shankar quantum mechanics solution

8/18/2019 Shankar Quantum Mechanics Solution

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SOLUTION MANUAL

COMPILED BY YEMI BUKKY+234(0)8057474928; +234(0)8064974071

( [email protected])

Department of Physics,

Federal University of Technology,

Minna, NG

Nigeria

mailto:[email protected]:[email protected]:[email protected]


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PRINCIPLES OF QUANTUM

MECHANICS

BY R. SHANKAR

SECOND EDITION

SOLUTIONSCOMPILED BY YEMI BUKKY ( [email protected])

Department of Physics,

Federal University of Technology, Minna, NG

Nigeria

+2348057474928

1

mailto:[email protected]:[email protected]:[email protected]


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Physics 710-712 due October 30, 2009

Problem Set 4

Problem 1: Do exercise 2.5.3 of the text.Solution: The problem asks us to get the equations of motion using the Hamiltonian method

for the system shown in Figure 1.5 (p. 46) of the text. Using x1 and x2 shown there as

the generalized coordinates, the kinetic energy and potential energy are

T = m

2 ẋ21

+ m

2 ẋ22

, V = k

2x21

+ k

2(x1 − x2)

2 + k

2x22

,

so the Lagrangian is L = T − V and thus the generalized momenta are

p1 = ∂ L

∂ ẋ1=

∂T

∂ ẋ1= mẋ1, ⇒ ẋ1 =

p1

m,

p2 = ∂ L

∂ ẋ2=

∂T

∂ ẋ2= mẋ2, ⇒ ẋ2 =

p2

m,

so the Hamiltonian is

H = T + V = p2

1

2m +

p22

2m +

k

2x21 +

k

2(x1 − x2)

2 + k

2x22

,

and Hamilton’s equations are

ẋ1 = ∂ H

∂p1=

p1

m, ˙ p1 = −

∂ H

∂x1= −2kx1 + kx2,

ẋ2 = ∂ H

∂p2=

p2

m, ˙ p2 = −

∂ H

∂x2= −2kx2 + kx1.

Problem 2: Do exercise 2.7.2 of the text.Solution: (i):

{q i, q j} :=k

∂q i

∂q k·

∂q j

∂pk−

∂q i

∂pk·

∂q j

∂q k

=k

∂q i

∂q k·0 − 0·

∂q j

∂q k

= 0

{ pi, pj} :=k

∂pi

∂q k·

∂pj

∂pk−

∂pi

∂pk·

∂pj

∂q k

=k

0·

∂pj

∂pk−

∂pi

∂pk·0

= 0

{q i, pj} :=k

∂q i

∂q k·

∂pj

∂pk−

∂q i

∂pk·

∂pj

∂q k

=k

(δ ikδ jk − 0·0) = δ ij,

and

{q i,H} :=k

∂q i∂q k ·

∂ H

∂pk −

∂q i

∂pk ·

∂ H

∂q k

=k

δ ik·

∂ H

∂pk − 0·

∂ H

∂q k

=

∂ H

∂pi = q̇ i,

{ pi,H} :=k

∂pi

∂q k·

∂ H

∂pk−

∂pi

∂pk·

∂ H

∂q k

=k

0·

∂ H

∂pk− δ ik·

∂ H

∂q k

= −

∂ H

∂q i= ˙ pi,

where in the last steps I used Hamilton’s equations.

(ii): The Hamiltonian given is H = p2x + p2

y + ax2 + by2. If a = b, H has a

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symmetry under simultaneous rotations in the x-y and px- py planes, under which z (the

generator) is conserved. Therefore {z,H} = 0.

We check this as follows:

{z,H} =k

∂z

∂q k·

∂ H

∂pk−

∂z

∂pk·

∂ H

∂q k

=

∂z

∂x·

∂ H

∂px+

∂z

∂y ·

∂ H

∂py−

∂z

∂px·

∂ H

∂x −

∂z

∂py·

∂ H

∂y .

But

∂ H

∂pk= 2 pk,

∂ H

∂xk=

∂ H

∂x ,

∂ H

∂y

= (2ax , 2by) ,

∂z

∂pk=

∂ (xpy − ypx)

∂pk=

∂z

∂px,

∂z

∂py

= (−y , x) ,

∂z

∂q k=

∂z

∂x ,

∂z

∂y

= ( py , − px) ,

so

{z , H} = py·2 px + (− px)·2 py − (−y)·2ax − x·2by = 2xy(a − b)

which vanishes if a = b.

Problem 3: Do exercise 2.8.1 of the text.Solution: Since g = p1 + p2, it generates the infinitesimal transformations

δx1 = + ∂ g

∂p1= +, δp1 = −

∂g

∂x1= 0,

δx2 = + ∂ g

∂p2= +, δp2 = −

∂g

∂x2= 0.

So, to order , these give the canonical transformations xi → x̄i(xj , pj) and pi → ¯ pi(xj , pj)

with

x̄1 = x1 + , ¯ p1 = p1,

x̄2 = x2 + , ¯ p2 = p2,

which is precisely a spatial translation of the whole system by an amount .

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Physics 710-711-712 November 16, 2009Problem Set 5

Problem 1: Do exercise 4.2.1 of the text.

(1)

Solution: The possible outcomes are Lz = {1, 0, −1}, which are the eigenvalues of Lz.(2)

Solution: Lz|ψ = 1 · |ψ implies

|ψ =10

0

.

(Note that I have normalized |ψ!) Then

Lx = ψ|Lx|ψ =

1 0 0

1√ 2

0 1 01 0 10 1 0

100

= 1√ 2

1 0 0

010

= 0.

L2x = ψ|L2x|ψ =

1 0 0 1

2

0 1 01 0 1

0 1 0

0 1 01 0 1

0 1 0

10

0

= 1

2

0 1 0

010

= 1

2.

∆Lx =

L2x − Lx2 =

1

2

− 02 = 1√

2.

(3)

Solution: The characteristic equation for Lx is

0 = det(Lx − λ) = det

−λ 1√ 2

01√ 2

−λ 1√ 2

0 1√ 2

−λ

= λ − λ3, ⇒ λ ∈ {1, 0, −1}.

