Graduate Quantum Mechanics Solution HW #6 ~HW...

Graduate Quantum MechanicsSolution

HW #6 ~HW #8(Sakurai Ch.2 #21~#37)

Solved by Sunghyon Kyeong([email protected])

Modfied by Chul Kim([email protected])

May 25, 2008Send me a mail. If you find any mistakes in this document

2. J.J. Sakurai 2.21

A particle in one dimension is trapped between two rigid walls:

V ={

0, for 0 < x < L∞, for x < 0, x > L

At t=0 it is known to be exactly at x=L/2 with certainty. What are the relative probabilities forthe particle to be found in various energy eigenstates? Write down the wave function for t ≥ 0. (Youneed not worry about absolute normalization, convergence, and other mathematical subtleties.)

(Solution)

1D-box problem is very well known. So, I will start the problem assuming we have already knownthe eigen ftn and eigen energy without calculating.

ψn =

√2L

sin(nπ

Lx)

En =π2h2n2

2mL2, where n ∈ Z.

At, t=0, particle is located at x = L2 which can be represented as follow:

δ(x − L

2

)=∑

n

cnψn

cn =

√2L

∫ L

0

dxδ(x − L

2

)sin

nπ

Lx

=

√2L

sinnπ

2

Therefore, δ-function can be expanded by complete set of eigen functions as follow:

δ(x − L

2

)=

2L

∞∑n=1

sinnπ

2sin

nπ

Lx

i. relative probability for the particle to be found in various energy eigenstates.

P (n) = |cn|2 =2L

sin2(nπ

2

)=

L

4(1 − cosnπ) =

L

2. for odd n

ii. time evolution for δ(x − 1

2

).

exp(− i

H

ht)δ(x − L

2

)=

∞∑n=0

( 2L

)exp

(− i

E2n+1

ht)sin

(2n + 1)π2

sin(2n + 1)π

Lx


Consider a particle in one dimension bound to a fixed center by a δ-function potential of the form

V (x) = −v0δ(x), (v0 real and positive).

Find the wave function and the binding energy of the ground state. Are there excited bound states?

(Solution)

We will start from the schrodinger equation

− h2

2m

d2

dx2ψ(x) − v0δ(x)ψ(x) = Eψ(x)

It yields bound state(E < 0)We’ll look first at region I (x < 0, V (x) = 0)

− h2

2m

d2

dx2ψ(x) = Eψ(x) =⇒ d2ψ(x)

dx2= −2mE

h2 ψ(x) = κ2ψ(x)

after solving above differential equation, we can get following:

ψ ={

ψI = Ae−κx + Beκx, for x < 0ψII = Ge−κx + Feκx, for x > 0

where κ2 = − 2mEh2 , κ > 0

We can get an appropriate solutions ψ (at both regions) by applying boundary conditions ψI(x →−∞) = 0 and ψII(x → ∞) = 0 with ψI(x = 0) = ψII(x = 0)

ψ ={

ψI = Beκx, for x < 0ψII = Be−κx, for x > 0

To find the remaining coefficients κ and B, we have to integrate the schrodinger equation from −εto +ε and then take the limit as |ε| → 0.

− h2

2m

∫ +ε

−ε

dxd2ψ(x)

dx2− v0

∫ +ε

−ε

dxδ(x)ψ(x) =∫ +ε

−ε

dxEψ(x)

(RHS) term is vanished by integrating process.

− h2

2m

(dψ(x)dx

∣∣∣+ε

− dψ(x)dx

∣∣∣−ε

)− v0ψ(0) = 0

=⇒ − h2

2m

(− κB − (−κB

)− v0B = 0

Finally, we can get the bound state energy

κ2 =m2v2

0

h4 = −2mE

h2 =⇒ E = −mv20

2h2

There exists only one bound state.∫ +∞

−∞dx|ψ|2 =

∫ 0

−∞dx|B|2e2κx +

∫ +∞

0

dx|B|2e−2κx =|B|2κ

= 1

Therefore, the last coefficient is B =√

κ and the final solution is given by.

ψ =

{ψI =

√mv0h2 e

mv0h2 x, for x < 0

ψII =√

mv0h2 e−

mv0h2 x, for x > 0

These two solution can be combined into one compact form as follow:

ψ(x) =√

mv0

h2 exp(− mv0

h2 |x|)


A particle of mass m in one dimension is bound to a fixed center by an attractive δ-function potential:

V (x) = −λδ(x), (λ > 0).

At t=0, the potential is suddenly switched off(that is, V = 0 for t> 0). Find the wave function fort> 0).(Be quantitative! But you need not attempt to evaluate an integral that may appear.)

(Solution)

I will start from schrodinger equation

− h2

2m

d2φ(x)dx2

− λδ(x)φ(x) = Eφ(x) (1)

The fourier transformation bewteen real space and momentum space is given as follow:

φ(x) =12π

∫ +∞

−∞dkφ(k)eikx (2)

We can get fourier transform of φ(x) as plugging equation(2) into equation(1) and by using theδ-function property.

