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I had a couple of questions regarding the coordinate Bethe ansatz after trying to follow along in some lecture notes.

  1. The first one regards showing that the $H_{XXX}$ Hamiltonian preserves down spins. This was claimed without proof. The way I figured to go about this was to show that $S^{z} = \sum_{i}S^{z}_{i}$ commutes with the Hamiltonian. We also note that $\left[S^{a}_{i}, S^{b}_{j} \right] = \delta_{ij} \epsilon^{abc}S^{c}_{i}$. My work is as follows:

$\left[S^{z}, H \right] = \left[S^{z},\sum_{i}J_{x}S_{i}^{x}S^{x}_{i+1} + J_{y}S_{i}^{y}S^{y}_{i+1} + J_{z}S_{i}^{z}S^{z}_{i+1} \right] = \sum_{i}J_{x}(\left[S^{z}_{i},S^{x}_{i} \right]S_{i+1}^{x} + S_{i}^{x}\left[S_{i}^{z},S_{i+1}^{x} \right]) + J_{y}(\left[S^{z}_{i},S^{y}_{i} \right]S_{i+1}^{y} + S_{i}^{y}\left[S_{i}^{z},S_{i+1}^{y} \right]) + J_{z}(\left[S^{z}_{i},S^{z}_{i} \right]S_{i+1}^{z} + S_{i}^{z}\left[S_{i}^{z},S_{i+1}^{z} \right]) = J_{x}\left[S_{i}^{z}, S_{i}^{x} \right]S^{x}_{i+1} + J_{y}\left[S_{i}^{z}, S^{y}_{i} \right]S^{z}_{i+1} = J(S^{y}_{i}S^{x}_{i+1} - S^{x}_{i}S^{y}_{i+1}).$

where I applied two commutator identities to go to the second equality, I applied the spin commutator to get to the third equality, and then applied it again to get to the fourth equality (also setting $J_{x}=J_{y}$ for XXX and XXZ). Am I allowed to re-define the indicies in the first term, i.e. $x \rightarrow y$, $y \rightarrow x$ so that

$J(S^{x}_{i}S^{y}_{i+1} - S^{x}_{i}S^{y}_{i+1}) = 0$? If so, what are the conditions by which I would know how one could rename indicies in this way?

  1. The kets for spin are given by: $S^{+}\vert \uparrow \rangle = 0, S^{-}\vert\uparrow\rangle = \vert\downarrow \rangle, S^{+}\vert\downarrow \rangle = \vert \uparrow \rangle, S^{-}\vert \downarrow \rangle = 0, S^{z}\vert\uparrow\rangle = \frac{1}{2}\vert\uparrow\rangle, S^{z}\vert\downarrow\rangle = -\frac{1}{2}\vert\downarrow\rangle $

I also note that $\vert n_{1}, n_{2}, ..., n_{N} \rangle = S^{-}_{n_{1}}S^{-}_{n_{2}}...S^{-}_{n_{N}}\vert\Omega \rangle$ and $\vert\Omega\rangle = \vert \uparrow \uparrow ...\rangle$ is the vacuum eigenvector of all spins up.

I was wondering if someone would be able to explain in a bit more detail how to compute

$\sum_{l=1}^{L}S^{+}_{l}S^{-}_{l+1}\vert n \rangle$,

$\sum_{l=1}^{L}S^{-}_{l}S^{+}_{l+1}\vert n \rangle$,

$\sum_{l=1}^{L}S^{z}_{l}S^{z}_{l+1}\vert n \rangle$.

I tried computed the first one and obtained $\vert n+1 \rangle$ by the first $S^{-}_{l+1}$ acting on the $\vert n \rangle$ turning a spin from up to down at the (n+1)-th site, and then the $S^{+}_{l}$ operator acting on the ket to flip the spin at the nth site from down to up. Hence we are left with an (n+1)-ket. When I tried doing the second one, it appeared that the whole ket was killed when I applied the first operator. I also didn't obtain the third value of $\frac{L-4}{4}$, so I am definitely missing something and would appreciate some help.

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  • $\begingroup$ \ket isn't a command in Mathjax, you can use \vert \uparrow \rangle: $\vert \uparrow \rangle$ $\endgroup$
    – Claudio
    Commented Sep 10 at 13:24
  • $\begingroup$ Claudio, thank you for the tip. I have fixed all of the kets! $\endgroup$ Commented Sep 11 at 5:23

1 Answer 1

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  1. You're missing some terms in the third equality: the sum disappeared and you've dropped the terms where you take the commutators with the second spin operator in each product.

  2. a) That's correct. b) The result is nonzero if $l+1=n$. Hint: first work out the commutators between $S^+_i, S^-_j, S^z_k$ and then do the computation more carefully. c) Hint: what is the eigenvalue of $S^z_i$ on $\vert \Omega\rangle$?

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    $\begingroup$ Hi Jules, thanks for the hints. I will try to get to them later today. As for 1, I thought the terms where I took the commutators with the second spin operator in each product were all zero due to i $\neq$ (i+1). This could be one of the points of my confusion. $\endgroup$ Commented Sep 11 at 17:42
  • $\begingroup$ It might help to replace the dummy index of one sum by $j$ and carefully use the commutation relations $\endgroup$ Commented Sep 11 at 18:48
  • $\begingroup$ I will definitely post my solutions when I have had some time to go through everything. Thanks! $\endgroup$ Commented Sep 12 at 7:33
  • $\begingroup$ I don't think that'd be needed, this is a very standard computation. You can just lmk if my hints were enough. $\endgroup$ Commented Sep 12 at 13:44
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    $\begingroup$ Hi Jules, I got them all now. Thanks for your hints. For 2b, I didn't need to use the commutators to obtain the right answer. I never worked with spins and eigenvalue sums before, and when I realized that I should be viewing the operations in 2. in pairs, it made the whole problem much more obvious. Thank you for giving hints and not the answer outright. $\endgroup$ Commented Sep 16 at 4:30

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