The corresponding eigenvectors, |λ, then satisfy

0 = (Lx − λ)|λ =

−λ 1√ 2

01√ 2

−λ 1√ 2

0 1√ 2

−λ

ab

c

=

−λa + b√ 2

a√ 2 − λb + c√

2b√ 2 − λa

where we have parameterized the components of |λ by (a b c). For λ = 1, we can solvefor b and c in terms of a, giving b = √ 2a and c = a. We then determine a by normalizing|λ = 1:

|λ = 1 = a√ 2a

a

, ⇒ 1 = λ = 1|λ = 1 = a∗ √ 2a∗ a∗

a√ 2a

a

= 4|a|2, ⇒ a = 1

2

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(where I have chosen the arbitrary phase to be 1). Thus, and doing the same thing for

λ = 0 and λ = −1, gives

|λ = 1 = 12

1√

2

1

, |λ = 0 = 1√

2

1

0

−1

, |λ = −1 = 1

2

1

−√ 21

.

(4)

Solution: The possible outcomes are Lx = {1, 0, −1}, which are the eigenvalues of Lx.|ψ is the normalized eigenstate of Lz with eigenvalue Lz = −1, which is

|ψ =00

1

.

So (here P stands for "probability of"):

P (Lx = 1) = |λ = 1|ψ|2

=

1

2

1 √ 2 10

01

2

=

1

4 ,

P (Lx = 0) = |λ = 0|ψ|2 =

1

2

1 0 −1

00

1

2

= 1

2,

P (Lx = −1) = |λ = −1|ψ|2 =

1

2

1 −√ 2 1

00

1

2

= 1

4.

(5)

Solution:

L2z =

1 0

1

, ⇒ the possible outcomes are L2z = {0, 1}.

An eigenbasis of the L2z = 1 eigenspace is {|a, |b} with

|a =10

0

, |b =

00

1

.

Therefore, upon measuring L2z = 1, the state collapses to

|ψ −→ |ψ = (

|a

a|

+|b

b|)|ψ|(|aa| + |bb|)|ψ| .

But

[|aa| + |bb|] |ψ =10

0

1 0 0+

00

1

0 0 1

1

2

11√

2

=

10

0

1

2+

00

1

1√

2=

1

2

10√

2

,

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has norm 12

1 0

√ 2 1

2

10√

2

=

√ 3

2 ,

so

|ψ = 1√ 3/2

12 10√

2

= 1√ 3

10√ 2

.The probability of L2z = +1 is

P (L2z = 1) = ψ| (|aa| + |bb|) |ψ = |a|ψ|2 + |b|ψ|2

=

1

2

1 0 0

11√ 2

2

+

1

2

0 0 1

11√ 2

2

= 1

4 +

1

2 =

3

4.

If we measured Lz the posible outcomes are the eigenvalues Lz, {0, ±1}, with probabilities

P (Lz = 1) = 1 0 0 |ψ

2

= 1

√ 3 1 0 01

0

√ 22

= 1

3

.

P (Lz = 0) =0 1 0 |ψ2 =

1√

3

0 1 0

10√ 2

2

= 0.

P (Lz = −1) =0 0 1 |ψ2 =

1√

3

0 0 1

10√ 2

2

= 2

3.

(6)

Solution: In the Lz eigenbasis

|Lz = 1 =10

0

, |Lz = 0 = 010

, |Lz = −1 = 001

,write the unknown state as

|ψ =ab

c

.

Then

P (Lz = 1) = 14

= |Lz = 1|ψ|2 =

1 0 0

abc

2

= |a|2,

P (Lz = 1) = 12

= |Lz = 0|ψ|2 =

0 1 0ab

c

2

= |b|2,

P (Lz = 1) = 14

= |Lz = −1|ψ|2 =

0 0 1ab

c

2

= |c|2.

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The most general solution to these three equations is then

a = 1

2eiδ1 , b =

1√ 2

eiδ2 , c = 1

2eiδ3 ,

for some arbitrary phases δ i, which gives the desired answer.

The δ i

phase factors are not irrelevant. For example

P (Lx = 0) = |λ = 0|ψ|2 =

1√ 2

1 0 −1 1

2

eiδ1√ 2eiδ2

eiδ3

2

=

eiδ12√ 2 − eiδ3

2√

2

2

= 1

8

eiδ1 − eiδ3 e−iδ1 − e−iδ3 = 1

8

1 − ei(δ3−δ1) − e−i(δ3−δ1) + 1

=

1

4 (1 − cos(δ 3 − δ 1)) ,

so something measurable (a probability) depends on the difference of the phases.

Problem 2: Do exercise 4.2.2 of the text.

Solution:

P = ψ|P |ψ = ∞−∞

dxψ|xx|P |ψ

=

∞−∞

dx ψ∗(x)

−i d

dx

ψ(x) = −i

∞−∞

dx ψ(x)dψ(x)

dx

= − i 2

∞−∞

dx d

dx

ψ(x)2

= −i

2 ψ2

∞−∞ = 0

if ψ → 0 as |x| → ∞.Alternatively, use the k-basis:

P = ψ|P |ψ = ∞

−∞dkψ|kk|P |ψ = ∞

−∞dk k ψ|kk|ψ = ∞

−∞dk k ψ∗(k)ψ(k).

But

ψ(k) = k|ψ = ∞−∞

dx k|xx|ψ = 1√ 2π

∞−∞

dx eikxψ(x),

therefore

ψ(−k) = 1√ 2π

∞−∞

dx e−ikxψ(x) = ψ∗(k)

since ψ(x) is real. So

P = ∞−∞

dk kψ∗(k)ψ(k) =

∞−∞

dk kψ(−k)ψ(k).

and under the change of variables k → −k, this becomes

P = ∞−∞

dk (−k)ψ(k)ψ(−k) = −P ,

and so P = 0.

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Problem 3: Do exercise 2.4.3 of the text.Solution:

eip0x/ψP eip0x/ψ = ∞

−∞dx eip0x/ψ|xx|P

eip0x/ψ = ∞−∞

dx

eip0x/ψ(x)∗

(−i ) ddx

eip0x/ψ(x)

= −i ∞

−∞dx ψ∗(x)e−ip0x/ ip0

e

ip0x/

ψ(x) + e

ip0x/dψ

dx

=

∞−∞

dx ψ∗(x) p0 ψ(x) − i ∞−∞

dx ψ∗(x)dψ

dx

= p0

∞−∞

dx ψ|xx|ψ

+

∞−∞

dx ψ|xx|P |ψ

= p0ψ|ψ + ψ|P |ψ = p0 + P .

