Feynman Simplified 3B: Quantum Mechanics Part Two [2 ed.]

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Feynman Simplified 3B: Quantum Mechanics Part Two Everyone’s Guide to the Feynman Lectures on Physics by Robert L. Piccioni, Ph.D.

Copyright © 2015 by Robert L. Piccioni Published by Real Science Publishing 3949 Freshwind Circle Westlake Village, CA 91361, USA Edited by Joan Piccioni

All rights reserved, including the right of reproduction in whole or in part, in any form. Visit our web site www.guidetothecosmos.com

Everyone’s Guide to the Feynman Lectures on Physics Feynman Simplified gives mere mortals access to the fabled Feynman Lectures on Physics.

This Book Feynman Simplified: 3B covers the second third of Volume 3 of The Feynman Lectures on Physics. The topics we explore include: Particle Exchange Model of Forces Barrier Penetration in Particle Exchange Quantum Operators & Matrices 2-State & N-State Systems Hyperfine Splitting & Spin-Spin Interactions Electrons in Crystals & Semiconductors Schrödinger’s Equation Symmetry & Conservation Laws

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Table of Contents Review: Summary of Part One Chapter 13: Molecules & Forces Chapter 14: Operators & Matrices Chapter 15: Two-State Spins Chapter 16: Hyperfine Splitting in Hydrogen Chapter 17: Spin ½ & Spin 1 Rotation Tables Chapter 18: Electrons in the Simplest Crystals Chapter 19: Electrons in Realistic Crystals Chapter 20: Semiconductors Chapter 21: Independent Particle Approximation Chapter 22: Schrödinger’s Equation Chapter 23: Symmetry & Conservation Laws Chapter 24: Review of Part Two

Summary of QM Part One

Particle-wave duality: Everything in our universe has both classical particle and classical wave properties. In particular, particles have wavelengths given by: λ=h/p. Uncertainty Principle: the wave properties of particles preclude simultaneously measuring both their position and momentum with unlimited precision, as quantified in four equations: Δx Δp ≥ ħ/2 Δy Δp ≥ ħ/2 Δz Δp ≥ ħ/2 Δt ΔE ≥ ħ/2 x y

z

where ħ=h/2π=1.055×10 joule-sec and h=6.626×10 joule-sec is Planck’s constant. –34

–34

A quantum state defines all the variable properties that an entity can have, such as position, linear momentum, energy, spin, and angular momentum. It does not define intrinsic properties, such as charge and mass. In general, different types of particles can be put into a given state, and can be moved from one state to another. It is beneficial to think of a quantum state as a vector that defines a location in the space of all possible properties. Spin is a quantized form of angular momentum that is an intrinsic and immutable property of each type of elementary particle. All fermions have half-integral spin: s=n/2, for some odd integer n>0. All elementary fermions have spin s=1/2, as do protons and neutrons. Bosons have integral spin: s=0, 1, or 2. There is an essential distinction between a particle’s spin and its component of spin along any axis. Like other forms of angular momentum, spin is a vector with magnitude and direction. The spin of every particle never changes, although the orientation of its spin may change. The primary fermions, those with spin 1/2, can have a component of spin along any selected axis of +ħ/2 or –ħ/2, which are called spin up and spin down. The Z and W bosons with spin 1 can have a component of spin along any selected axis of +ħ, 0, or –ħ. In general, a particle of spin s can have a component of spin along any selected axis of –sħ, –(s–1)ħ, …+(s–1)ħ, +sħ; the allowed values of spin components are always separated by integral multiples of ħ. Photons have spin 1, but cannot have a zero component of spin along any axis, because they are massless. A particle with spin s has a spin vector whose magnitude equals √{s(s+1)}; this magnitude is rarely mentioned, since it is so much easier to say “spin s.”

Measurement is a contact sport, more like boxing than dancing. Observation requires interaction. Substantial interactions alter the behavior of what is observed. Every measurement forces the observed entity into a definite state, one with an allowed value for that measurement. Particles have no memory of their prior history. If an atom’s spin is measured to be +1 along the z-axis, nothing else about its spin is knowable. Feynman’s General Principles of probabilistic quantum mechanics are: First: the probability P(y) of event y is proportional to the square of the magnitude of probability amplitude ø(y), which is a complex number. If event y can occur in only one way, P(y) is: P(y) = øø* = |ø|

2

In Dirac’s bra-ket notation, the amplitude that A results in, or goes to, B is . This is analogous to the dot product of normal vectors A and B. For example, in the two-slit experiment: ø(y) = ø(y) = Second: when event y can occur in N undistinguished ways, the amplitude of y equals the sum of the amplitudes for each separate way: = Σ , sum K=1…N K

We sum amplitudes only when each separate path results in exactly the same final state. As the two-slit experiment demonstrates, with no substantial disturbances, it is impossible — for both man and nature — to distinguish the paths by which particles travel to reach an event y. Paths are distinguished only when particle-waves along those paths are substantially disturbed. What matters is the magnitude of the disturbance, not whether it occurs naturally or is human-directed. But due to our scale, any human-directed interaction will almost certainly substantially disturb a quantum system. Third: when event y can occur in N distinguished ways, the probability of y equals the sum of the probabilities for each separate way: P(y) = Σ || , sum K=1…N 2

K

Fourth: the amplitude for a sequence of events equals the product of the amplitudes for each separate event. The amplitude for a particle to go from S to x and then from x to y is: . Fifth: For multiple particles, if ø is the amplitude of particle 1 going from S to F , and ø is the amplitude of particle 2 going from S to F , then the amplitude for both events to occur is ø × ø . 1

1

2

2

1

2

1

2

Sixth: When two intrinsically identical particles can enter, exit, or be in the same state, their amplitudes interfere. If the particles are bosons governed by Bose-Einstein statistics, their

amplitudes add. If the particles are fermions governed by Fermi-Dirac statistics, their amplitudes subtract. There is no third alternative. For two identical particles 1 and 2, and any two indistinguishable states A and B, the combined amplitude is: Fermions: – Bosons : + Particles with different spins are distinguishable. All fundamental particles of each type are intrinsically exactly identical. While not fundamental, all protons are exactly identical to one another as are all neutrons. The class of fermions includes protons, electrons, neutrons, neutrinos, and quarks. All particles that form material objects are fermions with spin 1/2. The class of bosons includes photons, gluons, W , Z , and the Higgs boson. All forces between elementary particles are carried by bosons. Photons are the force exchange particles of electromagnetism. Gluons mediate the strong force. The W and Z intermediate vector bosons mediate the weak force. See Higgs & Bosons & Fermions.... to further explore elementary particle physics. ±

0

±

0

The probability that N identical bosons are in a common state is larger by a factor of N! than the probability of N non-identical particles being in that state. The probability of one more boson entering a state occupied by N identical bosons is N+1 times greater than it would be if the state were empty. Bosons are groupies: the bigger the party, the greater the attraction. The gregarious behavior of bosons underlies Einstein’s laws of radiation and the operation of lasers. Fermions are antisocial; they never exist in the same state as another identical fermion. This is called the Pauli Exclusion Principle. Two fermions are not in the same state if their spins are different. Nature has room for both groupies and individualists. Similarity of Vectors & States In normal 3-D space, a vector v specifies any selected point within the space. In quantum mechanics, |S> specifies a state, any selected set of particle properties. In 3-D, we choose a complete set of orthonormal basis vectors e , e and e . x

y

z

In QM, we choose a complete set of orthonormal basis states |J>, J=1,…,N.

Orthonormal means for each J and K: e •e =δ in 3D; and =δ in QM. Here δ is the Kronecker delta, which equals 1 if J=K and equals 0 otherwise. J

K

JK

JK

JK

In QM, a basis is “complete” if every possible state is some linear combination of basis states — for any state |φ>: |φ> = Σ a |J> = a |1> + a |2> + … a |N> |φ> = Σ |J> J

J

1

2

N

J

where a = are complex numbers. The set of all a completely defines |φ>. J

J

Differences Between Vectors & States In 3-D, the dot product is commutative: A•B = B•A. But, in QM the product of two states is not commutative: = *, which is the complex conjugate of . The complex conjugate of a complex quantity is obtained by replacing i with –i. In general and are not the equal. Since = Σ for all states φ and ψ, Feynman writes the open equation: J

| = Σ |J> to identify states, for consistency. Operators transform one state vector into another. Any action that changes particle states is represented in quantum mechanics by an operator. As an example: = Σ JK

denotes the product of state χ with the vector resulting from A operating on state ψ, which is shown expanded in basis states. Knowing for all basis states J and K completely determines A. Operators are often represented as N×N matrices, where N is the number of basis states: matrix component A equals . If multiple operators act on a state sequentially, we express that using the product of those operators. Matrix products are not commutative, so the order of factors is KJ

essential. The first operator to act holds the right-most position, with the second immediately on its left, and continuing on to the last operator to act in the left-most position. For two operators, A followed by B, the expression is: = Σ

JKL



Waves in an Energy Potential: the probability amplitude for a particle of momentum p and energy E (kinetic plus mass), with potential energy V, is proportional to: exp{(E+V)t/iħ–p•r/iħ} The Hamiltonian equation governs the time evolution of states: iħ d|Ψ(t)>/dt = H |Ψ(t)>

Review of Two-State Systems For a system with two basis states |1> and |2>, any state |Ψ> can be expressed as a linear combinations of the basis states: |Ψ> = |1> + |2> |Ψ> = C |1> + C |2> 1

2

The Hamiltonian equation yields two independent differential equations: iħ d(C )/dt = H C + H C iħ d(C )/dt = H C + H C 1

11

1

12

2

2

21

1

22

2

We can also write this more compactly: iħ dC /dt = Σ H C , for any j, sum over k j

k

jk

k

Here H is the amplitude for state |2> to transition to state |1>, and H is the amplitude for state |1> to transition to state |2>. 12

21

Assuming the Hamiltonian components are constant in time, the Hamiltonian equation has two solutions that are stationary states, states of definite energy. These are: |ψ+> = |+> exp{E t/iħ}, with E = E +E* |ψ–> = |–> exp{E t/iħ}, with E = E –E* +

–

where E = (H +H )/2 0

11

22

+

–

0

0

E* = √[ (H –H ) /4 + H H ] 2

11

22

12

21

|+> = a |1> + a |2> |–> = a |1> + a |2> 1+

2+

1–

2–

In V3p10-2, Feynman says the coefficients of the basis states, the a’s in the above equation, are complex constants that satisfy: 1 = |a | + |a | 1 = |a | + |a | 2

1+

2

2+

2

1–

2

2–

a / a = H / (E – H ) a / a = H / (E – H ) 1+

2+

12

+

1–

2–

12

–

11

11

In the special case of a symmetric two-state system, where H =H and H =H =–E*: 11

a / a = –E* / (E – E ) = –1 a / a = –E* / (E – E ) = +1 1+

2+

+

1–

2–

–

0

0

|+> = { |1> – |2> } / √2 |–> = { |1> + |2> } / √2

22

12

21

Chapter 13 Molecules & Forces We begin Part Two of Feynman’s brilliant presentation of quantum mechanics by exploring more twostate systems and the forces that bind molecules and nuclei.

An Ionized Hydrogen Molecule A hydrogen molecule consists of two hydrogen atoms, and is denoted by: H . Since each hydrogen atom consists of one proton and one electron, a hydrogen molecule has two protons and two electrons, making it electrically neutral. 2

We consider here a hydrogen molecule that has been ionized by the removal of one electron, leaving two protons to share the lone remaining electron. Let’s examine how this electron can be shared. Two simple basis states present themselves: the electron could surround the left proton; or it could surround the right proton. These two states are symmetric and both effectively result in a neutral hydrogen atom with a lone proton nearby, as shown in Figure 13-1.

Figure 13-1 Two Basis States of Ionized H2 Dots are Protons, Gray Disks are Electrons

As Feynman says: “Of course, there are really many states of an electron near a proton, because the combination

can exist as any one of the excited states of the hydrogen atom. We are not interested in that variety of states now; we will consider only the situation in which the hydrogen atom is in the lowest state—its ground state—and we will, for the moment, disregard spin [entirely].” The energy required to remove an electron from a neutral hydrogen atom is 13.6 eV, which is defined to be 1 Rydberg. (1 eV denotes one electron-volt, the energy gained by an electron in traversing a potential of one volt.) On the atomic scale, 13.6 eV is substantial, about 6 times the energy of visible light. For the electron of an ionized hydrogen molecule to jump between states |1> and |2> it would have to overcome the binding energy of one proton before falling into the potential well of the other proton. This is a bit like a golf ball spontaneously disappearing from one cup and reappearing in another. No golfer has ever seen this happen, because in classical physics such actions are impossible. However, as we discovered in Chapter 9, barrier penetration does occur in the realm of quantum mechanics. Hence there is a non-zero amplitude H for the transition from |2> to |1>. Since 13.6 eV presents a large barrier, the amplitude H is small. By symmetry, states |1> and |2> must have the same energy (H =H ) and the same amplitude to transition from one to the other (H =H ). 12

12

11

22

12

21

As before, there are two stationary states of definite energy: |+> = {|1> – |2>}/√2 with E = E + E* |–> = {|1> + |2>}/√2 with E = E – E* +

0

–

0

where E =H =H –E* = H = H 0

11

22

12

21

If H = H = 0, if there were no transition amplitude between the basis states, there would be only one energy level: E . But because there is a non-zero transition amplitude, we say the single energy level is split into two levels: E and E . 12

21

0

–

+

Let’s be very clear about the quantum states just described. The sole electron can exist in either stationary state |+> or |–> indefinitely, with a precisely defined energy. In either of these states, the electron is at all times simultaneously around each proton. If we measure the electron’s position, it will always have a 50% probability of being found at the left proton and a 50% probability of being found at the right proton. Alternatively, the electron could be, at time t=0, entirely in one basis state, either |1> or |2>. But it will not remain there. The electron will transition back and forth between these two states. These states are therefore not stationary, and are not states of definite energy. If we measure the electron’s position at time t, the probability of finding it around the left proton will vary between 0 and 1, as given by sin (ωt+ø), while the probability of finding it around the right proton will be given by cos (ωt+ø). 2

2

In all cases, these are the states of a single electron, shared by two protons. Now let’s consider how the transition amplitudes H =H vary with the distance between the two protons. 12

21

In our prior study of barrier penetration, we learned that the penetration amplitude decreases exponentially as a function of the product of barrier height multiplied by barrier width. The same logic applies here to H . H has its maximum magnitude when the protons are adjacent, and decreases rapidly as the proton-proton separation D increases. 12

12

Since the energy level splitting is proportional to the transition amplitude, the splitting is also maximum at D=0, and decreases rapidly as D increases. This is illustrated in Figure 13-2.

Figure 13-2 Electron Energy vs. Proton Separation D for both Stationary Electron States

The energy of state |+> decreases with increasing D, creating a repulsive force that pushes the protons apart. Recall that force equals minus the gradient of potential energy; forces act to reduce potential energy. Conversely, the energy of state |–> energy decreases with decreasing D, creating an attractive force that pulls the protons together. But the electron’s energy is only part of the story. As two protons become closer, their electrostatic repulsive force increases in proportion to 1/D . Their electrostatic potential energy, E , is positive and proportional to 1/D, going to +∞ in the limit that D goes to zero. Adding the proton-proton potential energy, E , to the electron’s energy, we obtain the curves seen in Figure 13-3. 2

p

p

Figure 13-3 Total Energy vs. Proton Separation D for both Stationary Electron States

For |+>, adding E increases the repulsive force pushing the protons apart. While for |–>, adding E creates a minimum in the total energy at a proton-proton separation we will call D . At D , the proton-proton repulsive force is exactly balanced by the attractive force between the electron and each proton. Since the total energy is negative at D , the ionized hydrogen molecule has a stable bound state. The electron holds the two protons together, forming what chemists call a one-electron bond. p

p

min

min

min

Recall how the stationary states are formed from the basis states: |+> = {|1> – |2>}/√2 with E = E + E* |–> = {|1> + |2>}/√2 with E = E – E* +

0

+

0

Note that the lower energy state, the only bound state, is symmetric with equal amplitudes for the electron to be around each proton. The higher energy state is asymmetric, with opposite polarity amplitudes. In V3p10-4, Feynman provides physical insight into why the asymmetric state has a larger energy: “You remember that with a single proton the electron is “spread out” because of the uncertainty principle. It seeks a balance between having a low Coulomb potential energy and not getting confined into too small a space, which would make a high kinetic energy (because of the uncertainty relation ΔpΔx≈ħ). Now if there are two protons, there is more space where the electron can have a low potential energy. It can spread out—lowering its kinetic energy— without increasing its potential energy. The net result is a lower energy [for an ionic hydrogen molecule than for] a proton and a [separate] hydrogen atom. Then why does [|+>] have a higher energy? Notice that this state is [asymmetric and] must have zero amplitude to find the electron half-way between the two protons. This means that the electron is somewhat more confined, which leads to a larger energy.” In fact, the asymmetric state |+> excludes the electron from the energetically most favorable location: midway between two positive charges. Feynman cautions that our assumption that H is a two-state system breaks down at small values of D, requiring a more sophisticated quantum mechanical analysis than we are ready for. Our analysis is + 2

qualitatively, but not quantitatively, correct. The measured values for an ionized hydrogen molecule, denoted H , are: binding energy equals –2.8 eV at D = 1.06 angstroms (1.06×10 m). +

2

–10

min

The electron’s ability to be around one proton, or the other, or both simultaneously, results in a lower total energy for H than for a hydrogen atom and a separate proton. The electron provides an attractive force that holds the three particles together. +

2

Bridging a Negative-Energy Gap Let’s now examine more closely the barrier penetration phenomenon that makes the ionized hydrogen molecule stable. Barrier penetration occurs when a particle transitions across a region where its kinetic energy is negative. Figure 13-4 plots the potential energy of the H electron as a function of its distance x from the left proton. The dotted vertical lines mark the two proton positions. Classically, the electron’s potential energy is proportional to –1/r –1/r , where r and r are its distances to the left and right protons respectively. That sum approaches –∞ at the location of each proton. + 2

L

R

L

R

Figure 13-4 Electron Energy vs. Position Negative Kinetic Energy Region is Shaded

The H electron’s potential energy V is represented in Figure 13-4 by the solid curve, while its total energy (T+V) is represented by the horizontal dashed line. The electron is bound because T+V is negative. Midway between the protons, the potential energy V increases (becomes less negative), making T negative within the shaded region. Classically, an electron with the indicated total energy cannot scale the potential barrier and move from one proton to the other. + 2

But as we know, the wave properties of particles enable them to penetrate barriers in which their kinetic energy is negative. Since the energy involved here is always much smaller than an electron’s mass-energy, we employ non-relativistic equations to analyze barrier penetration.

T = mv /2 = p /2m p = √(2mT) 2

2

When T is negative, define ε = –T > 0. The equation for momentum then becomes: p = +i√(2mε) Now define Λ = iħ/p. In the negative-energy region, Λ is real and positive. Λ = iħ / [i√(2mε)] Λ = ħ / √(2mε) As we discovered in Feynman Simplified 1C, a particle-wave expanding outward from point P to point Q has an amplitude given by: exp{(i/ħ)p•r} / r where p is its momentum vector and r is the vector from P to Q. In most cases of barrier penetration, the distance z that the amplitude penetrates within the classically forbidden zone is generally much less than r, the distance to the source. The electron’s amplitude to transition from one proton to the other, which we are calling A, is proportional to: A ~ exp{(i/ħ) ∫pdz } A ~ exp{–∫(dz/Λ)} We write it with an integral because p and Λ are functions of z. Feynman explains why we ignored the negative imaginary solution of p=–i√2mε. That would result in A ~ exp{+∫(dz/Λ)}, which is physically unrealistic, corresponding to an amplitude within the barrier that increases exponentially toward infinity. We must be very clear about what quantum mechanics says about barrier penetration. It does not say a “real” electron can travel within a barrier in which its kinetic energy is negative. What penetrates the barrier is the electron’s probability amplitude, something that has no physical reality. An electron placed near the left proton has a small but non-zero probability amplitude to be at the right proton. If we repeatedly measure the electron’s position, we might almost always find it near the left proton. But eventually, we will find it near the right proton. Our classical intuition encourages the notion that the electron must have moved from left to right, passing through the impenetrable barrier. But that is not what quantum mechanics claims. Indeed, quantum mechanics says we will never observe the electron inside the barrier: no real particle can ever be observed to have a negative energy, negative mass, or imaginary momentum. What quantum mechanics says is that for each measurement, there is some probability of finding the electron near the left proton and another probability of finding it near the right proton. We may be able

to exactly predict these probabilities as functions of time, but regardless of how much we know, and regardless of what the electron’s prior history has been, its position is undetermined between measurements. To claim that an electron went from here to there, we would actually have to observe its position progressing step-by along that path.

Van Der Waals Effect In V3p10-5, Feynman adds a final note about the H molecule: the van der Waals effect. This effect is generally quite small, but at large separations it makes a contribution to the attractive force holding the three particles together. + 2

In an isolated hydrogen atom, the center of positive charge and the center of negative charge coincide with the atom’s center. If however, a second proton is nearby, the hydrogen atom’s electric charge centers shift, with the atom’s negative charge moving closer to and its positive charge moving farther from the second proton. This gives the atom an electric dipole moment µ. The dipole moment is proportional to the electric field E that the atom experiences from the second proton. The energy of a dipole µ in an electric field E is proportional to µE, which in this case is proportional to E . Recall that E is proportional to 1/r . Hence, at a proton-proton separation of r, the dipole energy is proportional to 1/r . Since force is the gradient of potential energy, the attractive van der Waals force is proportional to 1/r . 2

2

4

5

Typically, we can ignore terms like 1/r , but in this case the other contributing term is proportional to exp{–r}/r. At large enough distances, the van der Waals effect dominates. 4

Note that the van der Waals effect is attractive for both electron states, |+> and |–>, while the electron-exchange effect is repulsive for |+> and attractive for |–>.

Asymmetric Ionic Molecules V3p10-4 Now consider the combined system of two different atoms, each of which has a net electric charge of +1 (each is missing one electron). Feynman uses the example of a lone proton and a Li ion, but the analysis applies generally to any asymmetric objects with positive charge. +

In a non-symmetric system, H may be very different from H . For sake of discussion, assume: 11

H =H H >H (H – H ) >> H 12

21

11

22

11

Define:

22

12

22

–A = H E = (H + H )/2 Δ = (H – H ) 12

0

11

11

22

22

From the two-state equations given at the end of the Review of Part One, we have: E = E ± √[ Δ /4 + A ] 2

±

2

0

Above, we had Δ=0, resulting in E =E ±A. But here, for Δ>>A, this equation can be approximated by: ±

0

E ≈ E ± (Δ/2) {1 + 2(A/Δ) } 2

±

0

The energy difference between the two stationary states is then: E – E = Δ + 2A /Δ 2

+

–

Compared with the case of two identical ions, here the energy difference attributable to the electron’s state is reduced from 2A to 2A(A/Δ). Since we assumed A0. The total energy would then be mc plus T plus E , which must be greater than the initial energy mc , in violation of energy conservation. 2

γ

e

2

2

e

γ

Physicists typically resolve this apparent conflict by saying that the exchanged particle is a virtual photon, not a real photon. The virtual photon can never be directly observed, and is promptly absorbed by a real particle. Virtual particles, because they are unobservable, can have any energy, any mass, and any momentum. An alternative view is to invoke the uncertainty principle, and say that energy and momentum are not conserved at each vertex of a Feynman diagram, but are eventually conserved in the final states. The final states must be composed of real, observable particles that can be measured with very large values of Δt and Δx. Such measurements mandate very small values of ΔE and Δp, thereby ensuring energy and momentum conservation.

Both views result in the same math, and are thus experimentally indistinguishable. Choosing either, or both, is a matter of personal preference. If one accepts virtual particles, or their equivalent, the next question might be: which particles are real and which are virtual? Any particle could be real in one circumstance and virtual in another. Per the uncertainty principle, any particle that exists only for an infinitesimal time has, within its wave packet, values of energy, mass, and momentum that are spread over broad ranges. The longer the particle exists, the smaller those ranges become, and the less “virtual” and the more “real” the particle becomes. There is no sharp dividing line. It was Feynman who made this particle-exchange model work mathematically. In his scheme, to evaluate the total electromagnetic force between two electrons, one must sum over all possible virtual particles, emitted and absorbed in all possible ways. That might seem like an unattainable goal, but Feynman developed techniques to do such calculations systematically. In these summations, virtual particles that deviate only slightly from real particles make the greatest contributions. Virtual particles with large deviations (those we call “far off the mass-shell”) make insignificant contributions. Feynman, Tomonaga, and Schwinger were awarded the 1965 Nobel Prize for developing quantum electrodynamics (“QED”), the particle-exchange model of electromagnetism. Following the success of QED, physicists have tried to generalize the particle-exchange model to all other forces, with mixed results. These approaches are collectively called quantum field theories (“QFT”). The weak force is very successfully explained using the particle-exchange model. Shown in Figure 13-6 is the Feynman diagram for neutron decay. Neutrons are composite particles composed of one up-quark and two down-quarks (udd). One of the d-quarks decays to a u-quark plus a W boson, a carrier of the weak force. –

Figure 13-6 Neutron Decay to Proton, Electron & Antineutrino, Mediated by W – Boson

The new u-quark combines with the u- and d-quarks that remain from the original neutron, thus forming a proton. The W decays to an electron and an antineutrino. In the standard particle physics units in which c=1, some key masses are: 2.4 MeV = u-quark mass 4.8 MeV = d-quark mass 80.4 GeV = W boson mass When the d-quark changes into a u-quark, it releases 2.4 MeV of energy. That’s less than 1/30000th of the mass of the W. This means the virtual W is “far off the mass shell”, an extreme deviation from a real W. This is fundamentally why this decay is “weak”, and proceeds at such a low rate. The lifetime of a particle that can decay via the strong force is typically one trillionth of a trillionth of a second. The neutron lifetime is 15 minutes, 10 times longer. 27

Meson Exchange Model of Strong Force In the last 50 years, enormous progress has been achieved in our understanding of the strong nuclear force. We will first discuss what Feynman said in V3p10-6, and subsequently highlight later major developments. In 1935, Hideki Yukawa proposed that the force between nucleons (protons and neutrons) arose from their exchange of mesons, particles of intermediate mass, lighter than protons but heavier than electrons. By combining the uncertainty principle with the range of the strong force, Yukawa predicted that the meson’s mass would be about 200 times the electron’s mass. Let’s see how Yukawa estimated the meson’s mass. This analysis is intended only to approximate the range of masses that would satisfy Yukawa’s model. If a particle of mass m is spontaneously created from nothing, energy conservation is violated by the amount ΔE=mc . The uncertainty principle says this violation is undetectable (even by nature) for a time interval of up to Δt=ħ/(2ΔE). During time Δt, the maximum distance a particle can travel is d=cΔt. 2

If this particle is the exchange particle of the strong force, which has a maximum range of 1 fermi (10 m), then:

–

15

1 fermi = c Δt = c ħ/(2ΔE) 1 fermi = ħc / (2mc ) 1 fermi = 197 MeV-fermi / 2mc mc = 99 MeV = 193 × mass of electron 2

2

2

Here, I used ħc = 197 MeV-fermi, one of my favorite conversion factors. At the time Yukawa made this prediction, only one particle was known whose mass was between the electron mass and the proton mass. That particle was the muon, whose mass is 105.7 MeV,

tantalizingly close to Yukawa’s prediction. Muons are the most common particles in cosmic rays, due largely to their relatively long lifetime of 2.2 microseconds. But experiments from 1942 through 1945, by Marcello Conversi, Ettore Pancini, and Oreste Piccioni (my father), proved that the muon did not participate in the strong nuclear force. This meant the muon could not be the strong force exchange particle that Yukawa had predicted. The search continued. In 1947, physicists discovered two types of pions, one with charge +1 and the other with charge –1. In 1950, they found a third type of pion with zero charge. Charged pions have a mass of 139.6 MeV and a lifetime of 26 nanoseconds, while neutral pions have a mass of 135.0 MeV and a lifetime of 8.4×10 seconds. All pions are strongly interacting and composed of quarkantiquark pairs. The discovery of pions validated Yukawa’s prediction, and he was awarded the 1949 Nobel Prize. –17

In V3p10-6, Feynman applies the results of our analysis of the H molecule to the strong force. We considered the force that binds H to be the exchange of the sole electron between the two protons. We found the exchange amplitude was proportional to: + 2

+

2

A ~ exp{(i/ħ) ∫pdr } / r Recall that in a region between the protons, the electron momentum p is imaginary due to the potential energy barrier. Feynman says we have a similar situation in the strong force. When a proton or a neutron emits a virtual pion, that particle must have zero total energy. This means: E = 0 = p c + mc p = imc 2

2

2

4

Here, m is the pion mass. Using the same equation as for H , the pion exchange amplitude is: +

2

A ~ exp{(i/ħ) pr } / r A ~ exp{–(mc/ħ) r} / r This is a general result for any particle-exchange force model. Feynman notes that in electromagnetism, the exchange particle is the photon that has zero mass. In this case, the amplitude is proportional simply to 1/r. The force, being the gradient of energy, is then proportional to 1/r . Thus our particle-exchange model reduces to the famous inverse-squared law, as it logically must. 2

Quark Model of Strong Force Our understanding of the strong force has advanced greatly in the last 50 years. The modern understanding of elementary particles and the strong, electromagnetic, and weak forces is called the Standard Model of Particle Physics. It doesn’t explain everything, but it does correctly describe a very wide range of natural phenomena. The Standard Model says the strong force is the interaction of quarks with one another via the

exchange of gluons. Gluons are massless, spin 1 bosons. The force between nucleons, what we used to call the strong force in the good old days, is merely a residual effect of the much stronger interaction between the quarks within those nucleons. Sometimes physicists now use “strong force” to describe the quark-quark interaction and “nuclear force” to describe the residual effect between nucleons. The residual “nuclear force” does have a very limited range. Its strength drops precipitously at distances greater than 1 to 2 times the proton radius. But that’s not the most interesting aspect of the strong force. The strong force is much more complex than electromagnetism, and is not yet perfectly understood. Unlike all other forces, the strong force does not diminish with increasing distance. Indeed, the strength of the strong force as a function of quark-quark separation, D, is quite bizarre. Its strength seems to be very weak, perhaps zero, at D=0; it increases linearly with D up to a point; and it plateaus at an enormous value for all D ≥ the size of a proton. At that plateau, the strength of the strong force between two quarks is estimated to be 10,000 newtons, about the weight of a one-ton mass on Earth’s surface. Think about that: the strong force between two quarks equals the gravitational force between 10 quarks in Earth and 10 quarks in a one-ton mass — a ratio of 10 . 52

30

82

Any attempt to pull two quarks apart fails, because the energy required to overcome the strong force is sufficient to spontaneously create a quark-antiquark pair. This is shown schematically in Figure 137, where an up quark is pulled out of a proton. The field energy builds as the quarks separate, and eventually that energy converts into a down-antidown quark pair, increasing the number of quarks to five. Three of the five quarks combine to make a neutron, and the remaining two combine to make a pion.

Figure 13-7 Time Sequence Starting At Top Of Attempt To Pull U-Quark Out Of A Proton

This “explains” why individual quarks are never observed: they are too strongly attached to one another to ever be alone.

The Hydrogen Molecule We now consider the neutral hydrogen molecule composed of two protons and two electrons. To simplify a complex topic, we will discuss only the lowest energy states — the ground states — of the hydrogen molecule. Initially, we will treat the two electrons as distinguishable and call them electron A and electron B. We will later explore the effects of A and B being identical fermions. With electrons A and B assumed distinguishable, analyzing the neutral hydrogen molecule follows the same logic used for the ionized hydrogen molecule. We imagine two protons separated by a distance D and consider the possible electron states. Two electrons around one proton is clearly not the lowest energy state, since the electrons would repel one another and no advantage is being taken of the low potential around the other proton. The lowest energy basis states for electrons A and B are: |AB>: A at left proton; B at right proton |BA>: B at left proton; A at right proton By symmetry, these states have the same energy. But the electron stationary states of definite energy are: |+> = {|AB> – |BA>}/√2 with E = E + E* |–> = {|AB> + |BA>}/√2 with E = E – E* +

0

+

0

where –E* equals the amplitude for the transition from |AB> to |BA>, or vice versa. The total molecular energy must include the proton-proton electrostatic potential energy that acts to push the protons apart. The net effect is identical in form to the ionized hydrogen molecule, which we show again in Figure 13-8.

Figure 13-8 Energy of Hy drogen Molecule vs. Proton Separation D, For 2-Electron States

The energy curve for state |+>, with E=E +E , has no minimum and is unbound at all values of the proton-proton separation D. State |–>, with E=E +E , has a minimum at D=D and is bound. p

+

p

+

min

Now, we come to the issue that all electrons are identical fermions. The bound state |–> is: |–> = {|AB> + |BA>}/√2 If A and B are indistinguishable, |AB> and |BA> are also indistinguishable, and according to FermiDirac statistics: |BA> = –|AB> |–> = {|AB> – |BA>}/√2 = 0 This means the bound state |–> cannot exist if A and B are indistinguishable. Only one alternative remains to make the hydrogen molecule stable: A and B must have spins in opposite directions. Two electrons with opposite spin components are distinguishable: they would, for example, deflect in opposite directions in a Stern-Gerlach device. Therefore the hydrogen molecule is bound only when its electrons have spins in opposite directions. Two nearby hydrogen atoms will repel one another if their electrons spins are parallel. In V3p10-9, Feynman says: “…there appears to be an ‘interaction’ energy between two spins because the case of parallel spins has a higher energy than the opposite case. In a certain sense you could say that the spins try to reach an antiparallel condition and, in doing so, have the potential to liberate energy—not because there is a large magnetic force, but because of the exclusion principle.” As with almost any physics problem, we could have made this far more complex. We could have included higher-energy electron states and the multiplicity of proton spin directions. Our analysis isn’t complete, but it elucidates the key physics, which is our primary goal. There is a very important distinction between the one-electron two-state systems discussed earlier and two-electron two-state systems. Recall that for asymmetric nuclei, the one-electron bond is very weak because the lone electron’s energy is different at each nucleus (H is not equal to H ). 11

22

Conversely, two-electron bonds can be very strong even for asymmetric nuclei. This is because the basis states have the same electron energy (H =H ), since each basis state has one electron near each nucleus and it doesn’t matter which electron is near which nucleus. This means E is always substantially less than E , the energy of unbound atoms. 11

22

–

0

Indeed, two-electron bonds are the backbone of molecular chemistry, and the most common way in which atoms hold one another together.

