Tor KjellssonStockholm University

Chapter 4

4.15

Q. A hydrogen atom starts out in the following linear combination of the sta-tionary states n, l,m = 2, 1, 1 and n, l,m = 2, 1,−1:

Ψ(r, 0) =1√2

(ψ211 + ψ21−1) . (1)

a)Q. Construct Ψ(r, t) and simplify it as much as you can.

Sol:The normalized hydrogen wave functions are:

ψnlm =

√(2

na

)3(n− l − 1)!

2n [(n+ l)!]3 e−

rna

(2r

na

l)[L2l+1n−l−1(2r/na)

]Y ml (θ, φ) (2)

where L2l+1n−l−1(x) and Y ml (θ, φ) are the Laguerre polynomials and Spherical har-

monics. In Griffiths (second edition) you find the formal definitions in chapter4.2 and 4.1 respectively. A few examples have also been listed on p.139 andp.153 which we will refer the reader to.

What we do now is to simply put in our two sets of {n, l,m} into eq. (2):

ψ211 =

√(2

2a

)3(2− 1− 1)!

2 · 2 [(2 + 1)!]3 e−

r2a

(2r

2a

1)[L2·1+12−1−1(2r/2a)

]Y 11 (θ, φ)

ψ211 =

√1

a31

4 · 63e−

r2a

( ra

) [L30(r/a)

]Y 11 (θ, φ)

Now we look up the Laguerre polynomial on p.153 and the Spherical Harmonicon p.139 that we need:

L30(r/a) = 6 and Y 1

1 (θ, φ) = −√

3

8πsin(θ)eiφ.

Thus we have arrived at the following expression:

ψ211 = −√

1

a564πre−

r2a sin(θ)eiφ . (3)

1

The same procedure for n, l,m = 2, 1,−1 now gives:

ψ21−1 =

√1

a31

4 · 63e−

r2a

( ra

) [L30(r/a)

]Y −11 (θ, φ).

Checking the new Spherical harmonic on p.139 we now obtain:

ψ21−1 =

√1

a564πre−

r2a sin(θ)e−iφ

and thus:

Ψ(r, 0) =1√2

(ψ211 + ψ21−1)

Ψ(r, 0) =1√2

(−√

1

a564πre−

r2a sin(θ)eiφ +

√1

a564πre−

r2a sin(θ)e−iφ

)

Ψ(r, 0) = −√

1

2a564πre−

r2a sin(θ)

(eiφ − e−iφ

)Ψ(r, 0) = −

√1

a532πi re−

r2a sin(θ) sin(φ)

and to get the full (time-dependent) wave function we just multiply with theusual exponential:

Ψ(r, t) = −√

1

a532πi re−

r2a sin(θ) sin(φ) e−iE2t/~

which contains the time dependence and the energy of the state.

b)Q. Find the expectation value of the potential energy, 〈V 〉. Does it depend ont? Give both the formula and the actual number, in electron volts.

Sol:The expectation value is, like always, given by:

〈V 〉 = 〈Ψ|V |Ψ〉

and when the states are functions this is given by the integral (evaluated overthe space):

〈V 〉 =

∫Ψ∗VΨ dr =

∫Ψ∗VΨ dr (4)

we could drop the hat on V because this is a function and technically not anoperator. Otherwise you leave the hat on until you can present what it doesto the object it will act on. Also, note the bold integration letter - this is a

2

notation for a volume integral, which we of course must use because we mustconsider the whole space the wave functions are defined in.

Now we simplify the integrand:

Ψ∗VΨ =[√1

a532πi re−

r2a sin(θ) sin(φ) eiE2t/~

]V (r)

[−√

1

a532πi re−

r2a sin(θ) sin(φ) e−iE2t/~

]=

1

a532πr2e−

ra sin2(θ) sin2(φ)V (r)

Insert this now into eq. (4) and insert the potential V (r) = V (r) = − e2

4πε0r:

〈V 〉 =−1

a532π

∫r2e−

ra sin2(θ) sin2(φ)

e2

4πε0rdr.

Note that r = (x, y, z) is given in the cartesian coordinate system while ourfunctions are represented using spherical coordinates. Making the transition tospherical coordinates in the integral we get:

dr = dxdydz = r2 sin(θ)drdθdφ

〈V 〉 = − e2

128a5π2ε0

∫ ∞r=0

r3e−ra dr

∫ π

θ=0

sin3(θ) dθ

∫ 2π

φ=0

sin2(φ) dφ.

The last two integrals can be evaluated fairly easily but the first might most ofyou be unfamiliar with. You can evaluate it with the formula:∫ ∞

r=0

rne−ra dr = n! an+1

and you can find it in Physics Handbook as the Gamma function. Anyhow,evaluating the three integrals we get:

〈V 〉 = − e2

128a5π2ε0·[6a4]·[

4

3

]· [π] = − e2

16aπε0.

a is the Bohr radius and is given by the expression:

a ≡ 4πε0~2

me2(5)

so inserting this into our expectation value of the potential energy we obtain:

3

〈V 〉 = − e2

16πε0· me2

4πε0~2= −1

2· m

2~2

(e2

4πε0

)2

=1

2E1 = −6.8 eV

which is, as we see, independent of t.

4.18

Q. The raising and lowering operators change the value of m by one unit:

L±fml = (Aml ) fm±1l (6)

where Aml is some constant. What is this constant if the eigenfunctions fml areto be normalized?

Hint: First show that L∓ is the hermitian conjugate of L± (since Lx and Ly areobservables you may assume that they are hermitian but prove it if you like).Then use:

L2 = L±L∓ + L2z ∓ ~Lz (7)

to get the answer:

Aml = ~√l(l + 1)−m(m± 1) = ~

√(l ∓m)(l ±m+ 1). (8)

Sol:What do we know about fml ? They are the eigenfunctions of L2 and Lz, thatis:

L2fml = ~2l(l + 1)fml and Lzfml = ~mfml (9)

and since the states are normalized you get the eigenvalues by computing theexpectation value of the operators:

〈fml |L2fml 〉 = 〈fml |~2l(l + 1)fml 〉 = ~2l(l + 1)〈fml |fml 〉 = ~2l(l + 1) (10)

〈fml |Lzfml 〉 = 〈fml |~mfml 〉 = ~m〈fml |fml 〉 = ~m. (11)

Using eq. (7) we obtain:

〈fml |L2fml 〉 = 〈fml |L±L∓fml 〉+ 〈fml |L2zfml 〉+ 〈fml | ∓ ~Lzfml 〉. (12)

4

However, looking at the first term on the right hand side we note that it is thereversed order from what we want. That is, we want to compute the coefficientsAml when L± acts on fml , but we have L∓. We interchange the top andbottom cases to obtain:

〈fml |L2fml 〉 = 〈fml |L∓L±fml 〉+ 〈fml |L2zfml 〉+ 〈fml | ± ~Lzfml 〉. (13)

Note that the signs in front of ~Lz switched according to eq. (7). Now weinsert the expectation values that we earlier computed:

~2l(l + 1) = 〈fml |L∓L±fml 〉+ ~2m2 ± ~2m (14)

where the only (possible) question mark should come from the term ~2m2. Thisis however not strange, L2

z just means that you apply Lz twice in a row. Justshove one extra Lz into eq. (11) and see if you can get this result - it is a goodexercise! (If you get stuck, look in the footnote1.)

So, reshuffling our equation we get:

〈fml |L∓L±fml 〉 = ~2l(l + 1)− ~2m2 ∓ ~2m = ~2 (l(l + 1)−m(m± 1)) (15)

and since we know that the left hand side contains information about our stepoperator this looks promising. We now follow the hint of studying the hermitianconjugate of the step operators by inserting the explicit form of them. Here wedo this for two general functions g, h:

〈g|L±h〉 = 〈g| (Lx ± iLy)h〉 = 〈g|Lxh〉+ i〈g|Lyh〉 (16)

and since we know that Lx, Ly are observables they must be hermitian, so:

〈g|L±h〉 = 〈g|Lxh〉 ± i〈g|Lyh〉

= 〈Lxg|h〉 ± i〈Lyg|h〉

= 〈Lxg|h〉+ 〈∓iLyg|h〉

= 〈(Lx ∓ iLy) g|h〉

= 〈L∓g|h〉.

