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A quadratic polynomial of $2n$ variables is given as $$ H = \sum_{i,j=1}^{2n} A_{ij} x_i x_j = x^T A x, $$ where $A$ is a symmetric matrix. I am looking for a symplectic transformation of these variables into $y = Cx$--i.e., $C^TJC=J$ where $J=\begin{pmatrix}0 & I\\ -I & 0\end{pmatrix}$--such that $H$ becomes diagonal in $y$'s: $C^TAC = D$ for some diagonal matrix $D$.

It is clear that an orthogonal transformation doing the job always exists, but the question is about symplectic transformations. In addition I think $D$ cannot be the Jordan normal form of $A$, since in that case $C$ can (must?) be orthogonal and $C^TC=I$ is generically in conflict with $C^TJC=J$.

The question arises naturally if you want to use canonical transformations of classical mechanics to convert the most general quadratic Hamiltonian of a set of coordinates and momenta into non-interacting harmonic oscillators.

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I see you have an accessible book for this. Good. However, did you try the 2 by 2 case? What happened? –  Will Jagy Nov 16 '12 at 21:09
    
The 2x2 case is trivial. You can always write $ap^2+2bpq+cq^2 = a(p+bq/a)^2+(c-b^2/a)q^2$ and note that $\{q, p+bq/a\} = 1$. –  Mahdiyar Nov 17 '12 at 12:18
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In general, you cannot diagonalize a quadratic form on $\mathbb{R}^{2n}$ using a symplectic matrix. There is an analysis of all the possible canonical forms such a quadratic form might have, and it depends on the Jordan decomposition of the matrix $JA$. Check out Appendix 6 of Arnold's Mathematical Methods of Classical Mechanics for this analysis and a list of all possible normal forms of a quadratic Hamiltonian.

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Sounds like it's exactly what I want. I'm going to read it. Thanks. –  Mahdiyar Nov 16 '12 at 20:46
    
Alright, for the sake of completeness, let me say that according to Arnold there is a so-called Williamson's theorem which states that a non-trivial diagonalization exists iff the Jordan blocks of the matrix $JA$ are 1x1 and have purely imaginary eigenvalues. –  Mahdiyar Nov 19 '12 at 5:14
    
But let me also ask about a confusion of mine. Arnold says that for a pair of 1x1 blocks with eigenvalues $\pm ib$ we can turn $H$ into $H_1=\pm\frac12(b^2p^2+q^2)$; for a pair of 1x1 blocks with eigenvalues 0, into $H_2=0$, and for a 2x2 block with eigenvalue 0, into $H_3=\frac12 q^2$. I feel that the 2nd case should be a special case of the first one, but that would require $H_2$ and $H_3$ to be swapped! –  Mahdiyar Nov 19 '12 at 5:29
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Symplectically, under $(p,q) \mapsto (\frac{p}{b}, bq)$, $H_1$ is equivalent to $\pm \frac{b}{2} (p^2 + q^2)$, and when you take $b \rightarrow 0$, you indeed get $0$. Your shape of $H_1$ "hides" the division and so doesn't behave well under taking limits. –  levap Nov 19 '12 at 7:25
    
Got it. Thanks! [BTW, $H_3=\pm\frac12 q^2$ is correct in my previous comment.] –  Mahdiyar Nov 19 '12 at 7:44
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