+h2k2

2mφ(k) − λφ(x = 0) = Eφ(k)

From the above equation, we can get φ(k) and φ(x)as follows:

φ(k) =λφ(x = 0)

h2k2/2m − E=

2mλφ(0)/h2

k2 + κ2=

2κ32

k2 + κ2

φ(x) =12π

∫ +∞

−∞dk

κ32

k2 + κ2eikx

At t > 0, there exist only kinetic energy in hamiltonian

U(t) = exp(−i

H

ht

)= exp

(−i

1h

( p2

2m

)t

)= exp

(ih

2m

∂2

∂x2t

)We can get the time-dependent function by applying the above time-evolution operator.

φ(x, t) = Uφ(x, 0)

= exp(

ih

2m

d2

dx2t

)12π

∫ +∞

−∞dk

2κ32

k2 + κ2expikx

=12π

∫ +∞

−∞dk

2κ32

k2 + κ2exp

(−i

k2h

2mt + ikx

)You need not attempt to evaluate an integral that may appear.


A particle in one dimension(−∞ < x < ∞) is subjected to a constant force derivable from

V = λx, (λ > 0).

(a) Is the energy spectrum continuous or discrete? Write down an approximate expression for theenergy eigenfunction specified by E. Also sketch it crudely.

(Solution)

In the case under construction there is only a continuous spectrum and the eigenfunctions are nondegenerates. From the discussion on WKB approximation.region I : E > V (x)

ψI(x) =A

[E − V (x)]1/4exp

(+

i

h

∫ x2

x1

dx√

2m(E − V (x)))

=A

[E − V (x)]1/4exp

(√2mλ

h

∫ x2=E/λ

x1

dx√

(E/λ − x) − π

4

)

=A

[E − V (x)]1/4exp

(−2

3

(E

λ− x

)3/2√

2mλ

h− π

4

)

Similarly, we can easily find a wave function at the opposite region

region II : E < V (x)

ψII(x) =A2

[V (x) − E]1/4exp

(− 1

h

∫ x2

x1=E/λ

dx√

2m(V (x) − E)

)

=A2

[V (x) − E]1/4exp

(− 1

2mλh

∫ x2

x1=E/λ

d(2mλx)√

2m(V (x) − E)

)

=c2

[−q]1/4exp

[−2

3(−q)3/2

]

Where, q = α(

Eλ − x

)and α =

(2mλh2

)1/3

(b) Discuss briefly what changes are needed if V is replaced by

V = λ|x|.

(Solution)

We have bound states when we take the potential V = λ|x|. Therefore, the energy spectrum isdiscrete. So, in this case the energy eigenstates have to satisfy the consistency relation.∫ x2

x1

dx√

2m(E − λ|x|) =(

n +12

)πh

Where, n is non-negative integer.The turning points are x1 = −E

λ , x2 = Eλ

So, (n +

12

)πh =

∫ +E/λ

−E/λ

dx√

2m(E − λ|x|) = 2∫ +E/λ

0

dx√

2m(E − λ|x|)

= 2√

2mλ23

(E

λ

)3/2

We can finally get the discrete eigen energy for nth state as follow:

En =

[3(n + 1

2

)πhλ

4√

2m

]2/3


Consider an electron confined to the interior of a hollow cylinderical shell whose axis coincideswith the z-axis. The wave function is requared to vanish on the inner and outer wall, ρ = ρa and ρb,and also at the top and bottom, z = 0 and L.

(a) Find the energy eigenfunctions. (Do not bother with normalization.)Show that the energy eigenvalues are given by

Elmn =(

h2

2me

)[k2

mn +(

lπ

L

)2]

(l = 1, 2, 3, ..., m = 0, 1, 2, 3, ...),

where kmn is the nth root of the transcendental equation

Jm(kmnρb)Nm(kmnρa) − Nm(kmnρb)Jm(kmnρa) = 0.

(Solution)

Consider a wave function at the inner and outer wall.

− h2

2m2�2 ϕ = Eϕ

�2ϕ + k2ϕ = 0 where, k2 =2meE

h2

The schrodinger equation at the both inner and outer wall is(∂2

∂ρ2+

1ρ

∂

∂ρ+

1ρ2

∂2

∂φ2+

∂2

∂z2+ k2

)ϕ = 0 (3)

with the separation of coordinates

ϕ(ρ, φ, z) = R(ρ)Φ(φ)Z(z)

Then, equation (3) becomes

1R

(d2R

dρ2+

1ρ

dR

dρ

)+

1ρ2

1Φ

d2Φdφ2︸︷︷︸

−m2

+1Z

d2Z

dz2︸︷︷︸−k2

z

+k2 = 0

1R

(d2R

dρ2+

1ρ

dR

dρ

)+

−m2

ρ2− k2

z + k2 = 0

ρ2R′′(ρ) + ρR′(ρ) + ρ2

(k2

mn − m2

ρ2

)R(ρ) = 0 where, k2

mn = k2 − k2z

Where a prime denotes differentiation with respect to ρ. We are now in a position to apply boundaryconditions.

