1

A. Lim 3004

Calculate the expectation values of Lx and L2x for a state with angular momentum lh̄ and
a projection onto the z axis mh̄.
Solution:
The expectation value of Lx can be obtained using the commutation relations for com-
ponents of angular momentum: ih̄Lx = [Ly , Lz ]. Hence

1 1
hLx i = hlm| [Ly , Lz ]|lmi = hlm|Ly Lz − Lz Ly |lmi
ih̄ ih̄

We now use Lz |lmi = h̄m|lmi (and its conjugate) to find

1 1
hLx i = hlm|mh̄Ly |lmi − hlm|mh̄Ly |lmi = 0 .
ih̄ ih̄

Alternatively, we can make use of the rising and lowering operators for angular momentum

L+ = Lx + iLy L− = Lx − iLy .

Which gives Lx = 21 (L+ + L− ). As

p
L± |lmi = h̄ l(l + 1) − m(m ± 1)|lm ± 1i ,

we obtain
1 p
hlm|Lx |lmi = h̄ l(l + 1) − m(m + 1)hlm|lm + 1i
2
1 p
+ h̄ l(l + 1) − m(m − 1)hlm|lm − 1i
2
= 0,

as the basis of the |lmi states is orthonormal.

The expectation value of the square of the Lx operator can be obtained using the operator
of the magnitude of the momentum L2 and of the projection into the z axis. As L2 =
L2x + L2y + L2z and hL2x i = hL2y i, due to the symmetry of the problem, the expectation value
is hL2x i = 12 h(L2 − L2z )i. Using L2 |lmi = l(l + 1)h̄2 |lmi and Lz |lmi = h̄m|lmi, we obtain
hL2x i = 12 h̄2 (l(l + 1) − m2 ).
Using the raising and lowering operators, we have

1 1
L2x = (L+ + L− )(L+ + L− ) = (L+ L+ + L− L− + L− L+ + L+ L− )
4 4
 2

The expectation value of the squares of the raising or lowering operators is zero (as
hlm|L2+ |lmi = chlm|lm + 2i = 0) and only the mixed terms remain. They can be eval-
uated as
1
hL2x i = hlm|L− L+ + L+ L− |lmi
4
1 p 1 p
= h̄hlm|L− l(l + 1) − m(m + 1)|lm + 1i + h̄hlm|L+ l(l + 1) − m(m − 1)|lm − 1i
4 4
1 2 p p
= h̄ hlm| l(l + 1) − (m + 1)(m + 1 − 1) l(l + 1) − m(m + 1)|lmi
4
1 p p
+ h̄2 hlm| l(l + 1) − (m − 1)(m − 1 + 1) l(l + 1) − m(m − 1)|lmi
4
1 2 1
= h̄ (l(l + 1) − m(m + 1)) + h̄2 (l(l + 1) − m(m − 1))
4 4
1 2
= h̄ (2l(l + 1) − m(m + 1) − m(m − 1))
4
1
= h̄2 (l(l + 1) − m2 ) .
2
Which agrees with the previous result.

B. Lim 3007

A particle with spin S = 1 is in a state with an angular momentum of L = 2. A spin-orbit

Hamiltonian
H = AL · S

describes the interaction between the particles. What are the possible energies and their
degeneracies for this system.
Solution: The spin-orbit Hamiltonian does not commute with individual projections of
the spin and angular momentum, i.e. [Lz , L · S] 6= 0 and [Sz , L · S] 6= 0. The Hamiltonian
is, however, diagonal in the basis of the total momentum J = L + S and its projections
Jz = Lz + Sz .
The Hamiltonian can be rewritten using the operator of the magnitude of the total
momentum J 2 = L2 + S 2 + 2L · S, from which we find
1
L · S = (J 2 − L2 − S 2 ) .
2
The magnitude of the (L = 2) orbital momentum is L2 |lmi = l(l + 1)h̄2 |lmi = 6h̄2 |lmi and
for spin (S = 1) we have S 2 |ssz i = s(s + 1)h̄2 |ssz i = 2h̄2 |ssz i and are thus identical for all
the states.
 3

The rules of combination of angular momenta give possible values for the total momentum
J = L + S, . . . , |L − S|. Therefore, for L = 2 and S = 1 we have J = 3, 2, 1. The magnitude
of the total momentum is J 2 |JJz LSi = J(J + 1)h̄2 |JJz LSi, which, for the possible values of
J gives 12h̄2 , 6h̄2 , and 2h̄2 , respectively. As the expectation values of the Hamiltonian are

A
hHi = hJJz LS| (J 2 − L2 − S 2 )|JJz LSi ,
2

We get

E(J = 3) = 2Ah̄2
E(J = 2) = −Ah̄2
E(J = 1) = −3Ah̄2

which are 7, 5, and 3-fold degenerate, respectively.

