# 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. 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...

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Never mind, I got it figured out. For those interested:

Applying $H$ to $|\psi>$, the result can be written as $\sum_{1\leq n_1< n_2\leq L}^L \alpha(n_1,n_2) |n_1,n_2>+\sum_{n_1=1}^L\beta(n_1) |n_1,n_1+1>$,

where $\alpha$ and $\beta$ are functions containing "illegal" terms like $f(n_1,n_1)$. The next step would be demanding $\alpha(n_1,n_2)=Ef(n_1,n_2)$ and $\beta(n)=0$, so that $|\psi>$ is indeed an eigenvalue. Using the Bethe Ansatz for $f(n,n)$ as well then gives the desired result, but requires extending the definition of $f(n_1,n_2)$ to $n_1=n_2$.

Alternatively, when applying $H$ to $|\psi>$, the result can be written as $\sum_{1\leq n_1+1< n_2\leq L}^L \alpha'(n_1,n_2) |n_1,n_2>+\sum_{n_1=1}^L\beta'(n_1) |n_1,n_1+1>$.

In this case we need to require that $\alpha'(n_1,n_2)=Ef(n_1,n_2)$ (this time for $n_2>n_1+1$) and $\beta'(n)=Ef(n,n+1)$. These requirements contain no $f(n,n)$ terms. But to get the condition $2f(n_1,n_1+1)=f(n_1,n_1)+f(n_1+1,n_1+1)$,

we first need to insert the Bethe Ansatz in $\alpha'(n_1,n_2)=Ef(n_1,n_2)$ and calculate $E$, then notice that this equation also holds for $n_2=n_1+1$ (by inserting the found $E$) and then the condition follows from $\beta'(n)-\alpha'(n,n+1)=0$.

So, the answer is yes, the condition does depend on the form of the Bethe Ansatz!

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