PHYS 20602 Answers for Examples Class 2

PHYS 20602 Answers for Examples Class 2
1. Possible values of total angular momentum J = ˆL + Ŝ: J 2 = j(j + 1) 2 , L 2 =
l(l + 1) 2 , S 2 = s(s + 1) 2 . j, l, and s are all quantised and positive: l must
have integer values; j and s can also have half-integer values. Maximum value
of j is j max = l + s, minimum value j min = |l − s|. In between, values decrease
in steps of one. Electrons have s = 1 , so for l = 2,
2
and there are no intermediate values.
j max = 5/2, j min = 3/2
Possible values of m j run from j to −j in unit steps, hence
j 5/2 3/2 degeneracy
5/2 1
3/2 3/2 2
m j 1/2 1/2 2
−1/2 −1/2 2
−3/2 −3/2 2
−5/2 1
2. (i) Bra (operator acts to left on a bra to make a new bra).
(ii) Operator (outer products are operators; this is the projector to state |1〉).
(iii) Ket (direct product of two kets is a ket in the direct product space).
(iv) Complex numbers (inner products are complex numbers).
3. |←〉〈← | is the projection operator onto state |←〉; call it P. First basis ket is
|↑〉, second is |↓〉, so
P 11
= 〈↑ |P | ↑〉 = 〈↑ |←〉〈← |↑〉 = 1 〈↑ | (|↑〉 + |↓〉) (〈↑ | + 〈↓ |) |↑〉
2
= 1 2 (〈↑ |↑〉 + 〈↑ |↓〉) (〈↑ |↑〉 + 〈↓ |↑〉) = 1 2 (1 + 0)(1 + 0) = 1 2
using the orthonormality of {|↑〉, |↓〉}. Similarly:
P 12 = 〈↑ |P | ↓〉 = 〈↑ |←〉〈← |↓〉 = 1 2 (1 + 0)(0 + 1) = 1 2
P 21 = 〈↓ |P | ↑〉 = 〈↓ |←〉〈← |↑〉 = 1 2 (0 + 1)(1 + 0) = 1 2
P 22 = 〈↓ |P | ↓〉 = 〈↓ |←〉〈← |↓〉 = 1 2 (0 + 1)(0 + 1) = 1 2
1

So the matrix
[P] =
Eigenvalues and eigenvectors:
( )
P11 P 12
P 21 P 22
=
( 1
2
(a) Smart way: since P is a projector, it has eigenvalues 1 for eigenstate |←〉,
and 0 for the eigenspace orthogonal to |←〉. Since we are in a 2-D space, the
orthogonal eigenspace is 1-D and its basis ket is obviously |→〉. If you don’t
already know the expansion for |→〉 in terms of {|↑〉, |↓〉}, use the general rule
that the vector orthogonal to (a|1〉 + b|2〉) is (b ∗ |1〉 − a ∗ |2〉). In this case
1
2
1
2
1
2
)
.
|↓〉 =
|↑〉 − |↓〉
.
2
(b) Standard way: Solve for eigenvalues p of P:
1
2
∣
− p ∣ ( )
1 ∣∣∣ 2
2 1
1 1
− p =
2 2 2 − p − 1 4 = p2 − p = p(p − 1),
hence eigenvalues are p = 0, 1. Solving for the eigenvectors, starting with p = 1:
( 1 − p ) ( ) (
1
2 2 a −
a
1 1
− p =
+ ) ( )
b
2 2 0
a
b
− b =
0
2 2 2 2
i.e. a = b, so the eigenvector is
|p = 1〉 −→ S z
1 √2
( 1
1
)
.
Similarly for eigenvalue 0:
( a+b
2
a+b
2
) ( 0
=
0
)
,
4. (i)
i.e. a = −b so
|p = 0〉 −→ S z
1 √2
( 1
−1
ˆM = Ŝz2
+ 2
2
)
.
This is an operator equation so all terms should be operators. However, since
multiplying by the identity has no effect we often don’t bother writing Î; but to
be accurate we should say
ˆM = Ŝz2
+ 2 2 Î.
2

