quantum

Physics 751 & 752: Quantum Mechanics I & II

Russell Bloomer1

University of Virginia

Note: There is no guarantee that these are correct, and they should not be copied

1email: [email protected]

Contents

1 751: Problem Set 1 11.1 Problem 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.2 Problem 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21.3 Problem 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21.4 Problem 4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31.5 Problem 5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3

2 751: Problem Set 2 52.1 Problem 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52.2 Problem 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52.3 Problem 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72.4 Problem 4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 82.5 Problem 5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9

3 751: Problem Set 6 133.1 7.3.1 Principles of Quantum Mechanics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 133.2 7.3.2 Principles of Quantum Mechanics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 143.3 7.3.3 Principles of Quantum Mechanics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 143.4 7.3.4 Principles of Quantum Mechanics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 153.5 7.3.5 Principles of Quantum Mechanics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 153.6 7.3.6 Principles of Quantum Mechanics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 163.7 7.4.1 Principles of Quantum Mechanics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 163.8 7.4.2 Principles of Quantum Mechanics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 183.9 7.4.3 Principles of Quantum Mechanics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 193.10 7.4.4 Principles of Quantum Mechanics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 203.11 7.4.5 Principles of Quantum Mechanics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 203.12 7.4.6 Principles of Quantum Mechanics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22

4 752: Problem Set 6 254.1 Problem 1: Sakurai 5.13 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 254.2 Problem 2: Sakurai 5.15 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 254.3 Problem 3: Sakurai 5.28 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 264.4 Problem 4: Sakurai 5.29 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 274.5 Problem 5: Sakurai 5.30 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28

5 752: Problem Set 7 315.1 Problem 1: Shankar 18.2.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 315.2 Problem 2: Shankar 18.2.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 315.3 Problem 3: Shankar 18.2.4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 325.4 Problem 4: Shankar 18.2.5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 325.5 Problem 5: Shankar 18.2.6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33

i

5.6 Problem 6: Shankar 18.4.4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34

6 752: Problem Set 8 376.1 Problem 1: Sakuari 5.35 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 376.2 Problem 2: Sakuari 5.40 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 386.3 Problem 3: Shankar 18.5.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 396.4 Problem 4: Shankar 18.5.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 406.5 Problem 5: Shankar 19.3.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 416.6 Problem 6: Shankar 19.3.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41

A 511: Problem Set 5 43A.1 Merzbacher Exercise 10.9 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43A.2 Merzbacher Problem 10.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44A.3 Merzbacher Problem 14.6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46

B Special Harmonics 49B.1 Legendre Polynomials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49B.2 Associated Legendre Polynomials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50B.3 Spherical Harmonics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50

ii

Chapter 1

751: Problem Set 1

1.1 Problem 1

a

What is the rate an electron orbits a hydrogen atom in its circular Bohr orbit with n = 30?

In Bohr’s Model the electron has angular momentum of L = n~, where n is the first quantum number. The ki-netic energy of an orbiting electron is

1

2mv2 =

1

8πεo

e2

r(1.1)

The combination of the two equations and solve for r

1

2mv2 =

1

2m

(n~mr

)2

=1

8πεo

e2

r

rn =4πεo~2

me2n2

Lets define

ao =4πεo~2

me2

Look back to the definition of angular momentum, the orbital frequency can be found by dividing by mr2. Then

f =v

r=

n~mr2

→ ~ma2

on3

Entering the correct values for the constants

f = 1.53× 1012 1

sX

b

What is the frequency of a photon emitted from the transition from n = 30 to n = 29?

The energy level for state n is

En = − me4

32π2ε2o~2

1

n2(1.2)

The transition is then is ∆E, so the frequency is

ν =∆E

h→ me4

64π3ε2o~3

(1

n229

− 1

n230

)

1

Inserting the proper constants

ν =(9.109x10−31)(1.602x10−19)4

64π3(8.85x10−12)2(1.0546x10−34)3

(1

(29)2− 1

(30)2

)= 2.56x1011HzX

c

Comment on the answers from (a) and (b).

The two values are different. The value for part b classically should be the difference the rate at which the electronorbits the nucleus between n = 30 and n = 29. The difference is do to the electron does not actually orbit the nucleus.X

1.2 Problem 2

Does the inverse-square repulsion hold true if the α’s hit the gold nucleus? Find an approximate size of the nucleusand the minimum energy to show deviation.

No, it would not hold true. The amount of energy to hit the nucleus by the inverse-square repulsion is infinite.

The approximate size of the nucleus is

R = 1.2A1/3 × 10−15 m (1.3)

The size of the gold nucleus is then R = 6.98 fm. The path of closest approach is1

K =zZe2

4πεop(1.4)

where K is the kinetic energy and p is the point of closest approach. The distance would be the nucleus radius.

K =2(79)

(1.602× 10−19

)24π (8.85× 10−12) (6.98× 10−15)

= 32.6 MeVX

1.3 Problem 3

a

For a nucleus of charge Ze, what is the the n = 1 µ orbit compared to the electron orbit.

From equation 13.3.142, the orbit of the electron in the Hydrogen is

r = n2 ~2

me2(1.5)

To modify this for an electron in an atom with charge Ze is

re = n2 ~2

mZe2→ ~2

mZe2

For the muon, the electron mass has to be replaced by the reduced mass of the muon and the nucleus. For largernucleus, the reduce mass is approximately the mass of the reduce mass. In general though,

rµ = n2 ~2(207m+M)

207mMZe2→ ~2(207m+M)

207mMZe2

So rµ/re

rµre

=~2(207m+M)

207mMZe2

~2

mZe2

=(207m+M)

207X

1Williams. Nuclear and Particle Physics2Shankar Principles of Quantum Mechanics

2

b

What is the frequency of a photon emitted for the muon transition from n = 2 to n = 1

The transition is defined as

∆E = E2 − E1 = 2π~ν

The energy is defined as

En = − µZ2e4

3π22ε2o~2

1

n2

where µ is the reduced mass. For the transition

ν =E2 − E1

64π3ε2o~3=

µZ2e4

64π3ε2o~3

(1

12− 1

22

)=

3Z2e4

256π3ε2o~3µ =

3Z2e4

256π3ε2o~3

(207mM

207m+M

)X

c

What is the frequency of the photon in the transition n = 2 to n = 1 for a gold atom?

Using the result from the previous problem

ν =3Z2e4

256π3ε2o~3

(207mM

207m+M

)= 3.19× 1021HzX

1.4 Problem 4

What is the uncertainty in the measurement of the height of a helium atom at rest above the table?

The helium atom is at rest on the table. If the atom is at rest the momentum is at 0. This means that theuncertainty in the momentum measurement is 0. The Uncertainty Principle states that

∆x∆p ≥ ~2π

(1.6)

From this, the uncertainty in the measurement in the position is infinite. If the helium is not known to be at rest,then the position measurement can be more certain. X

1.5 Problem 5

How long would the Bohr atom last if it obeyed classical physics?

Classically the velocity of a particle is

v =√

2mK

where K is the kinetic energy. For the electron in the Bohr orbit, the kinetic energy is

K =1

8πεo

e2

r

The distance an object travels in a circular orbit is

d = 2πr

The the period of the rotation is

1

ν=

√16π3εomr3

e

3

The Bohr atom has allowed orbits of

rn =4πεo~2

me2n2

The classical period would be

1

νn=

32π3ε2o~3n3

me4

In the time of one period, the electron will decay out of orbit, so the time for the n = 1 orbit is

1

ν=

4(8.85× 10−12

)3 (6.626× 10−34

)31

(9.109× 10−31) (1.602× 10−19)2= 4.52× 10−16sX

4

Chapter 2

751: Problem Set 2

2.1 Problem 1

a

Define the derivative of the delta function by

δ′ (x) = lim∆ → 0d

dx

1

(4π∆2)1/2e−x

2/4∆2(2.1)

Sketch this function for small ∆. See Attached Sheet What is the value of < δ′(x− a)|ψ(x) >

The value for < δ′(x− a)|ψ(x) > is∫ ∞

−∞δ′(x− a)ψ(x)dx = δ(x− a)ψ(x)

∣∣∣∣∞−∞

−∫ ∞

−∞δ(x− a)ψ′(x)dx

by integration by parts. The value of ψ at infinity is zero by the boundary conditions, so∫ ∞

−∞δ′(x− a)ψ(x)dx = −

∫ ∞

−∞δ(x− a)ψ′(x)dx

= −ψ′(a)X

b

What function has the delta function for its derivative? Explain

The step function, which steps up, has the delta function as its derivative. The step function has the propertythat every where but one value has slope of 0. The one value is discontinuous at that point the derivative value thereis then infinite. X

2.2 Problem 2

On the interval (-1,1) the polynomials 1, x, x2, x3, . . . construct the first four orthonormal basis of these polynomials.

First the interproduct is

< f(x)|g(x) >=

∫ 1

−1

f(x)g(x)dx (2.2)

For this problem (and next) the following convention will be used: ui is the ith basis, ψi is the ith orthogonal basis,and ϕi is the ith orthonormal basis.

5

In Gram-Schmidt u1 ≡ ψ1, ψ1 = u1 = 1. Now to normalize this base

< ψ1|ψ1 >=

∫ 1

−1

dx = x|1−1 = 2 = a1

The orthonormal base is then

ϕ1 =ψ1

a1= c

The second basis is u2 = x. Now compare to ϕ1

< u2|ϕ1 >=

∫ 1

−1

xdx = x2|1−1 = 0

Then u2 = ψ2. Normalizing

< ψ2|ψ2 >=

∫ 1

−1

x2dx =x3

3|1−1 =

2

3= a1

Then

ϕ2 =ψ2√2/3

=x√2/3

=

√3

2x

The third base is u3 = x2. Now to create linear independence

< u3|ϕ1 >=

∫ 1

−1

x2

√2dx =

x3

3√

2|1−1 =

√2

3= b1

< u3|ϕ2 >=

∫ 1

−1

√3

2x3dx =

√3

2

x4

4|1−1 = 0

Then ψ3 = u3 − b1ϕ1. Normalizing

< ψ3|ψ3 > =

∫ 1

−1

(x2 − 1

3

)2

dx =

(x5

5− 2x3

9+x

9

)|1−1

=8

45= a1

Then

ϕ3 =ψ3√a1

=x2 − 1

3√845

=

√5

2

1

2

(3x2 − 1

)The fourth base is u4 = x3. Forming linear independent base

< u4|ϕ1 > =

∫ 1

−1

x3

√2dx =

x4

4√

2|1−1 = 0

< u4|ϕ2 > =

∫ 1

−1

√3

2x5dx =

√3

2

x4

5|1−1 =

2

5

√3

2= b2

< u4|ϕ3 > =

∫ 1

−1

√5

2

1

2

(3x5 − x3) dx

=

√5

2

1

2

(3/6x6 − 1/4x4) |1−1 = 0

So the base is ψ4 = u4 − b2ϕ2 = x3 − 35x. Normalizing

< ψ4|ψ4 > =

∫ 1

−1

(x3 − 3

5x

)2

dx

=

(x7

7− 6x5

25+

9x3

75

)|1−1 =

8

175= a1

6

So

ϕ4 =ψ4√a1

=x3 − 3

5x√

8175

=

√7

2

1

2

(5x3 − 3x

)The first four basis are

ϕ1 =1√2

(2.3)

ϕ2 =

√3

2x (2.4)

ϕ3 =1

2

(3x2 − 1

)(2.5)

ϕ4 =

√7

2

1

2

(5x3 − 3x

)X (2.6)

