I'm searching for funtions $f_n:\mathbb{R} \rightarrow \mathbb{R}$, for which $$ \sqrt{n}f_{n-1}(x) + \sqrt{n+1}f_{n+1}(x) = \sqrt{2} x f_n(x). $$ The closest set of functions I found are Hermite-polynomials, since $$ H_{n+1}(x) + 2nH_{n-1}(x) = 2xH_n (x). $$ I guess the solution will be some polynomials, similar to $H_n(x)$
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2$\begingroup$ It looks like you can define $f_0$ and $f_1$ however you like and that will lock in the rest of them. Is there some additional restriction on the functions you want? $\endgroup$– BlitzerCommented Jul 21, 2022 at 17:21
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$\begingroup$ For $f_0$, I have $f_0(x) = \pi^{-1/4}e^{-x^2/2}$ I can't calculate $f_1$. $\endgroup$– dnnagyCommented Jul 21, 2022 at 17:31
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1$\begingroup$ You can have an arbitrary scale factor in the $f_n$, so the $\pi^{-1/4}$ can be removed. $\endgroup$– marty cohenCommented Jul 21, 2022 at 18:12
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1 Answer
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No, the solution is not a polynomial for your initial condition in your comment.
There are systematic methods for solving homogeneous three term recursions, but in this problem, the answer is screaming at you! It is the harmonic oscillator in Fock space.
Recall $$ a|n\rangle +a^\dagger |n\rangle = (a+a^\dagger) |n\rangle ~~~\leadsto \\ (\sqrt{n} + \sqrt{n+1})\langle x|n\rangle= \sqrt{2} \langle x|\hat x|n\rangle ,\implies \\ f_n(x)=\langle x|n\rangle = \psi_n(x) = {1 \over \pi^{1/4} \sqrt{2^n n!}} \exp(-x^2 / 2)~ H_n(x) \\ = {1\over \pi^{1/4}\sqrt{ 2^n n!}} \exp(-x^2 / 2)~ \left(2x - \frac{d}{dx} \right)^n \cdot 1. $$ Hermite functions, indeed.
The normalization is arbitrary, by linearity, so you may adjust it as you wish... Further note the exponential may be taken away, again by linearity, so your recursion is also solved by the Hermite polynomials, properly normalized, $f_n=H_n/\sqrt{ 2^n n!}$; except you chose an initial condition disallowing that! But the two apparently different recursions in your question are essentially the same!
answered Jul 22, 2022 at 19:16