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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.

Ilmari Karonen
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Slava Kashcheyevs
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    It might be helpful if you use the expression $$\sum_{m=1}^n \prod_{\ell=0}^{m-1} \left(q^{k+1}-q^{k+\ell-n+1}\right)$$; for $n \leq k+1$, it's clear why they're polynomials. Why they're still polynomials when $n > k+1$ does not look so straightforward... – J. M. ain't a mathematician Oct 17 '11 at 10:04
  • ...and since you're in Mathematica anyway: QHypergeometricPFQ[{q^(-n), q}, {0}, q, q^(k + 1)] - 1 looks to be a more compact way of expressing your polynomials. – J. M. ain't a mathematician Oct 17 '11 at 10:19
  • @J.M. useful indeed, thanks! (you've flipped the overall minus sign) – Slava Kashcheyevs Oct 17 '11 at 10:31
  • Taking the limit of $n \to \infty$, $|n-k|$=finite would also be very useful. If $q \to 1$ on top of it, this must give the Fermi function. – Slava Kashcheyevs Oct 17 '11 at 10:34
  • Or you could rewrite the product as $q^{-nm}\prod(q^n-q^l)$ and expand it into $q^{-nm}$ times a polynomial in $q$ whose coefficients are alternating sums of special partition functions... – anon Oct 17 '11 at 10:37
  • @anon: would be very grateful if you elaborate a bit (perhaps submit an answer?): what are "special partition functions"? I'm asking since they seem to be closely related to physical canonical partition functions where this beast comes from... – Slava Kashcheyevs Oct 17 '11 at 10:43
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    Wait, doesn't point 4 imply that every negative power of $q$ in the function's expansion has coefficient zero? Though asking for a combinatorial or generatingfunctionological approach is intriguing. – anon Oct 17 '11 at 10:46
  • @anon: ineed, you've solved the primary question! Point (4.) comes from the ground state with first $n$ levels occupied having the dominating Gibbs weight in $T\to 0$ limit. But indeed, the coefficients of these polynomials seems to be intriguingly small and non-trivial... – Slava Kashcheyevs Oct 17 '11 at 10:52
  • What I was thinking of was $$\prod_{l=0}^{m-1}(q^n-q^l)= \sum_{a=m(m-1)/2}^{n(m-1)} \left(\sum_{b=0}^{m-1} (-1)^b p_b(a-bn,m)\right)q^a,$$ where $p_e(f,g)$ counts the number of ways to write $f$ as a sum of $e$ numbers $\le g$. This gets very very complicated, obviously. – anon Oct 17 '11 at 10:53
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    I have a suspicion that your polynomial is a special case of one of the $q$-Hahn class of orthogonal polynomials; I haven't fully browsed the usual database, but methinks it's in there somewhere... – J. M. ain't a mathematician Oct 17 '11 at 10:54
  • I don't think I have ever heard the phrase «combinatorial physics» :) – Mariano Suárez-Álvarez Oct 18 '11 at 02:49
  • @MarianoSuárez-Alvarez: it is also known as "statistical physics before thermodynamic limit" :) I've added a link from "combinatorial" to "physics" at the end of the question. – Slava Kashcheyevs Oct 18 '11 at 04:37

4 Answers4

7

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...

J. M. ain't a mathematician
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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.)
Community
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Slava Kashcheyevs
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6

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

Peter Taylor
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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.

Slava Kashcheyevs
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