2.3 Problem 3

Suppose the inner product for real functions is defined as

< f |g >=

∫ ∞

−∞f(x)g(x)e−x

2dx (2.7)

For the polynomials 1, x, x2, x3, . . ., find the first four orthonormal basis

Again u1 ≡ ψ1. Normalizing

< ψ1|ψ1 >=

∫ ∞

−∞e−x

2dx =

√pi = a1

Then

ϕ1 =ψ1√π

=1

π1/4

The second base is u2 = x. Forming linear independent base

< u2|ψ1 >=

∫ ∞

−∞

x

π1/4e−x

2dx = 0

So the orthogonal base is psi2 = u2 = x. Normalizing

< ψ2|ψ2 >=

∫ ∞

−∞x2e−x

2dx =

√pi

2= a1

Then

ϕ2 =ψ2√a1

=

√2

π1/4x =

2x√2π1/4

The third base is u3 = x2. Forming a linear independent base

< u3|ψ1 > =

∫ ∞

−∞

x2

π1/4e−x

2dx =

π1/4

2= b1

< u3|ψ2 > =

∫ ∞

−∞

2x3

√2π1/4

e−x2dx = 0

Then psi3 = u3 − b1ϕ1 = x2 − 12. Normalizing

< ψ3|ψ3 >=

∫ ∞

−∞

(x2 1

2

)2

e−x2dx =

2√pi

4= a1

7

Then

ϕ3 =ψ3√a1

=x2 1

2√2√π

4

=2

π1/4√

2

(x2 − 1

2

)

=1

π1/4√

8

(4x2 − 2

)The fourth base is u4 = x3. Forming a linear independent base

< u4|ψ1 > =

∫ ∞

−∞

x3

π1/4e−x

2dx = 0

< u4|ψ2 > =

∫ ∞

−∞

2x4

√2π1/4

e−x2dx =

3π1/4

2√

2= b2

< u4|ψ3 > =

∫ ∞

−∞

1

π1/4√

12

(4x5 − 2x3) e−x2

dx = 0

The orthogonal base is ψ4 = u4 − b2ϕ2 = x3 − 32x. Normalizing

< ψ4|ψ4 >=

∫ ∞

−∞

(x3 − 3

2x

)2

e−x2dx =

3√pi

4= a1

Then

ϕ4 =ψ4√a1

=x3 − 3

2x√

3√pi

4

=1

π1/4√

3

(2x3 − 3x

)=

1

π1/4√

12

(2x3 − 3x

)The four basis are

ϕ1 =1

π1/4(2.8)

ϕ2 =

√2

π1/4x =

2x

π1/4√

2(2.9)

ϕ3 =1

π1/4√

8

(4x2 − 2

)(2.10)

ϕ4 =1

π1/4√

12

(2x3 − 3x

)X (2.11)

2.4 Problem 4

Solve the time-independent Schrodinger equation in one dimension for an attractive potential V (x) = λδ(x) for abound state. Can there be more than one bound state? Explain.

The time-independent Schrodinger equation is

− ~2

2m

d2ψ

dx2− λδ(x)ψ = Eψ (2.12)

For a state to be bound E < 0. First consider the region for x < 0. By the definition of the delta function V (x) = 0.The Schrodinger equation reduces to

d2ψ

dx2= −2mE

~2ψ

This can be simplified by κ ≡√−2mE

~2 . The solution is then

ψ(x) = Ae−κx +Beκx

8

The wave equation must behave properly, so A = 0, because at x = −∞ blows up. Now when x > 0 needs to beconsidered. Again, V (x) = 0. The Schrodinger equation is again

d2ψ

dx2= −2mE

~2ψ

So the wave equations is then

ψ(x) = Ce−κx +Deκx

For the wave equation to behave properly, D = 0, because at x = ∞ it blows up. The wave equation must becontinuous, but because of the potential is infinite, the first derivative will be discontinuous at that point. To becontinuous B = C. Now integrate Schrodinger equation a small distance, ε, from 0

−∫ ε

−ε

~2

2m

d2ψ

dx2dx−

∫ ε

−ελδ(x)ψ(x)dx =

∫ ε

−εEψ(x)dx

The term on the right integrates to 0, when ε→ 0 and the first term is just the derivative of the wave function about0. In the middle term, apply the limit to find.

∆

(dψ

dx

)= −2mλ

~2ψ(0)

Because the derivative is discontinuous, it must be approached from both sides. On the left, dψdx

= Bκ and from the

right dψdx

= −Bκ. Now as the limit approaches 0, the left is subtracted from the right to yield

∆

(dψ

dx

)= −Bκ = −2mλ

~2B

Then the energy is

E = −~2κ2

2m→ −mλ

2

2~2

Lastly, normalizing the wave function.∫ ∞

−∞|ψ(x)|2dx = 1 → 2|B|2

∫ ∞

0

=|B|2

κ= 1

The wave equation is then

ψ(x) =

√mλ

~e−mλ|x|

~2 (2.13)

The allowed energy is again

E = −mλ2

2~2(2.14)

No, there is only one bound state. There is nothing that can be changed that does not changed the problem,such as the quantum number n. X

2.5 Problem 5

For a one-dimensional general time-dependent solution of the Schrodinger equation, prove:

a

d

dt

∫ ∞

−∞|Ψ(x, t)|2dx = 0 (2.15)

How would the result change if the limits were finite? Can you write an equation for this case?

9

So taking the derivative inside the integral

d

dt

∫ ∞

−∞|Ψ(x, t)|2dx =

∫ ∞

−∞

∂

∂t|Ψ(x, t)|2dx

the partial time derivative is

∂

∂t|Ψ|2 =

∂

∂x

i~2m

(Ψ∗ ∂Ψ

∂x− ∂Ψ∗

∂x

)Then

d

dt

∫ ∞

−∞|Ψ(x, t)|2dx =

i~2m

(Ψ∗ ∂Ψ

∂x− ∂Ψ∗

∂x

)|∞−∞

Because the functions must vanish at infinity to behave properly the integral is 0.

d

dt

∫ ∞

−∞|Ψ(x, t)|2dx = 0

The value would not be 0, but some finite value, because the particle could leave the space between a and b. Theequation would look like

d

dt

∫ b

a

|Ψ(x, t)|2dx =i~2m

(Ψ∗ ∂Ψ

∂x− ∂Ψ∗

∂x

)|baX

b

If < x >=∫x|Ψ|2dx, show that d

dt< x >= 

2m

The term on the left is

d

dt< x > =

∫x∂

∂x|Ψ|2dx

=i~2m

∫x∂

∂x

(Ψ∗ ∂Ψ

∂x− ∂Ψ∗

∂xΨ

)dx

= − i~2m

∫∂

∂x

(Ψ∗ ∂Ψ

∂x− ∂Ψ∗

∂xΨ

)dx

− x

(Ψ∗ ∂Ψ

∂x− ∂Ψ∗

∂xΨ

)|∞∞

Because the wave function must vanish at |x| = ∞, it reduces to

d

dt< x > = − i~

2m

∫∂

∂x

(Ψ∗ ∂Ψ

∂x− ∂Ψ∗

∂xΨ

)dx

= − i~m

∫Ψ∗ ∂Ψ

∂xdx

This is just < v >, which is < v >= m

, ie

d

dt< x >= − i~

m

∫Ψ∗ ∂Ψ

∂xdx =



mX

c

Prove

d

dt= −

⟨dV

dx

⟩(2.16)

Start by writing out ddt

d

dt= −i~

∫ (∂Ψ∗

∂t

∂Ψ

∂x+ Ψ∗ ∂

∂x

∂Ψ

∂t

)dx

10

Integrating by parts

d

dt = − ~2

2m

∫ (∂2Ψ∗

∂x2

∂Ψ

∂x−Ψ∗ ∂

2

∂x2

∂Ψ

∂x

)+

∫ (VΨ∗ ∂Ψ

∂x−Ψ∗ ∂(VΨ)

∂x

)dx

The first term vanishes because the function vanishes at 0. This reduces to

d

dt= −

∫Ψ∗ ∂V

∂xΨdx

Finally

d

dt= −

⟨dV

dx

⟩X

11

Chapter 3

751: Problem Set 6

3.1 7.3.1 Principles of Quantum Mechanics

Find the recursion relationship for 7.3.8

Equation 7.3.8

ψ′′ + (2ε− y2)ψ = 0 (3.1)

The power series to solve the differential equation used will be

u(y) =

∞∑n=0

Cnyn

The second derivative of the power series is

d2u

dy2=

∞∑n=0

n(n− 1)Cnyn−2 →

∞∑n=2

n(n− 1)Cnyn−2

The base can be change to start at 0

d2u

dy2=

∞∑n=0

(n+ 2)(n+ 1)Cn+2yn

The last term is

y2u(y) = y2∞∑n=0

Cnyn →

∞∑n=0

Cnyn+2

Reducing the power to just n

y2u(y) =

∞∑n=2

Cn−2yn

The differential becomes

∞∑n=0

(n+ 2)(n+ 1)Cn+2yn +

∞∑n=0

2εCnyn −

∞∑n=2

Cn−2yn

The y-terms can be collected by the same power. Then

(n+ 1)(n+ 2)Cn+2 + 2εCn − Cn−2 = 0

↓

Cn+2 =Cn−2 − 2εCn(n+ 1)(n+ 2)

So C0 and C2 are arbitrary. X

13


Verify that H3(y) and H4(y) obey the relation, Eq. (7.3.15)

Equation (7.3.15) states that

Cn+2 = Cn(2n+ 1− 2ε)

(n+ 2)(n+ 1)(3.2)

Lets consider H3(y) first. In this case, n = 3, so

Cn+2 = Cn(2n+ 1− 2ε)

(n+ 2)(n+ 1)→ C5 = C3

(7− 2ε)

(5)(4)

From Equation 7.3.18,

ε =2n+ 1

2(3.3)

where this n comes from the polynomial number. From this, C5 = 0. Now finding C3 in terms of C1

C3 = C1(3− 2ε)

(3)(2)= C1

(3− (2n+ 1))

(3)(2)= C1

−2

3

The coefficient C1 is the lowest, because there in no negative powers of y. So the recursion relationship becomes

u(y) = C1y + C3y3 + C5y

5 + . . . = C1

(y − 2

3y3 + 0

)Because C1 is arbitrary, it can be defined as C1 = −12. With that definition

u(y) = C1

(y − 2

3y3

)= −12

(y − 2

3y3 + 0

)= H3(y)

Now the case of H4(y). Again using equations 7.3.15 and 7.3.18 to find

C6 = C49− 2ε

(6)(5)= 0

Now to find C4 in terms of C2

C4 = C25− 2ε

(4)(3)= C2

−1

3

Again finding C2 in terms of C0

C2 = C01− 2ε

(2)(1)= C0(−4)

The recursion relationship becomes

u(y) = C0 + C2y2 + C4y

4 + C6y6 + . . .

= C0

(1− 4y2 +

4

3y4 + 0

)Here C0 is arbitrary, so define C0 = 12, then

u(y) = 12

(1− 4y2 +

4

3y4

)= H4(y)X


See Attached Sheets

14


See Attached Sheets


Using the symmetry arguments from exercise 7.3.3, show that⟨n∣∣X∣∣n⟩ =

⟨n∣∣P ∣∣big >= 0 and thus show that⟨

X2⟩

= (∆X)2 and⟨P 2⟩

= (∆P )2 in these states. Show that⟨1∣∣X2

∣∣1⟩ = 3~/2mω and⟨1∣∣P 2∣∣1⟩ = 3mω~/2. Show

that ψ(x) saturates the uncertainty bound ∆X ·∆P ≥ ~/2

For⟨n∣∣X∣∣n⟩, the coefficients can be set to 1 for simplicity of the argument below.

