Fundamentals of quantum mechanics: solutions manual

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Solutions Manual

Fundamentals of Quantum Mechanics: For Solid State Electronics and Optics C.L. Tang Cornell University Ithaca, N. Y.

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Chapter 2 2-1. (a) ψ(x) [in units of A]

4

2

−4

−2

0

4

2

x

6

(b) +∞

1

6

−∞

−4

1

1 = ∫ | Ψ ( x ) | 2 dx = | A | 2 [ ∫ ( 4 + x) 2 dx + ∫ ( 6 − x ) 2 dx ] =

∴ A=

(c)

1 5

3 10

= 1

250 | A |2 3

.

, by inspection.

Next, find σ2 first: 1

6

−4

1

σ 2 = | A |2 [ ∫ ( x − 1) 2 ( 4 + x) 2 dx + ∫ ( x − 1) 2 ( 6 − x) 2 dx ] =

therefore,

< x 2 > = σ2 + < x >2 =

7 2

2 - 1

.

5 2

;

(d)

The answer to this question is tricky due to the discontinuous change in the slope of the wave function at x = -4, 1, and 6. Taking this into account ,

< K. E. > = −

2-2.

2

2

h 3 3h (0 ⋅1−5 ⋅ 2+ 0 ⋅ 1) = 2m 250 50m

Given  2  sin(3 πx/a),for 0 < x < a Ψ (x ) =  a 0, for x ≤ 0 and x ≥ a. 

(a) < H >=−

2

h 2 2m a

∂

a

2

∫ sin( 3πx /a) ∂x

sin( 3πx /a) dx = 2

0

2

2

9π h 2ma 2

(b)

 h2 ∂2  ˆ H sin( 3πx / a ) =  − sin( 3πx / a) = E sin( 3πx / a) 2   2m ∂x 

∴ E =

2

2

9π h 2ma 2

.

(c) 9 π 2h

−i t 2 2 Ψ ( x , t )= sin( 3πx / a)e 2ma a

.

(d)

[

∆H = < H 2 > − < H > 2

]

2 - 2

.

.

2

h 2 =− 2m a

2

4  9π 2h 2  ∂ = 0 . sin( 3πx /a) sin( 3πx /a) dx  ∫ 2  ∂x 4  2ma  0 a

Yes, it is as expected. Since the given state is an eigen state of the Hamiltonian as shown in (b), the uncertainty in the total energy must be zero as shown in (2.5d).

2-3.

Prove the following commutation relationships: (a)

[ Aˆ + Bˆ , Cˆ ] = ( Aˆ + Bˆ ) Cˆ − Cˆ ( Aˆ + Bˆ ) = ( Aˆ Cˆ − Cˆ Aˆ ) + ( Bˆ Cˆ − Cˆ Bˆ ) , Q.E.D. = [ Aˆ , Cˆ ] + [ Bˆ , Cˆ ] (b) [ Aˆ , Bˆ Cˆ ] = Aˆ Bˆ Cˆ − Bˆ Cˆ Aˆ = Aˆ Bˆ Cˆ − ( Bˆ Aˆ Cˆ − Bˆ Aˆ Cˆ ) − Bˆ Cˆ Aˆ = ( Aˆ Bˆ Cˆ − Bˆ Aˆ Cˆ ) + ( Bˆ Aˆ Cˆ − Bˆ Cˆ Aˆ ) = [ Aˆ ,Bˆ ]Cˆ + Bˆ [ Aˆ , Cˆ ]

2-4.

.

Q.E.D.

Prove the following commutation relations: (a)

[ pˆ x ,xˆ n ] = − i hnxˆ n−1 .

Applying the left side to an arbitrary state function Ψ( x) gives:

[ pˆ x , xˆ n ] Ψ( x) = − i h ( = − ih n x

(b)

n −1

Ψ( x)

∂ n ∂ x − x n ) Ψ( x) ∂x ∂x .

Q.E.D.

[ xˆ , pˆ x2 ] = i 2h pˆ x .

Similar to (a) above: [ xˆ , pˆ 2x , ] Ψ ( x ) = − h 2 ( x = i 2 h pˆ x Ψ( x)

(c)

.

∂2 ∂2 ∂ − x ) Ψ( x) = h 2 [ 2 Ψ ( x ) ] 2 2 ∂x ∂x ∂x Q.E.D.

Not possible .

2 - 3

2-5.

0 1  0 − i 1 0  Consider the two-dimensional matrices σˆ x =   , σˆ y =   , and σˆ z =   , whose 1 0  i 0 0 −1

physical significance will be discussed later in Chapter VI. (a)

1 0  The eigen values σz of σˆ z =   are clearly +1 and -1.  0 − 1 For the eigen function corresponding to the eigen value σz = +1 :  a  1 0   a σˆ z   =     = + 1  b   0 − 1  b 

a   ; ∴ b

a   = b

 1   .  0

For the eigen function corresponding to the eigen value σz = -1 :  a  1 0   a σˆ z   =     = − 1  b   0 − 1  b 

a   ; ∴ b

a   = b

 0   .  1

0 1 The eigen values σx of σˆ x =   are clearly also +1 and -1. 1 0 For the eigen function corresponding to the eigen value σx = +1 :  a  0 1  a σˆ x   =     = + 1  b  1 0  b

 a   ; ∴ b

a   = b

1 2

1   . 1

For the eigen function corresponding to the eigen value σx = -1 :  a  0 1  a σˆ x   =     = − 1  b  1 0  b

a   ; ∴ b

a   = b

1 2

1   .  − 1

0 − i The eigen values σy of σˆ y =   are clearly also +1 and -1. i 0  For the eigen function corresponding to the eigen value σx = +1 :  a  0 − i   a  σˆ y   =     = + 1 b  i 0  b

 a   ; ∴  b

a   = b

1 2

1   . i 

For the eigen function corresponding to the eigen value σx = -1 :

2 - 4

 a  0 − i   a  σˆ y   =     = − 1 b  i 0  b

(b)

a   ; ∴ b

a   = b

1 2

1   . − i

These eigen states in the Dirac notation in the representation in which σˆ z is diagonal are as follows: The eigen functions corresponding to the eigen values σz = +1 and -1 are, respectively

| σ z = 1 , + >   1 | σ = − 1 , + >   0   =   , and  z  =   . | σ z = 1 , − >   0 | σ z = − 1 , − >   1 The eigen functions corresponding to the eigen values σx = +1 and -1 are, respectively

 | σ x = 1 , + >  1  1 | σ = − 1 , + >  1  1    =   , and  x  =   . 2  1 2  − 1 | σ x = 1 , − >  | σ x = − 1 , − >  The eigen functions corresponding to the eigen values σy = +1 and -1 are, respectively

 | σ y = 1 , + >  1 1 | σ y = − 1 , + >  1 1      | σ y = 1 , − >  = 2  i  , and  | σ y = − 1 , − >  = 2  − i  .         Note:

The + and – signs in the Dirac notation for the eigen states of the Pauli spinmatrices refer to the spin- up and spin-down states,

respectively,

in the

representation in which σˆ z is diagonal.

2-6.

Consider the Hamiltonian operator Hˆ with discrete eigen values. Suppose the Hamiltonian is a Hermitian operator which by definition satisfies the condition:

2 - 5

∫ Ψ ( x) Hˆ Φ( x)dx = ( ∫ Φ (x) Hˆ Ψ(x)dx ) *

(a)

*

*

.

