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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?)

Qmechanic
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physics
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1 Answers1

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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.

Toffomat
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