C. Two spins – Lim 3034

Consider a system with two non-interacting spins. The first is in a state sA

z = +1/2, the

second in a state sB
x = +1/2. What’s the probability that the total spin is zero?

Solution:
Two particles with spin one half lead to total spin one with three-fold degeneracy and a
non-degenerate spin zero state. For the total spin equal to zero, the state is (taking z as the
quantisation coordinate)

1
|S = 0, Sz = 0i = p (| ↑A ↓B i − | ↓A ↑B i) .
(2)

To be able to project on this state, we need to transform the sB B

x = +1/2 state into the sz

basis. The representation of the sB

x states can be found by diagonalisation of the sx operator

matrix 
h̄ 0 1
sx =   . (1)
2 1 0

The eigenvalues are ±1h̄/2 and the states are √12 11 for the +h̄/2 state and √1 1



2 −1
for the
−h̄/2 state. That is, |sx = +1/2i = √1 (|sz = +1/2i + |sz = −1/2i). Therefore,
2

1
|sA B A B A B
z = +1/2, sx = +1/2i = √ (|sz = +1/2, sz = +1/2i + |sz = +1/2, sz = −1/2i .
2
 4

The projection on the zero spin state is then (in a short-hand notation)
1
h00|sA B
z = +1/2, sx = +1/2i = (h↑A ↓B | − h↓A ↑B |) (| ↑A ↑B i + | ↑A ↓B i)
2
1
= (h↑A ↓B | ↑A ↑B i + h↑A ↓B | ↑A ↓B i − h↓A ↑B | ↑A ↑B i − h↓A ↑B | ↑A ↓B i)
2
1 1
= (0 + 1 + 0 + 0) =
2 2
(2)
The probability is
1
|h00|sA B 2
z = +1/2, sx = +1/2i| = .
4

D. Lim 3017

An electron is prepared with projection of the spin +h̄/2 along the z axis.

• What are the possible results of measurement of spin along the x axis?

• What is the probability of finding these results?

• If we measure the spin along axis restricted to the x − z plane and rotated by an angle
θ from the z axis, what are the probabilities of measuring the different results?

• What is the expectation value of spin measured along the rotated axis, given the initial
projection along z?

Solution:
The measured electron spin will be ±h̄/2 along any axis. This can be shown by finding
the eigenvalues of the sx operator that describes the act of measuring the spin along x
axis. Its eigenvalues are then the only possible results of measuring the spin (given an
isolated system). We find the eigenvalues by diagonalising the matrix representation of the
sx operator in the basis of the states corresponding to measurement along the z axis. The
sx operator is given as  
h̄ 0 1
sx =   .
2 1 0

The eigenvalues are found by calculating the determinant of

 
h̄  −λ 1
 = λ2 − 1,
2 1 −λ
 5

therefore the eigenvalues are indeed λ = ±h̄/2. The eigenvector corresponding to λ = h̄/2
is √12 11

E. Spin-spin Hamiltonian

Two particles (A and B) with spin 1/2 interact via Hamiltonian H = JsA · sB . Find the
eigenenergies of the Hamiltonian by rewriting it using the operators of the magnitudes of
the spin.
Solution:
The problem is analogous to the problem of spin-orbit interaction. The dot product in
the Hamiltonian does not commute with the operators of the projections of the individual
spins into the z axis. We therefore introduce the total spin S = sA ⊗ 1B + 1A ⊗ sB and
its projection Sz = sA B
z ⊗ 1B + 1A ⊗ sz , which communte with the Hamiltonian. By the
1 1
rules of combination of angular momenta the total spin can be either S = 2
+ 2
= 1 or
1 1
S= 2
− 2
= 0. In the first case, three projections are possible, in the latter, only one.
To find the new eigenvalues we rewrite the dot product using the operator of the magni-
tude of the total spin

