Homework 3 Solutions

AS.171.303: Quantum Mechanics IDue: Tuesday, October 8

1. (a) From equation (3.28), we have the following representations for the rotation gen-erators,

Jx → ~√2

0 1 01 0 10 1 0

Jy →~√2

0 −i 0i 0 −i0 i 0

Jz → ~

1 0 00 0 00 0 −1

Computing [Jx, Jy] is then simply matrix multiplication,

[Jx, Jy] = JxJy − JyJx

=~2

2

0 1 01 0 10 1 0

0 −i 0i 0 −i0 i 0

− ~2

2

0 −i 0i 0 −i0 i 0

0 1 01 0 10 1 0

= ~2

i 0 00 0 00 0 −i

= i~Jz

So we find the result [Jx, Jy] = i~Jz, which is the same result we found in Home-work 1. While this may not seem like an important result, this is a necessary

condition for these matrices to represent the generators of rotations in three-dimensions, and should hold for any basis and any value of j.

(b) This is also simply matrix multiplication, resulting in

1

J2 = J2

x + J2

y + J2

z

=~2

2

1 0 10 2 01 0 1

+~2

2

1 0 −10 2 0−1 0 1

+ ~2

1 0 00 0 00 0 1

= 2~2

1 0 00 1 00 0 1

= 2~2I

The fact that J2 is proportional to the identity matrix I tells us that any statewritten in this representation is an eigenstate of J2. This may seem surprising,but we stated at the very beginning of the problem that our particle had angularmomentum corresponding to j = 1, which automatically forces that particle tobe an eigenstate of J2 with eigenvalue ~

2j(j + 1) = 2~2.

(c) Since we found that J2 is simply proportional to I, we immediately know all threecommutators. The reason for this is that OI = IO = O for any operator O, soany commutator with J2 is simply

[O, J2] = [O, 2~2I] = 2~2[O, I] = 2~2(

OI − IO)

= 0

2. First, let’s find the z-basis representation for Jn. Based on the definition given in theproblem, we find

Jn = Jx sin θ cosφ+ Jy sin θ sinφ+ Jz cos θ

=~

2

((

0 sin θ cosφsin θ cosφ 0

)

+

(

0 −i sin θ sinφi sin θ sinφ 0

)

+

(

cos θ 00 − cos θ

))

=~

2

(

cos θ sin θ e−iφ

sin θ eiφ − cos θ

)

Now that we have this form, we can solve the eigenvalue equation, which takes theform

(

Jn − λ±I)

| ± n〉 =(

~

2cos θ − λ±

~

2sin θ e−iφ

~

2sin θ eiφ −~

2cos θ − λ±

)(

a±b±

)

= 0

Just like in the previous problem set, we can solve for λ± by knowing that the deter-minant of this matrix is zero,

det(

Jn − λ±I)

| ± n〉 =(

λ2± − ~2

4cos2 θ

)

− ~2

4sin2 θ = λ2± − ~

2

4= 0

→ λ± = ±~

2

2

The reason why we only have ±~

2, and not the more general eiδ ~

2, is because we know

that the eigenvalues of this Hermitian operator must be real.

Using these eigenvalues, we can now solve for the forms of | ±n〉. Starting with |+n〉,we find

(

Jn − λ+I)

|+ n〉 = ~

2

(

cos θ − 1 sin θ e−iφ

sin θ eiφ − cos θ − 1

)(

a+b+

)

= 0

This turns into the two equations,

(cos θ − 1)a+ = − sin θ e−iφb+

sin θ eiφa+ = (cos θ + 1)b+

While it may not look like it, these equations are actually redundant, as eigenvectorequations should be. We then have the relationship

b+ =sin θ

1 + cos θeiφa+

Requiring that 〈+n|+ n〉 = 1 gives us

|a+|2 + |b+|2 = |a+|2(

1 +sin2 θ

(1 + cos θ)2

)

= |a+|2(1 + cos θ)2 + sin2 θ

(1 + cos θ)2= 1

→ |a+|2 =(1 + cos θ)2

2(1 + cos θ)=

1 + cos θ

2= cos2

θ

2

→ a+ = eiδ cosθ

2

As in many problems in the past, the overall phase δ is arbitrary, and can be set tozero. This then gives us

a+ = cosθ

2

b+ = eiφsin θ

1 + cos θcos

θ

2= eiφ cos

θ

2

√

1− cos2 θ

(1 + cos θ)2= eiφ cos

θ

2

√

1− cos θ

1 + cos θ

= eiφ cosθ

2· sin

θ2

cos θ2

= eiφ sinθ

2

The process for determining | − n〉 is basically identical, so we’ll go through it ratherquickly,

