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I have been given the Hamiltonian of system (particle in a ring),

$${H = \sum_{0}^{N}\left[ − V_1(|g_n\rangle \langle g_{n+1}| + |g_{n+1}\rangle \langle g_n|) − V_2(|g_n\rangle \langle g_{n+2}| + |g_{n+2}\rangle \langle g_n|)\right]}$$

It's expressed in the basis ${\langle m\vert g_n\rangle = \delta_{mn}}$

How can I re-express this in the basis, ${|h_m\rangle: h_{m}(n) = e^{2\pi inm/N}}$ where, n=1,2,3.....N-1 and show that it is diagonal?

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  • $\begingroup$ Something simpler, but the same idea of a similarity transformation physics.stackexchange.com/questions/93884/… $\endgroup$
    – user108787
    Sep 18, 2016 at 2:18
  • $\begingroup$ I tried follow the same procedure! but I think I got a contradiction! could you please help me? $\endgroup$ Sep 18, 2016 at 2:21
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    $\begingroup$ Hi danula, if you show your attempt, you have a much better chance of an answer, other your question might be closed, sorry. It's 3.30am where I am, and there are much more experienced people than I, so I would strongly suggest that you expand your question to show your attempt, even on a small part of H. $\endgroup$
    – user108787
    Sep 18, 2016 at 2:28
  • $\begingroup$ Unrelated, but it's easy to do bras and kets in latex, just use the commands \rangle and \langle. e.g. |g_{n+1}\rangle -> $|g_{n+1}\rangle$ $\endgroup$ Sep 18, 2016 at 2:43
  • $\begingroup$ Your edit doesn't fix the problems brought up by @flippiefanus. It should almost certainly be $\langle g_m|g_n\rangle = \delta_{mn}$. And what does $|h_m\rangle = e^{(...)}$ mean? You can't set a basis vector equal to that expression, as it's meaningless. That should almost certainly be a projection $\langle g_n|h_m\rangle$ or $\langle h_m|g_n\rangle = e^{(...)}$. Where did you get this problem? Could you look back at the exact problem statement? $\endgroup$ Sep 18, 2016 at 13:11

2 Answers 2

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There are a number of things in the question that is a bit confusing. First, I assume the orthogonality relation actually reads: $$\langle g_m|g_n\rangle=\delta_{mn}$$ Second, one usually have explicit expression only when ket are contracted on bra's. In other words, I'd expect to see $$\langle h_m|?\rangle=\exp(i2\pi mn/N) . $$ Here I'll assume the $|?\rangle$ is $|g_n\rangle$. So then we have $$\tag{1} \langle h_m|g_n\rangle=\exp(i2\pi mn/N) . $$

To do the transformation one needs the completeness relation (also called the resolution of unity) $$\sum_m |h_m\rangle\langle h_m|={\cal I} , $$ where ${\cal I}$ is the identity. Then one inserts the identity on both sides of the Hamiltonian and expands them in terms of the completeness relations. Make sure to use different indices for the two new summations. Then one uses (1) and its Hermetian conjugate to convert the contractions between the two bases into exponentials. One can then evaluate the sum over $n$ (the original index in the Hamiltonian). Somehow this should produce Kronecker deltas with which you can remove one of the two new summations, thus ending up with one summation over diagonal terms.

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  • $\begingroup$ Both basis are orthogonal and complete! And I have done a mistake expressing ⟨hm|=e2πinm/N actually it's not a bra. it's the ket I have corrected that in the question. $\endgroup$ Sep 18, 2016 at 12:27
  • $\begingroup$ Both basis are orthogonal and complete! And I have done a mistake expressing ⟨hm|=e2πinm/N actually it's not a bra. it's the ket I have corrected that in the question. @Kyle Arean-Raines I'm confused about how did you handle kronika-deltas. <hm|gm> =e^(2πinm/N) δmn $\endgroup$ Sep 18, 2016 at 12:36
  • $\begingroup$ The way it is expressed still doesn't make sense to me. Where did you find this question? $\endgroup$ Sep 20, 2016 at 15:45
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Starting with your Hamiltonian in the $|g\rangle$ basis, you can use the completeness relation $\sum_{\alpha} |\alpha\rangle\langle \alpha| = 1$ (twice) to get

$$H = \sum_{m, m', n} \left[-V_1(|h_{m}\rangle\langle h_{m}|g_n\rangle\langle g_{n+1}|h_{m'}\rangle\langle h_{m'}|+|h_{m}\rangle\langle h_{m}|g_{n+1}\rangle\langle g_n|h_{m'}\rangle\langle h_{m'}|)-V_2(|h_{m}\rangle\langle h_{m}|g_n\rangle\langle g_{n+2}|h_{m'}\rangle\langle h_{m'}|+|h_{m}\rangle\langle h_{m}|g_{n+2}\rangle\langle g_n|h_{m'}\rangle\langle h_{m'}|) \right]$$

Then use the expression you have for the projection $\langle h | g \rangle$

$$H = \sum_{m,m',n} \left[-V_1 (e^{2\pi i n m / N}e^{-2\pi i (n+1)m' / N}|h_{m}\rangle\langle h_{m'}| + e^{2\pi i (n+1) m / N}e^{-2\pi i n m' / N}|h_{m}\rangle\langle h_{m'}|) - V_2(e^{2\pi i n m / N}e^{-2\pi i (n+2)m' / N}|h_{m}\rangle\langle h_{m'}| + e^{2\pi i (n+2) m / N}e^{-2\pi i n m' / N}|h_{m}\rangle\langle h_{m'}|) \right]$$

$$= \sum_{m,m',n}\left[-V_1 (e^{2\pi i/N(nm-nm'-m')} + e^{2\pi i/N (nm-nm'+m)}) - V_2(e^{2\pi i/N(nm-nm'-2m')} + e^{2\pi i/N (nm-nm'+2m)})\right]|h_{m}\rangle\langle h_{m'}|$$

It's helpful to look at the argument of each exponent, evaluate for a few inputs of $n$, $m$, and $m'$, and see how you can eliminate some of these combinations (what do I get if $n = m$, or $m = m'$, or $nm = N$, etc.). Once you've eliminated the sums over $n$ and $m'$ you're done.

Edit: @flippiefanus, I didn't see your answer before posting this.

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