Appendix E: Solutions
237
Exercise 5.10 Since both the interaction elements NH| ˆ
H|I are larger (t versus
t/
√
2 for triplet and singlet) and the denominators smaller (2 and 4K), one expects a
stronger energy lowering for the triplet energy than for the singlet. This means that
J TQ = (E T − E Q )/2 will be larger (in absolute value) than E ST = E T − E S .
Exercise 5.11 Assuming a square geometry with distance r 1 between neighboring
magnetic sites, the hole-particle contribution to the total energy of Φ α is q 1 · q 2 /r 1 =
−1/r.ForΦ γ , the same contribution arises. In the case of Φ β the hole (+1 charge)
and particle (−1 charge) are at r 2 =
√
2r 1 and the contribution becomes q 1 · q 2 /r 2 =
−1/
√
2r 1 , and hence, the energies of the ionic determinants are not strictly the same.
Exercise 5.12 |abcd|
t ad
− →| bcdd|
t cb
− →| bbdd|
t ba
− →| abdd|
t dc
− →| abcd| or
|abcd|
t ad
− →|bcdd|
t cb
− →|bbdd|
t dc
− →|bbcd|
t ba
− →|abcd|.
Exercise 5.13 I α
=| ϕ 1 ϕ 2 ϕ 2 ϕ 3 |, I β
=| ϕ 1 ϕ 3 ϕ 4 ϕ 4 |. Φ 4 | ˆ
H|I α =
ϕ 1 ϕ 2 ϕ 3 ϕ 4 | ˆ
H|ϕ 1 ϕ 2 ϕ 2 ϕ 3 =− − ϕ 1 ϕ 2 ϕ 3 ϕ 4 | ˆ
H|ϕ 1 ϕ 2 ϕ 3 ϕ 2 =− t 24 , Φ 4 | ˆ
H|I β =
−t 24 .
Exercise 5.14 −J 1 ( ˆ
S A ˆ
S B + ˆ
S C ˆ
S D )αααα =−J 1 (
1
4 αααα +
1
4 αααα) =−
1
2 J 1 αααα;
−J 2 ( ˆ
S A ˆ
S D + ˆ
S B ˆ
S C )αααα =− J 2 (
1
4 αααα +
1
4 αααα) =−
1
2 J 2 αααα; −J 3 ( ˆ
S A ˆ
S C +
ˆ
S B ˆ
S D )αααα =− J 3 (
1
4 αααα +
1
4 αααα) =−
1
2 J 3 αααα; J r ( ˆ
S A ˆ
S B )( ˆ
S C ˆ
S D )αααα =
J r ˆ
S A ˆ
S B
1
4 αααα =
1
16 J r αααα; J r ( ˆ
S A ˆ
S D )( ˆ
S B ˆ
S C )αααα = J r ˆ
S A ˆ
S D
1
4 αααα =
1
16 J r αα
αα; −J r ( ˆ
S A ˆ
S C )( ˆ
S B ˆ
S D )αααα =− J r ˆ
S A ˆ
S C
1
4 αααα =−
1
16 J r αααα ⇒⇒ αααα| ˆ
H|αα
αα=−
1
2 (J 1 + J 2 + J 3 ) +
1
16 J r .
Problem 5.1 The model space is reduced to a 2 × 2 matrix spanned by |ab| and |ba|.
The matrix representation is
|ab| ba
ab| 0 K ab
ba| K ab 0
The corresponding secular determinant leads to the equation E 2 − K 2
ab = 0, which
gives the eigenvalues E 1,2 =± K ab and the eigenfunctions Ψ 1,2 =| ab|±|ba|.The
energy difference is 2K ab and the ground state is the triplet, because K ab is positive.
Problem 5.2 (a) Ψ 1 = (|ab|+|ba|)/
√
2; Ψ 2 = (|ab|−|ba|)/
√
2; Ψ 3 = (|aa|+
|bb|)/
√
2; Ψ 4 = (|aa|−|bb|)/
√
2. (b) Ψ 1 | ˆ
H|Ψ 1 =
1
2 (ab| ˆ
H|ab++ab| ˆ
H|ba+
ba| ˆ
H|ab++ba| ˆ
H|ba) = h aa + h bb + J ab + K ab ; Ψ 2 | ˆ
H|Ψ 2 =
1
2 (ab| ˆ
H|ab−
ab| ˆ
H|ba−− ba| ˆ
H|ab++ ba| ˆ
H|ba) = h aa + h bb + J ab − K ab ; Ψ 3 | ˆ
H|Ψ 3 =
1
2 (aa| ˆ
H|aa++aa| ˆ
H|bb++bb| ˆ
H|aa++bb| ˆ
H|bb) = h aa +h bb +
1
2 (J aa +J bb )+K ab ;
Ψ 4 | ˆ
H|Ψ 4 =
1
2 (aa| ˆ
H|aa−−aa| ˆ
H|bb−−bb| ˆ
H|aa++bb| ˆ
H|bb) = h aa + h bb +
1
2 (J aa + J bb ) − K ab .(c)Ψ 1 = S g ; Ψ 2 = T u ; Ψ 3 = S g ; Ψ 4 = S u ; only Ψ 1 | ˆ
H|Ψ 3 is
non-zero. (d) Ψ 1 | ˆ
H|Ψ 3 =
1
2 (ab| ˆ
H|aa++ab| ˆ
H|bb++ba| ˆ
H|aa++ba| ˆ
H|bb) =