Orbital parity of simple bound states in atomic and particle physics

Question

The parity operator commutes with the Hydrogen atom Hamiltonian. The energy eigenfunctions are parity eigenstates with orbital parity $(-1)^\ell$ which follows from the fact that $Y_{\ell m}(\theta,\phi)$ is an eigenstate of parity with parity eigenvalue $(-1)^\ell$.

Question 1 What about the Helium (He) atom? The Hamiltonian of the He atom also commutes with parity but is not central. But I am not sure whether the energy eigenfuctions are still given by a product of a radial and an angular part with the angular part given by $Y_{\ell m}(\theta,\phi)$. So I cannot decide whether or not the He atom energy eigenstates also have the orbital parity $(-1)^\ell$?

Question 2 What about a meson $\bar{q}_1q_2$ i.e., a bound state of a quark and an antiquark? The strong interaction Hamiltonian commutes with parity. Mesons being strong interaction eigenstates (in absencce of weak interaction contamination) should also have definite parity. Can we say that mesons also carry an orbital parity $(-1)^\ell$? Apart from orbital parity there exists contribution from intrinsic parity which is not my concern here.

Answer 1

Basically, yes to both.

When you solve the Schroedinger equation you use separation of variables, getting two equations, one for $r$ and one for $(\theta,\phi)$. The first involves the potential $V(r)$ but the second does not. So you end up with the $Y_{\ell m}(\theta, \phi)$ spherical harmonics whatever the form of $V(r)$: Coulomb (as in hydrogen), screened Coulomb (as in Helium) or confining QCD (as in mesons). Provided the potential depends only on the distance and not on $\theta$ or $\phi$, it's always the same story as far as the angular dependence is concerned.

So for example the $a_1$ meson is a $u$ or $d$ quark and $\overline u$ or $\overline d$ antiquark, just like a $\pi$, but orbiting in an $\ell=1$ state and with their spins aligned. The total angular momentum $J$ is also 1 (as 1+1= 0, 1 or 2). The overall parity is +1,(in contrast to the $\pi$, for which it is$-1$) as the $-1$ from $-1^\ell$ multiplies the intrinsic parities, which are opposite for $q$ and $\overline q$.

s,p,d,f... are just code for $\ell=0,1,2,3... $. They stand for sharp, principal, diffuse and fine in old spectroscopists' notation.

Answer 2

In the central field approximation, the Hamiltonian of a many-electron atom is assumed to comprise of a large spherically symmetric part and a small spherically asymmetric component. Since the asymmetric part is small, it can be treated as a perturbation. The total angular momentum operator $\textbf{L}=\sum\limits_{i}\textbf{L}_i$ commutes with the spherically symmetric component $H_{\rm central}$ of the Hamiltonian. Same for the total spin $\textbf{L}=\sum\limits_{i}\textbf{S}_i$. Hence, it's possible to find simultaneous eigenfunctions of $$H_{\rm central},\textbf{L}^2,\textbf{S}^2,L_z,S_z$$ labelled as $$|\gamma LSM_LM_S\rangle$$ where $\gamma$ is additional quantum numbers and rest of them the usual meanings. Since $H_{\rm central}$ is the dominant component of the many-electron Hamiltonian, the total orbital angular momentum of any approximate eigenstate $|\gamma LSM_LM_S\rangle$ will have orbital parity $(-1)^L$ where $L$ can have values from $|\ell_1-\ell_2|$ to $(\ell_1+\ell_2)$ in steps of unity for the He atom.