⟨n∣∣X∣∣n⟩ =

∫ψ∗n(x)xψn(x)dx

=

∫xne−x/2x

(xne−x

2/2)dx

=

∫x2n+1e−x

2dx

From exercise 7.3.3, odd powers of x lead to∫

= 0. For⟨n∣∣P ∣∣n⟩, again, the coefficients are set to 1.

⟨n∣∣P ∣∣n⟩ =

∫ψ∗n(x)

d

dx(ψ(x)) dx

=

∫xne−x

2/2 d

dx

(xne−x

2/2)dx

=

∫xne−x

2/2(nxn−1e−x

2/2 − xn+1e−x2/2)dx

=

∫ (nx2n−1e−x

2− x2n+1e−x

2)dx

From exercise 7.3.3, odd powers of x are 0, so∫

= 0. So the (∆O)2 is defined as (∆O)2 =⟨O2⟩−⟨O⟩2

. Now for

both X and P the⟨⟩

= 0, then (∆X)2 =⟨X2⟩

and (∆P )2 =⟨P 2⟩. These are defined as

(∆X)2

=

(mω

π~22n (n!)2

)1/2 ∫x2

(Hn

[(mω~

)1/2

x

])2

e−mωx2/~

and

(∆P )2

=

(mω

π~22n (n!)2

)1/2 ∫~2 d

2

dx2

(Hn

[(mω~

)1/2

x

])2

e−mωx2/~

For⟨1∣∣X2

∣∣1⟩,⟨1∣∣X2

∣∣1⟩ =

(mω

π~22 (1!)2

)1/2 ∫x2

(2(mω

~

)1/2

x

)2

e−mωx2/~

=(mω

4π~

)1/2 (4mω

~

)∫x4e−mωx

2/~

=(mω

4π~

)1/2 (4mω

~

) 3π1/2

4

(~mω

)5/4

=3~

2mω

15

Now for⟨1∣∣P 2∣∣1⟩

⟨1∣∣P 2∣∣1⟩ = ~2

(mω

π~22 (1!)2

)1/2 (4(mω

~

))×∫

xd2

dx2xe−mωx

2/~dx

= ~2

(mω

π~22 (1!)2

)1/2 (4(mω

~

))×

∫ ((~mω

)2

x4 − 3~mω

x2

)e−mωx

2/~dx

=3

2mω~

Finally for ψ0(x). First⟨0∣∣X2

∣∣0⟩⟨0∣∣X2

∣∣0⟩ =(mωπ~

)1/2∫x2e−mωx

2/~dx

=(mωπ~

)1/2 ~2mω

(π~mω

)1/2

=~

2mω

Next⟨0∣∣P 2∣∣0⟩

⟨0∣∣P 2∣∣0⟩ = ~2

(mωπ~

)1/2∫e−mωx

2/2~ d2

dx2e−mωx

2/2~dx

= ~2(mωπ~

)1/2∫ ((mω

~

)2

x2 − mω

~

)e−mωx

2/~dx

=mω~

2

Finally ∆X ·∆P

(∆X)2 · (∆P )2 =~

2mω

mω~2

=~2

4

Then√

(∆X)2 · (∆P )2 = ~2

= ∆X ·∆P X


See Attached Sheets


Compute the matrix elements of X and P in the∣∣n⟩ basis compare with the results of 7.3.4

The matrix elements of X and P can be found from the raising and lowering matrices. These matrices can beconstructed from equations 7.4.26 and 7.4.27. Combining with equation 7.4.28 and 7.4.29, to find the matrix elements.From 7.4.26 ⟨

n′∣∣a∣∣n⟩ = n1/2δn,n−1 (3.4)

The lowering matrix can be found then to be

a =

0 11/2 0 0 . . .

0 0 21/2 0

0 0 0 31/2

.... . .

16

Now for the raising operator

a† =

0 0 0 . . .

11/2 0 0

0 21/2 0

0 0 31/2

.... . .

Equation 7.4.28 defines X =

( ~2mω

)1/2(a+ a†), then as a matrix

X =

(~

2mω

)1/2

0 11/2 0 0 . . .

11/2 0 21/2 0

0 21/2 0 31/2

0 31/2 0...

. . .

From 7.4.29, P = i

(mω~

2

)1/2(a† − a), then as a matrix

P = i

(mω~

2

)1/2

0 −11/2 0 0 . . .

11/2 0 −21/2 0

0 21/2 0 −31/2

0 31/2 0...

. . .

So⟨n′∣∣X∣∣n⟩ is then

(~

2mω

)1/2 [. . . n′ . . .

]·

0 11/2 0 0 . . .

11/2 0 21/2 0

0 21/2 0 31/2

0 31/2 0...

. . .

...n...

=

(~

2mω

)1/2 [(n+ 1)1/2δn′,n+1 + n1/2δn′,n−1

]Lastly for

⟨n′∣∣P ∣∣n⟩

(mω~

2

)1/2 [. . . n′ . . .

]·

0 −11/2 0 0 . . .

11/2 0 −21/2 0

0 21/2 0 −31/2

0 31/2 0...

. . .

...n...

= i

(mω~

2

)1/2 [(n+ 1)1/2δn′,n+1 − n1/2δn′,n−1

]This is identical to exercise 7.3.4 X

17


Find⟨X⟩,⟨P⟩,⟨X2⟩,⟨P 2⟩, and ∆X ·∆P in the state

∣∣n⟩The X and P operator are defined as the following, respectively.

X =

(~

2mω

)1/2 (a+ a†

)P = i

(mω~

2

)1/2 (a† − a

)So from this,

⟨n∣∣X∣∣n⟩ =

(~

2mω

)1/2 ⟨n∣∣(a+ a†)

∣∣n⟩=

(~

2mω

)1/2 (⟨n∣∣a†∣∣n⟩+

⟨n∣∣a†∣∣n⟩)

=

(~

2mω

)1/2 (n1/2⟨n∣∣n− 1

⟩+

(n+ 1)1/2⟨n∣∣n+ 1

⟩)By orthogonality of the

∣∣n⟩,⟨n∣∣X∣∣n⟩ =

(~

2mω

)1/2(n1/2δn,n−1+

(n+ 1)1/2 δn,n+1

)= 0

Now, for the momentum operator

⟨n∣∣P ∣∣n⟩ = i

(mω~

2

)1/2 ⟨n∣∣(a† − a)

∣∣n⟩= i

(mω~

2

)1/2 (⟨n∣∣a†∣∣n⟩− ⟨n∣∣a†∣∣n⟩)

= i

(mω~

2

)1/2 ((n+ 1)1/2

⟨n∣∣n+ 1

⟩−

n1/2⟨n∣∣n− 1⟩)

By orthogonality of the∣∣n⟩,⟨n∣∣P ∣∣n⟩ = i

(~

2mω

)1/2 ((n+ 1)1/2 δn,n+1 + n1/2δn,n−1

)= 0

Next, the square of the position operator⟨n∣∣X2

∣∣n⟩ =~

2mω

⟨n|(a+ a†

)2 ∣∣n⟩=

~2mω

⟨n|(a2 + aa† + a†a+

(a†)2) ∣∣n⟩

=~

2mω

(n1/2 (n− 1)1/2 δn+1,n−1 + (n+ 1)δn+1,n+1+

nδn−1,n−1 + n1/2 (n− 1)1/2 δn−1,n+1

)=

~2mω

(0 + n+ 1 + n+ 0) =~mω

(n+ 1/2)

18

Next, the square of the momentum operator⟨n∣∣P 2∣∣n⟩ = −mω~

2

⟨n|(a† − a

)2 ∣∣n⟩= −mω~

2

⟨n|(a2 − aa† − a†a+

(a†)2) ∣∣n⟩

= −mω~2

(n1/2 (n− 1)1/2 δn+1,n−1−

(n+1)δn+1,n+1 − nδn−1,n−1 + n1/2 (n− 1)1/2 δn−1,n+1

)= −mω~

2(0− n− 1− n+ 0) = mω~ (n+ 1/2)

By definition (∆φ)2 =⟨φ2⟩− (⟨φ⟩)2 where φ is some operator. Then (∆X)2 = ~

mω(n+ 1/2) − 0, so ∆X =√

~mω

(n+ 1/2). This can be repeated to so that ∆P =√mω~ (n+ 1/2). Then

∆X∆P =

√~mω

(n+ 1/2)√mω~ (n+ 1/2) = ~ (n+ 1/2) X


In classical mechanics, the viral theorem states that the average kinetic and potential energy, which is V (r) = ark inan orbit is related by

T = c(k)V (3.5)

where c(k) depends only on k. Show that c(k) = k/2. Using the previous exercise to show that for the oscillator⟨T⟩

=⟨U⟩

(3.6)

To show the viral theorem is true, there is no frictional force. The other important detail is this is in a circularorbit. In this case,

T =1

2F · r

=1

2∇V · r

=1

2∇rV r

Recalling that V = ark

T =1

2∇rV r

=1

2

(kark−1

)r

=k

2ark =

k

2V

Now, for the oscillator k = 2

⟨T⟩

=⟨U⟩→⟨P 2

2m

⟩=

⟨1

2mω2X2

⟩⟨mω~ (n+ 1/2)

2m

⟩=

⟨1

2mω2 ~

mω(n+ 1/2)

⟩⟨

~ω2

(n+ 1/2)

⟩=

⟨~ω2

(n+ 1/2)

⟩X

19


Show that

⟨n∣∣X4

∣∣n⟩ =

(~

2mω

)2

[3 + 6n(n+ 1)] (3.7)

First lets define X4,

X4 =

(~

2mω

)4/2

×(a4 + a3a† + a2a†a+ a2a†a† + aa†a2 + aa†aa†+

aa†a†a+ aa†a†a† + a†a3 + a†aaa† + a†aa†a+

a†aa†a† + a†a†aa+ a†a†aa† + a†a†a†a+ a†a†a†a†)

From the orthogonality, if there are not equal raising and lower operators, the expectation value would be 0. Then

⟨n∣∣X4

∣∣n⟩ =

(~

2mω

)2

×[⟨n∣∣(aaa†a† + aa†aa† + aa†a†a+ a†aaa†+

a†aa†a+ a†a†aa)∣∣n⟩]

Which reduces to

⟨n∣∣X4

∣∣n⟩ =

(~

2mω

)2 ((n+ 1)(n+ 2) + (n+ 1)2+

n(n+ 1) + n(n+ 1) + n2 + n(n− 1))

=

(~

2mω

)2

[3 + 6n(n+ 1)] X


At t = 0 a particle starts out in∣∣ψ(0)

⟩= 1/2(

∣∣0⟩+∣∣1⟩)

(1)

Find∣∣ψ(t)

⟩The time dependent part of the harmonic oscillator is e−iEnt/~. For n = 0, E0 = ~ω/2, and for n = 1, E1 = 3~ω/2

so

∣∣ψ(t)⟩

=1

21/2

(e−iE0t/~∣∣0⟩+ e−E1t/~∣∣1⟩)

=1

21/2

(e−iωt/2

∣∣0⟩+ e−3ωt/2∣∣1⟩)X

(2)

Find⟨X(0)

⟩=⟨ψ(0)

∣∣X∣∣ψ(0)⟩,⟨P (0)

⟩,⟨X(t)

⟩,⟨P (t)

⟩

20

For⟨X(0)

⟩⟨X(0)

⟩=⟨ψ(0)

∣∣X∣∣ψ(0)⟩

=1

21/2

(⟨0∣∣+ ⟨1∣∣)( ~

2mω

)1/2

(a+ a†)1

21/2

(∣∣0⟩+∣∣1⟩)