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Problem 2

In[3]:= Psi@x_, t_D := HPi H∆ ^ 2 + hbar^ 2 t^ 2 ê H m ^ 2 ∆^ 2LLL ^H−1 ê 2LExp@H−Hx − H p0 ê m L tL ^ 2L ê H∆^ 2 + hbar^ 2 t^ 2 ê H m ^ 2 ∆^ 2LLD;

PlotAEvaluate@Table@Psi@x, tD ê. 8 p0 → 1, ∆ → 1, hbar → 1, m → 1


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Physics 710-711-712 December 4, 2009Problem Set 7

Problem 1: Exercise 5.3.1

The Hamiltonian isH =

12m

P 2 + V r(X )− iV iwhere V r is a real function and V i a real constant. Therefore

H † = 1

2m(P †)2 + V r(X

†)− (i)∗V i = 12m

P 2 + V r(X ) + iV i = H,

so H is not Hermitian.Derivation of the continuity equation. Schrodinger’s equation and its complex con-

jugate in this case read

i ∂ψ∂t

= − 2

2m∇2ψ + V rψ − iV iψ,

−i ∂ψ∗

∂t = −

2

2m∇2ψ∗ + V rψ∗ + iV iψ∗.

Multiplying the first by ψ∗ and teh second by ψ and taking the difference, then dividingby i gives

∂P

∂t = − ∇· j − 2

V iP,

where, as before, P = |ψ|2 and j = (ψ∗ ∇ψ − ψ ∇)/(2mi) are the probability densityand current, respectively. Integrating this over all space, the

∇· j term vanishes (by

the divergence theorem, since we assume j → 0 at infinity), givingdP dt

= −2

V iP ,

where P = d3xP is the total probability. (I can pull V i out of the integral since it isassumed constant in the problem.) Integrating this differential equation gives

P (t) = P (0) e−2V it/.

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Problem 2: Exercise 5.3.4With primes denoting derivative with respect to x,

j :=

2mi [ψ∗ψ − ψ(ψ∗)]

=

2mi

(A∗

e−ixp/

+ B∗

eixp/

)(Aeixp/

+ Be−ixp/

)

− (Aeixp/ + Be−ixp/)(A∗e−ixp/ + B∗eixp/)=

1

2mi

(A∗e−ixp/ + B∗eixp/)(ipAeixp/− ipBe−ixp/)− (Aeixp/ + Be−ixp/)(−ipA∗e−ixp/ + ipB∗eixp/)

= p

2m

|A|2 + AB∗e2ixp/−A∗Be−2ixp/− |B|2+ |A|2 + A∗Be−2ixp/−AB∗e2ixp/− |B|2

= p

m|A|2 − |B|2

.


(a) For x = Ce

ikx + De−ikx.Scattering boundary conditions means we set D = 0 (no incoming particles fromx = +∞).

Now we need to figure out the boundary conditions at x = 0. Look at thetime-independent Schrodinger equation,

− 2

2m

ψ + V 0aδ (x)ψ = Eψ. (1)

Since the potential has an infinite jump in it, ψ will be continuous, but ψ mayhave a finite jump. To see how big the ψ jump is, integrate (1) from x = − tox = + to get

2

2m [ψ(−)− ψ()] + V 0aψ(0) = E

−

dxψ.

In the limit as → 0, the right hand side vanishes since ψ is continuous, fromwhich we learn that

ψ>(0)

−ψgives the two conditions

A + B = C

ikC − ikA + ikB = (2maV 0/ 2)C.

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Dividing through by A, and solving for B/A and C/A gives B/A = maV 0/(ik 2−

maV 0) and C/A = ik 2/(ik 2 − maV 0). Since R = |B/A|2 and T = |C/A|2, we

get

R = m2a2V 20

k2 4 + m2a2V 20, T =

k2 4

k2 4 + m2a2V 20.

(b) Call x < −a region I, |x| < a region II, and x > a region III. Then solving for theenergy eigenstates, ψ = 2m(E − V (x))ψ, of energy 0 < E ≤ V 0 in each regiongives ψI = Ae

ikx + Be−ikx, ψII = Ce−κx + Deκx, and ψII I = Ee

ikx + F e−ikx,with κ =

2m(V 0 −E ) and k =

√ 2mE . Scattering boundary conditions

means we set F = 0 (no incoming particles from x = +∞). The incoming wavehas amplitude A, the reflected has amplitude B, the transmitted amplitude E .Therefore R = |B/A|2 and T = |E/A|2, so we only need to solve for B/A andE/A.

The boundary conditions at x = ±a are that ψ and ψ are continuous, implying

Ae

−ika

+ Be

ika

= Ce

κa

+ De

−κa

,ikAe−ika − ikBeika = −κCeκa + κDe−κa,

Eeika = Ce−κa + Deκa,

ikEeika = −κCe−κa + κDeκa.

Dividing by A and eliminating C/A and D/A gives

B

A =

e−2iak(e4aκ − 1)(k2 + κ2)(e4aκ − 1)(k2 − κ2) + 2i(e4aκ + 1)kκ ,

E

A

= 4ie2a(κ−ik)kκ

(e4aκ

− 1)(k2

− κ2

) + 2i(e4aκ

+ 1)kκ

,

so

R = (e4aκ − 1)2(k2 + κ2)2

(e4aκ − 1)2(k2 − κ2)2 + 4(e4aκ + 1)2k2κ2 ,

and T = 1−R.

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Physics 711 January 15, 2010Problem Set 8


Plug the power series expansion ψ = ∞

n=0 cnyn

into the equation ψ

+(2ε−y2

)ψ = 0to get∞n=0

cnn(n− 1)yn−2 + (2ε− y2)yn = 0.

Shift n → n + 2 in the first term, and n → n− 2 in the third term to get∞n=0

yn [(n + 2)(n + 1)cn+2 + 2εcn − cn−2] = 0

with the convention that

c−2 = c−1 = 0.This implies

cn+2 = cn−2

(n + 1)(n + 2) − 2εcn

(n + 1)(n + 2)

for all n ≥ 0.

Problem 3: Exercise 7.3.7In the momentum basis |ψ → ψ( p), P → p, and X → i (d/dp), so the energy

eigenvalue equation

1

2mP 2 +

mω2

2 X 2

|E = E |E

becomes1

2m p2ψ( p) − mω

2

2

2ψ( p) = Eψ( p).

Compare this to the position-basis equation

mω2

2 x2ψ(x) − 1

2m 2ψ(x) = Eψ(x).

These are the same equations with the substitutions x ↔ p and m ↔ 1/(mω2

).