The eigen values of the Hamiltonian are all real: Let Ψ ( x) and Φ (x ) in the definition of a Hermitian operator be an eigen state of the Hamiltonian Hˆ corresponding to the eigen value Ei . * * ∫ Ψ ( x) Hˆ Φ ( x ) dx = ∫ ΨEi ( x) Hˆ ΨEi ( x) dx = Ei .

Similarly, ( ∫ Φ * ( x ) Hˆ Ψ ( x ) dx ) * = ( ∫ ΨE*i ( x ) Hˆ ΨEi ( x ) dx ) * = E i* . The condition of Hermiticity of the Hamiltonian leads to: E i = E *i . (b)

Let Ψ ( x) and Φ (x ) in the definition of a Hermitian operator be an eigen state of the Hamiltonian Hˆ corresponding to the eigen value Ei and Ej

, respectively.

The

Hermiticity condition gives: ( Ei − E j ) ∫ ΨE*i ( x) ΨE j ( x) dx = 0

Therefore,

if ( Ei − E j ) ≠ 0 ,

.

then ∫ ΨE*i ( x ) ΨE j ( x) dx = 0 ,

or the eigen functions

corresponding to different eigen values are necessarily orthogonal to each other.

2-7.

Consider a particle of mass m in a potential field V(x). (a)

On the basis of Heisenberg’s equation of motion, (2.49) , and the commutation relation (2.11a): pˆ 2 pˆ d i < xˆ >= < [ x + Vˆ ( xˆ ) , xˆ ] > = < x > . dt h 2m m 2 - 6

(b)

On the same basis, the time variation of the expectation value of the momentum is given by : pˆ 2 d i dV ( x ) < pˆ x >= < [ x + Vˆ ( xˆ ) , pˆ x ] > = − < > = Fx dt h 2m dx which is known as Ehrenfest’s theorem.

2 - 7

,

Chapter 3 3-1. ) λ(electron deBroglie =

(a)

proton ) λ(deBroglie =

(b)

h = 2mE

0.3 E(in eV )

12.3 E(in eV )

o

1.24 E (eV )

0.3 E(in eV )

o

A

μ

λDe Broglie

8

.

.

A

) ( proton ) λ(neutron deBroglie ≅ λdeBroglie = photon) λ(deBroglie =

o

A

(electron)

0.8

0.8

0.6

0.6

6 o

o

A

A 0.4

4 2

λ

3-2.

2

4

λphoton

0.4 0.2

0.2

(proton) De Broglie

0

λDe Broglie (neutron)

8 6 E(eV)

10

0

2

4

μ

6 E(eV)

8

10

Suppose we know that there is a free particle initially located in the range −a < x < a

with a

spatially uniform probability. (a)

The normalized state function Ψ( x ,t = 0 ) of the particle in the Schroedinger-

representation is, assuming the phase of the wave function is arbitrarily chosen to be zero:

ψ(x)

(2a)-1/2 x -a

(b)

The corresponding momentum representation is: Ψ( p) =

(c)

p h 1 sin( x a) π a px h

.

The corresponding state function at an arbitrary later time Ψ( x ,t > 0 ) in the integral form is : Ψ(x,t ) =

3-3.

+a

1

π

∫

+∞

1 2a

px a) i( p x x − p 2x t ) h e h 2mh dp x px

sin(

−∞

.

Consider a free particle with the initial state function in the form of: Ψ( x ,t = 0 ) = Ae − ax

(a)

2

+ ikx

.

To normalize this state function: 1= | A |

2

∫e

+∞ −∞

−2ax 2

dx = | A |

2

π 2a

;

⎛ 2a ⎞1/ 4 ∴ A =⎜ ⎟ ⎝π⎠

(b)

After a considerable amount of algebra by first completing the square of the exponential in Fourier-transform integral,

it can be shown that the corresponding momentum

representation of this state function is:

∫

+∞

1 Ψ( p) = 2πh

−∞

⎛ 2a ⎞ ⎜ ⎟ ⎝π⎠

1/ 4

e

−ax 2 + ikx −i p x x / h

where Δ px =

(c)

h =h a 2Δ x

,

− ⎡ 1 ⎤ e dx = ⎢ 2 ⎥ ⎣ 2π(Δpx ) ⎦ 1/ 4

< px > =k h

, Δx=

( p x −< p x >)2

1 4a

4 Δp x2

,

.

The corresponding state function Ψ( x ,t > 0 ) is: 1/ 4 1 ⎛ 1 ⎞ Ψ(x,t) = ⎟ ⎜ 2πh ⎝ 2π Δpx2 ⎠

∫e

+∞ − ( p x −< p x >) 4 Δp x2 −∞

⎛ 1 ⎞1/ 4 ⎡ 1 i t ⎤ + =⎜ ⎥ ⎢ 2 2⎟ ⎝ 2πh ⎠ ⎣ 4 Δ px h 2m ⎦

−1/ 4

e

2

e

−i

p p x2 t+i x x h 2mh

dpx

⎡ ⎤

⎢ ⎥ (x− x t )2 m ⎥ + i< p x > (x− < p x > t )+ iθ −⎢ 1 i t ⎥ ⎢ 2 h 2m 4 h ( + ) ⎢ ⎥ 4 Δ p x2 h 2m ⎦ ⎣

,

where θ is a time-dependent phase-shift of no physical consequence that goes to zero at t = 0. (d)

From c above, the expectation value and the corresponding uncertainty of the position for t > 0 are, respectively:

=

2 1 ⎡ ⎛ 2ha ⎞ ⎤ 2 Δx = 1+ t ⎢ ⎜ ⎟⎥ . 2 a⎣ ⎝ m ⎠⎦ 1

hk t m

and

Similarly, the expectation value and the corresponding uncertainty of the linear momentum for t > 0 are, respectively: < px > t = hk

(e)

and

Δ px (t) = Δ px (0) = h a

.

The uncertainty product of the position and momentum for this state is: 2 h ⎡ ⎛ 2ha ⎞ ⎤ 2 h t⎟ ⎥ ≥ (Δ x) (Δpx ) = ⎢1+ ⎜ 2 2⎣ ⎝ m ⎠ ⎦ 1

which satisfies Heisenberg’s uncertainty principle for all time t ≥ 0 .

,

Chapter 4 4-1. From Eq. (4-19):

e−ik3 d F = ⎤ A ⎡ k12 + k 22 cos k d − i sin k 2 d⎥ ⎢ 2 2k1k 2 ⎦ ⎣

.

The corresponding transmission coefficient is : 2 ⎡ ⎤ ⎛ k 2 + k 2 ⎞2 2 F 2 1 2 ⎢ T= = cos k2 d + ⎜ ⎟ sin k 2 d⎥ A ⎢⎣ ⎥⎦ ⎝ 2 k1k2 ⎠

⎫ ⎧ ⎡ (2 E − V ) 2 ⎤ 0 = ⎨1 + ⎢ − 1⎥ sin 2 k 2 d ⎬ ⎭ ⎩ ⎣ 4 E ( E − V0 ) ⎦ ⎡ ⎤ V = ⎢1 + sin 2 k 2 d⎥ ⎣ 4 E ( E − V0 ) ⎦ 2 0

−1

−1

−1

,

⎡ ⎤ −1 ΔV 2 2 = ⎢1 + sin k2 d⎥ ⎣ 4( E − VI )( E − VII ) ⎦

which is Eq. (4.20a). In the limit of (E − V0 ) → 0 , sin 2 k 2 d →

2m(E − V0 ) 2 d ; therefore, h2

−1 ⎡ ⎡ β2⎤ 2mV0 d 2 ⎤ T = ⎢1 + ⎥ = ⎢1 + ⎥ 4⎦ 4 h2 ⎦ ⎣ ⎣ −1

lim E →V0

-1-

.

