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 it is just a function. Also, note the boldintegration letter - this is a notation for a volume integral, which we use becausewe are working in R3.

2

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:

〈V 〉 = − e2

16πε0· me2

4πε0~2= −1

2· m

2~2

(e2

4πε0

)2

=1

2E1 = −6.8 eV

3

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:Recall that L± = Lx ± iLy. Following the hint we first show that (L+)

†= L−

by letting L+ work on a test function g:

〈g|L+g〉 = 〈g| (Lx + iLy) g〉 = 〈g|Lxg〉+ 〈g|iLyg〉 (9)

since Lx and Ly are observables they must be hermitian. It thus follows that:

〈g|L+g〉 = 〈g|Lxg〉+ 〈g|iLyg〉 = 〈Lxg|g〉+ 〈−iLyg|g〉 = 〈L−g|g〉. (10)

(You proved in problem 3.5 that you must complex conjugate any complex constant if

you move it from the ket to the bra.)

Continuing to follow the hint we now use eq.(7). Recall that to find a scalarfrom an operator we must compute some sort of inner product. So wecompute 〈fml |L2fml 〉:

〈fml |L2fml 〉︸ ︷︷ ︸(1)

= 〈fml |L±L∓fml 〉︸ ︷︷ ︸(2)

+ 〈fml |L2zfml 〉︸ ︷︷ ︸

(3)

+ 〈fml | ∓ ~Lzfml 〉︸ ︷︷ ︸(4)

. (11)

On the next page we evaluate the terms.

4

(1) :Recall that fml are eigenfunctions of L2:

L2fml = ~2l(l + 1)fml =⇒ 〈fml |L2fml 〉 = ~2l(l + 1) (12)

(2) :

Here we use (L+)†

= L−:

〈fml |L±L∓fml 〉 = 〈L∓fml |L∓fml 〉 = 〈Aml fm∓1l |Aml fm∓1l 〉 = |Aml |2 〈fm∓1l |fm∓1l 〉︸ ︷︷ ︸1

〈fml |L±L∓fml 〉 = |Aml |2 (13)

(3) :Recall that fml are eigenfunctions of Lz:

Lzfml = ~mfml =⇒ 〈fml |Lzfml 〉 = ~m. (14)

Now we use this to show that:

〈fml |L2zfml 〉 = 〈fml |Lz~mfml 〉 = 〈fml |~m · ~mfml 〉 = (~m)

2 〈fml |fml 〉︸ ︷︷ ︸1

〈fml |L2zfml 〉 = (~m)

2(15)

(4) :From eq.(14) get:

〈fml | ∓ ~Lzfml 〉 = ∓~2m (16)

Inserting the results from (1)-(4) into eq.(11) now gives:

〈fml |L2fml 〉︸ ︷︷ ︸(1)

= 〈fml |L±L∓fml 〉︸ ︷︷ ︸(2)

+ 〈fml |L2zfml 〉︸ ︷︷ ︸

(3)

+ 〈fml | ∓ ~Lzfml 〉︸ ︷︷ ︸(4)

.

~2l(l + 1) = |Aml |2 + (~m)2 ∓ ~2m. (17)

|Aml | = ~√l(l + 1)−m2 ±m = ~

√l(l + 1)−m(m∓ 1). (18)

The last algebraic step in eq.(8) is left out. A more important remark is thefact that eq.(18) does not determine the phase of Aml . The phases introducedby the ladder operators have no physical significance so we are free to set themto 0. However, in more advanced problems than we consider in this book youmust be careful with choosing the phase.

Also note that in eq.(11), term 2, we have used L∓ rather than L±. Therefore,eq.(18) have inverted signs.

