Undefined amplitudes in the Coordinate Bethe Ansatz for the XXX model?
Rather specific question for someone familiar with the Coordinate Bethe Ansatz... I am considering the Heisenberg XXX-model, consisting of a one-dimensional chain of L sites with a spin-1/2 particle at each site and periodic boundary conditions, i.e. $S_{n+L}=S_n$. The Hamiltonian is given by
$H=-\frac{J}{2} \sum_{n=1}^L S_n^+S_{n+1}^-+S_n^-S_{n+1}^+ +S_n^zS_{n+1}^z$ with $J$ a coupling constant.
Choosing the $z$-axis as a quantization axis, we can write $S^z = \frac{L}{2} - M $, where $M$ is the number spins down. 656
Due to conservation of $S^z$ we can find the eigenvectors of the Hamiltonian by looking at each value for $M$ separately.
For $M=2$, write a state as $|\psi> =\sum_{1\leq n_1< n_2\leq L}^L f(n_1,n_2) |n_1,n_2>$, where $|n_1,n_2>$ denotes the basis state where the spins at site $n_1$ and $n_2$ are down. The Coordinate Bethe Ansatz for the eigenvectors is
$f(n_1,n_2)=Ae^{i(k_1n_1+k_2n_2)}+Be^{i(k_2n_1+k_1n_2)}$ with A and B constants.
Applying the Hamiltonian to $|\psi>$, without using the Coordinate Bethe Ansatz, then yields an equation for the eigenvalue, as well as the following condition:
$2f(n_1,n_1+1) = f(n_1,n_1) + f(n_1+1,n_1+1)$.
Now the question: the condition above was derived without use of the Bethe Ansatz (see for example these notes, pages 62-63). It contains amplitudes $f(n_1,n_1)$, which are however not defined by the general expansion $|\psi>$ since there we have that $n_2>n_1$! Only by inserting the Bethe Ansatz afterwards, we yield the desired equations to solve for the spectrum of $H$. Can we make sense of this condition without using the Bethe Ansatz, i.e. why should it be well defined? Also, why does the Bethe Ansatz also have to hold for $n_2=n_1$ here? I could imagine just defining $f(n_1,n_1) = 0$ since the amplitude $f(n_1,n_1)$ doesn't appear in the expansion $|\psi>$ anyways.
I hope this is clear... 66