show/hide this revision's text 3 anticommuting case

A $2n\times 2n$ dimensional Hermitian matrix that can be diagonalized by a symplectic transformation can be viewed as an $n\times n$ matrix with elements consisting of $2\times 2$ blocks of the quaternion real form

${\bar{z}\;-\bar{w}}\choose{w\;\; z}$

so if you choose real $z$ and $w$ you have constructed a real symmetric matrix $M$ that can be diagonalized by a symplectic $S$; is S$.


@Federico: this what you need?is the general form for matrices that commute, $MT=TM$, with

$T=1_{N}\otimes$ ${0\; 1}\choose{-1\; 0}$ $K$

($K$ is the operator of complex conjugation); alternatively, one can take matrices that anticommute, $MT=-TM$; then the $2\times 2$ blocks have the form

${\bar{z}\;\bar{w}}\choose{w\;\; -z}$

and again, for a real $M$ one would choose real $w,z$. these two choices exhaust the possibilities.

In applications to physical systems, the matrix $M$ is a Hamiltonian and $T$ is the operator of time reversal. Then only commuting matrices, $MT=TM$, are permitted.

For a discussion in the physics context, see Section 1.4.2 of Forrester's book, online here:

http://www.ms.unimelb.edu.au/~matpjf/b1.ps

    Post Undeleted by Carlo Beenakker
show/hide this revision's text 2 new attempt at an answer

The condition is

A $2n\times 2n$ dimensional Hermitian matrix that the 2n eigenvalues can be diagonalized by a symplectic transformation can be viewed as an $n\times n$ matrix with elements consisting of M should come in n twofold degenerate pairs. Then $2\times 2$ blocks of the matrix S is both orthogonal quaternion real form

${\bar{z}\;-\bar{w}}\choose{w\;\; z}$

so if you choose real $z$ and $w$ you have constructed a real symmetric matrix $M$ that can be diagonalized by a symplectic .$S$; is this what you need?

    Post Deleted by Carlo Beenakker
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