5

4.19

Starting with the canonical commutation relations for position and momentum:

[ri, pj ] = i~δij , [ri, rj ] = [pi, pj ] = 0. (19)

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 (20)

[Lz, px] = i~py, [Lz, py] = −i~px, [Lz, pz] = 0 (21)

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

[Lz, x] = [xpy − ypx, x] = [xpy, x]− [ypx, x]. (22)

Focus on one commutator at a the time:

[xpy, x] = x[py, x] + [x, x]x = 0 + 0 (23)

where we applied:1) [AB,C] = A[B,C] + [A,C]B2) eq.(19) for [py, x] = 03) [x, x] = 0 for obvious reasons.

Turning now to the other commutator:

[ypx, x] = y[px, x] + [y, x]px = −i~y + 0 (24)

where we used eq.(19).

Inserting the results from eq.(23,24) into eq.(22) gives:

[Lz, x] = [xpy, x]− [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

For the remaining commutators [Lz, z], [Lz, px], [Lz, py], [Lz, pz] and from nowon we only write out non-zero commutators using the underbraces we show dothe same steps and mark if anything is not 0.

6

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

[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. (25)

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]

In the previous problem we proved that [Lz, p2x + p2y + p2z] = 0. This also

holds for Lx and Ly - just make the exchange in the previous problem andyou will see it. Since all components of L commute with p2 it follows that[L, p2x + p2y + p2z

]= 0 .

Now we will show that [V,L] = 0:

Assume that the potential is only a function of r =√x2 + y2 + z2, as we are

asked to do. Since V only depends on r it is useful to look what L looks like inspherical coordinates:

L = r× p =~ir× ~∇ (26)

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)

Note that all terms in eq.(28) commute with any function V (r) since the onlyoperator dependent on r is the scalar valued 1/r, which certainly commutes

with V (r). This concludes that [L, V (r)] = 0 and thus also that:

[H,L] =1

2m[p2,L] + [V,L] = 0 (29)

8

4.21

Q. Derive the following equation:

L+L− = −~2(∂2

∂θ2+ cot(θ)

∂

∂θ+ cot2(θ)

∂2

∂φ2+ i

∂

∂φ

)(30)

using the definition of the raising/lowering operators:

L± = ±~e±iφ(∂

∂θ± i cot(θ)

∂

∂φ

). (31)

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. (31) 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

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

]. (32)

B:

B = −~ ∂

∂φ

(e−iφ

∂g

∂θ

)+ i~ cot(θ)

∂

∂φ

(e−iφ

∂g

∂φ

)

9

= −~[∂

∂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

]. (33)

From the expression just above eq. (32):

L+L−g(r) = ~eiφ [A+ i cot(θ)B] (34)

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:

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

]. (35)

10

b)Q. Use the following three equations:

L+L− = −~2[∂2

∂θ2+ i

∂

∂φ+ cot(θ)

∂

∂θ+ cot2(θ)

∂2

∂φ2

](36)

Lz =~i

∂

∂φ(37)

L2 = L±L∓ + L2z ∓ ~Lz (38)

to derive:

L2 = −~2[

1

sin(θ)

∂

∂θ

(sin(θ)

∂

∂θ

)+

1

sin2(θ)

∂2

∂φ2

]. (39)

Sol:Now we let the ”upper” version of eq. (38) act on a test function g(r):

L2g(r) = L+L−g(r) + L2zg(r)− ~Lzg(r). (40)

We know the first term from the tedious calculation in the previous problem.The last two terms can be calculated by using eq. (37):

Lzg(r) =~i

∂

∂φg(r) =

~i

∂g

∂φ(41)

Lz (Lzg(r)) =~i

∂

∂φ

(~i

∂g

∂φ

)= −~2 ∂

2g

∂φ2. (42)

Now we insert this into eq. (40):

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

].

11

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

].

