# Demystification of the spin-sum for massive spin-1 particles

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1. Jul 29, 2015

### Dilatino

Assuming that a massive spin-1 particle has momentum only in the z-direction, the polarization vectors are given by

$$\varepsilon_{\mu}(J_z = +1) = (0,-\frac{1}{\sqrt{2}},-\frac{i}{\sqrt{2}},0 )$$

$$\varepsilon_{\mu}(J_z = 0) = (\frac{p}{m},0,0, \frac{E}{m})$$

$$\varepsilon_{\mu}(J_z = -1) = (0,\frac{1}{\sqrt{2}},-\frac{i}{\sqrt{2}},0 )$$

The so-called spin-sum is the claimed to be

$$\sum\limits_{J_z = -1,0,+1} \varepsilon_{\mu}\varepsilon_{\nu}^* = g_{\mu\nu} + \frac{p_{\mu}p_{\nu}}{m^2}$$

I absolutely dont understand how this spin-sum is evaluated.
What does $\varepsilon_{\mu}\varepsilon_{\nu}^*$ even exactly mean? Is it a scalar product between two of the three above polarization vectors, or is it a "tensor-product" between the components of a single polarization vector which results in a 4x4 matrix and one has finally to sum all such matrices for the three possible values of $J_z$?
I would really appreciate it if somebody can explain to me what this spin-sum exactly means and how it is evaluated step-by-step.

2. Jul 29, 2015

### Orodruin

Staff Emeritus
The latter option, it is a tensor product, which results in a rank two tensor (which may be represented by a matrix, it is not a matrix - it is a rank two tensor). The sum is taken over all possible spin states, i.e., you are summing several rank two tensors. The result is a new rank two tensor.

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