=

(~

8mω

)1/2 (⟨0∣∣+ ⟨1∣∣) (a+ a†

) (∣∣0⟩+∣∣1⟩)

=

(~

8mω

)1/2 (⟨0∣∣a∣∣0⟩+

⟨0∣∣a†∣∣0⟩+

⟨0∣∣a∣∣1⟩+⟨

0∣∣a†∣∣1⟩+

⟨1∣∣a∣∣0⟩+

⟨1∣∣a†∣∣0⟩+

⟨1∣∣a∣∣1⟩+

⟨1∣∣a†∣∣1⟩)

=

(~

8mω

)1/2

(0 + 0 + 1 + 0 + 0 + 1 + 0 + 0)

=

(~

2mω

)1/2

Likewise for⟨P (0)

⟩⟨P (0)

⟩=⟨ψ(0)

∣∣P ∣∣ψ(0)⟩

=1

21/2

(⟨0∣∣+ ⟨1∣∣) i(mω~

8

)1/2

(a† − a)1

21/2

(∣∣0⟩+∣∣1⟩)

= i

(mω~

8

)1/2 (⟨0∣∣+ ⟨1∣∣) (a† − a

) (∣∣0⟩+∣∣1⟩)

= i

(mω~

8

)1/2 (−⟨0∣∣a∣∣0⟩+

⟨0∣∣a†∣∣0⟩− ⟨0∣∣a∣∣1⟩+⟨

0∣∣a†∣∣1⟩− ⟨1∣∣a∣∣0⟩+

⟨1∣∣a†∣∣0⟩− ⟨1∣∣a∣∣1⟩+

⟨1∣∣a†∣∣1⟩)

= i

(mω~

8

)1/2

(0 + 0 + 1 + 0 + 0− 1 + 0 + 0) = 0

Next for⟨X(t)

⟩⟨X(t)

⟩=⟨ψ(t)

∣∣X∣∣ψ(t)⟩

=1

21/2

(eiωt

⟨0∣∣+ e3iωt/2

⟨1∣∣)( ~

2mω

)1/2

×

(a+ a†)1

21/2

(∣∣0⟩e−iωt + e−3iωt/2∣∣1⟩)

=

(~

8mω

)1/2 (⟨0∣∣a∣∣0⟩+

⟨0∣∣a†∣∣0⟩+⟨

0∣∣a∣∣1⟩e−iωt +

⟨0∣∣a†∣∣1⟩e−iωt +

⟨1∣∣a∣∣0⟩eiωt+⟨

1∣∣a†∣∣0⟩eiωt +

⟨1∣∣a∣∣1⟩+

⟨1∣∣a†∣∣1⟩)

=

(~

8mω

)1/2 (eiωt + e−iωt

)=

(~

2mω

)1/2

cosωt

21

Finally for⟨P (t)

⟩⟨P (t)

⟩=⟨ψ(t)

∣∣P ∣∣ψ(t)⟩

=1

21/2

(eiωt

⟨0∣∣+ e3iωt/2

⟨1∣∣) i(mω~

2

)1/2

×

(a† − a)1

21/2

(∣∣0⟩e−iωt + e−3iωt/2∣∣1⟩)

= i

(mω~

8

)1/2 (−⟨0∣∣a∣∣0⟩+

⟨0∣∣a†∣∣0⟩−⟨

0∣∣a∣∣1⟩e−iωt +

⟨0∣∣a†∣∣1⟩e−iωt − ⟨1∣∣a∣∣0⟩eiωt+⟨

1∣∣a†∣∣0⟩eiωt − ⟨1∣∣a∣∣1⟩+

⟨1∣∣a†∣∣1⟩)

= i

(mω~

8

)1/2 (−eiωt + e−iωt

)= −

(mω~

2

)1/2

sinωtX

(3)

Find⟨X(t)

⟩and

⟨P (t)

⟩using Ehrenfest’s theorem and solve for

⟨X(t)

⟩and

⟨P (t)

⟩and compare to part 2

For⟨X(t)

⟩,

⟨X(t)

⟩=

(−i~

)⟨[X,H]

⟩=

(−i~

)(⟨[X,

P 2

2m

] ⟩+⟨[X,V (x)]

⟩)=

(−i

4m2~

)(⟨P [X,P ]

⟩+⟨[X,P ]P

⟩)=

⟨P (t)

⟩m

For⟨P (t)

⟩⟨P (t)

⟩=

(−i~

)⟨[P,H]

⟩=

(−i~

)(⟨[P,

P 2

2m

] ⟩+⟨[P, V (x)]

⟩)= −i~

⟨dVdx

⟩= −mω2⟨X(t)

⟩For this to be true, the time dependent

⟨X(t)

⟩has to contain the momentum term. With that some initial value

needs to be chosen. In this case, t = 0. So

⟨X(t)

⟩=⟨X(t = 0)

⟩cosωt+

⟨P (t = 0)

m

⟩sinωt

Likewise ⟨P (t)

⟩=⟨P (t = 0)

⟩sinωt−mω2⟨X(t = 0)

⟩cosωt


Show that⟨a(t)

⟩= e−iωt

⟨a(0)

⟩and that

⟨a†(t)

⟩= eiωt

⟨a†(0)

⟩

22

First lets define how the operator a varies in time.

i~dadt

= [a(t), H] = i~(a(t)H −Ha(t)) =

~ω(a(t)(a†(t)a(t) +

1

2)− (a†(t)a(t) +

1

2)a(t)

)=

~ω[a(t), a†(t)

]a(t) = ~ωa(t)

Solving this differential equation

i~dadt

= ~ωa(t) → da

a(t)= −iωdt→ ln a(t) = −iωt+ C

ln a(t) = −iωt+ C → a(t) = C′e−iωt

In this case, C′ is an integration constant. So this initial condition is when a(t = 0) = a(0). Substituting that intothe previous equation to find

a(t) = C′e−iωt → a(t) = a(0)e−iωt

Now the expectation value would be ⟨ψ∣∣a(t)∣∣ψ⟩ =

⟨ψ∣∣a(0)e−iωt

∣∣ψ⟩Because the operator is changing with time, the basis are constant, so⟨

ψ∣∣a(0)e−iωt

∣∣ψ⟩ = e−iωt⟨ψ∣∣a(0)

∣∣ψ⟩ = e−iωt⟨a(0)

⟩For the raising operator, the fact that a† is the adjoint to a can be used.(

a(t) = a(0)e−iωt)†→ a†(t) = a†(0)eiωt

Here to the basis are constant with time⟨ψ∣∣a†(t)∣∣ψ⟩ =

⟨ψ∣∣a†(0)eiωt

∣∣ψ⟩ = eiωt⟨ψ∣∣a†(0)

∣∣ψ⟩= eiωt

⟨a†(0)

⟩X

23

Chapter 4

752: Problem Set 6

4.1 Problem 1: Sakurai 5.13

The Hamiltonian for the Stark effect isH = p2

2m− e2

r+ezε, where the last term is a perturbation. Now the Hamiltonian

matrix is

H =

(Eo2s + 〈s|ezε|s〉〈s|ezε|p〉〈p|ezε|s〉 Eo2p + 〈p|ezε|p〉

)From parity and Wigner-Eckart Theorem 〈s|z|s〉 = 〈p|z|p〉 = 0. When considering relativistic effects, Eo2s + δ = Eo2p.For convenience E = Eo2s = Eo2p. Then

H =

(E + δ 〈s|ezε|p〉〈p|ezε|s〉 E

)From the text, eε〈s|z|p〉 = eε〈p|z|s〉 =

√3eaoε. So the Hamiltonian becomes

H =

(E + δ

√3eaoε√

3eaoε E

)diag.⇒

∣∣∣∣ E + δ − λ√

3eaoε√3eaoε E − λ

∣∣∣∣ = 0 ⇒ (E − λ)2 + δ (E − λ)− 3e2a2oε

2 = 0

E − λ =−δ ±

√δ2 + 12e2a2

oε2

2⇒ λ = E +

δ

2±√δ2 + 12e2a2

oε2

2

The mean energy is E′ = E + δ/2 with shifts of ∆E = 12

√δ2 + 12e2a2

oε2. Now examining the shifts in energy for thetwo cases of δ eaoε and δ eaoε.The case of δ eaoε. Using the binomial expansion

∆E =δ

2

(1 +

12

2δ2e2a2

oε2

)⇒ ∆E =

δ

2

(1 +

6

δ2e2a2

oε2

)So in leading order in ε, it is quadratic.For the case δ eaoε

∆E =

√12

2eaoε

(1 +

δ2

24e2a2oε2

)⇒ ∆E =

√3eaoε

(1 +

δ2

24e2a2oε2

)So in leading order in ε, it is linear. As it has been shown in the text the unperturbed Hamiltonian are time invariant,but the dipole term 〈α′l′m′|z|αlm〉 = 0, so there is no time invariance for the dipole term from page 280. X


Suppose an electron has an intrinsic electric dipole moment µel, which is proportional to σ. This dipole causes asmall perturbative potential, −µel · E. Discuss qualitatively how the energy levels of a Na atom would be alteredin the absence of any external electromagnetic field. Are the energy shifts first or second order? State which statesget mixed with each other. Obtain an expression for the energy shifts of the lowest level that is affected by the

25

perturbation.

In the absence of an external electromagnetic field H = Ho+V , where V = −µel ·E. The electric field can be defined

as E = Eor. The potential becomes −µel · E = −Eoµel · r. The electric dipole can be changed to µel = ek√

4π3σ,

where k is some constant. Then

−µ ·E = −ekEo√

4π

3σ · r = −ekEo

√4π

3|σ||r| cos θ

From spherical harmonics −ekEo√

4π3|σ||r| cos θ = −ekEoz. This is similar to the Stark effect. The correction will be

first order, because z affects the states in first order. The valence electron in Na is in the 3s state, so then the states are|3, 0, 0〉, |3, 1, 0〉, |3, 2, 0〉, |3, 1,±1〉, |3, 2,±1〉, and |3, 2,±2〉. When the potential acts on these state, the selection rulesare ∆m = 0 and ∆l = ±1. The states |3, 2,±2〉 are unaffected. Now for the mixed states, 〈3, 2,±1|z|3, 1,±1〉 6= 0, sothe states |3, 2, 1〉 and |3, 1, 1〉 are mixed together, and |3, 2,−1〉 and |3, 1,−1〉 are mixed together. Now for the statem = 0 would be simple, but because the s-state can go to the p-state and the d-state can go down to the p-state, thethree have to be mixed together. The lowest state will be a m = 0 mixture. The matrix for m = 0 is1

V = ekEo

〈320|z|320〉〈320|z|310〉〈320|z|300〉〈310|z|320〉〈310|z|310〉〈310|z|300〉〈300|z|320〉〈300|z|310〉〈300|z|300〉

= ekEo

0 〈320|z|310〉 0〈310|z|320〉 0 〈310|z|300〉

0 〈300|z|310〉 0

= ekEo

0 −3√

3ao 0

−3√

3ao 0 −3√

6ao0 −3

√6ao 0

= −3√

3aoekEo

0 1 0

1 0√

2

0√

2 0

Now diagonalizing the matrix∣∣∣∣∣∣

−λ 1 0

1 −λ√

2

0√

2 −λ

∣∣∣∣∣∣ = 0 ⇒ −λ3 + 3λ = λ(λ2 − 3) = 0 ⇒ λ = 0,±√

3

The eigenvectors are the

λ = 0;1√3

√2

0−1

λ =√

3;1√6

1√3√2

λ = −√

3;1√6

1

−√

3√2

The energy shift of the lowest level, which is 1√

6

(|320〉+

√3|310〉+

√2|300〉

)is then

∆ = −3√

3aoekEo√

3 = −9aoekEo


For hydrogen atom in its ground state is placed in an uniform electric field

E =

0 t < 0

Eoe−t/τ t > 0

(4.1)

Find the probability for the transitions 2p (each state) and 2s.

