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

  1. Jul 29, 2015 #1
    Assuming that a massive spin-1 particle has momentum only in the z-direction, the polarization vectors are given by

    [tex]\varepsilon_{\mu}(J_z = +1) = (0,-\frac{1}{\sqrt{2}},-\frac{i}{\sqrt{2}},0 )[/tex]

    [tex]\varepsilon_{\mu}(J_z = 0) = (\frac{p}{m},0,0, \frac{E}{m})[/tex]

    [tex]\varepsilon_{\mu}(J_z = -1) = (0,\frac{1}{\sqrt{2}},-\frac{i}{\sqrt{2}},0 )[/tex]

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

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

    I absolutely dont understand how this spin-sum is evaluated.
    What does [itex]\varepsilon_{\mu}\varepsilon_{\nu}^*[/itex] 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 [itex]J_z[/itex]?
    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. jcsd
  3. Jul 29, 2015 #2

    Orodruin

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    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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