Deriving Rodrigues Formula and Generating function of Hermite Polynomial from $H_n(x)= e^{x^2/2}(x-\frac{d}{dx})^ne^{-x^2/2}$

Question

There are a variety of ways of first defining the Hermite Polynomials in a certain way and then to derive alternative representations of them. For example in Mary Boas' Mathemmatical methods (p. 607, 3rd edition) she starts with the differential equation $y''_n-x^2y_n = -(2n+1)y_n$. Arfken's Mathematical Methods (p. 817, 6th edition) starts with a generating function $g(x,t)=e^{2xt-t^2}=\sum_{n=0}^{\infty}\frac{t^n}{n!}H_n(x)$. Griffiths Intro to Q.M. (p.57,problem 2.17, 2end edition) starts with Rodrigues formula: $H_n(x)= (-1)^ne^{x^2}\frac{d^n}{dx^n}e^{-x^2}$.

I'd like to do the following: Start from the defintion $H_n(x)= e^{x^2/2}(x-\frac{d}{dx})^ne^{-x^2/2}$ and then derive both Rodrigues Formula $H_n(x)= (-1)^ne^{x^2}\frac{d^n}{dx^n}e^{-x^2}$ and then the generating function $g(x,t)=e^{2xt-t^2}=\sum_{n=0}^{\infty}\frac{t^n}{n!}H_n(x)$.

I'm not sure where to begin. My primary interest in this is quantum mechanics and how this Hermite Polynomial is used as a solution to the quantum Harmonic oscillator. I have never used Hermite Polynomials (or Laguerre, Legendre or Bessel functions)... So I'd be grateful for some advice on how to do this.

edit: Okay so here's my approach for deriving Rodrigues Formula $H_n(x)= (-1)^ne^{x^2}\frac{d^n}{dx^n}e^{-x^2}$: I use induction on $n$. For $n = 0$ both $H_n(x)= e^{x^2/2}(x-\frac{d}{dx})^ne^{-x^2/2}$ and Rodrigues give $1$. Now assume the equality holds for some $n$ and show it for $n+1$: but now I'm stuck again...

Answer 1

For a start, you can go from the Rodrigues Formula to the generating function by noting that the equality for the generating function holds if and only if $$\frac{d^n}{dt^n}|_{t=0} g(x,t) = H_n(x)$$ But the left hand side evaluates to $$\frac{d^n}{dt^n}|_{t=0} e^{2xt-t^2} = \frac{d^n}{dt^n}|_{t=0} e^{x^2} e^{-(x-t)^2} = e^{x^2}\frac{d^n}{dt^n}|_{t=0}e^{-(x-t)^2}$$ Now substitute $t=x-u$ in the expression on the right, so that it becomes $$e^{x^2}(-1)^n\frac{d^n}{du^n}|_{u=x}e^{-u^2}$$ which is exactly the Rodrigues formula.

Answer 2

There are different versions of the formula, but the arguments are the same.

Using integration by parts,

$\int_{-\infty}^\infty -2nH_mH_ne^{-x^2}dx=\int_{-\infty}^\infty -2m H_mH_ne^{-x^2}dx$

which implies Hermite Polynomials are orthogonal. See how?

The series solution implies all polynomials can be expressed as a linear combination of HPs. That means $xH_n$ is always some linear combination of HPs. By orthogonality, those HPs are $H_{n-1}$ and $H_{n+1}$.

$xH_n=A_nH_{n-1}+B_nH_{n+1}$

$H_n+xH_n'=A_nH_{n-1}'+B_nH_{n+1}'$

$2H_n'+xH_n''=A_nH_{n-1}''+B_nH_{n+1}''$

$2H_n'-2xH_n-2nxH_n=-2A_n(n-1)H_{n-1}-2B_n(n+1)H_{n+1}$

Now add $2(n+1)$ times the first equation to get:

$2H_n'=4A_nH_{n-1}\implies H_n'=2A_nH_{n-1}$

$\implies 2B_nH_{n+1}=2xH_n-H_n'$

$2B_nH_{n+1}e^{-x^2}=-(H_ne^{-x^2})'$

So up to a multiplicative constant, possibly depending on $n$, we have a Ladder operator that can generate the next HP from the previous, next being measured by the eigen value.

This suggests $\frac{d^{n+k}}{dx^{n+k}}(H_ne^{-x^2})=C_nH_{n+k}e^{-x^2}$

With some index juggling:

$H_n=C_n(-1)^ne^{x^2}\frac{d^n}{dx^n}(e^{-x^2})$

The constant is somewhat irrelevant since it depends on how we want to normalize the polynomials and there are multiple options for doing so. The math is simplified if $C_n=1$.

By inspection, it can be shown the formula produces the first few HPs. By induction, it can be shown it gives the rest.

The proof by induction requires Leibniz's Formula for derivative of products:

$\frac{d^n}{dx^n}(A(x)B(x))=\sum_{k=0}^n \binom{n}{k}\frac{d^kA}{dx^k}\frac{d^{n-k}B}{dx^{n-k}}$

Through a similar process, starting with orthogonality and the recurrence relationw with undetermiend coefficients, one can derive similar formulas for Legendre and Laguerre Polynomials. Both you see in a course in Quantum Mechanics.