Plots of T for β = 4:

and β = 10 :

4 -2.

A particle with energy E in a region of zero potential is incident on a potential well of depth Vo and width "d". From the expression for the probability of transmission T of the particle past the well given in (4.20a), the approximate values of E (in terms of h 2 /2md2) corresponding to the maxima and minima in T: (a)

for β = 10 are: En ≅

h2 [n 2 π 2 + 10 2 ] 2md 2

and

-2-

En ≅

h 2 (n + 1) 2 π 2 [ + 10 2 ] , respectively ; 2md 2 4

(b)

for β = 250 are:

En ≅

h2 [n 2 π 2 + 250 2 ] 2md 2

En ≅

and

h 2 (n + 1) 2 π 2 [ + 250 2 ] , respectively. 2md 2 4

4 –3. Consider a one-dimensional rectangular potential well structure such as that shown in Figure 4.9

below.

V = V1

for

V = 0

for -a < x < 0

V = V1/ 2

for

0 < x< a

V = V1

for

x> a

I

x < -a

III

II

IV

V1 V1/2

E

a V=0

-a

0

The wave functions in regions I through IV and the equations describing the boundary conditions on these wave functions for (a)

E > V1

are:

Ψ1 = e ik1 x + Ae−ik1 x Ψ2 = Be ik2 x + C e−ik2 x

Ψ3 = De ik3 x + F e−ik3 x Ψ4 = G e ik4 x

-3-

where

⎡ V1 ⎤ 2 ⎢ 2m (E − 2 ) ⎥ , k3 = ⎢ ⎥ , k4 = k1 . h2 ⎢ ⎥ ⎦ ⎣ 1

⎡ 2m ( E − V1) ⎤ 2 ⎡ 2mE ⎤ 2 k1 = ⎢ , k = 2 ⎢⎣ h 2 ⎥⎦ ⎥⎦ ⎣ h2 1

1

The boundary conditions (b. c.) at x = -a are:

e−ik1 a + Ae ik1 a = Be− ik2 x + C e ik2 a and

k1 [e−ik1 a − Ae−ik1 a ] = k 2 [Be− ik2 x − C e ik2 a ] .

The corresponding b.c. at x = 0 are:

B+C =D+ F

and

k2 (B − C) = k 3 (D − F)

.

The corresponding b.c. at x = a are:

De ik3 a + F e − ik3 a = G e ik4 x

(b)

and

k3 ( De ik3 a − F e − ik3 a ) = k 4 G e ik4 x

For V1 > E > V1 / 2 , the wave functions in the various regions are:

Ψ1 = Ae α1 x Ψ2 = Be ik2 x + C e−ik2 x Ψ3 = De ik3 x + F e−ik3 x

Ψ4 = G e−α 4 x

,

where

-4-

.

⎡2 m (V1 − E ) ⎤ 2 ⎡ 2mE ⎤ 2 α1 = ⎢ , k2 = ⎢ 2 ⎥ ⎥ 2 ⎣ ⎣ h ⎦ ⎦ h 1

1

⎡ V1 ⎤ 2 ⎢ 2m (E − 2 ) ⎥ , k3 = ⎢ ⎥ , α 4 = α1 . 2 h ⎢ ⎥ ⎦ ⎣ 1

The b. c. at at x = -a are:

Ae −α1 a = Be− ik2 x + C e ik2 a and

α1 Ae−α a = i k2 [ Be− ik2 x − C e ik2 a ] .

The corresponding b.c. at x = 0 are:

B+C =D+ F

and

k2 (B − C) = k 3 (D − F)

.

The corresponding b.c. at x = a are:

De ik3 a + F e − ik3 a = G e−α 4 a

(c)

and

i k3 ( De ik3 a − F e − ik3 a ) = − α 4 G e−α 4 x

For E < V1 / 2 , the wave functions in the various regions are:

Ψ1 = Ae α1 x Ψ2 = Be ik2 x + C e−ik2 x Ψ3 = De α 3 x + F e−α 3 x

Ψ4 = G e−α 4 x

,

where ⎡2 m (V1 − E ) ⎤ 2 ⎡ 2mE ⎤ 2 α1 = ⎢ , k2 = ⎢ 2 ⎥ ⎥ 2 ⎣ ⎣ h ⎦ ⎦ h 1

1

-5-

⎤2 ⎡ V1 ⎢2m ( 2 − E) ⎥ , α3 = ⎢ ⎥ , α 4 = α1 . 2 h ⎥ ⎢ ⎦ ⎣ 1

The b. c. at at x = -a are: Ae −α1 a = Be− ik2 x + C e ik2 a and

α1 Ae−α a = i k2 [Be− ik2 x − C e ik2 a ].

The corresponding b.c. at x = 0 are: B+C =D+ F

and

i k2 (B − C) = α 3 (D − F)

.

The corresponding b.c. at x = a are: De α 3 a + F e − α 3 a = G e−α 4 a

4 –4.

α 3 (De α 3 a − F e − α 3 a ) = − α 4 G e−α 4 x

and

Suppose the following wave function describe the state of an electron in an infinite square potential well, 0 < x < a, with V(x) = 0 inside the well:

( ) ( )

⎧⎪ Asin 32aπ x cos π2ax , for0≤ x≤ a Ψ(x)= ⎨ ⎪⎩0 elsewhere

(a)

.

One way to normalize the wave function is to expand the given wave function in terms of the normalized energy eigen functions from x = 0 to x = a: a⎡ 2 2πx 2 πx ⎤ sin( )+ sin( ) ] ⎥ . ⎢ a a a ⎦ 8⎣ a

Ψ( x ) = A

a a 2 1 = | A |2 [ + ] ; ∴ A = 8 8 a

(b)

.

Time-dependence of the normalized wave function is: Ψ ( x, t )=

πx −i 2 t 2πx −i ma 2 t 1 sin( )e sin( )e 2 ma + a a a a

1

π2h

2 π2h

-6-

(c)

If measurements of the energy of the electron are made, the values of the energy that will be measured and the corresponding absolute probabilities are: Energy

Probability

2 π2h 2 ma 2

1 2

π 2h2 2ma 2 4 -5.

1 2

Consider the one-dimensional potential of Figure 4.10: Region V= ∞

x= 0 , 2mω0

< n | pˆ x | n >=

hmω0 < n |( aˆ + + aˆ − ) | n >= 0 . 2

and

The expectation values of the potential energy and the kinetic energy (Tˆ ) are equal: p x2 hω hω 0 ˆ < n | T | n >=< n | | n >= 0 < n |( aˆ + + aˆ − )( aˆ + + aˆ − ) | n >= (2n + 1) , 2m 4 4

hω k h < n | Vˆ ( x ) | n >=− < n |( aˆ + − aˆ − )( aˆ + − aˆ − ) | n >= 0 (2n + 1)=< n | T | n > . 2 2mω0 4 (c)

The uncertainty product of

the position and momentum ∆x ∆ px is equal to

1 (n + )h : 2

∆ p2x = < p2x > − < px > 2 =

∆ x 2 = < x 2 > − < x >2 =

hmω 0 (2n + 1) 2

h (2n + 1) 2mω 0

5–1

,

;

1 ∴ ∆ px ∆x = (n + ) h 2

5-2.

.