(i) z-direction : general solution is Z(z) = Azsin(kzz) + Bzcos(kzz)

Z(z = 0) = Z(z = L) = 0 =⇒ Bz = 0

kz =lπ

Ll = 1, 2, 3, ...

(ii) φ-direction : general solution is Φ(φ) = Aφeimφ

and boundary condition in φ-direction is given as follow:

Φ(φ) = Φ(φ + 2π)

Therefore, m should be integer.(i.e. m ∈ Z)

(iii) ρ-direction : general solution at the finite region as following. ρz < ρ < ρb

R(ρ) = AρJm(kmnρ) + BρNm(kmnρ)R(ρ = ρa) = R(ρ = ρb) = 0(

Jm(kmnρa) Nm(kmnρa)Jm(kmnρb) Nm(kmnρb)

) (Aρ

Bρ

)= 0

To get a non-trivial solution below determinant is automatically satisfied for arbitrary Aρ and Bρ (Imean, determine Aρ and Bρ with the above matrix equation.)∣∣∣∣ Jm(kmnρa) Nm(kmnρa)

Jm(kmnρb) Nm(kmnρb)

∣∣∣∣ = 0

→ Jm(kmnρb)Nm(kmnρa) − Nm(kmnρb)Jm(kmnρa) = 0

Therefore, we can get the wave function.

ϕ(ρ, φ, z) = Cρ (Nm(kmnρb)Jm(kmnρ) − Jm(kmnρb)Nm(kmnρ)) sin(

lπ

Lz

)eimφ

Also, we can get the energy eigenvalue.

Enml =h2

2mek2 =

h2

2me

[k2

mn +(

lπ

L

)2]

where, l = 1, 2, 3, ...

(b) Repeat the same problem when there is a uniform magnetic field B = Bz for 0 < ρ < ρa. Notethat the energy eigenvalues are influenced by the magnetic field even tough the electron never“touches” the magnetic field.

(Solution)

The magnetic field is related to vector potential A through the relation B = ∇× A.We can find the a vector potential by using stokes’ theorem.

ΦB =∫

S

B · d S = πρ2aB =

∮C

A · d l = 2πρAφ

A =(

Bρ2a

2ρ

)φ =

(ΦB

2πρ

)φ

The Schrodinger equation in the presence of the magnetic field B can be written as follows

12me

[−ih∇− e

c A]·[−ih∇− e

c A]ϕ = Eϕ

− h2

2m2

[ρ

∂

∂ρ+ φ

1ρ

(∂

∂φ− ie

hc

ΦB

2π

)+ z

∂

∂z

]2

ϕ = Eϕ

We can see the change only in φ-component compare with B = 0 case.(ddφ

− ie

hc

ΦB

2π

)2

Φ(φ) = −m2Φ(φ) (4)

Trial solution of Φ(φ) can be written as follows:

Φ(φ) = CφeiMφ

By putting a trial solution into equation (4), we can get(ddφ

− ie

hc

ΦB

2π

)(iM − ie

hc

ΦB

2π

)CφeiMφ = −m2CφeiMφ

(iM − ie

hc

ΦB

2π

)(iM − ie

hc

ΦB

2π

)CφeiMφ = −m2CφeiMφ

(iM − ie

hc

ΦB

2π

)2

= −m2

So, we can get the relation between M and m.

M = ±m +ie

hc

ΦB

2π

Applying the boundary condition,

Φ(φ) = Φ(φ + 2π) =⇒ M = ±m +ie

hc

ΦB

2π∈ Z

This means that the energy eigenfunctions will

ϕ(ρ, φ, z) = Cρ (Nm(kmnρb)Jm(kmnρ) − Jm(kmnρb)Nm(kmnρ)) sin(

lπ

Lz

)eiMφ

But now m is not an integer. As a result the energy of the ground state will be

EnMl =h2

2mek2 =

h2

2me

[k2

mn +(

lπ

L

)2]

where, l = 1, 2, 3, ...

(c) Compare, in particular, the ground state of the B = 0 problem with that of the B = 0 problem.Show that if we require the ground-state energy to be unchanged in the presence of B, we obtain“flux quantization”

πρ2aB =

2πNhc

e, (N = 0,±1,±2, ...).

(Solution)

Now, m = M− iehc

ΦB

2π is not zero in general but it corresponds to M ∈ Z such that 0 ≤ M− iehc

ΦB

2π < 1.Notice also that if we require the ground state to be unchanged in the presence of B, we obtain “fluxquantization”

M − ie

hc

ΦB

2π= 0 =⇒ ΦB = πρ2

aB =2πMhc

ewhere, M ∈ Z.


Consider a particle moving in one dimension under the influence of a potential V (x). Suppose itswave function can be written as exp[iS(x, t)/h]. Prove that S(x, t) satisfies the classical Hamilton-Jacobi equation to the extent that h can be regarded as small in some sense.Show how one may obtain the correct wave function for a plane wave by starting with the solution ofthe classical Hamilton-Jacobi equation with V (x) set equal to zero. Why do we get the exact wavefunction in this particular case?

(Solution)

Hamilton-Jacobi equation in the classical limit is the following.