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(

Jn − λ+I)

| − n〉 = ~

2

(

cos θ + 1 sin θ e−iφ

sin θ eiφ − cos θ + 1

)(

a−b−

)

= 0

b− =sin θ

cos θ − 1eiφa−

|a−|2 =(cos θ − 1)2

(cos θ − 1)2 + sin2 θ=

1− cos θ

2= sin2

θ

2

a− = sinθ

2

b− =sin θ

cos θ − 1eiφ sin

θ

2= −eiφ sin θ

2

√

1− cos2 θ

(1− cos θ)2

= −eiφ cos θ2

Based on all of this, we have the final expression,

|+ n〉 = a+|+ z〉+ b+| − z〉 = cosθ

2|+ z〉+ eiφ sin

θ

2| − z〉

| − n〉 = a−|+ z〉+ b−| − z〉 = sinθ

2|+ z〉 − eiφ cos

θ

2| − z〉

3. (a) The expectation value of Jz can be found rather easily using this matrix repre-sentation,

〈Jz〉 = 〈ψ|Jz|ψ〉

=(√

1

5

√

2

5

√

2

5

)

~

1 0 00 0 00 0 −1

√

1

5√

2

5√

2

5

= −1

5~

The operator J2z can also be found easily by squaring the representation for Jz,

J2

z = ~2

1 0 00 0 00 0 −1

1 0 00 0 00 0 −1

= ~2

1 0 00 0 00 0 1

We can then calculate 〈J2z 〉 the same way as 〈Jz〉, finding

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〈J2

z 〉 = 〈ψ|J2

z |ψ〉

=(√

1

5

√

2

5

√

2

5

)

~2

1 0 00 0 00 0 1

√

1

5√

2

5√

2

5

=3

5~2

(b) Now that we have 〈Jz〉 and 〈J2z 〉, we can easily find the variance σ2

Jz,

σ2

Jz= 〈J2

z 〉 − 〈Jz〉2 =(

3

5~2

)

−(

−1

5~

)2

=14

25~2

(c) Since we know that J2 = 2~2I, we can actually calculate 〈J2〉 for any state |ψ〉,

〈J2〉 = 〈ψ|J2|ψ〉 = 〈ψ|2~2I|ψ〉 = 2~2〈ψ|ψ〉 = 2~2

This answer shouldn’t surprise us. The expectation value 〈J2〉 tells us the averagevalue we will get if we measure J2 for a very large number of particles all in thesame initial state |ψ〉. However, we know that every measurement we do will resultin 2~2, because we specified that our particle has angular momentum j = 1, sothe average value must also be 2~2.

To find σ2

J2, we must first calculate 〈(J2)2〉. This can also be done rather simply,

since we know that the operator (J2)2 must be

(J2)2 =(

2~2I)2

= 4~4I2 = 4~4I

We then have

〈(J2)2〉 = 〈ψ|(J2)2|ψ〉 = 〈ψ|4~4I|ψ〉 = 4~4

Based on these results, we can find the variance σ2

J2,

σ2

J2= 〈(J2)2〉 − 〈J2〉2 =

(

4~4)

−(

2~2)2

= 0

This answer should also not surprise us, because the variance is related to theuncertainty in our measurement of J2. Specifically, the uncertainty is the squareroot of the variance, or σJ2 . However, if every measurement of J2 for any stategives 2~2, then there is no uncertainty in our measurement. We know the valueof J2 for any state, since we already specified that j = 1.

