Problems and Solutions in Nonrelativistic Quantum Mechanics 9810246331, 9789810246334

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NONRELATIVISTIC QUANTUM MECHANICS

Anton Z. Capri *

1 i •

World Scientific

cpfo 6kwis Soíupons NONRELATIVISTIC QUANTUM MECHANICS

Anton Z. Capri Department of Physics University of Alberta, Canada

\ • m En > Em

2) induced from n - » m 3) induced from m —» n . At equilibrium, at a temperature T, the emission and absorption probabilities are given by Pmn

=

KNn[Bmnp(v)+Amn]

Pnm

=

KNm[Bnmp(v)]

emission

(1.7.32)

absorption.

(1.7.33)

Here, hu = En — Em, K is a proportionality constant, and Nn, Nm are respectively the number of atoms in the states n and m . The coefficients Amn, Bmn are known respectively as the "Einstein Coefficients" of spontaneous and induced emission, whereas the coefficient Bnm is known as the "Einstein Coefficient" of induced absorption. [1.2] Use these equations together with Planck's radiation law for the radiation density p(v) at equilibrium to show that 1) the Einstein coefficients of induced absorption and emission are the same, that is that Byim — Bum and t h a t 2) the Einstein coefficients of spontaneous and induced emission are related by Anm ~

có

Bnm

.

Solution In equilibrium, at a temperature T, if the number of atoms in the state n and m is given by Nn and Nrn respectively, we have t h a t Nn=Ne~E"/kBT

,

Nm = Ne~Bm/kBT

(1.7.34)

where N is the total number of atoms. Therefore, Nm = Nn e(En-Em)/kBT

_

N n ehv/kBT

(1.7.35)

Also at equilibrium the number of transitions from n —»• m equals the number of transitions from m —> n. Thus, we have NmPnm = NnPmn

(1.7.36)

or Nnehv'kBT

Brnnp{v)=Nn[Bnmp{v)

+ Anm}

.

(1.7.37)

Thus, solving for the radiation density p we get

.

2.4

(2.3.16)

(2.3.17)

Bohr-Sommerfeld: Particle in a B o x

Use Bohr-Sommerfeld quantization to calculate the energy levels of a particle confined to a box of length L. For simplicity assume this is a "one-dimensional box".

Solution As in the problem above we use Bohr-Sommerfeld quantization pdq = nh = 2nnh

.

(2.4.18)

Again we have that the momentum is related to the energy by p = ±V2mE

(2.4.19)

so that we have the particle moving to the right with constant momentum V¿mE and then returning to the left wit) constant momentum -y/2mE. Hence, we get (jipdq

=

nh

=

V2 mE

11: dx

=

\ZlmE

x 2L .

—

J

dxj (2.4.20)

Solving for E we find £

2.5

=

h2 s ^ "

!

I2-4'21)

-

Larmor Frequency

Suppose a gyroscope has a magnetic moment ¡2 proportional to its angular momentum j. according to ¡1 = ML . The potential energy due to placing the gyroscope in a magnetic field B is V = -¡I

B .

Assume that B is constant and derive the equation of motion for L. Show that the gyroscope precesses with the angular Larmor frequency u)L — MB .

Solution For a gyroscope with angular momentum L we have t h a t if the angular velocity is Q and the moment of inertia is 7 then L = Id

.

(2.5.22)

Thus, the kinetic energy is T=^IJ

2

= ^lé2.

(2.5.23)

Also, V =-p.

• B = -ML

• B = -M1Q

•B .

(2.5.24)

Then, the Lagrangian is L = \ie2 ^

- I' = l-lé2 + MI9B. z

(2.5.25)

The equation of motion is — lié + MIB1 dt L

= 0.

(2.5.26)

-I

This yields immediately a first integral 9 + MB = constant .

(2.5.27)

But, for no magnetic field (B = 0) we have the gyroscope at rest and we immediately get 0 = 0 for 5 = 0. So we see that the integration constant vanishes. Hence, 9 = -MB

2.6

= -lol

.