An extreme case of a two-electron bond with asymmetric nuclei arises when one atom attracts electrons much more forcefully than the other, as with NaCl. Here both electrons may be entirely captured by one atom, thereby giving both atoms a net charge. The atoms are then held together by the electrostatic attraction of opposite charges.

The Benzene Molecule A benzene molecule is composed of six hydrogen atoms and six carbon atoms arranged in a ring, as shown in Figure 13-9 in the representation used in organic chemistry. Here each line represents a single (two-electron) bond, while the double lines represent two bonds (four electrons being shared). Note that the six carbon atoms forming the ring are connected by alternating single and double bonds.

Figure 13-9 Diagram of Benzene Molecule

Carbon atoms have six electrons, two of which are in the innermost shell and are tightly bound to the nucleus. The remaining four electrons are valence electrons that are bound weakly enough to be easily shared with other atoms. In this diagram, each carbon atom has four lines connecting it to the atoms with which it is sharing its four valence electrons. A hydrogen atom has one electron that is easily shared, and in this diagram, one line connecting it to the atom with which it is sharing its electron. Chemists were originally unable to explain why benzene was so tightly bound — its binding energy was much larger than the sum of the bonds indicated in the diagram. In V3p10-11, Feynman highlights another historical dilemma with this molecule: some benzene-like molecules had only half the expected number of distinct combinations. Consider the example of orthodibromobenzene (names like that are one reason I didn’t become a chemist). This molecule is made by replacing two of benzene’s hydrogen atoms with bromine atoms. “Ortho” means the two bromine atoms are bonded to adjacent carbon atoms. The problem was: ortho-dibromobenzene does not have

the two distinct versions shown in Figure 13-10.

Figure 13-10 Two Versions of Ortho-dibromobenzene Single Bond (Left) & Double Bond (Right) At Arrow

The two versions are distinguished by the bond between the two carbon atoms to which the bromine atoms are attached. The left-hand version has a single bond, while the right-hand version has a double bond. Different chemical properties were expected for these two versions. But, experiments found no differences among ortho-dibromobenzene molecules. We can now solve these mysteries with our knowledge of two-state quantum systems. The ground state of benzene is a two-state system. We noted earlier that the six carbon atoms are connected by alternating single and double bonds. For each bond around the ring, there are two options: single bond or double bond. But having chosen one bond, all the others are determined. Thus there are two basis states, as shown in Figure 13-11.

Figure 13-11 Two Basis States of Benzene: |1> on the Left and |2> on the Right

By symmetry, the two basis states have the same energy. There is some amplitude for the electrons forming the ring bonds to flip from |1> to |2>, with the same amplitude to flip in the opposite direction. As we well know by now, this results in two stationary states with definite but different energy, one greater than and one less than the energy of the basis states. Only the lower energy stationary state exists at normal temperatures. The measured energy difference between the stationary states — the energy level splitting — is 1.5 eV, and the carbon bond lengths are 0.140 nm, nearly midway between 0.147 nm for the normal carbon-carbon single bond and 0.135 nm for the double bond. In the actual stationary state, each of the carbon-carbon bonds is effectively a one-and-a-half bond, being the quantum superposition of

equal parts single and double. The two expected versions of ortho-dibromobenzene also correspond to two basis states. But because of the transition amplitude between them, the basis states are not the states that exist in nature. The actual states, the stationary states of definite energy, are superpositions of both basis states, with only the lower energy stationary state existing at normal temperatures.

Dyes In V3p10-12, Feynman explores the theory of dyes, using magenta as his example. The molecular diagram we will now present is in a more abbreviated format that is more commonly used in organic chemistry. In our prior diagrams, very atom was labeled, but here most of the carbon and hydrogen labels are omitted, as those atoms are ubiquitous in organic molecules. In this abbreviated format, a carbon atom is assumed at each end of every line, unless otherwise noted. Also, hydrogen atoms are assumed wherever a carbon atom is shown with less than four bonds. With that in mind, Figure 13-12 shows the diagram of the magenta molecule.

Figure 13-12 Diagram of Magenta Extra Charge Can Be at Either Arrow

The magenta molecule is nearly symmetric. Its key feature is an unbalanced positive charge that can be located at either end of the molecule — the ends indicated by the two arrows in Figure 13-12. As shown in the figure, the positive charge is near the upper arrow; note the double bond at the nitrogen atom and two rather than three double bonds in the attached benzene ring. This is another two-state system. One basis state has the positive charge at the upper arrow, while the other basis state has it at the lower arrow. The stationary states are the usual symmetric and antisymmetric superpositions of basis states. The transition amplitude between the basis states is much smaller for magenta than for benzene,

because 12 bonds must change states to move the positive charge across the molecule. The energy splitting of the stationary states is therefore reduced to the energy of visible light. Another key property is magenta’s high dipole moment, due to the large distance between the two locations of the positive charge. This makes magenta an extremely efficient light absorber. Thus even a small amount of magenta produces a strong, vibrant color — the essential characteristic of a beautiful dye.

Chapter 13 Review: Key Ideas Quantum Field Theory (QFT), the Particle Exchange Model of Force, states that the electromagnetic, strong, and weak forces arise from fermions (particles of matter) exchanging other particles (most commonly bosons). Many attempts to include gravity within the QFT model have failed. These bosons are: Photons for the electromagnetic force W , W , Z for the weak force Gluons for the strong force +

–

0

The ionized hydrogen molecule, two protons sharing one electron, is a two-state system that demonstrates how forces arise from particle-exchange. Here, two positive charges are held together by exchanging an electron, thereby achieving a lower energy state than if they were separated. This is called a single-electron bond. The neutral hydrogen molecule is similarly held together by the exchange of two electrons. This is called a double-electron bond, the backbone of molecular chemistry. At the subatomic level, forces arise from the exchange of virtual particles. Virtual particles are not directly observable and may have exotic properties that can include negative energy, negative mass, and imaginary momentum. Feynman says: “We do not ‘see’ the photons inside the electrons before they are emitted or after they are absorbed, and their emission does not change the ‘nature’ of the electron.” The force between nucleons is merely a residual effect of the much stronger interaction between the quarks within those nucleons. We sometimes use “strong force” to describe the quark-quark interaction and “nuclear force” to describe the residual effect between nucleons. Barrier Penetration A particle’s probability amplitude A declines exponentially within a barrier in which its kinetic energy is negative and its momentum p is imaginary, according to: A ~ exp{(i/ħ) ∫pdz} Here z is the distance inside the barrier. What penetrates the barrier is not the “real” particle but

rather its probability amplitude, something that has no physical reality. The meaning of quantum barrier penetration is that if a particle starts on one side of a classically impenetrable barrier, there is a small but non-zero probability of finding it later on the opposite side of the barrier. Our classical intuition encourages the notion that the particle must have moved through the barrier. But that is not what quantum mechanics claims. Indeed, quantum mechanics says no real particle can ever be observed to have a negative energy, negative mass, or imaginary momentum. Complex Molecules, including benzene and magenta, are quantum superpositions of basis states when there is a non-zero amplitude to transition from one basis state to another. These basis states are often symmetric, or nearly so. In these cases, the molecule’s energy level is split by an amount proportional to the transition amplitude, resulting in stationary states of definite energy. The energy of one stationary state will be less than the energy of any basis state; this is the state found in nature.

Chapter 14 Operators & Matrices This chapter delves deeply into the mathematics of quantum systems and operators.

Quantum Operators V3p11-5 We begin by reviewing the concept of operators in quantum mechanics. Recall from Chapter 10 how we defined the Hamiltonian operator. We began by asking how does the state |Ψ(t)> evolve over time? More specifically, what is |Ψ(t+Δt)>? We defined the operator U(t,t+Δt) by defining each of its components relative to a set of basis states |j>: is the amplitude that state |k> will transition to state |j> in the time interval from t to t+Δt. Because the equations of quantum mechanics are linear, if we know how each basis state evolves, we can calculate the evolution of any state. Thus: = Σ k

It is essential that you are completely comfortable with this equation and its ilk. If you are, you can skip down to the *** below. To find out how Ψ changes, we break Ψ down into little pieces, with each piece being entirely in one basis state, call it |k>. We then take the amplitude for |k> to change into another basis state, call it |j>. We finally sum all those little changed pieces to find the amplitude for Ψ to change into |j>. From right to left, the equation reads: the amplitude of Ψ being in basis state |k>, multiplied by the amplitude of state |k> transitioning to state |j> between t and t+Δt, summed over all |k>, equals the amplitude for Ψ(t+Δt) being in state |j>. This logic is an example of a hallmark of Western science: reductionism, breaking problems down into smaller, and hopefully simpler, pieces. We could try to calculate how Ψ changes directly, but then we would have to repeat that difficult task from the beginning for every new state we encounter. With reductionism, we only have to do the difficult calculation once for each basis state. *** end skip In Chapter 10, we defined the Hamiltonian as the limit as Δt goes to zero of: = δ + H (t) Δt / (iħ) jk

jk

and we showed that the amplitudes C = are governed by these differential equations: j

iħ dC /dt = Σ H (t) C j

k

jk

k

We can rewrite this using the bra-ket expressions for C , and H : j

jk

iħ d()/dt = Σ k

Since the above equation is valid for any state |j>, it can be written as an open equation (see Chapter 10) as: iħ d(|Ψ>)/dt = Σ H |k> k

Feynman says we can describe this as: the amplitude of state |Ψ> being in state |k> multiplied by the Hamiltonian operator H operating on state |k> yields iħ multiplied by the time derivative of |Ψ>. Alternatively, in matrix notation, where |Ψ> is a vector and H is an operator, we write: iħ d(|Ψ>)/dt = H |Ψ> Feynman says we can describe this as: the Hamiltonian H operating on state |Ψ> yields iħ multiplied by the time derivative of |Ψ>. Either description is valid. Feynman carries the mathematical abstraction one step farther by writing: iħ d/dt = H This follows since the prior equation is valid for all |Ψ>. In V3p11-6, Feynman cautions that this equation: “is not a statement that the H operator is just the identical operator as d/dt. The equations are the dynamic laws of nature–the law of motion–for a quantum system.” This is the distinction between being equal and being identical: in Newtonian physics, F and ma are different entities, they are not identical, but their equality provides the law of motion. By contrast, E=mc identifies mass as a form of energy, with c being the factor that converts units of measure. 2

2

Now, “just to get some practice”, Feynman derives the above equations in another way. Start with |Ψ> represented in terms of its components and basis states: |Ψ> = Σ |k> k

To find how |Ψ> evolves, simply take its time derivative: d(|Ψ>)/dt = d/dt {Σ |k> } k

d(|Ψ>)/dt = Σ |k> d{}/dt k

Here we assumed that we have chosen basis states that do not change over time (d|k>/dt=0), as we assume unchanging coordinate axes when we analyze objects moving through space. We have the equation for the time derivative of from the definition of H. iħ d()/dt = Σ n

Plugging that into the prior equation yields: d(|Ψ>)/dt = Σ |k> Σ /(iħ) k

n

iħ d(|Ψ>)/dt = Σ |k> kn

iħ d(|Ψ>)/dt = Σ H |n> n

This is exactly the equation we derived in another way. As Feynman says: “So we have many ways of looking at the Hamiltonian. We can think of the set of coefficients H as just a bunch of numbers, or we can think of the ‘amplitudes’ , or we can think of the ‘matrix’ H , or we can think of the ‘operator’ H. It all means the same thing.” ij

ij

Matrix Arithmetic In quantum mechanics, operators are represented by matrices. We have touched upon the subject of matrices before (Chapters 7 and 8), but now we need to employ them more extensively. Matrix mathematics is an essential tool for analyzing complex quantum systems. Matrices are rectangular arrays of components laid out in rows and columns. In quantum mechanics, the components are typically amplitudes or numbers. An n×m (“n-by-m”) matrix has n rows, m columns, and n•m components. In quantum mechanics, most matrices have the same number of rows and columns; these are called square matrices. Let’s examine matrix arithmetic using the example of a 3×3 matrix M. The components of M are denoted M , where the first index i denotes the component’s row number and the second index j denotes the column number. Often we use M to denote the entire matrix, but sometimes we write M to emphasize that it is a matrix. The layout of matrix M is shown below. ij

ij

It is essential to remember that in matrix mathematics, M is not equal to M in general. In quantum mechanics, many matrices are Hermitian, which means M equals the complex conjugate of M : M = M *. ij

ji

ij

ji

ij

ji

Multiplication by a scalar is the simplest matrix operation. A scalar is a single entity that might be a number, a function, or an amplitude. You can think of a scalar being a 1×1 matrix. Multiplying matrix M by scalar s just means multiplying each component of M by s.

Matrix addition is defined only for matrices with the same number of rows and the same number of columns. If matrix A = matrix M plus matrix N, then A =M +N , for all i and j. ij

ij

ij

Matrix subtraction is simply multiplying one matrix by the scalar –1, followed by matrix addition. Matrix multiplication is a bit trickier. If matrix C equals the product of matrix A times matrix B: C = Σ A B for all i and j ij

k

ik

kj

This product is defined only when the number of columns in A equals the number of rows in B, which ensures that each term in the above sum over k is well defined. The product matrix C has the same number of rows as A and the same number of columns as B. The calculation of C is schematically illustrated below: the summation runs across row 3 of A and down column 2 of B. 32

Multiplication order is critical: A times B is not equal to B times A, in general. When several matrices are multiplied with one another, the sequence of multiplication is also critical. One must multiply matrices from right to left. To evaluate the matrix product ABC, multiply B times C, then multiply that result by A. Vectors with n components can be considered either n×1 matrices or 1×n matrices. Vectors and matrices can be multiplied together. For example, consider a 3×3 matrix R and two 3-component vectors ø and Ψ: Ψ =Σ R ø i

j

ij

j

Here Ψ and ø could be spin state vectors of a spin 1 particle, and R could be the rotation operator that transforms states from one coordinate basis to another. The particle’s state in one basis, ø, is transformed into its state in another basis, Ψ, by multiplying ø by R. The identity matrix, also called the unit matrix, is denoted by the Kronecker delta δ . ij

This may seem trivial, but the unit matrix is just as useful as the number 1. Matrix division is not defined. However, most matrices of interest in quantum mechanics have inverses. If A is the inverse of matrix A, then: –1

AA = A A= δ –1

–1

ij

Matrix A has an inverse only if it is a non-singular, n×n matrix, in which case its inverse is also a non-singular, n×n matrix. A matrix is singular if any of its rows equals a linear combination of its other rows, or if any of its columns equals a linear combination of its other columns. Simple examples of singular matrices are those whose components are all zero across an entire row or an entire column. Matrix equations can be valid only if every term is a matrix with the same number of rows and the same number of columns. The equation M=0*δ means that every component of matrix M is zero. In such equations, δ is sometimes not explicitly shown, but is implicitly assumed. ij

ij

Note that the sum of two Hermitian matrices is Hermitian, but their product is not, in general. The Hamiltonian equation for the time evolution of C and C , the coefficients for spin states |+z> and |–z> respectively, is then: 1

2

iħ dC /dt = Σ H C j

k

jk

k

Equivalently, treating C as a two-component vector, and suppressing the indices, we can write this more simply as: iħ dC/dt = H C

Pauli Spin Matrices Pauli spin matrices are very rarely presented to undergraduates, but in V3p11-2, Feynman does. He says: “They are used all the time by the professionals…anyone who is going to work in quantum physics really has to memorize them.” While this is true for professional theoretical physicists, the topic seems overly formal for undergraduates. Nonetheless, my mission is simplifying the Feynman Lectures, so here are the Pauli spin matrices Simplified. Suppressing the row and column indices for clarity, the three Pauli spin matrices are:

The products of pairs of Pauli spin matrices are easily calculated. Below is one example, with details of the matrix product calculation:

(σ σ ) (σ σ ) (σ σ ) (σ σ )

: row 1 col 1: 0•0 +1*i = i : row 1 col 2: 0•(–i) +1•0 = 0 : row 2 col 1: 1•0 +0*i = 0 : row 2 col 2: 1•(–i) +0•0 = –i

x

y 11

x

y 12

x

y 21

x

y 22

Try working out some of the other pairings yourself. The full set of products of pairs of Pauli spin matrices is listed at the end of this chapter. Now let’s examine the product of a Pauli spin matrix with a state vector, represented in z-spin basis states. Our example is:

Here the state vector Ψ is multiplied by σ . The process is the same as above except that the state vector has only one column. Here: y

p = q =

If p=1 and q=0, |Ψ>=|+z> and the above product shows that σ |+z>=–i|–z>. y

If p=0 and q=1, |Ψ>=|–z> and the above product shows that σ |–z>=+i|+z>. y

The full set of products of Pauli spin matrices with z-spin basis states is listed at the end of this chapter. Mathematically, we can think of the three Pauli spin matrices forming the three components of some sort of vector that we denote σ. This vector does not correspond to any physical entity; it is purely a mathematical definition of convenience. One feature of the Pauli spin matrices is that any 2×2 matrix can be expressed as a linear sum of δ and the three σ’s, possibly requiring complex constants.

ij

Two-States to N-States As we have often seen before, the most general Hamiltonian equations for two-state systems are: iħ dC /dt = H C + H C iħ dC /dt = H C + H C 1

11

1

12

2

2

22

2

21

1

We have also seen these equations expressed in the following two forms: iħ dC /dt = Σ H C iħ dC/dt = H C j

k

jk

k

In the last equation, C is a two-component state vector and H is a 2×2 Hamiltonian matrix. Expanding from two states to N states is conceptually straightforward. The number of basis states expands from 2 to N; C becomes an N-component state vector and H becomes a N×N Hamiltonian matrix. The last two equations still fully describe the physics, but there are now more components. This added algebraic burden is why we employ matrix mathematics. As before, we assume the Hamiltonian is not time-dependent, which means dH /dt=0 for all j and k. jk

Our approach to N-state systems follows the same logic that we employed for two-state systems. We find the stationary states of definite energy. Then from linear combinations of those, we can obtain all possible states. Any state |Ψ> can be represented as a linear sum of basis states: |Ψ> = Σ C |j> j

j

For |Ψ> to be a stationary state, its amplitudes to be in each basis state must all have the same time

dependence. We can therefore express |Ψ> with time-independent coefficients a and one common time factor: j

|Ψ> = exp{–iEt/ħ} Σ a j

j

Plugging |Ψ> into the Hamiltonian equation yields: iħ d|Ψ>/dt = H |Ψ> E |Ψ> = H |Ψ> Rearranging the last equation yields: 0 = (H – δ E) |Ψ> jk

jk

Matrix mathematics proves that the above equation has solutions only when the determinant of the matrix (H –δ E) is zero. A discussion of determinants and how to calculate them is provided at the end of this chapter. jk

jk

Setting the determinant equal to zero yields a polynomial equation, called the characteristic polynomial, with E as the independent variable. In general, this polynomial contains all powers of E from the zeroth power to the Nth power. As we know from the algebra of complex numbers, such equations have N roots, some of which might be equal. Each root corresponds to a solution of the Hamiltonian equation for some value of E; denote these N values E , n=1…N. These E are the energy levels of the N stationary states of our N-state system. n

n

For each energy E there is a state vector |n> such that: n

E |n> = H |n> n

For each n, the above expression produces a set of N equations with N unknowns; the unknowns are the amplitudes of state |n> being in basis state |j>, for j=1…N. Since the above expression is linear in |n>, it cannot determine the normalization of state |n>; 17|n> is as good a solution as 1|n>. Therefore we add the additional requirement that =1, which means the probability of |n> being in basis state |j>, the sum over all basis states, is 100%. Each stationary state is called an eigenstate, its energy is called an eigenvalue, and its state vector is called its eigenvector, from the German word “eigen” meaning an intrinsic characteristic. “Eigen” identifies the energies and states that are fundamental to the system, its stationary states, from which any other state can be obtained. Let’s see how this works with a simple two-state system that we already understand. We begin by calculating the eigenvalues by setting the determinant of (H –δ E) equal to zero. For a 2×2 matrix, the determinant is the product of the upper-left and lower-right components minus the product of the other two components. jk

jk

0 = Det |M| = (H –E)•(H –E) –H •H 0 = H •H –E•(H +H ) + E –H •H E = E ±√{E –H H +H H } 11

22

21

12

2

11

22

11

22

21

12

2

0

0

11

22

21

12

where we define: E =(H +H )/2. Now define the quantity in { }’s to be E* : 2

0

11

22

E* = E –H H +H H E* = (H + 2H H + H )/4 –H H +H H E* = (H – 2H H + H )/4 +H H E* = √{ (H – H ) /4 +H H } 2

2

0

11

2

22

21

2

11

2

11

22

11

22 2

2

11

11

22

12

2 22 2 22

21

11

22

21

12

21

12

12

Thus we can write the two eigenvalues as: E = E ± E* 0

Now let’s calculate the eigenvectors. For eigenvalue E: H |n> = E |n> Let the two components of |n> be p and q.

This matrix equation produces two equations, one from each row. We show both, but we actually only need one of them. H p+H q=Ep H p+H q=Eq 11

12

21

22

H q = (E – H ) p p/q = H / (E – H ) 12

11

12

11

For each of the two eigenvalues, the ratios of eigenvector components are: |+>: p/q = H / (E + E* – H ) |–>: p/q = H / (E – E* – H ) 12

0

11

12

0

11

In each case, we must normalize p and q such that |p| +|q| =1. 2

2

We can check the simple case where H =H =E and H =H =–A. 11

22

0

21

12

E* = √{(H –H ) /4 +H H } = A 2

11

22

21

12

E = E ± E* 0

|+>: p/q = –A / (E + A – E ) = –1 |–>: p/q = –A / (E – A – E ) = +1 0

0

0

0

|+>: p = +1/√2; q = –1/√2 |–>: p = +1/√2; q = +1/√2 |+> = (|1> – |2>)/√2 |–> = (|1> + |2>)/√2 This matches what we derived earlier without matrices.

Diagonal Matrices In the case of a two-state system, such as the ammonia molecule, we defined the basis states as: |1> for nitrogen above, and |2> for nitrogen below the hydrogen atoms. With that basis, the Hamiltonian and its eigenvectors and eigenvalues are:

Procedures exist to diagonalize a square matrix, thereby making all its off-diagonal components equal to zero: H =0 whenever j is not equal to k. Diagonal Hamiltonians are vastly easier to analyze; each eigenvector has only one non-zero component, and each diagonal component of the Hamiltonian is an eigenvalue. If we diagonalize the above Hamiltonian, the resulting matrix, eigenvectors and eigenvalues are: jk

When a matrix can be diagonalized it is usually well worth the effort. However diagonalizing a large matrix can be very laborious.

Matrix Determinants Determinants are defined only for N×N matrices. Determinants have geometric significance. Consider each column (or each row) to be an independent vector. The determinant of a 2×2 matrix equals ± the area of the parallelogram whose two sides are the two column vectors, with the sign determined by the order of the column vectors. The determinant of a 3×3 matrix equals ± the volume of the parallelepiped whose three sides are the three column vectors. The determinant of a N×N matrix equals ± the measure of the N-dimensional space enclosed by the N column vectors. The simplest way to calculate the determinant of a matrix is by iteration. The determinant of a 1×1 matrix simply equals its sole component. M = (M ) Det M = M 11

11

The determinant of a 2×2 matrix M equals the product of the upper-left and lower-right components minus the product of the other two components, as shown below:

Det |M | = ad – cb ij

The determinant of a 3×3 matrix M has three contributions. Begin by picking any row or any column; all choices yield the same determinant. In the example below, we chose row 1. We proceed to step across the row, sequentially selecting each component and evaluating its contribution. The first contribution equals the first component of row 1, M =a, multiplied by the determinant of the minor of a. The minor of a is the 2×2 matrix formed by deleting both the row and the column in which a appears. In the example below, the minor for each contribution is shown in bold type. 11

Det |M | = +a(ej–hf) –b(dj–gf) +c(dh–ge) ij

In this example, the minor of a is the matrix formed by components e, f, h, and j. The second contribution equals b (row 1, column 2) multiplied by the determinant of its minor, which is the matrix formed by d, f, g, and j. The third contribution equals c (row 1, column 3) multiplied by the determinant of its minor. Each of the three contributions must be summed with the proper sign: (–1) , where r is the row number and c is the column number of the selected component. For example, the first contribution selected row 1 column 1, which has sign (–1) =+1. The second contribution selected row 1 column 2, which has sign (–1) =–1. The signs alternate as one proceeds across the chosen row or column. r+c

1+1

1+2

For larger matrices, repeat this procedure iteratively. Pick any row (or column). Multiply each component in that row (or column) by the determinant of its minor and by its proper sign. Sum the contributions across the entire row (or column). An equivalent equation for the determinant of the N×N matrix M is: Det |M | = Σ Sign(abc…)•M •M •M •… ij

1a

2b

3c

Here, abc… is a permutation of the integers 1 through N. Each term in the sum is the product of N components, with one selected from each row and one selected from each column. The sum extends over all permutations of the integers 1 through N, and Sign(abc..) equals +1 for even permutations and –1 for odd permutations. A permutation is even (or odd) if it is obtained from the sequence 1, 2, 3, … N by an even (or odd) number of swaps of adjacent integers. For example, 1243 is obtained from 1234 with one swap, so Sign(1243)=–1, while 1423 requires two swaps, so Sign(1423)=+1.

Chapter 14 Review: Key Ideas 1. Matrices are arrays of components laid out in rows and columns. In quantum mechanics, the components are typically amplitudes or numbers. An n×m (“n-by-m”) matrix has n rows, m columns, and n•m components. For matrices, A, B, and C, and scalar s, basic matrix arithmetic includes: Multiplication by scalar s: A =sB Matrix addition: A =B +C Matrix multiplication: C =Σ A B Unit matrix δ has components: 1 if i=j, else zero If A is the inverse of A, A A=AA =δ . Only non-singular n×n matrices have inverses. ij

ij

ij

ij

ij

ij

k

ik

kj

ij

–1

–1

–1

ij

2. The Pauli spin matrices are:

3. The products of pairs of Pauli spin matrices are: σ =σ =σ =δ σ σ =–σ σ =iσ σ σ=–σ σ =iσ σ σ =–σ σ=iσ 2

2 y

x

2 z

ij

x

y

y

x

z

y

z

z

y

x

z

x

x

z

y

4. Pauli spin matrices operating on a vector yield:

σ σ σ σ σ σ

z z x x y y

|+z>= + |+z> |–z>= – |–z> |+z>= |–z> |–z>= |+z> |+z>= +i |–z> |–z>= –i |+z>

5. The general equation for the determinant of the N×N matrix M is: Det |M | = Σ Sign(abc…)•M •M •M •… ij

1a

2b

3c

Here, abc… is a permutation of the integers 1 through N. The sum extends over all permutations, and Sign(abc..) equals +1 for even permutations and –1 for odd permutations.

Chapter 15 Two-State Spins

Hamiltonian for Spin ½ in Magnetic Field In V3p10-12, Feynman says: “Some of what we will say has been covered in earlier chapters, but doing it again may help to make some of the puzzling points a little clearer.” Here’s a chance to catch up and master your spin-thinking (or is that spinning-thinking). Consider an electron in a magnetic field. What we discover about electrons will apply to any spin 1/2 particle. Our basis states are: |+> for spin in +z-direction |–> for spin in –z-direction Any state Ψ is a linear combination of the basis states. |Ψ> = C |+> + C |–> +

–

We now need the Hamiltonian matrix for an electron in a constant magnetic field B. If B is in the +z-direction, our basis states are already stationary states of definite energy: iħ dC /dt = E C = –µ B C iħ dC /dt = E C = +µ B C +

+

+

z

+

–

–

–

z

–

And the Hamiltonian is: H H H H

11 12 21 22

= –µ B =0 =0 = +µ B

z

z

Note that the magnetic moment µ of the electron is negative (it is antiparallel to the electron’s spin) because the electron has a negative electric charge. Here, we have chosen to measure energy relative to what the electron’s energy would be if there were no magnetic field. This means we set E =0. We can always shift the zero-point of energy 0

measurements, because adding a constant to each energy merely changes the phase of every amplitude by the same amount, which has no physical impact. Now, what happens if the magnetic field is in a direction other than +z? Here, Feynman states that the Hamiltonian must be linear with respect to B. More specifically, if H is the Hamiltonian for B and H is the Hamiltonian for B , then H +H is the Hamiltonian for B +B . He later says this assumption has been validated by experiment. This assumption is also consistent with classical physics, which is the macroscopic limit of quantum physics. In classical physics, the energy of a dipole in a magnetic field is –µB, with B = √(B +B +B ). 1

1

2

2

2

x

1

2 y

2

1

2

2 z

With linearity assumed, let’s derive the Hamiltonian for a constant magnetic field in any direction. Firstly, we know the energy of electrons whose spins are along the same axis as B. These are: E = –µB for spin up, parallel to B E = +µB for spin down, antiparallel to B up dn

So, the Hamiltonian must: 1. be linear in each component of B 2. have the same proportionality constant for each component of B 3. and yield the above energy for spin up and spin down, according to our now-familiar two-state equations. Recall the equations for energy levels of the stationary states of a two-state system. E = E – E* E = E + E* –

0

+

0

where, E = (H +H )/2 E* = √[ (H –H ) /4 + H H ] 0

11

22

2

11

22

12

21

Recall that the amplitude to transition from any state P to any state Q equals the complex conjugate of the amplitude to transition from Q to P. Also, recall our choice of our zero-point energy, E =0. This means: 0

H =H * H H = H H * = |H | 21

12

2

12

21

12

12

12

H = –H 11

22

E* = √[ H + |H | ] 2

11

2

12

In the special case of B entirely in the +z-direction B=(0,0,B ), we know what E* and H must be: z

11

E* = H = µB µB = √[ (µB ) + |H | ] |H | = 0 11

z

2

z

2

z

12

2

12

This means H can contain terms linear in B and B , but not in B . 12

x

y

z

In V3p10-14, Feynman says the only solutions are of the form: H = µ (B ± iB ) exp{iβ} 12

x

y

Let’s see why. For H to be linear in both B and B , with proportionality constant µ (like H = µB ), H must have the form: 12

x

y

11

z

12

H = exp{iδ} µ (B + B exp{iβ}) 12

x

y

The overall phase angle δ is irrelevant since it vanishes in |H | , which is: 2

12

|H | = µ (B + B exp{iβ}) (B + B exp{–iβ}) |H | = µ (B + B + B B [exp{iβ}+exp{–iβ}]) |H | = µ (B + B + B B [2 cosβ]) 2

2

2

2

2

2

12

x 2 x 2 x

12 12

y

x

y

2

y

x

y

x

y

2

y

To match classical physics, we must eliminate the cross term. This means β must be ±π/2, and exp{iβ} must be ±i. The choice of ± is arbitrary, because both have the same physical consequences. By convention we choose: H = –µ (B – iB ) 12

x

y

The complete Hamiltonian becomes: H H H H

11 12 21 22

= –µ B = –µ (B – iB ) = –µ (B + iB ) = +µ B z

x

y

x

y

z

We can write this Hamiltonian in matrix form (restoring the zero-point energy E ): 0

Feynman says this result is also valid in time-varying magnetic fields, as validated by experiments.

Electron in Magnetic Field

Now we will apply this Hamiltonian to some interesting situations. Given a constant magnetic field in the +z-direction, there are two stationary states: |–z> with energy +µB; and |+z> with energy –µB. An electron placed in either |–z> and |+z> will remain in that state indefinitely — that’s what we mean by “stationary” — so the transition amplitudes H and H are zero. 12

21

Next add a small B field in the +x-direction. From the above Hamiltonian, we see that H and H are no longer zero. That allows transitions between |+z> and |–z>, resulting in two new stationary states that are superpositions of |+z> and |–z>. The energy split E* increases, since E* =H +|H | . The energy levels of these new stationary states are therefore driven farther apart by B . 12

2

21

2

11

2

12

x

If B varies sinusoidally (B =cosωt), this situation is equivalent to the ammonia maser discussed in Chapter 11. There, the transition was driven by an oscillating electric field, here it is an oscillating magnetic field, but the physics and the math are the same. The oscillating field causes transitions between |+z> and |–z>, which are most substantial when the field frequency is near the dipole resonant frequency 2µB /ħ. x

x

z

Therefore, Feynman says, the two-spin-state electron, like ammonia, could be employed in a maser. Indeed, in 1976, John Madey built the first free-electron laser, the most versatile of all laser varieties, based on similar principles. Next let’s consider a magnetic field B in a direction tilted from +z by polar angle θ, and rotated about the z-axis by azimuthal angle ø, as shown in Figure 15-1.

