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I have seen in different posts (like this one) that, in case the potential $V(x)$ of a quantum system is symmetrical, you can always find a basis of eigenstates of the Hamiltonian that have definite even or odd parity.

Let such a basis of eigenstates be $\{\psi_n\}$. Can we know a priori for which $n$ they will be even and for which $n$ they will be odd?

For instance, in the one dimension infinite potential well centered at the origin, the eigenfunctions $\psi_n=\sqrt{\frac{2}{a}}\sin(\frac{n\pi}{a}(x-a/2))$ are even for odd $n$ and odd for even $n$. Could we say the same for a symmetric potential like $V(x)=kx^4$?

Quaerendo
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1 Answers1

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Could we say the same for a symmetric potential like $V(x)=kx^4$?

The $n^\text{th}$ eigenfunction of any Sturm-Liouville problem has exactly $n-1$ roots. (See Wikipedia for confirmation.) As you mentioned, for a symmetric potential the eigenfunctions are either even or odd. The first one has no roots so it must be even. The second has one root so it must be odd. Etc.

If $\psi_0$ is the ground state, the parity of $\psi_n$ is the parity of $n$. If $\psi_1$ is the ground state, the parities are opposite.

G. Smith
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