Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

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...

share|improve this question

1 Answer 1

up vote 3 down vote accepted

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!

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.