Student Seminar: Classical and Quantum Integrable Systems

Here in the first bracket we consider the terms with n 2 > n 1 +1, while the last bracket
represents the result of action of H on terms with n 2 = n 1 + 1. Using periodicity
conditions we are allowed to make shifts of the summation variables n 1 , n 2 in the
first bracket to bring all the states to the uniform expression |n 1 , n 2 〉. We therefore
get
{
H|ψ〉 = − J ∑
a(n 1 − 1, n 2 )|n 1 , n 2 〉 +
∑
a(n 1 , n 2 − 1)|n 1 , n 2 〉
2
n 2 >n 1
+ ∑
n 2 >n 1 +2
⎧
⎨
− J 2 ⎩
∑
1≤n 1≤L
n 2>n 1+2
a(n 1 + 1, n 2 )|n 1 , n 2 〉 + ∑
a(n 1 , n 2 + 1)|n 1 , n 2 〉 + L − 8
2
n 2>n 1
[
a(n 1 , n 1 + 1)
|n 1 , n 1 + 2〉 + |n 1 − 1, n 1 + 1〉 + L − 4
2
∑
n 2 >n 1 +1
] ⎫ ⎬
|n 1 , n 1 + 1〉
⎭ .
a(n 1 , n 2 )|n 1 , n 2 〉
Now we complete the sums in the first bracket to run the range n 2 > n 1 . This is
achieved by adding and subtracting the missing terms. As the result we will get
H|ψ〉 =
{ ∑
− J 2
−
− J 2 ⎩
n 2>n 1
(
∑
1≤n 1 ≤L
⎧
⎨
a(n 1 − 1, n 2 ) + a(n 1 , n 2 − 1) + a(n 1 + 1, n 2 ) + a(n 1 , n 2 + 1) + L − 8 )
a(n 1 , n 2 ) |n 1 , n 2 〉
2
(
a(n 1 , n 1 )|n 1 , n 1 + 1〉 + a(n 1 + 1, n 1 + 1)|n 1 , n 1 + 1〉 +
+ a(n 1 , n 1 + 1)|n 1 , n 1 + 2〉
} {{ } + a(n 1, n 1 + 2)|n 1 , n 1 + 2〉
} {{ } +L − 8
) }
a(n 1 , n 1 + 1)|n 1 , n 1 + 1〉
2
∑
1≤n 1≤L
[
a(n 1 , n 1 + 1) |n 1 , n 1 + 2〉 + |n 1 − 1, n 1 + 1〉
} {{ } +L − 4
] ⎫ ⎬
|n 1 , n 1 + 1〉
2
⎭ .
The underbraced terms cancel out and we finally get
H|ψ〉 =
{ ∑
− J 2
⎧
⎨
+ J 2 ⎩
n 2 >n 1
(
∑
1≤n 1 ≤L
}
a(n 1 − 1, n 2 ) + a(n 1 , n 2 − 1) + a(n 1 + 1, n 2 ) + a(n 1 , n 2 + 1) + L − 8 )
a(n 1 , n 2 ) |n 1 , n 2 〉
2
⎫
(
) ⎬
a(n 1 , n 1 ) + a(n 1 + 1, n 1 + 1) − 2a(n 1 , n 1 + 1) |n 1 , n 1 + 1〉
⎭ .
If we impose the requirement that
a(n 1 , n 1 ) + a(n 1 + 1, n 1 + 1) − 2a(n 1 , n 1 + 1) = 0 (5.1)
then the second bracket in the eigenvalue equation vanishes and the eigenvalue problem
reduces to the following equation
2(E − E 0 )a(n 1 , n 2 ) = J [ 4a(n 1 , n 2 ) − ∑
a(n 1 + σ, n 2 ) + a(n 1 , n 2 + σ) ] . (5.2)
σ=±1
Substituting in eq.(5.1) the Bethe ansatz for a(n 1 , n 2 ) we get
Ae (p 1+p 2 )n + Be i(p 1+p 2 )n + Ae (p 1+p 2 )(n+1) + Be i(p 1+p 2 )(n+1)
(
)
− 2 Ae i(p 1n+p 2 (n+1)) + Be i(p 2n+p 1 (n+1))
= 0 .
}
– 63 –