For a one-dimensional harmonic oscillator, in the basis in which the Hamiltonian is diagonal, the matrix representations of : (a)

the position and momentum operators xˆ and pˆ x are, respectively :

   h  xˆ = −i  2mω 0    

pˆ x =

(b)

   hmω 0   2    

0 − 1 0 0 1 0 − 2 0 0 2 0 − 3 0 0 3 0 • • • • • • • •

0 1 0 0 • •

1 0 2 0 • •

0 2 0 3 • •

0 0 3 0 • •

• • • • • •

• • • • • • •  • •  • •  •

the operator products aˆ + aˆ − and aˆ − aˆ + , respectively:

5–2

•  • •  • •  •

,

;

0  0 0 + − aˆ aˆ =  0  •  •

(c)

0 0 0 • •  1 0 0 • • 0 2 0 • •  , and 0 0 3 • •  • • • • •  • • • • •

0 0 0 • •  2 0 0 • • 0 3 0 • •  0 0 4 • •  • • • • •  • • • • •

.

Using the above matrices, it can be shown immediately that the commutator of aˆ − and aˆ + is : 1  0 0 − + + − aˆ aˆ − aˆ aˆ =  0  •  •

5-3.

1  0 0 − + aˆ aˆ =  0  •  •

0 0 0 • •  1 0 0 • • 0 1 0 • • =1 0 0 1 • •  • • • • •  • • • • •

.

Substituting the wave function of the form :

mω 0 2 − x mω 0 ΨEn ( x) = Cn H n ( x)e 2h h

into the Schroedinger-equation gives :

[−

.

− h2 d 2H n ( x) dH n (x) hω 0 + hω 0 x + H n (u) ] e 2 2m dx dx 2

Change the variable from x to u≡

h ω0 2h

x2

= E n Ψ n (x)

mω 0 x gives indeed Eq.(5.33), which defines the Hermit h

polynomials: 5–3

2

d H n (u) dH (u) E 1 − 2u n + 2( n − )H n (u) = 0 2 du du hω 0 2

5-4.

.

Suppose the harmonic oscillator is initially in a superposition state | Ψ(t = 0) > =

1 [ | 0 > + |1 >], 2

the expectation value of the position of the oscillator < x >t ≡< Ψ(t) | x | Ψ(t) > as a function of time is:

< x > t ≡< Ψ(t ) | x | Ψ (t ) > 1 h = [< 0 |+< 1 | e − iω 0 t ] [ −i (aˆ + −aˆ − )] [| 0 >+e i ω0 t | 1 >] 2 hmω 0 =

5-5.

h sin ω 0t . 2mω 0

r From Maxwell equations, (5.65 a-d), and the condition for transverse waves ∇⋅E = 0 that

r r 1 ∂2 E ∇ E − 2 2 =0 c ∂x 2

.

It is clear that

r r 2πhω k − −iω k t +i kz E ( r, t) ≡ E x (z,t) e x = i ak e − ak+e iω k t−ik z )e x ( L

(5.68a)

2

ω satisfies the above wave equation as long as k = 2k . Under the same condition, it can be c 2

shown that (5.68a) and

r r 2πhω k − −iω k t +ik z B( r, t) ≡ By (z,t) e y = i ak e − a+k e iω k t−ik z )e y ( L satisfy (5.65c).

5–4

(5.68b)

Substituting (5.68a) and (5.68b) into

1 8π

∫ [E L

0

2 x

(z,t) + By2(z, t) ] dz

and making use of the commutation rela tionship (5.69) after changing the a ± ’s into operators gives (5.70):

1 Hˆ = [ aˆ k+ aˆ −k + ] hω k 2 and

1 E k n = < n | Hˆ | n > = ( n + ) hω 0 2 5-6.

.

The Rayliegh-Jeans law and Planck’s law for black-body radiation as functions of wavelength and in units of energy per volume per wavelength- interval: Since ν=

c , Planck’s radiation law as a function of the wavelength is: λ

ρb (λ) d λ = ρ b (ν =

In the limit of

c ∂ν 8π hc 1 ) dλ = 5 ⋅ hc / λkB T dλ λ ∂λ λ e −1

.

h , of the z-component of the angular momentum of this particle is: 2 1 1 < Lˆ z > = h − h = − h 3 3 3

6-6.

.

The wave function of a particle of mass m moving in a potential well is, at a particular time t :

(a)

Ψ( x , y , z ) = ( x + y + z ) e − α

x2 + y2 + z 2

Ψ in the spherical coordinate system is:

Ψ ( x, y, z ) = ( x + y + z )e −α

x2 + y2 + z2

=[r sin θcos φ+rsin θsin φ+ r cos θ]e −α r

⎡⎛ − 1 + i ⎞ 8π 4π ⎤ − α r ⎛ 1 + i ⎞ 8π =⎢⎜ Y11 + ⎜ Y1−1 + Y10 ⎥ r e . ⎟ ⎟ 3 ⎝ 2 ⎠ 3 ⎣⎝ 2 ⎠ 3 ⎦ To normalize: ⎡ 2 2 8π 4π ⎤ 1 = ∫ | Ψ (θ, φ) | 2 sin θ dθ dφ = N 2 ⎢( + ) + ⎥ ; 3⎦ ⎣ 4 4 3

∴N =

1 4π

.

. The corresponding normalized wave function is: ⎛ −1+ i ⎞ ⎛1+ i ⎞ 1 Ψ (θ, φ) = ⎜⎜ ⎟⎟Y11 + ⎜⎜ ⎟⎟Y1−1 + Y10 . 3 ⎝ 6 ⎠ ⎝ 6 ⎠

(b)

The probability measurement of Lö2 and Lˆ z gives the values respectively, is:

6-4

2 h 2 and 0,

Pr obability =

6-7.

1 1/3 = 1/3 + 1/3 + 1/3 3

.

Consider a mixed state of hydrogen:

Ψ = R21(r)Y11(θ, φ ) + 2R32 (r)Y21(θ, φ ) (a)

The normalized Ψ is: Ψ=

(b)

1 5

R21 (r )Y11 (θ, φ) +

2 5

R32 (r )Y21 (θ, φ) .

Ψ is not an eigen function of Lö2 , but is an eigen function of Löz corresponding

to the eigen value h . (c)

The expectation value < Ψ | Lö2 | Ψ > is : 2 4 26 2 < Ψ | Lˆ2 | Ψ > = h 2 + 6 h 2 = h . 5 5 5

6-8.

(d)

The < Ψ | Löz | Ψ > is h .

(e)

1 1 4 1 < Ψ | Hö | Ψ > = − ( ⋅ + ⋅ ) 13.6 eV 5 4 5 9

.

Consider a hydrogen atom in the following mixed state at t=0: Ψ( r , θ, φ,t = 0 ) = 3R32 ( r )Y20 ( θ, φ ) + R21 ( r )Y11 ( θ, φ )

(a)

The normalized the wave function is:

6-5

Ψ (r , θ, φ, t = 0) =

(b)

3 10

R32 (r )Y20 (θ, φ) +

1 10

R21 (r )Y11 (θ, φ)

.

The atom is not in a stationary state, because it is in a mixed state of n=2 and n =3.

(c)

The expectation value of the energy for t > 0 is: 9 1 1 1 < Ψ | Hö | Ψ > = − ( ⋅ + ⋅ ) 13.6 eV 10 9 10 4

(d)

.

The expectation values are : 9 1 28 2 < Ψ | Lö2 | Ψ > = ( ⋅ 6 + ⋅ 2 ) h 2 = h 10 10 5 1 1 9 )h= h < Ψ | Löz | Ψ > = ( ⋅ 0 + 10 10 10

(e) The uncertainty of Löz in this state is: ΔL z = [< Ψ | Lˆ2z | Ψ > − < Ψ | Lˆ z | Ψ > 2 ] 2 = [ 1

1 1 1/ 2 3 − h . ] h= 10 100 10

.