Figure 15-1 Orientation of Magnetic Field B

From the figure, we see that the components of B are:

B = B cosθ B = B sinθ cosø B = B sinθ sinø z

x y

Let’s find the amplitudes for an electron that has its spin parallel to B to be in spin states |+z> and |– z>. We write the electron wave function as: |Ψ> = C |+z> + C |–z> 1

2

With its spin parallel to the magnetic field, the electron is in a pure stationary state of definite energy with E=–µB. This means the time dependence of Ψ must be exp{–iEt/ħ}, and that both C and C have that same time dependence. 1

2

C (t) = a exp{–iEt/ħ} C (t) = a exp{–iEt/ħ} 1

1+

2

2+

Note that we have no need to deal with a or a , because the electron’s spin is in a pure spin-up state relative to B. 1–

2–

From the two-state equations, we recall: 1 = |a | + |a | a / a = H / (E – H ) 2

2

1+

1+

2+

2+

12

+

11

From the Hamiltonian derived in the last section: H = –µ B = –µ B cosθ 11

z

H = –µ (B – iB ) H = –µ B sinθ (cosø – i sinø) H = –µ B sinθ exp{–iø} 12

x

y

12 12

Plugging these into the equation for a and a : 1+

2+

a / a = –µB sinθ exp{–iø} / (–µB + µB cosθ) a / a = sinθ exp{–iø} / (1 – cosθ) 1+

2+

1+

2+

The final step is applying the overall normalization condition 1 = |a | + |a | . Feynman has a trick to simplify this task: use the trig formulas for half-angles. 2

1+

1 – cosθ = 2 sin (θ/2) sinθ = 2 sin(θ/2) cos(θ/2) 2

We then have: a / a = 2 sin(θ/2) cos(θ/2) exp{–iø} / [2sin (θ/2)] 2

1+

2+

2

2+

a / a = cos(θ/2) exp{–iø} / sin(θ/2) |a / a | = cos (θ/2) / sin (θ/2) 1+

2+

2

1+

2

2

2+

1 = {|a / a | } |a | + |a | 1 = [{cos (θ/2) / sin (θ/2)} + 1] |a | 2

1+

2

2+

2

2+

2

2+

2

2

2+

1 = |a | {cos (θ/2) + sin (θ/2)} / sin (θ/2) sin (θ/2) = |a | a = sin(θ/2) exp{iδ} 2

2

2

2

2+

2

2

2+

2+

Here δ is an arbitrary phase angle. By convention, we chose δ=ø/2, which yields: a =cos(θ/2) exp{–iø/2} a = sin(θ/2) exp{+iø/2} 1+ 2+

C (t) =cos(θ/2) exp{–iø/2} exp{–iEt/ħ} C (t) = sin(θ/2) exp{+iø/2} exp{–iEt/ħ} 1 2

In V3p10-16, Feynman notes the interesting fact that none of the last four equations contains a B; the magnetic field has cancelled out of the equations. Our result remains true in the limit that B goes to zero. He says: “This means that we have answered in general the question of how to represent a particle whose spin is along an arbitrary axis. The amplitudes [above] are the projection amplitudes for spin one-half particles corresponding to the projection amplitudes we gave in Chapter [8]. We can now find the amplitudes for filtered beams of spin one-half particles to go through any particular Stern-Gerlach filter.” By “projection amplitudes” we mean that a and a are the amplitudes that relate spins along the zaxis with spins along another axis z*, whose orientation is defined by the polar angle θ and the azimuthal angle ø. For spin one-half particles, these amplitudes are: 1+

2+

= cos(θ/2) exp{–iø/2} = sin(θ/2) exp{+iø/2}

Electron Precession Consider an electron that at time t=0 is in a pure state of spin-up along axis z*. Using the results of the last section, we can express this state as the superposition of states |+z> and |–z>: |Ψ> = C (0) |+z> + C (0) |–z> C (0) = =cos(θ/2) exp{–iø/2} C (0) = = sin(θ/2) exp{+iø/2} 1

2

1 2

At t=0, we turn on a magnetic field in the +z-direction. At a later time t=T, what is the electron’s

state? States |+z> and |–z> are stationary states of definite energy. Since their energy levels are different, these states change differently over time, according to: C (t) = C (0) exp{+iµBt/ħ} C (t) = C (0) exp{–iµBt/ħ} 1

1

2

2

At t=T, these become: C (T) =cos(θ/2) exp{–iø/2} exp{+iµBT/ħ} C (T) = sin(θ/2) exp{+iø/2} exp{–iµBT/ħ} 1 2

Note that the exponents in the last two equations are identical, but with opposite signs. The exponent of C (t) can be rewritten as: 2

+iø/2 – iµBT/ħ = i {ø–2µBT/ħ} /2 = iβ/2 where we define β to be the expression in { }’s. With this substitution, we have: C (t) =cos(θ/2) exp{–iβ/2} C (t) = sin(θ/2) exp{+iβ/2} 1 2

This means the azimuthal angle of the electron’s spin was ø at t=0, and over time that angle rotated, becoming ø–2µBT/ħ at t=T. The azimuthal angle rotates about the z-axis with frequency ω=–2µB/ħ. (Recall that µ is negative for an electron.) This motion, shown in Figure 15-2, is called precession.

Figure 15-2 Precession of Electron Spin S

We have discussed precession before, but as Feynman says: “This result we discussed several times previously in a less complete and rigorous manner. Now we have obtained a complete and accurate quantum mechanical description of the precession of atomic magnets.”

In V3p10-17, Feynman says: “It is interesting that the mathematical ideas we have just gone over for the spinning electron in a magnetic field can be applied to any two-state system. That means that by making a mathematical analogy to the spinning electron, any problem about two-state systems can be solved by pure geometry. It works like this. First you shift the zero of energy so that (H +H ) is equal to zero so that H =–H . Then any two-state problem is formally the same as the electron in a magnetic field. All you have to do is identify −μB with H and −μ(B −iB ) with H . No matter what the physics is originally…you can translate it into a corresponding electron problem. So if we can solve the electron problem in general, we have solved all two-state problems.” 11

11

22

22

z

11

x

y

12

Employing Pauli Spin Matrices Using the Pauli spin matrices, the above Hamiltonian is: H = E – µ(σ B + σ B + σ B ) 0

x

x

y

y

z

z

As discussed above, assume the constant E is multiplied by the unit matrix. We can then write the Hamiltonian even more compactly: 0

H = E – µ σ•B 0

This is very reminiscent of the classical physics equation for the potential energy U of a dipole in a magnetic field. U = – µ•B Since the Hamiltonian is the energy matrix of quantum mechanics, this may feel like finally seeing a familiar lighthouse beacon through a thick quantum fog. We have introduced many mathematical concepts to reduce the size of our equations. It might seem that we are saving a few keystrokes at the expense of forcing you to learn too much exotic math. Your task most certainly is challenging, and is made more challenging with complex numbers, vectors, and matrices. But there is great value in this, value that is much more important than saving keystrokes. Our minds can only focus on so many things at once. We therefore want to avoid filling the screen with complicated entities expanded in their full glory. Instead, we want to focus on the essential physics in order to better understand the ideas. In this regard, consider Einstein’s most profound equations: E = mc E = hf G = 8π T 2

γ

Their brevity makes their statements bolder and clear.

Feynman went so far as to drop every 2π from his calculations. He said he had a scheme for figuring out how to restore them in the final equations. Regrettably, no one else managed to learn his scheme. In V3p11-4, Feynman offers this amusing and sage advice: “It is sometimes said that to each quantity in classical physics there corresponds a matrix in quantum mechanics. It is really more correct to say that the Hamiltonian matrix corresponds to the energy, and any quantity that can be defined via energy has a corresponding matrix. For example, the magnetic moment can be defined via energy by saying that the energy in an external field B is −μ•B. This defines the magnetic moment vector μ. Then we look at the formula for the Hamiltonian of a real (quantum) object in a magnetic field and try to identify whatever the matrices are that correspond to the various quantities in the classical formula. That's the trick by which sometimes classical quantities have their quantum counterparts. You may try, if you want, to understand how a classical vector is equal to a matrix μσ, and maybe you will discover something—but don't break your head on it. That's not the idea—they are not equal. Quantum mechanics is a different kind of a theory to represent the world. It just happens that there are certain correspondences … there is no mystery in the fact that in classical mechanics there is some shadow of quantum mechanical laws—which are [the true laws of nature]. To reconstruct the original object from the shadow is not possible in any direct way, but the shadow does help you to remember what the object looks like. [H=–µσ•B] is the truth, and [U=–µ•B] is the shadow. Because we learn classical mechanics first, we would like to be able to get the quantum formula from it, but there is no sure-fire scheme for doing that. We must always go back to the real world [to] discover the correct quantum mechanical equations… If the warnings above seem repetitious …, please excuse the conditioned reflexes of a professor who has usually taught quantum mechanics to students who hadn't heard about Pauli spin matrices until they were in graduate school. Then they always seemed to be hoping that, somehow, quantum mechanics could be seen to follow as a logical consequence of classical mechanics which they had learned thoroughly years before. (Perhaps they wanted to avoid having to learn something new.)” The Hamiltonian equation for an electron in a magnetic field can be written: iħ d|Ψ>/dt = –µ σ•B |Ψ> Now let’s write this as a matrix equation by replacing |Ψ> with state vector (C , C ) and replacing σ with the full Pauli matrices: +

–

Using the products of Pauli matrices and state vectors listed at the end of Chapter 14, the above matrix equation produces two component equations: iħ dC /dt = –µ { B C –B iC +B C } iħ dC /dt = –µ { B C +B iC –B C } +

x

–

y

–

z

+

–

x

+

y

+

z

–

iħ dC /dt = –µ B C –µ { B – iB } C iħ dC /dt = +µ B C –µ { B + iB } C +

z

+

x

y

–

–

z

–

x

y

+

This is the same set of equations, with different constants, that we found for the ammonia molecule in Chapter 10. The solution there, as here, has two stationary states of different energy that are each mixtures of the C and C spin states. +

–

Photon Polarization States Throughout this section, we orient the +z-axis along light’s direction of motion. Recall from Feynman Simplified 1C Chapter 36, that photons have two spin states. For basis states, we can choose: x- and y-polarizations, or left-handed and right-handed circular polarizations. Photons are spin 1 particles, which normally have three spin states, such as +1, 0, and –1. But spin 0 is not possible for photons, because they travel at the speed of light. We will ignore the variety of possible momentum states that photons may have, and consider only their two spin states. In V3p11-9, Feynman says: “In the classical theory, light can be described as having an electric field which oscillates horizontally or an electric field which oscillates vertically … “Now, however, suppose we have a single photon—just one. There is no electric field that we can discuss in the same way. All we have is one photon. But a photon has to have the analog of the classical phenomena of polarization. There must be at least two different kinds of photons. At first, you might think there should be an infinite variety—after all, the electric vector can point in all sorts of directions. We can, however, describe the polarization of a photon as a twostate system … [For a photon moving in the +z-direction] two base states |x> and |y> … are all

that are needed to describe any photon at all.” In that same chapter we discussed calcite and Polaroid filters. Calcite crystals, also called Icelandic spar, exhibit anomalous refraction because their refractive index is different along different axes. Calcite physically separates a beam of polarized photons into two beams, one with x-polarization and the other with y-polarization. This is entirely analogous to the Stern-Gerlach device that physically separates particle beams into beams of different spin states. This means all our discoveries about particle spin-states and Stern-Gerlach devices apply directly to photon spin states and calcite crystals. Let’s now examine Polaroid filters from a quantum perspective. Consider a Polaroid filter set to pass light in the x-direction. Photons with |y> polarization are absorbed by the filter, whereas those with |x> polarization pass freely through the filter and remain in the |x> state. What about photons polarized in another direction? Consider a photon polarized in the +x*-direction, where the x*-axis is obtained by rotating the x-axis by angle θ about the z-axis. In V3p11-10, Feynman says polarization in the x*-direction can be represented by a linear combination of polarizations in the x- and y-directions as: |x*> = cosθ |x> + sinθ |y> This means the amplitude for a photon entirely polarized in the x*-direction to pass through a Polaroid filter that passes x-polarization is: = cosθ which corresponds to a probability of || =cos θ. Feynman gives the example that for θ=30º, sinθ=1/2 and cosθ=√(3/4), so cos θ=3/4. This means the probability of a photon with polarization x* passing through a Polaroid filter set at 30º to x* is 75%, while the probability of that photon being absorbed in the filter is 25%. 2

2

2

In classical physics (1C Chapter 36), we say a beam of light whose electric field E is entirely in the x*-direction has a component of that field in the x-direction of Ecosθ, if x* is rotated by angle θ relative to x. The intensity of light (its energy) that passes through a Polaroid filter is E cos θ. 2

2

Both quantum physics and classical physics give the same pass ratio of cos θ — the two theories must agree in the classical limit. But the interpretation is different. As Feynman says in V3p11-11: 2

“The classical picture and the quantum picture give similar results. If you were to throw 10 billion photons at the [Polaroid], and the average probability of each one going through is, [quantum mechanically] 3/4, you would expect 3/4 of 10 billion would get through. Likewise, the energy that they would carry would be 3/4 of the energy that you attempted to put through. The classical theory says nothing about the statistics of the thing—it simply says that the energy that comes through will be precisely 3/4 of the energy which you were sending in. That is, of

course, impossible if there is only one photon. There is no such thing as 3/4 of a photon. It is either all there, or it isn't there at all. Quantum mechanics tells us it is all there 3/4 of the time. The relation of the two theories is clear.” Let’s now compare the quantum and classical views of circular polarization. Classically, right-hand circular (RHC) and left-hand circular (LHC) polarizations have equal x- and y-components, but the components are phase shifted. From 1C Chapter 36, these equations relate RHC and LHC to x- and y-polarizations: E E

RHC LHC

= (cos[ωt], +sin[ωt], 0)/√2 = (cos[ωt], –sin[ωt], 0)/√2

{E +E }/√2 = (cos[ωt], 0, 0)= x polarization {E –E }/√2 = (0, sin[ωt], 0) = y polarization RHC

LHC

RHC

LHC

Quantum mechanically, RHC and LHC have equal x- and y-amplitudes, but the amplitudes are phase shifted. |RHC> = (|x> + i|y>)/√2 |LHC> = (|x> – i|y>)/√2 |x> = (|RHC> + |LHC>)/√2 |y> = (|RHC> – |LHC>)/(i√2) Note that |RHC> and |LHC> are orthogonal: = |RHC>* |LHC> = {()/√2} = { –i –i –}/2 = {1 –0 –0 –1}/2 = 0 You might try to prove that |RHC> and |LHC> each have unit magnitude. In V3p11-11, Feynman explores what he considers “a curious point.” It seems that RHC and LHC polarized light should have no relation to the x- and y-axes. For light moving in the +z-direction, we can rotate the x- and y-axes by any angle about the z-axis without changing the meaning of right- and left-handedness. A right-hand screw remains a right-hand screw no matter how you turn it. This means one doesn’t need any specific set of xy-axes to define the handedness. Yet, the sum of |RHC> and |LHC> do define a specific x-axis. How is that possible if RHC and LHC are axis-independent? The answer is in the phase angles. Feynman demonstrates this by defining x* and y* coordinates that are rotated relative to x and y by angle θ about the z-axis, and by defining new circular polarization states |RHC*> and |LHC*> as:

|x*> = cosθ |x> + sinθ |y> |y*> = cosθ |y> – sinθ |x> |RHC*> = (|x*> + i|y*>)/√2 |LHC*> = (|x*> – i|y*>)/√2 |RHC*>=(cosθ|x>+sinθ|y>+icosθ|y>–isinθ|x>)/√2 |RHC*> = (|x>+i|y>)(cosθ–isinθ)/√2 |RHC*> = |RHC> exp{–iθ} |LHC*>=(cosθ|x>+sinθ|y>–icosθ|y>+isinθ|x>)/√2 |LHC*> = (|x>–i|y>)(cosθ+isinθ)/√2 |LHC*> = |LHC> exp{+iθ} What we find is that rotating xy-axes changes the phase angles but not the handedness. The phase angles preserve the identity of the xy-axes that defined them.

Chapter 15 Review: Key Ideas Using spin basis states |1>=|+z> and |2>=|–z>, the Hamiltonian for a particle with magnetic moment µ in a magnetic field B=(B ,B ,B ) is: x

H H H H

11 12 21 22

y

z

= –µ B = –µ (B – iB ) = –µ (B + iB ) = +µ B z

x

y

x

y

z

Feynman says this result is also valid even in time-varying magnetic fields. The projection amplitudes that relate spins along the z-axis with spins along another axis z*, whose orientation is defined by the polar angle θ and the azimuthal angle ø, are: = cos(θ/2) exp{–iø/2} = sin(θ/2) exp{+iø/2} A spin 1/2 particle, with magnetic moment µ and in a pure state of spin-up along the z*-axis at t=0, is exposed to a magnetic field B in the +z-direction. Let axis z* be rotated from axis z by polar angle θ and azimuthal angle ø. The particle’s state becomes the superposition of two states of definite energy, spin |+z> and |–z>: |Ψ> = C |+z> + C |–z> 1

with:

2

C (t) =cos(θ/2) exp{–iø/2} exp{+iµBt/ħ} C (t) = sin(θ/2) exp{+iø/2} exp{–iµBt/ħ} 1 2

The azimuthal angle precesses about the z-axis with frequency ω=–2µB/ħ. Magnetic moment µ is negative for negatively charge particles. The Hamiltonian equation for an electron in a magnetic field can be written in terms of Pauli spin matrices: iħ d |Ψ>/dt = –µ σ•B |Ψ>

Chapter 16 Hyperfine Splitting in Hydrogen

We will now extend our reach to a system with four spin states. As Feynman says, in V3p12-1, this is only “slightly more complicated” but “enough more complicated” that it can be generalized to “all kinds of problems.” A hydrogen atom consists of an electron bound to a proton by their mutual electrical attraction. In Chapters 3 and 6, we explored the electron orbits in hydrogen atoms. Recall that the electron orbits are grouped in shells denoted n=1, 2, 3…., where n is the principal quantum number. The energy of shell n, is: E = –e /2r = –13.61 eV / n 2

2

n

Here, E=0, the zero-point of the energy scale, is defined to be the energy of an electron and a proton that are infinitely far part. We call n=1 the ground state, and n=2 the first excited state. But our earlier discussions ignored the spin of both particles. Indeed, each orbital state is actually further subdivided into four states corresponding to the four combinations of: electron spin up and down; and proton spin up and down. These four spin states have slightly different energy levels, but these energy differences are minute compared with the energy differences between orbital states. For example, the separation between the n=1 and n=2 orbital states is about 10 eV, while the energy splitting between spin states is 100 million times less; hence the name hyperfine splitting. Hyperfine splitting results from the interaction of the electron and proton magnetic moments. To limit complexity, we will consider here only the four spin states of a hydrogen atom in the n=1 orbital ground state.

Basis States of Two Spin ½ Particles Step 1 in analyzing a quantum system is defining basis states, of which there will be four in this case. One might wonder: which of all possible choices of basis states is best? All are equally legitimate, but Feynman recommends the basis states that are “physically clearest.” He says: “It may not be the

solution to any problem, or may not have any direct importance, but it will generally make it easier to understand what is going on.” We therefore choose the most obvious basis states. With e denoting electron and p denoting proton, and up and down measured in the +z-direction, we define these basis states: |1> = |e+,p+>, e spin up, p spin up |2> = |e+,p–>, e spin up, p spin down |3> = |e–,p+>, e spin down, p spin up |4> = |e–,p–>, e spin down, p spin down With these basis states, any state |Ψ> is a linear combination of the form: |Ψ> = Σ C |j>, with C = j

j

j

In V3p12-2, Feynman addresses a common concern of students of quantum mechanics: “You may say, ‘But the particles interact, and maybe these aren't the right base states. It sounds as though you are considering the two particles independently.’ Yes, indeed! The interaction raises the problem: what is the Hamiltonian for the system, but the interaction is not involved in the question of how to describe the system. What we choose for the base states has nothing to do with what happens next. It may be that the atom cannot ever stay in one of these base states, even if it is started that way. That's another question. That's the question: How do the amplitudes change with time in a particular (fixed) base? In choosing the base states, we are just choosing the ‘unit vectors’ for our description.” Choosing basis states is equivalent to choosing coordinate axes in classical physics.

Basis States of More General Systems Here Feynman briefly pauses for an aside on choosing basis states for more general multi-particle systems. He says a spin 1/2 particle is completely described by the amplitudes to be in each of two basis states: | momentum p, spin up> | momentum p, spin down> Here we have two infinite sets of states, since there are an infinite number of values of momentum vector p. For multiple spin 1/2 particles, we expand the set of basis states in the obvious manner. For two spin 1/2 particles, #1 and #2: |#1: p , spin up; #2: p , spin up> |#1: p , spin up; #2: p , spin down> |#1: p , spin down; #2: p , spin up> 1

2

1

2

1

2

|#1: p , spin down; #2: p , spin down> 1

2

Hamiltonian of Four Spin States The Hamiltonian specifies how each amplitude C evolves over time, where: j

C = and |Ψ> = Σ C |j> j

j

j

Whereas choosing basis states is arbitrary, choosing the Hamiltonian is not. In V3p12-3, Feynman says there is no general procedure for determining a Hamiltonian that he can articulate at this stage of our understanding of quantum mechanics. The only proof that we have the right Hamiltonian is that experiments validate it. But, we can give some reasonable arguments justifying the Hamiltonian that Feynman presents. From our discussion of the ammonia maser, recall that two states split due to their potential energy difference in an electric field. That energy term is µ•E, where µ is ammonia’s dipole moment and E is the electric field. The only physical entities in our hydrogen atom are the magnetic moments of the electron and the proton, the vectors µ and µ , respectively. Since energy is a scalar quantity, we must look for a combination of two vectors that produces a scalar. e

p

Classically, the proton produces a magnetic field B proportional to its magnetic moment µ . The electron’s potential energy is therefore proportional to the dot product of B and its magnetic moment µ . This means the potential energy term must be proportional to µ •µ . p

e

e

p

Now recall what we discovered about the Hamiltonian of an electron in a magnetic field. Classically the electron’s energy is E +µ •B, where E is the energy that the electron would have with no magnetic field. We found that in quantum mechanics, the corresponding Hamiltonian is E +µ σ•B, where σ is the vector whose components are the Pauli spin matrices. 0

e

0

0

e

This suggests a Hamiltonian of the form: H = E δ + A σ •σ e

0

p

jk

where A is an undetermined constant, and σ and σ are the Pauli spin vector matrices for the electron and proton respectively, which are further described below. The potential energy of two dipoles depends in part on their spatial separation d, being proportional to 1/d . Yet we see no explicit distance dependence in this Hamiltonian. This is due to our assumption that the atom is in its orbital ground state, which specifies a distance distribution. The proportionality constant A reflects that average dipole separation, and also the magnetic moments of both the electron and the proton. e

p

3

In V3p12-5, Feynman cautions that the “classical qualitative picture” presented above may provide

insight and assist understanding, but such classical arguments cannot be relied upon in the quantum world. The proof of the Hamiltonian lies solely in its validation by experiment. He adds that in a complete quantum theory of the hydrogen atom, which we haven’t yet reached in this course, the constant A can be calculated to a precision of 30 parts per million. Each of σ and σ has exactly the same form as the Pauli spin vector matrix of Chapter 14, but each operates only on the spin of its designated particle. Thus σ does not alter proton spins, and acts on them like the identity operator δ . Similarly σ does not alter electron spins. e

p

e

p

jk

Each vector matrix has three components: σ , σ , and σ , as shown below. x

y

z

where the superscript F (F for fermion) is in this case either e for electron or p for proton. In Chapter 14, we found that the Pauli spin matrices operate on z-spin states as follows: σ σ σ σ σ σ

F z F z F x F x F y F y

|+z> = + |+z> |–z> = – |–z> |+z> = + |–z> |–z> = + |+z> |+z> = +i |–z> |–z> = –i |+z>

Any product of such operators, such as σ σ , is treated like any other operator product. The order of operation is from right to left. Thus: e

p

x

σ σ |e+,p–> = σ {σ |e+,p–>} σ σ |e+,p–> = σ (–1) |e+,p–> σ σ |e+,p–> = – |e–,p–> e

p

x

e

e

z

p

x

e

e

z

p

x

p

x

z

x

z

z

The first line states that the right-most operator acts first. The second line shows that σ returns the sign of the proton’s z-spin multiplied by the original state. The third line shows that σ flips the z-spin of the electron. Also note that the product of any matrix M (or vector v) by any scalar s is commutative: sM=Ms (sv=vs). p z

e

x

The following products of operator pairs will be most useful: σ σ σ σ

e

σ σ σ σ

e

σ σ σ σ

e

x e x e x e x

y e y e y e y

z e z e z e z

σ σ σ σ

p x

σ σ σ σ

p y

σ σ σ σ

p x p x p x

p y p y p y

p z p

|e+,p+> = + σ |e+,p–> = + σ |e–,p+> = + σ |e–,p–> = + σ

x

e x

e x

e x

|e+,p+> = +i σ |e+,p–> = –i σ |e–,p+> = +i σ |e–,p–> = –i σ

e

|e+,p+> = + σ |e+,p–> = – σ |e–,p+> = + σ |e–,p–> = – σ

|e+,p+> = + |e+,p+> |e+,p–> = – |e+,p–> |e–,p+> = – |e–,p+> |e–,p–> = + |e–,p–>

z z

y

e

e z z z

e

z

y

e

e

p

y

e

e

p

|e+,p–> = + |e–,p–> |e+,p+> = + |e–,p+> |e–,p–> = + |e+,p–> |e–,p+> = + |e+,p+>

e

z

y

|e+,p–> = – |e–,p–> |e+,p+> = + |e–,p+> |e–,p–> = + |e+,p–> |e–,p+> = – |e+,p+>

In V3p12-4, Feynman notes that the identity operator δ plus the three σ components makes a total of four operators, as do δ plus the three σ components. We can therefore form 16 different operator product pairs by multiplying one operator from the first group by another operator from the second group. That number matches the 16 components in the 4×4 Hamiltonian of a 4-state system. Feynman says any 4×4 matrix can be expressed as a linear sum of the 16 spin-matrix-product-pairs just enumerated. e

jk

p

jk

The Hamiltonian suggested above is: H = E δ + A σ •σ H = E δ + A {σ σ + σ σ + σ σ } e

0

jk

0

jk

p

e

p

x

e

x

p y

y

e

p

z

z

As before, we will reset the zero-point of our energy scale to make E =0, for now. This will simplify the math, and we can always restore E later. 0

0

Our next task is to calculate the 16 components of H. Ten of these components are independent, and the other 6 are determined by the relation H = H *. kj

jk

Let’s begin by calculating the effect of H operating on the first basis state. H |e+,p+> = A {σ σ + σ σ + σ σ } |e+,p+> H |e+,p+> = A {+|e–,p–> –|e–,p–> +|e+,p+>} H |e+,p+> = +A |e+,p+> e

p

x

e

x

p

y

e

y

p

z

z

This means: H H H H

11 12 13 14

= = +A = = 0 = = 0 = = 0

We continue with H operating on the second basis state. H |e+,p–> = A {σ σ + σ σ + σ σ } |e+,p–> H |e+,p–> = A {+|e–,p+> +|e–,p+> –|e+,p–>} H |e+,p–> = +2A |e–,p+> –A |e+,p–> e

p

x

H H H H

21 22 23 24

e

x

p

y

e

y

z

p z

= = 0 = = –A = = +2A = = 0

Next, is H operating on the third basis state. H |e–,p+> = A {σ σ + σ σ + σ σ } |e–,p+> H |e–,p+> = A {+|e+,p–> +|e+,p–> –|e–,p+>} H |e–,p+> = +2A |e+,p–> –A |e–,p+> e

p

x

H H H H

31 32 33 34

e

x

p

y

e

y

z

p z

= = 0 = = +2A = = –A = = 0

And finally, we have H operating on the fourth basis state. H |e–,p–> = A {σ σ + σ σ + σ σ } |e–,p–> H |e–,p–> = A {+|e+,p+> –|e+,p+> +|e–,p–>} H |e–,p–> = +A |e–,p–> e

p

x

H H H H

41 42 43 44

e

x

p

y

e

y

p

z

z

= = 0 = = 0 = = 0 = = +A

All this results in the Hamiltonian in matrix form:

Since iħ dΨ/dt = HΨ, this Hamiltonian matrix is exactly equivalent to these four equations: iħ dC /dt = +A C iħ dC /dt = –A C + 2A C iħ dC /dt = +2A C –A C iħ dC /dt = +A C 1

1

2

2

3

3

2

4

3

4

Solving the Hamiltonian Equations We can now calculate the stationary states and their energy levels — the eigenvectors and eigenvalues of the Hamiltonian. Let’s do this first using the matrix mathematics of Chapter 14, and then do it again with the component approach, for comparison. Recall that we define a matrix and set its determinant equal to zero, as follows: M = H – Eδ jk

jk

jk

Det M = 0 jk

The large number of zero components simplifies this determinant. We will proceed across the first row since it has only one non-zero component: M . The determinant of M equals M multiplied by the determinant of the minor of M . Proceeding iteratively, the determinant of the 3×3 minor of M equals M multiplied by the determinant of the 2×2 minor of M . This 2×2 minor is what remains of the original matrix after eliminating rows 1 and 4 and columns 1 and 4. The determinant of the 2×2 minor is M M –M M .The characteristic polynomial is therefore: 11

11

11

11

44

44

22

33

23

32

0 = (A–E)•(A–E)•{(–A–E) –4A } 0 = (A–E) {A +2AE +E –4A } 0 = (A–E) {E +2AE –3A } 0 = (E–A) {(E–A)(E+3A)} 2

2

2

2

2

2

2

2

2

2

The 4 eigenvalues are the 4 roots of this equation, which clearly are: E = A, A, A, and –3A Now we find the eigenvectors. Since the non-diagonal components are all zero in the first and fourth rows and columns, the first two eigenvectors are particularly easy. Let |Ψ> = (w,0,0,z). E |Ψ> = H |Ψ> E (w, 0, 0, z) = (Aw, 0, 0, Az) Two solutions are evident, both with energy E=A. w=1 and z=0, |Ψ> = (1, 0, 0, 0) w=0 and z=1, |Ψ> = (0, 0, 0, 1) Next we solve the eigenvector equations for the second and third rows, for an eigenvector of the form: |Ψ> = (0,x,y,0). E |Ψ> = H |Ψ> E (0, x, y, 0) = +A (0, –x+2y, 2x–y, 0) E (0, x, y, 0)= +A(0, –x+2y, 2x–y, 0) The first equation is solved by x=+y, which means: |Ψ> = (0,1,+1,0)/√2 with E=A. The second equation is solved by x=–y, which means: |Ψ> = (0,1,–1,0)/√2 with E=–3A. We will label the four eigenstates: a, b, c, and d. The four eigenvalues and their eigenvectors are thus: E E E E

a b c d

= +A; |a> = (+1, 0, 0, 0) exp{E t/iħ} = +A; |b> = ( 0, 0, 0,+1) exp{E t/iħ} = +A; |c> = ( 0, 1,+1, 0) exp{E t/iħ}/√2 =–3A;|d> = ( 0, 1,–1, 0) exp{E t/iħ}/√2 a

b

c

d

Note that the four eigenvectors are all orthonormal, as expected. All other solutions of the hydrogen atom can be obtained from linear sums of these stationary states. In the two-state case, the two energy levels were E +A and E –A, so it didn’t matter much if A was positive or negative. Here it does matter. In V3p12-12, Feynman says that A is definitely positive, as shown both by experiment and by the complete quantum theory. 0

0

Note also that the lowest energy state is the antisymmetric superposition of antiparallel spins. Since the electron and proton have opposite electric charges, when their spins are antiparallel their magnetic moments are parallel. Classically, that would be the higher energy state of two separated dipoles. But quantum mechanically, the particles are not separated. Their wave functions overlap and are concentric — the dipole moments are on top of one another. And, as we know from our macroscopic experience, magnetic dipoles arise from electric currents, and parallel electric currents attract one another. We have finished the matrix approach, now let’s compare that with the component approach. The first and fourth Hamiltonian equations are: iħ dC /dt = +A C iħ dC /dt = +A C a

a

d

d

Again, we see that C and C are both stationary states with energy E=A. a

d

Now examine the second and third Hamiltonian equations. iħ dC /dt = –A C + 2A C iħ dC /dt = +2A C –A C b

b

c

c

b

c

Add the first equation to the second equation. iħ d(C +C )/dt = +A C +A C b

c

b

c

This means C +C , properly normalized, is a stationary state with energy E=A. b

c

Now subtract the second equation from the first. iħ d(C –C )/dt = –3A C +3A C = –3A (C –C ) b

c

b

c

b

c

This means (C –C ), properly normalized, is a stationary state with energy E=–3A. b

c

The four eigenvalues and their eigenvectors are exactly what we derived in the matrix approach.

Pauli Spin Exchange Operator In V3p12-4, Feynman says he: “can’t resist telling you about a clever rule due to Dirac—it will make you feel really advanced—although we don’t need it for our work.” From the Hamiltonian equations above we have: σ •σ |e+,p+> = +|e+,p+> σ •σ |e+,p–> = 2|e–,p+> – |e+,p–> e

p

e

p

σ •σ |e–,p+> = 2|e+,p–> – |e–,p+> σ •σ |e–,p–> = +|e–,p–> e

p

e

p

Dirac rewrote the first and last equations, with this nicely symmetric result: σ •σ σ •σ σ •σ σ •σ e

p

e

p

e

p

e

p

|e+,p+> = 2|e+,p+> – |e+,p+> |e+,p–> = 2|e–,p+> – |e+,p–> |e–,p+> = 2|e+,p–> – |e–,p+> |e–,p–> = 2|e–,p–> – |e–,p–>

We therefore define the Pauli spin exchange operator: P , which exchanges the electron and proton spins. swap

P P P P

swap swap swap swap

|e+,p+> = |e+,p+> |e+,p–> = |e–,p+> |e–,p+> = |e+,p–> |e–,p–> = |e–,p–>

Using P

we can condense the prior four equations into one.

swap

σ •σ = 2P e

p

swap

–1

Feynman ends this section with this exuberant comment: “You see, you can do everything now. The gates are open.” Even if that hyperbole were true, I would still recommend reading the rest of Feynman Simplified, if only for fun.