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n|X |n = ∞

−∞

dxψ∗n(x)·x·ψn(x) = ∞

−∞

dx x·ψ2n(x) since ψn(x) is real

= 0 since x is odd and ψ2

n(x) is even.

n|P |n = ∞

−∞

dxψ∗n(x)(−i ) d

dxψn(x) = (−i )

∞

−∞

dxψnψ

n = −i

2

∞

−∞

dx (ψ2n)

= −i 2 ψ2n

∞−∞

= 0 since ψn → 0 as |x| → ∞.

1|X 2|1 = ∞

−∞

dxψ∗1x2ψ1 =

mω4π

1/2 ∞−∞

dx x2

2xmω

1/22e−mωx

2/

= 2

√ π mω

3/2

∞

−∞

dxx4e−mωx2/ =

2

√ π mω

3/2 3

√ π

4mω

−5/2

= 3

2mω.

1|P 2|1 = ∞

−∞

dxψ∗1(−i )2ψ1 = − 2 2√ π

mω

3/2 ∞−∞

dxxe−mωx2/2

xe−mωx

2/2

= − 2 2√ π

mω

3/2 ∞−∞

dxx2mω

mω

x2 − 3

e−mωx

2/

= − 2 2√ π

mω

5/2·(−1) 3

√ π

4

mω

−3/2

= 3mω

2 .

∆X 2 = 0|X 2|0 =

mω

π

∞

−∞

dxx2e−mωx2/ =

mω

1/2 √ π2

mω

−3/2

=

2mω.

∆P 2 = 0|P 2|0 = − 2

mω

π

∞

−∞

dx e−mωx2/2

e−mωx

2/2

= − 2

√ π

mω

3/2 ∞−∞

dxmω

x2 − 1

e−mωx

2/

=

−

2

√ π mω

3/2

·(

−1)

√ π

2mω

−1/2

= mω

2

.

∴ ∆X ∆P =

2mω·

mω

2 =

2.

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Physics 828: Homework Set No. 3

Due date: Friday, January 27, 2011, 1:00pm

in PRB M2043 (Biao Huang’s office)

Total point value of set: 60 points + 10 bonus points

Problem 1 (20 pts.): Exercise 10.3.5 (Shankar, p. 278)




Problem 5 (5 pts.): Exercise 11.4.2 (Shankar, p. 300). If you correctly derive in closedform the explicit expression for [ P̂ , Ĥ ] you receive 10 bonus points.



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Physics 828 Sketch of Solution to Set 2

Shankar 12.3.7

(1) The two-dimensional harmonic oscillator is obviously invariant under rotations

about the z -axis: the magnitude of the position and momenta are unaltered by rota-

tions around the z -axis. Therefore, the Hamiltonian commutes with the generator of

rotations about the z -axis, Lz.

(2) So we write ψ(ρ, φ) = eimφ REm(ρ) where m is an integer, positive or negative.

In two dimensions since we know the Laplacian we have

−h̄2

2µ

REm +

1

ρREm

+

h̄2m2

2µρ2 +

1

2µω2ρ2

REm(ρ) = EREm .

For small ρ assuming that REm(ρ) ∝ ρk (derivatives decrease the power by unity

increasing its importance for small ρ etc.) we can neglect the potential energy term

and the constant term on the right-hand side. Thus we have

REm + REm

ρ ∝

m2

ρ2 REm ⇒ [k(k − 1) + k] = m

2 ⇒ k2 = m2 .

For the wave function to be normalizable, for ∞

0 d ρ ρ R2Em(ρ) to be finite at the lower

limit k ≥ 0. Thus we have REm(ρ) ρ→0−→ ρ|m|.

(3) For large ρ, terms with inverse powers of ρ including the centrifugal term and

also the constant term on the right-hand side can be neglected. So we have

REm = µ2

ω2

ρ2

h̄2 REm ;

this is identical to the one-dimensional oscillator equation. See the careful analysis on

page 191. Up to powers of ρ the solution is1

REm(ρ) ρ→∞−→ e−

µω2h̄ ρ2 .

So we write as instructed1

REm = −µω

h̄ ρ e−

µω2h̄ρ2 and R Em =

µω

h̄

−1 +

µω

h̄ ρ2 e−

µω

2h̄ρ2 .

Thus we can neglect the constant term in R and recover the solution. One can use this to check

that terms we have neglected are indeed small compared to the terms we have retained. When you

neglect terms it is a good idea to substitute the solution you have obtained and check that the terms

you have neglected are indeed smaller.

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REm(ρ) = U m(ρ) ρ|m| e−

µω2h̄ ρ2 .

(4) Use the dimensionless variable = E h̄ω

and y2 = µωh̄

ρ2. Dividing the radial

equation by h̄ω we have dropping the annoying subscripts

−

1

2

d2

dy2 +

1

y

d

dy −

m2

y2

+

1

2y2

R(y) = R(y) .

(5) We do the substitution and do elementary calculus and obtain the result given:

U +

2|m|+ 1

y

− 2y

U + (2 − 2|m| − 2) U = 0 .

(6) We substitute U (y) = ∞

r=0 C ryr and collect the coefficient of yr. The second

derivative reduces the power by two and so we use the C r+2 term etc.

(r + 2)(r + 1)C r+2 + (2|m|+ 1)(r + 2)C r+2 − 2rC r + (2 − 2|m| − 2)C r = 0

yielding a two-term recursion relation.

(7) We write this as

C r+2C r

= − (2( − |m| − r − 1)

(r + 2)(2|m|+ r + 2) .

First if C 0 is given C 2 and the other even terms can be computed. As r →∞ we have

C r+2C r

→ 2

r .

This implies that U (y) grows as ey2

which overwhelms the e−y2/2 in R pushing it out

of the Hilbert space. So the series must terminate. Thus the boundary condition at

infinity leads as usual to energy quantization.

What about the odd terms? A series only with odd terms (set C 0 = 0 so that all even

terms vanish) is inconsistent since then U (y) ∼ y for small y and thus R(ρ) ∼ ρ|m|+1

inconsistent with our earlier result in (2). This appears to be suggested as an argu-

ment. What if one starts with C 0 and C 1 non-zero? Substituting into the equation for

U we find that the (1/y)(dU/dy) leads to the term C 1/y and there is no other sourceof y−1 terms. Thus C 1 = 0 and therefore, all odd terms vanish.

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Therefore, we have r = 2k and the termination of the series condition yields

= 1 + r + |m| = |m| + 2k + 1 ≡ (n + 1)

with k = 0, 1, , 2, · · ·.

