Schrodinger equation: If $V(x)=V(-x)$ then prove that $\psi(x)=\psi(-x)$ or $\psi(x)=-\psi(-x)$ [duplicate]

The title explains itself. If the potential is an even function then prove that wave function is either odd or even. I set $$-x$$ in Schrodinger equation and find out that $$\psi(-x)$$ is also a solution for the equation therefore any linear combination of $$\psi(x)$$ and $$\psi(-x)$$ is also a solution but I couldn't go any further from that. Any help is appreciated.

• The title statement (v3) is not correct. Possible duplicates: physics.stackexchange.com/q/44003/2451 , physics.stackexchange.com/q/13980/2451 and links therein. Commented Jul 13, 2019 at 17:55
• @Qmechanic I saw that post earlier but there is no solution in there and the question there is that there exist even and odd solutions for schrodinger equation which directly follows from linear combination of $\psi(x)+\psi(-x)$ and $\psi(x) - \psi(-x)$ Commented Jul 13, 2019 at 17:59
• Well, the post (v3) misquotes the actual statement. Commented Jul 13, 2019 at 18:03

By considering this basis instead of the plane wave basis, we have what you want - energy eigenstates which are also eigenstates of the parity operator (which flips the sign of $$x$$).
To help with your proof of this, note that if $$\psi(x)$$ is an energy eigenstate with energy $$E$$, then so is $$\phi(x)=\psi(-x)$$. From there, taking the even and odd linear combinations of these states yields energy eigenstates which are even and odd, respectively.