# Pi-meson question

1. Mar 25, 2005

### kakarukeys

I have a question about the neutral Pi-meson.

One of the decay modes of the neutral pi-meson is to 2 photons.

$$\pi^0 \rightarrow \gamma + \gamma$$

it seems to me that this electromagnetic process violates parity conservation.

L.H.S. parity is -1.
R.H.S. parity is (-1)(-1) = +1

Is there anything wrong?

2. Mar 26, 2005

### marlon

No,
The answer is situated in the photon spin and their direction with respect to the photon momentum : check out the bottom of : http://csm.jmu.edu/physics/giovanetti/particlePhysics/parity.htm [Broken]

Parity IS conserved

regards
marlon

Last edited by a moderator: May 2, 2017
3. Mar 26, 2005

### marlon

Clue of this story is the fact that in order for the two-photon state to have J = 0 (conservation of J and J=0 for the neutral pi meson because it is a superposition of two quark anti-quark states) 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, as is denoted in the text...

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

Last edited: Mar 26, 2005
4. Mar 26, 2005

### dextercioby

Could one of you please come up with the Feynman diagram...?I have a hunch this process typical SM-type (that is electroweak+quarks).

Daniel.

5. Mar 26, 2005

### marlon

The Feynman diagram is one page four

http://www.oup.co.uk/pdf/0-19-852913-9.pdf [Broken]

marlon

Last edited by a moderator: May 2, 2017
6. Mar 26, 2005

### dextercioby

Thanks,Marlon.I could have sworn it was a weak process,but it isn't...

Daniel.

7. Mar 26, 2005

### marlon

So did I...

As a matter of fact the OP asked a very good question and it took me a while to figure it out and find some good resources...

regards
marlon

8. Mar 26, 2005

### dextercioby

It definitely took you less than it took me to come up with all sorts of Z,W+-,quarks & photons diagrams...

Daniel.

9. Mar 26, 2005

### juvenal

It is a good question. I took a look at Sakurai's Invariance Principles and Elementary Particles, p. 44-45. The intrinsic parity of the photon is based on the behavior of the polarization vector under parity, and is negative. For the pi0 decay, he says:

"We consider the decay of a spin zero pion at rest into two photons. The final-state wave function must transform like a J=0 system, and it must be linear in each of the two photon polarization vectors. We have two possibilities (even and odd parity):
$$\vec{\epsilon_1} \cdot \vec{\epsilon_2}$$

or

$$(\vec{\epsilon_1} \times \vec{\epsilon_2} )\cdot k$$

(k is the unit relative momentum vector of the two-photon system). The first expression is seen to be invariant under P while the second changes sign. In general the final state is expected to be a linear combination of these two forms. If parity is conserved in the decay, however, only one of the two forms is allowed, and we could determine, by measuring the relative orientations of the polarization vectors whether the pi0 is described by a scalar or pseudoscalar field."

10. Mar 26, 2005

### marlon

Yes, juvenal, and the two formula's that you gave are the only possibilties because they are invariant under the transformations generated by J : ie the rotations

We could also take the tensorproduct of k with one of the polarzation tensors but this option is useless because the longitudinal mode of photons vanishes... only the transverse (two) photon-polarizations remain. In QFT, you can kick out one degree of freedom by imposing a gauge-condition for the A field $$\partial^{\mu} A_{\mu} = 0$$

The A field (which is a Lorentz-object with four degrees of freedom) now has three remaining degrees of freedom.

This is very good because spin 1 means that this object will transform as a vector under the 3-D rotation group. The fundamental representations of this group are (3*1)-column matrices with indeed three variables : ie the components.

The reason why there are 'initially', three degrees of freedom for the photon (ie the three polrizations) has to do with the fact that in QFT a photon is described as the excitations of a MASSIVE spin one-meson (denoted by a vectorfield) because this field needs to couple to the vectorcurrent J (this field is disturbed by the J-current which expresses nothing more then the charged particles that will make the spin 1 field vibrate).

The J-current is a Lorentz vector with 4 degrees of freedom (ie the components) but one degree is knocked out because of this equation : $$\partial_{\mu} J^{\mu} = 0$$ : ie conservation of charge.

This J-current now has three degrees of freedom. Then we let m evolve towards zero and 'hope' things will not get awkward...Well, they do...damn it...In order to solve these problems (the longitudinal polarization will yield a negative expectance value) and Gupta-Bleuler-theory teaches us how to get rid of this third variable : the longitudinal polarization : it will decople from the physics at hand.

marlon

Last edited: Mar 26, 2005
11. Mar 28, 2005

### dextercioby

Chapter 13,section 2 of Pokorski's book deals with chiral anomalies linked to [itex] \pi^{0}\rightarrow \gamma+\gamma [/tex] and a computation of the decay rate.

Daniel.