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Decay width average in the isospin invariant limit

+ 3 like - 0 dislike
26 views

Suppose we have the following experimental values for $\eta' \rightarrow \eta \pi \pi$ decay width:

$\Gamma_{\eta' \rightarrow \eta \pi^+ \pi^-} = 0.086 \pm 0.004$

$\Gamma_{\eta' \rightarrow \eta \pi^0 \pi^0} = 0.0430 \pm 0.0022$

We are investigating this decay in the isospin invariant limit and want to compare our results with experimental rates. So we should average over these 2 values to find the experimental decay width in this limit. It is written in one paper that we should average

$\Gamma_{\eta' \rightarrow \eta \pi^+ \pi^-} = 0.086 \pm 0.004$

2$\Gamma_{\eta' \rightarrow \eta \pi^0 \pi^0} = 0.0860 \pm 0.0044$

in a specific way (which is not my question). I don't know why we should twice the second value and then average?

This post imported from StackExchange Physics at 2014-05-04 11:26 (UCT), posted by SE-user soodeh
asked Apr 29, 2014 in Theoretical Physics by soodeh (15 points) [ no revision ]
Could you please provide a reference regarding the paper you mention?

This post imported from StackExchange Physics at 2014-05-04 11:26 (UCT), posted by SE-user Frederic Brünner
Yes, of course. Phys. Rev. D60, 034002 (page 3)

This post imported from StackExchange Physics at 2014-05-04 11:26 (UCT), posted by SE-user soodeh

1 Answer

+ 2 like - 0 dislike

The average is taken this way because their theoretical prediction does not distinguish between $\Gamma_{\eta' \rightarrow \eta \pi^+ \pi^-}$ and $2\Gamma_{\eta' \rightarrow \eta \pi^0 \pi^0}.$ See for example equation 2.1. The experimental value, however, is different for both decay channels. Therefore, in order to make a meaningful comparison, they choose to compare to the average of both values. Since the theoretical equivalence include the factor of $2$, the average should also include it.

This post imported from StackExchange Physics at 2014-05-04 11:27 (UCT), posted by SE-user Frederic Brünner
answered Apr 29, 2014 by Frederic Brünner (1,060 points) [ no revision ]
thank you so much.

This post imported from StackExchange Physics at 2014-05-04 11:27 (UCT), posted by SE-user soodeh

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