This means that the hermitian conjugate of L± is just L∓! Thus we get:

〈fml |L∓L±fml 〉 = ~2 (l(l + 1)−m(m± 1))

〈L±fml |L±fml 〉 = ~2 (l(l + 1)−m(m± 1))

〈Aml fm±1l |Aml fm±1l 〉 = ~2 (l(l + 1)−m(m± 1))

|Aml |2〈fm±1l |fm±1l 〉 = ~2 (l(l + 1)−m(m± 1))

|Aml |2 = ~2 (l(l + 1)−m(m± 1))

Aml = ~√l(l + 1)−m(m± 1)

1〈fml |L2

zfml 〉 = 〈fm

l |Lz~mfml 〉 = ~m〈fm

l |Lzfml 〉 = ~m〈fm

l |~mfml 〉 = ~2m2.

5

4.19

Starting with the canonical commutation relations for position and momentum:

[ri, pj ] = i~δij , [ri, rj ] = [pi, pj ] = 0. (17)

where (r1, r2, r3) = (x, y, z) and (p1, p2, p3) = (px, py, pz).

a)Q. Work out the following commutators:

[Lz, x] = i~y, [Lz, y] = −i~x, [Lz, z] = 0 (18)

[Lz, px] = i~py, [Lz, py] = −i~px, [Lz, pz] = 0 (19)

Sol:Insert Lz = xpy − ypx in each commutator and do the algebra. We will startby doing this in detail:

[Lz, x] = [xpy − ypx, x] = [xpy, x]− [ypx, x] (20)

where we used the distributive property of commutators. Now we apply anothercommutation relation that was derived in chapter 3:

[AB,C] = A[B,C] + [A,C]B (21)

and we focus on one commutator at time:

[xpy, x] = x[py, x] + [x, x]x = 0 + 0 (22)

where the first commutator is 0 by the canonical commutation relation and thesecond trivially is 0. Turning now to the other commutator:

[ypx, x] = y[px, x] + [y, x]px = −i~y + 0 (23)

where we used the canonical commutation relations on both commutators. In-serting these results back into our original equation we get:

[Lz, x] = [xpy − ypx, x] = 0− (−i~y) = i~y.

Similarly we get:

[Lz, y] = [xpy − ypx, y] = x [py, y]︸ ︷︷ ︸−i~

+ [x, y]︸ ︷︷ ︸0

py − y [px, y]︸ ︷︷ ︸0

− [y, y]︸︷︷︸0

px = −i~x

and from now on we only write out non-zero commutators using the underbraces.

[Lz, z] = [xpy − ypx, z] = x[py, z] + [x, z]py − y[px, z]− [y, z]px = 0

6

[Lz, px] = [xpy − ypx, px] = x[py, px] + [x, px]︸ ︷︷ ︸i~

py − y[px, px]− [y, px]px = i~py

[Lz, py] = [xpy − ypx, py] = x[py, py] + [x, py]py − y[px, py]− [y, py]︸ ︷︷ ︸i~

px = −i~px

[Lz, pz] = [xpy − ypx, pz] = x[py, pz] + [x, pz]py − y[px, pz]− [y, pz]px = 0

b)Q. Use the results from the previous problem to obtain [Lz, Lx] = i~Ly directlyfrom:

Lx = ypz − zpy, Ly = zpx − xpz, Lz = xpy − ypx. (24)

Sol:Insert the definition of Lx:

[Lz, Lx] = [Lz, ypz − zpy] = [Lz, ypz]− [Lz, zpy] = −[ypz, Lz] + [zpy, Lz] =

−(y [pz, Lz]︸ ︷︷ ︸

0

+ [y, Lz]︸ ︷︷ ︸i~x

pz

)+ z [py, Lz]︸ ︷︷ ︸

i~px

+ [z, Lz]︸ ︷︷ ︸0

py = −i~xpz + i~zpx = i~Ly

c)Q. Evaluate the commutators [Lz, x

2 + y2 + z2] and [Lz, p2x + p2y + p2z].

Sol:

[Lz, x2 + y2 + z2] = [Lz, x

2] + [Lz, y2] + [Lz, z

2]

= −(

[x2, Lz] + [y2, Lz] + [z2, Lz]

)= −

(x [x, Lz]︸ ︷︷ ︸−i~y

+ [x, Lz]︸ ︷︷ ︸−i~y

x+ y [y, Lz]︸ ︷︷ ︸i~x

+ [y, Lz]︸ ︷︷ ︸i~x

y + z [z, Lz]︸ ︷︷ ︸0

+ [z, Lz]︸ ︷︷ ︸0

z

)= 0

[Lz, p2x + p2y + p2z] = [Lz, p

2x] + [Lz, p

2y] + [Lz, p

2z]

= −(

[p2x, Lz] + [p2y, Lz] + [p2z, Lz]

)= −

(px [px, Lz]︸ ︷︷ ︸

−i~py

+ [px, Lz]︸ ︷︷ ︸−i~py

x+py [py, Lz]︸ ︷︷ ︸i~px

+ [py, Lz]︸ ︷︷ ︸i~px

py+pz [pz, Lz]︸ ︷︷ ︸0

+ [pz, Lz]︸ ︷︷ ︸0

pz

)= 0

7

d)Q. Show that the Hamiltonian H = p2/(2m) + V commutes with all threecomponents of L, provided that V only depends on r. (Thus H, L2 and Lz aremutually compatible observables.)

Sol:

[H,L] =1

2m[p2,L] + [V,L]

If we look carefully in the previous problem and especially at the rows withbig parenthesis you see2 that it still sums up to 0 even when you exchange Lzfor either Ly or Lx. Thus all components of and hence the angular momentumvector L commutes with both r2 and p2. So the first commutator is 0 and weturn our interest to the second commutator.

If we assume that the potential is only a function of r =√x2 + y2 + z2 we

can prove that the second commutator is 0 as well. One might claim that:

[L, r2] = 0 , [r2, f(r)] = 0 =⇒ [L, f(r)] = 0 (25)

which certainly is true for a lot of functions. It is however not a matter of coursethat this should be true for all functions3, so we test another thing.

Since the function is only dependent on r we can use spherical coordinates. Inspherical coordinates we have:

L = r× p =~ir× ~∇ (26)

which in spherical coordinates is:

L =~i

(r, 0, 0)×(∂

∂r,

1

r

∂

∂θ,

1

r sin(θ)

∂

∂φ

)(27)

L =

∣∣∣∣∣∣r θ φr 0 0∂∂r

1r∂∂θ

1r sin(θ)

∂∂φ

∣∣∣∣∣∣ = φ1

r

∂

∂θ− θ 1

sin(θ)

∂

∂φ(28)

so we see that (inserting a testfunction g(r)):

[L, f(r)]g(r) =(φ

1

r

∂

∂θ− θ 1

sin(θ)

∂

∂φ

)︸ ︷︷ ︸

L

f(r)g(r)− f(r)

(φ

1

r

∂

∂θ− θ 1

sin(θ)

∂

∂φ

)︸ ︷︷ ︸

L

g(r) (29)

2If not, exchange one of the angular momentum operators and explicitly check this. Re-gardless of which angular momentum operator you insert you will always get two terms thatare 0. The four other terms cancel each other pairwise.

3As an example using matrices, prove that it is not necessarily true that [A,B] = 0,[B,C] = 0 gives [A,C] = 0. Can you think of a simple counter example where this is nottrue? Hint: put B to the identity matrix.

8

but since the partial derivatives are with respect to θ, φ any function of only rwill be regardedas a constant. So the above equation now says:

=

f(r)

(φ

1

r

∂

∂θ− θ 1

sin(θ)

∂

∂φ

)︸ ︷︷ ︸

L

−f(r)

(φ

1

r

∂

∂θ− θ 1

sin(θ)

∂

∂φ

)︸ ︷︷ ︸

L

g(r) (30)

and this has to be 0 because the two terms in the square brackets are identicalup to a minus sign. Thus we have proved that L commutes with any functionthat is purely a function of the radial distance and hence [H,L] = 0.

