224
Appendix E: Solutions
Exercise 1.5 ˆ
S 2 |ϕ 1 ϕ 2 |=|ϕ 1 ϕ 2 |( ˆ
S + ˆ
S − − ˆ
S z + ˆ
S 2
z )αβ =|ϕ 1 ϕ 2 |
ˆ
S + ˆ
S − αβ − ˆ
S z αβ +
ˆ
S 2
z αβ
. With ˆ
S =ˆ s(1) +ˆ s(2) we arrive at |ϕ 1 ϕ 2 |
(αβ + βα) − (
1
2 αβ −
1
2 αβ) +
ˆ
s z (
1
2 αβ −
1
2 αβ)
. Since the last two terms are zero, we get |ϕ 1 ϕ 2 |(αβ + βα) =
|ϕ 1 ϕ 2 |+|ϕ 1 ϕ 2 |, which is the same result as obtained in Eq. 1.27 where the determinant
was fully expanded; ˆ
S 2
|ϕ 1 ϕ 2 |+| ϕ 1 ϕ 2 |
/
√
2 = (1/
√
2)|ϕ 1 ϕ 2 | ˆ
S 2 (αβ + βα) =
(1/
√
2)|ϕ 1 ϕ 2 |
(αβ +βα)+(βα +αβ)
= (2/
√
2)
|ϕ 1 ϕ 2 |+|ϕ 1 ϕ 2 |
; The expansion
of Φ 2 leads to (1/2)(ϕ 1 ϕ 2 −ϕ 2 ϕ 1 +ϕ 1 ϕ 2 −ϕ 2 ϕ 1 ) = (1/2)(ϕ 1 ϕ 2 −ϕ 2 ϕ 1 )(αβ +βα).
ˆ
S 2 Φ 2 = (1/2)(ϕ 1 ϕ 2 − ϕ 2 ϕ 1 ) ˆ
S 2 (αβ + βα) = (1/2)(ϕ 1 ϕ 2 − ϕ 2 ϕ 1 ) · 2(αβ + βα) =
ϕ 1 ϕ 2 − ϕ 2 ϕ 1 + ϕ 1 ϕ 2 − ϕ 2 ϕ 1 =
√
2
|ϕ 1 ϕ 2 |+|ϕ 2 ϕ 1 |).
Exercise 1.6 (a) Rewriting Eq. 1.23 gives |S, M S + 1=
ˆ
S + |S, M S /
√
S(S + 1) − M S (M S + 1) = ˆ
S +
|ab|+| ab|
/(
√
2(1(1 + 1) − 0(0 + 1)) =
|ab|+|ab|
/2 =|ab|. Similar for the M S −1 component: ˆ
S −
|ab|+|ab|
/
√
2(1(1+
1) − 0(0 − 1))
=
|ab|+|ab|
/2 =| ab|.( b )|S= ˆ
P 0 |ab|=( ˆ
S 2 − 2)|ab|=
|ab|+|ab|−2|ab|=|ab|−|ab|; |T = ˆ
P 1 |ab|=( ˆ
S 2 − 0)|ab|=|ab|+|ab|.
Exercise 1.7 |3/2, 3/2= ˆ
S +
|abc|+| abc|+| abc|
/(
√
3N) with N =
√
3/2(3/2 + 1) − 1(2(1/2 + 1) =
√
3 ⇒| 3/2, 3/2=(|abc|+| abc|+
|abc|)/3 =| abc|; |3/2, −1/2=
|abc|+| abc|+| abc|
/(
√
3N) with N =
√
3/2(3/2 + 1) − 1/2(1/2 − 1) = 2 ⇒| 3/2, −1/2=(|abc|+|abc|+|abc|+
|abc|+| abc|+| abc|)/2
√
3 = (|abc|+| abc|+| abc|)/
√
3; |3/2, −3/2=
ˆ
S − (|abc|+|abc|+|abc|)/(
√
3N) with N =
√
3/2(3/2 + 1) −−1/2(−1/2 − 1) =
√
3 ⇒|3/2, −3/2=(|abc|+|abc|+|abc|)/3 =|abc|.
Exercise 1.8 (a) Ψ A |Ψ B =(1/
√
12)abc − abc|2abc − abc − abc=(1/
√
12)
(−−abc|abc++abc|abc) = 0. (b) ˆ
S + ˆ
S − |abc|= ˆ
S + (|abc|+|abc|) =|abc|+|abc|+
|abc|+|abc|; ˆ
S + ˆ
S − |abc|= ˆ
S + (|abc|+|abc|) =| abc|+|abc|+|abc|+|abc|;
ˆ
S z (|abc|−|abc|) = (1/2+1/2−1/2)|abc|−(−1/2+1/2+1/2)|abc|⇒ ˆ
S 2
z (|abc|−
|abc|) = (1/4)(|abc|−|abc|). Collecting all the terms gives ˆ
S 2 (|abc|−|abc|)/
√
2 =
(1 − 1/2 + 1/4)(|abc|−|abc|)/
√
2, and hence, S(S + 1) = 3/4, S = 1/2.
Exercise 1.9 (a)
(b) Starting with Ψ(1, 1/2, ±1/2), the application of the formula in Eq. 1.43 gives
Ψ(2, 0, 0) = (−
√
0 − 0 + 1βα +
√
0 + 0 + 1αβ)/
√
2 · 0 + 2 = (αβ − βα)/
√
2.
Exercise 1.10 (a) ˆ
S + (ααββ − ββαα) = (αααβ + ααβα − αβαα − βααα), which
is equivalent to the function of Eq. 1.50b, except for the normalization factor. (b)
Starting with (αβ − βα)/
√
2, the singlet coupling for electron 1 and 2, Eq. 1.53