18
A. M. Korostil and M. M. Krupa
where
b 0 (k) =
|b R (k)| 2 + J 2 , cos γ (k) =
2B · b R (k)
|b R (k)| 2 + J 2 .
(33a)
The expansion (33) can be written more explicitly in the form
|b(k)| =
∞
n=0
n
p=0
B
p
n (k)(sinθ)
n
(cos ϕ)
p
(sinθ)
n− p
(34)
with expansion coefficients,
B
p
n (k) = (−1)
p 2
n
1
2
n
n
p
(J t
)
n
(sin k y )
p
(sin k x )
n− p
|b R (k)| 2 + J 2
n−1/2 .
(35)
The internal energy then has the corresponding expansion,
U (θ, ϕ) =
∞
n=0
n
p=0
U
p
n (sinθ)
n
(cos ϕ)
p
(sinϕ)
n− p
, U
p
n = −
d k
(2π )
2
B
p
n (k). (36)
Because the integrand is odd under k x → −k x and k y → −k y , coefficients
U
2 p+1
n
= 0, and U
k
2 p+1 = 0, i.e., only terms even in both p and n survive. In
combination with the symmetry of the binomial coefficients, this also leads to the
equality, U
2n−2 p
2n
= U
2 p
2n .
Then, the first terms in the expansion are
U (θ, ϕ) ≈ U
0
0 + U
0
2 sin
2
θ +
1
8
U
0
4 + U
2
4 +
2U
0
4 − U
2
4
cos 4ϕ
sin
4
ϕ
(37)
as in the phenomenological representation (20).
For the gapped half-filled case, it is consistent to expand the integrand in the
|t
| | J limit,
B
0
0 (k) ≈ J + r (k)
2J −
r (k)
J
, B
0
2 (k) ≈ −
(t
)
2
2J
sin
2 k x (1 − 3r (k)),
B
0
4 (k) ≈ −
(t
)
4
8J 3 sin
4 k x , B
2
4 (k) ≈ −
(t
)
4
4J 3 sin
2 k x sin
2 k y , (38)
where r (k) = |b R (k)|
2
/4J
2 .
Substituting (38) into (36) results in the expressions
A. M. Korostil and M. M. Krupa
where
b 0 (k) =
|b R (k)| 2 + J 2 , cos γ (k) =
2B · b R (k)
|b R (k)| 2 + J 2 .
(33a)
The expansion (33) can be written more explicitly in the form
|b(k)| =
∞
n=0
n
p=0
B
p
n (k)(sinθ)
n
(cos ϕ)
p
(sinθ)
n− p
(34)
with expansion coefficients,
B
p
n (k) = (−1)
p 2
n
1
2
n
n
p
(J t
)
n
(sin k y )
p
(sin k x )
n− p
|b R (k)| 2 + J 2
n−1/2 .
(35)
The internal energy then has the corresponding expansion,
U (θ, ϕ) =
∞
n=0
n
p=0
U
p
n (sinθ)
n
(cos ϕ)
p
(sinϕ)
n− p
, U
p
n = −
d k
(2π )
2
B
p
n (k). (36)
Because the integrand is odd under k x → −k x and k y → −k y , coefficients
U
2 p+1
n
= 0, and U
k
2 p+1 = 0, i.e., only terms even in both p and n survive. In
combination with the symmetry of the binomial coefficients, this also leads to the
equality, U
2n−2 p
2n
= U
2 p
2n .
Then, the first terms in the expansion are
U (θ, ϕ) ≈ U
0
0 + U
0
2 sin
2
θ +
1
8
U
0
4 + U
2
4 +
2U
0
4 − U
2
4
cos 4ϕ
sin
4
ϕ
(37)
as in the phenomenological representation (20).
For the gapped half-filled case, it is consistent to expand the integrand in the
|t
| | J limit,
B
0
0 (k) ≈ J + r (k)
2J −
r (k)
J
, B
0
2 (k) ≈ −
(t
)
2
2J
sin
2 k x (1 − 3r (k)),
B
0
4 (k) ≈ −
(t
)
4
8J 3 sin
4 k x , B
2
4 (k) ≈ −
(t
)
4
4J 3 sin
2 k x sin
2 k y , (38)
where r (k) = |b R (k)|
2
/4J
2 .
Substituting (38) into (36) results in the expressions