6.4. Bethe ansatz equation The solution of the six-vertex model consists in finding the eigenvalues of the transfer matrix T(p)ψ = (A(p)+D(p))ψ = λψ.

This problem can be solved by the algebraic Bethe ansatz, whose main steps are as follows: define the pseudo-vacuum φ, as the state annihilated by the operator C(p)
C(p)φ = 0, ∀p.

a. Prove that φ
{β} = N k=1 δβk,+ =↑ · · · ↑.

b. Prove that φ
{β} is an eigenstate of A and D with eigenvalues A(p)φ = aN(p)φ
D(p)φ = bN(p)φ.
However, applying B to φ results in neither an eigenvector nor zero, B(p)φ = φ,0. This suggests an eigenstate of the transfer matrix in the form ψ = B(p1) . . .B(pn)φ
where the parameters pi are to be determined.

c. Show that, applying A(p) and D(p) to ψ and pushing them through all the Bs by the commutation relations (6.4.71), (A(p)+D(p))ψ = (λA(p)+λB(p))ψ +unwanted terms where λA(p) = aN(p)
n k=1 a(pk −p)
b(pk −p)
, λB(p) = bN(p)
n k=1 a(pk −p)
b(pk −p)
The unwanted terms, coming from the second term in eqn. (6.4.71), contain a B(p)
and so they can never give a vector proportional to ψ, unless they vanish. Show that this happens if the Bethe ansatz equations hold b(pj)
a(pj)

N n k=1 a(pj −pk)
b(pj −pk)
b(pk −pj)
a(pk −bj)
= −1, j = 1, 2, . . .n.
Notice that the eigenvalue problem of the transfer matrix has been transformed in a set of transcendental equations above for the spectral parameters p1, . . .pn. A further elaboration of the solution of the Bethe ansatz equations leads to the expression (6.4.74)
of the free energy of the model.