The perturbation becomes V = −ex ·E = −eEz. So looking at the transition to the 2p states with magnetic momentsof ±1.

〈2, 1,±|V |1, 0, 0〉 = k〈2, 1,±1|z|1, 0, 0〉 = 0

1The integrals are found in Griffiths. Quantum. p 254

26

from parity and Wigner-Eckart Theorem. Because of this the probability of the transitions to |2, 1,±1〉 is 0. Firstorder time dependent theory for 2, 1, 0〉 state.

c(1)210(t) = − i

~

∫ t

0

eiω21t′〈2, 1, 0|V (t′)|1, 0, 0〉dt′ =

ieEo~〈2, 1, 0|z|1, 0, 0〉

∫ t

0

eiω21t′e−t

′/τdt′

= − ieEo~〈2, 1, 0|z|1, 0, 0〉1− eiω21t−t/τ

1/τ − iω21

So ω21 = (E2 − E1)/~ = 3e2/8ao~ and t τ , then

|c(1)210(t→∞)|2 =e2E2

o

~2/τ2 + (3e2/8ao)2∣∣〈2, 1, 0|z|1, 0, 0〉∣∣2

Now for the transition from the ground state to 2s state. This can be quick found by looking at 〈2, 0, 0|z|1, 0, 0〉.Under parity an even state has to go to an odd state for this potential. Therefore the probability is 0. X


For two spin 1/2 particle are in the following Hamiltonian t > 0 H = 4∆~2 S1 · S2 and at t < 0 H = 0. The particles

are initial state |+−〉

(a) Find the exact solution

First finding the product of S1 and S2.

S1 · S2 = S1xS2x + S1yS2y + S1zS2z =~2

4

[(|+〉〈−| + |−〉〈+|)1 (|+〉〈−| + |−〉〈+|)2 + i2 (−|+〉〈−| + |−〉〈+|)1 (−|+〉〈−| + |−〉〈+|)2

= (|+〉〈+| − |−〉〈−|)1 (|+〉〈+| − |−〉〈−|)2

]~2

4

[| + +〉〈− − | + | + −〉〈− + | + | − +〉〈+ − | + | − −〉〈+ + |

+ i2 (| + +〉〈− − | − | + −〉〈− + | − | − +〉〈+ − | + | − −〉〈− − |)

+ | + +〉〈+ + | − | + −〉〈+ − | − | − +〉〈− + | + | − −〉〈− − |]

So the Hamiltonian becomes

H = ∆

1 0 0 00 −1 2 00 2 −1 00 0 0 1

Only the center of the 2× 2 matrix has to be diagonalized∣∣∣∣ −1− λ 2

2 −1− λ

∣∣∣∣ = 0 ⇒ (−1− λ)2 − 4 = 0 ⇒ λ = 1,−3

For λ = 1 is the symmetric state 1√2

(|+−〉+ | −+〉). For λ = −3 is the anti-symmetric state 1√2

(|+−〉 − | −+〉).So in diagnonlized space the energy splitting becomes

∆ |+ +〉, | − −〉, 1√2

(|+−〉+ | −+〉)−3∆ 1√

2(|+−〉 − | −+〉)

Because the Hamiltonian is time-independent the time evolution operator is U(t, to) = e−i~H(t−to). Here to = 0.

Then the time evolution operator becomes U(t, 0)| + −〉 = 1√2U(t, 0) (|3〉+ |4〉), where |1〉 = | + +〉, |4〉 = | − −〉,

|3〉 = 1√2

(|+−〉+ | −+〉), and |4〉 = 1√2

(|+−〉 − | −+〉). Now the final state is

|f, t〉 = U(t, 0)|i, 0〉 =1√2e−

i~Ht (|3〉+ |4〉) =

1√2

[e−i∆t/~|3〉+ e3i∆t/~|4〉

]=

1

2

(e−i∆t/~ + e3i∆t/~

)|+−〉+

1

2

(e−i∆t/~ − e3i∆t/~

)| −+〉

27

To find the probability that the particle are in each of the final possible states

|〈+ + |f, t〉|2 = 0

|〈− − |f, t〉|2 = 0

|〈+− |f, t〉|2 =1

4

(2 + e4i∆t/~ + e−4i∆t/~

)=

1

2(1 + cos (4∆t/~))

|〈−+ |f, t〉|2 =1

4

(2− e4i∆t/~ − e−4i∆t/~

)=

1

2(1− cos (4∆t/~)) X

(b) Solve using first order time dependent perturbation theory.

For first order time dependent perturbation theory,

c(0)n = δni c(1)n (t) = − i

~

∫ t

0

dt′eiωnit′Vni(t

′)

So here Ho = 0 and V = 4∆~2 S1 · S2, and the initial state is |i〉 = |+−〉 and the final state is |f〉 = | −+〉. From the

undiagonalized matrix Vfi = 2∆, and the other two basis Vni = 0. Now ωfi =Ef−Ei

~ = 0. So

c(0)i = 1 c

(1)f = − i

~

∫ t

0

dt′2∆ = −2i∆t/~

Then the probability for each state is

P (|i〉) = |c(0)i |2 = 1

P (|f〉) = |c(1)f |2 = 4

(∆t

~

)2

This holds true for(

2∆t~)2 1 ⇒ ∆t

~ 1 ⇒ ∆ ~t

X


For the two level system E1 < E2. There is a time dependent potential that connects the two levels as follows:

V11 = V22 = 0 V12 = γeiωt V21 = γe−iωt (4.2)

and at t = 0, c1(0) = 1 and c2(0) = 0.

(a) Find |c1(t)|2 and |c2(t)|2 for t > 0 by exactly solving the coupled differential equations

i~ck =

2∑n=1

Vkn(t)eiωkntcn for n = 1, 2 (4.3)

The potential is

V = γ

(0 eiωt

e−iωt 0

)and ω21 = E2−E1

~ . So

i~c1 = γeiωte−iω21tc2; i~c2 = γe−iωteiω21tc1

i~c1 = γc2i(ω − ω21)ei(ω−ω21)t + γei(ω−ω21)tc2

Then

i~c1 = i(ω − ω21)[γc2e

i(ω−ω21)t]

+γ2

i~ei(ω−ω21)te−i(ω−ω21)tc1

c1 = −γ2

~2c1 + i(ω − ω21)c1 ⇒ c1(t) = ψei(ω−ω21)t/2

28

Then the differential equation becomes

ψ − (ω − ω21)2

4ψ + i(ω − ω21)ψ = −γ

2

~2ψ + i(ω − ω21)ψ −

(ω − ω21)2

2ψ

ψ =

[−γ

2

~2− (ω − ω21)

2

4

]ψ ⇒ ψ = A sin

√γ2

~2+

(ω − ω21)2

4t+B cos

√γ2

~2+

(ω − ω21)2

4t

For simplicity Ω ≡√γ2

~2+

(ω − ω21)2

4, so

c1(t) = ei(ω−ω21)t/2 [cos Ωt+B sinΩt]

c2(t) = e−i(ω−ω21)t/2A sinΩt

Then from the boundary conditions i~c1∣∣t=0

= 0 and i~c2∣∣t=0

= γ. From the second condition i~AΩ = γ →

A =γ

i~Ωand the first condition now i(ω−ω21)

2+BΩ = 0 → B =

−i(ω − ω21)

2Ω. Therefore

c1(t) = ei(ω−ω21)t/2

[cosΩt− i(ω − ω21)

2ΩsinΩt

]c2(t) = e−i(ω−ω21)t/2

[ γ

i~ΩsinΩt

]So

|c2(t)|2 =γ2

~2Ω2sin2 Ωt⇒ |c2(t)|2 =

γ2/~2

γ2/~2 + (ω − ω21)2/4sin2

(√γ2

~2+

(ω − ω21)2

4t

)X

And

|c1(t)|2 = cos2 Ωt+(ω − ω21)

2

4Ω2sin2 Ωt = 1− sin2 Ωt

Ω2

(Ω2 − (ω − ω21)

2

4

)= 1− sin2 Ωt

Ω2

(γ2

~2+

(ω − ω21)2

4− (ω − ω21)

2

4

)= 1− sin2 Ωt

Ω2

(γ2

~2

)⇒ |c1(t)|2 = 1− |c2(t)|2 X

(b) Now use time dependent perturbation theory for small values of γ and look at two cases (i) ω very different fromω21 and (ii) ω very close to ω21

The lowest order term would be first order term

c(1)2 = − iγ

~

∫ t

0

e−i(ω−ω21)t′dt′ =−iγ~

1

−i(ω − ω21)

(e−i(ω−ω21)t − 1

)=

γ2

~(ω − ω21)sin(ω − ω21

2t)

So the probability is

|c2(t)|2 =4γ2

~2(ω − ω21)2sin2

(ω − ω21

2t)

Ifγ is small and ω is very different from ω21, the term γ2/~2 + (ω − ω21)2/4 becomes (ω − ω21)

2/4, so the real termbecomes

|c2(t)|2 =γ2/~2

γ2/~2 + (ω − ω21)2/4sin2

(√γ2

~2+

(ω − ω21)2

4t

)≈ 4γ2

~2(ω − ω21)2sin2

(ω − ω21

2t)

So the |c2real(t)|2 ≈ |c2pert(t)|2. So in the case ∆ω/2 γ/~ is approximately true. In the second case of ω very closeto ω21, perturbation theory starts to breakdown. The ratio of the two is

|c2pert(t)|2

|c2real(t)|2=

4γ2

~2(ω−ω21)2sin2

(ω−ω21

2t)

γ2/~2

γ2/~2+(ω−ω21)2/4sin2

(√γ2

~2 + (ω−ω21)2

4t

) X

29

Chapter 5

752: Problem Set 7

5.1 Problem 1: Shankar 18.2.2

A hydrogen atom is in the ground state at t = −∞. An electric field E(t) = (kE)e−t2/τ2

is applied until t = ∞.Show that the probability that the atom ends up in any of the n = 2 states is, to first order,

P (n = 2) =

(eE~

)2(215a2

o

310

)πτ2e−ω

2τ2/2 (5.1)

where ω = (E2lm − E100)/~. Does the answer depend on whether or not we incorporate spin in the picture?

The perturbation is H1 = ezEe−t2/τ2

. Using the selection rules for the z operator, which are ∆l = ±1 and ∆m = 0,the only state that the ground state can transition to is |210〉. Now for the transition

eE〈210|z|100〉 = eE∞∫0

π∫0

2π∫0

1√

32πa3o

sin θ cos2θr4

ao

e−3r/2aodφdθdr

= eE1

√32πa3

o

(2π)

(2

3

)(4!

ao

(2ao

3

)5)= eE

215/2ao

35

Now for the time dependent term

i

~

∞∫−∞

et′2/τ2

e−iωt′dt′ =

i

~√πτ2e−ω

2τ2/4

The transition coefficient is

cf (t) =i

~√πτ2e−ω

2τ2/4eE 215/2ao35

The transition probability is the coefficient squared

|cf |2 = P (n = 2) =

(eE~

)2 215a2o

310πτ2e−ω

2τ2/2 X

For a changing electric field, there is an induced magnetic field. This magnetic field interacts with spin of the electronwhich would be Hm = −µ ·B


Consider a particle in the ground state of a box of length L. Argue on semiclassical grounds that the natural timeperiod associated with it is T ' mL2/~π. If the box expands symmetrically to double its size in time τ T what isthe probability of catching the particle in the ground state of the new box?

