So I was just reading back through Griffiths' "Introduction to Quantum Mechanics" and solving some of the problems for practice. There is a nice one (problem 2.1c for those playing at home) where you have to show that if the potential energy function $V(x)$ is an even function that $\psi(x)$ can always be taken to be even or odd. This is pretty straightforward, but it got me to thinking about the general restrictions on the wavefunction if $V(-x)=-V(x)$ (ie, it is odd). I decided just to use a one dimensional case. Starting with the time independent Schrodinger equation, and taking $x\rightarrow-x$ gives:
Now use the fact that $V(x)$ is odd, and put the minus sign in an illuminating location:
Now we see that the Schrodinger equation is satisfied if $\psi(-x)=-\psi(-x)=0$.
My suspicion: it is not consistent in the last line to have $V$ defined at $x$ and $\psi$ defined at $-x$. And this method, while it works nicely for the case of even potentials, is just meaningless for the case of odd potentials.
My question: Does the fact that this is meaningless mean that we can't have odd potentials in Quantum mechanics (This seems silly, we can just go into the lab and make some, right?) or does it (more likely) just mean that this method tells us nothing, and that the wavefunctions can be even, odd or neither depending on the problem? Thanks :)