Todd Trimble Completeness of polynomials

The Hermite functions $H_n$, which are important in the study of the quantum harmonic oscillator, may be defined in physics-speak as eigenstates of the Hamiltonian $H = p^2 + q^2$, where $p = -i\frac{d}{d x}$ and $q = x$ (multiplication by $x$). For the moment we will regard these as operators on Schwartz space $\mathcal{S}(\mathbb{R})$. If we introduce so-called annihilation and creation operators $a = q + i p$, $a^\dagger = q - i p$ respectively, then the ground state $H_0$ is a solution to $a(H_0) = 0$, and we easily find the solution $H_0(x) = e^{-x^2/2}$, up to a constant factor. Ignoring issues of normalization, the remaining $H_n$ (representing $n$-photon states, in physics-speak) are defined by $H_{n+1} = a^\dagger(H_n)$. One finds after a little calculation that each $H_n$ is a degree $n$ polynomial, say $P_n(x)$, times $e^{-x^2/2}$.

One claim that physicists frequently elide over is that the Hermite functions $H_n$ form an orthonormal basis or complete orthonormal system for $L^2(\mathbb{R}, d x)$. The orthonormality is the easy part; it falls out from the formalism of annihilation and creation operators. But how does one prove completeness of the orthonormal system? Sufficient amounts of spectral theory would probably get the job done, but I wanted something much simpler and more concrete.

Since $H_n(x) = P_n(x)e^{-x^2/2}$ where each $P_n$ is a degree $n$ polynomial, it’s really a matter of showing that the span of $x^n e^{-x^2/2}$ is dense in $L^2(\mathbb{R}, d x)$, or that polynomials are dense in the Hilbert space $\mathcal{H} = L^2(\mathbb{R}, w(x) d x)$, where $w(x)$ is the weight $e^{-x^2/2}$.

If the closure of the span of $x^n$ in $\mathcal{H}$ were a proper subspace $E \subset \mathcal{H}$, then we could find some $f \in \mathcal{H}$ for which $f - p_E(f) \neq 0$, where $p_E$ is the orthogonal projection onto the subspace $E$. Now

for all $e \in E$, by definition of $p_E$. Turning this around, if we prove that $g = 0$ whenever $\langle e, g \rangle = 0$ for all $e \in E$, or that $g = 0$ whenever $\langle x^n, g \rangle = 0$ for all $n$, then we will have proved completeness.

But the condition $\langle x^n, g \rangle = 0$ for all $n$ means that the entire function

vanishes identically. Setting $z = -i t$ in the last integral, this means that the Fourier transform of $g(x)w(x)$ vanishes identically. But this means that $g(x)w(x) = 0$ a.e. (since the Fourier transform is invertible), or that $g(x) = 0$ a.e., since $w(x)$ is nowhere zero. This completes the proof.

The same proof applies to any positive weighting function $w(x)$ which decays rapidly.