4.21

Q. Derive the following equation:

L+L− = −~2(∂2

∂θ2+ cot(θ)

∂

∂θ+ cot2(θ)

∂2

∂φ2+ i

∂

∂φ

)(31)

using the definition of the raising/lowering operators:

L± = ±~e±iφ(∂

∂θ± i cot(θ)

∂

∂φ

). (32)

Sol:We will now work with the explicit form of the operators and whenever yo dothat it is a good idea to use a test function. Our testfunction in this case willbe named g(r) and from eq. (32) we get:

L+L−g(r) =

[~eiφ

(∂

∂θ+ i cot(θ)

∂

∂φ

)][−~e−iφ

(∂

∂θ− i cot(θ)

∂

∂φ

)]g(r).

Now we first let L− act on the testfunction:

L+L−g(r) =

[~eiφ

(∂

∂θ+ i cot(θ)

∂

∂φ

)][−~e−iφ

(∂g

∂θ− i cot(θ)

∂g

∂φ

)]and then L+ shall act on L−g(r):

L+L−g(r) = ~eiφ

∂

∂θ

[−~e−iφ

(∂g

∂θ− i cot(θ)

∂g

∂φ

)]︸ ︷︷ ︸

A

+i cot(θ)∂

∂φ

[−~e−iφ

(∂g

∂θ− i cot(θ)

∂g

∂φ

)]︸ ︷︷ ︸

B

9

where we have named two parts to work on separtely in order to avoid algebraicerrors.

A:

A = −~e−iφ ∂2g

∂θ2+ ~e−iφi

∂

∂θ

(cot(θ)

∂g

∂φ

)

= −~e−iφ[∂2g

∂θ2− i ∂

∂θ(cot(θ)) · ∂g

∂φ− i cot(θ) · ∂

∂θ

(∂g

∂φ

)]

= −~e−iφ[∂2g

∂θ2+ i

1

sin2(θ)

∂g

∂φ− i cot(θ)

∂2g

∂θ∂φ

]. (33)

B:

B = −~ ∂

∂φ

(e−iφ

∂g

∂θ

)+ i~ cot(θ)

∂

∂φ

(e−iφ

∂g

∂φ

)

= −~[∂

∂φ

(e−iφ

)· ∂g∂θ

+ e−iφ · ∂∂φ

(∂g

∂θ

)]+i~ cot(θ)

[∂

∂φ

(e−iφ

)· ∂g∂φ

+ e−iφ · ∂∂φ

(∂g

∂φ

)]

= i~e−iφ∂g

∂θ− ~e−iφ

∂2g

∂φ∂θ+ ~e−iφ cot(θ)

∂g

∂φ+ i~e−iφ cot(θ)

∂2g

∂φ2

= ~e−iφ[i∂g

∂θ− ∂2g

∂φ∂θ+ cot(θ)

∂g

∂φ+ i cot(θ)

∂2g

∂φ2

]. (34)

From the expression just above eq. (33):

L+L−g(r) = ~eiφ [A+ i cot(θ)B] (35)

we see that we need to work out A+ i cot(θ)B:

A = −~e−iφ[∂2g

∂θ2+

i

sin2(θ)

∂g

∂φ− i cot(θ)

∂2g

∂θ∂φ

]

i cot(θ)B = i cot(θ) · ~e−iφ[i∂g

∂θ− ∂2g

∂φ∂θ+ cot(θ)

∂g

∂φ+ i cot(θ)

∂2g

∂φ2

].

A normal assumption about functions in physics is that partial derivatives with

respect to different variables commute. Hence ∂2g∂θ∂φ = ∂2g

∂φ∂θ so when we addthe two equations above the terms with mixed partial derivatives cancel eachother out:

10

A+i cot(θ)B = ~e−iφ[−∂

2g

∂θ2+i

−1︷ ︸︸ ︷(−1

sin2(θ)+ cot2(θ)

)∂g

∂φ−cot(θ)

∂g

∂θ−cot2(θ)

∂2g

∂φ2

]

= −~e−iφ[∂2g

∂θ2+ i

∂g

∂φ+ cot(θ)

∂g

∂θ+ cot2(θ)

∂2g

∂φ2

].

Now we are (finally!) done:

L+L−g(r) = ~eiφ (A+ i cot(θ)B)

L+L−g(r) = ~eiφ(−~e−iφ

[∂2g

∂θ2+ i

∂g

∂φ+ cot(θ)

∂g

∂θ+ cot2(θ)

∂2g

∂φ2

])

L+L−g(r) = −~2[∂2

∂θ2+ i

∂

∂φ+ cot(θ)

∂

∂θ+ cot2(θ)

∂2

∂φ2

]g(r)

L+L− = −~2[∂2

∂θ2+ i

∂

∂φ+ cot(θ)

∂

∂θ+ cot2(θ)

∂2

∂φ2

]. (36)

b)Q. Use the following three equations:

L+L− = −~2[∂2

∂θ2+ i

∂

∂φ+ cot(θ)

∂

∂θ+ cot2(θ)

∂2

∂φ2

](37)

Lz =~i

∂

∂φ(38)

L2 = L±L∓ + L2z ∓ ~Lz (39)

to derive:

L2 = −~2[

1

sin(θ)

∂

∂θ

(sin(θ)

∂

∂θ

)+

1

sin2(θ)

∂2

∂φ2

]. (40)

Sol:Now we let the ”upper” version of eq. (39) act on a test function g(r):

L2g(r) = L+L−g(r) + L2zg(r)− ~Lzg(r). (41)

We know the first term from the tedious calculation in the previous problem.The last two terms can be calculated by using eq. (38):

Lzg(r) =~i

∂

∂φg(r) =

~i

∂g

∂φ(42)

11

Lz (Lzg(r)) =~i

∂

∂φ

(~i

∂g

∂φ

)= −~2 ∂

2g

∂φ2. (43)

Now we insert this into eq. (41):

L2g(r) = L+L−g(r) + L2zg(r)− ~Lzg(r).

L2g(r) =

[−~2

(∂2

∂θ2+ i

∂

∂φ+ cot(θ)

∂

∂θ+ cot2(θ)

∂2

∂φ2

)]g(r)+

[−~2 ∂

2

∂φ2

]g(r)−~

[~i

∂

∂φ

]g(r)

so apparently:

L2 = −~2[∂2

∂θ2+ i

∂

∂φ+ cot(θ)

∂

∂θ+ cot2(θ)

∂2

∂φ2+

∂2

∂φ2− i ∂

∂φ

]

L2 = −~2[∂2

∂θ2+ cot(θ)

∂

∂θ+ cot2(θ)

∂2

∂φ2+

∂2

∂φ2

]

L2 = −~2[∂2

∂θ2+ cot(θ)

∂

∂θ+(cot2(θ) + 1

) ∂2

∂φ2

]

L2 = −~2[∂2

∂θ2+

cos(θ)

sin(θ)

∂

∂θ+

1

sin2(θ)

∂2

∂φ2

]

L2 = −~2[

1

sin(θ)

(sin(θ)

∂2

∂θ2+ cos(θ)

∂

∂θ

)+

1

sin2(θ)

∂2

∂φ2

].

Note now that you can re-write the first parenthesis:

sin(θ)∂2

∂θ2+ cos(θ)

∂

∂θ= sin(θ) · ∂

∂θ

(∂

∂θ

)+

∂

∂θ(sin(θ)) · ∂

∂θ

=∂

∂θ

(sin(θ)

∂

∂θ

)where the last manipulation follows from the differrentiation rule for a product.Hence we have arrived at the desired result:

L2 = −~2[

1

sin(θ)

∂

∂θ

(sin(θ)

∂

∂θ

)+

1

sin2(θ)

∂2

∂φ2

].

12

4.24

Q. Two particles of mass m are attached to the ends of a massless rigid rod oflength a. The system is free to rotate in three dimensions about the center (butthe center point itself is fixed).

a)Q. Show that the allowed energies of this rigid rotor are:

En =~2n(n+ 1)

ma2for n = 0, 1, 2, . . . (44)

Hint: first express the classical energy in terms of the total angular momentum.

Sol:Recall that the Hamiltonian operator H always tells you the energy of thesystem:

H|αn〉 = En|αn〉. (45)

You are used to the appearance:

H =p2

2m+ V (x) (46)

but this is just the operators for the kinetic and potential energies of the system!So what does these two operators look like in our current system?

