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I was hoping someone could clarify why the following decay does not occur:

$ \eta ^{'0} \rightarrow \pi ^{0} + \rho ^{0}$

The quark compositions and spin parity are as followed:

$ \eta ^{'0} : (u\bar{u}+d\bar{d}+s\bar{s}) / \sqrt{3} ;J^{P} = 0^{-} $

$ \pi ^{0} : (u\bar{u}-d\bar{d}) / \sqrt{2} ;J^{P} = 0^{-} $

$ \rho ^{0} : (u\bar{u}-d\bar{d}) / \sqrt{2} ;J^{P} = 1^{-} $

In order to conserve parity and angular momentum I thought that the two final particles states would have to be produced with angular momentum $l = 1$ between them (as parity of angular momentum 'part' is $ (-1)^{l}$ this would conserve parity and we can couple 0,1 and 1 to give 0 which conserves angular momentum). Does anyone know what is wrong this approach or alternatively a more straight forward reason why this does not occur.

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  • $\begingroup$ did you check for Isospin conservation? hyperphysics.phy-astr.gsu.edu/hbase/particles/parint.html $\endgroup$
    – anna v
    Commented May 22, 2015 at 19:34
  • $\begingroup$ In the question this is all I actually get so I have not looked at it - is there a way to determine if it occurs or not just given the information given? But thanks I will also have a look into that. $\endgroup$
    – msd27
    Commented May 22, 2015 at 19:44
  • $\begingroup$ There is G parity conservation as a generalization of the statement by Paganini en.wikipedia.org/wiki/G-parity $\endgroup$
    – anna v
    Commented May 23, 2015 at 4:41

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This decay (occurring via the strong interaction) violates the charge conjugation since $J^{PC}(\pi^0) = 0^{-+}, J^{PC}(\rho^0) = 1^{--}, J^{PC}(\eta'^0) = 0^{-+}$.

The charge conjugation transforms a particle in its anti-particle. In the case of the 3 particles involved in this decay, they are all their own anti-particle, and the effect of the charge conjugation operator $C$ is therefore (taking as an example the pion) $C|\pi^0> = \eta_C |\pi^0>$, meaning that the $\pi^0$ is eigenstate of the charge conjugation with eigenvalue $\eta_C = +1$. The $\rho^0$ has $\eta_C=-1$ and the $\eta'^0$, +1 (remark: $\eta_C$ is necessarily $\pm 1$ because when you apply twice the charge conjugation you should recover the initial state). The requirement of the charge conjugation conservation by the strong interaction would imposes: $\eta_C(\eta'^0) = \eta_C(\pi^0) \times \eta_C(\rho^0)$ which is not the case $+1 \ne (+1) \times (-1)$. Thus this reaction is forbidden.

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  • $\begingroup$ I am not familiar with charge conjugation - could you expand slightly? $\endgroup$
    – msd27
    Commented May 22, 2015 at 18:38
  • $\begingroup$ I have completed my answer to (try to) better explain. $\endgroup$
    – Paganini
    Commented May 22, 2015 at 20:56
  • $\begingroup$ Thanks that is really helpful, but why are the $\eta _{c}$ values for the pion, rho and eta, 1, -1 and 1 respectively? Is this just a property of the particles or can this be inferred from the information given? $\endgroup$
    – msd27
    Commented May 23, 2015 at 8:58
  • $\begingroup$ They are determined experimentally. The decay $\pi^0 \to \gamma \gamma$ is an electromagnetic decay (which respects as well the charge conjugation). The $\gamma$ is its own antiparticle and so eigenstate of $C$. Hence, we have $\eta_C(\pi^0) = \eta^2_C(\gamma) = 1$. Similarly $\eta' \to \gamma \gamma$ is seen and so $\eta_C(\eta') = 1$. Finally, $\eta' \to \rho^0 \gamma$ is observed, implying $1 = \eta_C(\eta') = \eta_C(\rho^0) \eta_C(\gamma) \rightarrow \eta_C(\rho^0) = \eta_C(\gamma)$. From the properties from the electromagnetic field, it can be shown that $\eta_C(\gamma)=-1$. $\endgroup$
    – Paganini
    Commented May 23, 2015 at 15:30
  • $\begingroup$ For the electromagnetic field $A^\mu$, since an E.M. interaction couples a current with $A^\mu$, a term $j_\mu A^\mu$, under charge conjugation the current reverts its sign (i.e. an electron becomes positron) and so the term remains invariant if $A^\mu$ becomes $-A^\mu$. Thus, the sign of $\eta_C(\gamma)=-1$. $\endgroup$
    – Paganini
    Commented May 23, 2015 at 15:52

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