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I have a question from Perkins:

I'm not sure I understand the line in boldface. If [itex]s_\pi = 1[/itex] then why is [itex]S_z = 0[/itex]? For neutral pions, the existence of the decay

[tex]\pi^{0} \longrightarrow 2\gamma[/tex]

proves that the pion spin must be integral (since Sy = 1) and that [itex]s_\pi \neq 1[/itex], from the following argument. It can be proved as a consequence of relativistic invariance that for any massless particle of spin s, there are only two possible spin substates, [itex]S_z = \pm s[/itex], where z is the direction of motion. Taking the common line of flight of the photons in the pion rest frame as the quantisation axis, the z-component of total photon spin in the above decay can thus have the values Sz = 0 or 2.Suppose [itex]s_\pi = 1[/itex]; then only [itex]S_z = 0[/itex] is possible, and the two-photon amplitude must behave under rotation like the polynomial [itex]P_{l}^{m}(\cos \theta)[/itex] with m = 0, where [itex]\theta[/itex] is the angle of the photon relative to the z-axis.

Thanks!

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# Spin of the pion

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