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This question already has an answer here:

I asked a similar question in the physics stack exchange, but realized my question is probably more suited here.

For any Hermitian matrix $H = H^{\dagger}$ we can write $H = P DP^{\dagger}$ where $P$ is the matrix containing the eigenvectors and $D$ is the matrix containing the eigenvalues on the diagonal. In this case $P^{-1} =P^{\dagger}$ and $P^{\dagger}$ means the Hermitian conjugate.

So I wanted to do this for a specific matrix of the form

h = {{x, -d}, {- d, -x}}

where $x$ and $d$ are both real.

I wrote

Eigensystem[h]

and obtained

{-(x+ Sqrt[d^2 + x^2])/d,1} and {(-x+ Sqrt[d^2 + x^2])/d,1}

as eigenvectors and

Sqrt[d^2 + x^2] and -Sqrt[d^2 + x^2]

as the corresponding eigenvalues.

However, if we now use the eigenvectors to form the matrix $P$

p = {
{-(x+ Sqrt[d^2 + x^2])/d,(-x+ Sqrt[d^2 + x^2])/d },
{1,1}
}

we see that $P$ does not satisfy $P^{-1} \neq P^{\dagger}$.

In physics I think this is a big deal, because it means that your eigenstates does not satisfy the necessary (anti)commutation relations.

My question is: How can I get mathematica to give me the eigenvectors that makes sure $P^{-1} = P^{\dagger}$?

I'm asking this question because I eventually want to generalize this procedure to a hermitian block diagonal $8 \times 8$ matrix.

In my case I think my normalized eigenvectors should be $(u,v)$ and $(v,u)$ where

$u^2 = \frac{1}{2}\left(1 + \frac{\xi}{\sqrt{\xi^2 + \Delta^2}} \right)$, $v^2 = \frac{1}{2}\left(1 - \frac{\xi}{\sqrt{\xi^2 + \Delta^2}} \right)$

However, I can't obtain this by just writing

FullSimplify[Normalize[{-(x+ Sqrt[d^2 + x^2])/d,1}]]

Is there an easy way to simplify this using Mathematica?

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marked as duplicate by Michael E2, AccidentalFourierTransform, mikado, Roman, m_goldberg Jun 25 at 22:42

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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    $\begingroup$ Normalize the eigenvectors. (This is done on approximate numeric matrices by Eigensystem[] but not on symbolic or exact matrices. See "Details and Options" of the documentation for Eigensystem, 2nd & 3rd points.) $\endgroup$ – Michael E2 Jun 25 at 17:51
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    $\begingroup$ Related: (72941), (79862), (139216), (151547) $\endgroup$ – Michael E2 Jun 25 at 18:01