If the particles are not moving at relativistic speed each has the kinetic energymv2/2. Since there are two of them the total kinetic energy of the system ismv2. Since there is no potential energy we thus get our Hamiltonian to be:

H = mv2. (47)

Following the hint given we can try to express this in terms of the angularmomentum of the system:

L = 2mrv = 2ma

2v = mav ⇐⇒ H =

L2

ma2(48)

but we know both the eigenvalues and eigenfunctions of L2! Given an eigen-function |fml 〉 we have:

L2|fml 〉 = ~2l(l + 1)|fml 〉 (49)

so the allowed energies for this system is (calling n = l):

En =~2n(n+ 1)

ma2for n = 0, 1, 2, . . . � (50)

13

b)Q. What are the normalized eigenfunctions for this system and what is the de-generacy of the n:th energy level?

Sol:The normalized eigenfunctions of L2 are the Spherical Harmonics: Y ml (θ, φ).for each l there are 2l+1 values that m can acquire so the degeneracy of the n:thenergy level (remember, we called n = l in the previous problem) is 2n+ 1 .

4.27

Q. An electron is in the spin state:

χ = A

(3i4

). (51)

a)Q. Determine the normalization constant A.

Sol:The normalization condition is χ†χ = 1, so:

χ†χ = A∗(−3i 4

)A

(3i4

)= |A|2 · 25 (52)

which gives:

1 = |A|2 · 25 ⇐⇒ A =1

5eiφ (53)

where the last exponential is called a phase. Now, since we only care aboutabsolute values you don’t need to include a phase, so you might as well put

φ = 0 (from now on, we will always do this). Hence A = 1/5 .

b)Q. Find the expectation values of Sx, Sy and Sz.

Sol:

〈Sx〉 = χ†Sxχ =1

25

(−3i 4

)( 0 ~/2~/2 0

)(3i4

)= 0 (54)

〈Sy〉 = χ†Syχ =1

25

(−3i 4

)( 0 −i~/2i~/2 0

)(3i4

)= −12

25~ (55)

14

〈Sz〉 = χ†Szχ =1

25

(−3i 4

)(~/2 00 −~/2

)(3i4

)= − 7

50~ (56)

c)Q. Find the ”uncertainties” σSx

, σSyand σSz

. (Note, these are not the Paulispin matrices!)

Sol:Recall that a variance can be written as:

σ2a = 〈a2〉 − 〈a〉2 (57)

so we see that we need to work out 〈S2i 〉 for i ∈ {x, y, z}.

S2x =

(0 ~/2

~/2 0

)(0 ~/2

~/2 0

)=

~2

4

(1 00 1

)(58)

S2y =

(0 −i~/2

i~/2 0

)(0 −i~/2

i~/2 0

)=

~2

4

(1 00 1

)(59)

S2z =

(~/2 00 −~/2

)(~/2 00 −~/2

)=

~2

4

(1 00 1

)(60)

which means that:

〈S2x〉 = 〈S2

y〉 = 〈S2z 〉 =

1

25

(−3i 4

) ~24

(1 00 1

)(3i4

)=

~2

4(61)

(this result will always be true since we are operating on normalized states us-

ing the identity matrix multiplied by ~2

4 .)

Inserting this result and our previous results into the equation about the variancewe get:

σ2Sx

= 〈S2x〉 − 〈Sx〉2 =

~2

4(62)

σ2Sy

= 〈S2y〉 − 〈Sy〉2 =

~2

4−(

12

25~)2

=49

2500~2 (63)

σ2Sx

= 〈S2z 〉 − 〈Sz〉2 =

~2

4

(7

50~)2

=576

2500~2 (64)

so:

σSx=

~2, σSy

=7

50~, σSz

=12

25~. (65)

15

d)Q. Confirm that your results are consistent with all three uncertainty principles:

σSxσSy ≥~2

∣∣〈Sz〉∣∣, σSzσSx≥ ~

2

∣∣〈Sy〉∣∣, σSyσSz≥ ~

2

∣∣〈Sx〉∣∣ (66)

Sol:Start by computing the left hand sides:

σSxσSy =7

100~2, σSzσSx =

6

25~2, σSyσSz =

7 · 12

50 · 25~2

and then look at the respective right hand sides:

~2

∣∣〈Sz〉∣∣ =7

100~2,

~2

∣∣〈Sy〉∣∣ =6

25~2,

~2

∣∣〈Sx〉∣∣ = 0 (67)

and in each case we see that the uncertainty principles are fulfilled. (Note thatthe two that are not trivial (where the right hand side is 0) hit the minimumvalues.)

4.29

a)Q. Find the eigenvalues and eigenspinors of Sy.

Sol:We know the matrix representation of this operator:

Sy =

(0 −i~/2

i~/2 0

)(68)

so we can directly solve for the eigenvalues and eigenspinors by solving theeigenvalue equation:

Sy

(ab

)= λ

(ab

)~2

(0 −ii 0

)(ab

)= λ

(ab

)which gives the characteristic equation:

det(Sy − λI) = 0

−λ · (−λ)−(−i~

2

)· i~

2= 0

λ = ±~2.

16

Inserting the eigenvalues into the eigenvalue equation we can then solve for theeigenspinors:

~2

(0 −ii 0

)(ab

)= ±~

2

(ab

)(69)

which gives the linear system of equations:

−ib = ±aia = ±b =⇒ b = ±ia.

note that the two equations are linearily dependent (which comes from the char-acteristic equation.). Also, note that it is important to keep track of the order ofthe plus and minus sign. That is, as I have written the results now, b = ±ia , Imean that for the positive eigenvalue the relation between the two componentsof the eigenspinor is b = ia. If I had gotten b = ∓ia then the componentsof the eigenspinor corresponding to the positive eigenvalue would have beenb = −ia!

Our normalized eigenspinors are thus:

χ(y)± =

1√2

(1±i

). (70)

b)Q. If you measured Sy on a particle in the general (normalized) state:

χ = aχ+ + bχ− (71)

what values might you get and with what probabilities? Check that the probabil-ities add up to 1. Remember that a and b need not be real!

Sol:Just a reminder on the notation of the χ±: if there are no superscripts thenit is the eigenspinors of Sz. Otherwise it is the eigenspinors of the superscriptthat is being used. In this problem I will for clarity include a superscript (z)where I feel it is needed, at some places I don’t include one. Whenever I don’tit still means that it is in the z-basis.

Now, we know that the eigenvalues of Sy (as well as the two other directions)are ±~

2 . To get the probabilities to obtain either of them we have to changethe basis from the z − basis to the y − basis.

I will solve this in two ways that are equivalent. In the first way we rewritethe eigenspinors in the z-basis as linear combinations of the eigenspinors in they-basis:

17

χ(z)+ =

(10

)=

1

2

(1i

)+

1

2

(1−i

)=

1√2

(χ(y)+ + χ

(y)−

)(72)

and

χ(z)− =

(01

)= − i

2

(1i

)+i

2

(1−i

)=−i√

2

(χ(y)+ − χ

(y)−

).

which then gives us:

χ(z) = aχ(z)+ + bχ

(z)− =

a√2

[χ(y)+ + χ

(y)−

]− ib√

2

[χ(y)+ − χ

(y)−

]so:

χ(y) =a− ib√

2χ(y)+ +

a+ ib√2χ(y)− . (73)

Before I answer the question about the probabilities I want to show you a (gen-erally) faster way to change basis. In this way you use the projection operator(see pages 122-123 in Griffiths) to change basis. Assume that you want to know

”how much of χ(z)+ lies along χ

(y)+ ”? You answer the question by using the

projection operator:

χ(y)+

(χ(y)+

)†︸ ︷︷ ︸

proj.oper.

χ(z)+ = c · χ(y)

+ (74)

and the complex constant c is here given by:

χ(y)+

(χ(y)+

)†χ(z)+︸ ︷︷ ︸

c

= χ(y)+ · c = c · χ(y)

+ . (75)

So for our state that is expressed in the z − basis we get:

χ(z) = aχ+ + bχ− =

(ab

)

χ(y) =

[χ(y)+

(χ(y)+

)†︸ ︷︷ ︸proj.oper. for +

+ χ(y)−

(χ(y)−

)†︸ ︷︷ ︸proj.oper. for −

](ab

)

χ(y) = χ(y)+

(χ(y)+

)†(ab

)︸ ︷︷ ︸

c

+χ(y)−

(χ(y)−

)†(ab

)︸ ︷︷ ︸

d

.

