 Normalization forces Hermite-series termination

For the Hermite differential equation $f^{\prime \prime }-2\xi f^{\prime }+(ϵ-1)f=0$ with real $ϵ$, the wavefunction $e^{-{\xi }^2/2}f$ is square integrable on the real line only if $f$ is a polynomial. To prove this without assuming that even and odd tails cannot cancel, project a square-integrable solution onto its even and odd parts first. Each projection remains square integrable and satisfies the same equation.

In either nonterminating parity sequence, $b_j=a_{2j+p}$ obeysThe tail has constant complex phase. For any fixed $0<\delta <1/2$, comparison with $(1-\delta )/(j+1)$ bounds its coefficient magnitudes below by $K(1-\delta {)}^j/j!$ for sufficiently large $j$. Thus on the positive real axis the tail dominates a positive multiple of ${\xi }^p{e}^{(1-\delta ){\xi }^2}$ after subtraction of finitely many terms. The Gaussian function factor leaves exponential growth $e^{(1/2-\delta ){\xi }^2}$, contradicting the square-integrable condition. Hence each nonzero parity sequence terminates. Its top coefficient forces $ϵ=2n+1$, giving Hermite oscillator quantization and a Hermite polynomial of degree $n$.