A problem from Griffith's Introduction to Quantum Mechanics asks to prove the following:
Given a symmetric potential $V(x)$ $(=V(-x))$, the solutions to the time-independent Schrödinger's equation can always be take to be either odd or even.
I understand that given a solution $\psi(x)$, $\psi(-x)$ is also a solution.
Now, $\psi_{even}(x) = \psi(x) + \psi(-x)$ and $\psi_{odd}(x) = \psi(x) - \psi(-x)$ are also solutions to the equation.
Thus $$\psi(x) = \frac {1}{2}(\psi_{even}(x) + \psi_{odd}(x))$$
But $$\psi_{even}(-x) + \psi_{odd}(-x) = \psi_{even}(x) - \psi_{odd}(x)$$
and hence $\psi(x)$ is neither even or odd!
I am unable to resolve this issue.
