The discussion centers on the decay of neutral pions, represented by the equation \(\pi^{0} \longrightarrow 2\gamma\), which confirms that the pion has integral spin. It is established that the spin of the pion, \(s_\pi\), cannot equal 1 due to the constraints of relativistic invariance and conservation of angular momentum. The argument presented indicates that if \(s_\pi = 1\), the only possible z-component of total photon spin, \(S_z\), would be 0, contradicting the requirement for angular momentum conservation in the decay process.
PREREQUISITESPhysicists, students of particle physics, and anyone interested in the fundamental properties of particles and their interactions, particularly in understanding spin and decay processes.
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.
xepma said:Conservation of angular momentum. two is larger than one!