Solution to Laguerre differential equation using generating function

Question

This is an exercise in Modern Quantum Mechanics by Sakurai and Napolitano.

Follow these steps to show that solutions to Kummer's equation (7.46) can be written in terms of Laguerre polynomials $L_n(x)$, which are defined according to a generating function as $$g(x, t) = \frac{e^{-xt/(1-t)}}{1-t} = \sum_{n=0}^\infty L_n(x) \frac{t^n}{n!}$$ where $0 < t < 1$. The discussion on generating functions for Hermite polynomials will be helpful.

(a) Prove that $L_n(0) = n!$ and $L_0(x) = 1$.

(b) Differentiate $g(x, t)$ with respect to $x$, show that $$L'_n(x) - nL'_{n-1}(x) = -nL_{n-1}(x),$$ and find the first few Laguerre polynomials.

(c) Differentiate $g(x, t)$ with respect to $t$ and show that $$L_{n+1}(x) - (2n+1-x)L_n(x) + n^2L_{n-1}(x) = 0.$$

(d) Now show that Kummer's Equation is solved by deriving $$xL''_n(x) + (1-x)L'_n(x) + nL_n(x) = 0,$$ and associate $n$ with the principal quantum number for the hydrogen atom.

I've done (a), (b), and (c), but I'm having trouble with (d). The difficulty seems to be that using the result in (d) to shift $n$ invariably introduces two new $n$ values, so I can never get it to be of use in simplifying what I get from (b). No matter what I do, I can't get an equation that only contains one $n$ value. But obviously I just haven't tried the right thing.

How can (d) be solved using (b) and (c)?

Answer 1

Replacing $n$ by $n+1$ in $L'_n(x) - nL'_{n-1}(x) = -nL_{n-1}(x)$, we get$$L_{n+1}'(x) = (n+1)L_n'(x) - (n+1)L_n(x).\tag*{(1)}$$Differentiating the above with respect to $x$,$$L_{n+1}''(x) = (n+1)L_n''(x) - (n+1)L_n'(x).\tag*{(2)}$$Replace $n$ by $n+1$ in the above equation to get$$L_{n+2}''(x) = (n+2)\left(L_{n+2}''(x) - L_{n+1}'(x)\right).$$Substitute $(n+1)L_n'(x) - (n+1)L_n(x)$ for $L_{n+1}'(x)$ to get$$L_{n+2}''(x) = (n+2)\left(L_{n+1}''(x) - (n+1)L_n'(x) - (n+2)L_n(x)\right)$$$$= (n+2)(n+1)\left(L_{n+1}''(x) - 2L_n'(x) + L_n(x)\right).\tag*{(3)}$$Differentiating equation $L_{n+1}(x) + (x - 2n - 1)L_n(x) + n^2L_{n-1}(x) = 0$ with respect to $x$,$$L_{n+1}'(x) + L_n(x) + (x - 2n - 1)L_n'(x) + n^2L_{n-1}'(x) = 0.$$Again differentiating,$$L_{n+1}''(x) + 2L_n'(x) + (x-2n-1)L_n''(x) + n^2L_{n-1}''(x) = 0.$$Replacing above equation with $n$ by $n+1$,$$L_{n+2}''(x) + xL_{n+2}''(x) + 2L_{n+1}'(x) - (2n+2)L_{n+1}''(x) + (n+1)^2L_n''(x) = 0.$$Substituting $(1)$, $(2)$, and $(3)$ in the above equation,$$\left((n+2)(n+1)\left(L_{n+1}''(x) - 2L_n'(x) + L_n(x)\right)\right) + x\left((n+1)L_n''(x) - (n+1)L_n'(x)\right) + 2\left((n+1)L_n''(x) - (n+1)L_n'(x)\right) - (2n+3)\left((n+1)L_n''(x) - (n+1)L_n'(x)\right) + (n+1)^2 L_n''(x) = 0.$$Simplifying the above equation, we get$$xL_n''(x) + (1-x)L_n'(x) + nL_n(x) = 0,$$which is what we wanted to show.

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Alternatively, we can prove this using the generator:$$xg'' + (1 - x)g' + ng = 0 - {{e^{{{-tx}\over{1-t}}}t(1-x)}\over{(1-t)^2}} + {{e^{{{-tx}\over{1-t}}}t^2x}\over{(1-t)^3}} + t\left({{e^{{{-tx}\over{1-t}}}}\over{(1-t)^2}} - {{e^{{{-tx}\over{1-t}}}}\over{(1-t)^2}} - {{e^{{{-tx}\over{1-t}}}}\over{(1-t)^3}}\right) = 0.$$