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4. (a) We’ll first compute the anticommutator of each Pauli matrix with itself, which is

{σi, σi} = σ2

i + σ2

i = 2σ2

i

{σ1, σ1} = 2

(

0 11 0

)(

0 11 0

)

= 2

(

1 00 1

)

= 2I

{σ2, σ2} = 2

(

0 −ii 0

)(

0 −ii 0

)

= 2

(

1 00 1

)

= 2I

{σ3, σ3} = 2

(

1 00 −1

)(

1 00 −1

)

= 2

(

1 00 1

)

= 2I

So we see that {σi, σi} = 2I = 2δiiI, exactly as we expected. Now let’s considerthe anticommutator of different Pauli matrices,

{σ1, σ2} = {σ2, σ1} =

(

0 11 0

)(

0 −ii 0

)

+

(

0 −ii 0

)(

0 11 0

)

=

(

i 00 −i

)

+

(

−i 00 i

)

= 0

{σ2, σ3} = {σ3, σ2} =

(

0 −ii 0

)(

1 00 −1

)

+

(

1 00 −1

)(

0 −ii 0

)

=

(

0 i

i 0

)

+

(

0 −i−i 0

)

= 0

{σ3, σ1} = {σ1, σ3} =

(

1 00 −1

)(

0 11 0

)

+

(

0 11 0

)(

1 00 −1

)

=

(

0 1−1 0

)

+

(

0 −11 0

)

= 0

We then see that {σi, σj} = {σj, σi} = 0 (for i 6= j). We then have the full

relationship, {σi, σj} = 2δij I.

(b) The cross product of two vectors yields a new vector, with a form given by

~σ × ~σ =

∣

∣

∣

∣

∣

∣

ı k

σ1 σ2 σ3σ1 σ2 σ3

∣

∣

∣

∣

∣

∣

= [σ2, σ3 ]ı− [σ1, σ3]+ [σ1, σ2]k

= 2iσ1ı+ 2iσ2+ 2iσ3k = 2i~σ

where we obtained the commutation relations from

[σi, σj] =

[

2

~Ji,

2

~Jj

]

=4

~2[Ji, Jj]

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(c) There is a vector identity which tells us that, for any four vectors ~a, ~b, ~c, and ~d,

(

~a×~b)

·(

~c× ~d)

=(

~a · ~c)(

~b · ~d)

−(

~a · ~d)(

~b · ~c)

This then tells us that, for any two vectors ~a and ~b,

(

~σ × ~σ)

·(

~a×~b)

=(

~σ · ~a)(

~σ ·~b)

−(

~σ ·~b)(

~σ · ~a)

Let’s look at the very last term in this equation. We can reverse the order ofthis term, but we need to be careful to take into commutation relations for the ~σterms (the vectors ~a and ~b are just numbers, so they commute with everything).To make this more evident, we can rewrite the dot products as sums,

(

~σ ·~b)(

~σ · ~a)

=(

∑

i

σibi

)(

∑

j

σjaj

)

=∑

i

∑

j

σiσjajbi

=∑

i

∑

j

(

{σi, σj} − σjσi

)

ajbi =∑

i

∑

j

(

2δij I − σjσi

)

ajbi

= 2I(

~a ·~b)

−(

~σ · ~a)(

~σ ·~b)

We can then plug this into the expression above, obtaining

(

~σ × ~σ)

·(

~a×~b)

=(

~σ · ~a)(

~σ ·~b)

− 2I(

~a ·~b)

+(

~σ · ~a)(

~σ ·~b)

= 2(

~σ · ~a)(

~σ ·~b)

− 2I(

~a ·~b)

Rearranging the equation, and using our result to part (b), we find

(

~σ · ~a)(

~σ ·~b)

=1

2

(

~σ × ~σ)

·(

~a×~b)

+ I(

~a ·~b)

= i~σ ·(

~a×~b)

+ I(

~a ·~b)

5. (a) First, let’s find the raising operator J+z. Based on the definition given in theproblem statement, we have

J+z = Jx + iJy =~√2

0 1 01 0 10 1 0

+ i~√2

0 −i 0i 0 −i0 i 0

= ~√2

0 1 00 0 10 0 0

Similarly, for the lowering operator J−z, we find

7

J−z = Jx − iJy = ~√2

0 0 01 0 00 1 0

(b) The representation of the state |j = 1,mz = −1〉 in this basis is simply

|j = 1,mz = −1〉 →

001

Acting with J±z then gives

J+z|j = 1,mz = −1〉 = ~√2

0 1 00 0 10 0 0

001

= ~√2

010

J−z|j = 1,mz = −1〉 = ~√2

0 0 01 0 00 1 0

001

=

000

We therefore see that the raising operator J+z turns the state |j = 1,mz = −1〉into the state |j = 1,mz = 0〉, multiplied an overall factor of ~

√2, and the

lowering operator J−z simply returns 0 (or ‘annihilates’ the state).

(c) We can find the representations of J±x the same way that we found J±z. Startingwith the raising operator J+x, we find

J+x = Jy + iJz =~√2

0 −i 0i 0 −i0 i 0

+ i~

1 0 00 0 00 0 −1

= i~√2

√2 −1 01 0 −1

0 1 −√2

Similarly, for the lowering operator J−x we find

J−x = Jy − iJz = i~√2

−√2 −1 0

1 0 −1

0 1√2

(d) Since |j = 1,mx = −1〉 should be an eigenstate of Jx with eigenvalue −~, weshould simply find Jx|j = 1,mx = −1〉 = −~|j = 1,mx = −1〉. Let’s check thiswith our representation,

8

Jx|j = 1,mx = −1〉 = ~√2

0 1 01 0 10 1 0

1

2

− 1√2

1

2

= −~

1

2

− 1√2

1

2

So far things look good. Now we also know that the lowering operator J−x shouldannihilate the state |j = 1,mx = −1〉, simply giving 0. Checking this, we find

J−x|j = 1,mx = −1〉 = i~√2

−√2 −1 0

1 0 −1

0 1√2

1

2

− 1√2

1

2

=

000

Both checks are satisfied, so our representation is fine.

(e) We know that acting with the raising operator J+x on the state |j = 1,mx = −1〉should give us something proportional to the state |j = 1,mx = 0〉. So to startwe should just calculate

J+x|j = 1,mx = −1〉 = i~√2

√2 −1 01 0 −1

0 1 −√2

1

2

− 1√2

1

2

= ~

i

0−i

We know that the state |j = 1,mx = 0〉 must be proportional to this answer. Inother words, we know that

J+x|j = 1,mx = −1〉 = n~|j = 1,mx = 0〉 = ~

i

0−i

where n is some unknown number that we need to calculate. All we have to dois solve for the value of n such that the state |j = 1,mx = 0〉 is normalized to 1.We can find this by calculating

(n∗〈j = 1,mx = 0|) (n|j = 1,mx = 0〉) = |n|2 =(

−i 0 i)

i

0−i

= 2

→ n = eiφ√2

We notice that we have an arbitrary phase, which we can just set to zero, givingus n =

√2. We therefore have

|j = 1,mx = 0〉 = 1

n~J+x|j = 1,mx = −1〉 =

i√2

0− i√

2

We can check this answer by acting with Jx, because we know that the state|j = 1,mx = 0〉 is an eigenstate of Jx with eigenvalue 0. Evaluating this, we find

9

Jx|j = 1,mx = 0〉 = ~√2

0 1 01 0 10 1 0

i√2

0− i√

2

=

000

So the state we found is, in fact, an eigenstate of Jx with eigenvalue 0.

Next, we need to find |j = 1,mx = +1〉. We can do this through exactly thesame process, except now we act with the raising operator J+x on the state wejust found, |j = 1,mx = 0〉.

J+x|j = 1,mx = 0〉 = i~√2

√2 −1 01 0 −1

0 1 −√2

i√2

0− i√

2

= ~

− 1√2

−1− 1√

2

We then simply need to normalize the state we found, finding

(n∗〈j = 1,mx = +1|) (n|j = 1,mx = +1〉) = |n|2

=(

− 1√2

−1 − 1√2

)

− 1√2

−1− 1√

2

= 2

→ n = eiφ√2

Just like before, we have an overall phase which we can set to zero, giving n =√2.

We therefore have

|j = 1,mx = +1〉 =

−1

2

− 1√2

−1

2

Just to be sure, let’s see if this is an eigenstate of Jx, with eigenvalue ~,

Jx|j = 1,mx = 0〉 = ~√2

0 1 01 0 10 1 0

−1

2

− 1√2

−1

2

= ~

−1

2

− 1√2

−1

2

While this process may have seemed a little long and inefficient, this is a verygeneral way of finding representations of states in more complicated systems. Wecreate raising and lowering operators out of combinations of other operators, findthe state which is annihilated by the lowering operator (the lowest possible state),then keep acting with the raising operator until we find the highest possible state(which is annihilated by the raising operator).