12m

| ∇S(x, t) |2 +V (x) +∂S(x, t)

∂t= 0

Use time dependent Schrodinger equation.

[−h2

2m∇2 + V (x)]ψ = ih

∂

∂tψ

· · ·ψ = exp[iS(x, t)

h]

· · · −h2

2m∇2exp[

iS(x, t)h

] =1

2m| ∇S(x, t) |2 exp[

iS(x, t)h

]

· · · ∂

∂texp[

iS(x, t)h

] =i

h

∂S(x, t)∂t

exp[iS(x, t)

h]

Then the time dependent Schrodinger equation becomes,

12m

| ∇S(x, t) |2 exp[iS(x, t)

h] + V (x)exp[

iS(x, t)h

] = −∂S(x, t)∂t

exp[iS(x, t)

h]

Divide both side with the exponential term then one can see that satisfies the Hamilton-Jacobiequation.

⇒ 12m

| ∇S(x, t) |2 +V (x) +∂S(x, t)

∂t= 0

If V(x) = 0 ,

12m

| ∇S(x, t) |2 +∂S(x, t)

∂t= 0 · · ·S(x, t) ≡ S1(x) + S2(t)

⇒ 12m

∂2S1

∂x2= −∂S2

∂t= C

Note that C is a constant. We can find the solution of S as following.

S1 =√

2mcx + const.S2 = −ct + const. ⇒ S =√

2mcx − ct

Insert this S into the exponential term of ψ,

⇒ ψ = exp[i(√

2mcx − ct)/h]

Now the wave function is a plane wave. Physically, if V(x) = 0 , then plane wave should come outbecause it is a free particle case. Mathematically, the second derivative of S is 0.


Using spherical coordinates, obtain an expression for j for the ground and excited states of thehydrogen atom. Show, in particular, that for ml = 0 states, there is a circulating flux in the sensethat j is in the direction of increasing or decreasing φ, depending on whether ml is positive ornegative.

(Solution)

General solution of 3D schrodinger equation is given by

ϕnlm = Rnl(r)Y lm(θ.φ) = Rnl(r)NlmP

|m|l (cosθ)eimφ

current density j can be expressed as follow.

j =ih

2me(ϕnlm∇ϕ∗

nlm − ϕ∗nlm∇ϕnlm)

and differential operator in spherical coordinate is given by

∇ =∂

∂rr +

1r

∂

∂θθ +

1rsinθ

∂

∂φφ

We can split into current density j by parts.

jr =ih

2me(ϕnlm∇rϕ

∗nlm − ϕ∗

nlm∇rϕnlm) = 0

jθ =ih

2me(ϕnlm∇θϕ

∗nlm − ϕ∗

nlm∇θϕnlm) = 0

These two term are automatically vanished because there are no imaginary components in the (r, θ)parts.

jφ =ih

2me(ϕnlm∇φϕ∗

nlm − ϕ∗nlm∇φϕnlm)

=ih

2meR2

nl(r)N2lm

1sinθ

p|m|l (cosθ)

(eimφ ∂

∂φe−imφ − e−imφ ∂

∂φeimφ

)

= 2mh

2me

1sinθ

|ϕnlm|2

=mh

mersinθ|ϕnlm|2

Finally, we can get the φ-direction current density as follow:

j = jφφ =mh

mersinθ|ϕnlm|2φ

There survives only φ-component, which means current flow in φ-direction depending on the sign ofmagnetic moment m


Derive (2.5.16) and obtain the three-dimensional generalization of (2.5.16).

(Solution)

K(x′′, t;x′, t0) = 〈x′′|exp(−i

H(t − t0)h

)|x′〉

=∫ +∞

−∞dp′〈x′′|p′〉〈p′|exp

(−i

p′2h2m

(t − t0))|x′〉

=1

2πh

∫ +∞

−∞dp′exp

(−i

p′2h2m

(t − t0))

exp(

ip′

h(x′′ − x′)

)

=1

2πh

∫ +∞

−∞dp′exp

[− i(t − t0)

2mh

[(p′ − x′′ − x′

t − t0m

)2

−(

x′′ − x′

t − t0m

)2]]

=1

2πhexp

[i

h

m(x′′ − x′)2

2(t − t0)

] ∫ +∞

−∞dp′exp

(− i(t − t0)

2mhp′2)

=1

2πhexp

(i

h

m(x′′ − x′)2(t − t0)

)√2mhπ

i(t − t0)

=√

m

2πih(t − t0)exp

[im(x′′ − x′)2h(t − t0)

]Generalization in three dimension

K(x′′, t;x′, t0) =(

m

2πih(t − t0)

)3/2

exp[im|x′′ − x′|2h(t − t0)

]


Define the partition function as

Z =∫

d3x′K(x′, t;x′, 0)|β=it/h,

as in (2.5.20)-(2.5.22). Show that the ground-state energy is obtained by taking

− 1Z

∂Z

∂β, (β → ∞).

(Solution)

Ground state energy ≡ EG.1.

