Functions satisfying this recurrence relation

Question

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 $H_n(x)$

Answer 1

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!