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The Feynman propagator is given by the expectation value of two time-ordered (scalar) field operators (evaluated in the vacuum): $$ G_\mathrm{F}(x,y) \equiv \langle 0 | \mathcal{T}\big( \hat{\phi}(x) \hat{\phi}(y) \big) | 0 \rangle $$

This is a Green's function for the Klein-Gordon operator in the sense that $$ \bigg( - \left( \tfrac{\partial}{\partial x^0} \right)^2 + \boldsymbol{\nabla}_{\mathbf{x}}^2 - m^2\bigg) G_\mathrm{F}(x,y) = - \delta^{(4)}(x-y) $$ See for example, Equation (6.2.17) in Weinberg's Volume 1.

Normally, mathematicians define a generic Green's function $G$ for an operator $\hat{L}$ as one satisfying; $$ \hat{L} G(x;y) = + \delta(x-y) $$ Note the difference in minus sign.

My question is, why do physicists have an extra minus sign for the Green's function they always use? Is it just a convention that caught on, or is there a practical reason?

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    $\begingroup$ The minus sign is not always there. It depends on conventions. It has no physical significance, and it doesn't change measurable results, as long as you are consistent. $\endgroup$ – AccidentalFourierTransform Aug 2 '18 at 19:09
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The rationale behind the convention is that in $d=1$, where there is only a time dimension, the partial differential equation (pde) reduces to $F=ma$. In fact, the pde with the wave equation can be viewed as a $f=ma$, where $ma=\ddot{\phi}$, the internal force density is given by $\nabla^2\phi-m^2\phi$, and the external force density is given by $\delta^{(4)}(x-y)$. So this sign convention is the same as saying external forces are directed parallel to the accelerations they cause. The opposite sign convention is allowed, of course, if you keep everything consistent, as @AccidentalFourierTransform notes.

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