6. We know that we can find the matrix representation for any operator O using therelationship

10

Oij = 〈i|O|j〉

Let’s start with Jz. We know that for j = 3

2there are four eigenstates, corresponding to

mz = +3

2,+1

2,−1

2,−3

2. Just like in the j = 1 case, we will take the first state to be the

one with the largest eigenvalue, |j,mz〉 = |32,+3

2〉. This state will then be represented

by

∣

∣

∣

∣

3

2,+

3

2

⟩

→

1000

Similarly, the next state (with 1 in the second entry) corresponds to |32,+1

2〉, and so

on. To find the first matrix entry in the first column for Jz, we then calculate

⟨

3

2,+

3

2

∣

∣

∣Jz

∣

∣

∣

3

2,+

3

2

⟩

=3

2~

Similarly, the first entry in the second column for Jz corresponds to

⟨

3

2,+

3

2

∣

∣

∣Jz

∣

∣

∣

3

2,+

1

2

⟩

= 0

In general, for any two states |j,mz〉 and |j,m′z〉, we have

〈j,m′z|Jz|j,mz〉 = mz~〈j,m′

z|j,mz〉

which is nonzero only when m′z = mz. Performing these calculations for the rest of the

matrix entries, we find the representation

Jz →1

2~

3 0 0 00 1 0 00 0 −1 00 0 0 −3

This form is very general. Whenever we are working in the z-basis, the matrix repre-senting Jz will simply consist of the eigenvalues on the diagonal (generally in descendingorder), and zero for all of the off-diagonal terms.

Next, let’s find the representations for J±. From equations (3.59) and (3.60) ofTownsend, we know that

J±z|j,mz〉 = ~

√

j(j + 1)−mz(mz ± 1)|j,mz ± 1〉

Similar to above, we then can write a general expression for the matrix entries,

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〈j,m′z|J±z|j,mz〉 = ~

√

j(j + 1)−mz(mz ± 1)〈j,m′z|j,mz ± 1〉

which is nonzero only whenm′z = mz±1. We can then calculate the full representations,

J+z → ~

0√3 0 0

0 0 2 0

0 0 0√3

0 0 0 0

J−z → ~

0 0 0 0√3 0 0 00 2 0 0

0 0√3 0

To get from these expressions to representations of Jx and Jy, we just need to realizethat

Jx =1

2

(

J+z + J−z

)

Jy = − i

2

(

J+z − J−z

)

We therefore have

Jx =~

2

0√3 0 0

0 0 2 0

0 0 0√3

0 0 0 0

+

0 0 0 0√3 0 0 00 2 0 0

0 0√3 0

= ~

0√3

20 0√

3

20 1 0

0 1 0√3

2

0 0√3

20

Jy = − i~2

0√3 0 0

0 0 2 0

0 0 0√3

0 0 0 0

−

0 0 0 0√3 0 0 00 2 0 0

0 0√3 0

= ~

0 −i√3

20 0

i√3

20 −i 0

0 i 0 −i√3

2

0 0 i√3

20

7. Since the Hamiltonian is proportional to the operator J2, any state with definite j isan eigenstate of the Hamiltonian, and the energy can be determined solely by knowingthe value of j. We can see this by seeing the action of H on any state |j,mz〉,

H|j,mz〉 =1

2IJ2|j,mz〉 =

1

2I~2j(j + 1)|j,mz〉

The energy of any state is then simply equal to the eigenvalue,

12

Ej =~2

2Ij(j + 1)

This confirms that the lowest possible energy is E = 0 and does correspond to a statewith j = 0. Since j is only allowed to be an integer, the three next lowest energiescorrespond to j = 1, 2, 3 with values

E1 =~2

I

E2 =3~2

I

E3 =6~2

I

Note that the energy is independent of the value ofmz, since it only depends on j. Thatmeans that there are multiple states which each have the same energy, correspondingto states with different mz values but the same j value.

It turns out that for any given j, there are 2j + 1 different possible values for mz,which means there are 2j+1 linearly independent eigenstates which all have the sameenergy. We therefore see that there is only one possible state with the lowest energy(mz = 0), but three possible states with the next lowest energy (mz = +1, 0,−1). Forthe three energies we considered, we then have

j = 1 → 3 states: mz = +1, 0,−1

j = 2 → 5 states: mz = +2,+1, 0,−1,−2

j = 3 → 7 states: mz = +3,+2,+1, 0,−1,−2,−3

j → 2j + 1 states: mz = +j,+(j − 1), ...,−(j − 1),−j

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