Hyperfine Splitting We found above that the energy levels of the four stationary spin states of the orbital ground state of the hydrogen atom are: E =A, E =A, E =A, and E =–3A. The average of these energies is zero, or more properly E if we restore the zero-point energy. a

b

c

d

0

A hydrogen atom with E=A can transition to the lowest state, E=–3A, by emitting a photon of energy E=4A. Conversely, a hydrogen atom with energy E=–3A can transition to one of the higher states by absorbing a photon of energy E=4A. Such a photon is in the microwave portion of the electromagnetic spectrum. Its frequency has been measured to a precision of about 10 parts per trillion. The experimental precision is about one million times better than the precision of the theoretical calculations. The measured values are: f = 1420.405,751,77 MHz λ = 21.106,114,051,3 cm 4A = 5.87433 micro-eV

This is the famous “21-cm” line, an extremely powerful tool for probing the universe. Its principal advantages are the ubiquity of hydrogen (92% of all atoms in the universe) and the extraordinary narrowness of the spectral line. With a lifetime of 10 million years, the line width is only 2×10 eV, for a resonance Q factor of over 30 trillion, trillion. A sharp line is very useful when measuring Doppler shifts due to relative motion or cosmic expansion. In turn, the redshift due to cosmic expansion determines the age and distance of everything in the universe that either emits or absorbs light; see Our Universe 5: How I Wonder Where You Are for a complete explanation. –30

Zeeman Splitting After we found the stationary states of the ammonia molecule in Chapter 10, we went on to study its interaction with an external electric field in Chapter 11. That led to an understanding of the ammonia maser. We will now do something similar: examine the interaction of the hyperfine states of hydrogen in a magnetic field. Recall from Chapter 15 that the Hamiltonian for an electron in a magnetic field is: H = E – µ σ•B 0

Since different types of energy combine linearly, it is reasonable that adding an external magnetic field simply adds another term to the Hamiltonian. With that assumption, the Hamiltonian for a hydrogen atom, in its orbital ground state, in an external magnetic field, is: H = A σ •σ – µ σ •B – µ σ •B e

p

e

p

e

p

Feynman says, in V3p12-9, that the proof this is the correct Hamiltonian is that it agrees with experiment. We will simplify the math by orienting the magnetic field in the +z-direction, and reduce clutter by defining: β = –(µ + µ ) B β = –(µ – µ ) B +

e

p

–

e

p

Note that –µ equals about 1000 µ , since magnetic moments are inversely proportional to mass (and several other factors). This means β and β are nearly equal, and as we have defined them, both are positive. e

p

+

–

With our choice of magnetic field direction, the added portion of the Hamiltonian has a particularly simple effect on the basis states. H = A σ •σ – (µ Bσ +µ Bσ ) e

p

e

e z

p

p z

–(µ Bσ +µ Bσ )|e+,p+> = (–µ –µ )B|e+,p+> –(µ Bσ +µ Bσ )|e+,p–> = (–µ +µ )B|e+,p–> e

e z

e

e z

p

p z

e

p

p

p z

e

p

–(µ Bσ +µ Bσ )|e–,p+> = (+µ –µ )B|e–,p+> –(µ Bσ +µ Bσ )|e–,p–> = (+µ +µ )B|e–,p–> e

e z

e

e z

–(µ Bσ –(µ Bσ –(µ Bσ –(µ Bσ e

e z

e

e z

e

e z

e

e z

p

p z

e

p

p

p z

e

p

+µ Bσ +µ Bσ +µ Bσ +µ Bσ p

p z

p

p z

p

p z

p

p z

) |e+,p+> = +β ) |e+,p–> = +β ) |e–,p+> = –β ) |e–,p–> = –β

|e+,p+> |e+,p–> |e–,p+> |e–,p–>

+

–

–

+

Note that the additional term never changes one basis state into another; it simply changes the coefficient of that state. (That’s why we picked B in the z-direction.) This means only the diagonal components of the Hamiltonian change. The new Hamiltonian is:

The characteristic equation becomes: 0=(A+β –E)•(A–β –E)•{ } +

+

where { } = { (–A+β –E)(–A–β –E) –4A )} 2

–

–

The first two roots are easy: E = A +β E = A –β a

+

b

+

Now the { }: 0 = ([A–β ]+E)([A+β ]+E) –4A 0 = A –β +E[2A] +E –4A 0 = –β +2AE +E –3A E = –A ± √(A + 3A +β )

2

–

2

– 2

2

2

–

2

2

2

–

2

2

2

–

The four eigenvalues are: E E E E

a b c d

= +A +β = +A –β = –A + A√(4 +β /A ) = –A – A√(4 +β /A ) +

+

2 – 2 –

2

2

Feynman suggests we check our results in the limit that B=0, where β =β =0. As B goes to zero… +

–

E E E E

a b c d

goes to +A goes to +A goes to –A +A√4 = +A goes to –A –A√4 = –3A

which checks with what we had before without a magnetic field. The four energy levels are plotted in Figure 16-1 as functions of the external magnetic field strength.

Figure 16-1 Energy Levels of Orbital Ground State of Hy drogen Atom vs. Magnetic Field Strength

This shifting of the energy levels as a function of the applied magnetic field is called Zeeman splitting. Without the magnetic field, there is only one emission/absorption spectral line — the 21-cm line at 1420 MHz., corresponding to an energy difference of 4A between state |d> and any of the other three states. With a magnetic field, the state energy levels of the first three states separate, resulting in four distinct energy levels, as Figure 16-1 shows. This means there are now six possible transitions, each with a different energy change and spectral frequency, as shown in Figure 16-2.

Figure 16-2 Transitions Between Energy Levels

Here, the four dashed horizontal lines indicate the four energy levels at a selected magnetic field value (the dashed vertical line). Each of the six possible combinations of two of four states has a transition with a unique energy and spectral frequency. In V3p12-12, Feynman says one can drive these transitions by applying an additional oscillating magnetic field in the xy-plane. This is the same physics that we explored with the ammonia maser; the only difference is that there we dealt with electric fields and here we have magnetic fields.

Eigenvectors & Quantum Numbers Let’s now further examine the four spin-states of the orbital ground state of hydrogen. We will focus on how the energy levels and the corresponding eigenvectors change as a function of the applied external magnetic field B. We can understand the energy relationships of the four states from the basic physics. At B=0, the eigenvalues E and eigenvectors are determined solely by the interaction of the electron and proton magnetic moments. But both change in an external magnetic field that is stronger than the dipole moments of the two particles. n

For very large B, where µB/A>>2, the four eigenvalues, the energy levels of the stationary states, become: E = +A +β E = +A –β E = –A +β a

+

b

+

c

–

E = –A –β d

–

Recall that β and β are nearly equal, that both are positive, and that: +

–

β = –(µ + µ ) B β = –(µ – µ ) B +

e

p

z

–

e

p

z

For large external fields, the electron and proton magnetic moments act independently in response to the field. The eigenstates and their eigenvector representation in the basis states change, as shown in the following table.

If the electron is spin up, parallel to the B field, its dipole moment is antiparallel to B (recall µ =|e+,p+> is the highest energy state. If the proton’s spin is flipped to down, its energy drops by about 2A, which corresponds to energy E ; thus at large B, |c> becomes |e+,p–>. e

c

If the electron is spin down, antiparallel to the B field, its dipole moment is parallel to B, which is the lowest energy orientation. If the proton is spin up, its dipole moment is parallel to the electron’s dipole moment, which decreases the state’s energy by A. Thus at large B, the lowest energy state is |e–,p+>, which is what |d> becomes. Finally, if the proton’s spin is flipped to down, its energy increases by about 2A, which corresponds to energy E . Thus at large B, |b> becomes |e–,p–>. b

We can show this effect analytically by calculating the eigenvectors of the mixed states. Let |Ψ> = (0,x,y,0). The eigenvector equation is: H |Ψ> = E |Ψ> (0, [–A+β ]x+2Ay, +2Ax–[A+β ]y, 0) = (0, Ex, Ey, 0) –

–

Examine the third component. 2Ax = (E+A+β )y x/y = (E+A+β ) / 2A –

–

For |c> with energy E =–A+A√(4 +β /A ): 2

c

–

2

x/y = {β + A√(4 +β /A )} / 2A 2

–

2

–

For B=0, x/y = 2A/2A = 1, meaning that |c> is a symmetric mixture of equal parts of |2> and |3>. But at very large B, x/y = 2β /2A, which goes to infinity, meaning that |c> becomes entirely |2>. –

For |d> with energy E =–A–A√(4 +β /A ): 2

d

2

–

x/y = {β – A√(4 +β /A )} / 2A 2

–

2

–

At B=0, x/y = –2A/2A = –1, meaning that |d> is an antisymmetric mixture of equal parts of |2> and |3>. At very large B, x/y goes to zero, meaning that |d> becomes entirely |3>. In V3p12-13, Feynman calls attention to the behavior of these states in a very weak magnetic field. Neglecting β and higher powers of the weak field, the energy levels of the eigenstates become: 2

E E E E

a b c d

= +A +β = +A –β = +A +β /8A = +A = –3A –β /8A = –3A +

+

2 –

2

–

Thus, E and E are approximately constant, while E increases with B and E decreases with B, as shown in Figure 16-3. c

d

a

b

Figure 16-3 Energy Levels vs. Magnetic Field for Weak Fields & Magnetic Quantum Number m

Now consider a beam of hydrogen atoms in their orbital ground state passing through a Stern-Gerlach device with a weak magnetic field. A beam of atoms selected to have energy E=–3A would pass undeflected. Since their energy is unaffected by a magnetic field, they experience no deflecting forces. Recall that force equals the negative gradient of potential energy. Conversely, a beam of atoms selected to have energy E=+A separates into three beams. State |a> deflects toward a weaker field, |b> deflects toward a stronger field, and |c> is undeflected. This behavior is exactly what we encountered in Chapter 7 for spin 1 particles.

We therefore say that the orbital ground state of hydrogen consists of a singlet state with spin 0 and a triplet state with spin 1. In this case, an atom composed of two fermions acts like a boson. Feynman describes these states using the same terminology used to describe the atomic orbits of electrons: n, the principal quantum number; j, the orbital quantum number; and m, the magnetic quantum number. All three of these quantum numbers must be integers. Modern convention employs a lower case l rather than j, but I will follow Feynman and use j to enhance eBook visibility. The principal quantum number n, also called the electron shell number, was described earlier in this chapter. It defines the average distance from the nucleus, which is the most important factor in determining the electron’s binding energy. The minimum value of n is 1, and the maximum is 7, for known elements. The orbital quantum number j, also called the angular momentum quantum number, specifies the orbit’s angular momentum L, with L =j(j+1)ħ ; j has an allowed range of j=0,1,2…(n–1). 2

2

The magnetic quantum number m, specifies the z-component of angular momentum L , with L =mħ; m has an allowed range of –j,…,0,…,+j. z

z

The electron ground state of any atom has quantum numbers n=1, j=0, m=0. Applying this terminology to an electron-proton pair is unconventional, but descriptive. The lowest energy state |d> with E=–3A has zero angular momentum, hence j=0 and m=0. If this atom were considered to be a single particle, its spin would be 0. Since a spin 0 particle has only one state, this is called a singlet state. The higher energy states with E=+A have z-components of spin equal to +1, 0, and –1, corresponding to a j=1 state with m=+1, 0, and –1 sub-states.

Chapter 16 Review: Key Ideas The orbital ground state of an electron in any atom has quantum numbers n=1, j=0, m=0, where n, is the principal quantum number; j, is the orbital quantum number; and m, is the magnetic quantum number. All three of these quantum numbers must be integers. Modern convention employs lower case l rather than j, but I follow Feynman’s notation to enhance eBook visibility. Each orbital electron state is further subdivided into spin states. The orbital ground state of hydrogen, for example, has four spin states corresponding to the four electron/proton up/down spin combinations. The Hamiltonian for the spin interaction is: H = A σ •σ e

p

Here, σ and σ are the Pauli spin vector matrices for the electron and proton respectively. Three of these hydrogen states form a spin 1 triplet, with m=+1, 0, and –1. The fourth is a spin 0 singlet state. e

p

With no external magnetic field, the triplet states have the same energy, which is 6 µeV above the singlet state energy. This hyperfine splitting of the energy levels enables transitions with an extremely sharp photon emission/absorption line of a frequency of 1420 MHz and a wavelength of

21-cm. This famous spectral line is a powerful tool in astronomy. When these spin states are exposed to an external magnetic field, the energy levels and composition of the stationary states change, which is called Zeeman splitting. The Hamiltonian for the spin interaction and the external magnetic field effect is: H = A σ •σ – (µ Bσ +µ Bσ ) e

p

e

e z

p

p z

As the external field increases, the electron and proton act ever more independently. The Pauli spin exchange operator, P , exchanges the electron and proton spins, allowing us to write: swap

σ •σ = 2P e

p

swap

–1

Chapter 17 Rotation for Spin ½ & Spin 1

This chapter derives rotation matrices for spin 1/2 and spin 1, all of which are tabulated at the end of this chapter. We will prove the important result that the amplitude of a spin s particle, with spin component +s along the z-axis, is multiplied by exp{isθ}, when rotated by angle θ about the z-axis. This fact was quoted in Chapter 8 to prove that spin 1/2 particles obey Fermi-Dirac statistics and spin 1 particles obey Bose-Einstein statistics. We will also prove that the amplitude for rotating N spin s particles equals the amplitude for rotating a single particle of spin Ns. This chapter is primarily a mathematical exercise; those who wish can skip to the next chapter without missing any new essential physics. We will complete here the derivation for the rotation matrices of spin 1/2 particles presented in Chapter 8, and provide a full derivation for the rotation matrices of spin 1 particles.

Arbitrary Spin ½ Rotations We didn’t show, in Chapter 8, the full derivation of the rotation of a spin 1/2 particle by arbitrary Euler angles. Let’s now complete that derivation, while reviewing what we discovered about spin 1/2 rotations. Recall that all possible rotations can be accomplished with a specific sequence of rotations by some set of three Euler angles. The rotation sequence, from coordinate system S to coordinate system T, is defined as: First, rotate about the z-axis by angle β. Second, rotate about the new x-axis by angle α. Third, rotate about the new z-axis by angle γ. This means we can calculate any rotation matrix from only two rotation matrices: R (θ), for a rotation z

about the z-axis by angle θ; and R (α), for a rotation about the x-axis by angle α. Here are the matrices for these rotations, for spin 1/2 particles, which are derived in component form in Chapter 8. x

Note, we labeled the rows and columns of R (θ) above. z

The most general rotation matrix for spin 1/2 is therefore: R(γ,α,β) = R (γ) R (α) R (β) z

x

z

Let’s calculate this first in component form. R =Σ R {Σ R R } jn

k

zjk

m

xkm

zmn

This sum is simplified by the fact that R is diagonal: R =0 unless j=k. The equation becomes: z

zjk

R =R R R jn

zjj

xjn

znn

From this, we can write down the components and define them as amplitudes a, b, c, and d. The most general equations for any rotation by Euler angles for spin 1/2 are: a = = R b = = R c = = R d = = R

11 21

12 22

= cos(α/2) exp{+i(β+γ)/2} = i sin(α/2) exp{+i(β–γ)/2} = i sin(α/2) exp{–i(β–γ)/2} = cos(α/2) exp{–i(β+γ)/2}

We can also calculate R by matrix multiplication. Starting from the right, we first multiply R (α) and R (β). jn

z

x

We next multiply that product by R (γ). z

Of course, both approaches give the same results.

Z-Rotation of Any Spin Consider now a particle of spin s, with its maximum possible spin, spin +s, along the z-axis. We want to show that rotating that particle by angle θ about the z-axis multiplies its amplitude by exp{isθ}. In the prior chapter, we discussed the orbital ground state of hydrogen, composed of two spin 1/2 fermions, one electron and one proton. We discovered that those two spin 1/2 particles can act like a single spin 1 particle. That result can be generalized to any spin: a group of 2s spin 1/2 particles can act like a single spin s particle. Let each of the 2s spin 1/2 particles have spin up along the z-axis. Since we know how to rotate a spin 1/2 particle in state |+S> through angle θ about the z-axis to state |+T>, we can calculate the result of rotating a spin s particle from |+S> through angle θ about the z-axis to state |+T>. The rotation of each spin 1/2 particle multiplies that particle’s amplitude by exp{iθ/2}. The rotation amplitude for all particles is the product of each particle’s individual amplitude. Thus, rotating all 2s spin 1/2 particles results in a total multiplicative factor of exp{isθ}, as stated in Chapter 8. This result explains why spin 1/2 particles obey Fermi-Dirac statistics and spin 1 particles obey Bose-Einstein statistics. Recall that for identical particles A and B in identical states |1> and |2>: = – for fermions = + for bosons We showed in Chapter 8 that, if A and B are adjacent, a rotation of 180º about the axis of symmetry is equivalent to exchanging the particles. This rotation multiplies the two-particle amplitude by exp{i(2s)π}, where s is the spin of A and B, and the factor of 2 arises because we rotated two

particles. For fermions, 2s is odd and exp{i(2s)π}=–1. For bosons, 2s is even and exp{i(2s)π}=+1.

Arbitrary Spin 1 Rotations We next derive the rotation matrices for the most general rotation of a spin 1 particle. We will use the rotation matrices for spin 1/2 particles just derived, and the fact that two spin 1/2 particles can act like a single spin 1 particle. In Chapter 7, we learned that a Stern-Gerlach machine S separates spin 1 particles into three spin states |+S>, |0S>, and |–S>. We also discussed the amplitudes for a particle in state |kS> to be in state |jT> with respect to a different Stern-Gerlach machine T, where j and k each have the values +1, 0, or –1. Define the axis of device S to be in the +z-direction, and the axis of device T to be in the +z*direction, with z* obtained by rotating z by angle ø about the common y-axis of both devices. With respect to device S, we can identify the three spin 1 states with the spin 1/2 triplet states of hydrogen as follows: |+S> = |e+S,p+S> |0S> = { |e+S,p–S> + |e–S,p+S> } /√2 |–S> = |e–S,p–S> Similar relationships hold for the three T-states. We now write the following expression equating two amplitudes: = The left-hand-side is the amplitude for a spin 1 object (electron+proton) in state |+S> to be in state |+T>. The right-hand-side is the product of two amplitudes: the amplitude for an electron in state |+S> to be in state |+T>; and the amplitude for a proton in state |+S> to be in state |+T>. The above expression says rotating two spin 1/2 particles is equivalent to rotating one spin 1 particle. All rotation amplitudes for electrons and protons are identical, and for this rotation both are equal to amplitude a, which we defined earlier. Therefore: = a

2

We similarly calculate these other amplitudes: = ab = ba = b 2

= ac = ad

= bc = bd = ca = cb = da = db = c = cd = dc = d 2

2

From these we can calculate the rotation amplitudes . These can be called projection amplitudes, since specifies how much of state |kS> projects onto state |jT>. = = a

2

= { + }/√2 = ab√2 = = b

2

= { + }/√2 = ac√2 = { + + + }/2 = ad+bc = { + }/√2 = bd√2 = = c

2

= { + }/√2 = cd√2 = = d

2

We next evaluate the above amplitudes in terms of the amplitudes required to rotate spin 1/2 particles by any set of Euler angles. Recall that the rotation matrix components are: a = = cos(α/2) exp{+i(β+γ)/2} b = = i sin(α/2) exp{+i(β–γ)/2} c = = i sin(α/2) exp{–i(β–γ)/2}

d = = cos(α/2) exp{–i(β+γ)/2} Before continuing, let’s calculate a quantity we will need later: ad–bc. Note that the exponents of amplitudes a and d sum to zero, as do the exponents of b and c. This means the exponents cancel one another in ad and bc. Hence: ad = cos(α/2) cos(α/2) exp{0} bc = isin(α/2) isin(α/2) exp{0} ad – bc = cos (α/2) + sin (α/2) ad – bc = 1 for any set of angles α, β, and γ 2

2

Now, let’s insert the spin 1/2 rotation amplitudes into the spin 1 amplitudes calculated above. = a = cos (α/2) exp{+i(β+γ)} = ab√2 = cos(α/2) exp{+i(β+γ)/2} × isin(α/2) exp{+i(β–γ)/2} √2 = b = –sin (α/2) exp{+i(β–γ)} 2

2

2

2

= ac√2 = cos(α/2) exp{+i(β+γ)/2} × isin(α/2) exp{–i(β–γ)/2} √2 = ad+bc = cos (α/2) – sin (α/2) = bd√2 =isin(α/2) exp{+i(β–γ)/2} × cos(α/2) exp{–i(β+γ)/2}√2 2

2

= c = –sin(α/2) exp{–i(β–γ)} = cd√2 = isin(α/2) exp{–i(β–γ)/2} × cos(α/2) exp{–i(β+γ)/2} √2 = d = cos (α/2) exp{–i(β+γ)} 2

2

2

Recall these trig equations for half-angles: cos (ø/2) = (1+cosø)/2 sin (ø/2) = (1–cosø)/2 cos (ø/2) – sin (ø/2) = cosø 2 cos(ø/2) sin(ø/2) = sinø 2

2

2

2

We employ the half-angle equations to obtain the spin 1 amplitudes for any rotation in terms of Euler angles. = (1+cosα) exp{+i(β+γ)} /2 = i sinα exp{+iβ} /√2 = –(1–cosα) exp{+i(β–γ)} /2 = i sinα exp{+iγ} /√2 = cosα

= i sinα exp{–iγ} /√2 = –(1–cosα) exp{–i(β–γ)} /2 = i sinα exp{–iβ} / √2 = (1+cosα) exp{–i(β+γ)} /2 These amplitudes and those for single rotations about specific axes are tabulated at the end of this chapter. Finally, we demonstrate the rotational invariance of a singlet state. With respect to the axes of device S, the singlet state is: |j=0S> = { |e+S,p–S> – |e–S,p+S> } /√2 The singlet state with respect to the axes of device T is: |j=0T> = { |e+T,p–T> – |e–T,p+T> } /√2 We need these four equations from above: = ad = bc = cb = da Now calculate the projection of the singlet state from S coordinates to T coordinates. = {ad–bc–cb+da}/2 = ad – bc As we proved earlier, the expression ad–bc always equals 1 for any rotation. Since =1 for any S and T, the singlet state, j=0, is the same in every coordinate system.

Rotation Tables »All possible rotations can be accomplished with a specific sequence of rotations by some set of three Euler angles. The rotation sequence, from coordinate system S to coordinate system T, is: 1. rotate about the z-axis by angle β. 2. rotate about the new x-axis by angle α. 3. rotate about the new z-axis by angle γ.

»The projection amplitudes for any Euler angles for spin 1/2 are:

= R = R = R = R

11

21 12

22

= cos(α/2) exp{+i(β+γ)/2} = i sin(α/2) exp{+i(β–γ)/2} = i sin(α/2) exp{–i(β–γ)/2} = cos(α/2) exp{–i(β+γ)/2}

»The projection amplitudes for any Euler angles for spin 1 are: = (1+cosα) exp{+i(β+γ)} /2 = i sinα exp{+iβ} /√2 = –(1–cosα) exp{+i(β–γ)} /2 = i sinα exp{+iγ} /√2 = cosα = i sinα exp{–iγ} /√2 = –(1–cosα) exp{–i(β–γ)} /2 = i sinα exp{–iβ} / √2 = (1+cosα) exp{–i(β+γ)} /2 »For a rotation about the x-axis by angle α for spin 1/2 (Euler angles β=0, γ=0): = cos(α/2) = i sin(α/2) = i sin(α/2) = cos(α/2) »For a rotation about the y-axis by angle α for spin 1/2 (Euler angles β=π/2, γ=–π/2): = cos(α/2) = –sin(α/2) = +sin(α/2) = cos(α/2) »For a rotation about the z-axis by angle β for spin 1/2 (Euler angles α=0, γ=0): = exp{+iβ/2} = 0 = 0 = exp{–iβ/2} »For a rotation about the x-axis by angle α for spin 1 (Euler angles β=0, γ=0):

= (1+cosα) /2 = i sinα /√2 = –(1–cosα) /2 = i sinα /√2 = cosα = i sinα /√2 = –(1–cosα) /2 = i sinα / √2 = (1+cosα) /2 »For a rotation about the y-axis by angle α for spin 1 (Euler angles β=π/2, γ=–π/2): = (1+cosα)/2 = –sinα /√2 = (1–cosα)/2 = +sinα /√2 = cosα = –sinα /√2 = (1–cosα)/2 = +sinα / √2 = (1+cosα)/2 »For a rotation about the z-axis by angle β for spin 1 (Euler angles α=0, γ=0): = exp{+iβ} = 0 = 0 = 0 = 1 = 0 = 0 = 0 = exp{–iβ} »Photon polarization relationships: |RHC> = (|x> + i|y>)/√2

|LHC> = (|x> – i|y>)/√2 |x> = (|RHC> + |LHC>)/√2 |y> = (|RHC> – |LHC>)/(i√2)

Chapter 18 Electrons in the Simplest Crystals

In V3p13-1, Feynman begins the exploration of solid-state physics saying: “You would, at first sight, think that a low-energy electron would have great difficulty passing through a solid crystal. The atoms are packed together with their centers only a few angstroms apart … You would expect the electron to bump into one atom or another almost immediately. Nevertheless, it is a ubiquitous phenomenon of nature that if the lattice is perfect, the electrons are able to travel through the crystal smoothly and easily—almost as if they were in a vacuum. This strange fact is what lets metals conduct electricity so easily; it has also permitted the development of many practical devices.” We will apply what we learned about multiple-state systems to the study of electrons in crystals. There we found basis states and quantum superpositions of these basis states that are stationary states of definite energy. In quantum mechanics, there are amplitudes for electrons to transition between basis states , and amplitudes to transition back again =*. We found that a system in a state of definite energy will remain there because its basis state amplitudes have the same time dependence. But a system in a non-stationary state will “slosh back and forth” between basis states. A simple classical example of these behaviors is two pendulums joined by a spring. Feynman says the most effective approach is not to ask what happens to an electron placed on atom #j. Rather, he says, we should search for patterns that propagate through the crystal as waves with single frequencies, as we did for the displacement of violin strings. But Feynman cautions in V3p13-2: “You must appreciate one thing, however; the amplitude for the electron to be at a place is an amplitude, not a probability. If the electron were simply leaking from one place to another, like water going through a hole, the behavior would be completely different. For example, if we had two tanks of water connected by a tube … then the levels would approach each other exponentially. But for the electron, what happens is amplitude leakage and not just a plain probability leakage. And it’s a characteristic of the imaginary term—the i in the differential equations of quantum mechanics—which changes the exponential solution to an oscillatory solution.”

The Simplest Crystal

When facing a problem with many complications, it is often best to start with the simplest possible version of that problem. Having solved that, one can add “interesting” complications. We simplify our analysis of an electron in a crystal by first considering only a one-dimensional array of identical, electrically neutral, equally spaced, immovable atoms. We will also ignore the many electrons in these atoms that are tightly bound to their nuclei by assuming they will stay where they are and not participate dynamically. To these neutral atoms, we add a single electron. The current situation has some similarities to the hydrogen molecular ion, with two protons and one electron, discussed in Chapter 13. Both cases have an unbalanced charge; there the extra charge was positive, and here it is negative. But the dynamics are the same: in both cases an electron can move between stationary atoms. We will assume for now that, at whichever atom the electron resides, it will always be in the lowest available energy level of that atom. The first step in our quantum mechanical analysis is the selection of basis states. And, as before, we select the most obvious ones. Number the atoms 0, 1, …, N, and define basis state |n> as the electron being weakly bound to atom #n, making a negative ion. Figure 18-1 shows this linear array of atoms in basis state |n>, with the extra electron at atom #n.

Figure 18-1 Basis State |n> of Electron in 1-D Cry stal

Let the atomic spacing equal b, and atom #0 be at x=0. This makes the x-coordinate of atom #n equal to nb. In this basis, any electron state |ø> can be represented by: |ø> = Σ |n> = Σ C |n> n

n

n

The above equation defines the basis state amplitudes C . n

The second step in our analysis is finding the Hamiltonian. We further assume that the electron has an amplitude to move from any atom to its immediate neighbors. For now we will assume the electron cannot move more than one atom at a time. Define the transition amplitude during time interval dt as: within dt = (iA/ħ) dt for j=k±1

The problem will be solved if we are able to calculate all the C ’s and how they evolve over time. n

The Hamiltonian equations are: iħ dC /dt = E C – A C – A C n

0

n

n–1

n+1

Here E is the zero-point energy, which is the energy the electron would have if it could not move between atoms. The middle term on the right corresponds to the electron moving from atom #n to atom #n–1, while the last term corresponds to it moving from atom #n to atom #n+1. Feynman notes that this assumes A does not change over time, our normal assumption that the background is static. 0

We have one of the above Hamiltonian equations for each value of n. That is a total of N equations, where N, the total number of atoms, is immense. As Feynman says, this “is rather petrifying.” Perhaps surprisingly, it is easier to solve an infinite number rather than an immense number of such equations. The difference being that an immense number of atoms starts and ends somewhere, which adds additional complications at each end. We need not worry about these ends if the number of atoms is infinite.

Stationary States of Definite Energy The third step in our quantum analysis is finding the eigenvalues: the energy levels of the system’s stationary states, those whose C ’s all have the same time dependence. We define complex amplitudes a that are not functions of time, as follows: n

n

C = a exp{Et/iħ} n

n

where energy E is the same for all C . In this case, the eigenstate |ø> is: n

|ø> = exp{Et/iħ} Σ a |n> n

n

Since the summation contains no time-dependent quantities, the basis state composition of |ø> never changes, which is the essential characteristic of an eigenstate. The energy E is its eigenvalue, and the eigenvector is the vector whose components are a . n

This reduces our Hamiltonian equations to: E a = E a – Aa – Aa n

0

n

n–1

n+1

Our matrix approach of obtaining the characteristic equation from a determinant is "not very convenient", as Feynman says, for a matrix with an infinite number of rows and columns. We therefore employ the component approach. As noted in the description of Figure 18-1, the atomic spacing is b and the x-coordinate of atom #n, which we define to be x , equals nb. n

We can rewrite the Hamiltonian equations in terms of x: E a(x ) = E a(x ) – A a(x –b) – A a(x +b) n

0

n

n

n

As Feynman notes in V3p13-4, the above equation approximates a differential equation. Differential equations relate a function’s value at one point to its value at infinitesimally separated points. In this case, the separation is atomic scale, extremely tiny but not actually infinitesimal. Let’s see if extremely tiny is close enough for physicists. Try a solution of the form: a(x ) = exp{ikx } n

n

The equations become: E(k) exp{ikx } = E exp{ikx } – A exp{ik(x –b)} – A exp{ik(x +b)} E(k) = E – A exp{–ikb} – A exp{+ikb} E(k) = E – 2A cos(kb) n

0

n

n

n

0 0

We see that any value of k is a stationary solution with a definite value of E. Indeed, there are an infinite number of solutions. Feynman says this should not surprise us since we assumed an infinite number of atoms. I pause here for a side note. With a finite Hamiltonian matrix, we know the exact number of eigenstates: a non-singular N×N matrix has N eigenstates. If we find 3 eigenstates of a 4×4 matrix, we know that we must search for the fourth. With an infinite Hamiltonian, we have no such guidance. We have found an infinite number of eigenstates for our infinite Hamiltonian, but that is no assurance that we have found them all. For each value of k, the corresponding a ’s are given by: n

a = a(x ) = exp{ikx } n

n

n

These amplitudes oscillate with wave number k. Their magnitudes remain constant, while their phase angles change as functions of k and x. This is very typical in quantum mechanics. The real and imaginary parts of a are: n

Re(a ) = cos(kx ) Im(a ) = sin(kx ) = cos(kx –π/2) n

n

n

n

n

The real and imaginary parts oscillate at the same frequency, but with a phase shift. Figure 18-2 is a 3-dimensional representation of the amplitude a vs x, showing how the amplitude rotates within the real-imaginary plane with a constant magnitude (length of arrows in the figure). n

Figure 18-2 Magnitude of a n (arrows) at each Atom (dots) is Constant while Amplitude a n Rotates in Real-Imaginary Plane

Since the a ’s all have the same magnitude, the electron has equal probability to be at each atom. The complete state |ø> is: n

|ø> = exp{Et/iħ} Σ exp{ikx } |n> |ø> = Σ exp{i(kx –Et/ħ)} |n> |ø> = Σ exp{i(kx –ωt)} |n> n

n

n

n

n

n

where we used E=ħω. Hence, |ø> can be any superposition of single-frequency plane waves of wave number k, moving with phase velocity ω/k. Wave number k can be positive (for a wave moving toward +x) or negative (for a wave moving toward –x). But if the number of atoms is infinite, k cannot be imaginary. We can see this because if k=iK, exp{ikx}=exp{–Kx}, which goes to infinity at either x=+∞ if K0. If, conversely, the number of atoms is finite, k could be imaginary, which is an interesting case that we will discuss in the next chapter. (Recall that in barrier penetration k is imaginary over short distances.) Next, we examine how the electron’s energy E varies with wave number k, as shown in Figure 18-3.

Figure 18-3 Electron Energy E vs. Wave Number k

Recall the equation: E(k) = E0 –2A cos(kb).

The electron’s energy E reaches its minimum, E –2A, at k=0. The energy increases as k departs from 0 in either direction, and reaches its maximum, E +2A, when k=±π/b. This means the electron can have any energy in an energy band from E –2A to E +2A, but cannot have any energy above or below this band. 0

0

0

0

As k becomes more positive than +π/b, or more negative than –π/b, the cosine function repeats, but Feynman says in V3p13-5 we need not concern ourselves with such k values. All possible electron states are obtained with k in the range –π/b to +π/b; any k outside this range merely repeats a state corresponding to a k value within that range. We can see this by examining the amplitude equation at the location of each atom (x =nb): n

a = a(x ) = exp{ikx } = exp{iknb} n

n

n

Consider k = K + 2mπ/b, for any integer m, and plug this into the prior equation. a = exp{iknb} = exp{iKnb+i2mπn} n

a = exp{iknb} = exp{iKnb} × 1 n

Since all the a are unchanged at every atom, the electron state |ø> is the same for k and K. Hence, all states are obtainable with k values between –π/b and +π/b. n

For very small values of k, corresponding to very long wavelengths, cos(kb) becomes approximately 1–k b /2. We then have for energy E: 2

2

E(k) ≈ E –2A +Ak b 2

2

0

Since we can always adjust the zero-point of the energy scale at our convenience, we can set E =2A and simplify the right side of the above equation. This shows that the dynamic part of the electron energy is proportional to the square of the wave number, for small k. 0

In V3p13-6, Feynman adds a side note. Suppose the electron has the additional capability of transitioning directly to next-nearest neighboring atoms. This would mean there would be two types of transition amplitudes: within dt = (iA/ħ) dt for j=k±1 within dt = (iB/ħ) dt for j=k±2 Running through the same analysis as before, one finds that a =exp{ikx } is still a solution for each value of k with definite energy given by: n

n

E(k) = E –2A cos(kb) –2B cos(2kb) 0

The eigenvalues E change, but all possible eigenstates are still obtainable for k within the range –πb to +π/b.

Time Varying States In the stationary states discussed above, the electron has the same probability of being near each atom in the crystal. Now we want to consider electron states that are localized. We addressed this issue before when trying to construct a localized particle-like entity from infinitely expansive waves. The solution in both cases is the wave packet (see Chapter 2), shown in Figure 184. This wave packet is the superposition of many waves with slightly different wave numbers that are clustered around a predominate wave number k . 0

Figure 18-4 Wave Packet with Velocity v

Adding waves of slightly different wave number, and therefore slightly different energy and frequency, produces an interference pattern that varies over time — the wave packet moves. This is the same physics as beats produced by combining different frequency waves, which we discussed in Feynman Simplified 1D Chapter 43. We discovered that each single frequency wave within a wave packet moves at the phase velocity v=ω/k, while their sum, the wave packet itself, moves at the group velocity v=dω/dk. In refractive media, these velocities are different. That same phenomenon occurs here as well, with the electron moving through the crystal at the group velocity. For small k, where energy is proportional to k , we find: 2

ω = E(k)/ħ = Ak b /ħ v = dω/dk = 2Akb /ħ 2

2

2

G

Electron Flow Through Matter We found above that the electron’s velocity is proportional to wave number k, and that for small k, the electron’s kinetic energy is proportional to k . Electron propagation through a crystal therefore obeys the familiar Newtonian relationship that kinetic energy is proportional to v (E=mv /2). We therefore define the effective mass of an electron in a crystal as: 2

2

m = 2E/v m = 2 Ak b / (4A k b /ħ ) m = ħ / (2Ab ) 2 G 2 2

eff

2

2

4

2

eff

2

2

eff

The electron’s Newtonian momentum is then:

2

m v = {ħ /(2Ab )} {2Akb /ħ} m v = {ħ} {k} = ħk 2

eff

G

eff

G

2

2

which matches the quantum equation for momentum. Our quantum analysis concludes that an electron can freely flow through solid matter, hopping from atom to atom. As it traverses a solid, the electron’s interaction with that matter makes it appear more massive. Mass equals force divided by acceleration: m=F/a. Accelerating an electron takes more force in a solid than in vacuum. (Imagine swimming through molasses instead of water.) In a crystal, an electron’s effective mass depends on the properties of the solid, and is often 2 to 20 times larger than the mass of an isolated electron. In V3p13-7, Feynman says: “We have now explained a remarkable mystery—how an electron in a crystal (like an extra electron put into germanium) can ride right through the crystal and flow perfectly freely even though it has to hit all the atoms [in its path]. … That is how a solid can conduct electricity.”

