# Photon Polarizations in Positronium Decay

1. Jun 4, 2015

### gobbles

1. The problem statement, all variables and given/known data
I am trying to solve problem 5.4(a) in Peskin & Schroeder. I am requested to calculate the decay rate of a $1S_0$ positronium state into two photons. Obviously, we have to sum over all photon polarizations eventually to get the total decay rate. However, I do not understand why the summation is only over circular polarizations, according to the solution given here
http://learn.tsinghua.edu.cn:8080/2010310800/Webpage/files/Peskin_Schroeder/Peskin_Chap05.pdf.

2. Relevant equations
For two outgoing photons with polarizations
$\epsilon_\nu^*(k)=(0, \alpha, \beta, 0), \epsilon_\mu^*(k^\prime)=(0, \alpha^\prime, \beta^\prime, 0),$
where $k,k^\prime$ are the 4-momenta of the photons, along $\hat{z}$ and $-\hat{z}$, respectively, I got the following expression for the amplitude of decay of $1S_0$ positronium into two photons:
$\mathcal{M}=\sqrt{2}2ie^2\big[\beta^\prime\alpha-\beta\alpha^\prime\big].$

3. The attempt at a solution
Obviously, if both photons are right circularly polarized or if both are left circularly polarized, this expression is not zero. But it's not zero also for, say,
$\epsilon_\mu^*(k^\prime)=(0, 1, 0, 0)$ and $\epsilon_\nu^*(k)=(0, 0, 1, 0)$. Why don't we sum over that polarization also? Am I missing something? Can a photon be anything but circularly polarized?

2. Jun 4, 2015

### Orodruin

Staff Emeritus
What else do you want to sum over? There are only two linearly independent polarisation states, which you can write either in the basis of circular polarisation or in the basis of linear polarisation. Which you use does not matter and is only a question of convenience.

3. Jun 4, 2015

### fzero

There's a physical reason why a single photon will always be measured in a circular polarization eigenstate. Once we fix the photon momentum to be in the $z$=direction, $k^\mu=(k,0,0,k)$, then the states are labeled by the eigenstates of the component of spin that leaves the momentum invariant: $S_z = \vec{S}\cdot \hat{k}$. In general these transformations are called the "little group" as a subgroup of the Lorentz group and the corresponding eigenvalue is called the helicity. In this 4-vector representation, we can take

$$S_z=\begin{pmatrix} 0 & 0 & 0 & 0 \\ 0 & 0 &-i & 0 \\ 0 & i & 0 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix},$$

and can verify that the circular polarizations are the eigenstates, not the linear ones. We can form a linear polarization by a superposition of circular polarizations, but a measurement of the helicity will always find the photon in a circular polarization state.

4. Jun 4, 2015

### Orodruin

Staff Emeritus
These photons are not necessarily being measured (if our measuring device was only sensitive to one helicity we would not sum over the states!). For the purposes of this problem, it does not matter if we use the circular or linear polarisation basis (as long as we sum over all states), the sum is the same.

5. Jun 4, 2015

### gobbles

I still do not understand. The final photons can be in various polarization states. For example, they can be right or left circularly polarized, they can be linearly polarized along the x or y axis or any other vector in the xy plane. So, to get the total decay rate, as I understand it, we would need to sum over the two linearly polarized photons along the x axis, along the y axis, along any other vector in the xy plane, along right and left circularly polarized photons, etc and that would give the final result. That is, we need to sum over all possible "kinds" of resulting photons. In the solution to this problem, the summation is only over the two basis vectors of polarizations. Maybe I don't quite understand what polarization means in the quantum sense, because in my head I treat it classically; a classical EM wave can be polarized along any direction.

6. Jun 4, 2015

### Orodruin

Staff Emeritus
Even classically, linearly polarised light is just a linear combination of circular polarised light (and vice versa). You can write any linearly polarised EM wave as a linear combination of circular polarised waves. This does not change in quantum mechanics. It is sufficient with two polarisation states to span the entire polarisation state space.

7. Jun 4, 2015

### vela

Staff Emeritus
You don't want to sum over all possible types of photons; you want to sum over all possible linearly independent types of photons. In other words, you want to sum over some basis.

As Orodruin noted, a linearly polarized wave is a linear combination of circularly polarized waves. If you summed over both linear and circular polarizations, you would be double counting. Whether you choose linear polarizations or circular polarizations is up to you, but you have to choose one or the other but not both.

8. Jun 5, 2015

### gobbles

Thank you, I think I understand now. The polarization vector $\epsilon_\mu$ of the Feynman rules of QED originates from the expansion of the vector potential $A_\mu$ into creation and annihilation operators, where the vectors that appear are the basis vectors of the polarization.

9. Jun 5, 2015

### Orodruin

Staff Emeritus
... and there are two linearly independent polarisations. Which ones you use is irrelevant as long as your selection is orthonormal.