The propagator for a spin 0 particle is (in momentum space, dropping $i\epsilon$ and other factors) $$\frac{1}{p^2-m^2}$$ which has the intuition "the particle likes to be on-shell". But the propagators for spin 1/2 and 1 are more complicated; they are $$\frac{\gamma^\mu p_\mu + m}{p^2 - m^2}$$ for spin 1/2 and $$\frac{\eta_{\mu \nu} - p_\mu p_\nu / p^2}{p^2-m^2}$$ Is there an intuitive explanation for what the extra terms in the numerators do? I've been given no explanation besides "this is what falls out of the theory".

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    $\begingroup$ I have no idea what you mean by "intuition" here. Why is the "intuition" for the scalar propagator "the particle likes to be on-shell"? Because it blows up to infinity there? (It does that for the other ones, too, and there are a bazillion other functions of $p$ and $m$ that also blow up near the shell values) $\endgroup$
    – ACuriousMind
    Sep 7, 2015 at 23:12
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    $\begingroup$ By intuition, I mean a couple of sentences that make me feel like I understand why these are the right expressions. You usually dismiss questions like this because you think they're trivial, but I don't find it trivial, so please tell me what you know. $\endgroup$
    – knzhou
    Sep 7, 2015 at 23:15
  • $\begingroup$ I don't think this question is trivial - I don't understand what it is asking for. I don't get why your claimed "intuition" for the scalar propagator would lead you to believe $\frac{1}{p^2 - m^2}$ is the right expression for the Fourier transform of the propagator (i.e. "the particle likes to be on-shell" doesn't lead me to that expression in any way, intuitive or not), so I can't begin to tell you what "intuition" might be behind the non-scalar ones. $\endgroup$
    – ACuriousMind
    Sep 7, 2015 at 23:17
  • $\begingroup$ One can deduce some properties without explicitly calculating the propagator. The pole location is determined by the particle mass, the functional form follows from respecting relativity and not violating unitarity. The tensor structure is determined by the gauge choice + we only have two tensors to 'build with' and so on. $\endgroup$
    – Winther
    Sep 8, 2015 at 0:25
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    $\begingroup$ $\frac{1}{p^2-m^2}$ is $(\Box+m^2)^{-1}$ in Fourier space, in that it solves $(\Box+m^2)\,\phi=j$ for $\phi$. The spin 1/2 case $\frac{\gamma\,p+m}{N}$ solves $(\gamma\,\partial-m)\,\psi=j$, and so on. $\endgroup$
    – Nikolaj-K
    Sep 9, 2015 at 22:37

1 Answer 1


To get a better intuition consider a field with a general spin, this can be written as \begin{align*} \psi_\ell &\propto \sum_\sigma \int d^3 p \left( u_\ell (\vec{p},\sigma) e^{i p\cdot x}a(\vec{p},\sigma) + v_\ell (\vec{p},\sigma) e^{-i p\cdot x}a^\dagger (\vec{p},\sigma) \right) \end{align*} where $\ell$ is the spin index, and $\sigma$ summed over all spin states.

then the propagator will have \begin{align*} \sum_\sigma u_\ell (\vec{p}, \sigma) u_m^*(\vec{p},\sigma) \end{align*} in its numerator (or equivalently $\sum vv^*$). So we are summing over $\textbf{polarizations}$

So now remembering that the propagator in a crude sense measures the correlation between disturbances of the field at different positions, you can understand that the numerator accounts for the contribution of different modes of polarization in propagating the disturbance, where you have to sum over all of them to get the total correlation. For a spin-less particle there is only one mode...


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