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The 3 leptons ($e,\mu,\tau$) of QED can form bound states. The positronium $e e^+$ can have spin 0 (para positronium) or spin 1 (ortho positronium). There is also muonium $e \mu^+$, true muonium $\mu \mu^+$, and similarly other bound states with $\tau$ as constituent.

What are the parities of each of these bound states? The spins and charge conjugation properties are rather clear, so in the $J^{PC}$ notation $J$ and $C$ are straightforward, but what about $P$?

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  • $\begingroup$ Same game. All elementary fermions have an intrinsic parity +1 while antifermions have -1. Then the spin and L combinations follow suit. Blind yourself to lepton flavor. $\endgroup$
    – Cosmas Zachos
    Commented Jan 7, 2022 at 18:36
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    $\begingroup$ Follow the rules in your text. P=- for the singlet para, and + for the triplet ortho. Look at the photon decays. $\endgroup$
    – Cosmas Zachos
    Commented Jan 7, 2022 at 19:26
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    $\begingroup$ Right. Sorry about the ortho.. it’s also -. $\endgroup$
    – Cosmas Zachos
    Commented Jan 7, 2022 at 19:44
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    $\begingroup$ Yes for the para, but the spin enters for the ortho, so $1^{-~-}$. $\endgroup$
    – Cosmas Zachos
    Commented Jan 7, 2022 at 19:51
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    $\begingroup$ Because of the standard formula $C=(-)^{L+S}$ your text should have derived. Ortho has S=1, L=0. $\endgroup$
    – Cosmas Zachos
    Commented Jan 7, 2022 at 20:02

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From comments of Cosmas Zachos:

$$C = (-1)^{L+S} \qquad \qquad P = (-1)^{L+1}$$

So the para positronium $(S=0, L=0)$ has $J^{PC} = 0^{-+}$ and the ortho positronium $(S=1, L=0)$ has $J^{PC} = 1^{--}$.

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