On the use of rational expressions in Combinatorial Physics: an example.

Gérard H. E. Duchamp and Christophe Tollu.^{∗}

Abstract In a joint work with J. Katriel, one of us gave the solution of the problem of matrix coefficients in the Fock space of carriers between two levels. We give here a complete proof of these continued fractions-type formulas and possible extensions of the method. ^{0}

1 Introduction

2 Brief review of some known formulas

In [1] was considered, as Fock space, a general vector space V over a field k with basis |e_{n}⟩ n = 0, 1, ⋯, equipped with its natural grading

V = ⊕_{n∈Z}V_{n} with V_{n} := ke_{n}; V_{−n−1} := (0) for n ≥ 0  (1)

and scalar product defined by ⟨e_{n}|e_{m}⟩ = δ_{n,m}. and f, g two linear operators in V of degrees −1, +1, respectively. Generically, they read

for n ≥ 0; f|e_{0}⟩ := 0; f|e_{k+1}⟩ = α_{k+1}|e_{k}⟩; g|e_{k}⟩ := β_{k+1}|e_{k+1}⟩  (2)

We consider the words in f, g:

w(f, g) := f^{p_{1}}g^{q_{1}}f^{p_{2}}g^{q_{2}} ⋯ f^{p_{n}}g^{q_{n}}  (3)

the degree (excess) of which is ∑_{k=1}^{n}(q_{k} − p_{k}) (this is, for algebraists, the degree of the graded operator f^{p_{1}}g^{q_{1}}f^{p_{2}}g^{q_{2}} ⋯ f^{p_{n}}g^{q_{n}}). This provides a representation μ of a two letters free monoid {b_{−}, b_{+}}^{∗} on V by μ(b_{−}) = f; μ(b_{+}) = g, which is graded for the weight on {b_{−}, b_{+}}^{∗}. In order to keep the reading of a word from left to right, one performs the the action on the right. Thus V becomes a {b_{−}, b_{+}}^{∗} right module by

e_{0}.b_{−} = 0 ; e_{n+1}.b_{−} = α_{n+1}e_{n} ; e_{n}.b_{+} = β_{n+1}e_{n+1}  (4)

one is interested by the matrix elements

⟨e_{n}.{b_{−}, b_{+}}^{i}|e_{m}⟩ = ω_{n→m}^{(i)} .  (5)

The following propsition gives the support i. e. the set of words w of length i such that ⟨e_{n}.w|e_{m}⟩ ≠ 0.

∗LIPN - UMR 7030 CNRS - Université Paris 13 F-93430 Villetaneuse, France

0Version compilée le 27-04-2008 23:04

Proposition 2.1 Define W_{n→m}^{(i)} as follows

W_{n→m}^{(i)} = {w ∈ {b_{−}, b_{+}}^{i} | (π_{e}(w) = m − n) and (w = uv ⟹ π_{e}(w) ≥ −n)} (6)

then

i) If the all the weights α_{n}; n ≥ 1, β_{n}, n ≥ 0 are not zero W_{n→m}^{(i)} is exactly the set of words of length i such that ⟨e_{n}.w|e_{m}⟩ ≠ 0. ii) In all cases the latter is a subset of W_{n→m}^{(i)} i. e.

{w ∈ {b_{−}, b_{+}}^{i} | ⟨e_{n}.w|e_{m}⟩ ≠ 0} ⊂ W_{n→m}^{(i)}  (7)

As a result,

μ(W_{m−n}^{(i)})|n⟩ = ω_{n→m}^{(i)}|m⟩ ; T_{n→n+k} := ∑_{i≥0} t^{i}ω_{n→n+k}^{(i)} .  (8)

One can expand T_{n→n+k} as a product of continued fractions. Let

F_{n}^{+} = \frac{1}{1 − \frac{t^{2}α_{n+1}β_{n+1}}{1 − \frac{t^{2}α_{n+2}β_{n+2}}{1 − \frac{t^{2}α_{n+3}β_{n+3}}{1 − ⋯}}}} = \frac{1}{1 − E_{n}^{+}}

F_{n}^{−} = \frac{1}{1 − \frac{t^{2}α_{n}β_{n}}{1 − \frac{t^{2}α_{n−1}β_{n−1}}{1 − \frac{t^{2}α_{n−2}β_{n−2}}{1 − ⋯}}}} = \frac{1}{1 − E_{n}^{−}}  (9)

and

F_{n} = \frac{1}{1 − E_{n}^{+} − E_{n}^{−}} .  (10)

Then, if k ≥ 0, we have

T_{n→n+k} = t^{k}F_{n+k} ∏_{i=0}^{k−1} F_{n+i}^{−} = t^{k}F_{n} ∏_{i=1}^{k} F_{n+i}^{+}  (11)

And, if k ≤ 0

T_{n→n+k} = t^{−k}F_{n+k} ∏_{i=0}^{−k−1} F_{n−i}^{+} = t^{−k}F_{n} ∏_{i=1}^{−k} F_{n−i}^{−}  (12)

References

References

[1] Katriel J., Duchamp G., Ordering relations for q-boson operators, continued fractions techniques, and the q-CBH enigma. Journal of Physics A 28 7209-7225 (1995).