(8) Since |m| = n − 2k for a given n (i.e., for a given energy) the maximum value

of m is n which occurs for k = 0, The azimuthal quantum number m decreases by stepsof 2 until m reaches the value −n. It is easy to se that there are n + 1 allowed values

of m yielding a degeneracy of n + 1. In Cartesian coordinates the energy is nx + ny + 1

in units of h̄ω. So the degeneracy corresponds to the number of ways in which we can

choose two non-negative integers to add up to n. We can choose nx to be any integer

from 0 to n and ny = n − nx. This yields the same degeneracy.

Shankar 12.6.1

(1) Since there is no angular dependence = 0.

(b) We have R(r) ∝ e−r/a0 and for large r (retaining only the dominant terms)

−h̄2

2mR = ER(r) .

Substituting the given form we obtain E = − h̄2

2ma20

.

(c) Clearly the R term and the energy term cancel for all r. If the equation is

valid for all r we must have

−h̄2

2m

2

r R + V (r) R(r) = 0 .

Substituting R(r) = e−r/a0 we obtain

h̄2

ma0r + V (r) = 0 ⇒ V (r) = −

h̄2

ma0r .

Shankar 13.1.1 and 13.1.3 You should be able to fill in the steps. Here are some

steps dropping some subscripts for notational simplicty.

v =∞k=0

C k ρk++1 .

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Substituting into

v − 2v +

e2λ

ρ −

( + 1)

ρ2

v = 0 (1)

we extract the coefficient of ρk+l carefully. Since two derivatives reduce the power of

ρ by two we should start from the term with ρk++2 with coefficient C k+1 for the first

term and similarly for the last term. For the second and third terms which reduce the

power of ρ by unity we can start with the ρk++1 term with coefficient C k. Thus we

have

(k + + 2)(k + + 1)C k+1 − 2(k + + 1)C k + e2λC k − ( + 1)C k+1 = 0 (2)

which yieldsC k+1

C k=

−q 2λ + 2(k + + 1)

(k + + 2)(k + + 1) − ( + 1) (3)

Since λ2 = 2mh̄2W

from 13.1.9. For the numerator to vanish we have

e4λ2 = −2me2

h̄2E = 4(k + + 1)2 .

This yields 13.1.14.

Mathematical aside: Here is a different representation using classical mathemat-

ical physics. Note that the function L(ρ) ≡ ∞

k=0 C k ρk obeys the equation

ρL(ρ) + [ 2( + 1) − 2ρ ]L(ρ) + (q 2λ − 2( + 1))L(ρ) = 0 .

Substituting q 2λ = 2n (where n is an integer eventually) we have

12

ρL + [ ( + 1) − ρ ]L + (n − ( + 1))L = 0 .

Let z = 2ρ we have

z d2L

dz 2 + [2( + 1) − z ]

dL

dz − ( + 1 − n)L = 0 .

We know (with a good mathematical methods course) that the general solution to

zw + (c− z ) w − aw = 0

is given by the confluent hypergeometric function w = 1F 1(a; c; z ). Thus we find that

the solution L(ρ) is 1F 1(+1−n; 2+2; 2ρ) . The particular terminating (for integer n)

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confluent hypergeometric function can be related to the associated Laguerre polynomial.

Shankar 13.3

We are considering the case n = 2 and = 1 and thus k = 0 in the notation of the

text. So we have W = me4

8h̄2 from 13.1.4 and form 13.1.6

ρ =

2mW

h̄2 r =

me2

2h̄2 r =

r

2a0

using 13.1.24. From 13.1.10 since k = 0 and = 1 we have v = C 0ρ2 and thus

R(r) = U (r)

r = C e

− r2a0 r

where C is an overall constant. We know that Y 10 =

34π

cos θ from 12.5.39 which is

normalized. So all we need is that Ce−r/(2a0)r is notmalized when integrated over the

radial coordinate. We have

C 2 ∞0

dr r2 r2 e− r

a0 = C 2 24 a50 .

Thus C =

124a3

0

× 1/a0 and including the normalization from the spherical harmonic

yields the quoted answer.

Shankar 13.5

Since we are asked to compute Ω for stationary states |nm its time derivativevanishes. Thus we have [Ω, H ] = 0 by Ehrenfests’ theorem. So we compute the

commutator for Ω = R · P as ordered. We calculate (using the summation convention)

R jP j,

P iP i2m

=

R j,

P iP i2m

P j = 2ih̄

P · P

2m = 2ih̄ T .

We have used

[R j, P iP i] = P i[R j, P i] + [R j, P i]P i = 2ih̄ P j .

We consider the potential energy term next:

[R jP j , V (R)] = R j[P j, V (R)] = R j

−ih̄

d

dR jV (R)

.

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Note that [P i, V (R)] can be evaluated in the coordinate representation (in Cartesian

coordinates) by acting on a function f (r):

[P i, V (r)]f (r) = −ih̄

∂

∂riV (r)f (r) − V (r)

∂

∂riV (r)f (r)

= −ih̄

∂

∂riV (r) = −ih̄ R· ∇V.

Thus we can write formally [P i, V (R)] = −ih̄∂V (R)/∂Ri. So we need to evaluate R · ∇V (R). In the coordinate representation using spherical coordinates for central

potentials this is just rV (r). For the Coulomb potential we obtain −V (r) and including

the factor of −ih̄ we obtain ih̄ V (r). Substituting into the basic relation we have

2T + V = 0

as asserted. If V (R) ∝ Rn, rV (r) = nV and thus we obtain T = n2V .

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Physics 710 March 12, 2010Problem Set 15

Problem 12.6.1: ψE = Ae−r/a0.

(1) No (θ, φ)-dependence implies ψE ∝ Y 00 , so we must have = 0 and m = 0.(2) Therefore ψE = RE,=0 =

1

rU E,0 which satisfies eqn. (12.6.5) with = 0:

(rψE ) +

2µ

2 [E − V (r)](rψE ) = 0. (1)

As r → ∞, V (r) → 0, which implies in this limit (rψE ) = A(1 − ra0 )e

−r/a0 ≈

−(A/a0)re−r/a0 , so (rψE )

≈ (A/a20)re−r/a0. Therefore, eqn. (1) reads in this

limitA

a20re−r/a0 = −

2µE

2 Are−r/a0,

from which it follows thatE = −

2

2µa20.

