I'm studying Merzbacher's Quantum Mechanics. In his treatment of harmonic oscillators, he introduces Hermite polynomials and their properties. He mentions that the Hermite polynomials (in the form such that $H_n$ has the coefficient of the highest power term equal to $2^n$) can be represented as
$$
H_n(x) = \frac{2^n}{\sqrt{\pi}} \int_{-\infty}^{\infty} (x+iu)^n e^{-u^2}\; du,
$$
for $n=0, 1, 2, \ldots$.
He justifies this by noting that the above satisfy, first, the relation
$$
H_n'(x) = 2nH_{n-1}(x)
$$
for all $n\ge1$, and, second, that their values agree with the Hermite polynomials at $x=0$.
Question: I'm unsure how the above two conditions are sufficient for any family of functions (indexed by $n=0, 1, 2, \ldots$) to be equal to the Hermite polynomials. I'm looking for a proof.
Best Answer
We proceed by induction. If $p_0(x)$ is a polynomials such that $p\,_0'(x)=0$ then $p_0(x)$ is a constant and since $p_0(0)=H_0(0)$ then $p_0(x)=H_0(x).$ Now suppose that $p_n(x)$ is a polynomial such that $p\,_n'(x) = 2nH_{n-1}(x).$ But also $H_n(x)$ satisfies that equation, thus they differ by a constant. The condition that $p_n(0)=H_n(0)$ implies that the constant is $0$ and thus $p_n(x)=H_n(x).$ By induction this is true for all $n$.