Consider a two parallel fermionic chains, $A$ and $B$, that “interact” with each other (i.e. imagine a ladder). The Hamiltonian I’m interested (seems to be exactly solvable) is given by $$ H = H^A_0+H^B_0 +H_I= J_A\sum_j^N (c^\dagger_{j,A}c_{j+1,A}+ h.c.)+ J_B\sum_j^N (c^\dagger_{j,B}c_{j+1,B}+ h.c.)+\lambda\sum_j^N (c^\dagger_{j,A}c_{j,B}+h.c.), $$ where as expected, $c_{j,A}$ annihilates a fermion at $j$ in ladder $A$. I have seen papers with a proper (non-quadratic) interaction which study this problem which naively makes me think the above Hamiltonian probably has an exact solution.

The problem is manifestly invariant under translations of the two ladders. I believe this is because fermions are indistinguishable particles and the interaction strength $\lambda$ is the same for all sites, this means that even the interaction term is invariant. This suggest (Bloch theo.) performing a Fourier transform by defining$^1$ $$ d_{k,A}\propto\sum_je^{ikj}c_{j,A} \implies c_{j,A}\propto\sum_ke^{-ikj}d_{k,A}, $$ which leads to $$ \sum_j^N (c^\dagger_{j,A}c_{j+1,A}+ h.c.)\propto \sum_j^N (\sum_ke^{ikj}d_{k,A}^\dagger \sum_{k’}e^{-ik’}e^{-ik’j}d_{k’,A} + h.c.)=\\ \sum_{k,k’} ( e^{-ik’} d_{k,A}^\dagger d_{k’,A} \sum_j e^{-ij(k’-k)}+ h.c.) \propto \sum_ke^{-ik}d^\dagger_{k,A}d_{k,A}+h.c.\propto \sum_k \cos(k)d^\dagger_{k,A}d_{k,A}, $$ where the lattice spacing is implicitly equal to $1$.

My problem is that the interaction term is not diagonal with this transformation. I believe the Hamiltonian now would read $$ H \sim \sum_k \{\cos(k) [J_A d^\dagger_{k,A}d_{k,A} + J_B d^\dagger_{k,B}d_{k,B}]+\lambda (d^\dagger_{k,A}d_{k,B}+d^\dagger_{k,B}d_{k,A})\}, $$ which is not diagonal. Does anyone know how to diagonalise $H$?

  1. For the other rookies like me, this should be because: $$ d_k\propto\sum_je^{ikj}c_j\propto \sum_je^{ikj}\sum_{k’}e^{-ik’j}d_{k’}\\= \sum_{k’} d_{k’}\sum_j e^{-ij(k’-k)}\propto \sum_{k’} d_{k’}\delta_{k,k’}=d_k \quad \text{ modulo conventions} $$

1 Answer 1


Your first step is correct (I didn't check the details, but as a step, it is correct; and also otherwise up to possibly prefactors). To fully diagonalize the Hamiltonian at the bottom of your post, you will have to make a new basis which is a linear combination of $d_{k,A}$ and $d_{k,B}$, i.e., define operators $$ a_k=\cos(\theta_k) d_{k,A} + \sin(\theta_k) d_{k,B} $$ and $$ b_k=\sin(\theta_k) d_{k,A} - \cos(\theta_k) d_{k,B}\ , $$ where you have to choose $\theta_k$ such that the Hamiltonian becomes diagonal, $$ H = \sum_k E_a(k)a_k^\dagger a_k + \sum_k E_b(k) b_k^\dagger b_k\ . $$ In essence, this amounts to diagonalizing a $2\times 2$ matrix.

Usually, one would call this a two-band model, where the $a_k$ and $b_k$ describe the two bands, with corresponding energies $E_a(k)$ and $E_b(k)$.

  • $\begingroup$ Thank you for your help as always! I dont see how your Hamiltonian form is diagonal having two different operator basis $a_k$ and $b_k$ in its form. I mean, I couldn't write your $H$ in matrix form because $a_k\neq b_k$ in general. Having said that, I guess the physical intuition is clear, one has two quasi-particles $a$ and $b$ living in a chain. One could imagine that we have decoupled the two chains and the modes within. I guess the spectrum is simply that of a product state of $a$ and $b$ fermions, i.e. $a^\dagger_1 b^\dagger_{32} |0 \rangle$ as they should share the same vacuum state. $\endgroup$ Jun 11, 2021 at 8:52
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    $\begingroup$ $a_k$ and $b_k$ act on different modes. There is no other difference between $a$ and $b$, they are all fermionic modes. In particular, they are no more different than two $a_k$ for different $k$ are different. If you prefer, call them $a_k$ and $a_{N+k}$. $\endgroup$ Jun 11, 2021 at 9:04
  • $\begingroup$ Of course! Always getting confused with the simplest things... thank you so much Norbert. $\endgroup$ Jun 11, 2021 at 10:46
  • $\begingroup$ Hi, I got that the Hamiltonian takes the form of $H=\sum_{k=1}^{2L}(J\cos (k)+\lambda)f_k^\dagger f_k$ as you explained. I was just wondering what is this of a two band model you were talking about. I mean, looking at the eigenvalues $E_k= J\cos(k)+\lambda$ where can I see this? $\endgroup$ Jun 14, 2021 at 10:55
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    $\begingroup$ If I would have to guess, I'd certainly say this (J cos(...) +- lambda) looks very reasonable. $\endgroup$ Jun 17, 2021 at 15:07

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