Momenta basis for a quantum system with configuration space $\mathbb{R}\mathrm{P}^2$

Let $$\mathcal{X}$$ be the configuration space of a (quantum) system.

• When $$\mathcal{X} = S^1 \simeq \mathrm{SO}(2)$$, a momenta basis is $$\{ |\ell\rangle : \ell \in \mathbb{Z} \}.$$

• When $$\mathcal{X} = \mathrm{SO}(3)$$, a momenta basis is $$\{ |\ell,m,n\rangle : \ell,m,n \in \mathbb{Z}, \ell \geq 0, -\ell \leq m,n \leq \ell \}.$$

• When $$\mathcal{X} = S^2 \simeq \mathrm{SO}(3)/\mathrm{SO}(2)$$, a momenta basis is $$\{ | \ell,m\rangle: \ell,m \in \mathbb{Z}, \ell \geq 0, -\ell \leq m \leq \ell \}.$$

What is a momenta basis when $$\mathcal{X} = \mathbb{R}\mathrm{P}^2 \simeq \mathrm{SO}(3)/\mathrm{O}(2)~?$$

• $RP^2$ is $S^2$ with antipodal points identified. So I would think you can just take the basis for $S^2$, and form states that are invariant under inversion e.g. $\frac{1}{\sqrt{2}}(|l, m\rangle+|l,-m\rangle)$. Feb 8, 2022 at 17:00
• Hi Eric Kubischta. Welcome to Phys.SE. I removed your last subquestion, which seemed too broad. Feb 8, 2022 at 17:02
• @MengCheng Are you talking about the time reversal operator? Doesn't the parity operator act like $\pi | \ell,m\rangle = (-1)^\ell |\ell,m \rangle$? In which case we would throw away all of the odd $\ell$'s and the momentum basis would be $\{ |\ell,m\rangle : \ell \in 2 \mathbb{Z}, m \in \mathbb{Z}, \ell \geq 0, -\ell \leq m \leq \ell\}$. Feb 11, 2022 at 15:38
• @EricKubischta You are right, I confused inversion with time-reversal. Then indeed only the even $l$ are kept. Feb 11, 2022 at 17:08