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  Evaluation of the spin-sum for a massive spin-1 particle?

+ 0 like - 0 dislike

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.869

asked Jul 29, 2015 in Theoretical Physics by anonymous [ no revision ]

It is a tensor-product, as you could have seen from the right hand side. Just evaluate the three matrices explicitly and sum them to get the result. It is pure linear algebra.

To add to Arnold, here might be a even straighter way for OP to see this is a elementary linear algebra calculation: Think of $\varepsilon_\mu(J_z)$ as a 4 by 3 matrix $\varepsilon$, rows labeled by $\mu$ and columns labeled by $J_z$, then the spin sum is nothing but $\varepsilon \varepsilon^\dagger$, which is a 4 by 4 matrix.

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