(2.5.28)

Applicability of Bohr-Sommerfeld Quantization

The system of quantization proposed by Bohr in 1913 and late gneralized by Sommerfeld is not applicable to all systems. To what general kinds of physical systems is Bohr's procedure applicable? For what kinds of systems is it not applicable.

Solution The system of quantization proposed by Bohr and Sommerfeld is applicable to systems with repetitive (periodic) motion. It is not applicable to systems with unbounded motion.

2.7

Schrodinger and Hamilton-Jacobi

Consider the time-dependent Schrodinger equation and put xjr = A e ' s / h where A = constant. Show t h a t in the limit as h —» 0 the equation h2 a 2 * 3* 2m dx2 + VV = ik- dt reduces to the Hamilton-Jacobi equation [2.1] 1

fdS\2 2m f e j

+ V { q )

, 8S -~dT-

Solution If we put V = AeiS'h in the Schrodinger equation we get

Now let H - » 0 then we get

"

+

-lV

(2.7.30)

dt 2m This is just the Hamilton-Jacobi equation.

2.8

W K B Approximation

In the problem above set S = W — Et and let W = Wo + hW\ + h¿W^ + • • • for the case of a one-dimensional Schrodinger equation. Find the equations for Wo and Wi and solve the equation for Wq . This is the so-called Wentzel-KramersBrillouin or W K B approximation.

Solution Clearly if in the problem above we set S = W — Et (before letting h —>• 0) we get (2.8.31) Letting W = Wo + hW\ + . . . and equating the coefficients of powers of h we get (2.8.32) So, (2.8.33) and (2.8.34) So we have an equation for W\

2.9

Dumbbell Molecule: Bohr-Sommerfeld

A dumbbell molecule of moment of inertia I is rotating about its centre of mass. a) Write the langrangian for this system and find the canonical momentum. b) Use Bohr-Sommerfeld quantization to find the energy levels of this molecule.

Solution a) If we take the plane in which the molecule is rotating as the plane z = 0 then the Lagrangian is just the kinetic energy L = iI

= exp(AV>) •

= BB"1

= 1 .

Solution In all cases, to see if the operator is linear, we have to check whether A(X\ipi + \2ip2)

=

^1 Aifii + \2A1fi2

(3.4.19)

where Aj and A2 are constants. The results are: a) Linear since J =

K(x,y)

Ai J

[XiAifii(y) + \2Aifi2(y)]

K(x,y)ifii(y)dy

+ A2 J

dy

I

" = TT = iS ' So we have to find an appropriate f(k) we use the initial condition that

/

and carry out the integral. To do this

oo J

k

k T ^ 2 -

(3.8.46)

•

(3-8.47)

This means that /(*) =

Therefore, gi[kx — hk2t/2m]

oo

/

J

k

k2

+

a2

•

(3.8.48)

Fortunately we have been told not to attempt the integral. Therefore, we are done.

3.9

Operator Hamiltonian

You are given the classical Hamiltonian for the motion of a particle in the form H = £-e~ax 2m where a is a constant. Find an acceptable Hamiltonian operator for this system. Notice that the answer is by no means unique.

Solution The classical Hamiltonian that we are given is ^classical =

2m

e~ax .

(3.9.49)

To make an operator out of this we must do two things: a) We must replace p by the operator (3-9-50)

Pop=-~ i ax

and b) rearrange the order of the operators P o p and xop so that the resultant Hamiltonian is hermitian. The easiest, and by no means unique, way of doing this is to symmetrize the operators. Therefore we propose •

i

3

-

9

-

5 1

»

CHAPTER

24

ó.

ELEMENTARY

SYSTEMS

Other possibilities are ¿Pope-axPop

Hop=

(3.9.52)

or any convex linear combination of (3.9.51) and (3.9.52).

3.10

Zero of Energy

In classical mechanics, the reference level for the potential energy is arbitrary. What are the effects on the wave function and energy of adding a constant potential in the time-dependent Schrodinger equation?

Solution Suppose we have the Hamiltonian H and we add a constant Vo to it to get the Hamiltonian H'. Then we have that ih—

= HV

(3.10.53)

as well as d^' ih—=

H'V

= {H+ V0)V'

.

(3.10.54)

If we try adding a constant phase —a/h to $ we have V = e-ia/n^ .