Z =∫

d3x′K(x′, t;x′, 0)

=∫

d3x′∑n

|〈x′|n〉|2exp(−βEn)

=∑

n

exp(−βEn)

− 1Z

∂Z

∂β= − 1

Z

∂

∂β[∑

n

exp(−βEn)] = − 1Z

[∑

n

(−En)exp(−βEn)]

=∑

n(En)exp(−βEn)∑n exp(−βEn)

In the limit of β → ∞, ∑n(En)exp(−βEn)∑

n exp(−βEn)⇒ EG (5)

2. In 1 − D box, Density of state is L/2π

Z =∑

n

exp(−βn2π2h2

2mL2)

⇒ − 1Z

∂Z

∂β=

π2h2

2mL2

∑n n2exp(−β n2π2h2

2mL2 )∑n exp(−β n2π2h2

2mL2 )

limπ2h2

2mL2

∑n n2exp(−β n2π2h2

2mL2 )∑n exp(−β n2π2h2

2mL2 )=

π2h2

2mL2= EG


The propagator in momentum space analogous to (2.5.26) is given by 〈p′′, t|p′, t0〉. Derive an explicitexpression for 〈p′′, t|p′, t0〉 for the free particle case.

(Solution)

Propagator in momentum space.

〈 p′′, t| p′, t0〉 = 〈 p′′|exp(−i

H

h(t − t0)

)| p′〉

= 〈 p′′|exp

(−i

p′2

2mh(t − t0)

)| p′〉

= exp

(−i

p′′

2mh(t − t0)

)δ3( p′′ − p′)

= K( p′′, t; p′, t0)


(a) Write down an expression for the classical action for a simple harmonic oscillator for a finitetime interval.

(Solution)

General expression of x(t) and x(t) for a simple harmonic oscillator is given by:

x(t) = Acosw(t − t1) + Bsinw(t − t1)x(t) = −wAsinw(t − t1) + wBcosw(t − t1)

The action can be written as follow:

S =∫ t2

t1

dt£(x, x; t) =∫ t2

t1

dt

(12mx2 − 1

2mw2x2

)

=mw2

2

∫ t2

t1

dt([−Asinw(t − t1) + Bcosw(t − t1)]2 − [Acosw(t − t1) + Bsinw(t − t1)]2

)To calculate the above action, we first find coefficients A and B by applying boundary conditions:

x(t = t1 − t1 = 0) = A

=⇒ A = x1

x(t = t2 − t1 = T ) = AcoswT + BsinwT = x1

=⇒ B =x2 − x1coswT

sinwT

We are now in a position to calculate the action as following procedure.

S =mw2

2

∫ t2

t1

dt([−Asinw(t − t1) + Bcosw(t − t1)]2 − [Acosw(t − t1) + Bsinw(t − t1)]2

)=

mw2

2

∫ T

0

dt([−Asinwt + Bcoswt]2 − [Acoswt + Bsinwt]2

)=

mw2

2

∫ T

0

dt[(A2 − B2)sin2wt − (A2 − B2)cos2wt − 4ABsinwtcoswt

]=

mw2

2

∫ T

0

dt[(B2 − A2)cos2wt − 2ABsin2wt

]=

mw2

2

[(B2 − A2)

12w

sin2wt + 2AB1

2wcos2wt

]T

0

=mw

4[(B2 − A2)sin2wT + 2AB(cos2wT − 1)

]=

mw

4[(B2 − A2)2sinwT coswT − 4ABsin2wT

]=

mw

2sinwT

[(B2 − A2)coswt − 2ABsinwT

]where coefficients are given by

B2 − A2 =(

x2 − x1coswT

sinwT− x2

1

)=

1sin2wT

[x2

2 + x21(cos2wT − sin2wT ) − 2x1x2coswT

]AB = x1

x2 − x1coswT

sinwT

We can finally get the action in terms of x1 and x2 by plugging in the coefficients as shown in above:

S =mw

2sinwT

[ coswT

sin2wT

(x2

2 + x21(cos2wT − sin2wT ) − 2x1x2coswT

)−2x1

x2 − x1coswT

sinwTsinwT

]=

mw

2sinwT

[(x2

2 + x21 −��

2x21sin

2wT − 2x1x2coswT )coswT

−2x1x2sin2wT +��2x2

1sin2wT coswT

]=

mw

2sinwT

((x2

2 + x21)coswT − 2x1x2

)

(b) Construct 〈xn, tn|xn−1, tn−1〉 for a simple harmonic oscillator using Feynman’s prescription fortn − tn−1 = Δt small. Keeping only terms up to order (Δt)2, show that it is in completeagreement with the t − t0 → 0 limit of the propagator given by (2.5.26).

(Solution)

S(n, n − 1) =∫ tn

tn−1

dt

[mx2

2− V (x)

]

= Δt

[m

2

(xn − xn−1

Δt

)2

− V

(xn + xn−1

2

)]

=m

2Δt

(x2

n − 2xnxn−1 + x2n−1 −

w2

2(xn + xn−1)2Δt

)

〈xn, tn|xn−1, tn−1〉 =√

m

2πihΔtexp

(i

h

∫dt£

)

=√

m

2πihΔtexp

[iΔt

h

(m

2

(xn − xn−1

Δt

)2

− m

2w2

(xn + xn−1

2

)2)]

Therefore, it agrees with the t-t0 ⇒ 0 limit of the propagator given by (2.5.26) and (2.5.18).


