A reasonable question to relate the eigenvalues between two matrices A and B is the following: If you have a specific homotopy, i.e. a continuous curve x(t) with x(0)=A and x(1)=B, then you can sometimes track eigenvalues along this curve. But the relation will depend on the choice of curve!
If you have no duplicate eigenvalues and A \approx B, i.e. if you are in the perturbative regime, then all short curves give the same matching, and this is why pertubation theory gives you a well-defined matching.
The basic test is to ensure that along your homotopy, you never have colliding eigenvalues. If you do have colliding eigenvalues, then you sometimes can still do stuff (cf bifurcation theory).
It sounds like you want to consider your off-diagonal coupling that way? I.e. consider x(t) = diag(A) + t *(A-diag(A)), assume that x(0) and x(1) both have non-degenerate spectra and want compatible permutations?
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I checked this paper and the paper OP mentioned, and surprisingly found they both make use of the same structure, i.e. commutator with an skew-symmetric/skew-Hermitian matrix. This generates a flow of matrices with the same eigenvalues, as the flow is driven by a sequence of unitary transformations.
It is always interesting to see similar ideas emerge in different areas independently. I am glad that Julia community can accommodate people with different background so such a interdisciplinary discussion can happen.
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I can see the similarities of the paper you mentioned. The spirit is probably exactly the same. But what I want to say is that I do need to solve those ODE problems (which is the flow equations), I was just hoping for a specific range (perturbative solution is well justified), my original basis would still be a good description. Also in the non-perturbative region, flow equation still works pretty well.
Yes, in fact I was a little bit afraid to use physical concepts in the beginning. But sometimes it do gives a simple picture. But as @mstewart mentioned I probably should work harder to try to convey my question using math and this will bring as much brilliant mind as possible to discuss and solve problems.
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Yes, I can see after solving the flow equation it will give you a formulae like this. But I am a little confused about the formulae you wrote, because in this case x(0)=diag(A) but x(1)=A which is kind like an opposite flow.