Intriguing polynomials coming from a combinatorial physics problem

Question

For real $0<q<1$, integer $n >0 $ and integer $k\ge 0$, define $$[k, n]_q \equiv -\sum_{m=1}^{n} q^{m(k+1)} (q^{-n}; q)_m = -\sum_{m=1}^{n} q^{m(k+1)} \prod_{l=0}^{m-1} (1-q^{l-n})$$

where $(\cdot\; ; q)_n$ is a $q$-Pochhammer symbol.

These functions express exact occupation numbers of $k$-th energy level in an ideal Fermi gas with equidistant spectrum and exactly $n$ fermions. (For physicists, $q$ is just $e^{-\Delta \epsilon/ kT}$).

My numerical experiments with Mathematica show so far :

1. All $[k, n]_q$ are polynomials in $q$.
2. $0<[k, n]_q<1$ for $0<q<1$.
3. $\lim\limits_{q\to 1} [k, n]_q = 0$.
4. $\lim\limits_{q\to 0} [k, n]_q = \begin{cases} 1, & k < n \\ 0, & k \ge n \end{cases}$.

Points (2.) to (4.) can be proven from the physics starting point, but I'm totally puzzled by (1.). The product from $l=0$ to $m-1$ contains negative powers of $q$ because of $n >l$, nevertheless, they conspire to cancel in the final sum.

Why are these functions polynomials? What would be the optimal way to compute their coefficients? Are there deeper mathematical properties to them?

Combinatorial context. The original "combinatorial physics" definition of these numbers can be written as

$$[k, n]_q =\frac{\sum_{ \{ \nu_k \} } \nu_k q^{\sum_k k \nu_k} \delta_{n, \sum_k \nu_k}}{\sum_{ \{ \nu_k \} }q^{\sum_k k \nu_k} \delta_{n, \sum_k \nu_k}}$$

where the summation indices run as $\nu_k=0,1$ for $k=0, 1, 2 \ldots$. Properties (2.)-(4.) follow easily from this definition. More physics context for the problem is being prepared for publication, see also a related post at Physics.SE.

Update: A physics paper describing these polynomials has been posted to arXiv, includes a reference to this question.

Answer 1

Some digging through Koekoek and Swarttouw's The Askey-scheme of hypergeometric orthogonal polynomials and its $q$-analogue reveals that $[k,n]_q$ is related to the Al-Salam-Carlitz I polynomials (see page 115 of Koekoek and Swarttouw). More precisely,

$$1-[k,n]_q={}_2\phi_1\left({{q^{-n},q}\atop{0}}; q, q^{k+1}\right)=(-1)^n q^{\frac{n}2(2k-n+3)}U_n^{(q^{-k-1})}\left(\frac1{q};q\right)$$

In particular, letting $S(n,k;q)=1-[k,n]_q$, there is the three-term recurrence

$$S(n+1,k;q)=(1+q^{k+1}-q^{k-n})S(n,k;q)-q^{k+1}(1-q^{-n})S(n-1,k;q)$$

with the initial conditions $S(0,k;q)=1,\quad S(1,k;q)=q^{k+1}-q^k+1$.

From the relation with the Al-Salam-Carlitz II polynomials (see page 116 of Koekoek and Swarttouw), we have another basic hypergeometric expression:

$$1-[k,n]_q={}_2\phi_0\left({{q^n,\frac1{q}}\atop{-}}; \frac1{q}, q^{k-n+1}\right)$$

(Mathematica note: unfortunately Mathematica doesn't have support for ${}_2\phi_0$... yet.)

One can also derive a "reversal" identity by reversing the summation order:

$$\begin{align*}1-[k,n]_q&=q^{n(k+1)}(q^{-n};q)_n\; {}_1 \phi_1\left({{q^{-n}}\atop{q^{-n}}};q,q^{-k}\right)\\&=(-1)^n q^{\frac{n}{2}(2k-n+1)}(q;q)_n\; {}_1 \phi_1\left({{q^{-n}}\atop{q^{-n}}};q,q^{-k}\right)\end{align*}$$

I'll update this post if I manage to dig up more information...

Answer 2

Hope this does not goes against the rules here, but I wanted to post a more permanent summary of the brainstorming that took place in the comments to the question.

• J.M. suggested writing

$$[k,n]_q = -\sum_{m=1}^n \prod_{\ell=0}^{m-1} \left(q^{k+1}-q^{k+\ell-n+1}\right)$$

and

$$[k,n]_q = 1-{}_2\phi_1\left({{q^{-n},q}\atop{0}}; q, q^{k+1}\right)$$

• anon noticed that point (4.) implies point (1.)

Answer 3

Not a full answer, but there's a recurrence $$[k,1]_q = q^{k}-q^{k+1}$$ $$[k,n+1]_q = \left(q^{k+1}-q^{k-n}\right) \left([k,n]_q -1\right)$$

One derivation is $$[k,n+1]_q = -\sum_{m=1}^{n+1} \prod_{\ell=0}^{m-1} \left(q^{k+1}-q^{k+\ell-n}\right)$$ $$ = -\left(q^{k+1}-q^{k-n}\right) \sum_{m=1}^{n+1} \prod_{\ell=1}^{m-1} \left(q^{k+1}-q^{k+\ell-n}\right)$$ Subst. $\ell^\prime = \ell - 1, \; m^\prime = m-1$ $$ = -\left(q^{k+1}-q^{k-n}\right) \sum_{m^\prime=0}^{n} \prod_{\ell^\prime=0}^{m^\prime-1} \left(q^{k+1}-q^{k+\ell^\prime-n+1}\right)$$ $$ = \left(q^{k+1}-q^{k-n}\right) \left(-1-\sum_{m^\prime=1}^{n} \prod_{\ell^\prime=0}^{m^\prime-1} \left(q^{k+1}-q^{k+\ell^\prime-n+1}\right)\right)$$ $$ = \left(q^{k+1}-q^{k-n}\right) \left([k,n]_q -1\right) $$

Answer 4

An alternative form, conjectured by my colleague:

$$[k,n]_q =1+\sum_{i=1}^{k+1}(-1)^i \frac{q^{(n-k)i+i(i-1)/2}}{1-q^{k+1}} (q^{k+1}\; ;q^{-1})_i= 1 - \sum_{i=1}^{k+1} q^{i(i+1)/2+n-k-1} \prod_{j=1}^{i-1}(q^{n-j}-q^{n-k-1}) $$

Checked for small $k$ and $n$, still need to prove. But this form explicitly proves that $[k,n]_q$ are polynomials for $k<n$. Also, the number of terms does not grow with $n$.

EDIT: Exploring the new form a bit more leads to a striking "reversal" relation:

$$[k,n]_q=1-q^{n-k} \sum_{m=0}^k q^{m(n+1)} (q^{-k} \;; q)_m=1-q^{n-k} \left (1 - [n,k]_q \right) $$

EDIT-2 Another empirical observation: $\frac{[k,n]_q}{(1-q^n)} q^{-k+(n-1)n/2}$ is a polynomial as well.