HS21 Physik II D-ITET Exercise 9

G. Scalari return on Friday 19.11.2021 by 13:00

1 Sequential Stern-Gerlach experiment

Figure 1: Setup of the sequential Stern-Gerlach experiment. The purple lines indicate beams
of atoms traveling from left to right through a series of magnets.

The effect of the Stern-Gerlach apparatus is to perform a measurement of Sz on the atoms.

After the measurement the atoms are in either the spin up or down state in the z-direction.
The basis of eigenstates to Ŝz is

~
| ↑iz with eigenvalue +
2
~
| ↓iz with eigenvalue − .
2

Consider now the sequential Stern-Gerlach experiment, shown in Fig. 1, where we block the
spin down atoms, and then perform the experiment again on the spin up atoms now with a
different measurement axis. The task here is, to find the probability of different outcomes
of spin measurements. Let {|Ψn i} be an eigenstate of the observable Ô with eigenvalue on .
The probability of a measurement on a system in state |Ψi to give a result on , is

Pon = |hΨn |Ψi|2 .

a) If we orient the second magnet in the z-direction, so that we measure Sz again, what is
the probability to measure spin up?

b) If we instead orient the apparatus in the x-direction, so that we measure Sx , the atoms
will be projected onto one of the eigenstates to the Ŝx operator:

~
| ↑ix with eigenvalue +
2
~
| ↓ix with eigenvalue − .
2

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Exercise 9 return on Friday 19.11.2021 by 13:00

The Sz basis states and the Sx basis states are related by

1
| ↑iz = √ (| ↑ix + | ↓ix )
2
1
| ↓iz = √ (| ↑ix − | ↓ix )
2
1
| ↑ix = √ (| ↑iz + | ↓iz )
2
1
| ↓ix = √ (| ↑iz − | ↓iz ).
2
What is the probability of measuring spin up in the x-direction?

c) Now block the spin down atoms in Sx -direction and put a third Stern-Gerlach apparatus
to measure Sz of the atoms in the | ↑ix state coming out of the Sx apparatus. What is
now the probability to measure spin up in the z-direction?

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HS21 Physik II D-ITET Exercise 9
G. Scalari return on Friday 19.11.2021 by 13:00

2 Angular Momentum Operators

In quantum mechanics, the angular momenta are given by operators:

~ˆ = ~rˆ × p~ˆ
L ⇒ (L̂x , L̂y , L̂z ) = (x̂, ŷ, ẑ) × (p̂x , p̂y , p̂z )
L̂x = ŷ p̂z − ẑ p̂y , L̂y = ẑ p̂x − x̂p̂z , L̂z = x̂p̂y − ŷ p̂x

a) Using the basic commutator relations, [x̂i , p̂j ] = i~δij , show that:

[L̂y , L̂z ] = i~L̂x , [L̂z , L̂x ] = i~L̂y , [L̂x , L̂y ] = i~L̂z .

b) Also evaluate the commutation relations [L̂z , r2 ] and [L̂z , p2 ], where r2 = x2 + y 2 + z 2 and
p2 = p2x + p2y + p2z . For that you need to use (maybe useful to derive them) the following
as well:
[L̂z , x̂] = i~ŷ, [L̂z , ŷ] = −i~x̂, [L̂z , ẑ] = 0,
and
[L̂z , p̂x ] = i~p̂y , [L̂z , p̂y ] = −i~p̂x , [L̂z , p̂z ] = 0.

c) Show that the Hamiltonian Ĥ = (p̂2 /2m) + V̂ commutes with all three components of
L̂, provided that V̂ depends only on r. Thus Ĥ, L̂2 , and L̂z are mutually compatible
observables. 1

d) Consider the operator L̂2 = L̂2x + L̂2y + L̂2z . Using your results above, show that:

[L̂z , L̂2 ] = 0.

Argue, without further calculation, that [L̂x , L̂2 ] and [L̂y , L̂2 ] must also vanish.

e) Consider the ”ladder” operators L̂− = L̂x − iL̂y and L̂+ = L̂x + iL̂y . Use the above
commutation relations to show that:

[L̂z , L̂± ] = ±~L̂± , [L̂2 , L̂± ] = 0.

f) Consider an eigenstate Ψm 2
l of both L̂ and L̂z such that

L̂2 Ψm 2 m
l = ~ l(l + 1)Ψl , L̂z Ψm m
l = ~mΨl .

