Sum involving Laguerre polynomials

Question

I am trying to prove the following equality: \begin{equation*} \sum_{k=0}^{\min(m,n)}\frac{(-1)^k}{(m-k)!(n-k)!k!}x^{\min(m,n)-k}=\frac{(-1)^{\max(m,n)}}{[\max(m,n)!]^2}L_{\max(m,n)}^{(|m-n|)}(x) \end{equation*} where $L_n^{(\alpha)}(x)$ denotes an associated Laguerre polynomial. I am not even confident of its correctness, as tests with Mathematica are inconsistent with the result. In any case, what is behind the idea of the author? In particular, I can't understand how to pass from a sum till $\min(m,n)$ to a sum to an upper or equal value $\max(m,n)$, as this is the boundary of the corresponding series expansion of the associated Laguerre polynomials \begin{equation*} L_n^{(\alpha)}(x)=\sum_{i=0}^n \binom{n+\alpha}{n-i}\frac {x^i}{i!} \end{equation*} The expression appears in eqn. (5.16) at page 457 of the paper I'm reading [Physica 12, 405 (1946); eprint] (please note that the author himself slips in the denomination of the Laguerre polynomials, referring to them as "Legendre").

The only relation that came to my mind that could help the issue is \begin{equation*} (-x)^i\frac 1 {i!}L_j^{(i-j)}(x)=(-x)^j\frac 1 {j!}L_i^{(j-i)}(x) \end{equation*} in combination with $\min(m,n)= \max(m,n)-|m-n|$.

Answer 1

I am not even confident of its correctness, as tests with Mathematica are inconsistent with the result.

Then those tests are sufficient to show that the result is wrong, and there's nothing else to say there.

In particular, I can't understand how to pass from a sum till $\min(m,n)$ to a sum to an upper or equal value $\max(m,n)$

You can't. The degree of the polynomial on the left is different to that on the right whenever $n\neq m$.

The expression appears in eqn. (5.16) at page 457 of the paper I'm reading [Physica 12, 405 (1946); eprint]

Then the paper is wrong, at least as regards this particular result. From a cursory reading of the subsequent text, it doesn't seem to be crucial to the rest of the arguments presented, so you should be able to fix any problems with it, if you really need to.

Answer 2

@Emilio Pisanty already answered your question, pointing out the degree of the l.h.s. of your target expression is min, but that of the r.h.s. is max.

Groenewold's expression (5.16) needs min(m,n) as a subscript of the Associated Laguerre, as, discounting the exponential, the highest power of the Wigner function involved must be $z^{(m+n)/2}$, where $z\equiv 2(x^2+p^2)/\hbar$. For the sake of illustration take max=M and min=n. Choose p=0 to efface the difference between left and right and make everything real.

You see then that Groenewold's expression, up to the exponential, starts with $z^{(M+n)/2}$, but concludes with $z^{(M-n)/2 +M}$; all is fixed if the Legendre subscript is n, the minimum, instead.

This is the canonical expression of the Bartlett & Moyal formula (2.5) in the comments, and (99) in our booklet, which eschews pesky summations of polynomials, by solving the resulting associated Laguerre equation from first principles, a naturalized phase space practice: The conceit is Schrodinger and the rest would have never been born!

---

I think, from here, you can chase down factors and signs. I strongly suspect the squaring of factorials in the denominator is another Groenewold typo. This is near the end of a history-changing, revolutionary (he virtually tore down his idol's, von Neumann, quantization recipe!) thesis, written, I assume you know, while Groenewold was on the lam from Nazis and local collaborators, hiding in the Polder, sleeping in barns, diving into ditches when hearing unfamiliar voices, etc. Three years later, Bartlett and Moyal had the leisure to smooth things out...