In my QFT class we have defined the Feynman propagator of a field $\phi^r$ (where $r$ could be a vector or spinor index, or even a multiindex if $\phi$ is a tensor field etc.) as $$ \Delta^{rs}_F(x - x') = \langle 0| \mathcal{T}\{\phi^r(x), \phi^s(x')\}|0\rangle $$

Suppose the equations of motion for the field(s) $\phi^r$ can be written as $L^{r s}\phi^s = 0$ for some differential operators $L^{rs}$. Then supposedly the Feynman propagator is a Green's function for $L^{rs}$ (up to sign), i.e. $$ L^{rs} \Delta^{s t}_F(x) = - \delta^4(x) \delta^{r t} $$ Q: Is this true, and if so, why?

It certainly is true for a Klein-Gordon field, but what about the general case? I can also see why we have $L^{r s} \Delta^{s t}_F(x) = 0$ for $t \neq 0$.

  • $\begingroup$ Well, you could take this as the definition of the propagator (see e.g. this PSE post, where I use the notation $\mathscr D$ instead of $L$). More precisely, you should let $\mathcal T$ be defined such that this condition is true (which sometimes agrees with the naive definition of time-ordering, but sometimes acquires extra contact terms, cf. this PSE post). $\endgroup$ Feb 10 '20 at 1:20
  • $\begingroup$ As a start, you could just consider the propagators as being Green functions of PDEs. You can do the same for the Dirac equation $(i\hbar\gamma^{\mu}\partial_{\mu} - m)\psi$ and define a fermion propagator. $\endgroup$
    – Judas503
    Feb 10 '20 at 5:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.