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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: \begin{equation}\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) \end{equation} whereas my first guess would've been \begin{equation}\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)} \end{equation} 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 $$ \underbrace{ \partial_\mu A_\nu(x) A_\rho(y)} $$

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1 Answer 1

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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

\begin{equation} \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)} \end{equation}

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

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

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

\begin{equation} \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} \end{equation}

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.

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