Ĵ = ˆL + Ŝ
J is an operator that acts on the combined (direct product) space of orbital
angular momentum and spin, but ˆL and Ŝ act only on one (each) of these factor
spaces. To be precise we should write
Ĵ = ˆL ⊗ Î + Î ⊗ Ŝ.
(ii) In the equation
ˆL z |m = 0〉 = 0,
the ‘0’ on the right-hand side is the zero vector (zero ket), since the LHS (operator
acting on ket) evaluates to a ket, not to a scalar.
5. (a) Given that we know H = AL 2 x, which is a function of L x , its eigenvalues
will be the same functions of the L x eigenvalues. L x must have the same
eigenvalues as L z , since x and z are just conventional labels. For l = 1, the
eigenvalues of L z and L x are , 0, and −. Hence we expect eigenvalues of
H to be
A 2 × (1, 0, 1)
Check by explicitly solving the matrix. For speed, solve the matrix without
the pre-factor (call it H ′ ) and then multiply the eigenvalues by A 2 /2:
1 − m 0 1
det(H ′ − mI) =
0 2 − m 0
∣ 1 0 1 − m ∣
= (1 − m)[(2 − m) × (1 − m) − 0] − 0 + 1 × [0 − (2 − m) × 1]
= (2 − m)[1 − 2m + m 2 − 1] = (2 − m)m(m − 2),
i.e. m = (2, 0, 2) and hence energy eigenvalues are E = A 2 (1, 0, 1) as
expected.
The ground state is the lowest energy value, in this case E = 0. Hence the
eigenvector satisfies:
⎛
⎝
1 0 1
0 2 0
1 0 1
⎞⎛
⎠⎝
a
b
c
⎞
⎠ = 0,
i.e. a + c = 0 and 2b = 0. Hence the normalised ground state ket is
⎛ ⎞
|E = 0〉 −→ 1
1
L z √2 ⎝ 0 ⎠ .
−1
3

(b) The state with m = −1 is an eigenstate of L z , hence one of our basis kets;
in fact it is the third basis ket as we always list the eigenvalues in the order
, 0, −. So
⎛ ⎞
|m = −1〉 −→ 0
L z
⎝ 0 ⎠
1
Per the question, write this as |m = −1〉 = a|E = 0〉 + |φ〉 where a and |φ〉
are unknown, but 〈φ|E = 0〉 = 0, i.e. |φ〉 is a vector in the subspace with
excited energy E = A 2 .
where
⎛
⎝
0
0
1
⎞
⎠ = a √
2
⎛
⎝
⎛
(1, 0, −1) ⎝
x
y
z
⎞
1
0
−1
⎞
⎛
⎠ + ⎝
x
y
z
⎞
⎠
⎠ = (x − z) = 0,
so the excited subspace contains any vector with x = z. The solution may
be obvious; formally we have to solve:
⎛ ⎞ ⎛
0 a/ √ ⎞
2 + x
⎝ 0 ⎠ = ⎝ y
1 −a/ √ ⎠ ,
2 + x
hence y = 0, x = −a/ √ 2, and − √ 2a = 1:
⎛ ⎞ ⎛ ⎞
0
⎝ 0 ⎠ = 1 1
⎝ 0 ⎠ − 1 2 2
1 1
The two terms have energies A 2 and 0, so we can apply the time evolution
operator to get the state at time t:
⎛ ⎞
|Ψ(t)〉 = e −iHt/ |m = −1〉 = e −iAt |φ〉−e 0 √ 1 |E = 0〉 −→ e
1
−iAt − 1
L z
⎝ 0 ⎠ .
2 2
e −iAt + 1
⎛
⎝
1
0
−1
This is a bit more illuminating if we pull out an overall phase factor:
⎛
⎞ ⎛
|Ψ(t)〉 −→ e −iAt/2 e −iAt/2 − e iAt/2 −i sin(At/2)
L z
⎝
0 ⎠ = e −iAt/2 ⎝ 0
2
e −iAt/2 + e iAt/2 cos(At/2)
The ion oscillates between L z = − and L z = , but without passing
through L z = 0.
⎞
⎠ .
⎞
⎠.
4