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, itโs 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

$\langle e, f - p_E(f) \rangle = 0$

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

$\array{
G(z) & = & \sum_{n \geq 0} \frac{z^n}{n!} \int_{\mathbb{R}} x^n g(x) w(x) d x \\
& = & \int_{\mathbb{R}} e^{x z} g(x)w(x) d x
}$

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.

Revised on October 1, 2011 at 18:30:08
by
Todd Trimble