# Finding the spectrum of a polynomial of the creation and annihilation operators

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)?

-

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 $V_n$, $n = 0, 1, . . .$

-
@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

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.

-
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