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?)

  • 1
    $\begingroup$ Hint: Does your potential $V$ possess a $\mathbb{Z}_2$-symmetry? $\endgroup$
    – Qmechanic
    May 9, 2017 at 9:43
  • $\begingroup$ 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? $\endgroup$
    – physics
    May 9, 2017 at 11:07
  • $\begingroup$ The thing is, that this linked potentials apparently are symmetrical, but I still can calculate a antisymmetric and symmetric solution. $\endgroup$
    – physics
    May 9, 2017 at 11:09
  • $\begingroup$ Possible duplicates: physics.stackexchange.com/q/13980/2451 , physics.stackexchange.com/q/44003/2451 and links therein. $\endgroup$
    – Qmechanic
    May 9, 2017 at 11:10
  • $\begingroup$ 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? $\endgroup$
    – physics
    May 9, 2017 at 11:39

1 Answer 1


The way I understand the question, it i smore about the motivation for definite-parity states, so I'll try to answer that one. The mathematics has been well covered in the answers linked to in the comments.

I asked myself the same question when I first solved the Schrödinger equation for very simple potentials such as one-dimensional piecewise-constant ones. In such cases, the equation can be solved in a number of ways, such as the one you've indicated, and you don't strictly have to use the symmetry properties of the potential. However, it is the smart thing to do and get into the habit now because you will soon encounter potentials which are (even just a bit) more complicated, and exploiting the symmetry of the Hamiltonian is essential to find the eigenstates.

  • $\begingroup$ Yes, I get what you mean. But I guess, when I have a given antisymmetrical square-potential I only my wave-function can only be solved by the odd parity/anti-symmetric way, right? $\endgroup$
    – physics
    May 10, 2017 at 19:21
  • $\begingroup$ And if I have a symmetric potential, you actually get two different ways with symmetrical and anti-symmetrical way. Right? $\endgroup$
    – physics
    May 10, 2017 at 19:22
  • $\begingroup$ Which means for symmetrical: $\psi(x)=\psi(-x)$ and for anti-symmetrical: $-\psi(x)= \psi(-x)$. $\endgroup$
    – physics
    May 10, 2017 at 19:23
  • $\begingroup$ @physics If the potential is antisymmetric, the Hamiltonian as a whole does not have a symmetry. Hence, I don't think you can say very much about the symmetries of the solutions in general. In this case, you can probably show that nontrivial solutions do not have definite parity, i.e. they are generically neither symmetric nor antisymmetric. $\endgroup$
    – Toffomat
    May 11, 2017 at 8:53

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