State the Schwinger action principle (see Finkelstein 1973, 155). Obtain the solution for 〈x2t2|x1t1〉by integrating the Schwinger principle and compare it with the corresponding Feynman expressionfor 〈x2t2|x1t1〉.Describe the classical limits of these two expressions.

(Solution)

i. Schiwinger action principleIn quantum mechanics we are interested in computing the transition amplitude 〈x2, t2|x1, t1〉where |x1, t1〉 represents the quantum state at time t1, and |x2, t2〉 represents the state at timet2 ≥ t1. These states are chosen to be eigenstates of the position operator xi:

xi|xα, tα〉 = xi(tα)|xα, tα〉 (α = 1, 2)

The Schwinger action principle states that

δ〈x2, t2|x1, t1〉 =i

h〈x2, t2|δS|x1, t1〉 (6)

where S represents the action obtained by the replacement of xi in the action for the classi-cal theory with xi, along with an operator ordering which leads to S being self-adjoint. δ inequation (6) represents any possible variation, including variations with respect to the timest1, t2, the dynamical variables qi, or the structure of the Lagrangian. The variations of thedynamical variables δqi will be chosen to be c-numbers, appropriate to bosonic theories.

ii. Obtain the solution for 〈x2, t2|x1, t1〉We have to integrate the equation from the statement of schwinger’s action principle referredfrom the Finkelstein’s Nonrelativistic Mechanics

δ〈x2, t2|x1, t1〉 =i

h〈x2, t2|δS|x1, t1〉

Let’s consider a action operator which acts from (x1, t1) to (x2, t2) sequently. We call it δS21.Then, we get

δ〈x2, t2|x1, t1〉 =i

h〈x2, t2|δS|x1, t1〉 =

i

h〈x2, t2|δS21|x1, t1〉

= δS′21

i

h〈x2, t2|x1, t1〉

=⇒ δ〈x2, t2|x1, t1〉〈x2, t2|x1, t1〉 =

i

hδS′

21

By integrating above equation, we get

〈x2, t2|x1, t1〉 = exp[

i

hS′

21

](7)

Let’s suppose that h can, in some sense, be regarded as a small quantity. Then S′21 which appears

in the equation (7) should ssatisfy the Hamilton-Jacobi equation in the classical mechanics.

12m

|∇ S′21( x, t)|2 + V ( x) +

∂ S′21

∂t= 0

iii. Compare the transition amplitude with the corresponding Feynman expressionFeynman expression for 〈x2, t2|x1, t1〉 is

〈x2, t2|x1, t1〉 =∑

all path

exp[

i

hS21

]

By taking the classical limit h → 0, then the dominant path is the only classical path. Therefore,transition amplitude 〈x2, t2|x1, t1〉 is exactly the same result for both of approaching.


Show that the wave-mechanics approach to the gravity-induced problem discussed in Section 2.6also leads to phase-difference expression (2.6.17).

(Solution)

Hamiltonian of this system is given by

H =p2

2mn+ mngz =

p2

2mn+ mngl2sinδ

Potential has different value at different position.

VAC = 0 if z = 0VBD = mngl2sinδ if z = l2sinδ

Then, we can calculate the phase difference as follow:

Δφ = − 1h

∫ lf

li

dl

v[VBD − VAC ] equation(2.6.9)

= − 1hv

mngl2sinδl1

= −mngl1l2sinδ

h

1v

= −mngl1l2sinδ

h

mnλ

h= −m2

ngl1l2λsinδ

h2

Where, λ is the deBroglie’s (matter) wave length which has to satisfy the following relation.

p = mnv =h

λ

v =h

mnλ


(a) Verity (2.6.25) and (2.6.27).

(Solution)

We can prove the equation (2.6.25) simply by using commutation relation.

Πi = pi − e

cAi

[Πi, Πj ] =[pi − e

cAi, pj − e

cAj

]=

[pi,−e

cAj

]+[−e

cAi, pj

]= −ih

∂

∂xi

(−e

cAj

)+ ih

∂

∂xj

(−e

cAi

)=

ihe

c

(∂

∂xiAj − ∂

∂xjAi

)

=ihe

cεijkBk

We can prove the equation (2.6.27) by using heisenberg equation of motion and commutation relation.

md2xi

dt2=

dΠi

dt=

1ih

[Πi,H]

=1ih

[Πi,

Π2

2m+ eφ(x)

]

=1ih

[Πi, eφ(x)] +1

2mih

⎡⎣Πi,

∑j

Π2j

⎤⎦

=1ih

[pi, eφ(x)] +1

2mih

∑j

([Πi, Πj ] Πj + Πj [Πi, Πj ])

= −e∂φ(x)∂xi

+1

2mih

(ihe

c

)(εijkBkΠj + εijkΠjBk)

= e

[Ei +

12c

(εijk

Πj

mBk − εikjBk

Πj

m

)]

= e

[ E +

12c

(d x

dt× B − B × d x

dt

)]i

(· · ·Πj =dxi

dt)

(b) Verity continuity equation (2.6.30) with j given by (2.6.31).