(3.10.55)

Then, th——=:e dt So, if we choose

-ia/h i t . t w ' + ih—— [ dt.

(3.10.56)

0 The relevant differential equations now are

(4.1.1) nn

Since the incoming beam is from the left we have only a transmitted beam on the right. Thus, the appropriate solutions are V>i(x) = iP2(x)

=

A(eik*'

+ Re~ikim)

x0

(4.1.2)

where k¡ = ^ f

,

k\ =

2m{E

~

Vo)

.

(4.1.3)

Using the continuity of the wavefunction and its derivative at x = 0 we get A(1 + R)=B

,

ikiA{l

- R) = ik2B .

(4.1.4)

Therefore, solving for R we get R =

k1

~ *2 . ki+k2

(4.1.5) K '

Cancelling a factor of 2 m / h 2 we get 2

\R\2 = VE + Vo-VE VE

+ V0 + VE

(4.1.6)

Regarding the car, the question is a red herring. The car going over a cliff has nothing to do with the problem we have just solved. For the car going over the cliff the potential is not a step function but varies smoothly from a potential that is 0 for x < 0 to a potential which is mgy for x > 0. This is a step function in the configuration of matter, not in the potential energy of the car. The car problem involves a potential that acts perpendicular to the car's original direction of motion. This explains why a probability of |ii| 2 = 0.84% is unreasonable, apart from the fact that we know that cars don't as a rule bounce away from unobstructed cliff edges. The problem of a car going over a cliff is considerably more complicated and is handled at the end of this chapter (problem 4.15).

4.2

D e e p Square Well

A particle moving in one dimension interacts with a potential of the form V(x) = 0 V

^

=

~^a

|ar| > a < 0

"

Find the equation determining the energy eigenvalues of this system. Solve it approximately assuming a is very small. What happens in the limit a —> 0?

Solution Here, x

^

_

/

0

for

for

I -f£

\x\ >a

l®l < a '

We define U = vo/2a. The problem now is exactly the same as the problem discussed in [4.2]. The solutions divide into positive and negative parity solutions and read

1

=:

ip-(x)

|r ^ I{

= =

AeKx B sin kx sinkx -Ae-Kx

x < —a |x| < a. x >a x < —a |x| < a x >a A o

iP+(x)

f AeKx 1 B cos kx { A e~Kx

(4.2.8)

where ,4.2.9) The equations that yield the energy eigenvalues are: Positive parity k tan ka = K

(4.2.10)

Negative parity kcotka

= —K .

(4.2.11)

Next, we solve for small a and use the explicit form of U k* =

2 m ( E

+ 'U°/2°) . h"

(4.2.12)

Thus, as a gets smaller the term vo/2a dominates so we can replace k'¿ by ImU/h2. This means that ka « 1 so we can set tan ¿a — ka (and cotA:a = 1/ka). The even parity solution now yields k2a = K. This gives that E

= -J¡T-

(4.2.13)

The odd parity solution requires that k/ka = —K. This leads to an energy that diverges as I/a2. Thus, we get only one bound state in the limit as a —> 0. The corresponding wavefunction is clearly

*+&={!?-*• itr

=^-"w.

Furthermore in this limit we find that U —>• oo. The result is a J-function potential.

4.3

Hydrogenic Wavefunction

The wavefunction of an electron in the ground state of an hydrogen-like atom is ip(r) = Ae

Zr a

!

, a = —-z

where Z is the charge on the nucleus. a) Determine the constant A, so that the wavefunction is normalized to unity. Remember that you are in three dimensions and r represents the radial variable in spherical coordinates. b) At what distance from the origin is the probability of finding the electron a maximum? c) Determine the average value of: the kinetic energy, the potential energy and the total energy. This verifies the virial theorem for the Coulomb potential.

Solution a) To normalize we simply integrate and impose 1

= J\ip\2dv.

