Todd Trimble
Completeness of polynomials

The Hermite functions H nH_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 2H = p^2 + q^2, where p=โˆ’iddxp = -i\frac{d}{d x} and q=xq = x (multiplication by xx). For the moment we will regard these as operators on Schwartz space ๐’ฎ(โ„)\mathcal{S}(\mathbb{R}). If we introduce so-called annihilation and creation operators a=q+ipa = q + i p, a โ€ =qโˆ’ipa^\dagger = q - i p respectively, then the ground state H 0H_0 is a solution to a(H 0)=0a(H_0) = 0, and we easily find the solution H 0(x)=e โˆ’x 2/2H_0(x) = e^{-x^2/2}, up to a constant factor. Ignoring issues of normalization, the remaining H nH_n (representing nn-photon states, in physics-speak) are defined by H n+1=a โ€ (H n)H_{n+1} = a^\dagger(H_n). One finds after a little calculation that each H nH_n is a degree nn polynomial, say P n(x)P_n(x), times e โˆ’x 2/2e^{-x^2/2}.

One claim that physicists frequently elide over is that the Hermite functions H nH_n form an orthonormal basis or complete orthonormal system for L 2(โ„,dx)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 โˆ’x 2/2H_n(x) = P_n(x)e^{-x^2/2} where each P nP_n is a degree nn polynomial, itโ€™s really a matter of showing that the span of x ne โˆ’x 2/2x^n e^{-x^2/2} is dense in L 2(โ„,dx)L^2(\mathbb{R}, d x), or that polynomials are dense in the Hilbert space โ„‹=L 2(โ„,w(x)dx)\mathcal{H} = L^2(\mathbb{R}, w(x) d x), where w(x)w(x) is the weight e โˆ’x 2/2e^{-x^2/2}.

If the closure of the span of x nx^n in โ„‹\mathcal{H} were a proper subspace EโŠ‚โ„‹E \subset \mathcal{H}, then we could find some fโˆˆโ„‹f \in \mathcal{H} for which fโˆ’p E(f)โ‰ 0f - p_E(f) \neq 0, where p Ep_E is the orthogonal projection onto the subspace EE. Now

โŸจe,fโˆ’p E(f)โŸฉ=0\langle e, f - p_E(f) \rangle = 0

for all eโˆˆEe \in E, by definition of p Ep_E. Turning this around, if we prove that g=0g = 0 whenever โŸจe,gโŸฉ=0\langle e, g \rangle = 0 for all eโˆˆEe \in E, or that g=0g = 0 whenever โŸจx n,gโŸฉ=0\langle x^n, g \rangle = 0 for all nn, then we will have proved completeness.

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

G(z) = โˆ‘ nโ‰ฅ0z nn!โˆซ โ„x ng(x)w(x)dx = โˆซ โ„e xzg(x)w(x)dx\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=โˆ’itz = -i t in the last integral, this means that the Fourier transform of g(x)w(x)g(x)w(x) vanishes identically. But this means that g(x)w(x)=0g(x)w(x) = 0 a.e. (since the Fourier transform is invertible), or that g(x)=0g(x) = 0 a.e., since w(x)w(x) is nowhere zero. This completes the proof.

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

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