(Solution)

The time-dependent schrodinger equations are

ih∂ψ

∂t= Hψ

−ih∂ψ∗

∂t= H∗ψ∗

Where, hamiltonian H is given by

H =1

2m

[−h2∇2 +

e

cih(∇ · A) + 2ih

e

c A · ∇ +

e2

c2A2

]+ eφ (8)

The derivation of ρ with respect to time t is

ih∂ρ

∂t= ih

∂ψ∗ψ∂t

= ihψ∗ ∂ψ

∂t+ ihψ

∂ψ∗

∂t= (ψ∗Hψ − ψHψ∗)

Multiplying ψ∗ and ψ at the left and right side of the hamiltonian as shown in the equation (8).

ψ∗Hψ =1

2m

[−h2ψ∗∇2ψ +

e

cih(∇ · A)|ψ|2 + 2ih

e

c A · ψ∗∇ψ +

e2

c2A2|ψ|2

]+ eφ|ψ|2

The complex conjugate of the above equation is

ψHψ∗ =1

2m

[−h2ψ∇2ψ∗ − e

cih(∇ · A)|ψ|2 − 2ih

e

c A · ψ∇ψ∗ +

e2

c2A2|ψ|2

]+ eφ|ψ|2

Then substracting the last two equations we get

− h2

2m

[ψ∗∇2ψ − ψ∇2ψ∗]+

( e

mc

)ih(∇ · A)|ψ|2

+( e

mc

)ih A · (ψ∗∇ψ − ψ∇ψ∗) = ih

(ψ∗ ∂

∂tψ + ψ

∂

∂tψ∗)

=⇒ − h2

2m∇ · [ψ∗∇ψ − ψ∇ψ∗] +

( e

mc

)∇ · ih A|ψ|2 = ih

∂

∂t|ψ|2

Dividing ih for both side of the last equation and rearranging process, we can get

∂

∂t|ψ|2 +

h

m∇ · Im(ψ∗∇ψ) −

( e

mc

)∇ · A|ψ|2 = 0

∂

∂t|ψ|2 + ∇ ·

(h

mIm(ψ∗∇ψ) −

( e

mc

)· A|ψ|2

)= 0

Finally, we get the continuity equation

∂ρ

∂t+ ∇ · j = 0

with j =(

hm

)Im(ψ∗∇ψ) −

(e

mc A|ψ|2

)and ρ = |ψ|2


Consider the Hamiltonian of a spinless particle of charge e. In the presence of a static magneticfield, the interaction terms can be generated by

poperator → poperator −eAc

,

where A is the appropriate vector potential. Suppose, for simplicity, that the magnetic field B isuniform in the positive z-direction. Prove that the above prescription indeed leads to the correctexpression for the interaction of the orbital magnetic moment (e/2mc)L with the magnetic fieldB. Show that there is also an extra term proportional to B2(x2 + y2), and comment briefly on itsphysical significance.

(Solution)

We have to choose an appropriate gauge to describe the system easily.

A = B(−y

2,x

2, 0)

=12

B × ρ

By using this gauge, we can construct the hamiltonian as follow:

H =1

2m

(p − e

c A)2

=1

2m

(p2 − e

c

(p · A + A · p

)+

e2

c2A2

)

=p2

2m− e

4mc

(p · ( B × ρ) + ( B × ρ) · p

)+

e2

2mc2

B2

4(x2 + y2)

=p2

2m− e

4mc

⎛⎜⎝ B · ( ρ × p)︸︷︷︸

�L

+( ρ × p)︸︷︷︸�L

· B

⎞⎟⎠+

e2

8mc2B2(x2 + y2)

=p2

2m− e

2mc B · L +

e2

8mc2B2(x2 + y2)

(comment)The hamiltonian can have maximum energy(unstable state) when the two vectors( B and L) areanti-parallel and minimum energy(stable state) when the two vectors are parallel.


An electron moves in the presence of a uniform magnetic field in the z-direction ( B = Bz).

(a) Evaluate[Πx, Πy] ,

whereΠx ≡ px − eAx

c, Πy ≡ py − eAy

c.

(Solution)

[Πx, Πy] =[px − e

cAx, py − e

cAy

]=

[px,−e

cAy

]+[−e

cAx, py

]= ih

e

c

(∂

∂xAy − ∂

∂yAx

)= ih

e

c

(∇× A

)z

= ihe

cB

Note that B only has z component.

(b) By comparing the Hamiltonian and the commutation relation obtained in (a) with those of theone-dimensional oscillator problem, show how we can immediately write the energy eigenvaluesas

Ek,n =h2k2

2m+( |eB|h

mc

)(n +

12

),

where hk is the continuous eigenvalue of the pz operator and n is a nonnegative integer includingzero.

