Appendix E: Solutions
231
ˆ
H(2ααββ + 2ββαα − αβαβ − αββα − βααβ − βαβα)/2
√
3 =−J · (1/2
√
3)[
1
2 (2 ·
0+2αββα +2αβαβ +2βαβα+2βααβ−ααββ−ααββ−ααββ−ααββ+2αββα +
2αβαβ + 2βαβα + 2βααβ + 2 · 0 − ββαα − ββαα − ββαα − ββαα) − 2ααββ −
2ββαα]=−J ·(1/2
√
3)[2αββα+2αβαβ +2βαβα+2βααβ−4ααββ−4ββαα]⇒
eigenvalue of singlet is 2J.
Exercise 3.10 Two-electrons: s 1 =
1
2 , s 2 =
1
2 , S a = 0, 1. Three electrons:
S a = 0, 1, s 3 =
1
2 , S b =
1
2 ,
1
2 ,
3
2 . Four electrons: S b =
1
2 ,
1
2 ,
3
2 , s 4 =
1
2 , S tot =
0, 1, 0, 1, 1, 2.
Exercise 3.11 (i) E(T 2,3 ) − E(Q) = J =− 129.7m e V ;E(S 2 ) − E(Q) = J =
−142.4meV;[E(T 1 ) − E(Q)]/2 =−130.0meV;[E(S 1 ) − E(Q)]/3 = J =−117.1
meV. (ii) [E(T 2,3 ) − E(S 2 )]×4 = J r = 50.7meV ;E(T 2,3 ) − E ( Q) = J =−129.7
meV; E(T 1 ) − E Q = 2J −
1
2 J r ⇒ J =
1
2 ∆E +
1
4 J r =−117.4meV ;E(S 1 ) − E Q =
3J +
3
4 J r ⇒ J =
1
3 ∆E −
1
4 J r =− 129.7 meV. (iii) E(S1) −
3
2 (E(T 1 ) − E Q ) =
3
2 J r ⇒ J r = 25.9meV ;E(T 1 ) − E Q = 2J −
1
2 J r ⇒ J =
1
2 (∆E +
1
2 J r =− 123.5
meV; E(T 2,3 ) − E Q = J + J 3 ⇒ J 3 = ∆E − J =−6.2 meV. These three parameters
exactly fit the energy difference E(S2) − E(Q) =−142.4meV= J + 2J 3 −
1
4 J r =
−123.5 − 2 × 6.2 −
1
4 × 25.9.
Exercise 3.12
ˆ
H =
ˆ
S x (1) ˆ
S y (1) ˆ
S z (1)
⎛
⎝
A xx A xy A xz
A yx A yy A yz
A zx A zy A zz
⎞
⎠
⎛
⎝
ˆ
S x (2)
ˆ
S y (2)
ˆ
S z (2)
⎞
⎠
=
A xx ˆ
S x (1) + A yx ˆ
S y (1) + A zx ˆ
S z (1), A xy ˆ
S x (1) + A yy ˆ
S y (1) + A zy ˆ
S z (1), A xz ˆ
S x (1) +
A yz ˆ
S y (1) + A zz ˆ
S z (1)
⎛
⎝
ˆ
S x (2)
ˆ
S y (2)
ˆ
S z (2)
⎞
⎠
=
A xx ˆ
S x (1) ˆ
S x (2) + A yx ˆ
S y (1) ˆ
S x (2) +
A zx ˆ
S z (1) ˆ
S x (2) + A xy ˆ
S x (1) ˆ
S y (2) + A yy ˆ
S y (1) ˆ
S y (2) + A zy ˆ
S z (1) ˆ
S y (2) + A xz ˆ
S x (1) ˆ
S z (2)
+ A yz ˆ
S y (1) ˆ
S z (2) + A zz ˆ
S z (1) ˆ
S z (2).
Exercise 3.13
⎛
⎜
⎜
⎜
⎝
10 0 0
0
1
√
2
1
√
2
0
00 0 1
0
1
√
2
−
1
√
2
0
⎞
⎟
⎟
⎟
⎠
⎛
⎜
⎜
⎝
H 11 H 12 H 13 H 14
H 21 H 22 H 23 H 24
H 31 H 32 H 33 H 34
H 41 H 42 H 43 H 44
⎞
⎟
⎟
⎠
⎛
⎜
⎜
⎜
⎝
100 0
0
1
√
2
0
1
√
2
0
1
√
2
0 −
1
√
2
001 0
⎞
⎟
⎟
⎟
⎠
=
⎛
⎜
⎜
⎜
⎝
H 11
H 12
H 13
H 14
H 21 +H 31
√
2
H 22 +H 32
√
2
H 23 +H 33
√
2
H 24 +H 34
√
2
H 41
H 42
H 43
H 44
H 21 −H 31
√
2
H 22 −H 32
√
2
H 23 −H 33
√
2
H 24 −H 34
√
2
⎞
⎟
⎟
⎟
⎠
⎛
⎜
⎜
⎜
⎝
100 0
0
1
√
2
0 −
1
√
2
0
1
√
2
0
1
√
2
001 0
⎞
⎟
⎟
⎟
⎠
231
ˆ
H(2ααββ + 2ββαα − αβαβ − αββα − βααβ − βαβα)/2
√
3 =−J · (1/2
√
3)[
1
2 (2 ·
0+2αββα +2αβαβ +2βαβα+2βααβ−ααββ−ααββ−ααββ−ααββ+2αββα +
2αβαβ + 2βαβα + 2βααβ + 2 · 0 − ββαα − ββαα − ββαα − ββαα) − 2ααββ −
2ββαα]=−J ·(1/2
√
3)[2αββα+2αβαβ +2βαβα+2βααβ−4ααββ−4ββαα]⇒
eigenvalue of singlet is 2J.