S 2 = (sA + sB )2 = s2A ⊗ 1B + 1A ⊗ s2B + 2sA · sB . (3)

Hence
1
sA · sB = (S 2 − s2A ⊗ 1B − 1A ⊗ s2B ) . (4)
2
The action of an operator of the magnitude of the spin is s2 |ssz i = h̄2 s(s + 1)|ssz i. For
particles with spin s = 21 , we obtain 34 h̄2 . For the triplet states with total spin S = 1, we
get 2h̄2 and for the singlet state with spin S = 0 we get 0.
The energies are then for triplet

J 3 3 J
E(T) = (2h̄2 − h̄2 − h̄2 ) = h̄2 (5)
2 4 4 4

and for the singlet state

J 3 3 3J
E(S) = (0 − h̄2 − h̄2 ) = − h̄2 . (6)
2 4 4 4

This agrees with the results obtained by exact diagonalisation.

 6

F. Spin-spin Hamiltonian in matrix representation

Two particles (A and B) with spin 1/2 interact via Hamiltonian H = JsA · sB . Write
the matrix representation of the Hamiltonian in the direct basis. Find the eigenvectors and
eigenvalues of the Hamiltonian.
Solution:
The Hamiltonian reads AsA · sB , where the dot product stands for

sA · sB = sA B A B A B
x ⊗ sx + sz ⊗ sz + sz ⊗ sz .

The matrix representation of the spin operators (sx etc.) in the standard quantisation along
z axis uses the Pauli matrices  
h̄ 0 1
sx =  
2 1 0
 
h̄ 0 −i
sy =  
2 i 0
 
h̄ 1 0
sz = 
2 0 −1
1 0



The basis functions are spin up and down along the z axis: | ↑i = 0
and | ↓i = 1
.
Calculating the expectation values of the spin components using these vectors gives zeros
h̄
for the x and y components and either 2
or − h̄2 for the z component, as expected.
The Hamiltonian is written as a direct product of the spin matrices, it can be thus written
as a 4 × 4 matrix. We will go from the direct product of two 2 × 2 matrices to the 4 × 4
matrix using the following scheme:
 
Aα Aβ Bα Bβ
     
A B α β
 Aγ Aδ Bγ Bδ 
 ⊗ =


 (7)
C D γ δ Cα Cβ Dα Dβ 
 

Cγ Cδ Dγ Dδ

This corresponds to a basis set | ↑↑i, | ↑↓i, | ↓↑i, and | ↓↓i, where the first spin is of the A
particle and the latter of the B particle.
 7

For example, the part of the Hamiltonian originating from the z components is
 
1 0
0 0
 
   
2 0 −1 0 0
h̄ 1 0  h̄ 1 0 h̄
J  ⊗ =J  (8)
 
2 0 −1 2 0 4 0 0 −1 0

−1  
0 0 0 1

This shows that for H = JsA B

z ⊗ sz there are two energy levels, one with aligned spins (H11
h̄2
for | ↑↑i state and H44 for | ↓↓i state). In this case the energy is E = 4
J. For the usual
ferromagnetic ordering, J < 0 and this will be the (degenerate) ground state. The states
with anti-parallel spins will be higher in energy. Anti-ferromagnetic order can be observed
as well in some materials and then J > 0 and the anti-parallel states have lower energy than
the states with parallel spins.
The parts of the Hamiltonian corresponding to the x and y components can be calculated
analogously to the z component, for x we obtain
 
0 0 0 1
     
2 0 0 1 0
h̄ 0 1 h̄ 0 1 h̄
J  ⊗  =J  (9)
 
2 1 0 2 1 4 0 1 0 0

0  
1 0 0 0

and for y
 
0 0 0 −1
     
2  0 0 1 0 
h̄ 0 −i h̄ 0 −i h̄
J  ⊗ =J  (10)
 
2 i 0 2 i 4 0 1 0 0

0  
−1 0 0 0
Overall, summing these three parts, we obtain the Hamiltonain
 
1 0 0 0
 
h̄2 0 −1 2 0
H=J (11)
 
4
 
0 2 −1 0
 
0 0 0 1

We see that the Hamiltonian is not diagonal in the direct product basis, specifically, the
states with anti-parallel spins are not eigenvectors of the new Hamiltonian. In contrast, the
states with aligned spins are still eigenvectors. To find the new states with anti-parallel
 8

spins we need to diagonalise the part of the Hamiltonian corresponding to these two states
 
2 −1 2
h̄
Hanti = J  . (12)
4 2 −1

For the eigenvalues we readily obtain 1,2 = −1 ± 2 corresponding to vectors √12 11 = √12 (| ↑↓

1 2
i + | ↓↑i) and √12 −1 = √12 (| ↑↓i − | ↓↑i). The energies of these two states are h̄4 J for the


2
first and − 3h̄4 J for the latter. Therefore, the state with symmetric spin wavefunction has
the same energy as the states with both spins up or both spins down. Together, they form
the triplet states with total spin S = 1 and projections Sz = 1, 0, −1. The state with anti-
symmetric spin wavefunction is the singlet state with S = 0 and only possible projection
Sz = 0. This can be verified by explicitly applying the operator of the magnitude of the
total spin S 2 (TODO) .

G. Coupling of momenta

A spin h̄/2 particle is bound in a spherically symmetric potential and is in a state with
orbital momentum l = 1h̄. What are the possible values of the total momentum and the
projections onto the z axis?
3h̄ h̄
The particle is in a state with j = 2
and j = 2
that can be written in the product basis
as
r r
2 1 1 1
|ψi = |j = 3/2, jz = 1/2, l, si = |l, lz = 0; s, sz = + i + |l, lz = 1; s, sz = − i .
3 2 3 2
• What are the respective results when the operators j 2 and jz act on the state |ψi?

• Verify, that the state |ψi is also an eigenstate of the operator jz = lz ⊗ 1 + 1 ⊗ sz in

the product basis of the original states.

• Rewrite the operator of the magnitude of the total momentum j 2 = (~l⊗1+1⊗~s)2 using
the operators l2 , lz , l− , l+ , s2 , sz , s− , s+ and verify that the state |ψi is an eigenstate of
j 2.

Solution:
The rules for combining angular momenta give two possible values for the total momen-
3h̄
tum: j = 2
and j = h̄2 . The possible projections are jz = 3h̄ h̄
2
, 2, − h̄2 , and − 3h̄
2
for the first
h̄
and jz = 2
and − h̄2 for the latter.
 9

3h̄
The state |ψi corresponds to j = 2
and jz = h̄2 , after acting with j 2 we obtain
3 3 15
j 2 |j = 3/2, jz = 1/2, l, si = ( +1)h̄2 |j = 3/2, jz = 1/2, l, si = h̄2 |j = 3/2, jz = 1/2, l, si .
2 2 4
For the projection onto the z axis we obtain
h̄
jz |j = 3/2, jz = 1/2, l, si = |j = 3/2, jz = 1/2, l, si .
2
We will now use the product basis where operators of the individual momenta act. For
the projection onto the z axis we obtain
r r
2 1 1 1
(lz ⊗ 1 + 1 ⊗ sz )[ |l, lz = 0; s, sz = + i + |l, lz = 1; s, sz = − i]
3 2 3 2
r r
2 1 2 1
= (lz ⊗ 1)|l, lz = 0; s, sz = + i + (sz ⊗ 1)|l, lz = 0; s, sz = + i
3 2 3 2
r r
1 1 1 1
+ (lz ⊗ 1)|l, lz = 1; s, sz = − i + (sz ⊗ 1)|l, lz = 1; s, sz = − i
3 2 3 2
r r
2 1 2 h̄ 1
= 0h̄|l, lz = 0; s, sz = + i + |l, lz = 0; s, sz = + i
3 2 32 2
r r
1 1 1 h̄ 1
+ h̄|l, lz = 1; s, sz = − i − |l, lz = 1; s, sz = − i
3 2 32 2
r r
h̄ 2 1 h̄ 1 1
= |l, lz = 0; s, sz = + i + |l, lz = 1; s, sz = − i
2 3 2 2 3 2
h̄
= |ψi .
2
That is, the results obatained in the product and coupled basis are identical, as they should.
The operator of the magnitude of the total momentum j 2 can be expressed as

j 2 = (l ⊗ 1 + 1 ⊗ s)2 = (l2 ⊗ 1 + 1 ⊗ s2 + 2lx ⊗ sx + 2ly ⊗ sy + 2lz ⊗ sz ) .