The time for a particle to move across the box can be found from the kinetic energy, E = 12mv2. Then

E =~2π2

2mL2=

1

2mv2 ⇒ v =

~πmL

31

The time to cross is v = L/T → T = mL2

~π . The initial state is ψ1 = 1√L

sin(πxL

). In the new volume ψ2 = 1√

Lcos(πxL

).

The new ground state is ψ = 1√2L

sin(πx2L

). Then

cf (t) =

3L/2∫L/2

1

L√

2sin(πx

2L

)cos(πxL

)dx =

8

3π

So the probability is

P = |cf |2 =

(8

3π

)2

X


In the β decay H3, the emitted electron has a kinetic energy of 16 keV. Argue that the sudden approximation maybe used to describe the response of an electron that is initially in the 1s state of H3. Show that the amplitude for itto be in the ground state of (He)+ is 16(2)1/2/27. What is the probability for it to be in the state

|n = 16, l = 3,m− 0〉 of (He)+? (5.2)

The time for the emitted electron, τ is much less than one orbit time, T , of electron in the ground state. So

τ ∼ ao/√

32000eVm

, where T ∼ 2ao/√

27.2eVm

∴ τ T . The ground state of H

|100〉 =

(1

πa3o

)1/2

e−r/ao

The ground state of (He)+

|100〉∗ =

(8

πa3o

)1/2

e−2r/ao

The amplitude is

〈100|100〉∗ =

∞∫0

π∫0

2π∫0

2√

2

πa3o

e−3r/aor2 sin θdφdθdr

=

∞∫0

8√

2

a3o

e−3r/aOr2dr =8√

2

a3o

2a3o

27=

16√

2

27

Now for the state |16, 3, 0〉 (He)+. The integral over spherical harmonics

π∫0

2π∫0

√7

8π

(−3 cos θ + 5 cos3 θ

)sin θdφdθ = 0

Because each term is multiplied be the spherical harmonics, the amplitude is 0 ∴ P (|16, 3, 0〉∗) = 0


An oscillator is in the ground state of H = Ho + H1, where the time-independent perturbation H1 is the linearpotential (−fx). If at t = 0, H1 is abruptly turned off, show that the probability that the system is in the ntheigenstate of Hois given by the Poisson distribution

P (n) =e−λλn

n!, where λ =

f2

2mω3~(5.3)

32

Because the perturbation is sudden, the particle is still in the initial state of H = Ho +H1. From Eq. 17.2.10

|n〉 = e−i(f/mω2~)P |no〉 (5.4)

Using λ = f2

2mω3~

e−i(f/mω2~)P |no〉 λ⇒ e

√λ(a†−a)|no〉

Using the hint and that the ground state is wanted

e√λ(a†−a)|0〉 = e−λ/2e

√λa†e−

√λa|0〉

Now let’s expand the exponential e−√λa = 1−

√λ+ . . .. Any lowering operator will be 0 when acting on the ground

state, so only the first term remains.

e−λ/2e√λa†e−

√λa|0〉 → e−λ/2e

√λa† |0〉

Using Eq. 21.1.115, Eq. 21.1.116, and Eq. 21.1.117

e√λa† =

∞∑n=0

(√λa†)n

n!(5.5)

So the state becomes

e−λ/2e√λa† |0〉 ⇒ e−λ/2

∞∑n=0

λn/2a†n

n!|0〉 ⇒ e−λ/2λn/2√

n!|n〉

The probability to be in the nth state

P (n) =

∣∣∣∣〈n|e−λ/2λn/2√n!

|n〉∣∣∣∣2 =

e−λλn

n!X


Consider a system subject to a perturbation H1(t) = H1δ(t). Show that if at t = 0−the system is in the state |io〉,the amplitude to be in a state |fo〉 at t = 0+ is, to first order,

df =−i~〈fo|H1|io〉 (f 6= i) (5.6)

From Eq. 18.2.9

df = δfi −i

~

0−∫−∞

〈fo|H1δ(t)|io〉eiωfitdt

= − i

~〈fo|0|io〉 = 0

Therefore there is no transition. Now for t = 0+

df = δfi −i

~

0+∫−∞

〈fo|H1δ(t)|io〉eiωfitdt

= − i

~〈fo|H1|io〉

33


(1) Write H for a particle in the potentials (A, φ).

The Lagrangian is

L =1

2mv2 − eφ+

e

cv ·A

The canonical momentum is

p =∂L∂v

= mv +e

cA

The Hamiltonian is

H = p · v− L =1

2mv2 + eφ

H =

(p− e

cA)2

2m+ eφ X

(2) Write down HΛ, the Hamiltonian obtained by gauge transforming the potentials.

The Lagrangian is

L =1

2mv2 − eφ− e

c

∂Λ

∂t+e

cv · (A−∇Λ)

The canonical momentum is

p =∂L∂v

= mv +e

c(A−∇Λ)

The Hamiltonian is

H = p · v− L =1

2mv2 + eφ+

e

c

∂Λ

∂t

H =

(p− e

c(A−∇Λ)

)22m

+ eφ+e

c

∂Λ

∂tX

34

(3) Show that if ψ(r, t) is a solution to Schrodinger’s equation with the Hamiltonian H, then ψΛ(r, t) given in Eq.(18.4.33) is the corresponding solution with H → HΛ

Let’s first look at the following relations (position, then momentum)

eieΛ/~cxe−ieΛ/~c = x

eieΛ/~cpe−ieΛ/~c = eieΛ/~c[p, e−ieΛ/~c

]+ p = −eieΛ/~ci~∇

[e−ieΛ/~c

]+ p = p +

e∇Λ

c(5.7)

Let’s look at the original Hamiltonian

i~ ddt|ψ〉 = H|ψ〉 =

[(p− e

cA)2

2m+ eφ

]|ψ〉

So for the second Hamiltonian

i~d

dt|ψΛ〉 = HΛ|ψΛ〉 =

[(p − e

c(A −∇Λ)

)2

2m+ eφ+

e

c

∂Λ

∂t

]|ψΛ〉

i~d

dt

[e−ieΛ/~c|ψ〉

]=e∂Λ

c∂te−ieΛ/~c|ψ〉 + e−ieΛ/~cH|ψ〉

Now the exponential part on the last term needs to be moved to the other side of the Hamiltonian. This does notcommute though. Using eq. 7 twice on p

i~d

dt

[e−ieΛ/~c|ψ〉

]=e∂Λ

c∂te−ieΛ/~c|ψ〉 + e−ieΛ/~cH|ψ〉 =

e∂Λ

c∂te−ieΛ/~c|ψ〉 +

[(p − e

c(A −∇Λ)

)2

2m+ eφ

]e−ieΛ/~c|ψ〉

=e∂Λ

c∂t|ψΛ〉 +

[(p − e

c(A −∇Λ)

)2

2m+ eφ

]|ψΛ〉 = HΛ|ψΛ〉 X

35

Chapter 6

752: Problem Set 8

6.1 Problem 1: Sakuari 5.35

The ground state of a hydrogen (n = 1, l = 0) is subjected to a time-dependent potential as follows:

V(x, t) = Vo cos(kz − ωt). (6.1)

Using time-dependent perturbation theory, obtain an expression for the transition rate at which the electron is emittedwith momentum p. Show, in particular, how you may compute the angular distribution of the ejected electron (interms of θ and φ defined with respect to the z-axis). Discuss briefly the similarities and the differences between thisproblem and the (more realistic) photoelectric effect. (Note: For the initial wave function see Problem 34. If youhave a normalization problem, the final wave function may be taken to be

ψf (x) =

(1

L3/2

)eip·x/~ (6.2)

with L very large, but you should be able to show that the observable effects are independent of L.)

The initial state is

ψi(x) =

(1

πa3o

)1/2

e−r/ao

The final state is

ψf (x) =

(1

L3/2

)eip·x/~

Now the perturbation in exponential form

V = Vo(eikz−iωt + e−ikz+iωt

)= Veiωt + V†e−iωt

From Equation 5.6.44 the transition rate is

wi→n =2π

~∣∣V†ni∣∣2δ(En − Ei − ~ω)

where∣∣V†ni∣∣2 =

V 2o4

∣∣〈n|eikz|i〉∣∣2. Now finding the matrix value

〈n|eikz|i〉 =

∫1

L3/2e−ikf ·xeikz

√1

πa3o

e−r/aod3r

=1

√πL3/2a

3/2o

∫e−i(kf ·x−kz)−r/aod3r

37

Now from the text k · x− kz = q, so the integral becomes

〈n|eikz|i〉 =8π

L3/2a5/2o

1

(1/a2o + q2)2

=8π

L3/2a5/2o

1(1/a2

o + (kf − kz)2)2

Using Equation 5.7.31 but converting to momentum space

n2dΩdn

dpfdpf =

L3p2f

(2π~)2dpfdΩ

The rate is

wi→pf =2π

~∣∣V†ni∣∣2δ(En − Ei − ~ω)

=2π

~V 2o

4

64π2

L3a5o

1(1/a2

o + (kf − kz)2)4 L3p2

f

(2π~)3dpfdΩ

=4V 2

o

~4a5o

p2fdpfdΩ(

1/a2o + (kf − kz)2

)4The angular dependence is in the denominator

(kf − kz)2 =[∣∣kf ∣∣ sin θ (cosϕex + sinϕey) +

(∣∣kf ∣∣ cos θ − k)ez]2

=∣∣kf ∣∣2 cos2 θ + k2 − 2k

∣∣kf ∣∣ cos θ + |kf |2 sin2 θ

= k2f + k2 − 2k

∣∣kf ∣∣ cos θ

The rate is

wi→pf =4V 2

o

~4a5o

p2fdpfdΩ(

1/a2o + k2

f + k2 − 2k∣∣kf ∣∣ cos θ

)4

There is no azimuthal dependence, but that is to be expected, because the perturbation is defined in the z direction.This is very similar to the photoelectric effect except that there is no polarization taken into account and no dipoleapproximation. X

6.2 Problem 2: Sakuari 5.40

Obtain an expression for τ(2p→ 1s) for the hydrogen atom. Verify that it is equal to 1.6× 10−9s.