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, . . . (43)

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〉. (44)

You are used to the appearance:

H =p2

2m+ V (x) (45)

but this is just the operators for the kinetic and potential energies of the system!So what do 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. (46)

12

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(47)

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

L2|fml 〉 = ~2l(l + 1)|fml 〉 (48)

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

En =~2n(n+ 1)

ma2for n = 0, 1, 2, . . . � (49)

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

). (50)

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 (51)

which gives:

1 = |A|2 · 25 ⇐⇒ A =1

5eiφ (52)

13

where the last exponential is called a phase. Now, like we discussed in thesolution to problem 4.18 the phase carries no physical significance, 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|Sx|χ〉 =1

25

(−3i 4

)( 0 ~/2~/2 0

)(3i4

)= 0 (53)

〈Sy〉 = 〈χ|Sy|χ〉 =1

25

(−3i 4

)( 0 −i~/2i~/2 0

)(3i4

)= −12

25~ (54)

〈Sz〉 = 〈χ|Sz|χ〉 =1

25

(−3i 4

)(~/2 00 −~/2

)(3i4

)= − 7

50~ (55)

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 (56)

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

)(57)

S2y =

(0 −i~/2

i~/2 0

)(0 −i~/2

i~/2 0

)=

~2

4

(1 00 1

)(58)

S2z =

(~/2 00 −~/2

)(~/2 00 −~/2

)=

~2

4

(1 00 1

)(59)

which means that:

〈S2x〉 = 〈S2

y〉 = 〈S2z 〉 =

1

25

(−3i 4

) ~24

(1 00 1

)(3i4

)=

~2

4(60)

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

ing the identity matrix multiplied by ~2

4 .)

14

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

σ2Sx

= 〈S2x〉 − 〈Sx〉2 =

~2

4(61)

σ2Sy

= 〈S2y〉 − 〈Sy〉2 =

~2

4−(

12

25~)2

=49

2500~2 (62)

σ2Sx

= 〈S2z 〉 − 〈Sz〉2 =

~2

4

(7

50~)2

=576

2500~2 (63)

so:

σSx=

~2, σSy

=7

50~, σSz

=12

25~. (64)

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

σSxσSy ≥~2

∣∣〈Sz〉∣∣, σSzσSx ≥~2

∣∣〈Sy〉∣∣, σSyσSz ≥~2

∣∣〈Sx〉∣∣ (65)

Sol:Start by computing the left hand sides:

σSxσSy =7

100~2, σSz

σSx=

6

25~2, σSy

σSz=

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 (66)

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.)

15

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

)(67)

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.

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

~2

(0 −ii 0

)(ab

)= ±~

2

(ab

)(68)

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 signs because they come from inserting two different eigen-values.

Our normalized eigenspinors are thus:

|χ(y)± 〉 =

1√2

(1±i

). (69)

16

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

|χ〉 = a|χ(z)+ 〉+ b|χ(z)

− 〉 (70)

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:The possible values are the eigenvalues of Sy and we know that they are ±~

2 .To get the probabilities we must express |χ〉 in the eigenvectors of Sy. This is

done by using the inner product of |χ〉 with each of the eigenvectors |χ(y)± 〉:

|χ〉 = 〈χ(y)+ |χ〉 |χ

(y)+ 〉+ 〈χ(y)

+ |χ〉 |χ(y)+ 〉 (71)

Working in the usual Sz-basis we have that:

|χ(z)+ 〉 =

(10

)|χ(z)− 〉 =

(01

)(72)

From the previous problem we know then know that the eigenvectors of Sy are:

|χ(y)+ 〉 =

1√2

(1i

), |χ(y)

− 〉 =1√2

(1−i

)(73)

which gives us:

〈χ(y)+ |χ〉 =

1√2

(1,−i)(ab

)=a− ib√

2. (74)

〈χ(y)− |χ〉 =

1√2

(1,+i)

(ab

)=a+ ib√

2. (75)

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

|χ〉 =a− ib√

2|χ(y)

+ 〉+a+ ib√

2|χ(y)− 〉 (76)

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 (77)

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

17

c)Q. If you measured S2

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

Sol:Recall from eq. (58) 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

)(78)

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.

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

1. 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〉1This is of course the same number as for Sy , Sx.

