230
Appendix E: Solutions
Exercise 3.4 Using a and b as shorthand notation for ψ a and ψ b ,
Ψ cov (0, 0)| ˆ
H|Ψ cov (0, 0)=
1
2 (ab| ˆ
H|ab++ab| ˆ
H|ba++ba| ˆ
H|ab++ba| ˆ
H|ba) =
1
2 (2a| ˆ
h|a+2b| ˆ
h|b++ ab|
1− ˆ
P 12
r 12
|ab++ ab|
1− ˆ
P 12
r 12
|ba++ ba|
1− ˆ
P 12
r 12
|ab+
ba|
1− ˆ
P 12
r 12
|ba) = h aa +h bb +J ab +K ab ; Ψ cov (1, 0)| ˆ
H|Ψ cov (1, 0)=
1
2 (ab| ˆ
H|ab−
ab| ˆ
H|ba−−ba| ˆ
H|ab++ba| ˆ
H|ba) =
1
2 (2a| ˆ
h|a+2b| ˆ
h|b++ab|
1− ˆ
P 12
r 12
|ab−
ab|
1− ˆ
P 12
r 12
|ba−−ba|
1− ˆ
P 12
r 12
|ab++ba|
1− ˆ
P 12
r 12
|ba) = h aa + h bb + J ab − K ab .T h e
energy difference is 2K ab .
Exercise 3.5 |φ a φ b |=
ψ a +νψ b
√
1+ν 2 ·
ψ b +νψ a
√
1+ν 2 = (ψ a ψ b +νψ b ψ b +νψ a ψ a +ν 2 ψ b ψ a )/
(1 + ν 2 ); |φ b φ a |=
ψ b +νψ a
√
1+ν 2 ·
ψ a +νψ b
√
1+ν 2 = (ψ b ψ a + νψ b ψ b + νψ a ψ a + ν 2 ψ a ψ b )/(1 +
ν 2 ). |φ a φ b |+|φ b φ a |=
1
1+ν 2 ((1 + ν 2 )(ψ a ψ b + ψ b ψ a ) + 2ν(ψ a ψ a + ψ b ψ b )) =
ψ a ψ b +ψ b ψ a +S ab (ψ a ψ a +ψ b ψ b ). Multiplying with 1/
√
2 + 2S 2 leads to (ψ a ψ b +
ψ b ψ a )/
√
2forS ab = 0.
Exercise 3.6 Maximum spin S max = S 1 + S 2 = 2S; E(S max ) =−
1
2 J
2S(2S + 1) −
2S(S + 1)
=− JS 2 ; Minimum spin S min = S 1 − S 2 = 0; E(S min ) =−
1
2 J
0 −
2S(S + 1)
= JS(S + 1).
Exercise 3.7
3 2 11
6
3
1 1
1 83
4
7 / 2 2 8
1731 3
49 / 2 7 / 2
4
Exercise 3.8 S min S S max , S = 0, 1 ⇒ 1(1 + 1)(2 · 1 + 1) exp(J · 1(1 +
1)/2kT )/(1 + 3exp(J · 1(1 + 1)/2kT )) = (6exp(J/kT ))(1 + 3exp(J/kT )) =
6/(3 + exp(−J/kT )). Multiplying with N A µ 2
B g 2
e /3kT one arrives at 2N A µ 2
B g 2
e /kT
(3 + exp(−J/kT )).
Exercise 3.9 (ˆ s
+
1 +ˆ s
+
2 )(ˆ s
−
3 +ˆ s
−
4 )ααββ = (ˆ s
+
1 +ˆ s
+
2 )(0 + 0) = 0; (ˆ s
+
1 +ˆ s
+
2 )(ˆ s
−
3 +
ˆ
s
−
4 )ββαα = (ˆ s
+
1 +ˆ s
+
2 )(βββα + ββαβ) = αββα + αβαβ + βαβα + βααβ; (ˆ s
+
1 +
ˆ
s
+
2 )(ˆ s
−
3 +ˆ s
−
4 )αβαβ = (ˆ s
+
1 +ˆ s
+
2 )(αβββ + 0) = ααββ; (ˆ s
+
1 +ˆ s
+
2 )(ˆ s
−
3 +ˆ s
−
4 )αββα =
(ˆ s
+
1 +ˆ s
+
2 )(0 + αβββ) = ααββ; (ˆ s
+
1 +ˆ s
+
2 )(ˆ s
−
3 +ˆ s
−
4 )βαβα = (ˆ s
+
1 +ˆ s
+
2 )(0 +
βαββ) = ααββ; (ˆ s
+
1 +ˆ s
+
2 )(ˆ s
−
3 +ˆ s
−
4 )βααβ = (ˆ s
+
1 +ˆ s
+
2 )(βαββ + 0) = ααββ;
(ˆ s
−
1 +ˆ s
−
2 )(ˆ s
+
3 +ˆ s
+
4 )ααββ = (ˆ s
−
1 +ˆ s
−
2 )(αααβ + ααβα) = βααβ + βαβα+ αβαβ +
αββα; (ˆ s
−
1 +ˆ s
−
2 )(ˆ s
+
3 +ˆ s
+
4 )ββαα = 0; (ˆ s
−
1 +ˆ s
−
2 )(ˆ s
+
3 +ˆ s
+
4 )αβαβ = (ˆ s
−
1 +ˆ s
−
2 )(0 +
αβαα) = ββαα; (ˆ s
−
1 +ˆ s
−
2 )(ˆ s
+
3 +ˆ s
+
4 )αββα = (ˆ s
−
1 +ˆ s
−
2 )(αβαα + 0) = ββαα;
(ˆ s
−
1 +ˆ s
−
2 )(ˆ s
+
3 +ˆ s
+
4 )βαβα = (ˆ s
−
1 +ˆ s
−
2 )(βααα + 0) = ββαα; (ˆ s
−
1 +ˆ s
−
2 )(ˆ s
+
3 +
ˆ
s
+
4 )βααβ = (ˆ s
−
1 +ˆ s
−
2 )(0 + βααα) = ββαα; (ˆ s z,1 +ˆ s z,2 )(ˆ s z,3 +ˆ s z,4 )ααββ =
(ˆ s z,1 +ˆ s z,2 )(−
1
2 −
1
2 )ααββ = (
1
2 +
1
2 )(−
1
2 −
1
2 )ααββ =−ααββ; (ˆ s z,1 +ˆ s z,2 )(ˆ s z,3 +
ˆ
s z,4 )ββαα = (−
1
2 −
1
2 )(
1
2 +
1
2 )ββαα =−ββαα. The action of (ˆ s z,1 +ˆ s z,2 )(ˆ s z,3 +ˆ s z,4 )
on all other determinants gives zero. Triplet: ˆ
H(ααββ − ββαα)/
√
2 =− J(0 −
1
2 (αββα+αβαβ +βαβα+βααβ)+
1
2 (βααβ +βαβα+αβαβ +αββα)−0−ααββ+
ββαα)/
√
2 =−J(−ααββ + ββαα)/
√
2 ⇒ eigenvalues of the triplet is J. Singlet: