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

+ 6 like - 0 dislike
118 views

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)?

This post imported from StackExchange Physics at 2016-10-04 13:43 (UTC), posted by SE-user Piotr Migdal

+ 6 like - 0 dislike

The subspaces $V_n = Span \{ (a_1^{\dagger})^{n_1}, . . . (a_d^{\dagger})^{n_d} |0>\}$, $n_i \ge 0$, $n_1 + . . . n_d = n$, constitute 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 $V_n$, $n = 0, 1, . . .$

This post imported from StackExchange Physics at 2016-10-04 13:43 (UTC), posted by SE-user David Bar Moshe

answered Jul 1, 2011 by (4,095 points)
edited Oct 4, 2016
@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).

This post imported from StackExchange Physics at 2016-10-04 13:43 (UTC), posted by SE-user Piotr Migdal
@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.

This post imported from StackExchange Physics at 2016-10-04 13:43 (UTC), posted by SE-user David Bar Moshe
+ 0 like - 1 dislike

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.

This post imported from StackExchange Physics at 2016-10-04 13:43 (UTC), posted by SE-user p_k
answered Jul 28, 2011 by (-10 points)
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?

This post imported from StackExchange Physics at 2016-10-04 13:43 (UTC), posted by SE-user Piotr Migdal

 Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead. To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL. Please consult the FAQ for as to how to format your post. This is the answer box; if you want to write a comment instead, please use the 'add comment' button. Live preview (may slow down editor)   Preview Your name to display (optional): Email me at this address if my answer is selected or commented on: Privacy: Your email address will only be used for sending these notifications. Anti-spam verification: If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:p$\hbar$ysic$\varnothing$OverflowThen drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds). To avoid this verification in future, please log in or register.