18

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 easily2 obtain Sz:

Sz = ~

1 0 00 0 00 0 −1

and now we proceed with the ladder operators. To find the matrix version ofan operator A, recall that the elements are given by Amn = 〈em|A|en〉:

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 check the different combinations of m and n. Start with theones 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:

2If you do not see this, pause for a second and breathe. Now look at what Sz does to thesekets.

19

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

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:

20

Sx =~√2

0 1 01 0 10 1 0

Sy = i~√2

0 −1 01 0 −10 1 0

4.33

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

B = B0 cos(ωt)z, (79)

where B0 and ω are constants.

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

Sol:

The Hamiltonian for an electron at rest in a magnetic field B is:

H = −µ ·B = −γS ·B (80)

where µ is the magnetic dipole moment and γ is the gyromagnetic ratio. In-serting B we obtain:

H = −γSz ·B0 cos(ωt) = −γB0 cos(ωt)~2

(1 00 −1

)(81)

since the magnetic field is pointing in the z-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

= H|χ〉 (82)

directly.

21

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

|χ(t)〉 =

(a(t)b(t)

)(83)

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

i~∂|χ〉∂t

= H|χ〉

i~∂

∂t

(a(t)b(t)

)= −γB0 cos(ωt)

~2

(1 00 −1

)(a(t)b(t)

)i~d

dt

(a(t)b(t)

)= −γB0 cos(ωt)

~2

(a(t)−b(t)

)(84)

which gives us a system of equations:

ida(t)

dt= −γB0 cos(ωt)

2a(t)

idb(t)

dt= γ

B0 cos(ωt)

2b(t).

Note that the two equations are almost identical. We can thus focus on one ofthem (say, the upper one) and later re-use the derivation.

ida

dt= −γB0 cos(ωt)

2a

now we separate the variables:

ida

a= −γB0 cos(ωt)

2dt

and integrate both sides:

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).

Following the same steps to find b(t) we obtain:

b(t) = Be−iγB02ω sin(ωt).

22

This gives us our state:

|χ(t)〉 =

(Aeγ

B02ω sin(ωt)

Be−iγB02ω sin(ωt)

)and from the initial value condition we can determine A and B :

|χ(0)〉 = |χ(x)+ 〉 =

1√2

(11

)=⇒ A = B =

1√2

(85)

giving:

|χ(t)〉 =1√2

(eiγ

B02ω sin(ωt)

e−iγB02ω sin(ωt)

). (86)

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

Sol:Recall that in order to find the probability of obtaining an eigenvalue in aspecific basis you must first express the state in that basis. Currently our stateis expressed in the z -basis3:

|χ(z)(t)〉 =1√2

(eiγ

B02ω sin(ωt)

e−iγB02ω sin(ωt)

).

Now we change basis by using the inner product of the state in our current basiswith the eigenstates of the new basis:

|χ(x)(t)〉 = 〈χ(x)+ |χ(z)(t)〉 |χ(x)

+ 〉+ 〈χ(x)− |χ(z)(t)〉 |χ(x)

− 〉. (87)

Note here that to answer the question we only need to compute the inner prod-

uct 〈χ(x)− |χ(z)(t)〉 because we are asked about −~/2.

In problem 4.29 you found the eigenstates of Sy. If you follow the same proce-dure for Sx you will find that:

|χ(x)− 〉 =

1√2

(1−1

)(88)

so:

〈χ(x)− |χ(z)(t)〉 =

1√2

(1 −1

) 1√2

(eiγ

B02ω sin(ωt)

e−iγB02ω sin(ωt)

)= i sin

(γB0

2ωsin(ωt)

)(89)

where we in the last step used one of Euler’s formulae.

3Because we got this state from working with Sz = ~2

(1 00 −1

). Since this matrix is diagonal,

it must be expressed in the z-basis.

