# Issue in Feynman propagator with derivative of a scalar field: which Feynman rule do I apply?

While solving a QFT exercise, I'm trying to calculate the Feynman propagator $$\underbrace{\partial_\mu \phi(x) \phi(y)} = \langle 0 \vert T { \partial_\mu \phi(x)\phi(y) } \vert 0\rangle$$ where the derivative acts over the $$x$$ variable.

In this task I encountered the problem of a relative minus sign between the two terms of the $$T$$-ordered product: $$$$\tag{1} \underbrace{\partial_\mu \phi(x)\phi(y)} = \int \frac{d^3\mathbf p}{(2\pi)^3 2E_\mathbf{p}} \left(-ip_\mu \right)\left( \theta (x^0 - y^0) e^{-ip(x-y)} - \theta (y^0 - x^0)e^{ip(x-y)}\right)$$$$ whereas my first guess would've been $$$$\tag{2} \langle 0|T{\partial_{\mu}\phi(x)\phi (y)}|0\rangle =\int \frac{d^4p}{(2\pi)^4} (-ip_{\mu}) \cdot \frac{ i }{p^2-m^2+i\epsilon} e^{-ip(x-y)}$$$$ leading to a Feynman contribution $$\frac{ p_\mu }{p^2-m^2+i\epsilon}$$ to the scattering amplitude.

My first question therefore reduces to: are $$(1)$$ and $$(2)$$ somehow equivalent, and if so, how?

Furthermore, accepting the validity of $$(2)$$, the scalar propagator does not recognize any orientation, thus $$p_\mu$$ seems to be ill-defined, again, due to a sign. Actually, another arbitrariness rises from the choice of deriving with respect to $$x$$ instead of $$y$$, when using Wick Theorem.

The same problem arises in computing the vector (e.g. photon) propagator $$\underbrace{ \partial_\mu A_\nu(x) A_\rho(y)}$$

Yes, (1) and (2) are equivalent. The issue of a relative minus sign is due to the derivative, assuming $$\mu\neq 0$$.

Eq. (1) is rather the intuitive expression for the time-ordered product of field operators $$\partial_\mu\phi(x)$$ and $$\phi(y)$$. (2) is a fancy way of writing the same thing, making use of contour integrals.

First, note that $$p^0=\pm E_p$$ is pole of the expression $$1/(p^2-m^2)=1/((p^{0})^2-\mathbf p^2-m^2)$$, so consider the integral

$$$$\int_{-\infty}^{\infty} \frac{dp^0}{2\pi} \frac{i}{p^2-m^2+i\epsilon}e^{-ip\cdot (x-y)} \equiv \int_{-\infty}^{\infty} \frac{dp^0}{2\pi} \frac{i}{(p^0 -E_p+i\epsilon)(p^0+E_p-i\epsilon)} e^{-ip\cdot(x-y)}$$$$

This can be converted into a contour integral with the semicircular arc at infinity in the lower half complex plane for $$x^0 - y^0> 0$$ and in the upper half complex plane for $$x^0 -y^0 < 0$$. In the first case, the pole lying inside the contour is $$p^0=E_p - i\epsilon$$, so, by the residue theorem, the integral equates to

$$$$\frac{e^{-iE_p(x^0-y^0)+i\mathbf p\cdot (\mathbf{x-y})}}{2E_p}$$$$

Note that the contour runs clockwise, so we incur an additional minus sign.

For the case $$x^0 - y^0<0$$, by the same procedure (contour anti-clockwise), we end up with

$$$$\frac{e^{iE_p(x^0-y^0) +i\mathbf p\cdot (\mathbf{x-y})}}{2E_p} =\frac{e^{-iE_p(y^0-x^0)-i\mathbf p\cdot (\mathbf{y-x})}}{2E_p}$$$$

To put this in the standard form, replace $$\mathbf p \rightarrow -\mathbf p$$. The spatial integral $$d^3p$$ remains unchanged, however, the derivative $$-ip_\mu$$ picks up a minus sign.