(3) Now plug E into (1) and use (rψE ) = A(− 2a0 +

ra20

)e−r/a0 to get

A

−

2

a0+

r

a20

e−r/a0 +

2µ

2

−

2

2µa20− V (r)

Are−r/a0 = 0,

which gives

V (r) = −

2

µa0r.

Problem 12.6.4:

(1) δ 3(r − r) is defined by the property that

d3rδ 3(r − r)f (r) = f (r). So simplycheck:

r2dr sin θdθdφ 1

r2 sin θδ (r − r)δ (θ − θ)δ (φ − φ)f (r,θ,φ)

=

drdθdφδ (r − r)δ (θ − θ)δ (φ − φ)f (r,θ,φ) = f (r, θ, φ).

(2) If r = 0 then ∇2(1r ) = 1

r2∂ ∂r

(r2 ∂ ∂r (1

r))+(angular parts) = 1

r2∂ ∂r

(r2(−1r2 )) = 1

r2∂ ∂r

(−1) =0. When r = 0 the above calculation breaks down since terms are singu-lar there. So consider an arbitrary continuous function f (r) and the integral

d3x∇2(1r )f (r) = lim→0 0

r2dr

dΩ∇2(1r )f (r), since ∇2(1r ) = 0 for r > 0.

Then, integrating by parts we get d3x∇2

1

r

f (r) = lim

→0

0

r2dr

dΩ

∇·

f (r) ∇

1

r

−

∇f (r)

· ∇

1

r

.

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Now recall that in spherical coordinates ∇ = r ∂ ∂r

+ θ 1r

∂ ∂θ

+ φ 1r sin θ

∂ ∂φ

, implying

that ∇(1/r) = r ∂ ∂r

(1/r) = − r/r2, so d3x∇2

1

r

f = lim

→0

0

r2dr

dΩ

∇·

−

r

f

r2

+

r· ∇f 1

r2

.

The second term on the right side vanishes, because as → 0, r· ∇f = (∂f/∂r)|r=0 =const., so:

lim→0

0

r2dr

dΩ

r· ∇f 1r2

= const.· lim→0

0

r2dr

dΩ

1

r2 = const.· lim

→04π = 0.

Therefore, d3x∇2

1

r

f = lim

→0

0

r2dr

dΩ ∇·

−

r

f

r2

= lim

→0

dΩr2

r·

−

r

f

r2

r=

= − lim→0

dΩ f |r= = −f (0) lim→0

dΩ = −f (0) lim→0

4π

= −4πf (0).

In the second step I used the divergence theorem which states R d

3x ∇·g = ∂R

d2a n·g where R is any region, ∂R is its boundary, n is the normal unit vectorto ∂R (pointing out of R) and d2a is the surface area element. In our caseR = {r < }, d2a = dΩ, n = r, and g = − r(f /r2). Thus we have shown that∇2(1/r) = 0 for r = 0 and for any f (r) that

d3x∇2(1/r)f = −4πf (0). This is

the definition of the delta function, so

∇21r = −4πδ 3(r).

Problem 12.6.9: Since = 0, ψ = R(r)Y 00 (θ, φ) = R(r). So the radial equationbecomes, with ψ(r) = (1/r)U (r),

d2

dr2 + k2

U in = 0 r ≤ r0, k ≡

2µ(E + V 0)

2 ,

d2

dr2

− κ2U out = 0 r ≥ r0, κ ≡ −2µE

2 ,

where k and κ are defined to be the positive root. The solutions of these equations are

U in = A sin kr + A cos kr,U out = Be

−κr + Be+κr.

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Physics 710-712 April 2, 2010Problem Set 16

Problem 13.3.1: To say the pion has a range of λ 10−5 angstroms is to say that

a single pion can be localized on this scale: ∆X ∼ λ. From the uncertainty principle∆X ∆P , we then deduce ∆P /λ. Since we assume λ is the smallest scaleon which we can localize the pion, it is plausible that the inequality is saturated, soP ∼ ∆P ∼ /λ. The relation between the energy and momentum is (from specialrelativity) E 2 = m2c4 + c2P 2 ∼ m2c4 + c2 2/λ2. Now, for the notion of a “singlepion” to exist, we must have E 2mc2. (See discussion in text on p. 363). So4m2c4 m2c4 + c2 2/λ2, or mc2 c /(

√ 3λ). Again, since λ is the smallest scale, it is

plausible that the inequality is saturated, giving

mc2 ∼ c √ 3λ∼ 2000 eV

◦

A

1.7× 10−5

◦

A

∼ 100 MeV.

(In reality mπc2 = 140 MeV.)

Problem 13.3.2: Since the kinetic energy is T = 200 eV 0.5 MeV mc2, theelectron is non-relativistic, so we can use T = p2/(2m) = ( p2c2)/(2mc2) which implies pc =

√ 2mc2T . Then

λ = 2π

p =

2π c

pc =

2π c√ 2mc2T

√

2π(2000 eV◦

A) (0.5× 106eV)(200 eV)

=

√ 2π

10

◦

A 1◦

A .

Problem 13.3.3: Recalling E n = −Ry/n2 −13eV/n2, we have

P (n = 2)

P (n = 1) = 4e−(E 2−E 1)/kT = 4e−[(−1/4)−(−1)]13 eV/(kBT ) 4e−105/(T/K )

where I used (kB/eV) ∼ 9 × 10−5K−1. So it is clear that we need T 105 K so thatthe exponent is not very small. For example, if T = 6000 K, then

P (n = 2)

P (n = 1) 4e−105/(6000) 4e−16 ∼ 2× 10−7 1.

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Physics 710-712 May 14, 2010Problem Set 21

Problem 17.2.1: With H 0 = 12m

P 2 + 12

mω2X 2 and H 1 = λX 4, and using that

X = /(2mω)(a + a†), we have:(1):

E 1n = n|H 1|n = λ 2

4m2ω2n|(a + a†)4|n

= λ 2

4m2ω2n|(a2a†2 + aa†aa† + a†a2a† + aa†2a + a†aa†a + a†2a2)|n

= λ 2

4m2ω2n|(6a2a†2 − 12aa† + 3)|n

= λ 2

4m2ω2{6n + 2|(n + 1)(n + 2)|n + 2 − 12n + 1|(n + 1)|n + 1 + 3n|n}

= λ 2

4m2ω2{6(n + 1)(n + 2) − 12(n + 1) + 3} = 3 2λ

4m2ω2(2n2 + 2n + 1),

where in the first line we dropped the zero superscripts from the unperturbed eigen-states; in the second line we kept only terms with equal numbers of a’s and a†’s; in thethird line we used [a, a†] = 1; and in the fourth line we used a†|n = √ n + 1|n + 1.