So let us now compute the constants c and d:

18

First we must remind ourselves about the eigenspinors of Sy:

χ(y)+ =

1√2

(1i

), χ

(y)− =

1√2

(1−i

)(76)

so:

c =(χ(y)+

)†(ab

)

=1√2

(1 −i

)(ab

)

=a√2− ib√

2

which gives us:

c =a− ib√

2. (77)

Similarily we now compute the constant d:

d =(χ(y)−

)†(ab

)

=1√2

(1 i

)(ab

)

=a√2

+ib√2

which gives us the state (expressed in the y-basis):

χ(y) =a− ib√

2χ(y)+ +

a+ ib√2χ(y)− (78)

just as before. The two procedures that we have just discussed are equivalentand you may choose whichever suits you the most. The last one, using theprojection operator, is generally faster since you don’t need to explicitly expressthe eigenspinors as linear combinations of the other basis - you just use theexpressions as they are!

Now that we have expressed the state in the right basis we get the probabilitiesby computing the magnitude squared of the coefficients:

P

(~2

)=

∣∣∣∣a− ib√2

∣∣∣∣2 =|a|2 − iba∗ + iab∗ + |b|2

2

P

(−~

2

)=

∣∣∣∣a+ ib√2

∣∣∣∣2 =|a|2 + iba∗ − iab∗ + |b|2

2

P

(~2

)+ P

(−~

2

)=

2(|a|2 + |b|2)

2= |a|2 + |b|2 = 1 (79)

19

where the last equality comes from the fact that the original state was normal-ized.

c)Q. If you measured S2

y , what values might you get and with what probabilities?

Sol:Recall from eq. (59) what S2

y looks like in matrix form:

S2y =

~2

4

(1 00 1

)We don’t need to find its eigenstates because they are just:(

10

)and

(01

)(80)

and have the same eigenvalue (the matrix is diagonal and the entries are thesame).

Since there is only one possible value to get, the answer is:~2

4with probability 1.

20

4.31

Q. Construct the spin matrices (Sx, Sy and Sz) for a particle of spin 1. Hint:How many eigenstates of Sz are there? Determine the action of Sz, S+ andS− on each of these states. Follow the procedure used in the text for spin 1/2.Also, study pages 173-174 carefully before attempting this problem.

Sol:First we need to determine the number of eigenstates of Sz

4. This is given bythe total number of different ms we can have: ms ∈ {−s,−s+ 1, · · · , 0, · · · , s}.For a spin 1 particle this is just ms ∈ {−1, 0, 1}. Let us call these states:

|s m〉 =|1 1〉|1 0〉|1 -1〉

By knowing the action of the operators on these states we can determine theirmatrix representation - precisely as was done in the book in section 4.4.1.

From eqns. (4.134-4.136) in the text book we read:

S2|1 m〉 = 2~2|1 m〉

Sz|1 m〉 = ~m|1 m〉

S±|1 m〉 = ~√

2−m(m± 1) |1 (m± 1)〉.

Since we are asked to find the operators on matrix form we make the naturalchoice:

|1 1〉 =

100

|1 0〉 =

010

|1 -1〉 =

001

With this choice, we easily5 obtain Sz:

Sz = ~

1 0 00 0 00 0 −1

and now we proceed with the ladder operators. For the purpose of being a kindtutor, I include some extra steps here:

4This is of course the same number as for Sy , Sx.5If you do not see this, pause for a second and breathe. Now look at what Sz does to these

kets.

21

Generally:Amn = 〈em|A|en〉

for some operator A. So:

S±:Note:

〈1 m|S±|1 n〉 = 〈1 n|~√

2− n(n± 1) |1 (n± 1)〉

= ~√

2− n(n± 1) · 〈1 m|1 (n± 1)〉 =⇒

〈1 m|S+|1 n〉 = ~√

2− n(n+ 1) · δm,(n+1)

〈1 m|S−|1 n〉 = ~√

2− n(n− 1) · δm,(n−1).

Now we need to investigate these matrix elements. Start with the ones for S+:

For m = 1 we will see that only n = 0 will give a contribution:

〈e1|S+|e1〉 = 〈1 1|S+|1 1〉 = 〈1 1|0 = 0

〈e1|S+|e2〉 = 〈1 1|S+|1 0〉 = 〈1 1|√

2~|1 1〉 =√

2~ 〈1 1|1 1〉︸ ︷︷ ︸1

=√

2~

〈e1|S+|e3〉 = 〈1 1|S+|1 -1〉 = 〈1 1|√

2~|1 0〉 =√

2~ 〈1 1|1 0〉︸ ︷︷ ︸0

= 0

so thus far we know that our matrix must have the form:

S+ = ~

0√

2 0

Continuing with the next row we obtain:

〈e2|S+|e1〉 = 〈1 0|S+|1 1〉 = 〈1 0|0 = 0

〈e2|S+|e2〉 = 〈1 0|S+|1 0〉 = 〈1 0|√

2~|1 1〉 =√

2~ 〈1 0|1 1〉︸ ︷︷ ︸0

= 0

〈e2|S+|e3〉 = 〈1 0|S+|1 -1〉 = 〈1 0|√

2~|1 0〉 =√

2~ 〈1 0|1 0〉︸ ︷︷ ︸1

=√

2~

and thus we now know the second row in the matrix:

S+ = ~

0√

2 0

0 0√

2

22

Finally, for the last row all matrix elements are 0:

〈e3|S+|e1〉 = 〈1 -1|S+|1 1〉 = 〈1 -1|0 = 0

〈e3|S+|e2〉 = 〈1 -1|S+|1 0〉 = 〈1 -1|√

2~|1 1〉 =√

2~ 〈1 -1|1 1〉︸ ︷︷ ︸0

= 0

〈e3|S+|e3〉 = 〈1 -1|S+|1 -1〉 = 〈1 -1|√

2~|1 0〉 =√

2~ 〈1 -1|1 0〉︸ ︷︷ ︸0

= 0

S+ = ~

0√

2 0

0 0√

20 0 0

.

The exact same procedure (do it!) for the lowering operator gives:

S− = ~

0 0 0√2 0 0

0√

2 0

.

Now we proceed to construct Sx and Sy. On p.174 we see that we get these as:

Sx =1

2(S+ + S−) Sy = − i

2(S+ − S−)

so:

Sx =~√2

0 1 01 0 10 1 0

Sy = i~√2

0 −1 01 0 −10 1 0

23

4.33

Q. An electron is at rest in an oscillating magnetic field:

B = B0 cos(ωt)k, (81)

where B0 and ω are constants.

a)Q. Construct the Hamiltonian matrix for this system.

Sol:

The Hamiltonian for an electron at rest is:

H = −µ ·B = −γS ·B (82)

where µ is the magnetic dipole moment and γ is the gyromagnetic ratio. So inthis case we have the Hamiltonian:

H = −γSz ·B0 cos(ωt) = −γB0 cos(ωt)~2

(1 00 −1

)(83)

since the magnetic field is pointing in the k-direction.

b)Q. The electron starts out (at t = 0) in the spin-up state with respect to the

x-axis. (that is: χ(0) = χ(x)+ ). Determine χ(t) at any subsequent time.

Beware: this is a time-dependent Hamiltonian, so you cannot get χ(t) in theusual way from stationary states. Fortunately, in this case you can solve thetime-dependent Schrodinger equation:

i~∂χ

∂t= Hχ (84)

directly.

Sol:First we express our state as a column vector:

χ(t) =

(a(t)b(t)

)(85)

and plug it in, together with our Hamiltonian, into the Schrodinger equation:

i~∂χ

∂t= Hχ

24

i~∂

∂t

(a(t)b(t)

)= −γB0 cos(ωt)

~2

(1 00 −1

)(a(t)b(t)

)

i~∂

∂t

(a(t)b(t)

)= −γB0 cos(ωt)

~2

(a(t)−b(t)

)(86)

which gives us a system of equations:

i∂a(t)

∂t= −γB0 cos(ωt)

2a(t)

i∂b(t)

∂t= γ

B0 cos(ωt)

2b(t).

First of all, the two equations are the same with the exception of a minus sign.We can thus focus on one of them (say, the upper one) and then re-use thesolution from that on the other:

ida

dt= −γB0 cos(ωt)

2a

and for this we can separate the variables:

ida

a= −γB0 cos(ωt)

2dt.