23

Computing the square of the absolute value of eq.(89) gives the probability:

P

(−~

2

)= sin2

(γB0

2ωsin(ωt)

). (90)

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 changed to |χ(x)− 〉. This would mean that:

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 time but itreaches maximum magnitude of ±1, so the minimum value of B0 (which occursat sin(ωt) = ±1) is:

B0(min) =π2ω

2γ=πω

γ.

24

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) The composite system can be written as a linear combination of the productof the individual systems:

|S M〉 =∑i

Ci|s1 m1〉|s2 m2〉 (91)

The coefficient Ci is called Clebsch-Gordan coefficient, but in this problemwe will not discuss these further.

iii) When s1 = s2 it is customary to only use m1,m2 for brevity:

|s m〉 =∑i

Ci|m1 m2〉. (92)

In this problem we consider s1 = s2 = 12 . 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:∣∣∣∣12 m1

⟩ ∣∣∣∣12 m2

⟩≡ |m1 m2〉 = |+ −〉 (93)

if m1 = + 12 and m2 = − 1

2 .

From the book (eq.(177) we know the following:

|S M〉 |m1 m2〉|1 1〉 = |+ +〉

|1 0〉 =1√2

[|+ −〉+ | − +〉] (94)

|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)− (95)

25

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

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 (96)

S−|−〉 = 0, S−|+〉 = ~|−〉. (97)

Now we use eq (94) 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 (94) 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

[|+ −〉 − | − +〉] (98)

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)+

)| − +〉

]

26

S+|0 0〉 =1√2

[(0 + ~|+ +〉)− (~|+ +〉+ 0)

]S+|0 0〉 = 0. (99)

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. (100)

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 tofulfil the following equation:

S2|S M〉 = ~2S(S + 1)|S M〉 (101)

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〉 = |+ +〉 in the individual systems:

S2|1 1〉 =(S(1) + S2)

)2|+ +〉 =

((S(1)

)2+(S(2)

)2+ 2S(1) · S2)

)|+ +〉

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〉 =(~2s1 · (s1 + 1)|+ +〉+ ~2s2 · (s2 + 1)|+ +〉+ 2S(1) · S(2)|+ +〉

)

S2|1 1〉 =

(~2

1

2· 3

2|+ +〉+ ~2

1

2· 3

2|+ +〉+ 2S(1) · S(2)|+ +〉

). (102)

27

The first two terms might have been easy but what about the last one? Notehere that:

S(1) · S(2) = S(1)x S(2)

x + S(1)y S(2)

y + S(1)z S(2)

z (103)

so:

S(1) · S(2)|+ +〉 = S(1)x S(2)

x |+ +〉+ S(1)y S(2)

y |+ +〉+ S(1)z S(2)

z |+ +〉. (104)

Recall now that we are using the z -basis so the last term is easy:

S(1)z S(2)

z |+ +〉 =~2· ~

2|+ +〉. (105)

For the S(1)x S

(2)x |+ +〉+ S

(1)y S

(2)y |+ +〉 we have to do some more work. The

fastest way is to use the matrix representation of Sx, Sy:

Sx =~2

(0 11 0

)Sy =

~2

(0 −ii 0

)(106)

Sx|+〉 =~2

(0 11 0

)(10

)=

~2

(01

)=

~2|−〉

Sy|+〉 =~2

(0 −ii 0

)(10

)= i

~2

(01

)= i

~2|−〉.

We let the reader verify the following using the information in eq(106):

S(1)x S(2)

x |+ +〉+ S(1)y S(2)

y |+ +〉 =

(~2

)2

| − −〉+

(i~2

)2

| − −〉 = 0. (107)

In total we therefore get:

S2|1 1〉 =(~2s1 · (s1 + 1)|+ +〉+ ~2s2 · (s2 + 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〉 .

As an exercise, follow the same steps to show that S2|1 -1〉 = 2~2|1 -1〉.

28

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 ~M = ~ · 1). The system is thus in the total 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 after a measurement of oneof the particles we can use the Clebsch-Gordan table.