Is L̂± Ψm 2
l an eigenfunction of L̂ or L̂z ? If yes, what would be the corresponding eigen-
value?

1
Commuting observables are called as well compatible observables since the measurement of one observ-
able has no effect on the result of measuring another observable.

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Exercise 9 return on Friday 19.11.2021 by 13:00

3 Eigenfunctions of L̂2 and L̂z

In this problem we will use the explicit coordinate representations of the angular momenta
in spherical coordinates to study the eigenfunctions, Ylm (θ, φ), of the angular momentum
operators.

Working in spherical coordinates, L̂2 and L̂z take the form,

1 ∂2
   
2 2 1 ∂ ∂ ∂
L̂ = −~ sin θ + , L̂z = −i~ .
sin θ ∂θ ∂θ sin2 θ ∂φ2 ∂φ

and the first few spherical harmonics Yl,m take the form,
r r r
1 3 3
Y00 = , Y10 = cos θ, Y1±1 =∓ sin θe±iφ
4π 4π 8π
 
a) (Optional) Derive the expression for L̂2 using the expression for L̂± = ±~e±iφ ∂
∂θ
∂
± i cot θ ∂φ ,
and the relation L̂+ L̂− = L̂2 − L̂2z + ~L̂z . (Hint: In order to find the expression for L̂+ L̂−
use a test function, otherwise you are likely to drop some terms!)

b) Show that the functions Y00 , Y10 and Y1±1 are properly 2
normalized and orthogonal to
one another.

c) Show that these functions are eigenfunctions of both L̂2 and L̂z , and compute the corre-
sponding eigenvalues.

d) What is L̂+ Yll ? (No calculation allowed!)

e) Use the results of the last part, together with expression for L̂± and the fact that L̂z Yll =
~lYll , to determine Yll (θ, φ), up to a normalization constant.

We will now study linear combinations on the example of the hydrogen wavefunctions. For
this part you need the first radial wave functions Rnl (r):

−3/2 1 −3/2  r
R10 = 2a exp(−r/a) and R20 =√ a 1− exp(−r/2a) (1)
2 2a

1 −3/2  r 
R21 = √ a exp(−r/2a) (2)
2 6 a
2
Remember that we normalized the total wavefunction in spherical coordinates by normalizing separately
the radial part and angular part.

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HS21 Physik II D-ITET Exercise 9
G. Scalari return on Friday 19.11.2021 by 13:00

f) Consider the linear combination of the stationary states n = 1, l = 0, m = 0 and n = 2,

l = 0, m = 0:
1
Ψ1 (~r, 0) = √ (ψ100 + ψ200 ). (3)
2
Write down the wavefunction Ψ1 (~r, t) and simplify it as much as you can. What do you
notice?

g) Now, consider the linear combination of the stationary states n = 2, l = 1, m = 1 and

n = 2, l = 1, m = −1:
1
Ψ2 (~r, 0) = √ (ψ211 + ψ21−1 ). (4)
2
Construct Ψ2 (~r, t) using the radial equations and the spherical harmonics and simplify
the result as much as you can.

h) Find the expectation value of the potential energy, hV i 3 . Does it depend on t? Give
both the formula and the actual number in electron volts.

After this small detour, let’s go back to spherical harmonics. From the last results and from
[L2 , Lx ] = [L2 , Ly ] = [L2 , Lz ] = 0, we can see that choosing the quantization axis z for the
angular momentum states is arbitrary, as also linear superpositions of spherical harmonics
can be stationary. Let’s for now choose x as our prefered axis instead.

i) Find the eigenfunction, Ψx of the operators L2 and Lx that gives eigenvalues 2~2 and
~, respectively. (Hint: You can find the solution by rotation.)
You should find r
3
Ψx = − (− cos θ + i sin θ sin φ) (5)
8π

k) Express the Ψx just found as a linear combination of eigenfunctions of L2 and Lz .

Rb Rb
The following integrals might be handy: a sin2 cxdx = [ x2 − 4c
3 1
sin 2cx]ba , a sin3 cxdx = [ cos12c
3cx
− 3 cos cx b
4c ]a
R b n cx n n!
and a x e dx = [ecx i=0 (−1)n−i i!cn−i+1 xi ]ba
P