Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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
up vote 3 down vote accepted

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.

share|cite|improve this answer
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
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.