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 individualm-values. The state is thus4:

|J M〉 =∑

m1+m2=M

Cs1,s2,Jm1,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〉 (108)

From eq.(108) we see that the possible m-values of the spin 2 particle are 2, 1, 0.The possible results and probabilities are thus:

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

4You need to practice this by yourself - there are no shortcuts! .

29

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:Note that this problem takes the opposite perspective from the previous prob-lem. In the previous problem you knew the total angular momentum and had tofind out information about the individual systems. In this problem you actuallyhave information about the individual systems:

|L ml〉|s ms〉 = |1 0〉| 12 - 12 〉

and want to find information about the total system |JM〉:

|L ml〉|s ms〉 =∑J

CL,s,Jm1,m2,m|J M〉

|1 0〉| 12 - 12 〉 =∑J

CL,s,Jm1,m2,m|J M〉.

To find the coefficients CL,s,Jm1,m2,m we go to the Clebsch-Gordan (1×1/2) table.In the previous problem you read off a column - in this case we instead read offa row. We are specifically looking for the row (0 - 12 ). Once you find it you willsee:

|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

30

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 inserted to the Schrodinger equation 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 X,Y,Z we mustdeduce that:

i~1

T

∂T

∂t= E (113)

where E is a constant5. (If it was not a constant, then T would depend onX,Y,Z which contradicts the assumption.)

5Try 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.

31

Now we can insert this back to obtain the time-independent Schrodinger equa-tion:

Eψ(x, y, z) =

[− ~2

2m

(∂2

∂x2+

∂2

∂y2+

∂2

∂z2

)+

1

2mω2

(x2 + y2 + z2

)]ψ(x, y, z)

which of course something familiar to us all.

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

E XY Z =

[− ~2

2m

(∂2

∂x2+

∂2

∂y2+

∂2

∂z2

)+

1

2mω2

(x2 + y2 + z2

)]XY Z

E XY Z = − ~2

2m

(Y Z

d2X

dx2+XZ

d2Y

dy2+XY

d2Z

dz2

)+

1

2mω2

(x2 + y2 + z2

)XY Z

and now divide by XY Z:

− ~2

2m

(1

X

d2X

dx2+

1

Y

d2Y

dy2+

1

Z

d2Z

dz2

)+

1

2mω2

(x2 + y2 + z2

)= E.

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

(− ~2

2m

1

X

d2X

dx2+

1

2mω2x2

)+

(− ~2

2m

1

Y

d2Y

dy2+

1

2mω2y2

)+

(− ~2

2m

1

Z

d2Z

dz2+

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

d2X

dx2+

1

2mω2x2︸ ︷︷ ︸

Ex

− ~2

2m

1

Y

d2Y

dy2+

1

2mω2y2︸ ︷︷ ︸

Ey

− ~2

2m

1

Z

d2Z

dz2+

1

2mω2z2︸ ︷︷ ︸

Ez

= E

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

− ~2

2m

d2X

dx2+

1

2mω2x2X = ExX (114)

− ~2

2m

d2Y

dy2+

1

2mω2y2Y = EyY (115)

− ~2

2m

d2Z

dz2+

1

2mω2z2Z = EzZ. (116)

32

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)

and in our case it applies in each direction i ∈ {x, y, z}:

Ei(ni) =

(ni +

1

2

)~ω (118)

with:

En = Ex(nx) + Ey(ny) + Ez(nz) =

(n+ 3 · 1

2

)~ω (119)

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. (120)

33

4.40

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

2〈T 〉 = 〈r · ~∇V 〉 (121)

for stationary states.

Sol:In problem 3.31 we used eq.(3.71):

d

dt〈Q〉 =

i

~〈[H, Q]〉+

⟨∂Q

∂t

⟩(122)

to show that

d

dt〈xp〉 = 2〈T 〉 −

⟨xdV

dx

⟩(123)

and since all expectation values are constant in time for stationary states weobtained the one dimensional Virial theorem:

2〈T 〉 =

⟨xdV

dx

⟩.