6-9.

This problem is somewhat like the finite square-well potential problem considered in the text, Sect. 4.4. The Hamiltonian of a particle of mass m in a finite spherical potential well: ⎧ 0, V( r ) = ⎨ ⎩V0 ,

if r ≤ a

if r ≥ a

6-6

is: h2 Hˆ = − ∇ + V (r ) 2m

.

For l = 0 , it is: h2 1 ∂ 2 ∂ [ (r ) ] + V (r ) Hˆ = − 2m r 2 ∂ r ∂r

.

The corresponding Schroedinger’s equation is

and

⎫ ⎧ h2 1 ∂ 2 ∂ [ 2 (r ) ] + V (r )⎬ Rno (r ) = E n Rno (r ) , ⎨− ∂r ⎭ ⎩ 2m r ∂ r

for

r≤a ,

⎧ h2 1 ∂ 2 ∂ ⎫ [ 2 (r ) ]⎬ Rno (r ) = E n Rno (r ) ⎨− ∂r ⎭ ⎩ 2m r ∂ r

for

r≥a .

,

The equation for r ≤ a can be converted to: 2mE d2 U (r ) = − 2 U (r ) 2 dr h

,

where U(r) = r R(r) . The general solution of this equation is: U (r ) = A cos kr + B sin kr where k =

2mE . To satisfy the boundary condition that Rn0(r) must be finite at r =0, h

A must be equal to 0, or

6-7

U (r ) = B sin kr Similarly,

where α =

,

for

r ≤ a.

for r ≥ a ,

2 m ( V0 − E )

to zero and:

h

U (r ) = C e α r + D e − α r .

,

For U(r) or Rn0(r) to be finite at r → ∞ , C must be equal U (r ) = D e − α r

,

for

r≥a .

Continuity of the wave function Rn0(r) and its derivative at r = a leads to the secular equation:

− k cot ka = α

.

Defining ξ = ka , the above equation is of exactly the same form as that corresponding to the antisymmetric solution of the finite square potential-well problem: − ξ cot ξ = β 2 − ξ 2 ,

where β 2 =

Just like in that problem, there is no solution, if β
=

−C 21 e = 2 1+ C12 ≅

8.6

−e < 100 | z | 210 >C12 + complex conjugate 2 1 + C12

∞

∫R

10

π

2

r R21 r d r

0

∫Y

00

cosθ Y10 sin θ d θ + C. C.

0

−C 21 e 1.5 a0 + C.C. 2 1+ C12

.

An electron in the n = 3, l = 0, m = 0 state of hydrogen decays by a sequence of (electric dipole) transitions to the ground state. (a)

The decay routes open to it are: |300> → |210 >

(b)

→

|100 >

→ |21 ± 1> →

|100 >

.

The allowed transitions from the 5d states of hydrogen to the lower states are: s 5 4 3

2

1

p

d

f

g

8.7.

Assume a Lorentzian fluorescence linewidth of 10 Ghz. The stimulated emission cross-section (in cm2 ) defined in connection with (8.31) for a hypothetical hydrogen laser with linearly polarized emission at 121.56 nm (Lyman-α line)is: 2

2

4π e ν 2 4 σ st = x12 g f (ν) ≅ 7.1x10 x12 hc

.

2

Using the value of the dipole moment found in Problem 8-5, x12 ≈ 0.62x10−16 cm 2, ∴

σ st ≈ 4.4 x10−12 cm 2

.

Assuming all the degenerate states in the 2p level are equally populated, the corresponding spatial gain coefficient (in cm-1 ) is: g = ( N 2 − N 1 ) σ st ≈ 4.4 x10−2 cm −1

,

if the total population inversion between the 1s and 2p levels of hydrogen in the gaseous medium is 1010 cm-3 .

Chapter 9 9-1.

The spin-orbit interaction in hydrogen is of the form, (6.62) : r r  Ze 2  ˆr ˆr ˆ  L ⋅ S ≡ ζ( r ) Lˆ ⋅ Sˆ H s−o =   2m 2 c 2 r 3   

The corresponding matrix for l = 1 in the representation in which Lˆ2 , Lˆ z , Sˆ 2, Sˆ z are diagonal is a 6x6 matrix.

To diagonalize this matrix within the manifold of degenerate states

|n,l = 1, ml , s = 1/2, ms > , the columns and rows corresponding to the pairs of (ml , ms ) values are arranged in a particular order:

( ml , ms )

( − 1, −

1 2

) ( − 1, +

1 2

) ( 0, −

1 2

) ( 0, +

1 2

) ( + 1, −

1 2

) ( + 1, +

1 2

)

(−1,−1/ 2) 1/2 0 0 0 0 0   (−1, +1 / 2 )  0 −1/2 1/ 2 0 0 0 ( 0,−1/ 2)  0 1/ 2 0 0 0 0 2   ⋅ ζ nl h . ( 0, +1 / 2 )  0 0 0 0 1/ 2 0  ( +1,−1 / 2 )  0 0 0 1/ 2 −1/2 0    (+1, +1 / 2)  0 0 0 0 0 1/2 

This matrix breaks down into two 2x2 and two 1x1 matrices which can be easily diagonalized . Doing so according to the degenerate perturbation theory yields two new eigen values: ζ nl h2 / 2 and −ζ nl h2 . These correspond to the two new sets of 4-fold ( j= ( j=

3 1 3 , mj = ± , ± ) and 2- fold 2 2 2

1 1 , mj = ± ) degenerate levels split from the original 6-fold degenerate level in the absence 2 2

of spin-orbit interaction as given in Sect. 6.5. The two sets of new eige n states correspond to the spin-orbit coupled The

diagonization

j = 3 / 2 , m j = ± 3 / 2, ±1 / 2

procedure

gives

also

9 -1

and

j =1/ 2 , m j = ± 1/ 2 hydrogenic states.

the

relevant

vector-coupling

coefficients

< lm l sm s | jm jls >

defined in (6.59) for the eigen functions for this particula r case. For

example, the vector-coupling coefficients: < j m j l s |l m l s m s > =< 3/ 2,±3/2,1,1/2 |1,± 1,1/2,± 1/ 2> = 1 ,

< j m j l s |l ml s ms > =< 3/2,−1/2,1,1/2|1,−1,1/2,1/2 > =

1 , 3

< j m j l s |l ml s ms > =< 3/2,−1/2,1,1/2|1,0,1/2,−1/2 > =

2 , 3

etc. This is the procedure for calculating vector-coupling coefficients in general.

9-2.

The perturbation theory for the covalent bonded homo-nuclear diatomic molecule can be extend to the case of hetero-nuclear diatomic molecules: E  A  Hˆ BA

Hˆ AB   EB 

 CAγ    = Eγ  CBγ 

 CAγ     CBγ 

where

E A ≡ < A | Hˆ | A > ≠ E B ≡ < B | Hˆ | B >, HAB ≡ < A | Hˆ | B > = H*BA ≡ < B | Hˆ | A >* Setting the corresponding secular determinant to zero gives: 2

Eγ 2 − (E A + EB ) Eγ − H AB + E A EB = 0

,

which gives the bonding and anti-bonding levels of the heteronuclear molecule:

9 -2

.

