Spin of the Pion: Neutral Decay Proves Integral Spin

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In summary, the conversation discusses the existence of the decay \pi^{0} \longrightarrow 2\gamma for neutral pions and how it proves that the pion spin must be integral and s_\pi \neq 1. The argument is based on relativistic invariance and shows that for a massless particle of spin s, there are only two possible spin substates, S_z = \pm s. The conversation also addresses the question of why S_z = 0 when s_\pi = 1, and the answer is that it conserves angular momentum, as two is larger than one. However, further clarification is needed to fully understand how Sz = 0 explicitly conserves angular momentum.
  • #1
maverick280857
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Hi,

I have a question from Perkins:

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.

I'm not sure I understand the line in boldface. If [itex]s_\pi = 1[/itex] then why is [itex]S_z = 0[/itex]?

Thanks!
 
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  • #2
Conservation of angular momentum. two is larger than one!
 
  • #3
xepma said:
Conservation of angular momentum. two is larger than one!

Yes. Right, but does Sz = 0 explicitly conserve angular momentum? How do I convince myself?
 

Related to Spin of the Pion: Neutral Decay Proves Integral Spin

1. What is the spin of a pion?

The spin of a pion is a quantum mechanical property that describes its intrinsic angular momentum. It is a fundamental property of subatomic particles and is measured in units of Planck's constant (h-bar).

2. How is the spin of a pion determined?

The spin of a pion can be determined through experiments, such as measuring its decay products. In the case of neutral pions, their spin can be determined by analyzing their decay into two photons, which is only possible for particles with integral spin.

3. What does it mean for a particle to have integral spin?

A particle with integral spin means that its spin value is a whole number, such as 0, 1, 2, etc. This is in contrast to particles with half-integral spin, which have spin values of 1/2, 3/2, 5/2, etc. Integral spin particles follow different rules and behaviors compared to half-integral spin particles.

4. How does neutral decay of pions prove integral spin?

The neutral decay of pions into two photons is only possible if the pion has integral spin. This is because the conservation of angular momentum requires that the total spin of the decay products must be equal to the spin of the parent particle. Since photons have a spin of 1, the spin of the pion must also be 1 in order for the decay to be possible.

5. Why is the spin of a pion important to study?

The spin of a pion, and other subatomic particles, provides important information about their fundamental properties and interactions. It is also a key factor in understanding the structure of matter and the fundamental forces of the universe. By studying the spin of pions, scientists can gain a better understanding of the underlying principles of the universe.

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