Here is a problem to solve: diagonalize the following hamiltonian using quantum fourier transform. The hamiltonian reads: $$ \sum_{i,j=1}^N e^{-\theta_{ij}} c_i^\dagger c_j + h.c. $$ Where $c_j$ are bosonic operators and summation is done on a ring. Using Fourier transform of operators from coordinate space to momentum space $$ c_j = \frac{1}{\sqrt{N}} \sum_{n=1}^N e^{i \frac{2 \pi n}{N}j} b_n $$ and changing $i$ to $j+1$ which corresponds to ring positioning, I obtain $$ \frac{1}{N} \sum_{m,n,j=1}^N e^{-\theta_i} e^{2 \pi i (m-n) \frac{j}{N}} e^{2 \pi i \frac{m}{n}} b_n^{\dagger} b_m + h.c. $$ I believe it should be simplified by series definition of Kronecker's symbol, but I still have thetas that ruin my sum. Any thoughts?


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Basically, this can't be done for the generality implied by the hamiltonian you've given. Fourier-transform methods are useful when you have translation invariance in some meaningful way, but with your hamiltonian

  • the notion of translation invariance doesn't even make sense, as the connectivity graph is complete instead of local, which means that you cannot even say how many dimensions your system lives in and, moreover,
  • even if you could, the $(i,j)$-dependent coupling $e^{-\theta_{ij}}$ completely breaks the translation invariance.

And, in addition, it's important to note that your manipulations are incorrect: there's no way to turn the $e^{-\theta_{ij}}$ in your original hamiltonian into a $j$-independent term $e^{-\theta_{i}}$ without additional assumptions, so your sum should read $$ \frac{1}{N} \sum_{m,n,j=1}^N e^{-\theta_{ij}} e^{2 \pi i (m-n) \frac{j}{N}} e^{2 \pi i \frac{m}{n}} b_n^{\dagger} b_m + \rm h.c. $$ at best.

If you introduce additional hypotheses about what $\theta_{ij}$ is, then this might be solvable, but without that then this method just isn't going to work.

  • $\begingroup$ I just implied the fact that if the particles are on a ring, then particle $j$ interacts only with particle $j+1$ and introduced periodic boundary conditions. What hypotheses about $\theta_{ij}$ might make it solvable? $\endgroup$ – Aleksandr Berezutskii Jul 6 '18 at 11:39
  • $\begingroup$ I don't think anything short of $\theta_{ij} = f(i-j)$ will make FT methods useful in this context; even then, unless $e^{-f(\Delta i)}$ (or, generally, the coupling) is linear or a very-low-order polynomial in $\Delta i$, it's unlikely to help much. $\endgroup$ – Emilio Pisanty Jul 6 '18 at 12:06
  • $\begingroup$ @AleksandrBerezutskii This is not a check-my-work site. $\endgroup$ – Emilio Pisanty Jul 9 '18 at 13:38

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