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.
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$\begingroup$ The title statement (v3) is not correct. Possible duplicates: physics.stackexchange.com/q/44003/2451 , physics.stackexchange.com/q/13980/2451 and links therein. $\endgroup$– Qmechanic ♦Commented Jul 13, 2019 at 17:55
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$\begingroup$ @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 $\psi(x) - \psi(-x)$ $\endgroup$– lifeistodCommented Jul 13, 2019 at 17:59
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$\begingroup$ Well, the post (v3) misquotes the actual statement. $\endgroup$– Qmechanic ♦Commented Jul 13, 2019 at 18:03
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As stated, that statement is false. Consider as a counterexample the infinite potential well centered at the origin with periodic boundary conditions. A plane wave with the appropriate momentum is an eigenstate of the Hamiltonian, but is neither even nor odd.
Notice, however, that each allowed energy level is twofold degenerate - you can have a plane wave propagating to the left or to the right, and these states are linearly independent. Therefore, rather than taking the plane waves as a basis, we can consider the even and odd linear combinations of them at each momentum (coordinating to cosines and sines).
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 $x$).
This is the key point - it's not every possible energy eigenstate is even or odd, but rather that you can always find a basis of energy eigenstates which are.
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.
answered Jul 13, 2019 at 18:12