Families of quadratic Hamiltonians - MathOverflow most recent 30 from http://mathoverflow.net2013-05-26T06:44:30Zhttp://mathoverflow.net/feeds/question/76763http://www.creativecommons.org/licenses/by-nc/2.5/rdfhttp://mathoverflow.net/questions/76763/families-of-quadratic-hamiltoniansFamilies of quadratic Hamiltonianszoltan2011-09-29T14:15:57Z2011-09-30T10:19:41Z
<p>Hi. What type of 2n dimensional real symmetric matrices can be diagonalized with symplectic transformations (meaning M->SMS^T, S^T means transpose and S is an element of the 2n dimensional real symplectic group. Usually normal forms of the literature are given as representatives of orthogonal group orbits, but I need to know the symplectic version. Thanks for any help, recommendation of literature etc.
Zoltan</p>
http://mathoverflow.net/questions/76763/families-of-quadratic-hamiltonians/76765#76765Answer by Carlo Beenakker for Families of quadratic HamiltoniansCarlo Beenakker2011-09-29T14:26:24Z2011-09-30T10:19:41Z<p>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 </p>
<p>${\bar{z}\;-\bar{w}}\choose{w\;\; z}$</p>
<p>so if you choose real $z$ and $w$ you have constructed a real symmetric matrix $M$ that can be diagonalized by a symplectic $S$.</p>
<hr>
<p>@Federico: this is the general form for matrices that commute, $MT=TM$, with</p>
<p>$T=1_{N}\otimes$ ${0\; 1}\choose{-1\; 0}$ $K$</p>
<p>($K$ is the operator of complex conjugation); alternatively, one can take matrices that anticommute, $MT=-TM$; then the $2\times 2$ blocks have the form</p>
<p>${\bar{z}\;\bar{w}}\choose{w\;\; -z}$</p>
<p>and again, for a real $M$ one would choose real $w,z$.
these two choices exhaust the possibilities.</p>
<p>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.</p>
<p>For a discussion in the physics context, see Section 1.4.2 of Forrester's book, online here:</p>
<p><a href="http://www.ms.unimelb.edu.au/~matpjf/b1.ps" rel="nofollow">http://www.ms.unimelb.edu.au/~matpjf/b1.ps</a></p>
http://mathoverflow.net/questions/76763/families-of-quadratic-hamiltonians/76786#76786Answer by Federico Poloni for Families of quadratic HamiltoniansFederico Poloni2011-09-29T18:19:38Z2011-09-29T18:19:38Z<p>I don't have a prompt answer (though I would guess "all those that are normal with respect to the scalar product induced by $J$"), but I suggest you to take a look at <em>Indefinite linear algebra and applications</em>, Gohberg and Lancaster.</p>