(2): For any finite value of λ, as n gets large

E 1n∆E

∼ 2λn2/(m2ω2)

ω ∼ n2 λ

m2ω3 1.

Physically, at large x, no matter how small λ is,

V (x) = mω2

2 x2 + λx4 ∼ λx4

for λ = 0.Problem 17.2.2: H = − µ· B = −γ S · B = H 0 + H 1 with H 0 = −γB0S z andH 1 = −γBS x. The H 0 eigenvalues are E 0± = ∓γB0 /2 with eigenstates |±0, the S zeigenstates. Then

E 1± = ±|0H 1|±0 = −γB±|0S x|±0 = −γB

2 ±|0(S + + S −)|±0 = 0,

and

E 2± =m

|±|0H 1|m0|2E 0± − E 0m

= |±|0H 1|∓0|2

E 0± − E 0∓=

γ 2B2

4

|±|0(S + + S −)|∓0|2∓γB0

= ∓ γB2

4B0 |±|0S ±|∓0|2 = ∓ γB

2

4B0 |±|0

(12 ∓ (∓ 1

2))(3

2 ± (∓ 1

2))|∓0|2 = ∓γ B

2

4B0,

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Problem 17.3.2: H = AS 2z + B(S 2x − S 2y) on the spin-1 Hilbert space. In the S z

basis, |m (m = ±1, 0), we have

S z =

1 0 00 0 0

0 0 −1

, S x = √

2

0 1 01 0 1

0 1 0

, S y = i √

2

0 −1 01 0 −1

0 1 0

(see p. 328 of the text). This implies

H = 2

A 0 B0 0 0B 0 A

. (2)Clearly

|m = |0 =

010

is the eigenvector with E = 0. So we only need to look at the |m = |±1 subspacewhere H = H 0 + H 1 with

H 0 = 2A

1 00 1

, H 1 = 2B

0 11 0

on the |m = {|±1} subspace.

H 0 is degenerate: H 0|m = 2A|m for m = ±1. The basis stable under H 1 is the onewhich diagonalizes H 1. Since

0 11 0

11

=

11

and

0 11 0

1−1

= −

1−1

,

the eigenvectors of H

1

are1 ≡ 1√ 2

(|1 + |−1) and−1 ≡ 1√

2(|1− |−1) .

To order O(B), the energy shifts of 1 and −1 are

E 11

=

1H 11 = 2B 1√

2

1 1

0 11 0

11

1√

2= 2B,

E 1−1

=−1

H 1−1 = 2B 1√

2

1 −10 1

1 0

1−1

1√ 2

= − 2B.

So, the eigenvalues, to O(B), areE 1 =

2(A + B), E −1 = 2(A − B), E 0 = 0.

Compare this to the exact eigenvalues of (2) given by 0 = det(H −λ) = λ( 2(B−A) +λ)( 2(B + A)−λ), which implies λ = {0, 2(A−B), 2(A + B)}. So, the O(B) resultsare exact.

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Physics 710-712 May 26, 2010Problem Set 22

Problem 18.2.2: To first order, the amplitude d2m(∞) for the atom to be in the|n = 2, ,m state is

d2m(∞) = − i

∞

−∞

2m|(eE Z )|100e−t2/τ 2eiω21tdt

since the potential H 1(t) = − µ· E = eZ E e−t2/τ 2. Here ω21 = (E 2 − E 1)/ , which fromnow on we’ll just call ω. So we need to evaluate 2m|Z |100. Z is a componentof a vector irreducible tensor operator, Z = T 01 , so by angular momentum selectionrules, only the 210|Z |100 matrix element is non-vanishing. This matrix element iseasily evaluated using the wave functions for the the |210 and |100 states, and thatZ = r cos θ:

210|Z |100 = ∞

0

r2dr

dΩ

1

25πa30

1/2 r

a0e−r/2a0 cos θ

(r cos θ)

1

πa30

1/2

e−r/a0

= 2π

1−1

d(cos θ)cos2 θ

·

∞

0

dr r4e−3r/2a0· 125/2πa40

= 2π·23·

2a03

5 ∞0

dρ ρ4e−ρ· 125/2πa40

= 215/2·3−5·a0.

Therefore

d2m(

∞) = δ 1δ m0

−i eE·

215/2

·3−5

·a0

∞

−∞

e−t2/τ 2eiωtdt

= δ 1δ m0

−ieE

215/2·3−5·a0

√ πτ 2e−ω

2τ 2/2,

and so the probability for the transition is

P (n=2) =,m

|d2m|2 =eE

2215a20

310

πτ 2e−ω

2τ 2/2.

This answer does not depend on the electron spin: since S z is conserved by H 0 and

H 1, there is still only a single final state it can go into.

Problem 18.2.4: The kinetic energy of the emitted electron is 16 keV = 12mv2e =

12mc

2(ve/c)2 = 12(511keV)(ve/c)

2, which implies that ve/c ∼ 1/4. Therefore the time,τ , for emission is τ ∼ a0/ve = (a0/c)/(ve/c) ∼ 4a0/c, since the typical size of the1s electron orbit is r ∼ a0, the Bohr radius. In comparison, the characteristic timescale of the 1s electron, T , is T ∼ a0/vs = (a0/c)/(vs/c) = (a0/c)/α ∼ 140a0/c, since

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the typical velocity of an electron bound in the hydrogen atom is vs/c = α, the finestructure constant. Therefore, T τ , and the sudden approximation is appropriate.

In the sudden approximation, right after emission the 1s electron will be in the samestate, the |100(Z =1) state of hydrogen. Therefore, the amplitude for the electron to bein the

|100

(Z =2) state of (He

3)+ is given by the overlap

(Z =2)100|100(Z =1) = ∞

0

r2dr

dΩ

Z 3

πa30

1/2e−rZ/a0·

1

πa30

1/2e−r/a0

= 4π·23/2

πa30

∞

0

r2dre−3r/a0 = 27/2

a30

a03

3 ∞0

ρ2dρe−ρ = 29/2·3−3,

where in the first line Z = 2 and I used the fact that under changing the hydrogennucleus charge from 1 to Z , all that changes is a0 → a0/Z .