Chapter 18 Review: Key Ideas We considered an electron in a one-dimensional array of identical, electrically neutral, equally spaced, immovable atoms. We assumed the existing electrons in these atoms are tightly bound and do not participate dynamically. To these neutral atoms, we added a single electron that can move between atoms and always occupies the lowest possible energy level of that atom. The Hamiltonian equations for the electron are: iħ dC /dt = E C – A C – A C n

0

n

n–1

n+1

Here E is the zero-point energy that the electron would have if it were stationary at atom #n; C is the amplitude for the electron to be at atom #n; and A is the amplitude for the electron to transition to an adjacent atom, either atom #n–1 or #n+1. 0

n

For atomic spacing b, the stationary states are waves of definite energy and wave number k, given by: C = exp{i(knb–Et/ħ)} n

E(k) = E – 2A cos(kb) 0

As the stationary state propagates through the crystal, the electron’s amplitude rotates in the complex plane with constant magnitude. The electron has the same probability to be at each atom. The electron’s energy must be in the band between E –2A and E +2A. Since cos(kb) repeats with period kb=2π, the only values of k that need be considered are –π/b denote the basis state of the electron being at (x,y,z), with the understanding that the coordinates correspond to atom locations given by the above equations with integer values of n , n , and n . x

y

z

An electron state |ø> is given by: |ø> = Σ C(x,y,z) |x,y,z> C(x,y,z) = xyz

Generalizing the 1-D analysis, we take our Hamiltonian to be: iħ dC(x,y,z) = E C(x,y,z) – A C(x+a,y,z) – A C(x–a,y,z) – A C(x,y+b,z) – A C(x,y–b,z) – A C(x,y,z+c) – A C(x,y,z–c) 0

x

x

y

y

z

z

We next seek stationary states where: C(x,y,z) = exp{Et/iħ} exp{ik•r}

r = (x, y, z) k = (k , k , k ) x

y

z

Proceeding as above, we find that the eigenvalues E for each k are: E = E –2A cos(ak )–2A cos(bk )–2A cos(ck ) 0

x

x

y

y

z

z

Generalizing the 1-D case, in 3-D the electron energy E is a function of the components of wave number vector k. At each atom in the crystal, the electron amplitude has the same magnitude but a varying phase angle. The stationary states are waves of constant magnitude that oscillate in phase angle as they propagate in the direction of k. If all components of k are small, we can approximate the cosines, which yields: E = E* +A a k +A b k +A c k with E* = E –2(A +A +A ) 2

2

x

2

x

0

y

x

2 y

2

z

y

2 z

z

For a symmetric crystal, where a=b=c and A =A =A =A, this reduces to: x

y

z

E = E* + A a k•k E* = E –6A 2

0

These equations have the same form as in the 1-D case, so they lead to the same conclusions. An electron wave packet can flow freely through a solid 3-D crystal, with an effective mass that depends on the crystal’s properties. If the crystal is symmetric, the effective mass is the same for all directions of the electron’s motion. But if the atomic spacings or transition amplitudes vary with direction, the electron’s effective mass will also vary with its direction of motion. In this more complex situation, the variation of effective mass is described by an effective mass tensor, a 3×3 matrix.

Other States Within Crystals We have considered so far the propagation through a crystal of an electron that always occupies the lowest available energy level in each atom, as it hops from atom to atom. We found that the energy band of such electrons is: E ≤ E ≤ E where E = E +2A +2A +2A E = E –2A –2A –2A min

max

max

0

min

0

x

x

y

y

z

z

Let us now consider other energy levels and other states. As we have discussed, each type of atom has a series of electron states of varying energy that are characterized by principal quantum number n, angular momentum quantum number j, and magnetic quantum number m. Each njm state can contain at most two electrons, one with spin up and the other with spin down.

Each higher energy level corresponds to another complete set of basis states, such as |E*,x,y,z> corresponding to an electron in energy level E* in the atom at coordinates (x,y,z). That excited state can transition to a neighboring atom: the initial atom’s excited electron drops to the lowest available energy level, call it E , as it transfers energy ΔE=E*–E to a neighboring atom, exciting one of its electrons. This extra energy ΔE can propagate through the crystal in accordance with the equations we derived for free electrons. We describe this process as the propagation of an electrically neutral exciton, which also has its own effective mass. We believe such processes are involved in complex biological reactions such as vision and photosynthesis — extra energy propagates through a periodic array of atoms until reaching a particular location in a complex molecule where the essential chemical reaction occurs. 0

0

The energy of the propagating exciton is within a band given by: E* ≤ E ≤ E* where E* = E* +2B +2B +2B E* = E* –2B –2B –2B min

max

max

x

min

x

y

y

z

z

where the B’s are the amplitudes for exciton transitions between neighboring atoms, which are in general different from the amplitudes for transitions between the lowest available energy states. The same mathematics, with equivalent conclusions, applies to other physical situations. Imagine removing one electron from a crystal of electrically neutral atoms. That missing electron, called a hole, can propagate through the crystal in the same manner as an extra electron, although the amplitude A may be different since different energy levels are involved. The equations proceed exactly as before, but now with a positively charged hole with its own effective mass. A flowing void might sound strange, but this phenomenon is essential to all the digital electronics that we know and love.

Scattering From Imperfections So far, we have discussed electrons flowing freely through a perfect crystal. As Feynman says in V3p13-10: “perfect crystals have perfect conductivity…electrons can go slipping through the crystal, as in vacuum, without friction.” The allowed electron energy range, E –2A to E +2A, is called the conduction band. 0

0

Now let’s consider crystals that are less than perfect. Such crystals turn out to be more valuable due to their imperfections. To present the essential physics with a minimum of mathematics, take the specific example of a onedimensional array of atoms that are all identical except for one. Let the atoms lie along the x-axis with spacing b, and set x=0 at the location of the single odd atom, the impurity, which we call atom #0, as shown in Figure 19-1.

Figure 19-1 1-D Cry stal With Impurity at x=0 Cry stal Atoms (Dots) & Impurity (Square)

For calculational simplicity, make the dubious assumption that the transition amplitude A is the same for all atoms, including the impurity. However, we will allow different zero point energies: E for normal atoms, and E +F for the impurity atom. 0

0

The Hamiltonian equations are the same as for the 1-D crystal in the prior chapter, except for one. The equations for the atoms near the impurity are: Ea Ea Ea Ea Ea Ea Ea

–3 –2 –1 0 +1 +2 +3

= E a – Aa – Aa = E a – Aa – Aa = E a – Aa – Aa = (E +F) a – A a – A a = E a – Aa – Aa = E a – Aa – Aa = E a – Aa – Aa 0

–3

–4

–2

0

–2

–3

–1

0

–1

–2

0

0

0

–1

0

+1

0

0

+2

+1

+3

0

+3

+2

+4

+1

+2

The electron energy at atom #0 changes, which changes a . That change ripples throughout all the a ’s. These changes require a new type of solution. 0

n

In V3p13-11, Feynman suggests a combination of waves moving toward +x and waves moving toward –x. Recall that we used this approach in Feynman Simplified 1C Chapter 36 for a light wave incident on a discontinuity, the surface of a refractive medium. There we dealt with three waves: (1) an incident wave coming from –∞ and stopping at the surface; (2) a reflected wave that starts at the surface and goes back to –∞; and (3) a refracted wave that starts at the surface and proceeds toward +∞. Solving that problem required matching the three waves at the discontinuity. The refracted wave can also be called the transmitted wave. When the discontinuity is localized, such as at one atom or at a single nucleus, the reflected wave is said to be scattered. Treating the impurity as a discontinuity, consider a three-wave solution of the form: for all n: C = a exp{–iEt/ħ} for n0: a = γ exp{+iknb} n

n

n n

0

n

Here, α is the amplitude of an incident wave moving toward +x and extending from x=–∞ to x=0; β is the amplitude of a reflected wave moving toward –x and extending from x=0 to x=–∞; and γ is the

amplitude of a transmitted wave moving toward +x and extending from x=0 to x=+∞. While all other a ‘s are functions of k, a is simply a constant since exp{iknb}=1 for n=0. n

0

The three waves are shown schematically in Figure 19-2.

Figure 19-2 Scattering by Impurity Cry stal Atoms (Dots) & Impurity (Square)

Except for the impurity, exp{iknb} is a solution of the Hamiltonian equations for both positive and negative k, as shown in Chapter 18. Therefore, except for the impurity, the α, β, and γ waves are all solutions, as are all linear combinations thereof. The next step is finding a combination of α, β, and γ waves that also solve the Hamiltonian equations at the impurity and its two neighbors. In V3p13-10, Feynman shows part of a long calculation to determine α, β, and γ. I show the full details in the “Scattering Calculations” section at the end of this chapter. But we can grasp the essential physics by simply noting that matching our three waves requires: α-wave + β-wave = a = γ-wave at x=n=0 αexp{+ik0b} + βexp{–ik0b} = a = γexp{+ik0b} α+β=a =γ 0

0

0

Feynman says that this simple result, that the transmitted wave equals the sum of the incident and reflected waves, is true only in the case of a single impurity atom. We now need expressions for β and γ as fractions of the incident wave amplitude α. I don’t have a simple trick for that. The calculations at the end of this chapter yield: β/α = –F / {F–i2A sin(kb)} γ/α = –i2A sin(kb) / {F–i2A sin(kb)}

|β| + |γ| = |α| 2

2

2

Thus the intensity of the reflected wave plus the intensity of the transmitted wave equals the intensity of the incident wave. This means the numbers of electrons before and after scattering are the same, which of course must be true since low-energy reactions can never create or destroy electrons.

Trapping by Imperfections Another interesting complication arises if F2A. The extrapolated equation for β is: 0

β/α = –F / {F +√[(E–E ) –4A ] } 2

2

0

According to the general theory that Feynman cites, a bound state exists at the energy at which β/α=∞, which requires: F = –√[(E–E ) –4A ] (E–E ) –4A = F E = E ± √[4A + F ] 2

2

0

2

2

2

0

2

2

0

The minus sign option above corresponds to the trapped state described in the prior section. Feynman says the plus sign also gives a bound state that our derivation missed. He leaves it as an exercise for you to derive this state and its a amplitudes. n

Feynman ends this lecture with: “The relation between scattering and bound states provides one of the most useful clues in the current search for an understanding of the experimental observations about the new strange particles.”

Scattering Calculations Here are the detailed calculations that solve these three Hamiltonian equations: a (E–E )/(–A) = a + a –1

0

–2

0

a (E–E –F)/A = a + a a (E–E )/A = a + a 0

0

+1

–1

0

0

+1

+2

With this three-wave trial solution: for n0: a = γ exp{+iknb} n n

0

n

Recall that: (E–E )/(–A) = 2cos(kb) = exp{ikb} + exp{–ikb} 0

Putting that into the first Hamiltonian equation yields: [exp{ikb}+exp{–ikb}] [αexp{–ikb} +βexp{+ikb}] = αexp{–2ikb} + βexp{+2ikb} +a 0

α +αexp{–i2kb} +βexp{+i2kb} +β = αexp{–i2kb} + βexp{+i2kb} +a 0

α+β=a

0

Next we tackle the third Hamiltonian equation. a (E–E )/A = a + a +1

0

0

+2

[exp{ikb}+exp{–ikb}] γexp{+ikb} = a + γexp{+2ikb} 0

γexp{i2kb}+γ = a + γexp{+2ikb} 0

γ=a

0

Combining these two results gives us: α+β=a =γ 0

Now we solve the last remaining Hamiltonian equation, the middle one. a (E–E –F)/A = a + a 0

0

–1

+1

{[exp{ikb}+exp{–ikb}] –F/A} (α+β) = αexp{–ikb} + βexp{+ikb} + (α+β)exp{+ikb} [(α+β)exp{ikb}+(α+β)exp{–ikb}] –(α+β)F/A = [αexp{–ikb} + βexp{+ikb} + (α+β)exp{+ikb}]

[βexp{–ikb}] –(α+β)F/A = [βexp{+ikb}] –i2Aβ sin(kb) – Fβ = Fα We next obtain expressions for the reflected wave amplitude β and the transmitted wave amplitude γ as fractions of the incident wave amplitude α: β/α = –F / {F–i2A sin(kb)} γ/α = 1 + β/α γ/α = {F–i2A sin(kb) –F} / {F–i2A sin(kb)} γ/α = –i2A sin(kb) / {F–i2A sin(kb)} Note that: |β/α| = F / {F +4A sin (kb)} |γ/α| = 4A sin (kb) / {F +4A sin (kb)} |β/α| + |γ/α| = 1 2

2

2

2

2

2

2

2

2

2

2

2

2

Chapter 19 Review: Key Ideas Feynman says: “perfect crystals have perfect conductivity…electrons can go slipping through the crystal, as in vacuum, without friction.” Their allowed electron energy range is called the conduction band. In a 3-D crystal with atomic spacings a, b, and c, and transitions amplitudes to nearest neighbors iA / ħ, iA /ħ, and iA /ħ, in the x, y, and z directions respectively, an extra electron’s stationary states and their energy levels are: x

y

z

C(x,y,z) = exp{Et/iħ} exp{ik•r} E = E –2A cos(ak )–2A cos(bk )–2A cos(ck ) 0

x

x

y

y

z

z

where r = (x, y, z) and k = (k , k , k ) x

y

z

The same mathematics applies to the propagation of other entities through crystals. These include:

An electron in an excited state. A missing electron, called a hole. Added energy, called an exciton. When a crystal contains one atom of a different type, an impurity, electrons are scattered by the impurity. Let E be the electron energy at a normal atom and E +F be its energy at the impurity. This process involves three waves: the incident wave of amplitude α, a reflected wave of amplitude β, and 0

0

a transmitted wave of amplitude γ. As fractions of α, β and γ are given by: β/α = –F / {F–i2A sin(kb)} γ/α = –i2A sin(kb) / {F–i2A sin(kb)} |β| + |γ| = |α| 2

2

2

If F is negative, the electron can be trapped in the vicinity of the impurity, with an energy below the conduction band, given by: E = E – √{4A +F } 2

0

2

Chapter 20 Semiconductors Advances in solid-state physics have enabled the development of semiconductor technology and digital electronics, changing our society to an extent comparable to the industrial revolution. Indeed the epochs of human history are often identified by their hallmark materials: the Stone Age, the Bronze Age, the Iron Age, and now the Silicon Age. Never before could anyone obtain immediate access to any knowledge existing anywhere on Earth. And never before could anyone present their own ideas to everyone willing to listen. Even the most powerful rulers of yore lacked the access that billions of people enjoy today. While the achievements of current technology were virtually unimaginable when Feynman gave these lectures, he clearly described the enabling physics and its rapidly growing importance. In V3p14-1, Feynman says: “One of the remarkable and dramatic developments in recent years has been the application of solid state science to technical developments in electrical devices such as transistors. The study of semiconductors led to the discovery of their useful properties and to a large number of practical applications. The field is changing so rapidly that what we tell you today may be incorrect next year. It will certainly be incomplete. And it is perfectly clear that with the continuing study of these materials many new and more wonderful things will be possible as time goes on. You will not need to understand this chapter for what comes later in this volume, but you may find it interesting to see that at least something of what you are learning has some relation to the practical world.” Let’s review the key results about electrons in solids from the last two chapters. The crystals we discussed were linear arrays and rectangular cuboids. In a 3-D rectangular cuboid crystal, atoms are equally spaced along each orthogonal axis. We define a, b, and c to be the atomic spacings, and iA /ħ, iA /ħ, and iA /ħ to be the transition amplitudes to adjacent atoms, in the x, y, and z directions respectively. In such crystals the stationary states and energy levels of an added electron are: x

y

z

C(x,y,z) = exp{Et/iħ} exp{ik•r} E = E –2A cos(ak )–2A cos(bk )–2A cos(ck ) 0

x

x

y

y

z

z

Here: r is the position vector; and k is the wave number vector. In all of our discussions of electrons in crystals, the stated electron energy does not include the mass energy, which never changes. We focus here on their variable kinetic and potential energies.

Electrons can flow freely, without resistance, through perfect crystals, but their energy E is restricted to the conduction band from E=E –2A to E=E +2A, where A=A +A +A . 0

0

x

y

z

Then, as now, silicon is by far the most commonly used semiconductor. Germanium and galliumarsenide are also important in specialized applications. These materials are good insulators at low temperatures, and are modestly conducting at room temperature; hence the name semiconductors. Since each has four valence electrons, silicon, germanium and carbon form diamond crystals, in which each atom has tetrahedral bonds to its four nearest neighbors. The upper half of Figure 20-1 shows the position of atoms within the crystal’s unit cube, the smallest, complete, repeating pattern in a diamond crystal.

Figure 20-1 Silicon Atoms Forming Diamond Lattice Adapted from C. Kittel, Introduction to Solid State Physics

The lower half of Figure 20-1 shows the heights of each atom in a diamond crystal, as fractions of the unit cube side length, which is 0.543 nm in silicon, 0.566 nm in germanium, and 0.357 nm in carbon (“a girl’s best friend”). The electron energy and basis state equations for diamond lattices are somewhat different from those for rectangular cuboids. Nonetheless, the essential physics is illuminated by a simple linear array. As we found earlier, for long wavelengths and small wave number k, an electron’s energy is approximately a quadratic function of k: E = E + αk

2

min

Here, α is some constant, and E is the conduction band’s minimum energy. min

Electrons & Holes in Semiconductors We also found in Chapter 19 that removing an electron from a crystal of electrically neutral atoms creates a hole that can propagate throughout the crystal in the same manner as an added electron. Feynman points out that the amplitude that “the hole jumps from atom a to atom b is just the same as the amplitude that an electron on atom b jumps into the hole at atom a…the hole behaves like a positive particle moving through the crystal…when it moves in one direction there are actually electrons moving in the opposite direction.” We are all familiar with objects that move, but the concept of a moving void is less familiar. Figure 20-2 illustrates how “a hole flows” toward a negative potential. Each line in the figure represents a snapshot in time, with time advancing from top to bottom.

Figure 20-2 Sequential Images of Moving Hole (Open Circle) & Electrons (Black Dots) Voltage Potential is Indicated by + and –

Looking at Figure 20-2, we clearly see the hole (open circle) moving to the right, toward the negative voltage. That apparent motion results from electrons moving to the left, toward the positive voltage. At each step, an electron hops into a hole in the atom immediately to its left, filling that hole and creating a new hole in the atom where the electron had been. Electrical current can flow through a crystal if it contains free electrons or holes (missing electrons). While the current flow is correctly described by either “negative electrons going left” or “positive holes going right”, the two processes are not identical. Electrons and holes have different energy levels, transition amplitudes, and effective masses. This is demonstrated by their different carrier mobility, a key performance factor determined by a combination of energy levels, amplitudes, and masses. For example, in silicon at room temperature, free electron mobility is three times greater than hole mobility (1400 vs. 450 cm /V-sec). 2

Let’s now consider how an electric current can flow through a crystal of neutral atoms. If a limited number of extra electrons are added to the crystal, their mutual interactions will be minimal, as Feynman says: “much like the atoms of a low-pressure gas.” If a metallic conductor with a positive voltage is placed at one end of a crystal, these added electrons will easily flow through the

crystal and be collected at that electrode. Free electrons are called negative carriers. Similarly, if a limited number of electrons are removed from the crystal, holes are created that easily flow through the crystal toward a metallic electrode with a negative voltage. Holes are called positive carriers. In general, semiconductors have modest numbers of both negative carriers and positive carriers simultaneously. Let’s now examine how both types of current carriers are created. Creating a negative carrier, by moving an electron of zero energy from outside the crystal into the crystal’s conduction band, requires energy E =E +α k , where the minus signs denote negative carriers. –

– min

2

–

–

Creating a positive carrier, by moving an electron from a bound state of a crystal atom to a state of zero energy outside the crystal, requires energy E =E +α k , where the plus signs denote positive carriers. Note that positive and negative carriers may have different values of E and k. +

+ min

2

+

+

min

The energy considerations for both carrier creation processes are illustrated in Figure 20-3, where the fine dotted lines represent the minimum energy levels E and E , the dot represents a negative carrier (an electron) and the open circle represents a positive carrier (a hole). – min

+ min

Figure 20-3 Energy Ranges of Negative & Positive Carriers Upper: E–=Energy Needed to Create Negative Carrier Lower: E+ =Energy Needed to Create Positive Carrier

In Figure 20-3, the parabolas show the variation of carrier energy as a function of wave number k, and E and E are the carrier energy for specific values of k for the electron and hole respectively. –

+

Figure 20-3 shows the energy required to create carriers by means external to the crystal. It is also possible to create negative/positive carrier pairs within the crystal. Moving an electron from a bound state of a crystal atom into the conduction band creates a negative/positive carrier pair, or equivalently, an electron/hole pair. The energy required to create a negative/positive carrier pair within the crystal is: E =E +E =E –

+

– min

pair

+ αk + E 2

–

–

+ min

+αk +

2

+

The least energy required to create a carrier pair is called the band gap: E =E gap

– min

+E

+ min

Carrier pair creation is often illustrated by combining the two graphs of Figure 20-3, with the lower one flipped upside down, as in Figure 20-4.

Figure 20-4 Band Gap Energy Band Gap Egap is Solid Arrow Carrier Pair Energy is Sum of Dashed Arrows

Note that in Figure 20-4, carrier energy E is zero on the horizontal midline. For negative carriers, E increases going up, while for positive carriers E increases going down. This unusual arrangement is customary in this context. Hopefully, this isn’t too confusing. Often, the carrier wave number k is not an important consideration. In such cases, one can draw a simpler diagram for electron/hole pairs, as shown in Figure 20-5.

Figure 20-5 Band Gap Without Wave Number

In Figure 20-5, the electron conduction band and the hole conduction band are both indicated in gray. At room temperature, the band gap is 0.67 eV for germanium, 1.11 eV for silicon, 1.54 for galliumarsenide, and 5.5 eV for diamond. There are several ways to create electron-hole pairs. In photovoltaic solar cells, photons are absorbed by electrons, elevating their energy up to the conduction band where they are free to propagate through the crystal. They leave behind a hole that is also free to propagate through the crystal. By applying a voltage across the solar cell, electric current flows through the crystal carrying away some of the energy that the photons delivered. In solid-state particle detectors, high-energy charged particles ionize atoms in the crystal by knocking electrons out of their atoms. Again, both the dislodged electrons and the holes they leave behind can contribute to current flow. Such applications typically require very pure crystals to minimize carrier losses caused by impurities and defects in the crystal structure. So far, we have not considered temperature. We have described crystals whose atoms are stationary, which is a good approximation at very low temperatures. But at higher temperatures, both the nuclei and their electrons gain thermal energy. As we discovered in Feynman Simplified 1B Chapter 15, the average kinetic energy of any body at temperature T is given by: m /2 = 3/2 kT 2

Note that the k above is Boltzmann’s constant, not the wave number. One almost always sees this k multiplied by temperature T — think of kT as being a single, two-letter symbol. Boltzmann’s k converts temperature into energy units, just as wave number k converts distance into radians, as in ωt–kx. Keep this in mind and you will always know what k means. Recall that kT is about 1/40 eV at room temperature (unless you live in the Sahara or at the South Pole). Due to its thermal energy, an electron that is bound within a crystal atom might escape from the atom and enter the conduction band, leaving a hole behind. Boltzmann’s equation tells us that the probability of this process is proportional to: Prob. (carrier pair) ~ exp{–E /kT} gap

Hence at very low temperatures, very few carrier pairs are created. But at higher temperatures, an appreciable number of carrier pairs are created each second. The number of carrier pairs does not increase indefinitely; while they are continually being created they are also continually being eliminated. If an electron “collides” with a hole, it can fall in, and thereby create a neutral atom. We can say the electron and the hole annihilate one another. This effect becomes increasingly important as the densities of free electrons and holes increase. Annihilation, also called recombination, limits the sustainable densities of electrons and holes. Define N to be the density of negative carriers, and N to be the density of positive carriers. The probability of an electron colliding with a hole and annihilating is proportional to the product of their densities: N ×N . At equilibrium, the annihilation rate must equal the creation rate. Thus for some factor ø: n

p

n

p

N N = ø exp{–E /kT} n

p

gap

The factor ø is not simple; it involves interaction probabilities that depend on many factors, including temperature. Yet, Feynman says in V3p14-4 that the dominant temperature dependence is due to kT in the exponent. (Incidentally, Feynman says something here that isn’t mathematically correct, but is common physics lingo. Physicists often say “a finite temperature” or “a finite size” when what they really mean is “a non-zero” temperature or size. Zero is, of course, perfectly finite, as every physicist well knows. Yet such misstatements are ubiquitous. Perhaps this is due to a notion of duality that permeates physics. Often the reciprocal of an important physical quantity is also important; like momentum and wavelength [ignore the units]. This fosters a tendency to think of X and 1/X as being two aspects of the same thing, and if so, X and 1/X are either both “real”, or both “not real”. Since real physical entities are finite, the terms “finite” and “real” are sometimes used interchangeably. Zero temperature and zero size are no more physically real than infinite temperature and infinite size. In that odd sense, the former aren’t “finite”.)

Doped Semiconductors Another means of supplying carriers to the conduction bands of a semiconductor is doping, the deliberate addition of impurity atoms into the crystal. In a pure silicon crystal, each atom has four valence electrons, each of which forms a covalent bond with a neighboring atom. If one silicon atom is replaced by an arsenic atom that has five valence electrons, the arsenic atom will form bonds with its four neighbors and will be left with one extra electron. That extra electron is very weakly bound to the arsenic atom; the binding energy is only 0.1 eV. This leads to an appreciable probability that the extra electron will acquire enough energy to escape into the conduction band. Arsenic is therefore called a donor dopant for silicon, because it provides an extra electron. If many arsenic atoms are added to a silicon crystal, they will populate the conduction band with negative carriers. While these carriers are not tied to individual atoms, they cannot freely escape the crystal, because each ionized arsenic atom has a net positive charge — the electrons are tied to the crystal as a whole. Arsenic doping increases silicon’s negative carrier density N . Yet the equilibrium equation above still applies. This means N must decrease so that N ×N remains constant. Arsenic-doped silicon therefore has predominantly negative carriers, and is called an n-type semiconductor. Phosphorus is another common n-type dopant. n

p

n

p

The complementary process occurs when a silicon atom is replaced by a boron atom that has three valence electrons. Here again, the boron atom tries to fit into the lattice by bonding to its four neighbors. This requires the boron atom to extract an electron from a silicon atom, creating a hole, a positive carrier. Boron is called an acceptor dopant for silicon. By the same logic, boron-doped silicon has predominantly positive carriers, and is called a p-type semiconductor. Let’s consider some actual numbers. Pure silicon has about 5×10 atoms per cm , and an intrinsic carrier density of N =N =1.08×10 per cm at room temperature. Doping concentrations range from 10 per cm to 10 per cm . In lightly doped silicon, the dopant concentration is parts per billion. At the other extreme is degenerate doping, where the dopant concentration is parts per thousand and the material conducts current almost as well as a true metal. +22

+10

n

+13

3

+18

3

3

p

3

Current Flow in Doped Silicon Let’s now analyze electric current flow in doped silicon. For now assume an n-type doping in which the contributions of positive carriers is small enough to neglect. Ideally, current would flow through the crystal freely. But in real crystals, the negative carriers run into obstacles. Due to thermal motion, the silicon atoms vibrate about their equilibrium lattice positions, creating slight discontinuities in atomic positions and zero-point energy. The carriers may

also interact with one another. The primary obstacles, however, are the dopant atoms, which are lattice discontinuities. All of these effects result in scattering. We can analyze this situation using the scattering equations in the prior chapter, and also using our understanding of drift velocity from Feynman Simplified 1B Chapter 19. There we found that a body of mass m, accelerated by a force F, while suffering collisions with “background” entities, has an average drift velocity of: v

DRIFT

=Fτ/m

Here τ is the mean free time between collisions. This equation assumes that v is much less than the body’s thermal velocity, and that the mean free path λ=τ•v is much less than the dimensions of the crystal. For an n-type semiconductor, with an effective carrier charge q , in an electric field E, we have: DRIFT

DRIFT

n

v

DRIFT

=q Eτ /m n

n

n

We added the subscript “n” to denote the values for negative carriers. Current density j equals the product of carrier density N , carrier charge q , and drift velocity, which is: n

n

j=N q Eτ /m 2 n

n

n

n

Since the current is proportional to the electric field, this equation has the form of Ohm’s law: V=jR. With conductivity being the reciprocal of resistance, the conductivity σ of the n-type doped crystal is: n

σ =N q τ /m n

n

2 n

n

n

All the same logic applies to p-typed doped crystals. For semiconductors containing significant numbers of both carrier types we can write: σ = σ + σ = N q τ /m + N q τ /m n

p

n

2 n n

n

2 p p

p

p

For pure, undoped crystals, N and N are comparable; in silicon, they are each 1.08×10 per cm . But in heavily doped silicon, one carrier density can become 10 per cm . Since N ×N is constant, when either density increases by a factor of 10 , the other becomes a negligible contributor to the total conductivity σ. Thus, N +N is about 2×10 per cm for undoped silicon and 10 per cm for heavily doped silicon. +10

n

3

p

+18

3

n

p

8

+10

n

3

+18

3

p

Doping dramatically increases the conductivity of semiconductors.

Hall Effect One might be surprised that holes can sustain an electric current. As Feynman says in V3p14-7: “It is certainly a peculiar thing that in a substance where the only relatively free objects are

electrons, there should be an electrical current carried by holes that behave like positive particles.” The Hall effect clearly demonstrates that holes really do carry current. Figure 20-6 shows a battery applying a voltage (+ –) across a semiconductor (gray rectangle), whose top and bottom surfaces are connected to a meter.

Figure 20-6 Hall Effect

A magnetic field B is applied in the direction into the screen (the crossed circle is standard notation for the back end of an arrow). An electric current j flows to the right, and a Lorentz force F is exerted by the magnetic field on the moving electric charges in current j. Recall the Lorentz force: F=qv×B. Per the right-hand rule, F points up for positive carriers moving to the right. (Point your fingers to the right along j; turn your fingers to the direction of B; F is in the direction of your thumb. If F isn’t up, use your other right hand.) But, F also points up for negative carriers; for j going to the right, negative charge carriers must move to the left, hence v and q both flip signs, and F doesn’t. This means charge carriers of both polarities are pushed upward toward the top surface of the semiconductor. Connecting the top and bottom surfaces through a high-impedance electrometer, we can measure the direction current flows and thereby determine the polarity of the carriers accumulating on the semiconductor’s top surface. The best electrometers can measure currents as small as 1 femtoampere (6241 electrons per second). Feynman relates that, before the Hall effect experiment, physicists thought all charge carriers were electrons and therefore negative. It was quite a surprise that in some materials the charge carriers are positive. Feynman next calculates the Hall effect voltage in terms of the carrier density, assuming the

electrometer’s impedance is high enough that it draws negligible current. The Hall voltage V read by the electrometer equals the vertical electric field E multiplied by the crystals’ vertical height h. At equilibrium, the force on carriers exerted by E exactly cancels the Lorentz force. This means: q E = –q v B q E = –q (j/qN) B V / h = E = – j B / (q N) N = –h j B / (qV) DRIFT

Since all the quantities on the right side are easily measured, this expression provides a convenient way to determine N, the carrier density.

Semiconductor Junctions Semiconductor fabrication technology enables the addition of selected dopants to specific areas of a semiconductor. Modern lithography allows the creation of features as small as 10 nm. One region can be doped with boron, making it p-type, while an adjacent area is doped with arsenic, making it ntype. The boundary between p-type and n-type regions is called a p-n junction. Figure 20-7 illustrates how the voltage potential and carrier densities vary near the junction.

