# 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?)

## marked as duplicate by Qmechanic♦ quantum-mechanics StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Jul 13 at 17:57

• Hint: Does your potential $V$ possess a $\mathbb{Z}_2$-symmetry? – Qmechanic May 9 '17 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? – physics May 9 '17 at 11:07
• The thing is, that this linked potentials apparently are symmetrical, but I still can calculate a antisymmetric and symmetric solution. – physics May 9 '17 at 11:09
• Possible duplicates: physics.stackexchange.com/q/13980/2451 , physics.stackexchange.com/q/44003/2451 and links therein. – Qmechanic May 9 '17 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? – physics May 9 '17 at 11:39

• Which means for symmetrical: $\psi(x)=\psi(-x)$ and for anti-symmetrical: $-\psi(x)= \psi(-x)$. – physics May 10 '17 at 19:23