Hamiltonian diagonalisation using quantum Fourier transform [closed]

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?

closed as off-topic by sammy gerbil, Jon Custer, heather, Kyle Kanos, By SymmetryJul 5 '18 at 11:19

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• 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.
• 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? – Aleksandr Berezutskii Jul 6 '18 at 11:39
• 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. – Emilio Pisanty Jul 6 '18 at 12:06