Now we exchange Q = xp to the three-dimensional analogy Q = rp in eq.(122):

d

dt〈r · p〉 =

i

~〈[H, r · p]〉+

⟨∂r · p∂t

⟩. (124)

For a stationary state we know that the left hand side must be 0. Also, the lastterm on the right hand side is 0 because there is no explicit time dependence inneither r or p. Therefore:

0 =i

~〈[H, r · p]〉. (125)

Now we expand commutator in eq.(125):

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

= −(

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

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

)

34

Inserting H = p2

2m + V (r) gives:

[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

).

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.

Any component of p will commute with p2 so:

[pj ,p2

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

p2

2m]︸ ︷︷ ︸

0

+[pj , V (r)]

which simplifies some of the terms in our hideous expression 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

).

Note now that any function f(r) must commute with the coordinate variablesx, y, z ( r1, r2, r3):

[rj , V (r)] = rjV (r)− V (r)rj = rjV (r)− rjV (r) = 0 (126)

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(127)

35

so inserting that we get:

[H, r · p] = −(r1

(−ih∂V (r)

∂r1

)+ [r1,

p2

2m]p1 + r2

(−ih∂V (r)

∂r2

))+

−(

[r2,p2

2m]p2 + r3

(−ih∂V (r)

∂r3

)+ [r3,

p2

2m]p3

).

Now we only have three more commutators and we will evaluate them all atonce:

[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

(pk(−i~) + (−i~)pk) δkj =i~m

which we now insert into our commutator and then simplify the expression:

[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

)

[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.

Referring back to back into eq. (125) we get:

i

~〈[H, r · p]〉 = 0

i

~

⟨ih · r · ~∇V − i~ · p

2

m

⟩= 0

〈r · ~∇V 〉 =

⟨p2

m

⟩and since 〈T 〉 = 〈 p

2

2m 〉 we have finally arrived at:

〈r · ~∇V 〉 = 2〈T 〉.

36

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

〈T 〉 = −En; 〈V 〉 = 2En. (128)

Sol:

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

For Hydrogen we have the potential V (r) = − e2

4πε0rwhich gives:

~∇V =∂V (r)

∂rr +

1

r

∂V (r)

∂θθ +

1

r sin(θ)

∂V (r)

∂φφ =

e2

4πε0r2r

=⇒ 〈r · ~∇V 〉 =

⟨rr · e2

4πε0r2r

⟩=

⟨e2

4πε0r

⟩= −〈V 〉.

Exchanging the left hand side in the Virial theorem with this gives:

−〈V 〉 = 2〈T 〉. (129)

Lastly, recall Ehrenfest’s theorem - expectation values obey classical laws. So:

〈T 〉+ 〈V 〉 = En. (130)

Combining eq.(129) and eq.(130) we obtain:

〈T 〉 = −En〈V 〉 = 2En.

(131)

37

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. (132)

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

〈r · ~∇V 〉 = 2〈T 〉.

For 3D harmonic oscillator we have the potential V (r) = 12mω

2r2 giving:

~∇V =∂V (r)

∂rr +

1

r

∂V (r)

∂θθ +

1

r sin(θ)

∂V (r)

∂φφ = mω2rr

=⇒ 〈r · ~∇V 〉 = 〈rr ·mω2rr〉 = 〈mω2r2〉 = 2〈V 〉.

Exchanging the left hand side in the Virial theorem with this gives:

2〈V 〉 = 2〈T 〉 ⇐⇒ 〈V 〉 = 〈T 〉. (133)

Lastly, recall Ehrenfest’s theorem - expectation values obey classical laws:

〈T 〉+ 〈V 〉 = En. (134)

Combining eq.(133) and eq.(134) gives:

2〈T 〉 = 2〈V 〉 = En (135)

which is what we wanted to show.

38