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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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2 Answers 2

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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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Much later than the OP, but I just came across your question and think I can add another perspective to your answer. It's a somewhat long story, but I hope that (for the patient reader) it makes clear what is happening on a more conceptual level; maybe this will be useful for someone at some point.

Let us first consider a sector with any (fixed) number $M$ of magnons. It has (coordinate) basis that I will denote by $$\lvert n_1,\dots,n_M\rangle\!\rangle \equiv \sigma^-_{n_1} \, \cdots \sigma^-_{n_M} \lvert \uparrow \cdots \uparrow\rangle, \qquad 1\leqslant n_1 < \dots < n_M \leqslant L \, ,$$ where the restriction on the positions $n_m$ of the $\downarrow$s avoids overcounting. The wave function of any vector in this sector is $f(n_1,\dots,n_M) \equiv \langle\!\langle n_1,\dots,n_M \vert \psi\rangle$.

The spin chain is translationally invariant, where the (cyclic) translation (shift) operator acts by $G \, \lvert n_1,\dots,n_M\rangle\!\rangle = \lvert n_1 - 1,\dots,n_M -1\rangle\!\rangle$ if $n_1 >1$ and $G \, \lvert 1,n_2,\dots,n_M\rangle\!\rangle = \lvert n_2 -1, \dots, n_M-1,L\rangle\!\rangle$. Here I was careful to write the result for the latter in terms of the coordinate basis (with allowed range of $n_m$). It's a little annoying that we have to treat the case $n_1 = 1$ separately, as in the other answer. So let us instead use the periodicity $\sigma^\alpha_{L+1} = \sigma^\alpha_1$ (periodic boundary conditions of the local spin operators) to extend the coordinate basis to $\lvert n_1,\dots, n_M\rangle\!\rangle$ where we now allow for any $n_1 < \dots < n_M$ with the condition $n_M - n_1 < L$, and we identify basis vectors according to $\lvert n_1,\dots,n_M\rangle\!\rangle \equiv \lvert n_2, \dots, n_M, n_1 + L\rangle\!\rangle$ to avoid overcounting. This just amounts to different choices of where we cut open the circle to write down our basis on an interval $\{n,n+1,\dots,n+L\}$ to avoiding overcounting. In terms of eigenvectors we thus extended the wave function $f(n_1,\dots,n_M)$ to any arguments such that $n_1 < \dots < n_M$ with $n_M - n_1 < L$ and we now have to impose the periodic boundary conditions in the form $$f(n_1,\dots,n_M) = f(n_2, \dots, n_M, n_1 + L). \tag{PBC} \label{PBC}$$

So far we only use properties of the coordinate basis, and translation invariance of our system.

Now we focus on $M=2$ magnons as in the question. From the Heisenberg Hamiltonian (let me set $J=1$ to simplify notation) we compute what the eigenequation looks like in terms of the coordinate basis by applying $\langle\!\langle n_1,n_2|$ to $H\,\lvert\psi\rangle = E \, \lvert\psi\rangle$ to get the (difference) equations $$f(n_1 - 1, n_2) + f(n_1 +1,n_2 + 1) + f(n_1, n_2 - 1) + f(n_1,n_2 + 1) - 4\,f(n_1,n_2) = E \, f(n_1,n_2) \tag{I} \label{one}$$ for the ('well separated') cases where $n_1,n_2$ are not adjacent on the circle, and $$f(n_1 - 1, n_2) + f(n_1,n_2 + 1) - 2\,f(n_1,n_2) = E \, f(n_1,n_2) \tag{II} \label{two}$$ for the case when $n_2 = n_1 + 1$. Because we extended our coordinate basis and wave function the latter automatically includes the case $n_1 = 1$, $n_2 = L$. So far we have used the tranlational invariance and the explicit form of the Heisenberg hamiltonian, and the extension of the wave function by periodicity.

At this point it is tempting to use \eqref{one} to simplify \eqref{two}, but we are not allowed to do this because that would violate the allowed range of $n_1 < \dots <n_M$. To this end we (formally) extend the domain of the wave function a little further: we assume that $f$ can be extended to any $n_1\leqslant \dots \leqslant n_M$ with $n_M - n_1 < L$. Note that $f(\dots,n,n,\dots)$ are non-physical values of the wave function: since $(\sigma^-_n)^2 = 0$ the corresponding coordinate basis vector $\lvert \dots,n,n,\dots \rangle\!\rangle =0$ vanishes. So in principle we could set such values of $f$ to whatever value we like; but of course we have to make sure everything is consistent. Now we can use \eqref{one} with $n_1 = n$ and $n_2 = n+1$ to get the equation from the OP, $$ f(n,n) + f(n+1,n+1) - 2\,f(n,n+1) = 0 \, , \tag{I'} \label{three}$$ and either of the previous two equations, say \eqref{two}, now extended to all $n_1 \leqslant n_2$ with $n_2 - n_1 < L$ including the well-separated case. Just like \eqref{PBC} cuts back the space of all possible wave functions $f$ solving our wave equation \eqref{two} to those wave functions that are compatible with periodicity, \eqref{three} cuts it back further to be compatible with our second extension of the domain of $f$.

So far we have used translational invariance to extend the wave function once, and assumed it can be (formally) extended to (nonphysical) cases with coinciding arguments. This still does not rely on the Bethe ansatz!

What the Bethe ansatz finally does, is actually solve the wave equation as well as the two constraints related to our extensions. This gives the additive energies $E(k_1,k_2) = E(k_1) + E(k_2)$ from \eqref{two}, determines ratio $A/B$ of the coefficients in the Bethe ansatz as a function of $k_1,k_2$ by \eqref{three}, and yields the Bethe-ansatz equations from \eqref{PBC}.

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