The energy operator for infinite statistics

Abstract

We construct the energy operator for particles obeying infinite statistics defined by aq-deformation of the Heisenberg algebra.

The aim of this paper is to construct the energy operator for particles which obey the so-called infinite statistics defined by theq-deformation of the Heisenberg algebra. This topic was studied in the previous article [1], where a conjecture was formulated concerning the form of the energy operator. Our main result is a proof of this conjecture in a slightly modified form (cf. Remark 1).

We will essentially use the same notations as in [1]. Thus,T _1k will denote the particular elements of$\mathfrak{S}_n$ which send [1, 2,...,n] to [k, 1,...,k−1,k+1,...,n], ie

$$T_{1k} (i) = \left\{ {\begin{array}{*{20}c} {k, if i = 1;} \\ {i - 1, if 1< i \leqslant k;} \\ {i, if k< i \leqslant n;} \\ \end{array} } \right.$$

and

$$T_{1k} (i) = \left\{ {\begin{array}{*{20}c} {k, if i = 1;} \\ {i - 1, if 1< i \leqslant k;} \\ {i, if k< i \leqslant n;} \\ \end{array} } \right.$$

represent the following subsets of$\mathfrak{S}_n$:

$$\mathfrak{S}_{n,p} = \{ \sigma \in \mathfrak{S}_{n,} with \sigma = T_{1k_1 } T_{1k2} ,...,T_{1k_p } ,1< k_1< ...< k_p \leqslant n\} ,$$

for 1≤p≤n−1 and$\mathfrak{S}_{n,0} = \{ 1\}$. (This differs from the definition of$\mathfrak{S}_{n,p}$ in [1].) In [1] ann!×n! matrix$A_n (\pi ,\sigma ), \pi ,\sigma \in \mathfrak{S}_n ,$, with coefficients in ℤ[q] was studied and shown to be invertible for |q|<1. As in [1], we will work with the group algebra ℂ[G _n ] rather than its matrix representation, so we have elements

$$\alpha _n = \sum\limits_{\varrho \in \mathfrak{S}n} {A_n (\varrho ,1)\sigma } = \sum\limits_{\varrho \in \mathfrak{S}n} {q^{I(\varrho )} \varrho , } \alpha _n^{ - 1} = \sum\limits_{\varrho \in \mathfrak{S}n} {A_n^{ - 1} (\varrho ,1)\varrho . } $$

((1))

Let ℰ be the energy operator of particles obeying infinite statistics, defined by the commutation relation (1) in [1]. ℰ acts on ℋ(q) and eachx ₁ is an eigenvector of ℰ satisfying the eigenvalue equation

$$\mathcal{E}a^\dag (l_n )...a^\dag (l_1 )|0\rangle = \sum\limits_{i = 1}^n {E(l_i )a^\dag (l_n )...a^\dag (l_1 )|0\rangle ,} $$

((2))

whereE(l _i) is the energy of a particle with momentuml _i.