Approximate diagonalisation of (3,3) hermitian matrix

  • Thread starter SUSY
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  • #1
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Hi,

I have a 3 by 3 hermitian matrix K that I need to diagonalise. More accurately, I am searching for a unitary matrix S such that [tex]S^{\dagger} K S[/tex] is diagonal.

The problem is that K is very complicated and the expression for S in mathematica takes up quiiiiiet a lot of space.

Is it possible to find approximate expressions for the entries of S in terms of entries in K? What I am looking for is something like (as an example) [tex]S_{ij}=K_{ij} + K_{ji} bla bla[/tex]
Or is it maybe possible to decompose S into matrices which I can then approximate by such expressions (they can, of course, be more complicated than the example given above).

Since the entries in K are given as a power series of some small parameter ε [tex]K_{ij}=a_0 + a_1 \epsilon + ...[/tex] and I am only interested in the lowest non-vanishing order anyway, it would be nice to have expressions for the entries in S in terms of the entries in K. Then, I could easily evaluate the entries in S to lowest non-vanishing order in ε (so I am only interested in an approximate S anyway).

Does anybody know of such a method?

Thanks,
Susy
 

Answers and Replies

  • #2
AlephZero
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If you can write ##K = K_0 + \epsilon K_\epsilon##, can you diagonalize ##K_0## and ##K_\epsilon## simultaneously? This is routinely done numerically, as a generalized eigenproblem, but I don't know about doing it symbolically.
 
  • #3
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Yes, you should be able to diagonalize both simultaneously. In all cases I can think of, $K_0$ will already be a diagonal matrix and $K_{\epsilon}$ is a correction ($K_{\epsilon}$ can in principle be an arbitrary hermitian matrix).
 

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