Generating Functions and Associated Laguerre Polynomials

Question

To give you context, I am currently attempting to derive the radial wavefunctions for a hydrogenic atom, from scratch. B.H. Bransden, C.J. Joachain - Physics of Atoms and Molecules states:

$$U_{p}(\rho, s) = \frac{(-s)^{p} e^{-\rho s /(1-s)}}{(1-s)^{p+1}} = \sum^{\infty}_{q=p} \frac{L^{p}_{q}(\rho)}{q!} s^q ; |s|<1$$ An explicit expression for $L_{n+l}^{2l+1} (\rho)$ is given by $$L_{n+l}^{2l+1} (\rho) = \sum^{n-l-1}_{k=0} (-1)^{k+1} \frac{[(n+l)!]^{2} \rho^{k}}{(n-l-1-k)! (2l+1+k)! k!} $$ and is easily verified by substitution into the above with $q = n + l$ and $p=2l+1$.

I am stuck on how to verify this by substitution. Perhaps this is a naive approach but here is what I have attempted so far:

$$ U_{p}(\rho, s) = \sum^{\infty}_{n-l-1=0} \frac{L_{n+l}^{2l+1}(\rho)}{(n+l)!} s^{n+l} $$ Any tips on how to progress would be very helpful.

Ultimately, I would also like to obtain $L^{2l+1}_{n-l-1}({\rho})$.

I found this question that is somewhat related to mine, but I would like to use the expressions provided in Joachim and Bransden for the moment.

Edit: Since I asked the question, I have also attempted the following: I first plugged in the taylor expansion of $e^{\rho s/(1-s)}$ and the binomial expansion of $(1-s)^{p+1}$ into the expression, giving:

$$ \sum^{2l+1}_{b=0} \sum^{\infty}_{a=0} \frac{(-s)^{2l+1-b+a} \rho^a b! (2l+1-b)!}{a!(2l+1)!(1-s)^a} $$

Then I tried to shift the summation index from $2l+1$ to $n-l-1$, introducing a new variable $k = b + (n-3l-2)$ but I am not sure what to do from this point. I also set $ a = n-l-1 $.

Answer 1

Here is a solution for a nonnegative integer $p$. We can write $U_p(\rho, s)$ as a Riordan array: $$ U_p(\rho, s)=\left(\frac{1}{1-s},-\frac{s}{1-s}\right)*(s^p e^{\rho s}). $$ The left-hand side is the column vector of the coefficients of the power series expansion of $U_p(\rho, s)$. The right-hand side is a product of a matrix and a column vector. The left factor is the lower-triangular matrix filled with the Pascal triangle with columns alternating signs. In other words, $$ \left(\frac{1}{1-s},-\frac{s}{1-s}\right)=\left[(-1)^k\binom{n}{k}\right]_{n\ge k\ge 0}, $$ where $n$ is the row index and $k$ is the column index. The right factor is the column vector of the coefficients of $s^p e^{\rho s}$, i.e. $$ \left[\frac{\rho^{k-p}\binom{k}{p}p!}{k!}\right]_{k\ge 0}. $$

Taking the product, we see that $$ [x^n] U_p(\rho, s)=\sum_{k=0}^{n}(-1)^k\binom{n}{k}\frac{\rho^{k-p}\binom{k}{p}p!}{k!}=\sum_{k=0}^{n-p}(-1)^k\binom{n}{k+p}\frac{\rho^{k}}{k!}, $$ i.e. $$ L^{p}_{n}(\rho)=\left[\frac{x^n}{n!}\right]U_p(\rho, s)=\sum_{k=0}^{n-p}(-1)^k\binom{n}{k+p}\frac{n!}{k!}\rho^{k}. $$ Now make the substitution $n\gets n+l$, $p\gets 2l+1$.