0

I have been trying to prove the following orthogonal relation which is used for the normalization of the hydrogenic radial wave function, $$\int_{0}^{\infty}e^{-\rho}\rho^{2l+2}[L^{2l+1}_{n+l}]^2d\rho=\frac{2n[(n+l)!]^{3}}{(n-l-1)!} \tag{1}$$

using the generating function $$U_p(\rho,s)=\frac{(-s)^pe^{\frac{-\rho s}{1-s}}}{(1-s)^{p+1}}=\sum_{q=p}^{\infty}\frac{L^{p}_{q}(\rho)}{q!}s^{q} \tag{2}$$

plugging (2) into (1) and relabeling $k:=2l+1$ and $m:=n+l$ I get $$\frac{(-1)^{2k}(st)^{k}}{[(1-s)(1-t)]^{k+1}}\int^{\infty}_{0}\rho^{k+1}e^{-\rho\frac{1-st}{(1-s)(1-t)}}d\rho=\sum_{v}^{\infty}\sum_{w}^{\infty}\frac{s^{v}}{v!}\frac{t^{w}}{w!}\int_{0}^{\infty}e^{-\rho}\rho^{k+1}L^{k}_vL^{k}_{w}d\rho \tag{3}$$

Now using integration by parts iteratively on the LHS of $(3)$ I get

$$\frac{(-1)^{2k}(st)^{k}}{[(1-s)(1-t)]^{k+1}}\frac{(-1)^{k+2} [(1-s)(1-t)]^{k+2}}{(1-st)^{k+2}}\Gamma[k+2]=\frac{(-1)^{k+2}(st)^{k}(1-s)(1-t)}{(1-st)^{k+2}}\Gamma[k+2] \tag{4}$$

I then used the binomial expansion of the denominator on the RHS of $(4)$.

$$ (1-st)^{-k-2}=\sum^{\infty}_{z=0}\binom{-k-2}{z}(-1)^z(st)^z \tag{5}$$

Looking at the binomial coefficient,

$$\binom{-k-2}{z} = \frac{(-k-2)(-k-2-1)...(-k-2-z+1)}{z!}=\frac{(-1)^{z}(k+z+1)!}{z!(k+1)!} \tag{6}$$

plugging $(6)$ and $(5)$ into $(4)$ and expanding the powers of $s$ and $t$ I get the following,

$$\begin{eqnarray}(4) &=& (-1)^{k+2} \Gamma[k+2] \sum^{\infty}_{z=0} \frac{(k+z+1)!}{z!(k+1)!} [-s^{k+1+z}t^{k+z}-s^{k+z}t^{k+1+z}+(st)^{k+z+1}+(st)^{k+z}] \\ &=& \sum_{v}^{\infty}\sum_{w}^{\infty}\frac{s^{v}}{v!}\frac{t^{w}}{w!}\int_{0}^{\infty}e^{-\rho}\rho^{k+1}L^{k}_vL^{k}_{w}d\rho\end{eqnarray} \tag{7}$$

Here is where I get stuck I don't really know where to go from here. Looking at the RHS of $(7)$ if $ v\neq w$ the integral is zero since Laguerre polynomials are orthogonal functions. Ive tried comparing coefficients of $st$ for example $k+z+1=2v$ since $v=n+l$ and $k=2l+1$ solving for $z$ I got $z=2n-2$ plugging that into the coefficient I got $\frac{n(n+l)!}{n!}$ which isn't right, there is also the issue of that $(-1)^{k+2}$ factor out front which I cant get rid of. Any help would be appreciated ive spent a long time trying to figure this out.

seraphimk
  • 573

1 Answers1

1

So I managed to figure it out and thought I would leave the answer for anyone elese who might be stuck as I was.

Firstly, I will address the factor $(-1)^{k+2}$ which is simply an my error when calculating the integral on the LHS of $(3)$. The integral actually gives $$\int^{\infty}_0\rho^{k+1}e^{-\rho \frac{1-st}{(1-s)(1-t)} } d\rho = \frac{[(1-s)(1-t)]^{k+2}}{(1-st)^{k+2}}$$

Secondly, when looking at the sum on the LHS of $(7)$ and comparing the coefficients of s and t in particular for when they have the same coefficients there is actually two contributions one from $(st)^{k+z}$ and another from $(st)^{k+(z-1)+1}$.

So again comparing coefficients $k+z=v$ so that $v=v-k=(n+l)-(2l+1)=(n-l-1)$.and looking at the binomial coefficients for the two $st$ contributions (i.e. for $v$ and $v-1$) we get.

$$\frac{(k+z+1)!}{z!(k+1)!}+\frac{(k+z)!}{(z-1)!(k+1)!}$$

which when inserting $k=2l+1$ and $z=n-l-1$ we get.

$$(n+l+1)\frac{(n+l)!}{(n-l-1)!(2l+2)!}+(n-l-1)\frac{(n+l)!}{(n-l-1)!(2l+2)!}=2n\frac{(n+l)!}{(n-l-1)!(2l+2)!}$$

Finally, We end up with

$$2n\frac{(n+l)!}{(n-l-1)!(2l+2)!}\Gamma[k+2]= \frac{1}{[(n+l)!]^{2}}\int^{\infty}_{0}e^{-\rho}\rho^{k+1}[L_{v}^{k}]^{2} d\rho$$

Which simply cancelling the gamma function and rearranging some terms gives the final answer. $$\int^{\infty}_{0}e^{-\rho}\rho^{k+1}[L_{v}^{k}]^{2} d\rho = 2n\frac{[(n+l)!]^{3}}{(n-l-1)!}$$

seraphimk
  • 573