Now we integrate both sides and simplify the result:

i

∫da

a= −γB0

2

∫cos(ωt)dt

i log |a|+ C = −γB0

2

1

ωsin(ωt) +D

log |a| = iγB0

2

1

ωsin(ωt) + i(C −D)

a(t) = AeγB02ω sin(ωt)

where A = ei(C−D). Similarly (note the minus sign):

b(t) = Be−iγB02ω sin(ωt).

This gives us our state:

χ(t) =

(Aeγ

B02ω sin(ωt)

Be−iγB02ω sin(ωt)

)

25

and from the initial value we get conditions on A and B :

χ(0) = χ(x)+ =

1√2

(11

)=⇒ A = B =

1√2

(87)

giving:

χ(t) =1√2

(eiγ

B02ω sin(ωt)

e−iγB02ω sin(ωt)

). (88)

c)Q. Find the probability of getting −~/2 if you measure Sx. (Note: this is ofcourse at time t.)

Sol:Recall that we need to express the state in the x-basis in order to answer thisquestion. The coefficient in front of the eigenspinor corresponding to this eigen-

value (χ(x)− ) contains information of the probability to obtain this eigenvalue.

From problem 4.29 we know that we can change basis by projecting the currentstate onto the eigenspinors of the new basis. Thus:

χ(x)(t) = f(t)χ(x)+ + g(t)χ

(x)− (89)

where g(t) is obtained by the projection:

g(t) =(χ(x)−

)†χ(z)(t) =

1√2

(1 − 1)1√2

(eiγ

B02ω sin(ωt)

e−iγB02ω sin(ωt)

).

g(t) =1

2

(eiγ

B02ω sin(ωt) − e−iγ

B02ω sin(ωt)

)g(t) = i sin

(γB0

2ωsin(ωt)

)(90)

which gives the probability:

P

(−~

2

)= |g(t)|2 = sin2

(γB0

2ωsin(ωt)

). (91)

d)Q. What is the minimum field (B0) required to force a complete flip in Sx?

Sol:The system starts out in χ

(x)+ and we want to estimate the minimum value of

B0 so that the system is only in χ(x)− . This would mean that:

26

P

(−~

2

)= 1

sin2

(γB0

2ωsin(ωt)

)= 1

sin

(γB0

2ωsin(ωt)

)= ±1

γB0

2ωsin(ωt) = ±nπ

2∀n ∈ Z+.

Now, since we want the minimum value of B0 we set n = 1 (for larger nthen B0 gets larger.). The term sin(ωt) is of course dependent on the time butit reaches magnitude of ±1, so the minimum value of B0 (which occurs whensin(ωt) = ±1) is:

B0(min) =π2ω

2γ=πω

γ.

27

4.34

a)Q. Apply S− to |1 0〉 and confirm that you get

√2~|1 -1〉.

Sol:Let us first start to clarify the notation that we will use:

i) For the composite system we will write |s m〉 where s can go from |s1 − s2|up to s1 + s2 in integer steps. The m value can then range in integer steps from−s to s.

ii) We could write the composite system in terms of linear combinations of theindividual systems:

|s m〉 =∑i

ci|s1 m1〉|s2 m2〉 (92)

but when s1 = s2 we will instead use the notation:

|s m〉 =∑i

ci|m1 m2〉. (93)

In this problem we consider s1 = s2 = 12 and from now on this is always implied.

The first thing to note is that the total spin can either be 0 or 1.

The values mi can only be ± 12 and we will simply refer to them as ±. Thus the

following is true in our notation:

|12m1〉|

1

2m2〉 = |m1 m2〉 = |+ −〉 (94)

if m1 = + 12 and m2 = − 1

2 .

From the book we know the following:

|s m〉 |m1 m2〉|1 1〉 = |+ +〉

|1 0〉 =1√2

[|+ −〉+ | − +〉] (95)

|1 -1〉 = | − −〉

In the composite system of particle 1 and particle 2, our spin operator is thesum of the respective spin operators:

S− = S(1)− + S

(2)− (96)

in the sense that they each act on ”their own” particle.

28

From the book (eq. 4.136) we know the ladder operators to a general spin state:

S(j)± |sj m〉 = ~

√s(s+ 1)−m(m± 1)|sj (m± 1)〉

so in our case:

S+|−〉 = ~|+〉, S+|+〉 = 0 (97)

S−|−〉 = 0, S−|+〉 = ~|−〉. (98)

Now we use eq (95) and these to get:

S−|1 0〉 =(S(1)− + S

(2)−

) 1√2

[|+ −〉+ | − +〉]

S−|1 0〉 =1√2

[(S(1)− + S

(2)−

)|+ −〉+

(S(1)− + S

(2)−

)| − +〉

]

S−|1 0〉 =1√2

[(~| − −〉+ 0) + (0 + ~| − −〉)

]S−|1 0〉 =

1√2

[2~| − −〉

]S−|1 0〉 =

√2~| − −〉

which we see from eq (95) is:

S−|1 0〉 =√

2~|1 − 1〉.

b)Q. Apply S± to |0 0〉 and confirm that you get 0.

Sol:From the book we see:

|0 0〉 =1√2

[|+ −〉 − | − +〉] (99)

So applying the ladder operators (one at a time):

S+|0 0〉 =(S(1)+ + S

(2)+

) 1√2

[|+ −〉 − | − +〉]

S+|0 0〉 =1√2

[(S(1)+ + S

(2)+

)|+ −〉 −

(S(1)+ + S

(2)+

)| − +〉

]

29

S+|0 0〉 =1√2

[(0 + ~|+ +〉)− (~|+ +〉+ 0)

]S+|0 0〉 = 0. (100)

Similarly:

S−|0 0〉 =(S(1)− + S

(2)−

) 1√2

[|+ −〉 − | − +〉]

S−|0 0〉 =1√2

[(S(1)− + S

(2)−

)|+ −〉 −

(S(1)− + S

(2)−

)| − +〉

]

S−|0 0〉 =1√2

[(~| − −〉+ 0)− (0 + ~| − −〉)

]S±|0 0〉 = 0. (101)

c)Q. Show that |1 1〉 and |1 −1〉 are eigenstates of S2, with the appropriate eigen-values.

Sol:For a state |s m〉 to be an eigenstate with the proper eigenvalue it needs to fulfilthe following equation:

S2|s m〉 = ~2s(s+ 1)|s m〉 (102)

so we need to show the the right hand side is indeed what we get if we startwith the left hand side.

Starting with |1 1〉:

S2|1 1〉 =(S(1) + S2)

)2|1 1〉 =

((S(1)

)2+(S(2)

)2+ 2S(1) · S2)

)|1 1〉

S2|1 1〉 =

((S(1)

)2+(S(2)

)2+ 2S(1) · S2)

)|+ +〉

S2|1 1〉 =

((S(1)

)2+(S(2)

)2+ 2S(1) · S2)

)|+ +〉

S2|1 1〉 =(~2s · (s+ 1)|+ +〉+ ~2s · (s+ 1)|+ +〉+ 2S(1) · S(2)|+ +〉

)

30

S2|1 1〉 =

(~2

1

2· 3

2|+ +〉+ ~2

1

2· 3

2|+ +〉+ 2S(1) · S(2)|+ +〉

). (103)

Remember, |+ +〉 = | 12 +〉| 12 +〉 which explains the first two terms. But whatis the last term? Well:

S(1) · S(2) = S(1)x S(2)

x + S(1)y S(2)

y + S(1)z S(2)

z (104)

so:

S(1) · S(2)|+ +〉 = S(1)x S(2)

x |+ +〉+ S(1)y S(2)

y |+ +〉+ S(1)z S(2)

z |+ +〉. (105)

Recall now that the z-basis is the one used by default, so:

S(1)z S(2)

z |+ +〉 =~2· ~

2|+ +〉. (106)

For the S(1)x S

(2)x |+ +〉+ S

(1)y S

(2)y |+ +〉 we now make use of the matrix rep-

resentation of the operators Sx, Sy:

Sx =~2

(0 11 0

)Sy =

~2

(0 −ii 0

)(107)

Sx|+〉 =~2

(0 11 0

)(10

)=

~2

(01

)=

~2|−〉

Sy|+〉 =~2

(0 −ii 0

)(10

)= i

~2

(01

)= i

~2|−〉

this gives us:

S(1)x S(2)

x |+ +〉+ S(1)y S(2)

y |+ +〉 =

(~2

)2

| − −〉+

(i~2

)2

| − −〉 = 0. (108)

In total we thus get:

S2|1 1〉 =(~2s · (s+ 1)|+ +〉+ ~2s · (s+ 1)|+ +〉+ 2S(1) · S(2)|+ +〉

)S2|1 1〉 =

(3~2

4|+ +〉+

3~2

4|+ +〉+ 2 · ~

2

4|+ +〉

)S2|1 1〉 = 2~2|+ +〉 = 2~2|1 1〉 .

