# Polarizations with respect to the scattering plane

In one problem in a QFT book the author asks to compute the cross section for a Compton scattering scalar QED process $\gamma\phi\to \gamma\phi$ with the incoming photon polarization $\epsilon^{\text{in}}_\mu$ polarized in the plane of scattering and polarized transverse to the plane of scattering.

Now I don't really know what he is asking here because he didn't use this terminology before.

He introduced the polarization vectors as part of the solution to the equations of motion in Fourier space.

In the end he derived a basis for polarization that is

$$\epsilon^1_\mu=(0,1,0,0),\quad\epsilon_\mu^2=(0,0,1,0)$$

so that a general polarization is $\epsilon_\mu = \sum c_j \epsilon_\mu^j$.

These are fixed to a plane, in particular, the plane $x,y$. I can't make this be sitting on the plane of scattering no matter how I choose the coefficients.

Furthermore, what is this plane of scattering by the way? How is it defined? Because if we use the refrence frame of the center of mass and set the axes so that the photon and electron are on the $z$ axis, the scattering would occur in any plane that contains the $z$-axis right?

So: what is this scattering plane, and what is the mathematical way to use the fact that $\epsilon_\mu^\text{in}$ is polarized inthe plane of scattering or transverse to it?

• I believe that the scattering plane is the plane containing both the incident vector and the scattered vector. – Floris Jun 5 '17 at 12:35

The easiest way to think about the scattering surface is to work in the Lab frame, where you have a particle that you scatter from a fixed target. In this frame, the scattering surface is the $2D$ surface that contains both the incoming and outgoing particle trajectories.

The polarization vectors represent the fact that photons only have two degrees of freedom (being massless spin 1 fields) and can thus be written as a superposition of any two basis vectors. Restricting polarization to the $x,y$ plane means we can assume that the photon is travelling in the $z$ direction, since polarization is always perpendicular to the direction of motion.

After thinking about the issue I believe to have found the point. We consider the scattering process in the center of mass reference frame.

Now we define the axes of the frame. The particles are moving towards each other along a line. We take this line as defining the $z$ axis and complete one orthonormal cartesian system with other axes $x'$,$y'$.

Thus we have necessarily $p_1^\mu = (E_1,0,0,k)$ and $p_2^\mu = (E_2,0,0,-k)$ the components of their momenta with respect to these axes in the center of mass frame. After the scattering, i.e., in the outgoing states, the momenta change to $p_3^\mu = (E_3,\mathbf{p})$ and $p_4^\mu = (E_4,-\mathbf{p})$.

Now together the pair $(\mathbf{p},\mathbf{e}_3$) spans a plane. This is the scattering plane. By virtue of a rotation around the $z$ axis, we can rotate the $x',y'$ axes to get new $x,y$ axes orthogonal to $z$, so that $\mathbf{p}$ lies in the $zy$ plane. With this done, we have one orthonormal basis $e_\mu$ such that

$$p_1^\mu=(E_1,0,0,k),\quad p_2^\mu=(E_2,0,0,-k)$$

$$p_3^\mu=(E_3,0,p\sin\theta,p\cos\theta),\quad p_4^\mu=(E_4,0,-p\sin\theta,-p\cos\theta)$$

where we defined $\theta$ the angle between $\mathbf{e}_3$ and $\mathbf{p}$.

For this case we suppose particle $1$ is an incoming photon. Since its momentum is $p_1^\mu = (E_1,0,0,k)$ with respect to the basis we chose, the two polarizations are:

$$\epsilon_1=(0,1,0,0),\quad \epsilon_2=(0,0,1,0)$$

as they satisfy the required condition $p_1^\mu \epsilon_{i,\mu}=0$.

We then observe that: $\epsilon_1$ is orthogonal to the scattering plane which is the $yz$ plane. If the photon is polarized like this it is polarized transverse to the scattering plane. In the same way, $\epsilon_2$ is obviously contained in the scattering plane, so if the photon is polarized like this it is polarized in the scattering plane.