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 $$\mathcal{X} = \mathbb{R}\mathrm{P}^2 \simeq \mathrm{SO}(3)/\mathrm{O}(2)~?$$

  • 1
    $\begingroup$ $RP^2$ 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. $\frac{1}{\sqrt{2}}(|l, m\rangle+|l,-m\rangle)$. $\endgroup$
    – Meng Cheng
    Feb 8, 2022 at 17:00
  • $\begingroup$ Hi Eric Kubischta. Welcome to Phys.SE. I removed your last subquestion, which seemed too broad. $\endgroup$
    – Qmechanic
    Feb 8, 2022 at 17:02
  • $\begingroup$ @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 $\{ |\ell,m\rangle : \ell \in 2 \mathbb{Z}, m \in \mathbb{Z}, \ell \geq 0, -\ell \leq m \leq \ell\}$. $\endgroup$ Feb 11, 2022 at 15:38
  • 1
    $\begingroup$ @EricKubischta You are right, I confused inversion with time-reversal. Then indeed only the even $l$ are kept. $\endgroup$
    – Meng Cheng
    Feb 11, 2022 at 17:08


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