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I want to diagonalise the following operator $$ \mathcal{L}= 2 \sum_k^N\epsilon_k(c^\dagger_{2k-1}c_{2k}-c_{2k}^\dagger c_{2k-1})+2iA\sum_k^N c^\dagger_{2k-1}c^\dagger_{2k}-B \sum^{2N}_kc^\dagger_kc_k, $$ that arose in some non-unitary evolution I’m studying. I have tried to attack it with a Bogoliubov transformation although $\mathcal{L}$ is not hermitian (I don’t know if that is a problem).

How does one normally go about attacking these kind of problems?

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One can always think of an operator as a matrix and refer to the corresponding matrix methods. Since the operator in question is non-hermitian, its right and left eigenvectors are generally not the same, and the diagonalizing transformation is generally not a unitary one, i.e., one cannot restrict it to matrices satisfying $S^\dagger=S$.

The more general approach to diagonalizing matrices is singular value decomposition - rarely encountered in quantum mechnaics, but in fact rather well studied.

To quote the Wikipedia article linked above:

In linear algebra, the singular value decomposition (SVD) is a factorization of a real or complex matrix. It generalizes the eigendecomposition of a square normal matrix with an orthonormal eigenbasis to any m × n matrix.

$$M=U\Sigma V^\dagger,$$ where $\Sigma$ is a diagonal matrix, whereas $U,V$ are unitary matrices.

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    $\begingroup$ I'm guessing then that the Bogoliubov transformation will not diagonalise this operator. $\endgroup$ Sep 8, 2021 at 14:56
  • $\begingroup$ certainly not with the usual constraints on the coefficients. Also, it looks like a tight-binding Hamiltonian. $\endgroup$ Sep 8, 2021 at 15:01
  • $\begingroup$ That being said, how are you certain that a regular Bog. transf. will not work? Generically, it will not, but for this particular example, it might. It is not obvious to me that it doesn't. Is there something that I am missing? $\endgroup$ Sep 8, 2021 at 15:20
  • $\begingroup$ @AccidentalFourierTransform I am glad that we understood each other, and it improved the answer. It depends on what exactly is called Bogoliubov transformation here - if you mean that the transformation is a sum of creation and annihilation operators, then it is likely enough. But it requires some work. $\endgroup$ Sep 8, 2021 at 15:37

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