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Playing around with different approaches to solve the radial part of the Schrodinger equation for the hydrogen-like atom, I have obtained the following expression ($l$ and $n$ are non-negative integers) $$ \tilde L^{(2l+1)}_n(x) = \frac{1}{n!} x^{-(l+1)}e^{x/2} \left(x \frac{d}{dx}-x \frac{d^2}{dx^2} + \frac{l(l+1)}{x} -\frac{x}{4} \right)^n \bigl(x^{l+1} e^{-x/2} \bigr) $$ which I believe is a representation of the generalized Laguerre polynomials.

The latter are conventionally defined as via de Rodrigues formula $$L_n^{(2l+1)}(x) = \frac{1}{n!} x^{-(2l+1)} e^{x} \frac{d^n}{dx^n} \Bigl(x^{2l+1+n} e^{-x} \Bigr)\,. $$

I have the feeling that there is a simple way to show that the two expressions are the same with $L= \tilde L$. Can somebody point me in the right direction?

Fabian
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1 Answers1

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Denoting \begin{equation} h(x)=x^{l+1}e^{-x/2};\quad \varphi(x)=x^{2l+1}e^{-x}; \quad\Theta=x \frac{d}{dx}-x \frac{d^2}{dx^2} + \frac{l(l+1)}{x} -\frac{x}{4} \end{equation} both polynomial definitions can be written as \begin{align} \tilde{L}^{(2l+1)}_n(x) &= \frac{1}{n!h(x)}\Theta^n\left[h(x)\right]\\ L^{(2l+1)}_n(x) &= \frac{1}{n!\varphi(x)}\frac{d^n}{x^n}\left[x^n\varphi(x)\right] \end{align} We introduce the function $$F(x)=x^{-l}e^{x/2}$$ It can easily be checked that $\Theta\left[F(x)\right]=0$. Then, for any function $z(x)$, \begin{equation} \Theta\left[F(x)z(x)\right]=xF(x)z'(x)-xF(x)z''(x)-2xF'(x)z'(x) \end{equation} But $F'(x)=\frac{1}{2}F(x)-\frac{l}{x}F(x)$, hence \begin{align} \Theta\left[F(x)z(x)\right]&=F(x)\left[2lz'(x)-xz''(x)\right]\\ &=F(x)\frac{d}{dx}\left[ 2l+1 -x\frac{d}{dx}\right]z(x) \end{align} Denoting \begin{equation} M= 2l+1 -x\frac{d}{dx} \end{equation} we have shown that \begin{equation} \Theta\left[F(x)z(x)\right]=F(x)\frac{d}{dx}M\left[z(x)\right] \end{equation} and thus, for $n=1,2,3\ldots$, \begin{equation} \Theta^n\left[F(x)z(x)\right]=F(x)\left( \frac{d}{dx}M \right)^n\left[z(x)\right] \end{equation} Choosing $z(x)=\varphi(x)=x^{2l+1}e^{-x}$, we have $F(x)\varphi(x)=h(x)$ and then \begin{equation} \Theta^n\left[h(x)\right]=F(x)\left( \frac{d}{dx}M \right)^n\left[\varphi(x)\right] \end{equation} or, dividing both sides by $n!h(x)$, \begin{equation} \frac{1}{n!h(x)}\Theta^n\left[h(x)\right]=\frac{1}{n!\varphi(x)}\left( \frac{d}{dx}M \right)^n\left[\varphi(x)\right] \end{equation} It can be checked that $M\left[\varphi(x)\right]=x\varphi(x)$ which gives, for $n=1$, \begin{equation} \frac{1}{h(x)}\Theta\left[h(x)\right]=\frac{1}{\varphi(x)}\frac{d}{dx}\left( x\varphi(x) \right) \end{equation} and thus $\tilde{L}^{(2l+1)}_1(x)=L^{(2l+1)}_1(x)$. To go further, we use induction. Suppose that for some $n$ \begin{equation} \left( \frac{d}{dx}M \right)^n\left[\varphi(x)\right]=\frac{d^n}{dx^n}\left( x^n\varphi(x) \right) \end{equation} then \begin{equation} \left( \frac{d}{dx}M \right)^{n+1}\left[\varphi(x)\right]= \frac{d}{dx}M \frac{d^n}{dx^n}\left( x^n\varphi(x) \right) \end{equation} For any function $z(x)$, \begin{align} M\frac{d}{dx}z(x)&=\left( 2l+1 -x\frac{d}{dx}\right)\frac{d}{dx}z(x)\\ &=\frac{d}{dx}\left(2l+2-x\frac{d}{dx}\right)z(x)\\ &=\frac{d}{dx}\left( M+1 \right)z(x) \end{align} More generally, \begin{equation} M\frac{d^n}{dx^n}=\frac{d^n}{dx^n}\left( M+n \right) \end{equation} Then \begin{align} \left( \frac{d}{dx}M \right)^{n+1}\left[\varphi(x)\right]&=\frac{d^{n+1}}{dx^{n+1}}\left( M+n \right)\left( x^n\varphi(x) \right)\\ &=\frac{d^{n+1}}{dx^{n+1}}\left( 2l+n+1-x\frac{d}{dx} \right)\left(x^{2l+n+1}e^{-x} \right)\\ &=\frac{d^{n+1}}{dx^{n+1}}\left( x^{2l+n+2}e^{-x} \right)\\ &=\frac{d^{n+1}}{dx^{n+1}}\left( x^{n+1}\varphi(x) \right)\\ \end{align} As the induction hypothesis is true for $n=1$, we have shown that \begin{equation} \frac{1}{n!h(x)}\Theta^n\left[h(x)\right]=\frac{1}{n!\varphi(x)} \frac{d^n}{dx^n}\left(x^n\varphi(x)\right) \end{equation} and thus $\tilde{L}^{(2l+1)}_n(x)=L^{(2l+1)}_n(x)$, both series of polynomials are identical.

Paul Enta
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