Ea = b

[

( EA + EB ) 1 2 ± ( E A − E B ) 2 + 4 H AB 2 2

]

1/ 2

2

H AB ≈ EA ± ( EA − E B ) B

for ( E A − E B ) >> H AB

2

,

and EA > EB. The corresponding wave functions of the bonding and

antibonding orbitals of the mole cule are: | b > = C (Ab ) | A > + C (Bb ) | B >

and

| a > = C A(a ) | A > + C (Ba ) | B > ,

where ( E A − E a,b )CA(a,b ) + HAB CA(a,b) = 0 . More specifically, they are:

,b ) C(a = A

[

H AB 2

H AB + (E A − E a,b ) 2

]

and

,b ) C(a = A

[

E A − E a,b 2

H AB + (E A − E a,b ) 2

]

.

[ Note: H AB = − H AB .]

9-3.

Suppose the un-normalized molecular orbital of a diatomic homo-nuclear diatomic molecule is:

Ψmo = CA | A > +CB | B > where | A > and | B > are the normalized atomic orbitals. (a)

The normalized molecular orbital is:

Ψm.o. =

1 | CA | + | CB |2 + 2SCA CB 2

9 -3

[ CA | A > + CB | B > ]

,

where S ≡< A | B > is the overlap integral between the atomic orbitals and CA and CB are assumed to be real.

(b)

The corresponding molecular energy is: 2

2

| C | E A + | CB | E B + 2CA CB H AB E m ≡ < Ψm | Hˆ | Ψm > = A | CA |2 + | CB |2 +2SCA CB

,

and

E m [| CA |2 + | CB |2 + 2S CA CB ] = | CA |2 EA + | CB |2 EB + 2CA CB H AB

.

Following the basic concept of Coulson’s molecular-orbital theory, differentiate the above equation against variations in CA gives:

∂E m | CA |2 + | CB |2 + 2SCA CB ]+ 2Em [CA + S C B ] = 2 [CA EA + CB H AB ] [ ∂CA Minimizing the molecular energy against variations in CA, or setting ∂E m /∂C A = 0 , yields one condition that Em , CA, and CB must satisfy:

(E A − Em ) CA + ( H AB − EmS ) CB = 0 Similarly,

.

by minimizing the molecular energy against variations in CB,

∂E m /∂C B = 0 , yields another condition Em , CA, and CB must satisfy:

(HBA − E m S ) CA + ( E B − E m ) CB = 0

.

The secular determinant of these two homogeneous equations must be zero:

9 -4

or setting

EA − Em H BA

H AB ≅ 0 EB − Em

,

assuming the overlap integral is negligible or S ≈ 0 . This result is the same as that obtained in Problem 9-2 above according to degenerate perturbation theory.

9-4.

The number of atoms per cubic cell of volume a3 in such a lattice:

N 1 1 8 = 2⋅[( 8 ⋅ ) + 6 ⋅ ] = 3 3 a 8 2 a

.

The number of valence electrons per conventional unit cell of diamond lattice = 4 ⋅

9-5.

9-6.

8 . a3

The primitive translational vectors for; SCC:

r r r a = a ex , b = a ey , c = a ez

FCC:

r a r a r a a = (e x + e y ) , b = (e y + e z ) , a = (e x + e z ) 2 2 2

;

1 1 1 Diamond lattice = FCC with 2 atoms per basis at ( 0, 0, 0) and  , ,  . It is, therefore, 4 4 4 equivalent to two inter- laced FCC lattice displaced one quarter the distance along the body diagonal of the FCC.

9-7.

The C-C bond length in the diamond structure =

9 -5

3 a. 4

9 -6

Chapter 10

10-1. For a two-dimensional electron gas, the density-of-state is independent of the energy; therefore, the Fermi energy is directly proportional to the electron density:

EF

Ne =

∫D

(2)

(E) d =

0

m EF π h2

.

10-2. (a)

The chemical potential of a free-electron gas in two dimensions is given can be found from Eq.(10.29):

Ne =

∞

∫ 0

m k BT 1 d E m kB T µ/k T = ln [e B + 1 ] 2 (E −µ)/ k B T 2 πh e +1 k B T πh πh 2 N e

µ (T) = kB Tln[e mk T −1]

∴

B

,

for Ne electrons per unit area. (b)

Plot µ(T ) / E F as a function of kT / E F as in Figure 10.6(b): µ E

F

1.00 0.95 k T B

0

0.1

0.2

10 - 1

E

F

.

;

10-3. For a typical 1-D energy band, sketch graphs of the relationships between the wave vector, k, of an electron and its: (a)

energy,

E

k 0

(b)

group velocity,

Vg

k 0

(c)

and effective mass.

m*

k 0

10 - 2

d.

The approximate density-of-states D(1) (E) ) for the energy band of part a aboveis

(1)

D (E)

k 0

10-4. The E(k x) vs. k x dependence for an electron in the conduction band of a one-dimensional semiconductor crystal with lattice constant a = 4 Å is given by: E (kx ) = E 2 −( E 2 − E1 )cos 2[k x a /2] ; E 2 > E1

(a)

.

The E(k x) for this band in the reduced and periodic zone schemes. E

E2

E1 − 2π

−π 0

π

reduced zone periodic zone

(b)

The group velocity of an electron in this band is:

10 - 3

2π

k xa

 1 ∂E  ( E2 − E1 ) a vg =  sin k x a = 2h  h ∂k  and is sketched below as a function of k x: Vg − π

(c)

k xa

π

0

The effective mass of an electron in this band as a function of k x is: 2  ∂2 E  −1   −1 2 ( E 2 − E1 ) a m*= h  2  = h  cos kx a 2 ∂ k    2

and is sketched below it in the reduced- zone scheme:

m*e

0

−π

π

k xa

A uniform electric field Ex is applied in the x-direction, the motion of the electron is as follows:

10 - 4

kx a

vg

m*e

Acceleration

0.2 π

>0

>0

-x direction

0.5 π

>0

→∞

=0

0.9 π

>0

0

>0

+x direction

0.5 π

>0

→∞

=0

0.9 π

>0

= N Trace  11   ρ 21 ρ22 

 0 1 h  Nh ( ρ12 + ρ21 )   =  1 0 2  2

,

,

11-2. An electrical charged particle with a spin angular momentum will have a magnetization proportional to the spin angular momentum. Suppose the averaged expectation value of the magnetization of r r M = N Trace [ ρˆ (γ Sˆ ) ] . (a)

the

medium

considered

in

Problem

11-1

above

is

The three Cartesian components of the magnetization in terms of the appropriate density- matrix elements as in Problem 11-1 above are:

Mz =

N γh ( ρ11 − ρ22 ) 2

,

My =

i N γh (ρ12 − ρ 21 ) 2

,

11 - 1

Mx =

(b)

N γh (ρ12 + ρ 21 ) 2

.

The Hamiltonian of the spin-1/2 particles in the presence of a static magnetic field r r r r H = H x x + H y y + H z z , but in the absence of any relaxation processes is: γh  H z Hˆ = −  2  H x + iγ H

H x - iγ H y  y

-H

z

 

From the results of Part a above, on the basis of the density- matrix equation (11.16), the dynamic equations describing the precession of the magnetization r M around such a magnetic field are:

d i (ρ11 − ρ22 ) = − 2 [H12 ρ 21 − ρ12 H21] = γ [i H x (−ρ12 + ρ21)+ H y ( ρ12 + ρ 21)] dt h

,

which can be shown to be

[

r r d M z = γ [−H x M y + H y M x ] = γ M × H dt

]

,

z

making use of the results in (a) above. Similarly for the x- and y-components of r r r r dM M, ∴ = γ M × H , just like in classical mechanics. dt (c)

r Suppose a magnetic field consisting of a static component in the z -direction and a r weak oscillating component in the plane perpendicular to the z -axis is applied to r r r r r the medium: H = H 0 z + H x x ≡ H 0 z + H 1cosω 0t x . The corresponding Hamiltonian is:

11 - 2

H 1 cosω 0 t  γh  H 0 Hˆ = −   H0 2  H 1 cosω 0t 

.