Finally, (Z =2)16, 3, 0|100(Z =1) = 0 since = 3 states are orthogonal to = 0 ones(and the radial part, and therefore Z , does not affect this).

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Problem S18.4.3 (Shankar, page 494):

(1) Show that a gauge transformation on potentials 0φ and 0 A using

( )0, ( , )

t

r t c r t dt φ −∞

′ ′Λ = −

∫ (1.1)

gives ( )1 , 0r t φ = and 1 0 0( , ) A r t A= − ∇Λ

Solution: This is problem 1 of HW8.

(2) Show that if we transform once more to 2φ and 2 A using

311

( , )1

4

A r t d r

r r π

′ ′∇′Λ = −

′−∫ i

(1.2)

then 2( , ) 0 A r t

∇ =i .

Solution: Using equation (18.4.12) from Shankar ( A A′ = − ∇Λ ),we have

312 1

( , )1( , ) ( , )

4

A r t A r t A r t d r

r r π

⎡ ⎤′ ′∇′= + ∇ ⎢ ⎥

′−⎣ ⎦∫

i

(1.3)

Note that the integrand is the gradient of a scalar function, which produces a vector. If

we now take the divergence of this, we get

2 312 1

( , )1( , ) ( , )

4

A r t A r t A r t d r

r r π

⎡ ⎤′ ′∇′∇ = ∇ + ∇ ⎢ ⎥

′−⎣ ⎦∫

i

i i (1.4)

Since the only r-dependence in the integrand comes from the denominator, we can usethe identity

( ) ( )2 31/ 4r r r r πδ ′ ′∇ − = − − (1.5)

Substituting this into (1.4) and doing the integrals over 3d r gives

[ ]2 1 1 1 11

( , ) ( , ) 4 ( , ) ( , ) ( , ) 04 r r

A r t A r t A r t A r t A r t π π ′=

′ ′∇ = ∇ + − ∇ = ∇ − ∇ =i i i i i (1.6)

(3) Verify that 2φ is also zero using 0 0 E ∇ =i .

Solution:

312 1 1

( , )1 1( , )

4

A r t r t d r

c t c t r r φ φ φ

π

⎧ ⎫′ ′∇∂Λ ∂ ⎪ ⎪′= + = − ⎨ ⎬′∂ ∂ −⎪ ⎪⎩ ⎭

∫ i

(1.7)

Incorporating the results of part (1), this becomes

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Physics 215C Homework #1 Solutions

Richard Eager

Department of Physics

University of California; Santa Barbara, CA 93106

Shankar 20.1.1

Derive the continuity equation

∂P

∂t + ∇ · j = 0

where P = ψ†ψ and j = cψ†αψ

The Dirac equation takes the equivalent forms

i ∂ |ψ

∂t =

cα · P + βmc2

|ψ

∂ψ

∂t =

−cα · ∇ −

i

βmc2

ψ

The conjugate equation is

∂ψ†

∂t = −cα · ∇ + i

βmc2ψ†

∂P

∂t = ψ†

∂ψ

∂t +

∂ ψ†

∂t ψ

= −c

ψ†α · ∇ψ + (α · ∇ψ†)ψ

∇ · j = c∇ψ†αψ + ψ†α · ∇ψ

Adding both equations together yields the desired result

∂P

∂t + ∇ · j = 0

Both terms in the Hamiltonian, cα · P and βmc2

are Hermitian, so where didthe relative minus sign come from? To show that P is a Hermitian operator youneed to integrate by parts. If you are working with the L2 inner product youcan drop boundary terms, but when working locally to derive the continuityequation we don’t integrate by parts and get a relative minus sign.

1

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2

Show that the probability current j of the previous exercise reduces in the non-relativistic limit to Eq.(5.3.8) [which is the same as Sakurai Eq.(2.4.16)].

The probability current j = cψ†αψ and α =

0 σσ 0

We can write the wave

function ψ =

χΦ

in terms of its relativistic and non-relativistic components,

Φ and χ respectively.

j = c

χ† Φ†0 σ

σ 0

χΦ

(0.1)

= c

χ†σΦ + Φ†σχ

(0.2)

In the non-relativistic limit (20.2.13)

Φ ≈ σ · π

2mc

j = χ† ˆ p

2mχ + (

ˆ p

2mχ†)χ

which is the non-relativistic current (5.3.8)

j =

2mi (ψ∗∇ψ − ψ∇ψ∗)

from the identification ˆ p = −i ∇.

Shankar 20.2.1

Show that

π × π = iq

c B

where π = P − qAc

.

π × π = P × P − qA

c × P − P ×

qA

c +

qA

c ×

qA

c

= −q

c (A × P + P × A)

= iq

c (A ×∇ + ∇× A)

The simplest way to manipulate operators is to act on a test function ψ

iq

c

(A ×∇ + ∇× A) ψ = iq

c

(A ×∇ψ + ∇× (ψA))

= iq

c (A ×∇ψ + (∇ψ)A + (∇× A)ψ)

= iq

c (∇× A) ψ

= iq

c Bψ

Therefore

π × π = iq

c B

2

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Shankar 20.1.1

Solve for the 4 spinors w that satisfy Shankar Eq. 20.3.3. You may assume thatthe 3- momentum p is along the z-axis. Normalize them to unity, and show that

they are mutually orthogonal.

Equation (20.3.3) isEw = (α · p + βm)w

In terms of the relativistic and non-relativistic components,

E − m −σ · p−σ · p E + m

χΦ

=

00

The solutions are given in equations (20.3.7) and (20.3.8). Choosing the basis10

and

01

for Φ and letting p be in the z direction,

w1,3 =

p/(±E − m)010

w2,4 =

0

p/(±E − m)

01

Orthogonality of the spinors is easy to see using E 2 = p2 + m2.

5

The five terms in 20.2.28 are

P 2

2m Hermitian

V Hermitian

− P 4

8m3c2 Hermitian

iσ · P × [P, V ]

4m2c2 Hermitian

P · [P, V ]

4m2c2 anti-Hermitian

Recall that P̂ is a Hermitian operator, the potential V is assumed to beHermitian (for conservation of probability) and the Pauli matrices σ are Her-mitian. The commutator [X, Y ] of two Hermitian operators is anti-Hermitiansince [X, Y ]† = (XY )† − (Y X )† = Y †X † − X †Y † = Y X − XY = −[X, Y ].The cross product of two Hermitian (vector) operators is again Hermitian since(X × Y )† = X † × Y † = X × Y.

3

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shankar quantum mechanics solution

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