= Solution
The <Time-dependent Schrödinger equation> for the one-dimensional <Quantum harmonic oscillator> is
$$
i\hbar\frac{\partial\Psi}{\partial t}
=\left(-\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2}
+\frac12m\omega^2x^2\right)\Psi.
$$
A <stationary state> has $\Psi(x,t)=\psi(x)e^{-iEt/\hbar}$, so cancellation of the common time factor gives
$$
-\frac{\hbar^2}{2m}\psi''+\frac12m\omega^2x^2\psi=E\psi.
$$
Set
$$
y=\sqrt{\frac{m\omega}{\hbar}}x,
\qquad
\epsilon=\frac{2E}{\hbar\omega}.
$$
The equation becomes
$$
\boxed{-\frac{d^2\psi}{dy^2}+y^2\psi=\epsilon\psi}.
$$
With $\psi=f(y)e^{-y^2/2}$,
$$
\psi''=\left[f''-2yf'+(y^2-1)f\right]e^{-y^2/2},
$$
and hence $f$ obeys the <Hermite differential equation>
$$
\boxed{f''-2yf'+(\epsilon-1)f=0}.
$$
If $f$ is monic of degree $N$, the coefficient of $y^N$ in this equation is $\epsilon-1-2N$. It must vanish, so
$$
\boxed{\epsilon=2N+1,
\qquad E=\hbar\omega\left(N+\frac12\right)}.
$$
Writing $f=\sum_{k=0}^Na_ky^k$ gives the recurrence
$$
a_{k+2}=\frac{2(k-N)}{(k+2)(k+1)}a_k.
$$
The coefficient of $y^{N-1}$ first gives $a_{N-1}=0$, and the recurrence preserves parity. If $N$ is even, every odd coefficient vanishes, so $f$ is an <even function>. The Gaussian factor is also even; therefore the stationary state $\psi$ has even <parity>.
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