The initial state is |2lm〉 ⊗ |0〉 and the final state is |100〉 ⊗ |k~〉. Now Fermi’s golden rule

Ri→f =2π

~

∣∣∣∣〈fo|e

mcA ·P|io〉

∣∣∣∣2δ(E100 + ~ω − E2lm)

Because of the spin of the photon, ∆l = 1. The matrix term

〈100|〈~k|A|0〉 ·P|21m〉 =

(~c2

4π2ω

)1/2 ∫ψ∗100e

k·rεψ21md

3r

Applying the dipole approximation

〈fo|A ·P|io〉 =

(~c2

4π2ω

)imω

∫ψ∗100ε · rψ21md

3r

where

ε · r = −ε11r−11 + ε01r

01 − ε−1

1 r11

38

The integral becomes

(4π

3

)1/2 ∫R10rR21r

2dr

∫Y

00

(−ε11Y

−11 + ε

01Y

01 − ε

−11 Y

11

)Y

m1 dΩ

The radial integral

2√24a7

o

∫r4e−3r/2aodr =

128

81

√2

3ao

Using orthogonality for the spherical harmonics the integral becomes

〈fo|A ·P|i〉 =256

243

1√2ao(ε11δm,1 + ε01δm,0 + ε−1

1 δm,−1

)Then

∣∣〈fo|A ·A|io〉∣∣2 =

215

310a2o

(∣∣ε11∣∣2δm,−1 +∣∣ε01∣∣2δm,0 +

∣∣ε−11

∣∣2δm,1

)Averaging all three polarizations

∣∣〈fo|A ·A|io〉∣∣2 =

215

311a2o

The rate becomes

Ri→f =2π

~

( e

mc

)2 ~2c4

4π2ωm2ω2a2

o215

311δ (E100 + ~ω − E21m)

The momentum of the outgoing photon∫δ (E100 + ~ω − E21m) k2dkdΩ =

4πk2

~c

where k = E21m−E100~c = e2

2ao~c(1− 1

4

)= 3e2

8ao~c . Then the rate is

Ri→f =2π

~

( e

mc

)2 ~2c2

4π2ωm2ω2a2

o215

311

(3e2

8ao~c

)24π

~c

=

(2

3

)8(e2

mc

)5mc2

~⇒ Ri→f =

(2

3

)8

α5mc2

~

The rate is Ri→f =(

23

)8 ( 1137

)5( (.511×106)(3×1018)1973.3

)= .61× 109 s−1. The lifetime is

τ = 1/R =1

.61× 109= 1.6× 10−9 s X


(1) By going through the derivation, argue that we can take the eik·r factor into account exactly, by replacing pfby pf − ~k in Eq. 18.5.19

Looking at Eq. 18.5.5

H1fi =

e

2mc

1

(2π~)3/2

√1

πa3o

∫e−ipf ·r/~

eik·r

Ao · (−i~∇) e−r/aod

3r

=e

2mc

1

(2π~)3/2

√1

πa3o

∫e−i(pf−~k

)·r/~


3r

=e

2mc

1

(2π~)3/2

√1

πa3o

∫e−ip′f ·r/~


3r

39

where p′f = pf − ~k. Because of the spherical symmetry of the system and all of the vectors are defined with respectto pf (so p′f would be a rotation, which has no affect on this system) all of the following steps are the same so

H1fi = N

∫e−i(pf−~k)·r/~Ao · (i~∇) e−r/aod3r

where N =(

e2mc

)1

(2π~)3/2

√1πa3o

X

(2) Verify the claim made above about the electron momentum distribution.

Integrating by parts of the previous problem (remembering that p′f = pf − ~k)

H1fi = NAo · p′f

∫e−ip

′f ·r/~e−r/aod3r

= NAo · p′f∫ ∞

0

∫ 1

−1

∫ 2π

0

e−ip′f ·r/~e−r/aor2drd(cos θ)dφ

= 2πNAo · p′f∫ ∞

0

(e−ip

′f ·r/~ − eip

′f ·r/~

−ipfr/~

)e−r/aor2dr

=8π/ao(

(1/ao)2 + (p′f/~)2)2NAo · p′f

Now for the δ-function

p2f

2m= E

oi + ~ω ⇒ δ

(p2f

2m− E

oi − ~ω

)=m

pf

δ(pf − (2mE

oi + ~ω)

1/2)

The rate becomes

Ri→dΩ =2π

~

∣∣Ao · p′f∣∣2 (8π/ao)

2((1/ao)2 + (p′f/~)2

)4mpf( e

2mc

)2 1

(2π~)31

πa3o

=4a3oe

2pf∣∣Ao ·

(pf − ~k

) ∣∣2mπ~4c2

[1 +

(ao~)2 (

pf − ~k)2]4 X


(1) Estimate the photoelectric cross section when the ejected electron has a kinetic energy of 10 Ry. Compare it tothe atom’s geometric cross section ' πa2

o.

Looking at Eq. 18.5.30 and pf =√

2mE1

σ =128π

3

1

~ω

e2

c~

p2f c

m~c

11 +p2

fc2

~2c2a2

o

4

=128π

3

1

E + 13.6α√

8E3/2 m3/2c

m

1

~c

a3o(

1 + 2mc2Ea2o

1~2c2

)4

=128π

√8

3

(136)3/2

149.6

1

137

(.511 × 106

)1/2

1973.3

(.511)3(1 + 2(.511×106)(136)(.511)2

(1973.3)2

)4

= 1.4 × 10−4A2

This is much smaller than the geometrical cross section of the atom. X

(2) Show that if we consider photoemission from the 1s state of a charge Z atom, σ ∝ Z5, in the limit pfao/Z~ 1.

The Bohr radius is inversely proportional to Z, then a = ~2

Zme2= ao

Z. Looking at Eq. 18.5.30

σ =128πa3e2p3

f

3m~3ωcp6fa

6/~6∝ 1

a5=Z5

a5o

∝ Z5 X

1All values from Shankar p. 669

40


Show that

σY ukawa = 16πr2o

(gµro~2

)2 1

1 + 4k2r2o(6.3)

where ro = 1µo is the range. Compare σ to the geometrical cross section associated with this range.Looking at Eq. 19.3.11

dσ =4µ2g2

~4

dΩ

((1/ro)2 + 4k2 sin2 (θ/2))2

σ =4µ2g2

~4

∫ π

0

∫ 2π

0

sin θdθdφ

((1/ro)2 + 4k2 sin2 (θ/2))2

=8πµ2g2

~4

∫ 1

−1

dx

((1/ro)2 + 2k2 − 2k2x)2

=16πµ2g2

~4

r4o1 + 4k2r2o

⇒ σ = 16πr2o

(gµro

~2

)2 1

1 + 4k2r2o

In comparison with the geometrical cross section, which is roughly σ ≈ 4πr2o = 4πfm2,

σY ≈ 4g2µ2r2o

~4

(4πr2o

)= 4r2o

e4m2c4

~4c4~4 (4πr2o)

= 4(10−5)2 α2

(109)2

(2000)2~4 (4πr2o)

= 5× 10−3~4 (4πr2o)So the Yukawa cross section is much smaller the the geometrical cross section. X


(1) Show that if V (r) = −Voθ(ro − r),

dσ

dΩ= 4r2o

(µVor

2o

~2

)2(sin qro − qro cos qro)

2

(qro)6 (6.4)

Looking at Eq. 19.3.8

f(θ, φ) = − µ

2π~2

∫e−iqr

′ cos θ′V (r′)d(cos θ′)dφ′r2dr′

=2µVo~2

∫ ro

0

r′ sin qr′

qdr′

=2µVo~2

(sin qro − qro cos qro

q3

)= 2ro

µVor2o

~2

(sin qro − qro cos qro)

(qro)3

Finally dσdΩ

= |f |2

dσ

dΩ=

(2ro

µVor2o

~2

(sin qro − qro cos qro)

(qro)3

)2

= 4r2o

(µVor

2o

~2


2

(qro)6X

(2) Show that as kro → 0, the scattering becomes isotropic and

41

σ ∼=16πr2o

9

(µVor

2o

~2

)2

(6.5)

From Eq. 19.3.6 q = 2k sin(θ/2)

σ =

∫ π

0

∫ 2π

0

4r2o

(µVor

2o

~2


2

(qro)6sin θdθdφ

=

∫ π

0

8πr2o

(µVor

2o

~2


2

(qro)6sin θdθ

So expanding using small angle approximation sin qro − qro cos qro ≈ qro − (qro)3

6− qro + (qro)3

2= (qro)3

3. Then

σ = 8πr2o

(µVor

2o

~2

)2 ∫ π

0

((qro)3

3

)2

(qro)6sin θdθ︸︷︷︸

2/9

=16π

9r2o

(µVor

2o

~2

)2

X

42

Appendix A

511: Problem Set 5

A.1 Merzbacher Exercise 10.9

Find the eigenvalues for kinetic and potential energy for the harmonic oscillator. Explain why they do not add up tothe energy eigenvalues for the total energy.The energy eigenvalues for total energy :

En = ~ω(n + 1

2

)The Hamiltonian for the harmonic oscillator :

H =p2

2m+

1

2mω2q2

If we define a new operator as

a =

√mω

2~

(q + i

p

mω

)(A.1)

It then follows that

H = ~ω(a†a+

1

2

)where a† is the Hermitian adjoint of a. Now we can use this to solve for the p and q.

q =

√~

2mω

(a† + a

)(A.2)

p = i

√m~ω

2

(a† − a

)(A.3)

We can now apply kinetic and potential operators to |n〉.

K|n〉 =p2

2m|n〉

Now we can apply Equation 3 and replace p.

p2

2m|n〉 =

(i√

m~ω2

(a† − a

))2

2m|n〉

= −~ω4

(a† − a

)(a† − a

)|n〉

= −~ω4

(a†2 + a2 − a†a− aa†

)|n〉

= −~ω4

[(a†2 + a2

)|n〉 −

(a†a+ aa†

)|n〉]

43

From Equation 10.75 in Merzbacher, we know a†a|n〉 = λo|n〉, then:

= −~ω4

[(a†2 + a2

)|n〉 − aa†|n〉 − λo|n〉

]

From Equation 10.72 in Merzbacher, we know aa† = a†a+ 1

= −~ω4

[(a†2 + a2

)|n〉 − 1|n〉 − 2λo|n〉

]

K|n〉 = −~ω4

(a†2 + a2

)|n〉+ ~ω

(n

2+

1

4

)(A.4)

Now we can proceed to the potential energy.

U |n〉 =1

2mω2q2|n〉

Using the same mathematical procedure, we find:

U |n〉 =~ω4

(a†2 + a2

)|n〉+ ~ω

(n

2+

1

4

)(A.5)

It would appear that we can add (4) and (5) to find the total energy eigenvalues, but they are not eigenvalues forthe kinetic and potential energies so we must find those eigenvalues instead.

K|p〉 =p2

2m|p〉 (A.6)

U |q〉 =1

2mω2q2|q〉 (A.7)

The total energy is the kinetic energy plus the potential energy, so we think that we may add them as:

K|p〉+ U |q〉 =p2

2m|p〉+

1

2mω2q2|q〉 =

(p2

2m+

1

2mω2q2

)|p, q〉

That is true, if the p and q commuted when in fact we know that they do not commute.

p2

2m|p〉+

1

2mω2q2|q〉 6=

(p2

2m+

1

2mω2q2

)|p, q〉

Therefore,

K|p〉+ U |q〉 6= H|n〉X

A.2 Merzbacher Problem 10.3

Find ∆x∆p for a linear harmonic oscillator nth energy eigenstate.

We know

(∆x)2 = 〈n|x2|n〉 − 〈n|x|n〉2

(∆p)2 = 〈n|p2|n〉 − 〈n|p|n〉2

44

Let us begin with 〈n|x|n〉 :We recall from the previous problem,

x =

√~

2mω

(a† + a

)We replace x with this new definition.

〈n|√

~2mω

(a† + a

)|n〉 =

√~

2mω〈n|(a† + a

)|n〉

=

√~

2mω

(〈n|a†|n〉+ 〈n|a|n〉

)=

√~

2mω(〈n|n+ 1〉+ 〈n|n− 1〉)

Because eigenstates n, n-1 and n+1 are orthonormal

〈n|n+ 1〉 = 0

〈n|n− 1〉 = 0

Then 〈n|x|n〉 = 0.Now for the term 〈n|p|n〉 :We recall from the previous problem,

p = i

√m~ω

2

(a† − a

)We replace p with this new definition.

