Expressing Feynman Green's function as a 4-momentum integral

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I am a bit confused on how we can just say that (z',p) form a 4-vector. In my head, four vectors are sacred objects that are Lorentz covariant, but now we introduced some new variable and say it forms a 4-vector with momentum. I understand that these are just integration variables but I still do not see how this is okay. The interpretation of z' now is different.
 
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realanswers said:
I am a bit confused on how we can just say that (z',p) form a 4-vector. In my head, four vectors are sacred objects that are Lorentz covariant, but now we introduced some new variable and say it forms a 4-vector with momentum. I understand that these are just integration variables but I still do not see how this is okay. The interpretation of z' now is different.
But where did it come from? ##\theta(x^0 - y^0)##, right? The restricted Lorentz group (identity-connected part) preserves time orientation, so it's ok to use ##\theta(x^0 - y^0)## in that circumstance. The rest just involves expressing it in momentum space.
 
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It's given a gas of particles all identical which has T fixed and spin S. Let's ##g(\epsilon)## the density of orbital states and ##g(\epsilon) = g_0## for ##\forall \epsilon \in [\epsilon_0, \epsilon_1]##, zero otherwise. How to compute the number of accessible quantum states of one particle? This is my attempt, and I suspect that is not good. Let S=0 and then bosons in a system. Simply, if we have the density of orbitals we have to integrate ##g(\epsilon)## and we have...
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