Зайцевский А.В. Релятивистская теория электронного строения молекул
Подождите немного. Документ загружается.
0
+
− 0
−
T
0
G s
(2s + 1)
SU(2)
G T
0
G
T
0
× T
s
T
s
G
T
s
j = s
C
2v
s = 1
I I C
2
, C
2
I σ
v
, σ
v
I σ
v
′
, σ
v
′
I
T
s
A
2
B
1
B
2
A
1
A
2
B
1
B
2
3
A
2
−→ A
1
, B
1
, B
2
3
B
1
−→ A
1
, A
2
, B
2
3
B
2
−→ A
1
, A
1
, B
1
T
s
T
s
= 0
+
0
−
) ⊕ 1 ⊕2 ⊕ . . . ⊕ |s|
T
s
= 1/2 ⊕ 3/2 ⊕ . . . ⊕ |s|
g
m
s
= 0
π
• s |s, m
s
= 0i 0
+
• 0
−
3
Σ
−
u
0
−
u
D
∞h
0
−
g
s 1
g
3
Σ
−
u
−→ 0
−
u
× (0
−
g
⊕ 1
g
) = 0
+
u
⊕ 1
u
.
C
1
I I
G
g
k
K
tr
a
k
=
K
tr
g
k
G
Ψ
µ
G
T(a
k
) : Ψ
µ
a
k
−→
X
ν
Ψ
ν
T(a
k
)
νµ
, T (g
k
) : Ψ
µ
g
k
−→
X
ν
Ψ
ν
T(g
k
)
νµ
.
T(g
k
), T(a
k
)
a
k
g
l
Ψ
µ
= g
k
K
tr
g
l
Ψ
µ
= g
k
X
ν
K
tr
Ψ
ν
T(g
l
)
νµ
= g
k
X
ν
T(g
l
)
∗
νµ
K
tr
Ψ
ν
=
X
λν
T(g
l
)
∗
νµ
T(a
k
)
λν
Ψ
λ
,
T(a
k
g
l
) = T(a
k
)T(g
l
)
∗
T(g
k
), T(a
k
)
G
{Ψ
µ
} −→ {Ψ
′µ
=
X
ν
Ψ
ν
U
νµ
}
T(g
k
) −→ U
−1
T(g
k
)U
T(a
k
) −→ U
−1
T(a
k
)U
∗
.
G
|m|
C
2v
E C
2v
{Ξ
k
}
Ψ =
X
k
C
k
Ξ
k
.
{Ξ
k
}
ψ
L
=
X
K
C
LK
ξ
LK
, ψ
S
=
X
k
C
Sk
ξ
Sk
ξ
LK
ξ
Sk
1p
ψ
L
{Ξ
k
} {
ξ
LK
0
}
{
0
ξ
Sk
}
hΨ|H
D
|Ψi
hΨ|Ψi = 1
H
D
C = SCE
H
D
S hΞ
k
|H
D
|Ξ
k
′
i
hΞ
k
|Ξ
k
′
i
C C
k
C
{C
LK
} {C
Sk
} C =
C
L
C
S
V
LL
C
L
+ c(σ, p)
LS
C
S
= ES
LL
C
L
c(σ, p)
SL
C
L
+ V
SS
C
S
= (E + 2c
2
)S
SS
C
S
A
A
LL
= khξ
LK
|A|ξ
LK
′
ik
A
LS
= khξ
LK
|A|ξ
Sk
ik
S
LL
S
SS
2c
2
V
LL
C
L
+ (σ, p)
LS
C
S
= EC
L
(σ, p)
SL
C
L
− 2C
S
= 0
C
S
=
V
LL
+
1
2
(σ, p)
LS
(σ, p)
SL
!
C
L
= EC
L
.
V
LL
+
1
2
(p
2
)
LL
!
C
L
= EC
L
,
1
2
(p
2
)
LL
=
1
2
((σ, p)(σ, p))
LL
O
{ξ
Sk
} (1−O)
1 = O + (1 − O)
1
2
(p
2
)
LL
=
1
2
((σ, p)O(σ, p))
LL
+
1
2
((σ, p)(1 − O)(σ, p))
LL
1
2
N
X
k=1
(σ, p)
Kk
(σ, p)
kK
′
N {ξ
Sk
}
{ξ
Sk
} {ξ
Sn
} n = N + 1, N + 2, . . . ∞
(1 − O) =
∞
X
n=N+1
|ξ
Sn
ihξ
Sn
|
1
2
∞
X
n=N+1
(σ, p)
Kn
(σ, p)
nK
′
{ξ
Sk
}
{ξ
LK
}
{ξ
Sk
}
(σ, p)ξ
LK
O(σ, p)ξ
LK
= (σ, p)ξ
LK
, (1 − O)(σ, p)ξ
LK
= 0
c → ∞ c
c
−1
Ψ
HF
=
1
√
n!
det
Ψ
1
(1)Ψ
2
(2)...Ψ
n
(n)
Ψ
u
u = 1, . . . , n
g
np
12
E
HF
=
occ
X
u
hΨ
u
| H
D
|Ψ
u
i +
1
2
occ
X
u,v
( hΨ
u
Ψ
v
| g
np
12
|Ψ
u
Ψ
v
i − hΨ
u
Ψ
v
| g
np
12
|Ψ
v
Ψ
u
i )
hΨ
u
| H
D
|Ψ
u
i =
Z
1
Ψ
u
(1)
†
H
D
(1)Ψ
u
(1)d1,
hΨ
u
Ψ
v
| g
np
12
|Ψ
u
Ψ
v
i =
Z
1
Z
2
Ψ
u
(1)
†
Ψ
v
(2)
†
g
np
12
Ψ
u
(1)Ψ
v
(2) d2 d1
hΨ
u
Ψ
v
| g
np
12
|Ψ
v
Ψ
u
i =
Z
1
Z
2
Ψ
u
(1)
†
Ψ
v
(2)
†
g
np
12
Ψ
v
(1)Ψ
u
(2) d2 d1
Ψ
u †
AΨ
v
≡
P
4
i=1
ψ
u ∗
i
Aψ
v
i
Λ
+
(1)Ψ
u
(1) = Ψ
u
(1), hΨ
u
| (Λ
+
(1) − 1) |Ψ
u
i = 0
g
np
12
f
+
Ψ
v
= ε
v
Ψ
v
, f
+
= Λ
+
fΛ
+
= Λ
+
(H
D
+ J − K) Λ
+
,
J K
J =
X
v
Z
2
Ψ
v
(2)
†
g
12
Ψ
v
(2)d2, K : KΨ
u
(1) =
X
v
Ψ
v
(1)
Z
2
Ψ
v
(2)
†
g
12
Ψ
u
(2)d2