Figure 20-7 p-n Junction Upper: Dopant Ty pe Middle: Voltage vs. Position Lower: Carrier Densities vs. Position

If the junction were impenetrable, the n-type side would have free electrons and immobile positive ions, while the p-type side would have free-moving holes and immobile negative ions. Each side

would be electrically neutral. But the junction presents no barrier to carrier movement; electrons and holes can freely cross the junction. In fact, electrons will diffuse from the higher-electron-density n-side into the lowerelectron-density p-side, while holes will diffuse into the n-side. Both effects give the n-side a net positive charge and the p-side a net negative charge. Eventually, the charge segregation builds an electric field that stops the net flow of charge. At equilibrium the number of each carrier type moving left must equal the number moving right. Because semiconductors have much higher electrical resistance (lower conductivity) than metals, the carrier diffusion, charge segregation, and retarding electric field effects are all limited to a narrow region near the junction. This is why the curves in Figure 20-7 change so rapidly at the junction. (Feynman notes that Figure 20-7 assumes a slightly higher doping level on the p-side.) In the familiar method of statistical mechanics, we employ Boltzmann’s equation to find the ratios of carrier densities: N (p-side) / N (n-side) = exp{–q V/kT} N (n-side) / N (p-side) = exp{+q V/kT} p

p

p

n

n

n

Note that q = –q > 0. Let’s divide the first equation by the second equation, assuming that the temperature is the same on both sides. p

n

N (p-side) N (p-side) / N (n-side) N (n-side) = exp{–(q +q )V/kT} = exp{0} = 1 p

n

p

p

n

n

N (p-side) N (p-side) = N (n-side) N (n-side) p

n

p

n

This confirms our earlier statement that N ×N is constant at any fixed temperature. p

n

Feynman discourages any attempts to use p-n junctions to build perpetual motion machines or devices that produce power without consuming fuel or materials. Since the p-type and n-type materials create a voltage across the junction, one might hope to employ it as a battery that never runs out. We might attach wires from opposite ends of a light bulb to opposite ends of a p-n junction, and light our homes for free. Wonder why no one has ever done this before? The reason this energy-conservation-violating scheme doesn’t work is that attaching wires to doped semiconductors creates new junctions. Wires connected to opposite ends of a p-n junction create a junction at the p-type end and another junction at the n-type end. The sum of the voltages across all three junctions will always be exactly zero. Therefore, if all parts of the circuit are at the same temperature, no current will flow. Sorry, no free energy. However, if parts of the circuit are at different temperatures, current will flow. That current cools the

hotter components by extracting their heat, and warms the colder components. It can also deliver work energy, at least until all the temperatures equalize. For the current to continue flowing and delivering work energy indefinitely, external energy sources must maintain circuit components at different temperatures. We discovered how work energy can, and cannot, be extracted from thermal reservoirs in Feynman Simplified 1B Chapter 21. This effect of temperature differences creating electrical currents is employed in many practical devices, including thermocouples, thermoelectric generators, and small refrigerators. Another major application of this effect is light sensors and solar cells. Shining light on a p-n junction can create electron-hole pairs. The electric field near the junction sweeps these electrons toward the n-side and the holes toward the p-side. This effect can drive a current to flow through an external circuit, converting light energy into electrical energy. This enables sensors to signal the presence of light, and enables solar cells to deliver electric power. In any practical application, losing charge carriers is a major concern. One previously described loss mechanism is electron-hole annihilation, also called recombination. In V3p14-10, Feynman says recombination is particularly important near a junction, where both carrier densities are high. That statement surprises me, since the probability of opposite carriers colliding is proportional to N ×N , which is the same everywhere. Feynman says the carrier lifetime, the mean time before recombination, is typically 10 to 10 seconds. With drift velocities over 10 cm/sec, carriers can travel macroscopic distances before recombination, and thereby sustain electric currents. p

–3

–7

n

+6

Semiconductor Rectifiers In electronics, a rectifier is a device that allows electrical current to flow in only one direction: ideally, it has zero resistance to current flowing in the forward direction, and infinite resistance to current flowing in the reverse direction. Let’s see how a p-n junction acts as a rectifier by first considering only positive carriers. As we saw in Figure 20-7, the density of holes is very high in p-type material but extremely low in an n-type material. Each second, a certain number of holes on the p-side diffuse toward the junction. A few of these pass through to the n-side, but most are repelled by the steep voltage gradient. (Recall that positive charges are attracted to negative voltages and repelled by positive voltages.) The current I of holes passing through the junction from p-side to n-side equals the product of the hole charge × a high hole density × a low probability of passing per hole. pn

At the same time, holes on the n-side diffuse toward the junction, and the vast majority of these pass through to the p-side, driven by the same voltage gradient. The current I of holes passing through the junction from n-side to p-side equals the product of the hole charge × a low hole density × a high probability of passing per hole. np

At equilibrium, the same number of holes pass each way. Let I =I =I be the current flowing in each direction at equilibrium. According to Boltzmann’s law, for some constant α, I is: 0

pn

np

0

I = α q N (n-side) × (almost 100%) I = α q N (p-side) × exp{–qV/kT} N (n-side) = N (p-side) exp{–qV/kT} 0

p

0

p

p

p

This is the same relationship that we had above, but expressed slightly differently. Now, imagine attaching a battery of voltage +ΔV to the p-side. This reduces the voltage difference across the junction to V–ΔV, thereby changing the current. Let I be the new hole current flowing from the p-side to the n-side. We now have: pn

I = α q N (p-side) exp{–q(V–ΔV)/kT} I = I exp{+qΔV/kT} pn pn

p

0

The number of holes flowing from the n-side to the p-side is virtually unchanged, since their individual probability of crossing the junction is still nearly 100%. Their current remains I . The net current flow I from the p-side to the n-side is: 0

I = I – I = I [ exp{+qΔV/kT} –1 ] pn

0

0

This is the current that the battery must provide to maintain a voltage difference of V–ΔV. Figure 20-8 shows a graph of current I versus battery voltage ΔV. At ΔV=0, the forward (p-side to the n-side) and backward currents are equal, making the net current zero.

Figure 20-8 Current vs. Voltage for p-n Junction Rectifier

For ΔV0, the device is said to be forward biased; the forward current increases exponentially, which means the device’s forward resistance decreases exponentially toward zero. 0

Our analysis was for positive carriers, but it applies equally for negative carriers. The direction of current flow will be the same, with the electrons moving in the opposite direction of the holes.

Combining carriers of both polarities doubles I , I , and I, resulting in the same final equation. 0

pn

Feynman notes that we obtained the same rectification curve in a mechanical system: the ratchet and pawl in Feynman Simplified 1B Chapter 23.

The Transistor A digital transistor is an electronically controlled switch that turns an output on or off, in accordance with an input signal. The transistor is, without any question, the most important semiconductor device, and one of the most important inventions of the 20th century. William Shockley, John Bardeen, and Walter Brattain made the first working transistor at Bell Labs in 1947, and received the 1956 Nobel Prize for their achievement. I had the pleasure of seeing that first transistor; it was about the size of a tennis ball. Transistors have come a long way in the following six decades. The smallest commercial transistors are now 14 nm wide, and single computer chips can contain a billion transistors. At the leading edge of research, a Korean team produced a 3 nm-wide transistor in 2006, and Australian physicists produced a 0.36 nm-wide transistor made from a single atom of phosphorus in 2012. Many different transistor designs have been developed, but they all employ common principles. The simplest design is a sandwich of three regions of alternating carrier polarities. Let’s consider a type of transistor called p-n-p. As illustrated in Figure 20-9, a p-n-p transistor is basically two p-n junctions put back-to-back, with electrodes attached to each region. In the figure, a thin base region in the middle separates the emitter on the left and the collector on the right. The base is doped n-type, while the emitter and collector are both doped p-type. For simplicity, assume the p-type regions are equally doped.

Figure 20-9 p-n-p Transistor Top, From Left: Emitter, Base, & Collector Middle: Potential Without External Voltages Lower: Potential With Applied Voltages

Without applied external voltages, the n-type region has a higher voltage potential than the p-type regions, just as we described above for the p-n junction. This is represented by the curve in the middle of Figure 20-9. In typical operation, the collector voltage is set to zero and the emitter voltage is positive, often connected to +5 volts through a resistor. The base voltage controls the transistor’s operation, and is never greater than the emitter voltage nor less than the collector voltage. In a p-n-p transistor, current is predominantly due to positive carriers. Since the voltage applied to the emitter is higher than that applied to the base, that p-n junction is forward biased, as described in the last section, leading to a large flow of positive carriers into the ntype base. The vast majority of this current flow passes through the base and into the collector, with only a modest amount flowing out through the base electrode. Over 99% of the emitter current can reach the collector. There are four reasons for such high efficiency. Firstly, the collector has a lower applied voltage than the base, making the collector more attractive to positive carriers than the base. Secondly, the intrinsic voltage potential is lower in the ptype collector than in the n-type base; the large negative potential gradient near the base-collector junction creates a strong electric field pulling positive carriers toward the collector. Thirdly, transistor bases are fabricated to be as thin as possible to enhance the probability of positive carriers passing straight through without recombining with negative carriers that dominate in the base. And fourthly, bases are typically lightly doped, further reducing the recombination rate of positive carriers. As stated above, the emitter-base junction is forward biased, which means that the p-n junction is

operating on the steep part of the rectifier curve shown in Figure 20-8. Even small changes in the applied base voltage therefore result in large changes in current flowing from emitter to collector. This makes transistors superb amplifiers — base voltage changes produce collector current changes that are up to 200 times larger. Electron flow is undesirable in a p-n-p transistor, and is reduced by minimizing the n-type doping of the base. Transistors are also fabricated with the opposite doping scheme. An n-p-n transistor has a thin p-type base region sandwiched between an n-type emitter and an n-type collector. In an n-p-n transistor, it is the negative carriers that carry most of the current. In addition, there are many other types of transistors that optimize various performance features for a wide variety of applications.

Chapter 20 Review: Key Ideas Silicon is the most commonly used semiconductor, with germanium and gallium-arsenide being important in specialized applications. These materials are good insulators at low temperatures, and are modestly conducting at room temperature; hence the name semiconductors. Electric current in a semiconductor is carried by both negative carriers (free electrons) and positive carriers (holes, vacant electron orbits in crystal atoms). The minimum energy required to create a negative/positive carrier pair within a semiconductor is called its band gap, E , which at room temperature is 1.11 eV for silicon, 0.67 eV for germanium, and 1.54 for gallium-arsenide. gap

There are several ways to create electron-hole pairs, including photon absorption in solar cells, ionization by high-energy particles in solid-state detectors, and conversion of thermal energy at sufficient temperatures, where the probability of carrier pair creation is proportional to exp{– E /kT}. gap

In a pure crystal, the negative carrier density N and the positive carrier density N are equal. Adding dopants, atoms of other elements, increases one carrier density while reducing the other. Their product is always proportional to exp{–E /kT}. n

p

gap

For pure silicon at room temperature, N and N are 1.08×10 per cm . In silicon, boron is a p-type dopant that increases N , while arsenic and phosphorus are n-type dopants that increase N . In heavily doped silicon, one carrier density can become 10 per cm . Doping dramatically increases the conductivity of semiconductors. +10

n

3

p

p

n

+18

3

In the Hall effect, a semiconductor carrying a current j is subject to a traverse magnetic field B. This forces carriers of both polarities to accumulate on the same side of the crystal. If one carrier polarity dominates, the Hall effect creates an electric field E, with E=–jB/qN, where N is the dominant carrier

density, and q is the carrier unit charge. A p-n junction is the interface between a p-type region and an n-type region within a semiconductor. Without an externally applied voltage, electrons diffuse into the lower-electron-density p-side, while holes diffuse into the n-side. Both effects give the n-side a net positive charge and the p-side a net negative charge. Eventually, charge segregation builds an electric field that stops the net flow of charge. At equilibrium, the same current I flows in both directions. Attaching a bias voltage of +ΔV to the p-type region creates a rectifier, which allows positive current to flow easily from p to n and inhibits current flow from n to p. Net current flow from p to n is: 0

I = I [ exp{+qΔV/kT} –1 ] 0

A p-n-p transistor has a thin n-type base region sandwiched between a p-type emitter and a p-type collector. Positive carriers carry most of the current in a p-n-p transistor. Typically, the collector voltage is set to zero and the emitter voltage is connected to +5 volts through a resistor. The base voltage is between the other voltages, and controls the transistor’s operation. The emitter-base junction is forward biased, leading to a large flow of positive carriers into the base. Typically over 99% of this current flow passes through the base and into the collector. Even small changes in the applied base voltage result in large changes in current flowing from emitter to collector — base voltage changes produce collector current changes up to 200 times larger. An n-p-n transistor has a thin p-type base region sandwiched between an n-type emitter and an ntype collector. In an n-p-n transistor, it is the negative carriers that carry most of the current.

Chapter 21 Independent Particle Approximation Natural phenomena are often complicated convolutions of many different factors. Earth’s motion, for example, is dominated by our Sun’s gravity. But, we are also affected by the gravity of the moon, of the other planets, of the other stars in the Milky Way, and on and on. No one could cope with all that complexity at once. Our success in understanding nature often relies on clever approximations — identifying the most important factors and neglecting, at least initially, the others. This approach is essential both in advancing science and in learning science. It is hard enough to learn all the major principles without having to simultaneously understand the innumerable potential complications. But these lesser influences should not be neglected indefinitely nor go without notice. Sometimes significant discoveries are made by examining discrepancies caused by annoying complications. An example of reducing complex phenomena to manageable proportions is the independent particle approximation stealthily employed in the last three chapters. We examined free electrons in crystals and focused on how these electrons interact with immobile atoms. But we did not address the interactions of these electrons with one another — we approximated each as an independent particle. Starting in V3p15-1, Feynman examines new phenomena with and without the independent particle approximation.

Ferromagnetism We will first consider a ferromagnetic crystal. Ferromagnetism is described in classical terms starting in V2p36-1. But in fact, ferromagnetism is an entirely quantum mechanical phenomenon. Very briefly, here are the highlights of ferromagnetism. Elementary particles with spin have magnetic dipole moments. In classical physics, a spinning electric charge creates a circulating electric current that creates a magnetic field. Electrons are the primary source of magnetism, because their dipole moments are 1000x greater than those of protons or neutrons. Fully occupied electron orbits contain two electrons with opposite spins, so their magnetic moments cancel. Hence, an atom’s net magnetic field comes from unpaired electrons in outer valence orbits. In ferromagnetic materials, unpaired electrons in neighboring atoms tend to align themselves parallel to one another. Summed over

countless atoms, this results in a macroscopically substantial magnetic field. This effect is contrary to the normal rule that dipoles align antiparallel to an electromagnetic field. However, in atoms, electrons with parallel spins must be farther apart, due to the Pauli exclusion principle. That greater separation reduces electrostatic repulsion, and makes parallel spin alignment the lowest energy state. We begin by analyzing a crystal in which each atom has one unpaired valence electron. We make three simplifying approximations: (1) the crystal is a one-dimensional array of identical atoms; (2) each unpaired electron is localized at one atom at a time; and (3) there are spin interactions between adjacent pairs of electrons but not between non-adjacent pairs. As in our discussion of the hyperfine splitting of hydrogen (Chapter 16), the interaction energy for each adjacent electron pair is: E = –σ •σ A/2, for m=n+1 nm

n

m

Stipulating m=n+1 ensures we count each pair exactly once. For A>0, total energy is reduced when spins are parallel. The factor of 2 is included to provide consistency with Chapter 17. For N+1 atoms in a linear array, there are N adjacent pairs. At each lattice site, there is one electron whose spin is either up or down. Specifying the spins at every lattice site provides a complete description of the crystal’s state. The Hamiltonian for the entire system is: H = – (A/2) Σ σ •σ n

n

n+1

The lowest energy level corresponds to all spins being parallel along any arbitrary direction. For convenience, let’s call that direction “up.” Recall from Chapter 16, the Pauli spin exchange operator P , and its action on a state with two spins: swap

P P P P

swap swap swap swap

|+ +> = |+ +> |+ –> = |– +> |– +> = |+ –> |– –> = |– –>

σ •σ = 2P n

m

swap

–1

This allows us to rewrite the Hamiltonian in a simpler form. For better ereading clarity, we replace the subscript “swap” on P with the numbers of the two electrons whose spins are being exchanged. We then have: H = – A Σ (P n

n,n+1

–1/2)

When P operates on a state with spin up at both lattice sites n and n+1, it simply returns the original state, which is equivalent to multiplying the state by +1. Hence: n,n+1

(P

n,n+1

–1/2) |+ +> = (1/2) |+ +>

The state of all-spins-up, |++…++>, is called the ground state. Because operator P multiplies the allspins-up state by a constant, the ground state is an eigenstate with energy E , where: 0

E E E E

|++…++> = H |++…++> |++…++> = –A Σ (1/2) |++…++> |++…++> = –A (N/2) |++…++> = – AN/2

0 0

n

0 0

Recall that N is the total number of adjacent pairs of atoms. For convenience, we rewrite the Hamiltonian factoring out E . 0

H = E – A Σ (P 0

n

n,n+1

–1)

Let’s double check: (P

n,n+1

–1) operating on |++…++> is zero for all n, so the energy of that state is E . 0

Single Spin Waves Now, let’s consider the energy of states with exactly one spin down, as sketched in Figure 21-1.

Figure 21-1 All Spins Up Except n = 5

Define |n> to be the state with all spins up except the spin at atom #n. The set of states |0> through |N> is a complete basis set for a crystal with only one spin down. How does the Hamiltonian operate on these basis states? Take state |5> for example. The operator P exchanges the spins at atoms 7 and 8, which are both spin up in state |5>; hence P makes no change to |5>.

7,8

7,8

P |5> = |5> (P -1) |5> = 0 7,8

7,8

This means all P’s without a “5” subscript make zero contribution to the Hamiltonian. But when n or n+1 equals 5, P does change state |5>. Specifically: n,n+1

(P –1) |5> = |4> – |5> 4,5

(P –1) |5> = |6> – |5> 5,6

The Hamiltonian operating on |5> is therefore: H |5> = E –A (|4> + |6> – 2|5>) 0

In general, the Hamiltonian operating on state |n> is: H |n> = E –A (|n–1> + |n+1> – 2|n>) 0

In V3p15-3, Feynman points out that the spin-spin interaction σ •σ exchanges spins but never changes the sum of the spins of all atoms. Total spin must be conserved with this Hamiltonian because that is the only form of angular momentum that we included in our analysis. n

n+1

The components of the Hamiltonian matrix are therefore all zero except: H = E + 2A H = H = –A n,n

0

n,n–1

n,n+1

This is the same Hamiltonian we solved for a free electron in a crystal in Chapter 18, with a slightly different definition of E . As before, we look for stationary states given by: 0

|ø> = Σ C |n> n

n

C = a exp{Et/iħ} n

n

Here all a ’s, the amplitudes to be in each basis state, are constant over time. Since these amplitudes don’t change, |ø> is always the same mixture of basis states, making it a stationary state of definite energy E. n

Define b to be the atomic spacing, which makes x , the x-coordinate of atom #n, equal to nb. n

For a trial solution of the form: a (k) = exp{ikx } = exp{iknb} n

n

the Hamiltonian equations become: E(k) exp{iknb} = (E +2A)exp{iknb} – Aexp{ik(n–1)b} – Aexp{ik(n+1)b} 0

E(k) = (E +2A) – Aexp{–ikb} – Aexp{+ikb} E(k) = (E +2A) – 2A cos(kb) 0 0

The eigenstates of definite energy are waves that propagate through the crystal flipping adjacent electron spins: at each lattice site, one spin flips down while its neighbor flips up. These are called spin waves, which clearly conserve angular momentum.

As before, for long wavelengths (small k values), the cosine approximates a parabola. If we also suppress the zero-point energy, the result is: E(k) = + A k b 2

2

This has the form of the Newtonian equation for kinetic energy. To see this, substitute mv = p =ħk. m v /2 = A b (m v / ħ ) m = ħ / (2 A b ) 2

2

2

2

2

2

2

The down spin travels through the crystal as if it were a particle of mass m. We call such nonmaterial propagating effects quasiparticles, and a quasiparticle of this spin-flipping type is called a magnon.

Double Spin Waves Next we consider states in which all electrons have spin up, except two, as shown in Figure 21-2. We define |n,m> to be the basis state of all spins up except at atom #n and atom #m. Note that |m,n> denotes the same state as |n,m>; there are two labels, but only one such state. Also, there is no state corresponding to |n,n>, since the electron at atom #n cannot be spin down twice. A simple way to eliminate these labeling issues is to require m>n.

Figure 21-2 All Spins Up Except n= 2 & 5

Stationary states have the form: |ø> = Σ Σ n

m>n

C |n,m> n,m

C = a exp{Et/iħ} n,m

n,m

Here all a ’s, the amplitudes to be in each basis state, are constant over time, and the sums are over all n, and all m>n. n,m

In V3p15-4, Feynman says: “The complications which now arise are not complications of ideas—they are merely

complexities in bookkeeping. (One of the complexities of quantum mechanics is just the bookkeeping. With more and more down spins, the notation becomes more and more elaborate with lots of indices and the equations always look very horrifying; but the ideas are not necessarily more complicated than in the simplest case.)” In the single spin down case discussed above, the Hamiltonian is: H = – (A/2) Σ σ •σ H = E – A Σ (P –1) H |n> = E –A (|n–1> + |n+1> – 2|n>) n

0

n

n

n+1

n,n+1

0

This leads to the following being the only non-zero Hamiltonian components: H = E + 2A H = H = –A n,n

0

n,n–1

n,n+1

Here, the “bookkeeping” expands into new dimensions: the C’s and a’s now have two indices, making them N×N matrices; and H now has four indices, making it a N×N×N×N 4-dimensional matrix. And adding to our enjoyment, N is so enormously large that it is easier to assume it is infinite. Let’s see how Feynman handles bookkeeping in four infinite dimensions. The Hamiltonian is: H = – (A/2) Σ σ •σ – (A/2) Σ σ •σ n

Define H

n,n+1,m,m+1

n

n+1

m

m

m+1

=

If atoms n and m are far apart, their actions are independent and we just have twice as much of everything as before. H

n,n+1,m,m+1

= 2E –A(P 0

n,n+1

–1) –A(P

m,m+1

–1)

H

= 2E –A(|n-–1,m> +|n+1,m> –2|n,m>) –A(|n,m–1> +|n,m+1> –2|n,m>)

H

= 2E –A(|n–1,m> +|n+1,m> |n,m–1> +|n,m+1> –4|n,m>)

n,n+1,m,m+1

n,n+1,m,m+1

0

0

Now, what happens when n and m are not far apart? If for example, m=n+1, the two down spins are adjacent. If n flips to spin up, n+1 cannot become more spin down than it already is, and hence angular momentum cannot be conserved. For m=n+1, the states |n+1,m> and |n,m–1> become |n+1,n+1> and |n,n> neither of which is legitimate. To be precise, we must eliminate these terms from the Hamiltonian, which greatly complicates the bookkeeping. Feynman’s solution: ignore this complication — make the approximation that of the “doubly” infinite combinations of n and m, we can neglect cases like m=n+1, which are only “singly” infinite.

With that approximation, solutions have the form: a = exp{ik nb} exp{ik mb} n,m

1

2

E = 2E + 4A –2A cos(k b) –2A cos(k b) 0

1

2

This is just what we would get if we assumed the two down spins were completely independent. We would then have a quasiparticle “n” with: a = exp{ik nb} E = E + 2A – 2A cos(k b) n

n

1

0

1

and a quasiparticle “m” with: a = exp{ik mb} E = E + 2A – 2A cos(k b) m

m

2

0

2

The rule for independent actions is to multiply the amplitudes and sum the energies. This rule yields exactly what we derived above, which means the assumption in our derivation is equivalent to the independent particle approximation. This isn’t very surprising, since we assumed we could ignore what happens when the particles come together.

An Exact Two Spin Solution As a side note, Feynman says that, for the case of precisely two spin downs, the exact solution including spin-spin interactions is calculable. It is: a = exp{ik (n+m)b} sin{k (n–m)b} n,m

+

–

where k = (k +k )/2 and k = (k –k )/2. +

1

2

–

1

2

This solution has the same energy as above. In V3p15-6, Feynman says: “This solution includes the ‘interaction’ of the two spins. It describes the fact that when the spins come together there is a certain chance of scattering. The spins act very much like particles with an interaction. But the detailed theory of their scattering goes beyond what we want to talk about here.” No exact solution is presented for three or more down spins.

Independent Particles

The approximate solution derived above for two down spins is: a = exp{ik nb} exp{ik mb} n,m

1

2

While this is indeed a solution, Feynman says it is not the most complete solution because it lacks the symmetry of the natural phenomenon we are trying to describe. In all branches of physics, symmetry is very important, but it is particularly important in quantum mechanics due to the special role of identical particle statistics (bosons and fermions). We said earlier that |n,m> and |m,n> refer to exactly the same state, of which there exists only one, not two. Yet, because k and k are different in general, if we swap n and m, our solution changes, which it should not. Feynman notes that another valid solution is: 1

2

a = exp{ik nb} exp{ik mb} n,m

2

1

Any linear combination of the first and second solutions is also a solution. The symmetric solution we seek is: a = exp{ik nb} exp{ik mb} + exp{ik nb} exp{ik mb} n,m

1

2

2

1

This amplitude represents one combined state that is the superposition of two quasiparticle magnons, one with wave number k and the other with wave number k . This is the same type of superposition state that we discovered in analyzing the hyperfine splitting of hydrogen in Chapter 16, There the electron and proton spins of one eigenstate is: 1

2

|up, down> + |down, up> We know that the spins are opposite, but we cannot know which particle has which spin. As we learned there, this spin combination belongs to a triplet of spin 1 states, making it a boson. Here also, this combined two-down-spin state is a boson with spin 1. In other situations, two solutions might be superposed with a minus sign, making the combination a fermion. But that is inappropriate here, because |n,m> and |m,n> are not two identical states, but are actually the same single state; therefore, switching n and m must not change the amplitudes at all, not even by a minus sign. Feynman says: “In general, quasi particles … may act like either Bose particles or Fermi particles.... The ‘magnon’ stands for a spin-up electron [being] turned over. The change in spin is one. The magnon has an integral spin, and is a boson.” The essential point of the two-spin-down phenomenon is the behavior of independent particles. When two or more particles do not interact amongst themselves, their amplitudes multiply and their energies add. For example, if: α exp{E t/iħ} = amplitude particle 1 does A 1

β exp{E t/iħ} = amplitude particle 2 does B 2

then the amplitude that particle 1 does action A and particle 2 does action B is: αβ exp{(E +E )t/iħ} 1

2

Since the amplitudes multiply, clearly the energies in the exponent must add. Whatever actions independent particles can do individually they can also do in combination. Feynman adds this final note: “However, you must remember that if they are identical particles, they [will] behave either as Bose or as Fermi particles depending upon the problem. Two extra electrons added to a crystal, for instance, would have to behave like Fermi particles. When the positions of two electrons are interchanged, the amplitude must reverse sign. In the equation [combining the two solutions] there would have to be a minus sign between the two [solutions]. As a consequence, two Fermi particles cannot be in exactly the same condition—with equal spins and equal k’s. The amplitude for this state is zero.”

Ethylene Molecule Ethylene contains two carbon atoms and four hydrogen atoms, with a covalent double bond between the two carbon atoms, as shown in Figure 21-3.

Figure 21-3 Ethy lene Molecule

In addition to four valence electrons, a carbon atom has two electrons in its innermost shell that are very tightly bound. We will assume throughout that such inner shell electrons never change states, and need not concern us. Let’s imagine removing one electron from the central double bond making a singly-ionized ethylene molecule. This creates a two-state system in which the two carbon atoms have a single bond between them and one extra electron. The two electrons in all single bonds, including both the carbon-carbon

bonds and the carbon-hydrogen bonds, are sufficiently tightly bound that we can ignore them and assume their states never change. That leaves only the extra electron to consider. It has two basis states: |1> electron on the upper carbon atom; and |2> electron on the lower carbon atom. (Remember how much simpler two-state systems are? No 4-D infinite matrices, etc.) The two stationary states and energy levels are: E =E +A: (|1>–|2>)/2 E =E –A: (|1>+|2>)/2 +

0

–

0

Here –A is the amplitude to transition from |1> to |2>. At moderate temperatures, the electron will occupy the lowest energy state with E=E –A. 0

Now, imagine replacing the electron that we removed earlier. Into what state will this second electron go? Neglecting interactions between these two electrons (the independent particle approximation), the second electron will enter the same state as the first, but will have the opposite spin. (Pauli excludes them from being in the same state with the same spin.) With our independent particle approximation, the energy of the second electron equals that of the first electron, and their total energy is: E = 2(E –A) 0

We discover that the energy of the double bond in ethylene is 2(E –A). 0

Let’s next apply these ideas to more complex molecules.

Benzene Molecule All the principles and rules needed to determine the behavior of molecules are provided by quantum mechanics. However, the equations of quantum mechanics are sufficiently complex that they have been solved exactly for only the simplest atoms: hydrogen and helium. Let’s see to what extent clever approximations can provide useful, if inexact, solutions. In Chapter 13, we described the benzene molecule: six carbon atoms in a ring, with six hydrogen atoms outside the ring, one attached to each carbon atom. Among the six carbon-carbon bonds forming the ring, three are double bonds and three are single bonds. We used this as a example of a two-state system, with the two basis states shown in Figure 21-4.

Figure 21-4 Two Basis States of Benzene

We found that the stationary states are mixtures of the basis states, with the lowest energy being E –A, where E is the energy of either basis state and –A is the transition amplitude between them. That analysis assumed only two basis states; we made the approximation that we could neglect all the excited states of benzene. 0

0

Feynman now reanalyzes this molecule with a different approach, making different approximations. Different approaches, with different approximations, can lead to different results. If someone were able to entirely eliminate the approximations made in either approach, they would get the exact description of benzene. Absent that, Feynman says: “you should not be surprised if the two descriptions do not agree exactly.” Our new approach is to remove six electrons from the three carbon-carbon double bonds. All six ring bonds are now single bonds, and the ionized molecule has a net charge of +6. Again, we assume all electrons in single bonds can be ignored. We next add back one electron. It could attach itself to any carbon atom, giving us six basis states. It can also transition from one carbon atom to its neighbor. This is now a six-state system that we could solve in the usual manner. Or, we could use one of Feynman’s tricks. In Chapter 18, we solved the infinite-state system of an electron in an infinite, linear array of crystal atoms. Six is somewhat less than infinity. Nonetheless, we can adapt the infinite solution to a finite ring of any length, say N atoms. In a ring, equally spaced atoms go on forever in both directions, with the added feature that the ring repeats every N atoms. The stationary state solution and its energy that we derived for an infinite linear array are: a (k) = exp{iknb} n

E = E –2A cos(kb) 0

We now add the requirement that the amplitude repeat exactly every N atoms. exp{ik(N+n)b} = exp{iknb} exp{ikNb} =1 kNb = 2πm, for some integer m In the infinite case, we showed that all solutions correspond to k values in the range:

–π/b < k ≤ +π/b. This means all ring solutions correspond to k values in the range: –N/2 < m ≤ +N/2 The N-atom ring has N stationary states of definite energy. Each value of m defines one state with k equal to: k = (2π/Nb) m m

The energy levels of these states are: E = E –2A cos(k b) E = E –2A cosθ θ = (2π/N) m 0

m

0

The lowest energy is E –2A, which occurs for m=0. Each increment of m increases θ by 2π/N. 0

The energy levels for benzene (N=6) are displayed in Figure 21-4 in the graphic form Feynman suggests. Here, the eigenvalues correspond to points on the circle with 60º (π/3 radians) angles between each value of m.

Figure 21-4 Electron Energy Levels in Benzene

For benzene, the six stationary states have four distinct energy levels: for m=+3, E = E for m=±2, E = E for m=±1, E = E for m = 0, E = E

0 0 0

0

+2A +A –A –2A

The eigenvectors are given by:

a (k) = exp{iknb} a (k) = exp{i(π/3)nm} n n

for m=+3, (–1,+1,–1,+1,–1,+1)/√6 for m=±2, (–1,–1,+2,–1,–1,+2)/√12 for m=±1, (+1,–1,–2,–1,+1,+2)/√12 for m = 0, (+1,+1,+1,+1,+1,+1)/√6 States m=–1 and m=+1 are said to be degenerate because they have the same energy. States m=–2 and m=+2 are also degenerate. Each value of m identifies an individual state that can accommodate two electrons, one spin up and the other spin down. The first two of benzene’s six outer electrons fill the lowest energy state, m=0. The next four fill the next lowest energy states, m=–1 and m=+1. The total energy of these six electrons is: E = 2(E –2A) + 4(E –A) = 6E –8A 0

0

0

This is less than 6(E –A), the energy of three double bonds that we derived above for ethylene. This means benzene is more tightly bound and more stable than ethylene. This tighter bonding means the valence electrons are more efficiently shared by the carbon atoms than they would be in any threedouble-plus-three-single bond configuration. The eigenvectors of this Hamiltonian show that the six carbon atoms share their valence electrons equally: in effect, benzene has six 1.5 bonds. 0

Exposure to light can excite benzene’s outer electrons and elevate them to the m=±2 or m=3 states. A photon of energy E=hf=2A can elevate a benzene electron from state m=±1 to m=±2. A photon with E=3A can elevate an electron from m=0 to m=±2, or from m=±1 to m=3, and with E=4A the m=0 to m=3 transition is possible. Measurements of the binding energy of ethylene and benzene indicate A=0.8 eV. However, measurements of the photon absorption spectrum of benzene indicate A is between 1.4 and 2.4 eV. In V3p15-9, Feynman says the substantial difference between these two different measurements of A demonstrates the limitations of neglecting electron-electron interactions.

Butadiene Molecule Organic chemistry is a fascinating field unto itself; but that isn’t Feynman's area of expertise, nor is it mine. I will, therefore, abbreviate Feynman’s further discussion of organic chemistry and organic chemists, and focus on those aspects that illuminate quantum physics. We have examined a two-state molecule (ethylene) and a ring molecule (benzene). Now let’s consider a linear molecule. As shown in Figure 21-5, butadiene consists of a line of four carbon atoms, with two carbon-carbon double bonds and six hydrogen atoms. The sole single bond can be between any pair of adjacent carbon atoms, leading to three possible combinations that we can select as basis states.

Figure 21-5 One Basis State of Butadiene

Just as we did with ethylene and benzene, we will analyze butadiene by imagining that we remove four electrons from the two double bonds of butadiene, and then add them back one by one, assuming the independent particle approximation. We could run through all the Hamiltonian equations for the case of four atoms in a row, but it is easier to use another of Feynman’s tricks. We can easily derive the equations for a four-atom line from our prior solution to an infinite line of atoms. Figure 21-6 shows a row of carbon atoms that we might imagine extending to infinity in each direction. Now consider how the Hamiltonian changes when the line length changes from infinity to four.