From (11.19), (th)

(th)

d ( ρ − ρ ) − ( ρ11 − ρ 22 ) ( ρ11 − ρ22) = − 11 22 + iγ H x (−ρ12 + ρ21) dt T1

.

Also, M ± =Mx ± iMy . Therefore, M − Mz d Mz = − z dt T1

(th)

+i

γH1 ( M + − M − ) cosω 0 t 2

.

Similarly, for the other components:

d M M ± = − ± ± iγ H 0 M m m iγ H 1 M z cosω ot dt T2

.

These are the well-known Bloch equations in the literature on magnetic resonance phenomena.

11-3. (a)

From the dispersion relation for light waves,

k 2 = εω 2 /c 2, and the definitions

k ≡ β+ iα and ε≡ ε'+iε" = ε 0 + iε" , β ≅ ε0 ω 0 /c

and

α ≅ ε"ω 20 / β c 2 .

Therefore, on the basis of (11.44) and near the resonance, ω 0 ≈ ω 21 and :

11 - 3

2 2 ε"ω 02 ε"ω 0 4 π ( N1 − N 2 )ν 0 e z12 α≅ = = β c2 ε0 c ε0 hc

where

ωp ≡

4πNe2 ≈ m

2

2

g f (ν 0 ) =

ωp f 4 ε0 c

g f (ν 0 )

,

4π(N1 − N2 )e 2 , if most of atoms are in the ground m

state, is known as the “plasma frequency” and f z ≡

2mω 21 2 z 21 is known as the h

“oscillator strength”. (b)

To compare the result obtained in Part a above with the classical result based on a damped harmonic oscillator model instead of the two-level atom model: Suppose the equation of mo tion of the harmonic oscillator is of the form: 2

1/ 2

d d f e ˜ −iω 0 t ˜ * iω 0 t 2 z(t) + Γ z(t) + ω 21 z(t) = − ( Ez e + Eze ) 2 dt dt m

which describes the oscillating motion of a particle of mass m and negative charge of the magnitude f 1/ 2e bound to a fixed point in space similar to the oscillator shown in Figure 5.1.

The spring constant of the harmonic oscillator is equal to

2 mω 21 ; the damping constant is Γ ; and the deviation of the particle from its

equilibrium position in the absence of any electric field Ez is z(t). For

the

classical

result,

assume ω 0 ≈ ω 21 >> Γ − 1

so

that

2 ω 02 −ω 21 ≅ 2ω 0 (ω 0 − ω 21) . Solving the above equation for a damped harmonic

oscillator:

z= −

~ f eE 1 g (ν) m 4ω0 f

;

and

Q

 N f e z ε = ε' + i ε" = 4 π [ χ'+ i χ"] = 4 π χ' − i 4 π   , E˜  

11 - 4

therefore, the classical model gives also : ε"ω 0 ε0 c

=

ω 2p f 4 ε0 c

g f (ν 0 ) = α

,

which is the same as the result obtained in (a) above on the basis of the quantum mechanic density-matrix equation. (c)

Since the complex dielectric constant based on the oscillator strength f using either the quantum mechanical model or the classical harmonic oscillator model gives the same result for ε” , it is obvious that the same should be true for ε’. The classical harmonic oscillator model, therefore, can be used to characterize the dispersion and absorption characteristics of linear optical media with only three phenomenological parameters: the oscillator strength f, that characterizes the strength of the charge e2 , the resonance frequency ω21 , and the damping constant Γ associated with the bound particle in the harmonic oscillator model.

11-4. Differentiating (11.51) ( th ) ρ mn  i ρ mn (t ) = ∫  + ∑ [ρ mm' (t ' )V (t ' ) m'n − V ( t ' ) mm' ρ m 'n (t ' )]e ( iωmn +1/ Tmn )( t'− t ) dt ' h m' −∞  Tmn  t

gives: ) ρ(th d i 1 mn ρmn = + ∑[ρ mm'V (t) m' n − V (t) mm' ρm 'n ] − ( iω mn + ) ρmn dt Tmn h m' Tmn

which is Eq.(11.27). (11.51), therefore, satisfies and is a solution of (11.27).

11 - 5

,

11.5

The second-order nonlinear optical susceptibility χ˜ (2 ) (ω1 + ω 2 = ω 3 ) relates the induced macroscopic polarization component Pi (ω 3 ) to the applied electric field components E j (ω1) and E k (ω 2 ) in the medium:

Pi ( ω3 ) = ∑ χ˜˜ (2) ijk (ω1 + ω 2 = ω 3 ) E j (ω1 ) E k (ω 2 )

.

j ,k

For any such medium with inversion symmetry, inverting the coordinate axes leaves χ˜ (2) ijk (ω1 + ω 2 = ω 3 ) invariant but changes the signs of all the vector components:

− Pi ( ω 3 ) = ∑ χ˜˜ (2) ijk (ω1 + ω 2 = ω 3 ) [− E j (ω1 )][−E k (ω 2 )] = Pi (ω 3 ) . j,k

Therefore, χ˜ (2 ) (ω1 + ω 2 = ω 3 ) must vanish. 11-6. Consider a laser with the following parameters: T1 ~ 10-9 sec, T2 ~ 10-12 sec, Tph ~ 5x10-12 sec , Rpump ~ 1027 /cm3 - sec , Bhν 0 g f (v 0 ) ~ 6x10−7 cm3/sec . The corresponding laser rate equations are:

    

d ( N − N1 ) = −109 ( N 2 − N1 ) −1.2 ⋅10−6 ( N2 − N1 ) N ph +10 27 dt 2 d N ph =− 2 ⋅1011 N ph + 6 ⋅ 10−7 ( N 2 − N1 ) N ph + 0(N (spont) ) ph dt

Changing the scales:

t → 10 τ, ( N 2 − N1) →1015 n, N ph →1014 N so that the

numbers are more manageable in the numerical computation:

11 - 6

    

d n = −10 n −1.2 n N + 10 4 dτ .

d N =− 2 ⋅ 103 N + 6 n N + 0(N (spont) ) dτ

The steady-state solutions of these equations are: Changing to normalized parameters:

nss = 333.3

n ≡ y and 333.3

and

Nss ≈16.6 .

N ≡ z , the above rate 16.6

equations become:

    

d y = −10 y (1+ 2 z ) + 30 dτ d z =2 ⋅10 3 z ( y −1) + 0( z(spont) ) dτ

.

The turn-on dynamics of such a laser can be calculated numerically on the basis of these normalized laser rate equations using, for example, the Mathematica program:

NDSolve[{y'[x] == -20 y[x] Abs[z[x]] - 10 y[x] + 30, z'[x] == -2000 z[x] + 2000 y[x] z[x] + 0.001, y[0] == 0, z[0] == 0}, {y, z}, {x, 0, 2}] g = 0/0 Plot[Evaluate[ y[x] /. g],{x, 0, 0.5},PlotRange-> {0, 2}, AxesOrigin->{0, 0}, AxesLabel->{"t", "n(t)/n(s.s)"}] Plot[Evaluate[ z[x] /. g], {x, 0, 0.5}, PlotRange->{0, 10}, AxesLabel->{"t", "N(t)/N(s.s.)"}] The resulting calculated dynamics for the normalized population inversion and intracavity intensity are shown in the figures ( t in 10-7 sec) below. These results show a pattern of laser relaxation oscillations with the frequency in the range of a few tenths of a Ghz and a damping time on the order of tens of nsec, numbers characteristic of a semiconductor laser.

