maverick280857
Feb5-10, 06:41 AM
Hi,
I have a question from Perkins:
For neutral pions, the existence of the decay
\pi^{0} \longrightarrow 2\gamma
proves that the pion spin must be integral (since Sy = 1) and that s_\pi \neq 1, 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, S_z = \pm s, 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 s_\pi = 1; then only S_z = 0 is possible, and the two-photon amplitude must behave under rotation like the polynomial P_{l}^{m}(\cos \theta) with m = 0, where \theta is the angle of the photon relative to the z-axis.
I'm not sure I understand the line in boldface. If s_\pi = 1 then why is S_z = 0?
Thanks!
I have a question from Perkins:
For neutral pions, the existence of the decay
\pi^{0} \longrightarrow 2\gamma
proves that the pion spin must be integral (since Sy = 1) and that s_\pi \neq 1, 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, S_z = \pm s, 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 s_\pi = 1; then only S_z = 0 is possible, and the two-photon amplitude must behave under rotation like the polynomial P_{l}^{m}(\cos \theta) with m = 0, where \theta is the angle of the photon relative to the z-axis.
I'm not sure I understand the line in boldface. If s_\pi = 1 then why is S_z = 0?
Thanks!