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I have the following $1$D fermionic Hamiltonian $H$, given by $$ H = H^A_0+H _0^B+H_I^{AB}=\sum_{jk\in A} H_{jk}^Ac^\dagger_j c_k + \sum_{jk\in B} H_{jk}^Bc^\dagger_j c_k + \lambda \sum_{j\in A, \ k \in B} H^{AB}_{jk} (c^\dagger_j c_k+ c_k^\dagger c_j), $$ i.e. two independent fermionic subsystems “interacting” via some $H_I^{AB}$, the $H_{jk}$’s could be whatever, I’m just interested in the above form. I know I could gather everything together into one big $H_{jk}$, diagonalise it, define some new $c_j=\sum_n \phi^n_j d_n$ such that $H=\sum_n \epsilon_n d^\dagger_n d_n$ and call it a day. The problem is that I would loose track of $\lambda$ , a parameter I want to tweak later on as it modulates the strength of the subsystem’s interaction.

How can one diagonalise $H$ keeping track of $\lambda$?

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    $\begingroup$ The Hamiltonian is one-particle, but written in second quantization. It is not clear what you mean by loosing track of $\lambda$, since it will enter the new energies and the coefficients $\phi_j^n$. $\endgroup$
    – Roger V.
    Commented Jun 8, 2021 at 11:49
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    $\begingroup$ Are you interested in perturbation theory in $\lambda$ or specifically in exact solutions? $\endgroup$
    – jacob1729
    Commented Jun 8, 2021 at 11:58
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    $\begingroup$ @FriendlyLagrangian It depends on the structure of your Hamiltonian - you have to diagonamize a block-diagonal matrix with non-diagonal blocks proportional to $\lambda$. It is hard to be more specific than that without additional constraints on $H_{ij}^\alpha$. $\endgroup$
    – Roger V.
    Commented Jun 8, 2021 at 14:08
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    $\begingroup$ @FriendlyLagrangian The simplest example is a two level system: each subsystem has one level and they are coupled with strength $\lambda$ $\endgroup$
    – Roger V.
    Commented Jun 8, 2021 at 14:58
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    $\begingroup$ Roger Vadim's comments seem spot on. If there is some structure (e.g. if H_{ij}s are translation invariant so that you can Fourier transform and reduce the matrix dimension) you may get somewhere, otherwise it's probably most promising to just study the system numerically for different values of $\lambda$. $\endgroup$
    – Anyon
    Commented Jun 8, 2021 at 19:00

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The Hamiltonian you give is completely general. There is no restriction whatsoever, even if you set $\lambda=1$. Thus, there cannot be any special structure in the way you diagonalize it (or on $\lambda$, as you can absorb $\lambda$ in $H^{AB}$), unless you give exttra information.

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