11 - 7

Figure 11.1 - Examples of the transient dynamics of a semiconductor laser.

11 - 8

Errata

Chapter 4 - Page 48, Eeq.(419) should read : F e −ik3d = A ⎡ ⎤ k12 + k 22 − k d i cos sin k 2 d ⎥ ⎢ 2 2 k1 k 2 ⎣ ⎦

,

not :

e−ik 3 d F = ⎤ A ⎡ k12 + k 2 cos k d − i sin k2 d ⎥ ⎢ 2 2k1k2 ⎣ ⎦ . - Page 55, line 7 should read:

ˆ Pˆ ∑ C Ψ ( x)=∑ (-1) n +1 C E Ψ ( x)=Pˆ H ˆ ∑ C Ψ ( x) H n En n n En n En n

n

,

n

not :

ˆ Pˆ H

∑C Ψ n

n

En

ˆ ∑ C Ψ (x) (x) = ± ∑ C n E n ΨE n (x) = Pˆ H n En n

n

- Page 60, Problem 4.1, line 4 should read:

for β= 4 and 10....... …..

not :

,

.

for β = 2 and 6 .......…..

.

- Page 61, Problem 4.4, equation should read:

Ψ ( x) = ............

,

not : V(x) = ……… .

- Page 62, Problem 4.6 (c), should read:

… for the lowest two bound states…..

,

for the lowest three bound states …..

.

not :

Chapter 5 - Page 84, line 12 should read:

Δn= < α | (aˆ + aˆ − ) 2 | α >−(< α | aˆ + aˆ − | α >) 2 =| α |=(n ) 2

,

Δ n = < α | ( aˆ + aˆ ) 2 | α > − (< α | aˆ + aˆ | α >) 2 = | α | = (n ) 2

.

1

not

- Page 82, third line from the bottom should read:

1

2 | α |2n 2 = < 0 | α > e |α | = < 0 |α > ∑ n! n

2

not :

= < 0 |α >

2

∑ n

α 2n n!

= < 0 | α > e |α | 2

2

,

.

Chapter 6 – - Equation (6.3) should read :

⎧ h2 ⎡ 1 ∂ ⎛ 2 ∂ ⎞ 1 1 ∂ 2 ⎤ e2 ⎫ ∂ ⎞ ∂ ⎛ sin θ + r + ⎜ ⎟ ⎜ ⎟ ⎨− ⎢ 2 ⎥ − ⎬ΨE (r , θ, φ) 2 ∂θ ⎠ r 2 sin 2 θ ∂φ 2 ⎦ r ⎭ ⎩ 2m ⎣ r ∂r ⎝ ∂r ⎠ r sin θ ∂θ ⎝ = EΨE (r , θ, φ) ,

not :

⎧ h2 ⎡ 1 ∂ ⎛ 2 ∂ ⎞ ∂ ⎛ ∂ 2 ⎤ e2 ⎫ 1 1 ∂⎞ ⎨− ⎟+ 2 2 ⎜r ⎜sin θ ⎟ + 2 2 ⎢ 2 ⎥ − ⎬ Ψ (r,θ, φ ) ∂θ ⎠ r sin θ ∂φ 2 ⎦ r ⎭ E ⎩ 2m ⎣r ∂r ⎝ ∂r ⎠ r sin θ ∂θ ⎝ = EΨE (r,θ, φ )

.

- Equation (6.36) should read: ⎧ h 2 ⎡ 1 ∂ ⎛ 2 ∂ ⎞ l(l + 1) ⎤ e 2 ⎫ − ⎬ R El (r ) = E l R El (r ) , ⎟− ⎜r ⎨− ⎢ 2 r 2 ⎥⎦ r ⎭ ⎩ 2m ⎣ r ∂r ⎝ ∂r ⎠

not :

⎧ h 2 ⎡ 1 ∂ ⎛ 2 ∂ ⎞ ∂ l(l + 1) ⎤ e 2 ⎫ ⎨− ⎟ − ⎢ 2 ⎜r ⎥ − ⎬ REl (r) = El REl (r) r2 ⎦ r ⎭ ⎩ 2m ⎣r ∂r ⎝ ∂r ⎠ ∂r

.

- Page 91, Eq.(6.31) should read :

< lml | Lˆ ± | l' ml, >= [(l m ml, )(l ± ml, + 1)] 2 hδ ll ' δ m ,( m' ±1) 1

l

,

l

- not :

< lml | Lˆ ± | l' ml, >= [(l m ml )(l ± ml + 1)] 2 hδ ll ' δ m ,( m' ±1) 1

l

l

.

- Page 99, equation in should read :

2 ππ

2 2 ∫ ∫ R10 (r ) Y00 r dr sin θdθdφ = R10 (r ) r dr , 2

2

00

not : 2 ππ

2 2 ∫ ∫ R10 (r ) r dr sin θdθdφ = 4π R10 (r ) r dr , 2

2

00

- Page 99, Equation (6.38) should read :

………….. and

not :

| p y >=

i (| p+1 > + | p−1 >) . 2

(6.38) ,

…………….. and | p y >= i (| p+1 > + | p−1 >).

(6.38) .

- Page 103, Eq.(6.49) should read : < jm j | Jˆ ± | j ' m 'j >= [( j m m 'j )( j ± m 'j + 1)] 2 hδ jj ' δ m 1

,

' j , ( m j ±1)

- not : < jm j | Jˆ± | j' m 'j >= [( j m m j )( j ± m j + 1)] 2 hδ jj'δm 1

- Page 109, Problem 6.8(e) should read :

ΔLz = [< Ψ | Lˆ2z | Ψ > − < Ψ | Lˆ z | Ψ > 2 ] 2

,

ΔLz = [< Ψ | Lˆz | Ψ > − < Ψ | Lˆz | Ψ > 2 ] 2

.

1

not : 1

- Page 108, Problem 6.8(b) should read :

……measurement of L2 and Lz that gives …..,

j

,(m 'j ±1)

.

not :

…….measurement of L2 and Lz gives …..,

.

Chapter 7 – - line 18 should read : r r r r r r r r Ψa1a2 ....a N −1a N (r1 , r2 ,..., rN −1 , rN ) = Ψa2 (r1 )Ψa1 (r2 )...Ψa N −1 (rN −1 )Ψa N (rN )

,

not:

r r r r r r r r Ψa1a2 ....a N −1a N (r1 , r2 ,..., rN −1 , rN ) = Ψa2 (r1 )Ψa1 (r2 )...ΨnN −1l N −1ml N −1 s N −1ms N −1 (rN −1 )ΨnN l N ml N s N ms N (rN )

Chapter 8 – - Equation (8.25) should read:

| E f >=| 100 > e

i − E1t h

+ C12(1) | 210 > e

i − E2t h

,

not : | E f >=|100 > e

Chapter 10 – - Eq.(10.46) should read:

i − E1 t h

+ C12(1) | 210 >,e

i − E2t h

.

3/2 π ⎛ me* k B T ⎞ NC = , ⎜ ⎟ 2 ⎝ π 2 h2 ⎠

not :

1 NC = 2

⎛ me* k B T ⎞ ⎜ 2 2 ⎟ ⎝π h ⎠

not :

1 NV = 2

⎛ mh* k B T ⎞ ⎜ 2 2 ⎟ ⎝π h ⎠

3/2

.

- Eq.(10.47) should read : π ⎛ mh* k B T ⎞ ⎜ 2 2 ⎟ NV = 2 ⎜⎝ π h ⎟⎠

3/ 2

,

3/2

.