# Alternative expression for Bethe vectors of su(2) XXX chain.

+ 3 like - 0 dislike
297 views

As is well known, the eigenvectors of the transfer matrix for the su(2) XXX spin chain of length $L$ are given by acting on the ground state $|\uparrow^L\rangle$ with all spins point up with the operator $B(u)$ of the monodramy matrix  Then, a state $B(u_1)\dots B(u_M)|\uparrow^L\rangle$ is an eigenvector of the transfer matrix provided that the rapidities $u_1,\dots,u_M$ satisfy the Bethe equations $\frac{(u_j+i/2)^L}{(u_j-i/2)^L}=\displaystyle\prod_{k\neq j}^M\frac{(u_j-u_k+i)}{(u_j-u_k-i)}$ for $j=1,\dots,M$.

This paper claims that it is possible to write eigenvectors in terms of the $C$ operators acting on the state with all spins pointing down, $|\downarrow^L\rangle$. However, they do not explicitly give the proportionality factor. They do give some arguments for it's validation, such as conservation of spin, etc. This paper, which I studied extensively as part of my Bachelor's thesis, claims that the exchange is given by

$B(u_1)\dots B(u_M)|\uparrow^L\rangle=\frac{(-1)^M}{(L-2M)!}(S^+)^{L-2M}C(u_1)\dots C(u_M)|\downarrow^L\rangle$ provided that the rapidities satisfy the Bethe equations. Here $S^+$ is the global raising operator which satisfies various commutation relations with the entries of the monodromy matrix, which can be found in review articles such as that of Faddeev.

However, this constant of proportionality has been something which has eluded me a bit, I tried to prove it using a brute force approach. For $M=1$ I have the following result (the rapidities can be off-shell)

$B(u)|\uparrow^L\rangle=\left(\frac{(S^+)^{L-1}}{(L-1)!}(D(u)-A(u))-\frac{(S^+)^{L-2}}{(L-2)!}C(u)\right)|\downarrow^L\rangle$

which can be shown easily by noting that $|\uparrow^L\rangle=\frac{(S^+)^{L}}{L!}|\downarrow^L\rangle$ and then performing induction on the number of sites. For $M=1$ the Bethe equations reduce to $(D(u)-A(u))|\uparrow^L\rangle=0$ and we do indeed obtain the desired prefactor.

Now for $M=2$ we need to act on the above expression with $B(v)$ where now $u$ and $v$ are on-shell. We need to use the full expression since we can no longer state that  $(D(u)-A(u))|\uparrow^L\rangle=0$ (what we actually have is $(A(u)A(v)-D(u)D(v))|\uparrow^L\rangle=0$)

After some work we obtain an expression with six terms; one which vanishes identically, another which is our desired on shell result, $\frac{(S^+)^{L-4}}{(L-4)!}C(v)C(u)|\downarrow^L\rangle$, and four other terms which can be grouped into pairs of powers of $S^+$. However, from what I can tell, the terms do not vanish. Even by solving the Bethe equations for the case $L=4,M=2$ and plugging in the values they are still non-zero. I also tried appealing to the symmetry of the vectors in $u$ and $v$ but this didn't achieve anything. It is also not possible to proceed by straight forward induction on $M$ since we cannot state anything such as "if $B(u_1)\dots B(u_k)|\uparrow^L\rangle$ is an eigenvector then so is $B(u_1)\dots B(u_k)B(u_{k+1})|\uparrow^L\rangle$".

I will post the non-vanishing terms later when I get time but in the mean time I would be interested in hearing some opinions on the matter. Specifically, where am I screwing up?

asked Oct 26, 2015
edited Oct 26, 2015

## Your answer

 Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead. To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL. Please consult the FAQ for as to how to format your post. This is the answer box; if you want to write a comment instead, please use the 'add comment' button. Live preview (may slow down editor)   Preview Your name to display (optional): Email me at this address if my answer is selected or commented on: Privacy: Your email address will only be used for sending these notifications. Anti-spam verification: If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:$\varnothing\hbar$ysicsOverflowThen drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds). To avoid this verification in future, please log in or register.