Using lx = 21 (l+ + l− ) a lx = 1
(l
2i +
− l− ) and similarly for the spin component, we rewrite
the x and y components as

lx ⊗ sx + ly ⊗ sy =
1 1
= (l+ + l− ) ⊗ (s+ + s− ) − (l+ − l− ) ⊗ (s+ − s− )
4 4
1
= (l− ⊗ s+ + l+ ⊗ s− ) .
2
Overall, we have

j 2 = (l2 ⊗ 1 + 1 ⊗ s2 + l− ⊗ s+ + l+ ⊗ s− + 2lz ⊗ sz ) .
 10

3h̄2
The state |ψi is an eigenstate of both l2 and s2 , with eigenvalues 2h̄2 and 4
, respectively.
The remaining contributions can be found by explicitly applying the operators on the states.
For the first remaining term
r r
2 1 1 1
l− ⊗ s+ [ |l, lz = 0; s, sz = + i + |l, lz = 1; s, sz = − i] ,
3 2 3 2
the action on the first component will give zero, s+ |s, sz = + 21 i = 0, as the spin projection
onto the z axis is already maximal. We then have to evaluate
r
1 1
l− ⊗ s+ |l, lz = 1; s, sz = − i
3 2
r r
2 1p 1 1 1 1 1
= h̄ 1(1 + 1) − 1(1 − 1) ( + 1) − (− )(− + 1)|l, lz = 0; s, sz = i
3 2 2 2 2 2
√ √
r
1 1
= h̄2 2 1|l, lz = 0; s, sz = i
3 2
r
2 1
= h̄2 |l, lz = 0; s, sz = i
3 2
In the term
r r
2 1 1 1
l+ ⊗ s− [ |l, lz = 0; s, sz = + i + |l, lz = 1; s, sz = − i] ,
3 2 3 2
the second contribution will be zero as the l projection is the largest and the spin projection
is the lowest possible. We therefore need to evaluate
r
2 1
l+ ⊗ s− |l, lz = 0; s, sz = i
3 2
r r
2 p 1 1 1 1 1
= h̄2 1(1 + 1) − 0(0 + 1) ( + 1) − ( )( − 1)|l, lz = 1; s, sz = − i
3 2 2 2 2 2
2√ √
r
1
= h̄2 2 1|l, lz = 1; s, sz = − i
3 2
2 2 1
= h̄ √ |l, lz = 1; s, sz = − i .
3 2
Finally, the term containing the projections onto the z axis becomes
r r
2 1 1 1
2lz ⊗ sz [ |l, lz = 0; s, sz = + i + |l, lz = 1; s, sz = − i]
3 2 3 2
r r
h̄ 2 1 h̄ 1 1
= 2(0h̄)( ) |l, lz = 0; s, sz = + i + 2(h̄)(− ) |l, lz = 1; s, sz = − i
2 3 2 2 3 2
r
1 1
= −h̄2 |l, lz = 1; s, sz = − i .
3 2
 11

Summing the last three results we find

(l− ⊗ s+ + l+ ⊗ s− + 2lz ⊗ sz )|ψi =

r
2 2 1 2 1
=h̄ |l, lz = 0; s, sz = i + h̄2 √ |l, lz = 1; s, sz = − i
3 2 3 2
r
1 1
− h̄2 |l, lz = 1; s, sz = − i
3 2
r r
2 2 1 2 1 1
=h̄ |l, lz = 0; s, sz = i + h̄ |l, lz = 1; s, sz = − i
3 2 3 2
r r
2 1 1 1
=h̄2 [ |l, lz = 0; s, sz = i + |l, lz = 1; s, sz = − i]
3 2 3 2
=h̄2 |ψi .

Therefore, |ψi is an eigenstate of the operator 2l · s with an eigenvalue of h̄2 . Together with
the contributions of l2 and s2 the total eigenvalue of j 2 is (2 + 34 + 1)h̄2 = 15 2
4
h̄ , as it should
for j = 32 h̄.