# Parity of the eigenfunctions of Hamiltonian for a symmetrical potential

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 $$V(x)=kx^4$$?

• How is the quantum number $n$ assigned? Commented Aug 28, 2020 at 19:43
• Let $\psi_0$ be the ground state, $\psi_1$ the first excited state, and so on. Commented Aug 28, 2020 at 19:49
• 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. Commented Aug 28, 2020 at 20:13
• Thanks, I have corrected the errata in the question Commented Aug 31, 2020 at 19:56

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.
• 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? Commented Aug 28, 2020 at 19:57
• Ok, I just wanted to make sure that your answer was for general even potentials, not only for the example of $V(x)=kx^4$ Commented Aug 29, 2020 at 8:14