The decay $\omega \rightarrow e^+e^-$ has a partial width that is around 10 times smaller than the partial width for the decay $\rho \rightarrow e^+e^-$. Why is this the case?

The only difference I have found between the $\omega$ and $\rho^0$ mesons is that $\omega$ is an isospin singlet and $\rho^0$ is an isospin triplet. I can't see how this should have any effect on the partial widths for electron positron decay. The context of the question is helicity surpression and the Wu experiment if that helps.

My second question is that I don't understand how both an isospin singlet (the $\omega$) and isospin triplet (the $\rho^0$) can exist with otherwise identical quantum numbers - surely this would imply that one of the $\omega$ or the $\rho^0$ must have even exchange symmetry overall, violating the pauli exclusion principle.


1 Answer 1


You might profit from this old talk by Nichitiu. The "theoretical" ratio of coupling widths to the photon is, in fact, 9 to 1.

The respective flavor wavefunctions of light vector mesons are, schematically, skipping γs, color, spin, etc... $$ \rho^0\sim \frac{\bar u u -\bar dd}{\sqrt{2}}; ~~\omega\sim \frac{\bar u u +\bar dd}{\sqrt{2}};~~ \phi \sim \bar s s. $$ The VDM EM couplings to the photon, then, are the overlaps of the EM current with the above wavefunctions, respectively: effective charges. Squared for the width, they then lead to the ratios $$ \Gamma_{ee} (\rho)~: \Gamma_{ee} (\omega)~:\Gamma_{ee} (\phi) \\ = \left ({2/3+1/3\over\sqrt{2}}\right )^2~: \left ({2/3-1/3\over\sqrt{2}}\right )^2~: (-1/3)^2 \\ = 9:1:2 ~. $$

I'm not sure what your second question might mean: the point of the generalized Pauli principle is that if you wrote a wave-function with the wrong (anti)symmetry, it would vanish. But you see manifestly and explicitly that the two wavefunctions above, contrasting the isovector to the isosinglet, need not vanish... how could they? (The analogous thinking would, by contrast lead to zeros in baryon wavefunctions.)

  • 2
    $\begingroup$ Thank you for this, it puzzled me. $\endgroup$
    – anna v
    Dec 16, 2021 at 5:08
  • $\begingroup$ In what sense does a wavefunction with the wrong exchange symmetry vanish? $\endgroup$ Dec 16, 2021 at 9:15
  • $\begingroup$ In the absence of color, spin, etc, uu vanishes but $\bar u u$ not, for fermions. $\endgroup$ Dec 16, 2021 at 11:40

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