Now I urge you to confirm, in the same manner, that S2|1 -1〉 = 2~2|1 -1〉.

31

4.36

a)

Q. A particle of spin 1 and a particle of spin 2 are at rest in a configurationsuch that the total spin is 3, and its z component is ~. If you measured the zcomponent of the angular momentum of the spin-2 particle what values mightyou get and what is the probability of each one?

Sol:

When you combine angular momentum J1 and J2, the possible results are:Jtot = |J1−J2|, |J1−J2|+ 1, . . . , J1 +J2 and the new m-value (the projectionon an axis) can range from −Jtot to Jtot in integer steps.

In our case we are told that Jtot = 3 and that M = 1 (since the z componentis 1 · ~. cf eq.(4.135 in the text book)). We thus know that the system is in thetotal state |3 1〉.

Recall that this system is built up by eigenstates of the two individual particles.To find the probability of obtaining a certain value for a measurement of one ofthe particles we must first find out how the total system is built up and this isdone using the Clebsch-Gordan table on p.188 in the text book.

From the table labelled 2 × 1 we find the column 3+1 . Then we see that there

are three rows with fractions and to the left of these one can see the m-valuesfor the individual particles. The state is thus6:

|J M〉 =∑

m1+m2=M

Cs1,s2,sm1,m2,m|s1 m1〉|s2 m2〉

|3 1〉 =

√1

15|2 2〉|1 -1〉+

√8

15|2 1〉|1 0〉+

√6

15|2 0〉|1 1〉

So now we can answer the question! A measurement of the z component of thespin 2 particle can yield m · ~, which gives the answers:

1) |2 2〉, → 2 · ~ with P (2~) =1

15

2) |2 1〉, → 1 · ~ with P (~) =8

15

3) |2 0〉, → 0 · ~ with P (0) =6

15

6You need to practice this by yourself - there is no fast way in learning how to use thesetables.

32

b)

Q. An electron with spin down in the state ψ510 of the hydrogen atom. If youcould measure the total angular momentum squared of the electron alone (notincluding the proton spin), what values might you get and what is the probabilityof each?

Sol:

In this case we actually know what the constituent parts of the total system is:

|L ml〉|s ms〉 =∑s

Cs1,s2,sm1,m2,m|J M〉

|1 0〉| 12 - 12 〉 =∑s

Cs1,s2,sm1,m2,m|J M〉 =

why do we know this? It is in the problem statement, try to find it!

This means that this problem is the inverse of the previous problem (there weknew the total system but not the individual parts). Here we now ask ourselveshow we must combine possible total states to obtain the specific combination ofthe individual states |1 0〉| 12 - 12 〉.

To find this out, go to the Clebsch-Gordan (1 × 1/2) table. In the previousproblem you read off a column - in this case we instead read off a row. We arespecifically looking for the row (0 - 12 ). Once you find it you will see:

|1 0〉| 12 - 12 〉 =

√2

3| 32 - 12 〉+

√1

3| 12 - 12 〉

So now we can answer the question of the possible values we might obtain andwith what probabilities when we measure the total angular momentum squared:

1)

∣∣∣∣32 -1

2

⟩→ ~2

3

2

(3

2+ 1

)=

15

4~2 with P

(15

4~2)

=2

3

2)

∣∣∣∣12 -1

2

⟩→ ~2

1

2

(1

2+ 1

)=

3

4~2 with P

(3

4~2)

=1

3

33

4.38

Q. Consider the three-dimensional harmonic oscillator, for which the po-tential is:

V (r) =1

2mω2r2. (109)

a)Q. Show that separation of variables in cartesian coordinates turns this intothree one-dimensional oscillators, and exploit your knowledge of the latter todetermine the allowed energies.

Sol:Consider the Schrodinger equation:

i~∂Ψ(x, t)

∂t= HΨ(x, t) (110)

where

H =p2

2m+ V (x) = − ~2

2m

(∂2

∂x2+

∂2

∂y2+

∂2

∂z2

)+

1

2mω2

(x2 + y2 + z2

). (111)

Assume now a separable solution:

Ψ(x, t) = ψ(x, y, z)T (t) (112)

which gives:

i~ψ(x, y, z)∂T

∂t=

[− ~2

2m

(∂2

∂x2+

∂2

∂y2+

∂2

∂z2

)+

1

2mω2

(x2 + y2 + z2

)]ψ(x, y, z)T

i~1

T

∂T

∂t=

1

ψ(x, y, z)

[− ~2

2m

(∂2

∂x2+

∂2

∂y2+

∂2

∂z2

)+

1

2mω2

(x2 + y2 + z2

)]ψ(x, y, z)

now, from the assumption about separable variables it is implied that the vari-ables must be independent of each other. Seeing that the left hand side is onlya function of T and the right hand side is only a function of spatial parts (youmay not ”divide away” ψ(x, y, z), why7?) we thus deduce that:

i~1

T

∂T

∂t= E (113)

where E is a constant8. Then we re-write the Schrodinger equation on a time-independent form:

7Answer: Because the one on the very far right is acted upon by the operators. You mustfirst let them act before you can try to do something else.

8Try to work out the dimension on the left hand side and you’ll see the appropriateness ofthe letter E. Hint: T (t) should give you a quantity measured in a time unit.

34

Eψ(x, y, z) =

[− ~2

2m

(∂2

∂x2+

∂2

∂y2+

∂2

∂z2

)+

1

2mω2

(x2 + y2 + z2

)]ψ(x, y, z)

which of course is what you all are used to (the time-independent Sch. Eq.)

Assume further now, in our separation of variables, that ψ(x, y, z) = X(x)Y (y)Z(z):

[− ~2

2m

(∂2

∂x2+

∂2

∂y2+

∂2

∂z2

)+

1

2mω2

(x2 + y2 + z2

)]XY Z = EXY Z

− ~2

2m

(Y Z

∂2X

∂x2+XZ

∂2Y

∂y2+XY

∂2Z

∂z2

)+

1

2mω2

(x2 + y2 + z2

)XY Z = EXY Z

and now divide by XY Z:

− ~2

2m

(1

X

∂2X

∂x2+

1

Y

∂2Y

∂y2+

1

Z

∂2Z

∂z2

)+

1

2mω2

(x2 + y2 + z2

)= E.

Lastly, group together terms that are dependent on the same variable:

(− ~2

2m

1

X

∂2X

∂x2+

1

2mω2x2

)+

(− ~2

2m

1

Y

∂2Y

∂y2+

1

2mω2y2

)+

(− ~2

2m

1

Z

∂2Z

∂z2+

1

2mω2z2

)= E.

Now, each of the terms in the round brackets are only dependent on that re-spective variable and our assumption is that they are independent of each other.Since the right hand side is constant it follows that each of these terms mustbe constant! Let us call these constants for Ex, Ey, Ez:

− ~2

2m

1

X

∂2X

∂x2+

1

2mω2x2︸ ︷︷ ︸

Ex

− ~2

2m

1

Y

∂2Y

∂y2+

1

2mω2y2︸ ︷︷ ︸

Ey

− ~2

2m

1

Z

∂2Z

∂z2+

1

2mω2z2︸ ︷︷ ︸

Ez

= E

(where Ex +Ey +Ez = E) so effectively we now have one second order differ-ential equation per variable:

− ~2

2m

∂2X

∂x2+

1

2mω2x2X = ExX (114)

− ~2

2m

∂2Y

∂y2+

1

2mω2y2Y = EyY (115)

− ~2

2m

∂2Z

∂z2+

1

2mω2z2Z = EzZ. (116)

These three differential equations are each equivalent to the one dimensionalHarmonic Oscillator from section 2.3 in the text book! The allowed energies inthat section (for one dimension) are derived to be:

En =

(n+

1

2

)~ω (117)

35

so it applies in each direction in our case:

Ex(n) =

(nx +

1

2

)~ω (118)

Ey(n) =

(ny +

1

2

)~ω (119)

Ez(n) =

(nz +

1

2

)~ω (120)

with:

En = Ex(n) + Ey(n) + Ez(n) = Ez(n) =

(n+

3

2

)~ω (121)

where n = nx + ny + nz and nx, ny, nz goes from 0,1,2....

b)Q. Determine the degeneracy d(n) of En.