( 4 . 3 . 1 5 )

Since the wavefunction is spherically symmetric we have dv = 4irr2dr and get oo

I = 4n Í

Jo

, \ i p ( r ) \ 2u rJJdr rOO

=

47r|^4|2 /

=

a' 4tt|¿|-*4^3 '

Jo

e~2Zr/a

r2dr

3

(4-3-16)

So, after choosing the phase of A

"=v5b) Let the probability of finding the electron between r and r + dr be P(r) dr. Let the probability of finding it in the volume between-!) and v + dv be pdv. Due to the spherical symmetry we again have dv = 4irr2 dr and so the two probabilities describe the same thing. P(r) dr = 4np(r) r2 dr .

( 4 . 3 . 1 8 )

Hence, P =

|V>|24ttR

2

=

—-J-

e~2Zr'a

4ttr2 .

( 4 . 3 . 1 9 )

4.4.

t f U U i y u s I'A'IHJ

RAVCRUIV^KLILV,

O U Í T I T W I , MUMCJVI UM

33

We now maximize P. dP n ^ = 0.

(4.3.20)

This yields 2 r e - 2 Z r / a [1 - Zr/a\ = 0 .

(4.3.21)

So, the answer is r = a/Z. c) We know that the average energy in this state equals the energy in this state since we have an eigenstate of the Hamiltonian. Therefore, Z2e2 = --2^-

M

(13.22)

The average potential energy is given by (V) = J

\ip\2V(r)dv

(V)

\ Jo

.

(4.3.23)

So, =

e -Zr!a

- Z e

2

r2dr

r

Ze2 a

(4.3.24)

Then, the average kinetic energy (T) is given by = (H) - (V) = -7^-

4.4

•

(4-3.25)

Bound State Wavefunction, Current, Momentum

a) Show that the wavefunction for a particle in a bound state may always be chosen to be real. b) By computing the current density give an explanation of the physical meaning of this result. c) Use the results obtained to show that for a bound state the expectation value (p) of the momentum vanishes.

Solution a) To obtain the wavefunction for a bound state we have to solve a differential equation with all the coefficients real and real boundary conditions. As a consequence, we may always choose the normalization constant so that the solution is real. This is never the case for scattering problems where the boundary conditions describe travelling waves and are of necessity complex.

34

*-XJ ± Jl

b) Now, suppose that we have a real wavefunction V- The probability current in this case is, as always, 7

J

—

=

2im

-r

K

>

=

3{V-*VV>} •

(4.4.26)

l m

Since V is real, this current vanishes. This means, using the equation of continuity, that the probability density remains constant in time. Thus, a particle in a certain region of space, determined by this wavefunction, remains in that region of space. We have a bound state. c) Writing out the expression for the current we have 7=o = ^ - ( r ^ - ^ r ) 2im Therefore, =

(p)

\ J ^

x

= ^ - [ r ^ + ( p r ) i ] lm

.

Wi»l> + (pV'*)^] = 0 .

(4.4.27)

(4.4.28)

Thus, the average momentum of a particle in a bound state vanishes. On the average the particle remains where it is.

4.5

T i m e Evolution for Particle in a B o x

For a particle in a box with sides at x — ± a the eigenfunctions and eigenvalues are ip+}n = 4

TTX ti2 7T2 cos[(n + 1/2).—] , E+ }n = (n + 1/2) 2 a ¿ma¿

„ . , TTX. ip->n = Bn sin n — J a

,

E-

n

= n

,

h2 7T2 t . 2ma¿

If we have a particle in such a box and its wavefunction at time t = 0 is given by 7TT

\P(0,£) = -7=sin[5—] y/a a find \P(t, x) .

Solution We are given the wavefunction at time t = 0 as ^(0,2;) = -^= sin(57ra;/a) . y/a

(4.5.29)

This is an energy eigenfunction with energy £

=

5 1

£

•

Vo is incident from the left on a potential Í 0 V(x) = i v ó \ 0

x < 0 0 Vo is incident from the right. Calculate the total particle current travelling to the right. [4.1] Hint: The waves from the left and right are completely independent and their scattering from the potential can be handled independently.

Solution To solve this problem we simply use the results of problem 4.8. Then the total particle current to the right is given by the transmitted part of the current incident from the left plus the reflected part of the current incident from the right. Corresponding to these two currents we have a transmission amplitude TL for the current from the left given by 2ikLKLe^k--2K