Parity in the \eta to two photon decay

Click For Summary
SUMMARY

The discussion centers on the conservation of parity in the decay process \(\eta \rightarrow \gamma \gamma\). The \(\eta\) meson has odd parity, while the two photons produced have an even parity when considered collectively. The key insight is that the parity does not add linearly due to the superposition of photon states, which requires the application of Clebsch-Gordan coefficients to analyze the combined system. The presence of orbital angular momentum \(l=1\) introduces a factor of \((-1)^l\), which is crucial for understanding the overall parity of the decay process.

PREREQUISITES
  • Understanding of quantum mechanics and particle physics
  • Familiarity with parity conservation laws in electromagnetic interactions
  • Knowledge of Clebsch-Gordan coefficients for angular momentum coupling
  • Basic concepts of photon polarization and angular momentum addition
NEXT STEPS
  • Study the application of Clebsch-Gordan coefficients in quantum mechanics
  • Research the implications of parity violation in particle decays
  • Explore the role of angular momentum in photon interactions
  • Examine historical experiments on \(\eta\) and \(\pi^0\) decays, particularly those by Yang
USEFUL FOR

Physicists, particularly those specializing in particle physics and quantum mechanics, as well as students seeking to deepen their understanding of parity conservation in particle decays.

fliptomato
Messages
78
Reaction score
0
Greetings, I'm curious about parity conservation in the decay[tex]\eta \rightarrow \gamma \gamma[/tex]. The [tex]\eta[/tex] has odd parity, while the product of the two photon parities (each is odd) is even. Now, parity is conserved in the EM interactions, so there must be a factor of (-1) coming in from orbital angular momentum factors--but the two photon final state has no orbital angular momentum. What am I missing here?
 
Physics news on Phys.org
IN SHORT : PARITY DOES NOT ADD UP LINEARLY...

I mean : in order for the two-photon state to have J = 0 (conservation of J we must write it as a superposition of two states A and B. Each state has the two photons with anti-parallel spins and in state A the photonspin is aligned with the photon-momentum, in B photon spin is opposite wrt photon momentum. You can derive these states by using the Clebsch-Gordan coefficients.

A photon has indeed parity -1 but since we are working with a superposition of TWO photonstates, the parity is relative. If you had just a single two-photon state then parity would be -1 * -1 but because of the superposition, it is the parity from state A with respect to state B that determins the actual parity of the entire wavefunction...

regards
marlon

More to be found here
 
fliptomato said:
Greetings, I'm curious about parity conservation in the decay[tex]\eta \rightarrow \gamma \gamma[/tex]. The [tex]\eta[/tex] has odd parity, while the product of the two photon parities (each is odd) is even. Now, parity is conserved in the EM interactions, so there must be a factor of (-1) coming in from orbital angular momentum factors--but the two photon final state has no orbital angular momentum. What am I missing here?

In \eta-->2\gamma, there is orbital angular momentum l=1, which gives the factor (-1)^l. The angular momentum addition for the photon spins must be S=1+1=1, and then l+S=1+1=0.
The eta decay is the same as the pi^0 decay. The parity of the pi^0, and of the eta^0, were determined by the relative plane polarizations of the two photons, which can be found from the spin addition 1+1=1.
This was first done by Yang.
 

Similar threads

Replies
4
Views
2K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 10 ·
Replies
10
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
2
Views
2K
  • · Replies 13 ·
Replies
13
Views
8K
  • · Replies 6 ·
Replies
6
Views
5K