In deriving the Feynman propagator in Timo Weigand's 2014 QFT2 notes, at the top of page 37, (equation 1.170), we use Cauchy's integral formula:

$$g(z_0)=\frac{1}{2\pi i}\oint_{C_1}\frac{g(z)}{z-z_0}\mathrm dz \tag{1.169},$$

to rewrite the equation 1.168 as a four-momentum integral:

$$\Theta(x^0-y^0)\frac{1}{2E_p}e^{-iE_p(x^0-y^0)}=-\Theta(x^0-y^0)\frac{1}{2\pi i}\oint_{C_1}\mathrm dp^0\frac{e^{-ip^0(x^0-y^0)}}{(p^0-E_p)(p^0+E_p)} \tag{1.170}.$$

In the above derivation the $+E_p$ is enclosed by the integral and the negative sign is due to the clockwise direction of the integral, we close the contour in the lower half of the $\Bbb C$-plane.

However, a straight application of equation 1.169 appears to give:

$$\Theta(x^0-y^0)\frac{1}{2E_p}e^{-iE_p(x^0-y^0)}=-\Theta(x^0-y^0)\frac{1}{2\pi i}\frac{1}{2}\oint_{C_1}\mathrm dp^0\frac{e^{-ip^0(x_0-y_0)}}{p^0(p^0-E_p)},$$

where have I gone wrong here?

Sidenote: I understand some people might believe this to be a mathematical question not a physics one, I am happy to move it if that's appropriate, but the language and notation is definitely physics and might be confusing.


1 Answer 1


(Here I use $z$ instead of $p_0$ for clarity). In your first equation, we take $$g(z) := \frac{e^{-iz(x_0-y_0)}}{z+E_p}$$

In particular, at $E_p$, $g(z)$ is equivalent to the advanced propagator: $$ g(E_p) = \frac{e^{-iE_p(x_0-y_0)}}{2E_p} $$ In principle, you could take any function $g(z)$ such that $g(E_p)$ is the advanced propagator, but this choice is particularly nice since we end up with a manifestly Lorentz-invariant result.

Now invoke Cauchy's integral formula: $$ g(E_p) = \frac1{2\pi i}\oint_C \frac{g(z)}{z-E_p} \mathrm{d}z \\ =\frac1{2\pi i}\oint_C \frac{e^{-iz(x_0-y_0)}}{(z+E_p)(z-E_p)} \mathrm{d}z $$


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