# Spin polarization 4-vector

1. Jan 4, 2008

### jdstokes

Can someone explain to me why the spin polarizations of a particle can be represented by the four unit 4-vectors, ie partial derivative vector fields with respect to each coordinate function?

I also do not understand why the probability of a particle to be created or absorbed with spin polarization in any given direction is proportional to the spin polarization vector in that direction. If spin polarization is a 4-vector, how can it be related to a probability (scalar)?

In Zee's textbook on QFT (p. 32), he claims that the total probability of a particle to be created and absorbed with momentum $k$ is

$\sum_a \varepsilon^{(a)}_\nu (k) \varepsilon_\lambda ^{(a)}(k)$.

Why does he only sum over the orthogonal polarization vector fields $a$?

Thanks.

2. May 17, 2010

### bobloblaw1

Hey jdstokes,

I too was confused by this comment in Zee and have put some thought into it.
In classical EM,ie Griffiths sec 9.2.2, it is shown how the polarization is a 3 vector and points along the direction of the electric field(by convention). Written in the language of SR the polarization must be represented by a 4 vector or some tensor. It turns out that it can indeed be represented by a four vector as follows.

First consider a plane wave solution to classical EM without sources. The vector potential can be written as $A_\mu = C_\mu e^{i k_\nu x^\nu}$. Then the field strength tensor can be written as $$F = \nabla \wedge A = k \wedge A = k \wedge C e^{i k_\nu x^\nu}$$. From this you can see that the four vector C is serving as a polarization vector since the F tensor is "orthogonal" to both k and C. If you write what the F tensor looks like in 3+1 notation you will see that the spatial component of C is essentially serving as the standard polarization 3 vector. Since polarization can be represented by a four vector the space of all possible polarizations therefore has a basis of four 4-vectors.

To answer the second part of your question about probability, firstly I must say that I think he is abusing the word probability a lot. What he is really is constructing is an operator built out of quantum fields. The calculation he is performing is somewhat like computing the propagator in momentum space. Recall for electrons the propagator is $$D_{ab}(x-y) =\left<0\right| \psi_a(x)\bar{\psi}_b(y)\left|0\right>$$. And just like this case we interpret this as the probability amplitude for a particle to be created at y and be destroyed at x. Likewise the fourier transform coefficients of $$\psi$$ create an annhilate particles of definite momentum. Finally since the polarization vector C is the fourier transform coefficient it should create photons with definite momentum when we quantize the EM field. So he's really thinking of the polarization vectors as operators that are creating particles and he's imaging sandwiching the operators between kets and thats why he's talking about "probability" (amplitude). So the product of two polarization vectors is envisaged the same way the product of the two spinor fields in the Dirac propagator. The sum is a sum over all possible polarizations, we are determining all the possible ways the particle can propagate from x to y. A similar sum also appears in the dirac propagator where the sum is over spin up and down states in the product of the fourier transform coefficients.