〈n|i√m~ω

2

(a† − a

)|n〉 = i

√m~ω

2〈n|(a† − a

)|n〉

= i

√m~ω

2

(〈n|a†|n〉 − 〈n|a|n〉

)= i

√m~ω

2(〈n|n+ 1〉 − 〈n|n− 1〉)


〈n|n+ 1〉 = 0

〈n|n− 1〉 = 0

Then 〈n|p|n〉 = 0.We move to the term 〈n|x2|n〉 :

x2 =~

2mω

(a2 + a†2 + aa† + a†a

)So

〈n|x2|n〉 = 〈n| ~2mω

(a2 + a†2 + aa† + a†a

)|n〉

=~

2mω

[〈n|a2|n〉+ 〈n|a†2|n〉+ 〈n|

(aa† + a†a

)|n〉]

=~

2mω

[〈n|n− 2〉+ 〈n|n+ 2〉+ 〈n|

(1 + 2a†a

)|n〉]

45


=~

2mω

[0 + 0 + 〈n|

(1 + 2a†a

)|n〉]

=~mω

[〈n|a†a+

1

2|n〉]

〈n|x2|n〉 =~mω

(n+

1

2

)(A.8)

Finally for the last term 〈n|p2|n〉 :

p2 = −m~ω2

(a2 + a†2 − aa† − a†a

)So

〈n|p2|n〉 = 〈n| − m~ω2

(a2 + a†2 − aa† − a†a

)|n〉

= −m~ω2

[〈n|a2|n〉+ 〈n|a†2|n〉 − 〈n|

(aa† + a†a

)|n〉]

= −m~ω2

[〈n|n− 2〉+ 〈n|n+ 2〉 − 〈n|

(1 + 2a†a

)|n〉]


= −m~ω2

[0 + 0〈n|

(1 + 2a†a

)|n〉]

= m~ω[〈n|a†a+

1

2|n〉]

〈n|p2|n〉 = m~ω(n+

1

2

)(A.9)

Then

(∆x)2 = 〈n|x2|n〉 − 〈n|x|n〉2 =~mω

(n+

1

2

)(A.10)

(∆p)2 = 〈n|p2|n〉 − 〈n|p|n〉2 = m~ω(n+

1

2

)(A.11)

Then the uncertainty for the linear harmonic oscillator

∆x∆p = ~(n+

1

2

)X

A.3 Merzbacher Problem 14.6

A linear harmonic oscillator, with energy eigenstates |n〉, is acted upon by a time dependent interaction between the|0〉 and |1〉 :

V (t) = F (t) |1〉〈0|+ F ∗ (t) |0〉〈1| (A.12)

a.

46

Find the coupled equations of motion for the probability amplitudes 〈n|Ψ (t)〉We know

i~d|Ψ (t)〉dt

= H|Ψ (t)〉

where

H = Ho + V (t)

= Ho + F (t) |1〉〈0|+ F ∗ (t) |0〉〈1|

with this substitution

i~d|Ψ (t)〉dt

= (Ho + F (t) |1〉〈0|+ F ∗ (t) |0〉〈1|) |Ψ (t)〉

so then 〈n|Ψ (t)〉 would be follow.

〈n|i~d|Ψ (t)〉dt

= 〈n| (Ho + F (t) |1〉〈0|+ F ∗ (t) |0〉〈1|) |Ψ (t)〉

i~ ddt〈n|Ψ (t)〉 = 〈n|Ho|Ψ (t)〉+ 〈n|F (t) |1〉〈0|Ψ (t)〉+ 〈n|F ∗ (t) |0〉〈0|Ψ (t)〉X

b.

Find the energy eigenvalues and stationary states for t>0, if F (t) =√

2~ωη (t)

En|n〉 = H|n〉= Ho|n〉+ V |n〉

We know from the earlier problem that

Ho|n〉 = ~ω(n+

1

2

)|n〉

For the definition of the Heaviside function η (t) = 1 for t>0. Then

V |n〉 = F (t) |1〉〈0|n〉+ F ∗ (t) |0〉〈1|n〉

=√

2~ω|1〉〈0|n〉+√

2~ω|0〉〈1|n〉

For the interaction term then

=√

2~ω (|1〉δ0n + |0〉δ1n)

Therefore

En|n〉 = ~ω(n+

1

2

)|n〉+

√2~ω (|1〉δ0n + |0〉δ1n) (A.13)

This means that the stable states are form when n>1.X

c.

Find 〈n|Ψ (t)〉 for t>0 with the initial condition of it in the ground state.

i~d|Ψ (t)〉dt

= Ho|Ψ (t)〉+(√

2~ω|1〉〈0|+√

2~ω|0〉〈1|)|Ψ (t)〉

i~d|Ψ (t)〉|Ψ (t)〉 =

(Ho +

√2~ω|1〉〈0|+

√2~ω|0〉〈1|

)dt

Now integrate both sides

i~∫d|Ψ (t)〉|Ψ (t)〉 =

∫ (Ho +

√2~ω|1〉〈0|+

√2~ω|0〉〈1|

)dt

i~ln (|Ψ (t)〉) =(Ho +

√2~ω|1〉〈0|+

√2~ω|0〉〈1|

)t + C

ln (|Ψ (t)〉) = − i

~

(Ho +

√2~ω|1〉〈0|+

√2~ω|0〉〈1|

)t + C

|Ψ (t)〉 = Ce−i~ (Ho+

√2~ω|1〉〈0|+

√2~ω|0〉〈1|)t

47

Now we apply the condition that the function is in the ground state at t=0.

|Ψ (0)〉 = C = |0〉

So

|Ψ (t)〉 = |0〉e−i~Hoe−i

√2ω(|1〉〈0|+|0〉〈1|)t (A.14)

Therefore

〈n|Ψ (t)〉 = 〈n|0〉e−i~Hoe−i

√2ω(|1〉〈0|+|0〉〈1|)t (A.15)

If this is a probability the sum over 〈n| should be 1. We notice that coefficient will be 0 other than n = 0, where itwill equal 1. This means that the exponent must equal 1 there, too. The first term is the transpose of the unitarymatrix. We must check if the second term is a unitary matrix.If

e−i√

2ω(|1〉〈0|+|0〉〈1|)t = U (t)

then

U† (t) = ei√

2ω(|1〉〈0|+|0〉〈1|)t

and

UU† = 1

So

UU† = e−i√

2ω(|1〉〈0|+|0〉〈1|)tei√

2ω(|1〉〈0|+|0〉〈1|)t

= e−i√

2ω(|1〉〈0|+|0〉〈1|)t+i√

2ω(|1〉〈0|+|0〉〈1|)t

= e0 = 1

Since this term is a unitary matrix and the other term was a transpose of a unitary matrix, the probability willalways equal 1 and equation (15) is correct.X

48

Appendix B

Special Harmonics

B.1 Legendre Polynomials

The spherical harmonics are developed from the associated Legendre Polynomials. Why do we care? The simple caseof electrostatic potentials:

ϕ =q

4πεo

1

(r2 + a2 − 2ar cos θ)1/2=

q

4πεo

∞∑n=0

Pn (cos θ)(ar

)n(B.1)

where Pn (cos θ) are the Legendre Polynomials. What are these polynomials, they are a set of orthogonal polynomials.1

The first nine terms are

P0(x) = 1P1(x) = xP2(x) = 1

2

(3x2 − 1

)P3(x) = 1

2

(5x3 − 3x

)P4(x) = 1

8

(35x4 − 30x2 + 3

)P5(x) = 1

8

(63x5 − 70x3 + 15x

)P6(x) = 1

16

(231x6 − 315x4 + 105x2 − 5

)P7(x) = 1

16

(429x7 − 693x5 + 315x3 − 35x

)P8(x) = 1

128

(6435x8 − 12012x6 + 6930x4 − 1260x2 + 35

)Now the recurrence relation

(2n+ 1)xPn(x) = (n+ 1)Pn+1(x) + nPn−1(x) (B.2)

Some of the differential relations

(2n+ 1)Pn(x) = P ′n+1(x)− P ′

n−1 (B.3)

P ′n+1(x) = (n+ 1)Pn(x) + xP ′

n(x) (B.4)

P ′n−1(x) = xP ′

n(x)− nPn(x) (B.5)(1− x2)P ′

n(x) = nPn−1(x)− nxPn(x) (B.6)(1− x2)P ′

n(x) = (n+ 1)xPn(x)− (n+ 1)Pn+1(x) (B.7)

The parity is

Pn(−x) = (−1)nPn(x) (B.8)

As same as before, these functions are orthogonal, so∫ 1

−1

Pn(x)Pm(x)dx = δn,m (B.9)∫ π

0

Pn(cos θ)Pm(cos θ) sin θdθ = δn,m (B.10)

1See Arfken and Weber. Mathematical Methods for Physicists 5th ed. 2001. pg. 596. on Gram-Schmidt Orthogonalization

49

B.2 Associated Legendre Polynomials

Now for the Associated Legendre Polynomials. The first ten are in the following polynomials

P 11 (x) =

(1− x2

)1/2= sin θ

P 12 (x) = 3x

(1− x2

)1/2= 3 cos θ sin θ

P 22 (x) = 3

(1− x2

)= 3 sin2 θ

P 13 (x) = 3

2

(5x2 − 1

) (1− x2

)1/2= 3

2

(5 cos2 θ − 1

)sin θ

P 23 (x) = 15x

(1− x2

)= 15 cos θ sin2 θ

P 33 (x) = 15

(1− x2

)3/2= 15 sin3 θ

P 14 (x) = 5

2

(7x3 − 3x

) (1− x2

)1/2= 5

2

(7 cos3 θ − 3 cos θ

)sin θ

P 24 (x) = 15

2

(7x2 − 1

) (1− x2

)= 15

2

(7 cos2 θ − 1

)sin2 θ

P 34 (x) = 105x

(1− x2

)3/2= 105 cos θ sin3 θ

P 44 (x) = 105

(1− x2

)2= 105 sin4 θ

The recurrence relations here are

(2n+ 1)xPmn = (n+m)Pmn−1 + (n−m+ 1)Pmn+1 (B.11)(1− x2)1/2 Pm′n =

1

2Pm+1n − 1

2(n+m)(n−m+ 1)Pm−1

n (B.12)

(2n+ 1)(1− x2)1/2 Pmn = Pm+1

n+1 − Pm+1n−1 (B.13)

The parity of the associated polynomials

Pmn (−x) = (−1)n+mPmn (x) (B.14)

The orthogonality of the functions

n(n−m!)

(n+m)!

∫ 1

−1

Pmn (x)P kn (x) (1− x)−1 dx = δm,k (B.15)

B.3 Spherical Harmonics

Now combining the Associated Legendre Polynomials with the azimuthal dependence to find the Spherical Harmonics.Spherical Harmonics are defined as

Y mn (θ, ϕ) = (−1)m

√2n+ 1

4π

(n−m)!

(n+m)!Pmn (cos θ)eimϕ (B.16)

The first nine are

Y 00 (θ, ϕ) = 1√

4π

Y 11 (θ, ϕ) = −

√38π

sin θeiϕ

Y 01 (θ, ϕ) =

√34π

cos θ

Y −11 (θ, ϕ) =

√38π

sin θe−iϕ

Y 22 (θ, ϕ) =

√5

96π3 sin2 θe2iϕ

Y 12 (θ, ϕ) = −

√5

24π3 sin θ cos θeiϕ

Y 02 (θ, ϕ) =

√54π

(32

cos2 θ − 12

)Y −1

2 (θ, ϕ) =√

524π

3 sin θ cos θe−iϕ

Y −22 (θ, ϕ) =

√5

96π3 sin2 θe−2iϕ

The orthogonality function is ∫ 2π

0

∫ π

0

Y m1∗n1 (θ, ϕ)Y m2

n2 (θ, ϕ) sin θdθdϕ = δn1,n2δm1,m2 (B.17)

50

Index

dipole moment, 25

Stark Effect, 25

51

quantum

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