Sol:Given the criteria on nx, ny, nz, the degeneracy is given by how many ways wecan get the same value n. The easiest way is to look for some sort of structure:

We start by choosing nx, then look at the possibilities left:

If nx = n then ny = 0 and nz = 0, this gives one way.

If nx = n− 1 then ny = 0, 1 and nz = 1, 0, this gives two ways.

If nx = n− 2 then ny = 0, 1, 2 and nz = 2, 1, 0, this gives three ways.

Since we can continue with this down to:

If nx = 0 then ny = 0, 1, 2, . . . n and nz = n, n − 1, n − 2, . . . , 0, this gives n+1ways.

Summing all these possibilities we obtain:

d(n) =

n+1∑i

i =(n+ 1)(n+ 2)

2. (122)

36

4.40

a)Q. Prove the three-dimensional Virial theorem:

2〈T 〉 = 〈r · ~∇V 〉 (123)

for stationary states.

Sol:In problem 3.31 we proved the one-dimensional Virial theorem:

d

dt〈xp〉 = 2〈T 〉 −

⟨xdV

dx

⟩.

using:

d

dt〈Q〉 =

i

~〈[H, Q]〉+

⟨∂Q

∂t

⟩. (124)

The three-dimensional analogy is now:

d

dt〈r · p〉 =

i

~〈[H, r · p]〉+

⟨∂r · p∂t︸ ︷︷ ︸0

⟩. (125)

Also, since we are considering stationary states any expectation value must beconstant in time:

0 =i

~〈[H, r · p]〉 (126)

so our next step is naturally to study the commutator.

[H, r · p] = [H, r1p1 + r2p2 + r3p3] = −[r1p1 + r2p2 + r3p3, H]

[H, r · p] = −(

[r1p1, H] + [r2p2, H] + [r3p3, H])

[H, r·p] = −(r1[p1, H] + [r1, H]p1 + r2[p2, H] + [r2, H]p2 + r3[p3, H] + [r3, H]p3

)but H = p2

2m + V (r) (omitting hats on p, since we have not included them onthe other components of p):

[H, r · p] = −(r1[p1,

p2

2m+ V (r)] + [r1,

p2

2m+ V (r)]p1 + r2[p2,

p2

2m+ V (r)]

)+

−(

[r2,p2

2m+ V (r)]p2 + r3[p3,

p2

2m+ V (r)] + [r3,

p2

2m+ V (r)]p3

).

37

In the equation above we have 12 commutators if we expand each of the six intotwo commutators. However, a lot of them will actually be 0 and now we huntthose down. Recall problem 4.19 and the canonical commutators:

[rj , pk] = i~δjk, [rj , rk] = [pj , pk] = 0.

From above one sees that any component of p will commute with p2 so:

[pj ,p2

2m+ V (r)] = [pj ,

p2

2m]︸ ︷︷ ︸

0

+[pj , V (r)]

which simplifies two terms in the upper row and one in the lower in our hideousexpression of [H, r · p]:

[H, r · p] = −(r1[p1, V (r)] + [r1,

p2

2m+ V (r)]p1 + r2[p2, V (r)]

)+

−(

[r2,p2

2m+ V (r)]p2 + r3[p3, V (r)] + [r3,

p2

2m+ V (r)]p3

).

Now, the other three commutators can be simplified as well. Any function f(r)must commute with the coordinate variables x, y, z ( r1, r2, r3):

[rj , V (r)] = rjV (r)− V (r)rj = rjV (r)− rjV (r) = 0 (127)

where we can change the order to the factors because they are not operatorsbut pure scalars. Thus:

[rj ,p2

2m+ V (r)] = [rj ,

p2

2m] + [rj , V (r)]︸ ︷︷ ︸

0

which then gives:

[H, r · p] = −(r1[p1, V (r)] + [r1,

p2

2m]p1 + r2[p2, V (r)]

)+

−(

[r2,p2

2m]p2 + r3[p3, V (r)] + [r3,

p2

2m]p3

).

From problem 3.13c) we know that:

[pj , V (r)] = −ih∂V (r)

∂rj(128)

so inserting that we get:

[H, r · p] = −(r1

(−ih∂V (r)

∂r1

)+ [r1,

p2

2m]p1 + r2

(−ih∂V (r)

∂r2

))+

38

−(

[r2,p2

2m]p2 + r3

(−ih∂V (r)

∂r3

)+ [r3,

p2

2m]p3

).

To finish this we now look at:

[rj ,p2

2m] =

1

2m

([rj , p

21 + p22 + p23]

)=

1

2m

3∑k=1

[rj , p2k] = − 1

2m

3∑k=1

([p2k, rj ]

)=

= − 1

2m

3∑k=1

(pk[pk, rj ] + [pk, rj ]pk) = − 1

2m

3∑k=1

(−2i~ · pkδkj)

which we now insert into our commutator:

[H, r · p] = −(r1

(−ih∂V (r)

∂r1

)+

(i~m

)p21 + r2

(−ih∂V (r)

∂r2

))+

−((

i~m

)p22 + r3

(−ih∂V (r)

∂r3

)+

(i~m

)p23

).

which simplifies to:

[H, r · p] = ih

(r1∂V (r)

∂r1+ r2

∂V (r)

∂r2+ r3

∂V (r)

∂r3

)− i~m

(p21 + p22 + p23

)[H, r · p] = ih · r · ~∇V − i~ · p

2

m.

Inserting this back into eq. (126) we get:

i

~〈[H, r · p]〉 = 0

i

~

⟨ih · r · ~∇V − i~ · p

2

m

⟩= 0

which gives:

〈r · ~∇V 〉 =

⟨p2

m

⟩= 2〈T 〉.

39

b)Q. Apply the Virial theorem to the case of Hydrogen and show that:

〈T 〉 = −En; 〈V 〉 = 2En. (129)

Sol:

The Virial theorem states:〈r · ~∇V 〉 = 2〈T 〉.

For Hydrogen we have the potential:

V (r) = − e2

4πε0r

~∇V =∂V (r)

∂rr +

1

r

∂V (r)

∂θθ +

1

r sin(θ)

∂V (r)

∂φφ =

e2

4πε0r2r

〈r · ~∇V 〉 = 〈 e2

4πε0r〉 = −〈V 〉.

Using the Virial theorem we thus get:

−〈V 〉 = 2〈T 〉. (130)

but recall Ehrenfest’s theorem: expectation values obey classical laws. So:

〈T 〉+ 〈V 〉 = En. (131)

Combining the equations we then get:

〈T 〉 − En = 2〈T 〉 ⇐⇒ 〈T 〉 = −En (132)

and also:

〈T 〉+ 〈V 〉 = −En + 〈V 〉 = En =⇒ 〈V 〉 = 2En. (133)

40

c)Q. Apply the Virial theorem to the three-dimensional harmonic oscillator (Prob-lem 4.38) and show that in this case:

〈T 〉 = 〈V 〉 = En/2. (134)

Sol:The same steps as the previous problem:The Virial theorem states:

〈r · ~∇V 〉 = 2〈T 〉.

For 3-D harmonic oscillator we have the potential:

V (r) =1

2mω2r2

~∇V =∂V (r)

∂rr +

1

r

∂V (r)

∂θθ +

1

r sin(θ)

∂V (r)

∂φφ = mω2rr

〈r · ~∇V 〉 = 〈mω2r2〉 = 2〈V 〉.

Using the Virial theorem we thus get:

2〈V 〉 = 2〈T 〉 ⇐⇒ 〈V 〉 = 〈T 〉. (135)

Recall Ehrenfest’s theorem: expectation values obey classical laws. So:

〈T 〉+ 〈V 〉 = En. (136)

Combining the equations we then get:

2〈T 〉 = 2〈V 〉 = En (137)

which is what we wanted to show.

41