(Solution)

Hamiltonian for this system is given by

H =1

2m

(p − e

c A)2

=1

2mΠ2

x +1

2mΠ2

y +1

2mΠ2

z

=p2

z

2m+

(Π2

x

2m+

Π2y

2m

)= Hfree + HB−field

Where, HB−field =(

Π2x

2m + Π2y

2m

)and Hfree = p2

z

2m

Let’s check a commutator relation between Hfree and HB−field.

[Hfree, HB−field] =[

12m

(px − e

cAx

)2

+1

2m

(py − e

cAy

)2

,p2

z

2m

]= 0 (9)

Where, we use basic commutator relation as follow:

[pi, pj ] = 0[pi, A ( x)

]= −ih

∂

∂xi

A( x)

Since, Hfree and HB−field are commute as shown in equation (9). There exits a set of simulta-neous eigenkets |k, n〉 of the operators (Hfree and HB−field).

If hk is the continuous eigenvalue of the operator pz and its eigenkets |k, n〉, we will have

Hfree|k, n〉 =h2k2

2m|k, n〉

On the other hand HB−field is similar to the hamiltonian of the one-dimensional harmonicoscillator problem which is given by

H =p2

2m+

12mw2x2

with [x, p] = ih. In order to get the similar form with the eigen energy of the 1 − D harmonicoscillator En = hw

(n + 1

2

)we should have the same commutator relation between the squared

operators in the hamiltonian. From (a) we have

[Πx, Πy] = ihe

cB =⇒

[cΠx

eB, Πy

]= ih

Considering the commutator relation between p and x, [p,x] and comparing it with the abovecommutator relation, one can write HB−field as the following form.

HB−field =1

2mΠ2

y +12mw2

(eΠx

|eB|)2

Where, the frequency is given w = |eB|mc By combining two separate hamiltonian, we get

H|k, n〉 = HB−field|k, n〉 + Hfree|k, n〉

=|eB|mc

(n +

12

)|k, n〉 +

h2k2

2m|k, n〉

=[ |eB|

mc

(n +

12

)+

h2k2

2m

]|k, n〉


Consider the neutron interferometer.

Prove that the difference in the magnetic fields that produce two successive maxima in the countingrates is given by

ΔB =4πhc

|e|gnλl

where gn(=-1.91) is the neutron magnetic moment in units of −eh/2mnc. [If you had solved thisproblem in 1967, you could have published it in Physical Review Letters!]

(Solution)

First, we have to know the fact that

“A nearly monoenergetic beam of thermal neutrons is split into two parts - path A and path B. Path Aalways goes through a magnetic-field-free region; in contrast, path B enters a small region where a staticmagnetic field is present.”

Now, Let’s consider a hamiltonian for neutron with spin angular momentum in z-direction

H = − μ · B = wSz

Where w is the spin-precession frequency

w = gn|eB|mnc

for the neutron with a magnetic moment of gneh/2mnc

We now consider a simultaneous eigenket of energy and spin angular momentum satisfying belowproperties.

H|Sz;±〉 = ± hw

2|Sz;±〉

Sz|Sz;±〉 = ± h

2|Sz;±〉

We are now in a position to find the time evolution operator.

U(T ) = U(T ) = exp(−i

H

hT

)= exp

(−i

wT

hSz

)= exp

(−i

wT

2σz

)

= 1 +−iwT

2σz +

(−iwT/2)2

2!+

(−iwT/2)3

3!σz + ...

= cos(wT

2) − iσzsin(

wT

2) =

(cos(wT

2 ) − isin(wT2 ) 0

0 cos(wT2 ) + isin(wT

2 )

)

=(

exp(−iwT2 ) 0

0 exp(+iwT2 )

)

=

(exp(−iwlmnλ

2h ) 00 exp(+iwlmnλ

2h )

)

Where T is the time spent in the B = 0 region and it is given by

T =l

v=

l

p/mn=

lmnλ

h

We now have to think about the state ket at the interference region. It is combined with two ketswhich went through different path. Only two of one ket is affected by the magnetic field.

So, we get|β〉 = C (|α〉 + U(T )α〉) = C[I + U(T )]|α〉

To find maxima in the counting rates, we have to calculate norm of |β〉 and check its behaviors.

Norm of |β〉

〈β|β〉 = C2

⎡⎣I2 + U + U† + UU†︸︷︷︸

I

⎤⎦

= C2(2 + U + U†) = C2

[2 +

(2cos(wT

2 ) 00 2cos(wT

2 )

) ]

= C2

[2 + 2cos

(wT

2

)]= 4C2cos2

(wT

4

)Counting rate is related to the frequency change as follow:

Δw = gn|e|ΔB

mnc

There are two solutions in the range of 0 ≤ ΔwT4 < 2π

ΔwT

4= 0, π

However, 0 gives a trivial solution. Therefore, the counting rates is calculated as follow:

π =ΔwT

4=

14

(gn

|e|ΔB

mnc

)T

ΔB =4πmnc

|e|gnT=

4πhc

|e|gnλl

Graduate Quantum Mechanics Solution HW #6 ~HW...

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Transcript of Graduate Quantum Mechanics Solution HW #6 ~HW...