Figure 21-6 Row of Carbon Atoms

In an infinite line, there are electron transitions from atom #0 to atom #1, and also from atom #5 to atom #4. However, these transitions cannot occur in a four-atom line. As far as atoms 1 though 4 are concerned, that is the only difference between an infinite line and a four-atom line. We can eliminate that difference by modifying the infinite solution. We force the electron amplitudes to be zero at atom #0 and at atom #5. If electrons can never be at atom #0 or at atom #5, then the fouratom line is isolated and it makes no difference whether the rest of the infinite line exists or not. How do we zero the amplitudes at atom #0 and atom #5? Recall that our solutions to the infinite line include both positive and negative wave numbers. A valid solution can be any linear combination of the form: a (k) = exp{+iknb} – exp{–iknb} a (k) ~ sin(knb) n n

Recall that n is the atom number and b is the atomic spacing. Since we explicitly include a minus sign in the second exponent, only positive values of k are required to span all solutions. Note that this is a stationary state, because it contains only one value of k. Subtracting the two exponentials ensures that a=0 at n=0. We can force a=0 at atom #N+1 by requiring:

k (N+1) b = π m, for integer m kb = π m / (N+1) The only values of m that correspond to stationary states are: 0 < m < N+1 For m=0, a (k) is zero at every atom, while for m>N, the same N solutions are simply repeated. For butadiene, N=4 and m has the values +1, +2, +3, and +4. The stationary state wave numbers and energy levels are: n

E = E –2A cos(kb) 0

m = 4, kb=4π/5, E = E m = 3, kb=3π/5, E = E m = 2, kb=2π/5, E = E m = 1, kb = π/5, E = E

0 0 0 0

+1.618A +0.618A –0.618A –1.618A

Note that ø = 2cos(π/5) = 1.618… is called the golden ratio. The special properties of ø were first discovered by the ancient Greeks. Innumerable geometry problems and natural phenomena exhibit this ratio, including the chambers of the nautilus, and the arrangement of sunflower seeds. Artists consider this ratio to be inherently appealing; the height and width of Mona Lisa’s face are in the golden ratio. (I’ll bet you didn’t know what the Mona Lisa and butadiene have in common.) The golden ratio has the amusing property that ø–1=1/ø. Two of butadiene’s valence electrons occupy the m=1 state and the other two occupy the m=2 state, making their total binding energy: E = 2(E –1.618A) + 2(E –0.618A) E = 4(E –A) – 0.472A 0

0

0

Molecular Stability The independent particle approximation is not adequate to precisely calculate energy levels, but it does properly describe qualitative trends. In all the molecules examined above, we found a series of energy levels, each of which can accommodate two electrons of opposite spin. This means that as electrons are added to a highly ionized molecule, pairs of electrons fill each state, starting with the lowest energy state and progressively filling states of higher and higher energy. Absent degeneracy (if none of the stationary states have the same energy), each group of two electrons has a different binding energy. If some states are degenerate, such as the m=±1 and m=±2 states of benzene, the groups of electrons with the same binding energy become larger (>2). All such groups, also called energy shells, always have an even number of electrons, half with spin up and half with

spin down. Quantum mechanics clearly identifies the group sizes for each type of molecule. Measurements show that molecules are most stable when their least-bound (highest energy) electrons are members of a completely filled group. For example, the group sizes of benzene, from lowest energy to highest energy, are: 2, 4, 4, and 2, corresponding to m=0, ±1, ±2, and 3. Benzene is more stable with 2 electrons rather than 1 or 3, and is more stable with 6 electrons (2+4), rather than 5 or 7. Consider another example: a molecule with a ring of three atoms. Using the N-atom-ring solutions derived above, the N=3 stationary state energy levels are: E = E –2A cos(θ) θ = (2π/3) m, for m = –1, 0, +1 0

for m=±1, E = E +A for m = 0, E = E –2A 0

0

For this molecule, the groups are: 2 and 4. It is most stable with two electrons filling the m=0 state, and much less stable with the addition of a 3rd, 4th, or 5th electron. This means the stable state of this molecule has one fewer electron than ring atoms, giving it a net charge of +1. For a molecule with a ring of five atoms, the energy levels are: m=±2, E = E +1.618A m=±1, E = E –0.618A m = 0, E = E –2A 0 0

0

The groups are: 2, 4, and 4. Such a molecule is more stable with 6 electrons (2+4) than 5, giving it a net charge of –1.

Concluding Comments In V3p15-12, Feynman says the independent particle approximation is qualitatively useful in understanding: (1) the arrangement of electron orbits in atoms, which leads to the structure of the Periodic Table; and (2) the arrangement of proton and neutron states in nuclei, which results in enhanced stability of nuclei with magic numbers of particles. The magic numbers are 2, 8, 20, 28, 50, and 82. I think it makes sense to defer both topics, since we will present a more complete theory of particle orbits after discussing the quantization of angular momentum. Feynman ends this lecture saying: “The independent particle approximation has been found useful in a wide range of subjects—

from solid-state physics, to chemistry, to biology, to nuclear physics. It is often only a crude approximation, but is able to give an understanding of why there are especially stable conditions —in shells. Since it omits all of the complexity of the interactions between the individual particles, we should not be surprised that it often fails completely to give correctly many important details.”

Chapter 21 Review: Key Ideas Natural phenomena are often complicated convolutions of many different factors. Our success in understanding nature often relies on clever approximations — identifying the most important factors and neglecting, at least initially, the others. One means of reducing complex phenomena to manageable proportions is the independent particle approximation. When a few “special” particles are subject to a pervasive, unchanging “background”, assuming that special particles act independently of one another, can provide useful insights, if not precise numerical results. »For a 1-D crystal whose atoms each have one valence electron, the Hamiltonian for the spin-spin interactions of adjacent electrons is: H = –(A/2) Σ σ •σ = E –A Σ (P n

n

n+1

0

n

n,n+1

–1)

Here: E = – AN/2 is the energy of the ground state, the state of all spins up; N+1 is the total number of atoms; and P is the Pauli spin exchange operator that swaps the spins of the electron of atom #n with the electron of atom #n+1. 0

n,n+1

»For a crystal with only one down spin, each eigenvector has the form: |ø> = exp{Et/iħ} Σ a |n>, where |n> is the basis state of all spins up except at atom #n. n

n

The eigenstates are spin waves, with wave number k, that propagate through the crystal flipping adjacent electron spins. Define b to be the atomic spacing. Each value of k in the range, –π/b < k ≤ +π/b, corresponds to an eigenstate, whose components and energy are: a (k) = exp{iknb} E(k) = (E +2A) – 2A cos(kb) n

0

The single down-spin travels through the crystal as if it were a particle of mass m=ħ /(2Ab ). We call such non-material propagating effects quasiparticles, and a quasiparticle of this spin-flipping type is called a magnon. 2

2

»For a crystal with exactly two down spins, each eigenvector has the form: |ø> = exp{Et/iħ} Σ Σ n

m>n

a |n,m>, where |n,m> is the state of all spins up except at atom #n and atom #m. The properly symmetric amplitudes and energy levels of the stationary states are: n,m

a = exp{ik nb} exp{ik mb} + exp{ik nb} exp{ik mb} n,m

1

2

2

1

E = 2E +4A –2Acos(k b) –2Acos(k b) 0

1

2

»The solution for an N-atom ring molecule is obtained from the solution for an infinite line of atoms by requiring that the solution repeat every N atoms. The eigenstates are (for integer m): a (k) = exp{iknb}, for n=1…N k = (2π/Nb)m, for –N/2 and another set of conditions might be state |ψ>. A state describes all relevant extrinsic properties that the particle can have, including momentum, spin, and position. A state does not describe a particle’s intrinsic properties, including charge and mass, because these never change and are thus not part of the dynamics of interest. States are mathematically similar to vectors. The amplitude for state |ø> to also be in state |ψ> is written and is often called a projection amplitude, because it projects one state onto another, just as the dot product projects one vector onto another. Amplitudes are generally complex quantities of the form: αexp{iβ}. The probability that a particle in state |ø> is in state |ψ> equals || = × *, where || is the magnitude of the amplitude squared, z* denotes the complex conjugate of z for any z, and * for any ø. States are generally normalized so that =1. 2

2

To analyze a quantum system, we begin by defining a set of basis states that might be labeled |1>, |2>, …|N>. Basis states are the equivalent of coordinate axes, and must be orthonormal and complete. Orthonormal means: = δ = 1 if j=k and 0 otherwise jk

Complete means any state |ø> equals some linear combination of basis states according to: |ø> = Σ |n> n

Here is the amplitude that |ø> is in basis state |n>, and the sum is over all basis states. Lastly, we can express as: = Σ n

Chapter Roadmap Feynman provides an overview of what will be accomplished in this chapter — a roadmap of our coming journey. Recall what we learned about electrons in a 1-D crystal, a linear array of identical atoms that are equally spaced with separation b. Let iA/ħ be the amplitude for an electron to jump from one atom to its neighbor. We found that the states of definite energy are waves that propagate through the crystal, with the electron hopping from atom to atom. For long wavelengths (small wave number k), the wave energy E is approximately: E = (E –2A) + Ak b 2

2

0

These waves have properties that are mathematically similar to those of real particles. They are called quasiparticles and each has an effective mass given by: m = ħ /(2Ab ) 2

2

eff

Now we are ready to look down the road ahead. In V3p16-2, Feynman suggests we imagine the atomic spacing b becoming smaller and smaller. In the limit that b goes to zero, the atom positions become points along a continuous line, and the electron can be anywhere along that line. If we next remove all the atoms, Feynman says: “This would be one way to describe the motion of an electron in a vacuum. In other words, if we imagine that space can be labeled by an infinity of points all very close together and we can work out the equations that relate the amplitudes at one point to the amplitudes at neighboring points, we will have the quantum mechanical laws of motion of an electron in space.” Let’s see how this works. Define a set of basis states in which |n> denotes the state of the electron being at atom #n. Any state |ψ> can be represented as: |ψ> = Σ C |n>, where C = n

n

n

The Hamiltonian equation for a 1-D crystal (see Chapter 18) yields: iħ ∂C /∂t = E C –AC –AC n

0

n

n+1

n–1

The right-most term represents the transition of atom #n–1 to atom #n, while the preceding term represents the transition from atom #n+1 to atom #n. We employ the partial derivative ∂/∂t since C is a function of x and t and we only want the change in C over time. The solutions of this equation for the states of definite energy are: C = exp{–iEt/ħ} exp{iknb} n

We now replace C by C(x), where we define x=nb. The Hamiltonian equation becomes: n

iħ ∂C(x)/∂t = E C(x) –AC(x+b) –AC(x–b) 0

Next reset the zero-point energy so that E =2A. 0

iħ ∂C(x)/∂t = A (2C(x) –C(x+b) –C(x-b)) We will show below that, in the limit that b goes to zero, this becomes: iħ ∂C(x)/∂t = – Ab ∂ C(x)/∂x 2

2

2

Recall that Ab = ħ / 2m . Since the crystal’s atoms were removed, we can identify m as m, the mass of a free electron. 2

2

eff

eff

What results is the Schrödinger equation for an electron moving in one dimension in vacuum: iħ ∂C(x)/∂t = – (ħ /2m) ∂ C(x)/∂x 2

2

2

Feynman advises that this not a rigorous derivation, but only a demonstration of plausibility. Feynman says this equation describes: “…the diffusion of a probability amplitude from one point to the next along the line. That is, if an electron has a certain amplitude to be at one point, it will, a little time later, have some amplitude to be at neighboring points.” The hallmark distinction between the equation for the diffusion of gas molecules and the equation for the diffusion of probability amplitudes is ‘iħ”, the imaginary coefficient of the time derivative. Gas diffusion has real coefficients that result in exponentials with real exponents. In stark contrast, in quantum mechanics, the imaginary coefficients result in oscillatory complex waves. Here is the derivation of the second derivative employed above. I will show this for the normal derivative, but it applies equally to the partial derivative. Let C(x) be any smooth function of x, and let y(x)=dC(x)/dx be the first derivative of C. By the definition of a derivative, the following equations hold in the limit that dx goes to zero: y(x) = dC(x)/dx = {C(x+dx) – C(x)} / dx y(x+dx) = dC(x+dx)/dx

y(x+dx) = {C(x+2dx) – C(x+dx)} / dx dy/dx = {y(x+dx) – y(x)} / dx dy/dx = { [C(x+2dx)–C(x+dx)/dx – [C(x+dx)–C(x)]/dx} / dx dy/dx = {C(x+2dx) –2C(x+dx) +C(x)}/dx d C(x)/dx = {C(x+2dx) –2C(x+dx) +C(x)}/dx 2

2

2

2

Wave Functions With this roadmap in mind, we begin our voyage to describe the motion of an electron in free space. We will treat space as being continuous, which necessitates continuous functions and an infinite set of basis states. Define the basis state |x> to be the state of an electron being precisely at position x. Any state |ψ> is a linear superposition of basis states. The amplitude of |ψ> being in basis state |x> equals , which we define to be C(x). Thus: C(x) = Feynman offers the comforting thought that continuously varying amplitudes are not new to us. In Chapter 9, we discovered that a particle with definite momentum p, and therefore a definite energy E, has an amplitude to be at position x that is proportional to: exp{ipx/ħ} Since that chapter, we have often used the equivalent expression: exp{ikx}, recall that p=ħk There are then at least two options for basis states: states of definite position; or states of definite momentum. Either choice provides an equally valid description of a quantum system; sometimes one is more convenient than the other. It is advantageous to be comfortable with both approaches. In V3p16-5, Feynman alerts us to a standard physics notation that can be confusing. We have consistently (I hope) used C(x) to denote an amplitude of a state of interest being in basis state |x>. This works well when only one particle is being considered. But, since physicists often deal with multiple particles and their interactions, we will need to be more explicit about to which state C(x) refers. One could add superscripts or subscripts to C, but the standard notation, used in virtually every physics book and research paper, is simply to replace “C” by “ψ”, as in: ψ(x) = For multiple particles, we assign one symbol to each particle, and then proceed with the calculations.

Your challenge is to realize that ψ(x) is the amplitude for a particle to be at position x (in basis state |x>), whereas |ψ> is the particle state that might be spread across all x. Since all physicists use this notation, one must learn to deal with it. Life is full of notational challenges, such as “read” being either present tense or past tense. Hopefully, the “| >” and “( )” are distinctive enough to clarify that | ψ> is a state, and ψ(x) is an amplitude that is a function of x. In fact, ψ(x) is called the wave function since it incorporates oscillatory terms like exp{(Et–px)/iħ}. Since we are dealing with continuous functions and infinitely many basis states, we must adapt some earlier definitions that assume small numbers of basis states. For example, it is sensible to calculate the probability of an electron being at a specific carbon atom in benzene; a likely answer is 1/6. But it is not sensible to calculate the Prob(x), the probability that an electron is precisely at point x, since any line segment has an infinite number of points, making the probability at a single point = 1/∞ = 0. Instead, we must calculate the probability Prob(x)Δx, the probability that an electron is between x and x+Δx, for some small Δx. We therefore write: Prob(x) Δx = |ψ(x)| Δx 2

If we take the limit as Δx goes to zero, we find the relative probability that the particle is at x as compared with it being somewhere else. To make this distinction, Prob(x) is properly called the probability density of finding a particle near x. In everyday communication, physicists often leave this distinction implicit. Additionally, with continuous functions and infinite basis states, we must replace summations with integrals. For example, the equation for a finite number of position states |x> is: = Σ x

With an infinite number of position states, the replacement equation is: = ∫ dx = ∫ * ψ(x) dx = ∫ø*(x) ψ(x) dx because ø(x) = , and = *. We also need to address how continuous states are normalized.

Normalizing Momentum States Assume an electron is in state |ψ> with probability density: Prob(x) = |ψ(x)|

2

Define |p> to be the state of definite momentum p. By our usual rules, the amplitude for the electron to have momentum p is: = ∫ dx, over all x In V3p16-7, Feynman says there are several options for normalizing the momentum amplitudes; he chooses: = * = exp{–ipx/ħ} The probability density of p is then: Prob(p) = || / h (not ħ) 2

Frankly, I would not choose this normalization because it creates an exception to the general rule that probabilities equal the squared magnitudes of amplitudes. But restating all the remaining equations in Volume 3 would be more confusing than just accepting his choice, which is what I will do. With Feynman’s normalization choice, the momentum amplitude becomes: = ∫ exp{–ipx/ħ} dx, over all x Let’s consider an example with an electron wave function given by: ψ(x) = K exp{–x /4σ } 2

2

Prob(x) = K exp{–x /2σ } 2

2

2

To find K, we require the probability of the electron being somewhere to be 100%, which means: 1 = ∫ Prob(x) dx, over all x Integral tables list ∫exp{–u /2}du = √(2π), when integrated over all u, from u=–∞ to u=+∞. With u=x/ σ this integral becomes: 2

∫exp{–x /2σ }dx/σ = √(2π) ∫exp{–x /2σ }dx = σ√(2π) 2

2

2

2

Comparing this equation with that for Prob(x), it is evident that: K = 1/√(2πσ ) K = (2πσ ) 2

2

2 –1/4

Prob(x) = 1/√(2πσ ) exp{–x /2σ } 2

2

2

This probability density is a Gaussian distribution with standard deviation σ. For a distribution centered at x=0, the standard deviation is defined to be:

σ = 1/√(2πσ ) ∫ x exp{–x /2σ } dx 2

2

2

2

2

Now let’s calculate the momentum amplitude (initially keeping K to reduce clutter): = K ∫ exp{–ipx/ħ} exp{–x /4σ } dx 2

2

Feynman then substitutes: u = x + 2ipσ /ħ u = x +4ixpσ /ħ –4p σ /ħ u /4σ = x /4σ +ixp/ħ –p σ /ħ –x /4σ –ixp/ħ = –u /4σ –p σ /ħ 2

2

2

2

2

2

2

2

2

4

2

2

2

2

2

2

2

2

2

2

2

= K exp(–p σ /ħ ) ∫exp{–u /4σ ) du 2

2

2

2

2

Now define v = σ√2. = K exp(–p σ /ħ ) ∫exp{–u /2v ) du = K exp(–p σ /ħ ) v √(2π) = (2πσ ) exp(–p σ /ħ ) σ√2 √(2π) = (2 σ π /2πσ ) exp(–p σ /ħ ) = (8πσ ) exp(–p σ /ħ ) 2

2

2

2

2

2

2

2 –1/4

4

4

2

2

2

2

2 1/4

2 1/4

2

2

2

2

2

2

2

Prob(p) = || / h Prob(p) = (8πσ /h ) exp(–2p σ /ħ ) Prob(p) = (2σ /πħ ) exp(–2p σ /ħ ) 2

2

2

2 1/2

2

2

2

2 1/2

2

2

2

Now define η = ħ/2σ, which means σ = ħ/2η. Prob(p) = (2ħ /4η πħ ) exp(–p /2η ) Prob(p) = (2πη ) exp(–p /2η ) 2

2

2 –1/2

2 1/2

2

2

2

2

This probability density is also a Gaussian distribution with standard deviation η=ħ/2σ. The product of the standard deviations of the x-distribution and the p-distribution are: σ × ħ/2σ = ħ/2 Feynman says that if we equate the standard deviations of these Gaussian distributions with the uncertainties in the values of x and p, we obtain the Heisenberg Uncertainty Principle. Δx × Δp = ħ/2

Normalizing Position States With position becoming a continuous variable with an infinite number of basis states, we must address the normalization of position states.

When the number of position basis states is finite, the requirement that the basis states be orthonormal is given by: = δ = 1 if j=k, else 0 jk

When the number of position basis states is infinite, the requirement becomes: = ∫ ψ(x’) ψ(x) dx = 1 if x’=x, else 0 To evaluate this, recall the equation: = ∫ dx Replace can be defined by giving all of the amplitudes to find the two particles at r and r . This generalized amplitude is therefore a function of the two sets of coordinates r and r . You see that such a function is not a wave in the sense of an oscillation that moves along in three dimensions. Neither is it generally simply a product of two individual waves, one for each particle. It is, in general, some kind of a wave in the six dimensions defined by r and r . If there are two particles in nature which are interacting, there is no way of describing what happens to one of the particles by trying to write down a wave function for it alone.” 1

2

1

1

1

2

2

2

Schrödinger’s Equation Schrödinger’s equation describes dynamics in quantum mechanics, much as F=ma describes dynamics in Newtonian physics. Recall that dynamics is the study of forces and their effect on motion. To describe dynamics, our wave functions must become functions of time as well as space. The coordinate representation of a wave function is written: ψ(r, t) = ψ(x, y, z, t) We will later consider other representations, including the momentum representation. Recall from Chapter 10, the equation relating the time evolution of amplitude C to all other amplitudes, each multiplied by a component H of the Hamiltonian matrix: j

jk

iħ dC /dt = Σ H C j

k

jk

k

and its equivalent form: iħ d/dt = Σ k

We now seek to transform this equation to the realm of infinite basis states, and continuous space and time. We must replace the summation by an integral and replace the finite basis states |j> and |k> with continuous basis states |x> and |x’>. This yields: iħ d/dt = ∫ dx’ Define H(x,x’) = , which is clearly a function of two variables. Combining this with our notation ψ(z)= yields: iħ d/dt ψ(x) = ∫ H(x,x’) ψ(x’) dx’ H(x,x’) is the amplitude per unit time that a particle will transition from x’ to x. Feynman says: “It turns out in nature, however, that this amplitude is zero except for point x’ very close to x.” This means we can express H(x,x’) in terms of ψ(x) and the derivatives of ψ at x. As we saw at the start of this chapter, the equation for a particle moving freely in one dimension is: ∫ H(x,x’) ψ(x’) dx’ = –ħ /2m ∂ /∂x ψ(x) 2

2

2

iħ d/dt ψ(x) = –ħ /2m ∂ /∂x ψ(x) 2

2

2

For a particle subject to a force that is described by a potential V(x), Schrödinger’s equation becomes: iħ dψ/dt = –ħ /2m ∂ ψ/∂x + V(x) ψ 2

2

2

The three-dimensional generalization of this is the full Schrödinger equation: iħ dψ/dt = –ħ /2m ☐ψ + V ψ 2

where ☐ = ∂ /∂x + ∂ /∂y + ∂ /∂z 2

2

2

2

2

2

As with all quantum mechanical equations in this course, this equation is valid only for nonrelativistic velocities. Relativistic quantum mechanics is a challenge for a subsequent course. (Please note that conventional notation for the sum of the three second-order partial derivatives is the gradient, an inverted Δ, squared. Since that symbol is not generally available on ereaders, I am borrowing the d’Alembertian ☐ from 4-D differential geometry, where it customarily includes the time derivative as well.) We have not derived the Schrödinger equation, because it cannot be derived from anything presented in this course.

In V3p16-12, V3p16-13, and V3p16-4, Feynman explains the origins and importance of Schrödinger’s equation: “It came out of the mind of Schrödinger, invented in his struggle to find an understanding of the experimental observations of the real world.” “It was the first quantum-mechanical equation ever known. It was written down by Schrödinger before any of the other quantum equations we have described in this book were discovered.” “When Schrödinger first wrote it down, he gave a kind of derivation based on some heuristic arguments and some brilliant intuitive guesses. Some of the arguments he used were even false, but that does not matter; the only important thing is that the ultimate equation gives a correct description of nature.” “The great historical moment marking the birth of the quantum mechanical description of matter occurred when Schrödinger first wrote down his equation in 1926. For many years the internal atomic structure of matter had been a great mystery. No one had been able to understand what held matter together, why there was chemical binding, and especially how it could be that atoms could be stable. …Schrödinger’s discovery of the proper equations of motion for electrons on an atomic scale provided a theory from which atomic phenomena could be calculated quantitatively, accurately, and in detail. In principle, Schrödinger’s equation is capable of explaining all atomic phenomena except those involving magnetism and relativity. It explains the energy levels of an atom, and all the facts of chemical binding. This is, however, true only in principle—the mathematics soon becomes too complicated to solve exactly [in] any but the simplest problems.” To encompass multiple particles, we can expand this equation by replacing ψ and V with: ψ(t, r , r , …, r ) 1

2

n

V(r , r , …, r ) 1

2

n

If the only forces present are the electrostatic interactions of these particles, V is given by: V(r , r , …, r ) = Σ Z Z e / r 2

1

2

n

j>k

j

k

jk

Here: we sum each pair of particles once, the Z’s are the particle charges measured relative to the proton charge q ; and e =q /4πε . 2

p

2 p

0

Quantized Energy Levels In the final portion of this lecture, Feynman seeks to demonstrate how a: “…most remarkable consequence of Schrödinger’s equation comes about—namely, the surprising fact that a differential equation involving only continuous functions of continuous

variables in space can give rise to quantum effects such as the discrete energy levels in an atom. The essential fact to understand is how it can be that an electron which is confined to a certain region of space by some kind of a potential “well” must necessarily have only one or another of a certain well-defined set of discrete energies.” For simplicity, we consider Schrödinger’s equation with a potential V in one dimension. –ħ /2m ∂ /∂x ψ(x) + V(x) ψ(x) = iħ d/dt ψ(x) 2

2

2

We seek the stationary states of definite energy E of an electron whose wave function ψ is: ψ = a(x) exp{–iEt/ħ} Plugging this into Schrödinger’s equation yields: –ħ /2m d a/dx exp{–iEt/ħ} + V a exp{–iEt/ħ} = iħ (–iE/ħ) a exp{–iEt/ħ} 2

2

2

d a/dx = –(2m/ħ ) [E–V] a = –β a with β = (2m/ħ ) [E–V] and a(x) = A exp{±ixβ} 2

2

2

2

2

2

Note that in any physically realistic situation, both a(x) and β are finite, and therefore so is d a/dx . This means a(x) and da/dx have no discontinuities; they are smooth functions. 2

2

2

Feynman presents a long, heuristic, and frankly unsatisfying discussion of different shaped curves and piecing them together. Instead of going through all that, here is a rigorous proof of Feynman’s statement that Schrödinger’s equation leads to discrete solutions in atoms. If β > 0, define k = √β . Wave number k is real and the solutions are sinusoids, with the most general solution being: 2

2

a(x) = A cos{kx+ø} da(x)/dx = – kA sin{kx+ø} if β < 0, define K=+√(–β ). Here β is imaginary and the solutions are exponentials, with the most general solutions being of two types: 2

2

a(x) = C exp{+Kx}, for x0 We encountered such solutions when analyzing barrier penetration in Chapter 9. Since any physically realistic solution must be finite everywhere, each solution is possible only in the indicated range.

In atoms and similar bound systems, the electron is bound because the potential V is greater than E far from the center of the attractive force. But V must be less than E near the center, or the electron could not exist there. In the central region, a(x) oscillates sinusoidally, but far away a(x) decreases exponentially. The essential point is that the sinusoidal and exponential pieces of a(x) must match where they meet at V=E. To properly match, both the values a(x) and the slopes da/dx on each side of V=E must be equal, because Schrödinger’s equation tells us that a(x) and da/dx have no discontinuities. To illuminate the physics with the simplest possible math, examine a square well potential of width 2d, as shown in Figure 22-1 and defined by: V(x) = 0, for |x| < d V(x)=V , for |x| > d and V > E 0

0

Figure 22-1 Square Well Potential V E is Energy of Bound Electron

In the central region, a(x) is sinusoidal with the same wave number k throughout. In the classically forbidden regions, x+d, a(x) decreases exponentially as |x| increases; both regions have the same value of K throughout, but with possibly different amplitudes. Recall that: k = √{(2m/ħ ) [E]} K = √{(2m/ħ ) [V –E]} 2

2

0

Let’s start at the far left, where x, the state of an electron being at x; and |p>, the state of an electron having momentum p. The wave function ψ is: ψ(x,t) = Here ψ(x,t), also written ψ(x), is the amplitude for a particle to be at position x at time t, whereas |ψ> is a particle state that might be spread across all x. Other restated equations are: = ∫ dx = ∫ * ψ(x) dx = ∫ø*(x) ψ(x) dx because ø(x) = , and = *.

The probability that an electron is between x and x+Δx, for small Δx, is: Prob(x) Δx = |ψ(x)| Δx 2

In the limit that Δx goes to zero, Prob(x) is the probability density of finding the particle near x. Feynman chooses this normalization for momentum amplitudes: = * = exp{–ipx/ħ} Prob(p) = || / h (not ħ) 2

The Dirac delta function is defined as: ∫ δ(x) dx = 1, integrated over all x δ(x) = limit of exp{–x /ε }/(ε√π) as ε goes to 0 2

2

In this limit, δ(x) becomes infinitely narrow, but δ(0) becomes infinitely large. For two position states |x> and |x’>: ψ(x’) = ∫ ψ(x) dx ψ(x’) = ∫ δ(x’–x) ψ(x) dx = δ(x’–x) = 0 if x’ is not equal to x The extension to three dimensions, with r=(x,y,z), is: = ∫ dx dy dz = ∫ ø*(r) ψ(r) dx dy dz = δ(x’–x) δ(y’–y) δ(z’–z) »For a particle subject to a force described by a potential V, Schrödinger’s equation for nonrelativistic velocities is: In 1D: iħ dψ/dt = –ħ /2m ∂ ψ/∂x + V ψ 2

2

2

In 3D: iħ dψ/dt = –ħ /2m ☐ψ + V ψ 2

where ☐ = ∂ /∂x + ∂ /∂y + ∂ /∂z 2

2

2

2

2

2

»Feynman says a “most remarkable consequence of Schrödinger’s equation… [is] that a differential equation involving only continuous functions of continuous variables … [gives rise to] discrete energy levels in an atom.”

Chapter 23 Symmetry & Conservation Laws The symmetries of natural laws have profound consequences. As we have often discussed, many fundamental principles of physics are stated as conservation laws, such as the conservation of energy and of momentum. These conservation laws are intimately associated with symmetry properties. Feynman Simplified 1D Chapter 49 presents a complete list of such symmetries, which we repeat here. The following symmetry operations are universally valid. Translation in Space Translation in Time Spatial Rotation Motion at Constant Velocity Exchange of Identical Particles Change of Quantum Phase CPT Symmetry The following symmetry operations are valid for all but the weak interaction. Spatial Reflection / Inversion Time Reversal Reversal of Electric Charge Polarity Exchange of Matter & Antimatter Physicists are awed by the beautiful relationship between symmetry and conservation laws. This profound connection, proven by Amalie Emmy Noether is called Noether’s Theorem. Her theorem applies to all of physics, including to Newton's laws. But in quantum mechanics, Noether’s Theorem becomes more expansive. Proving Noether’s Theorem isn’t overly complex, but requires a deeper understanding of mathematics than we presume here. Feynman expresses our fascination with symmetry in V1p52-3: “A fact that most physicists still find somewhat staggering, a most profound and beautiful thing, is that, in quantum mechanics, for each of the rules of symmetry there is a corresponding conservation law.”

And also in V3p17-1: “The most beautiful thing of quantum mechanics is that the conservation … laws are very deeply related to the principle of superposition of amplitudes, and to the symmetry of physical systems under various changes.”

Reflection Symmetry of H2 Ion We begin with a very simple example: the reflection symmetry of a singly-ionized hydrogen molecule. In Chapter 13, we analyzed this molecule that consists of two protons and one electron. We defined two basis states: |1> electron attached to proton A; and |2> electron attached to proton B. If the protons’ properties are identical, including momentum and spin, states |1> and |2> are symmetric about a plane that is perpendicular to and halfway along the line joining them, as shown in the upper half of Figure 23-1.

Figure 23-1 Upper: Two Basis States of H2+ Lower: States Reflected in Sy mmetry Plane AB Dots are Protons; Gray Disk is Electron

The lower half of Figure 23-1 shows the basis states reflected in that symmetry plane, effectively swapping left and right. Mathematically, if the horizontal axis is x and x=0 on the symmetry plane, reflection is the operation of inverting the sign of every x-coordinate. The reflected states are labeled P|1> and P|2>, and it is evident from the figure that: P|1> = |2> P|2> = |1>

We treat P as an operator that changes quantum states, and like all operators in quantum mechanics it is represented as a matrix. The matrix elements are easy to calculate: P P P P

11 12 21 22

= = = 0 = = = 1 = = = 1 = = = 0

In matrix form this is:

In quantum mechanics, Feynman says the words “operator” and “matrix” are as interrelated as the words “numeral” and “number” — the distinction he says: “is something pedantic that we don’t need to worry about.” Feynman despised pedantry; he thought ideas were important, and labels were not. While we focus here on symmetry, we should realize that many things in nature are not symmetric. A molecule like H need not be symmetric; it might, for example, have something else nearby that breaks its symmetry. Also, one proton might be spin up and the other spin down, which would change the symmetry property: the molecule would not be left-right symmetric, but would be symmetric under the simultaneous exchange of both left-right and up-down. + 2

Feynman stresses that a system that is symmetric has a special time-evolution property. We defined the operator U(Δt) to represent the effect of time advancing from the current time t to time t+Δt. The Hamiltonian operator H is defined in terms of U(Δt): U(Δt) = 1 + H/iħ, for infinitesimal Δt, or H = limit of iħ{U(dt)–1} as dt goes to 0 Now suppose that a wave function was entirely in state |1> at time t=0, and that at time t=Δt, the wave function evolved to have amplitude 0.8 to be in state |1> and amplitude 0.6i to be in state |2>, which we write: U(0) |1> = |1> U(Δt) |1> = 0.8 |1> + 0.6i |2> Recall that the vector (A,B) represents a state with amplitude A to be in state |1> and amplitude B to be in state |2>. We can write the prior equation in vector form: U(Δt) (1, 0) = (0.8, 0.6i)

If the laws of nature are symmetric under reflection, and if states |1> and |2> are symmetric with respect to reflection, then nature must treat state |2> as it does state |1>. (We bolded the if’s to emphasize that this must be established experimentally for every operator and every set of physical phenomena. Not all forces are symmetric under reflection, for example.) If so: U(Δt) |2> = 0.8 |2> + 0.6i |1> or in vector form: U(Δt) (0, 1) = (0.6i, 0.8) Now, we bring all these pieces together and compare different combinations of time-evolution and reflection operators. In vector form: ψ(t=0) = (1, 0) P ψ(t=0) = P (1, 0) = (0, 1) U(Δt) (1, 0) = (0.8, 0.6i) U(Δt) (0, 1) = (0.6i, 0.8) P U(Δt) ψ(0) = P (0.8, 0.6i) = (0.6i, 0.8) U(Δt) P ψ(0) = U(Δt) (0,1) = (0.6i, 0.8) Thus: P U(Δt) = U(Δt) P Reflection×Evolution = Evolution×Reflection The key conclusion is that time-evolution and reflection commute. Here, unlike the general case, the order of the two operators makes no difference. If we know how any system evolves, we also know how the reflected version of that system evolves. This is because reflection doesn’t change the fundamental nature of this system: the system is symmetry under reflection. We can generalize this invariance of operator order. (Feynman says making principles more general necessarily means making them more abstract.) Let |ψ > be some initial state, and let Q be a symmetry operator that leaves the basic physics of any system unchanged. And let |ψ > evolve into |ψ >, according to the time-evolution operator U: 1

1

2

|ψ > = U |ψ > 2

1

Now if the laws of nature are symmetric under Q, nature must treat the symmetric states |ψ > and Q| ψ > equally. If the ifs are satisfied, the evolution of Q|ψ > and the evolution of |ψ > must be symmetric: the evolved version of Q|ψ > must equal the symmetric version of the evolved |ψ >. This means: 1

1

1

1

U Q |ψ > = Q |ψ > U Q |ψ > = Q U |ψ > Thus: Q U = U Q 1 1

2

1

1

1

Feynman says this is the “mathematical statement of symmetry.” If the last equation is true, operators Q and U are said to commute. Feynman defines symmetry as: “A physical system is symmetric with respect to the operation Q when Q commutes with U, the operation of the passage of time.” If we now let Δt go to zero, we obtain: QU= UQ Q (1+H/iħ) = (1+H/iħ) Q QH=HQ This means symmetry is equally well defined by Q commuting with H, which is a much more prominent operator than U. I recommend remembering the symmetry definition as QH=HQ. Examples of symmetry operators are: reflection, discussed above; interchange of identical particles; rotation by any angle about any axis of a spherical system; and translation in space or time. Some symmetries are universal, while others apply in limited circumstances. Non-spherical objects can be symmetric under rotations of some angles, about some axes. A cube, for example, has three symmetry axes about which any 90º rotation leaves the cube invariant. The ammonia molecule, discussed in Chapter 10, has a symmetry axis that is perpendicular to the hydrogen atom plane; it is symmetric under a rotation of any angle about that axis.

