In Griffiths, he writes the following:
For normalizable solutions,the power series of h(z), where phi = h(z)e^(-z^2)/2, must terminate. For some highest subscript, call it n, $a_{n+2}=0.$ This will truncate either the odd or the even series, the other must be zero from the start.
I am with him until we get to this last part. Why is this necessarily true?
If $\psi(x)$ solves $$ -\frac{\hbar^2}{2m}\frac{d^2}{dx^2}\psi(x)+V(x)\psi(x)=E\psi(x) \tag{1} $$ then $\psi(-x)$ solves $$ -\frac{\hbar^2}{2m}\frac{d^2}{dx^2}\psi(-x)+V(-x)\psi(-x)=E\psi(-x)\tag{2} $$ since the double derivative w/r to $x$ does not pick up a sign when $x\to -x$, i.e. $$ \frac{d^2}{d(-x)^2}=\frac{d^2}{dx^2}\, . $$ Now, if $V(x)=V(-x)$, we have that $\psi(-x)$ also solves (1) in addition to solving (2), and thus the linear combinations of solutions with the same $E$: $$ \psi_\pm (x) = \psi(x)\pm \psi(-x) $$ also solve (1) (or (2)).
The function $\psi_+(x)$ has positive parity since $\psi_+(-x)=+\psi_+(x)$. The function $\psi_-(x)$ has in turn negative parity.
Hence in solving for (1) we can do “half” the work by restricting the search to even or odd functions. In the case of the harmonic oscillator, this means we can restrict to Gaussians which multiply polynomials in even or odd powers only. In the case of a finite well, the boundary conditions at one end of the well - say $x=-a$ - are easily related to the boundary conditions at $x=+a$ so you only need to look at one boundary rather than 2.
This is because if your potential is an even function (which is true for the harmonic oscillator) then you can show with the Shrodinger equation that the wavefunction must be even or odd.
Therefore, since the series solution can be broken up into an even series and an odd series, we must take one of them to be 0 for this to be true. I believe this is also exploited in other parts of the book (I think for the finite square well potential, but I could be wrong since I currently don't have the book with me).