Exercise 3.10 Two-electrons: s 1 =
1
2 , s 2 =
1
2 , S a = 0, 1. Three electrons:
S a = 0, 1, s 3 =
1
2 , S b =
1
2 ,
1
2 ,
3
2 . Four electrons: S b =
1
2 ,
1
2 ,
3
2 , s 4 =
1
2 , S tot =
0, 1, 0, 1, 1, 2.
Exercise 3.11 (i) E(T 2,3 ) − E(Q) = J =− 129.7m e V ;E(S 2 ) − E(Q) = J =
−142.4meV;[E(T 1 ) − E(Q)]/2 =−130.0meV;[E(S 1 ) − E(Q)]/3 = J =−117.1
meV. (ii) [E(T 2,3 ) − E(S 2 )]×4 = J r = 50.7meV ;E(T 2,3 ) − E ( Q) = J =−129.7
meV; E(T 1 ) − E Q = 2J −
1
2 J r ⇒ J =
1
2 ∆E +
1
4 J r =−117.4meV ;E(S 1 ) − E Q =
3J +
3
4 J r ⇒ J =
1
3 ∆E −
1
4 J r =− 129.7 meV. (iii) E(S1) −
3
2 (E(T 1 ) − E Q ) =
3
2 J r ⇒ J r = 25.9meV ;E(T 1 ) − E Q = 2J −
1
2 J r ⇒ J =
1
2 (∆E +
1
2 J r =− 123.5
meV; E(T 2,3 ) − E Q = J + J 3 ⇒ J 3 = ∆E − J =−6.2 meV. These three parameters
exactly fit the energy difference E(S2) − E(Q) =−142.4meV= J + 2J 3 −
1
4 J r =
−123.5 − 2 × 6.2 −
1
4 × 25.9.
Exercise 3.12
ˆ
H =
ˆ
S x (1) ˆ
S y (1) ˆ
S z (1)
⎛
⎝
A xx A xy A xz
A yx A yy A yz
A zx A zy A zz
⎞
⎠
⎛
⎝
ˆ
S x (2)
ˆ
S y (2)
ˆ
S z (2)
⎞
⎠
=
A xx ˆ
S x (1) + A yx ˆ
S y (1) + A zx ˆ
S z (1), A xy ˆ
S x (1) + A yy ˆ
S y (1) + A zy ˆ
S z (1), A xz ˆ
S x (1) +
A yz ˆ
S y (1) + A zz ˆ
S z (1)
⎛
⎝
ˆ
S x (2)
ˆ
S y (2)
ˆ
S z (2)
⎞
⎠
=
A xx ˆ
S x (1) ˆ
S x (2) + A yx ˆ
S y (1) ˆ
S x (2) +
A zx ˆ
S z (1) ˆ
S x (2) + A xy ˆ
S x (1) ˆ
S y (2) + A yy ˆ
S y (1) ˆ
S y (2) + A zy ˆ
S z (1) ˆ
S y (2) + A xz ˆ
S x (1) ˆ
S z (2)
+ A yz ˆ
S y (1) ˆ
S z (2) + A zz ˆ
S z (1) ˆ
S z (2).
Exercise 3.13
⎛
⎜
⎜
⎜
⎝
10 0 0
0
1
√
2
1
√
2
0
00 0 1
0
1
√
2
−
1
√
2
0
⎞
⎟
⎟
⎟
⎠
⎛
⎜
⎜
⎝
H 11 H 12 H 13 H 14
H 21 H 22 H 23 H 24
H 31 H 32 H 33 H 34
H 41 H 42 H 43 H 44
⎞
⎟
⎟
⎠
⎛
⎜
⎜
⎜
⎝
100 0
0
1
√
2
0
1
√
2
0
1
√
2
0 −
1
√
2
001 0
⎞
⎟
⎟
⎟
⎠
=
⎛
⎜
⎜
⎜
⎝
H 11
H 12
H 13
H 14
H 21 +H 31
√
2
H 22 +H 32
√
2
H 23 +H 33
√
2
H 24 +H 34
√
2
H 41
H 42
H 43
H 44
H 21 −H 31
√
2
H 22 −H 32
√
2
H 23 −H 33
√
2
H 24 −H 34
√
2
⎞
⎟
⎟
⎟
⎠
⎛
⎜
⎜
⎜
⎝
100 0
0
1
√
2
0 −
1
√
2
0
1
√
2
0
1
√
2
001 0
⎞
⎟
⎟
⎟
⎠