Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

Is there a general algorithm to find the spectrum of $S S^\dagger$, where $S$ is a homogenous polynomial (of degree $n$) of the annihilation operators (of $d$ variables)?

share|improve this question
add comment

2 Answers

up vote 7 down vote accepted

The supspaces $V_n = Span \{ (a_1^{\dagger})^{n_1}, . . . (a_d^{\dagger})^{n_d} |0>\}$, $n_i \ge 0$, $ n_1 + . . . n_d = n$, constitute of invariant subspaces of the operator $ S S^{\dagger}$ action. The dimension of $V_n$ is $ \frac{(d+n-1)!}{(d-1)! n!}$. Thus the operator can be represented on each of these subspaces as a square matrix of size $ \frac{(d+n-1)!}{(d-1)! n!}$ for which the spectrum can be found by elementary linear algebra. The spectrum on the whole of the Fock space is the union of the spectra over $V_n$, $ n = 0, 1, . . .$

share|improve this answer
    
@Moshe, did you lose your old account? –  lurscher Jul 1 '11 at 16:50
    
@Moshe, thank you a lot. Pushing in further - for a polynomial of degree $k$ does it suffice to know eigenvalues in the first $i$ subspaces (i.e. $V_0,\ldots, V_i$) to predict all the other? Like for $n=1$ it suffices to know the eigenvalue for $i=1$ (all others are their multipilicites). –  Piotr Migdal Jul 1 '11 at 19:56
    
@David Bar Moshe: The formula (v1) for $\dim V_n$ must contain an error/typo(?), which can e.g. be seen by putting $d=1$. –  Qmechanic Jul 27 '11 at 16:12
    
@Qmechanic - corrected thanks. –  David Bar Moshe Jul 28 '11 at 6:39
    
@Piotr - sorry for the error, of course the dimension is equal to the number of ordered partitions of $n$ into at most $d$ pieces, or equivalently the dimension of the fully symmetric $n$-tensorial representation of $SU(d)$, which can be calculated for example by using the hook length formula. By the way I am trying to think occasionally on your interesting suggestion in your last comment, but I haven't reached an answer yet. –  David Bar Moshe Jul 28 '11 at 6:39
add comment

One can always reorder the operators in your polynomial to make it a polynomial of individual particle number operators. E.g. $a^+_k a^+_k a_k a_k = \pm n_k^2+n_k$ (the sign depends on the statistics of your particles). Since the particle number operators for different modes commute, the calculation of the spectrum is straightforward.

share|improve this answer
    
I am afraid it is not that simple. Even for the simplest non-trivial case $S=\frac{\alpha}{\sqrt{2}} a_1^2+\beta a_1 a_2+\frac{\gamma}{\sqrt{2}} a_2^2$ you get cross-terms in $S S^\dagger$, e.g. $\frac{\alpha \beta^*}{\sqrt{2}} a_1^2 a_1^\dagger a_2^\dagger$. Do you know a general algorithm to 'diagonalize' it, so there are not any cross-terms? –  Piotr Migdal Jul 29 '11 at 8:46
add comment

Your Answer

 
discard

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.