# Solutions to time-independent Schrödinger's equation with symmetrical (even) potential [duplicate]

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.

• Nobody told you that $\psi(x)$ is even or odd. Griffith is telling you that you can construct a solution from a solution, that is neither even nor odd. Jul 16, 2015 at 12:15
• @gonenc: Later in problem 2.27, the author uses this fact to find solutions to the double delta potential. To do so, he assumes first even solutions, and then odd solutions, and finds the number of solutions thus possible. So I suppose the author is implying that the solution to an even potential has to be either odd or even. That's the only way you can enumerate all (bound state) solutions to the double delta potential.
– Sidd
Jul 16, 2015 at 12:19
• No, he implies that they CAN be chosen to be that. If you CAN choose it, then it simplifies to consider them separately.
– Omry
Jul 16, 2015 at 12:20
• @Omry: But then how can we find all bound states to the double delta potential $\delta(x-a) + \delta(x+a)$? The author does so by first finding all even solutions possible, and then all odd solutions possible. But wouldn't that mean that the only solutions are either even of odd?
– Sidd
Jul 16, 2015 at 12:24
• No, because if two solutions have the same energy, then their sum/difference is also a solution.
– Omry
Jul 16, 2015 at 12:33