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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 $V(x)=kx^4$?

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  • $\begingroup$ How is the quantum number $n$ assigned? $\endgroup$
    – Qmechanic
    Aug 28, 2020 at 19:43
  • $\begingroup$ Let $\psi_0$ be the ground state, $\psi_1$ the first excited state, and so on. $\endgroup$
    – Quaerendo
    Aug 28, 2020 at 19:49
  • $\begingroup$ In your example, the ground state is $\psi_1$ and is even, not odd. $\psi_0$ is zero everywhere and thus is not a wavefunction. $\endgroup$
    – G. Smith
    Aug 28, 2020 at 20:13
  • $\begingroup$ Thanks, I have corrected the errata in the question $\endgroup$
    – Quaerendo
    Aug 31, 2020 at 19:56

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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.

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  • $\begingroup$ Ok, thanks. And, in general, for a potential $V(x)$ that has even or odd symmetry, could we know a priory the symmetry of the eigenfunctions? $\endgroup$
    – Quaerendo
    Aug 28, 2020 at 19:57
  • $\begingroup$ This question is about symmetric (i.e., even) potentials. Please don’t use comments to ask a new question about odd potentials. (Hint: Does the logic in the post you linked to apply to an odd potential?) $\endgroup$
    – G. Smith
    Aug 28, 2020 at 20:22
  • $\begingroup$ Ok, I just wanted to make sure that your answer was for general even potentials, not only for the example of $V(x)=kx^4$ $\endgroup$
    – Quaerendo
    Aug 29, 2020 at 8:14
  • $\begingroup$ @Quaerendo Yes, it’s for general even potentials. $\endgroup$
    – G. Smith
    Aug 29, 2020 at 16:12

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