# Why do you need symmetric and antisymmetric solutions of the time-independent Schrödinger Equation by a given potential $V(x)$? [duplicate]

I've calculated many symmetric and antisymmetric solutions of the time-independent Schrödinger Euqation by a given square potentials $V(x)$. Just for practice etc., but honestly I do not understand why I have to calculate the symmetric and antisymmetric solution of the time-independent Schrödinger Equation.

I.e. for every $i$-th constant potential part $V_i$ in $V(x)$ you can compute a time-indepedent wave-function $\psi_i(x)$, which solves the time-independent Schrödinger Euqation $$\psi_i(x)'' + k_i^2 \psi_i(x)'=0$$ with $k_i=\frac{\sqrt{2m(V_i-E)}}{\hbar}$ and in the end you got $\psi(x)$ solved in a symmetric way $(\psi(x)=\psi(-x)$, even parity$)$ and solved in an anti-symmetric way$(\psi(-x) = -\psi(x)$, odd parity$)$ with respect to continium conditions between each $\psi(x)_i$-boundary of course.

So what meaning does the symmetric and antisymmetric solution have for a particle with mass m in a certain (square) potential in Quantum Mechanics?

(I tried to find a solution for my problem, but I neither really found it in Albert Messiah books nor here. But I read that post here: Definite Parity of Solutions to a Schrödinger Equation with even Potential?)

• Hint: Does your potential $V$ possess a $\mathbb{Z}_2$-symmetry? May 9, 2017 at 9:43
• E.g. potentials like this: upload.wikimedia.org/wikipedia/commons/thumb/3/3a/… where V(x)=V(-x) or this s3.amazonaws.com/answer-board-image/… . I guess you meant that by Z2-Symmetry? May 9, 2017 at 11:07
• The thing is, that this linked potentials apparently are symmetrical, but I still can calculate a antisymmetric and symmetric solution. May 9, 2017 at 11:09
• Possible duplicates: physics.stackexchange.com/q/13980/2451 , physics.stackexchange.com/q/44003/2451 and links therein. May 9, 2017 at 11:10
• Okay so if we have an even potential $V(x)$, so we can also say that $\psi(x) =\psi(-x)$(symmetric solution) because of $V(x)=V(-x)$. Right? But I couldn't read the antisymmetric solution out of those links. So why this wave function also have odd parity? May 9, 2017 at 11:39

• Which means for symmetrical: $\psi(x)=\psi(-x)$ and for anti-symmetrical: $-\psi(x)= \psi(-x)$. May 10, 2017 at 19:23