Special Symmetry Considerations A number of special symmetry properties are worth exploring. Sometimes, a symmetry operator Q operating on a state |ψ> returns that state with no changes whatsoever: |ψ> = Q |ψ>. In other cases, Q |ψ> is not exactly the same state, but is a physically equivalent state that differs from |ψ> by only a phase angle. This means: Q |ψ> = exp{iβ} |ψ> for some phase angle β. Such symmetry operators are called unitary, because they don’t change the magnitude of |ψ> and thus never change the number of particles in any system. Feynman avoids saying this here, but all the symmetry operators we are discussing are unitary. Consider again the example of the H molecule. We found that the stationary states of definite energy are mixtures of the basis states, given by: + 2

|+> = {|1> – |2>}/√2 |–> = {|1> + |2>}/√2

Now let’s operate on the stationary states with the reflection operator P. P |+> = {|2> – |1>}/√2 = – |+> P |–> = {|2> + |1>}/√2 = + |–> The operator P shifts the phase of |+> by 180º, but does not shift the phase of |–>. Note that |+> and |–> are eigenstates of operator P, because P operating on either state returns that state multiplied by a constant. In contrast, |1> and |2> are not eigenstates of P, because P operating on either state returns a different state. Another example is the symmetry of a circularly polarized photon under a rotation about its direction of motion. As we previously worked out, rotating an object with spin up along the z-axis by angle θ about the z-axis, multiplies the state by exp{isθ}, where s is the particle’s spin. For photons, s=1. The next consideration is appreciating that if Q changes state |ψ > by only a phase angle at any one moment in time, it must also produce the same change at all other times. Let’s see what this means. This time let’s employ the Hamiltonian H as the time evolution operator: 1

H |ψ > = |ψ > 1

2

Q |ψ > = exp{iδ} |ψ > H Q |ψ > = exp{iδ} H |ψ > H Q |ψ > = exp{iδ} |ψ > 1

1

1

1

1

2

Note that all numbers, like exp{iδ}, commute with all matrices. Since H Q = Q H: Q H |ψ > = exp{iδ} |ψ > Q |ψ > = exp{iδ} |ψ > 1

2

2

2

Hence Q produces the same result on the evolved state |ψ > as it did on the original state |ψ >. 2

1

Therefore, if a Hamiltonian H has a symmetry whose operator is Q, and Q has an eigenstate |ψ>, then Q|ψ> yields the same result at every time t. This, Feynman says, establishes a conservation law: Q|ψ> never changes. Feynman says, even if we know nothing about the mechanisms by which states evolve, and even if we don’t care to know, there is one thing that we do know. After the passage of any amount of time, Q|ψ> remains constant. In V3p17-4, Feynman says: “That's the basis of all the conservation laws of quantum mechanics.” The next special consideration relates to what we called the reflection operator P, which we used to invert the sign of one axis. The convention amongst professional physicists is that P inverts the sign of all three spatial axes: every x becomes –x, every y becomes –y, and every z becomes –z. We do this because it treats all axes equally, which is more symmetric. One can accomplish this: (1) in three

steps with three mirrors; or (2) with one mirror followed by a 180º rotation; or (3) by inverting the polarity of all axes.. When using the reflection operator, we must specify the reflection plane. When using the inversion operator, we must specify the origin point through which all coordinates invert. Changing P from the reflection operator to the inversion operator does not invalidate anything we said before. P is also called the parity operator. If a system is symmetric with respect to the inversion operator P, we have: P |ψ> = exp{iδ} |ψ> Applying P twice yields: P [P |ψ>] = P [exp{iδ} |ψ>] = exp{i2δ} |ψ> But, operating with P twice brings all coordinates back to their original values, therefore: P [P |ψ>] = |ψ> P P = +1 exp{i2δ} = +1 exp{iδ} = ±1 In classical physics, when P inverts a symmetric object, the result exactly equals the original. But, in quantum mechanics, there are two options: exactly the original, or exactly –1 times the original. The former is called even parity and the later is called odd parity. All elementary particles have definite intrinsic parity, either even or odd. The next consideration regards nomenclature and sign conventions. One can think of rotations and axes inversions in two ways. One can rotate or invert the physical objects being considered, or one can rotate or invert the coordinate systems. The two approaches are equivalent, except for polarities. Rotating objects about the x-axis by angle θ is equivalent to rotating the coordinate system about the x-axis by angle –θ. Feynman says that in these lectures he usually rotates or inverts the axes, not the objects. Some care is therefore required when comparing equations from different sources since they may use different conventions. The final special consideration relates to non-degenerate stationary states. Recall that a state of definite energy is non-degenerate if no other eigenstate of that system has the same energy. [In determining degeneracy, do not consider spin. Every eigenstate describing how a wave function is distributed among basis states exists in multiple spin combinations. If spin interactions are not included in the Hamiltonian, all these spin combinations will have the same energy. Nonetheless these are not called degenerate.] We wish to prove a new principle by considering a non-degenerate eigenstate |ψ> of definite energy E and any symmetry operator Q of that system. We know that a system’s symmetry operators commute

with its Hamiltonian. This means: H Q |ψ> = Q H |ψ> = Q E |ψ> Regrouping terms for clarity (E is a constant that commutes with Q): H { Q|ψ> } = E { Q|ψ> } This means {Q|ψ>} is an eigenstate of the Hamiltonian with eigenvalue E, the same energy as state | ψ>. But, since |ψ> is not degenerate, it is the only eigenstate with energy E. Hence Q|ψ> and |ψ> must be equal within a phase angle. This means: Q |ψ> = exp{iδ} |ψ> Symmetry operators can change a non-degenerate state only by shifting their phase, not by changing the magnitudes or changing them into other states. It is worth exploring why this argument does not apply to degenerate states. In Chapter 21, we found that the E=–A eigenstate of benzene is degenerate. Both of the following eigenvectors corresponds to that energy: |ψ+> = (+1,–1,–2,–1,+1,+2)/√6 |ψ–> = (+2,+1,–1,–2,–1,+1)/√6 Benzene is symmetric under a rotation of 60º within the ring plane. Let's call that 60º-rotation operation Q. We see that |ψ–> = Q |ψ+>. In this case, the symmetry operator Q changes one degenerate state into another, which is a greater change than simply shifting a phase angle.

Symmetries & Forces Some symmetries are valid properties of all of nature’s forces, but others are not. The best current evidence suggests that all the symmetries listed at the start of this chapter are valid properties of the electromagnetic force, of the strong nuclear force, and of gravity. The weak force is remarkably different. Weak interactions are not symmetric with respect to spatial inversion (parity, P), reversal of electric charge polarity (charge conjugation, C), time reversal (T), and matter-antimatter exchange (CP). The weak force Hamiltonian does not commute with the operators P, C, T, and CP. Parity is a particularly interesting example. Absent weak interactions, an isolated state of definite parity will maintain the same parity forever, however complex the state or its internal interactions. If an atom with even parity emits a photon, the combined system of that atom and photon will always retain its even parity. In addition, absent weak interactions, any non-degenerate state of definite energy will also be a state

of definite parity. We proved above that any symmetry operator Q can change a non-degenerate eigenstate |ψ> of definite energy E by at most a phase angle. In the case of parity, that phase angle can only be ±1, because applying P twice restores the original state exactly. A non-degenerate stationary state must therefore be a state of definite parity, either even or odd. Only the weak force can change that parity.

Conservation of Angular Momentum Consider the operator R (ø) that performs a rotation about the z-axis, rotating the coordinate system by angle +ø, or equivalently rotating all physical objects by angle –ø. We assume no external influences are altered by the rotation. If any external influences do exist, they must act entirely parallel to the zaxis. z

In such circumstances, there are particular states for which the rotation changes those states by only a phase angle. One such state is spin up along the z-axis. We will repeat a prior proof (Chapters 7 and 8) that this phase angle must be proportional to the rotation angle ø. Let one such state be |ψ> and define angle δ by: R (ø) |ψ> = exp{iδ} |ψ> z

If we perform this rotation twice, rotating about the z-axis by a total angle of 2ø, we obtain: R (ø) R (ø) |ψ> = exp{iδ} R (ø) |ψ> R (2ø) |ψ> = exp{i2δ} |ψ> z

z

z

z

Evidently, if we perform this rotation N times, we obtain: R (Nø) |ψ> = exp{iNδ} |ψ> z

This equation holds for any angle ø with some corresponding angle δ. We can therefore make ø as small as we wish, and make the integer N as large as we wish, so that any angle θ of any magnitude can be equated to Nø. Making that substitution, we get: θ = Nø R (θ) |ψ> = exp{iθ(δ/ø)} |ψ> = exp{imθ} z

Here m = δ/ø is a real constant that is a property of the system, and is unrelated to any specific value of rotation angle θ. This proves that the phase angle change to state |ψ> must be proportional to the rotation angle. I pause here for a brief side comment. Physicists are often accused, not always without justification, of having a cavalier attitude toward some of the finer points of mathematics. Someone might suggest

that the logic just employed could “prove” that sin(x) equals x. After all, for infinitesimal x, we routinely approximate sin(x)=x. Simply multiplying x by some large N, “proves” that sin(Nx)=Nx for any Nx, which is of course wrong. Approximations requires finesse. The essential difference is: sin(x) is not exactly equal to x, but R (ø)|ψ> is exactly equal to exp{iδ}|ψ>. z

Now recall a general property of symmetry operators: if a system is symmetric under R (ø) and if R (ø) changes state |ψ> only by a phase angle, that phase angle will never change. In this case, it is m that never changes. Absent external off-z-axis influences, we can say that m is conserved, or equivalently that m is a constant of the motion. In quantum mechanics, mħ equals the angular momentum of the system about the z-axis. And in the limit of large systems, the sum of all mħ’s equals the classical angular momentum of the total system. z

z

Thus the principle of the conservation of angular momentum derives from rotation operators changing states only by phase angles that are proportional to the rotation angle.

Conservation of Momentum & Energy Similar connections exist with the conservation of momentum and energy. Suppose a physical system is invariant under translation — that it can be moved from one location to another without changing any of its internal properties. Examples of such situations include moving a pot of water 10 cm across an empty lab bench, or moving a rocket 1 km in intergalactic space. This invariance implies that the system’s Hamiltonian cannot depend on any absolute coordinate values, but only on their differences. Define D (Δ) to be the translation operator that moves the u-axis by the amount Δ in the +u-direction, which is equivalent to decreasing all u-coordinate values by Δ. D (Δ) is also called the displacement operator. u

u

Again, there are particular states which when operated on by D (Δ), the x-translation operator, are changed only by a phase angle. As above, this phase angle must be proportional to Δ, the distance translated. You might try proving that as an exercise. x

For one such state |ψ> this means: D (Δ) |ψ> = exp{ik Δ} |ψ> x

x

Here, k is a real constant that is a property of the system, and is unrelated to any specific value of Δ. Again, k is a conserved quantity that never changes, absent external forces. In quantum mechanics, k ħ equals the x-component of linear momentum of the system. And in the limit of large systems, the sum of all k ħ’s equals the classical x-momentum of the total system. x

x

x

x

As Feynman says in V3p17-8: “If the Hamiltonian is unchanged when the system is displaced, and if the state starts with a

definite momentum in the x-direction, then the momentum in the x-direction will remain the same as time goes on. The total momentum of a system before and after collisions—or after explosions or what not—will be the same.” Now consider displacements in time, with the operator D (Δ) advancing all clocks by the amount Δ, which is equivalent to decreasing all measured times by Δ. To increase all measured times by Δ, use the operator D (–Δ). Take the example of a roller coaster ride releasing cars at the top of track, where: t

t

|ψ> is the state of a car released at t=0 D (–60)|ψ> = “ “ “ released at t=60 sec. D (–120)|ψ> = “ “ released at t=120 sec. D (–180)|ψ> = “ “ released at t=180 sec. t t t

Once again, if the Hamiltonian is symmetric under time displacements, and if the above states are identical except for phase angles, then D Δ)|ψ> must change |ψ> by only a phase angle that is proportional to Δ, according to: t

D (–Δ) |ψ> = exp{–iωΔ} |ψ> t

Here, ω is a real constant that is a property of the system, and is unrelated to any specific value of Δ. Again, ω is a conserved quantity that never changes, absent external forces. In quantum mechanics, ωħ equals the energy of the system. And in the limit of large systems, the sum of all ωħ’s equals the classical energy of the total system. Note that –Δ corresponds to advancing time, which conforms to our standard sign convention of writing exp{–iEt/ħ). Finally, some notes on connecting quantum and classical quantities. »For an infinitesimal time interval Δ: D (–Δ) |ψ> = exp{–iωΔ} |ψ> = (1–iωΔ) |ψ> D (–Δ) |ψ> = U(Δ) |ψ> = (1–iHΔ/ħ) |ψ> H |ψ> = E |ψ> t t

Thus, E = ω/ħ »For an infinitesimal spatial displacement dx: D (dx) |ψ> = exp{ik dx} |ψ> = (1 +ik dx) |ψ> ∂ψ/∂x = (D (dx) |ψ> – |ψ>)/dx = ik |ψ> x

x

x

x

x

For |ψ> = Aexp{i(–Et+p•r)/ħ), ∂ψ/∂x = (p /ħ)ψ x

Thus, p = k /ħ is the x-momentum operator. x

x

»For an infinitesimal rotation θ, physicists define J , the z-component of angular momentum, as: z

R (θ) = 1 + iJ θ/ħ z

z

From above: R (θ) |ψ> = exp{imθ} |ψ> = (1 + imθ) |ψ> z

Thus, J is the z-angular momentum operator, and J |ψ> will always equal mħ|ψ>. z

z

Polarized Light To compare the classical and quantum concepts of angular momentum conservation, Feynman examines polarized light. In Feynman Simplified 1D Chapter 49, we discussed light polarization and rotation, using the standard physics sign conventions for polarization orientation and angle of rotation. With those conventions, the amplitude of right hand circularly (RHC) polarized light, moving in the +z-direction, is multiplied by exp{+iθ} when operated on by R (θ), which rotates the x-y axes by angle +θ. Feynman uses standard physics conventions in this lecture, but as he acknowledges in V3p17-9, his earlier lecture used the sign conventions of chemistry, which has the opposite polarity. We use standard physics conventions throughout. Natural laws do not depend on how our thumbs are oriented, but understanding physics equations is much easier if we use consistent conventions. z

We found above that R (θ) multiplies states of z-spin m by exp{imθ}. This means that RHC photons have spin +1, consistent with the right hand rule. A beam of RHC photons, therefore, carries angular momentum. This would be expected classically, as we now demonstrate. z

In Feynman Simplified 1C Chapter 36, we found that, for light moving in the +z-direction, the electric field vector of RHC light rotates counterclockwise at frequency ω, within the xy-plane, according to: E

RHC

= (cos[ωt], +sin[ωt], 0)/√2

If circularly polarized light is incident on a material object, the light’s electric field will accelerate electrons in the atoms of that object. We can, as before in classical physics, treat these electrons as harmonic oscillators. In this case, driven by the photons’ electric field, the electrons rotate within the xy-plane about the centers of their atoms. The motion of one such electron and the electric field driving it are shown in Figure 23-2.

Figure 23-2 Electric Field E of RHC Light & Position of an Electron e Both Rotate Within xy -Plane

The electron rotates in the same direction and at the same frequency as the driving electric field, but with a phase delay. In this figure the rotation is counterclockwise, and the delay is apparent from the greater rotation angle of the electric field. The reason for this delay is apparent in the expanded view in Figure 23-3: only E , the transverse component of E, drives the electron’s rotation. t

Figure 23-3 Expanded View of Electron e Driven By Transverse Component of Electric Field E

Classically, the longitudinal component of E slightly displaces the electron from its equilibrium position, by the amount r, while the transverse component of E drives its rotation. Without a phase delay, there would be no transverse component. Since the electric field of every photon of RHC light is turning counterclockwise, all electrons are driven to rotate counterclockwise as well. Recall the classical equations:

transverse velocity v = ω r work W = force × distance Power P = dW/dt = force × velocity torque τ = force × lever arm τ = d/dt (angular momentum J) Here, in appropriate units, we have: force = E q P = dW/dt = E q v τ = dJ /dt = E q r (dJ /dt) / (dW/dt) = r / v dJ /dW = 1/ω t

t

z

t

z

z

Classically, electrons, driven by the RHC light’s electric field, absorb the light’s energy and angular momentum in the ratio: J /W=1/ω. z

Compare this classical analysis with a quantum analysis of the same phenomenon. If the total energy of the absorbed light equals W, and if each photon has energy ħω, the number N of absorbed photons is: N = W / ħω Since each photon of RHC light has its spin parallel to its direction of motion, +z in this case, the angular momenta of one photon and of N photons are: One photon: J = +ħ All photons: J = N ħ z

z

Recall that when physicists say “spin 1” or “spin 1/2”, there is always an implicit ħ: “spin +1” means spin +1ħ, and “spin –1/2” means spin –ħ/2. The angular momentum to energy ratio is: J / W = Nħ / (Nħω) = 1/ω z

Classical and quantum concepts agree in the limit of vast numbers of particles, as they must. All that pertained to RHC polarization. For left hand circularly (LHC) polarized light, the same analysis applies with only an overall sign change: LHC photons have spin –1 in their direction of motion, making J negative. z

What about other polarizations? Again from Feynman Simplified 1C Chapter 36, the classical equations for the electric fields of +x

and +y polarizations for light moving in the +z-direction are: +x polar.: {E +E }/√2 = (cos[ωt], 0, 0) +y polar.: {E –E }/√2 = (0, sin[ωt], 0) RHC

LHC

RHC

LHC

For a vast number of photons, the total electric field experienced by electrons equals the average of the cosine and sine functions, which is zero. Additionally, these fields each oscillate along one axis only; they move electrons up and down, or left and right, but both motions pass through x=y=0, which means the angular momentum of such motions is zero. Quantum mechanically, a photon with spin in the +x-direction has equal amplitudes to be RHC with zspin +1 and LHC with z-spin –1. This means the expectation value of z-spin is zero. The same is true for photons with spin in the +y-direction.

Spin of Zero Mass Particles In general, a spin 1 particle can have a component of spin angular momentum along any selected axis equal to +1, 0, or –1 (all times ħ). Massless particles, however, are different. Photons are the only confirmed particles that are absolutely massless. (To be precise, the photon mass is measured to be less than 10 times the electron mass, the lightest particle with a well-measured, non-zero mass.) I state without proof that, because photons are massless, they cannot have a zero spin component in any direction. A photon’s spin component along any selected axis can only be +1 or –1. –24

Neutrinos are different as well. It was long known that the mass of the neutrino was imperceptible, very much less than the mass of the electron. For many decades, it was universally believed that neutrinos had zero mass, which is what Feynman says in his lectures. But observations published in 2005 and 2006, from experiments in Japan and the U.S. respectively, definitively proved that the three types of neutrinos are not massless. While their masses are not exactly zero, cosmological measurements limit neutrino masses to extremely small values, less than one millionth of the electron mass. (Remarkably, the most precise limit of neutrino masses is obtained by measuring the mass of the observable universe and subtracting everything else.) Since the energy of neutrinos emitted in typical reactions is at least one million times larger than their mass, neutrino velocities are at least 0.999,999,5 c. No real detector is ever going to travel fast enough to observe neutrinos in their rest frame. In the reference frames in which neutrinos are emitted, and in the reference frames of our laboratories, we observe that neutrinos are always LHC polarized and antineutrinos are always RHC polarized.

Chapter 23 Review: Key Ideas

»Feynman says: “for each of the rules of symmetry there is a corresponding conservation law.” This is called Noether’s Theorem. Feynman says: “A physical system is symmetric with respect to the operation Q when Q commutes with U [or H], the [operators for] the passage of time.” Mathematically, this means: QH=HQ. »The following symmetry operations are universally valid. Translation in Space Translation in Time Spatial Rotation Motion at Constant Velocity Exchange of Identical Particles Change of Quantum Phase CPT Symmetry The following symmetry operations are valid for all but the weak interaction. Spatial Reflection Time Reversal Reversal of Electric Charge Polarity Exchange of Matter & Antimatter »If a Hamiltonian H has a symmetry whose operator is Q, and Q has an eigenstate |ψ>, then Q |ψ> yields the same result forever, absent external interference. This establishes a conservation law: Q|ψ> is conserved. »A symmetry operator operating a non-degenerate eigenstate can only change its phase angle. »A translation in time, represented by the operator D (Δ), increases all clock readings by Δ and decreases the time of all physical events by Δ. Symmetry under time translation implies the conservation of energy. t

A translation of the x-axis, represented by the operator D (Δ), decreases the x-coordinates of all physical events by Δ. Symmetry under x-axis translation implies the conservation of linear momentum in the x-direction. The x-momentum operator is p =k /ħ, with ∂ψ/∂x = (p /ħ)ψ. The same is true for y and z. x

x

x

x

A rotation of the xy-axes about the z-axis by angle θ is represented by the operator: R (θ). Symmetry under z-axis rotation implies the conservation of angular momentum in the z-direction. Spin m in the z-direction is equivalent to angular momentum J =mħ. The same is true for x and y. z

z

»A photon’s spin along its direction of motion can be +1 or –1, but never 0, because photons are massless. Neutrinos are definitely not massless, but their masses are extremely small, less than one millionth of the electron mass. In the reference frames in which neutrinos are emitted, and in the reference frames of our laboratories, neutrinos are always LHC polarized and antineutrinos are always RHC polarized.

Chapter 24 Review of Part Two Quantum Field Theory (QFT), the Particle Exchange Model of Force, states that the electromagnetic, strong, and weak forces arise from fermions (particles of matter) exchanging other particles (most commonly bosons). Gravity has not been successfully incorporated within the QFT model. The force-exchange bosons are:

Photons for the electromagnetic force W , W , Z for the weak force Gluons for the strong force +

–

0

An ionized hydrogen molecule is a two-state system in which two protons are held together by exchanging an electron, thereby reducing their energy and forming a single-electron bond. A neutral hydrogen molecule is similarly held together by the exchange of two electrons forming a doubleelectron bond, the backbone of molecular chemistry. At the subatomic level, forces arise from the exchange of virtual particles that are not directly observable and may have exotic properties, including negative energy and imaginary momentum. The force between nucleons (protons and neutrons) is a residual effect of the much stronger interaction between the quarks within those nucleons. Barrier Penetration A particle’s probability amplitude A declines exponentially within a barrier in which its kinetic energy is negative and its momentum p is imaginary, according to: A ~ exp{(i/ħ) ∫p dz} Here z is the distance inside the barrier. What penetrates the barrier is the particle’s probability amplitude, which has no physical reality. If a particle starts on one side of a classically impenetrable barrier, there is a small but non-zero probability of finding it later on the opposite side of the barrier. But quantum mechanics says no real particle can ever be observed to have negative energy, negative mass, or imaginary momentum; hence particles can never be observed within such barriers. »Complex Molecules, including benzene and magenta, are quantum superpositions of basis states

when there is a non-zero amplitude to transition from one basis state to another. These basis states are often symmetric, or nearly so. In these cases, the molecule’s energy level is split by an amount proportional to the transition amplitude, resulting in stationary states of definite energy. The energy of one stationary state will be less than the energy of any basis state; this is the state found in nature. »Matrices are arrays of components laid out in rows and columns. In quantum mechanics, the components are typically amplitudes or numbers. An n×m (“n-by-m”) matrix has n rows, m columns, and n•m components. For matrices, A, B, and C, and scalar s, basic matrix arithmetic includes: 1. 2. 3. 4. 5. 6.

Multiplication by scalar s: A =sB Matrix addition: A =B +C Matrix multiplication: C =Σ A B Unit matrix δ has components: 1 if i=j, else zero If A is the inverse of A, A A=AA =δ Only non-singular n×n matrices have inverses. ij

ij

ij

ij

ij

ij

k

ik

kj

ij

–1

–1

–1

ij

The general equation for the determinant of an N×N matrix M is: Det |M | = Σ Sign(abc…)•M •M •M •… ij

1a

2b

3c

Here, abc… is a permutation of the integers 1 through N. The sum extends over all permutations, and Sign(abc..) equals +1 for even permutations and –1 for odd permutations. »The Hamiltonian equation for an electron in a magnetic field can be written in terms of Pauli spin matrices: iħ d |Ψ>/dt = –µ σ•B |Ψ> A spin 1/2 particle, with magnetic moment µ and spin up along the z*-axis, is exposed to a magnetic field B in the +z-direction. Let axis z* be rotated from axis z by polar angle θ. The particle’s state is the superposition of two states of definite energy, spin |+z> and |–z>. Its azimuthal angle precesses about the z-axis with frequency ω=–2µB/ħ. Magnetic moment µ is negative for negatively charge particles. »The orbital states of electrons in atoms have three integer quantum numbers: n is the principal quantum number; j is the angular momentum quantum number; and m is the magnetic quantum number, which is the component of j along any selected axis. Modern convention employs lower case l rather than j, but I follow Feynman’s notation to enhance eBook readability. Each orbital electron state is further subdivided into spin states. The orbital ground state of hydrogen has four spin states corresponding to the four electron/proton up/down spin combinations. The Hamiltonian for the spin interaction is:

H = A σ •σ e

p

Here, σ and σ are the Pauli spin vector matrices for the electron and proton respectively. Three of these hydrogen states form a spin 1 triplet, with m=+1, 0, and –1. The fourth is a spin 0 singlet state. e

p

With no external magnetic field, the triplet states have the same energy: 6 µeV above the singlet state. This hyperfine splitting enables transitions with an extremely sharp photon emission/absorption line of frequency 1420 MHz and wavelength 21-cm, a famous spectral line in astronomy. When exposed to an external magnetic field, the energy levels and composition of the stationary states change, which is called Zeeman splitting. »In a one-dimensional array of identical, electrically neutral, equally spaced, immovable atoms, the existing electrons are tightly bound and do not participate dynamically. Feynman says an additional electron “can ride right through the crystal and flow perfectly freely even though it has to hit all the atoms [in its path]. … That is how a solid can conduct electricity.” The Hamiltonian equations for the electron are: iħ dC /dt = E C – A C – A C n

0

n

n–1

n+1

Here E is the zero-point energy of a stationary electron; C is the amplitude for the electron to be at atom #n; and A is the amplitude for the electron to transition to an adjacent atom. For atomic spacing b, the stationary states are given by: 0

n

C = exp{i(knb–Et/ħ)} E(k) = E – 2A cos(kb) n

0

The electron’s energy is in the conduction band between E –2A and E +2A, with –π/b basis representation, the wave function ψ is: ψ(x,t) = ψ(x) = Here ψ(x,t), or ψ(x), is the amplitude for a particle to be at position x at time t, whereas |ψ> is a particle state that might be spread across all x. Other restated equations are: = ∫ dx = ∫ * ψ(x) dx = ∫ø*(x) ψ(x) dx The probability that an electron is between x and x+Δx, for small Δx, is: Prob(x) Δx = |ψ(x)| Δx 2

In the limit that Δx goes to zero, Prob(x) is the probability density of finding the particle near x. Feynman chooses this normalization for momentum amplitudes:

= * = exp{–ipx/ħ} Prob(p) = || / h (not ħ) 2

The Dirac delta function is defined as the limit as ε goes to 0 of: δ(x) = exp{–x /ε }/(ε√π) ∫ δ(x) dx = 1, over all x 2

2

For two position states |x> and |x’>: ψ(x’) = ∫ ψ(x) dx ψ(x’) = ∫ δ(x’–x) ψ(x) dx = 0 if x’ is not equal to x The extension to three dimensions, with r = (x,y,z), is: = ∫ dx dy dz = ∫ ø*(r) ψ(r) dx dy dz = δ(x’–x) δ(y’–y) δ(z’–z) »For a particle subject to a force described by a potential V, Schrödinger’s equation for nonrelativistic velocities is: 1D: iħ dψ/dt = –ħ /2m ∂ ψ/∂x + V ψ 3D: iħ dψ/dt = –ħ /2m ☐ψ + V ψ 2

2

2

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where ☐ = ∂ /∂x + ∂ /∂y + ∂ /∂z 2

2

2

2

2

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»Feynman says a “most remarkable consequence of Schrödinger’s equation… [is] that a differential equation involving only continuous functions of continuous variables … [gives rise to] discrete energy levels in an atom.” »Feynman says: “for each of the rules of symmetry there is a corresponding conservation law.” This is called Noether’s Theorem. Feynman says: “A physical system is symmetric with respect to the operation Q when Q commutes with U [or H], the [operators for] the passage of time.” Mathematically, this means: QH=HQ. »The following symmetry operations are universally valid. Translation in Space Translation in Time

Spatial Rotation Motion at Constant Velocity Exchange of Identical Particles Change of Quantum Phase CPT Symmetry The following symmetry operations are valid for all but the weak interaction. Spatial Reflection Time Reversal Reversal of Electric Charge Polarity Exchange of Matter & Antimatter »If a Hamiltonian H has a symmetry whose operator is Q, and Q has an eigenstate |ψ>, then Q|ψ> yields the same result forever, absent external interference. This establishes a conservation law: Q|ψ> is conserved.

»A symmetry operator operating a non-degenerate eigenstate can only change its phase angle. »A translation in time, represented by the operator D (Δ), increases all clock readings by Δ and decreases the time of all physical events by Δ. Symmetry under time translation implies the conservation of energy. t

A translation of the x-axis, represented by the operator D (Δ), decreases the x-coordinates of all physical events by Δ. Symmetry under x-axis translation implies the conservation of linear momentum in the x-direction. The x-momentum operator is p =k /ħ, with ∂ψ/∂x = (p /ħ)ψ. The same is true for y and z. x

x

x

x

A rotation of the xy-axes about the z-axis by angle θ is represented by the operator: R (θ). Symmetry under z-axis rotation implies the conservation of angular momentum in the z-direction. Spin m in the z-direction is equivalent to angular momentum J =mħ. The same is true for x and y. z

z

»A photon’s spin along its direction of motion can be +1 or –1, but never 0, because photons are massless. Neutrinos are definitely not massless, but their masses are extremely small, less than one millionth of the electron mass. In the reference frames in which neutrinos are emitted, and in the reference frames of our laboratories, neutrinos are always LHC polarized and antineutrinos are always RHC polarized.

Meet The Author Congratulations and thank you for reading my book. I know your time is valuable, and I sincerely hope you enjoyed this experience. I’d like to tell you something about myself and share some stories. First, the obligatory bio (as if 3 “tweets”-worth can define anyone): I have a B.S. in physics from Caltech, a Ph.D. in high-energy particle physics from Stanford University, and was on the faculty of Harvard University. Now “retired,” I teach at the Osher Institutes at UCLA and CSUCI, where students honored me as “Teacher of the Year.” In between, I ran eight high-tech companies and hold patents in medical, semiconductor, and energy technologies. My goal is to help more people appreciate and enjoy science. We all know one doesn’t have to be a world-class musician to appreciate great music — all of us can do that. I believe the same is true for science — everyone can enjoy the exciting discoveries and intriguing mysteries of our universe. I’ve given 400+ presentations to general audiences of all ages and backgrounds, and have written 3 printed books and 29 eBooks. My books have won national and international competitions, and are among the highest rated physics books on Amazon.com. I’m delighted that two of these recently became the 2 and 3 best sellers in their fields. nd

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Richard Feynman was a friend and colleague of my father, Oreste Piccioni, so I knew him well before entering Caltech. On several occasions, Feynman drove from Pasadena to San Diego to sail on our small boat and have dinner at our home. Feynman, my father, my brother and I once went to the movies to see “Dr. Strangelove or: How I Learned to Stop Worrying and Love the Bomb.” It was particularly poignant watching this movie next to one of the Manhattan Project’s key physicists. At Caltech I was privileged to learn physics directly from this greatest scientist of our age. I absorbed all I could. His style and enthusiasm were as important as the facts and equations. Top professors typically teach only upper-level graduate classes. But Feynman realized traditional introductory physics didn’t well prepare students for modern physics. He thought even beginners should be exposed to relativity, quantum mechanics, and particles physics. So he created a whole new curriculum and personally taught freshman and sophomore physics in the academic years 1961-62 and 1962-63. The best students thrived on a cornucopia of exciting frontier science, but many others did not. Although Caltech may be the world’s most selective science school, about half its elite and eager students drowned in Feynman’s class. Even a classmate, who decades later received the Nobel Prize in Physics, struggled in this class. Feynman once told me that students sometimes gave him the “stink eye” — he added: “Me thinks he didn’t understand angular momentum.” Some mundane factors made the class very tough: Feynman’s book wasn’t written yet; class notes

came out many weeks late; and traditional helpers (teaching assistants and upper classmen) didn’t understand physics the way Feynman taught it. But the biggest problem was that so much challenging material flew by so quickly. Like most elite scientists, Feynman’s teaching mission was to inspire the one or two students who might become leading physicists of the next generation. He said in his preface that he was surprised and delighted that 10% of the class did very well. My goal is to reach the other 90%. It’s a great shame that so many had so much difficulty with the original course — there is so much great science to enjoy. I hope to help change that and bring Feynman’s genius to a wider audience. Please let me know how I can make Feynman Simplified even better — contact me through my WEBSITE. While you’re there, check out my other books and sign-up for my newsletters. Printed Books, each top-rated by Amazon readers:

Everyone's Guide to Atoms, Einstein, and the Universe Can Life Be Merely An Accident? A World Without Einstein

The Everyone's Guide Series of Short eBooks

Einstein: His Struggles, and Ultimate Success, plus Special Relativity: 3 Volumes, A to Z General Relativity: 4 Volumes, from Introduction to Differential Topology Quantum Mechanics: 5 Volumes, from Introduction to Entanglement Higgs, Bosons, & Fermions… Introduction to Particle Physics Cosmology Our Universe: 5 Volumes, everything under the Sun Our Place in the Universe: a gentle overview Black Holes, Supernovae & More We are Stardust Searching for Earth 2.0

Smarter Energy Timeless Atoms Science & Faith

Table of Contents Feynman Simplified 3B: Quantum Mechanics Part Two Summary of QM Part One Chapter 13 Molecules & Forces Chapter 14 Operators & Matrices Chapter 15 Two-State Spins Chapter 16 Hyperfine Splitting in Hydrogen Chapter 17 Rotation for Spin ½ & Spin 1 Chapter 18 Electrons in the Simplest Crystals Chapter 19 Electrons in Realistic Crystals Chapter 20 Semiconductors Chapter 21 Independent Particle Approximation Chapter 22 Schrödinger’s Equation Chapter 23 Symmetry & Conservation Laws Chapter 24 Review of Part Two