Evaluation of Feynman propagator in position space

[spaghetti3451]

Homework Statement

Compute ##\displaystyle{\int\ \frac{d^{4}p}{(2\pi)^{4}}} \frac{i}{p^{2}-m^{2}+i\epsilon}e^{-ip \cdot{(x-y)}}## in terms of the invariant interval.

Interpret your answer in the limit of small and large invariant intervals and for zero mass.

Homework Equations

The Attempt at a Solution

##\displaystyle{\int\ \frac{d^{4}p}{(2\pi)^{4}}} \frac{i}{p^{2}-m^{2}+i\epsilon}e^{-ip \cdot{(x-y)}}##

##=\displaystyle{\int\ \frac{d^{4}p}{(2\pi)^{4}}} \frac{i}{(p^{0})^{2}-(E_{\vec{p}})^{2}+i\epsilon}e^{-ip \cdot{(x-y)}}##

##=\displaystyle{\int\ \frac{d^{4}p}{(2\pi)^{4}}} \frac{i}{(p^{0})^{2}-(E_{\vec{p}}-i\epsilon)^{2}}e^{-ip \cdot{(x-y)}}##

##=\displaystyle{\int\ \frac{d^{4}p}{(2\pi)^{4}}} \frac{i}{(p^{0}-(E_{\vec{p}}-i\epsilon))(p^{0}+(E_{\vec{p}}-i\epsilon))}e^{-ip \cdot{(x-y)}}##

Therefore, the contour prescription ##i\epsilon## reminds us to integrate under the pole at ##p^{0}=-E_{\vec{p}}## and over the pole at ##p^{0}=E_{\vec{p}}##. So, let's now drop the ##i\epsilon## term and obtain

##=\displaystyle{\int\ \frac{d^{4}p}{(2\pi)^{4}}} \frac{i}{(p^{0}-E_{\vec{p}})(p^{0}+E_{\vec{p}})}e^{-ip \cdot{(x-y)}}##

##=\displaystyle{\int\ \frac{d^{3}p}{(2\pi)^{3}}}\ \bigg[ \int\ \frac{dp^{0}}{2\pi} \frac{i}{(p^{0}-E_{\vec{p}})(p^{0}+E_{\vec{p}})}e^{-ip^{0}(x^{0}-y^{0})} \bigg]\ e^{-ip^{i}(x_{i}-y_{i})}##

For ##x^{0}>y^{0}##, we close the contour below (clockwise sense introduces a negative factor) and for ##x^{0}<y^{0}##, we close the contour above (anti-clockwise sense introduces a positive factor), so that

##=\displaystyle{\int\ \frac{d^{3}p}{(2\pi)^{3}}}\ \bigg[ \int\ \frac{dp^{0}}{2\pi} \frac{i}{(p^{0}-E_{\vec{p}})(p^{0}+E_{\vec{p}})}e^{-ip^{0}(x^{0}-y^{0})} \bigg]\ e^{-ip^{i}(x_{i}-y_{i})}##

##=\displaystyle{\int\ \frac{d^{3}p}{(2\pi)^{3}}}\ \bigg[ \theta(x^{0}-y^{0})\frac{1}{2E_{\vec{p}}}e^{-iE_{\vec{p}}(x^{0}-y^{0})} +\theta(y^{0}-x^{0})\frac{1}{-2E_{\vec{p}}}e^{iE_{\vec{p}}(x^{0}-y^{0})} \bigg]\ e^{-ip^{i}(x_{i}-y_{i})}##

Am I correct so far?

 

[Fightfish]

##=\displaystyle{\int\ \frac{d^{4}p}{(2\pi)^{4}}} \frac{i}{(p^{0}-E_{\vec{p}})(p^{0}+E_{\vec{p}})}e^{-ip \cdot{(x-y)}}##

##=\displaystyle{\int\ \frac{d^{3}p}{(2\pi)^{3}}}\ \bigg[ \int\ \frac{dp^{0}}{2\pi} \frac{i}{(p^{0}-E_{\vec{p}})(p^{0}+E_{\vec{p}})}e^{-ip^{0}(x^{0}-y^{0})} \bigg]\ e^{-ip^{i}(x_{i}-y_{i})}##

There's probably a sign problem here (unless you have somehow absorbed it into the indices) - the temporal and spatial portions of your dot product should have different signs.

 

[spaghetti3451]

Well,

##-ip\cdot{(x-y)}=-i[p^{\mu}(x-y)_{\mu}]=-i[p^{0}(x-y)_{0}+p^{i}(x-y)_{i}]##

Now, using the mostly negative signature,

##-i[p^{0}(x-y)_{0}+p^{i}(x-y)_{i}]=-i[p^{0}(x-y)^{0}+p^{i}(x-y)_{i}]=-ip^{0}(x-y)^{0}-ip^{i}(x-y)_{i}]##.

Right?

 

[Fightfish]

Ah okay, it's somewhat confusing but your negative signs are buried within the subscripts - so in that case, I think whatever you've done seems correct. I think though that its generally cleaner to switch to three-vector notation i.e. ##p \cdot x = E_{\vec{p}} t - \vec{p} \cdot \vec{x}##

 

[spaghetti3451]

Ok, so

##=\displaystyle{\int\ \frac{d^{4}p}{(2\pi)^{4}}} \frac{i}{(p^{0}-E_{\vec{p}})(p^{0}+E_{\vec{p}})}e^{-ip \cdot{(x-y)}}##

##=\displaystyle{\int\ \frac{d^{3}p}{(2\pi)^{3}}}\ \bigg[ \int\ \frac{dp^{0}}{2\pi} \frac{i}{(p^{0}-E_{\vec{p}})(p^{0}+E_{\vec{p}})}e^{-ip^{0}(x^{0}-y^{0})} \bigg]\ e^{i\vec{p}(\vec{x}-\vec{y})}##

For ##x^{0}>y^{0}##, we close the contour below (clockwise sense introduces a negative factor) and for ##x^{0}<y^{0}##, we close the contour above (anti-clockwise sense introduces a positive factor), so that

##=\displaystyle{\int\ \frac{d^{3}p}{(2\pi)^{3}}}\ \bigg[ \int\ \frac{dp^{0}}{2\pi} \frac{i}{(p^{0}-E_{\vec{p}})(p^{0}+E_{\vec{p}})}e^{-ip^{0}(x^{0}-y^{0})} \bigg]\ e^{i\vec{p}(\vec{x}-\vec{y})}##

##=\displaystyle{\int\ \frac{d^{3}p}{(2\pi)^{3}}}\ \bigg[ \theta(x^{0}-y^{0})\frac{1}{2E_{\vec{p}}}e^{-iE_{\vec{p}}(x^{0}-y^{0})} +\theta(y^{0}-x^{0})\frac{1}{2E_{\vec{p}}}e^{iE_{\vec{p}}(x^{0}-y^{0})} \bigg]\ e^{i\vec{p}(\vec{x}-\vec{y})}##

##=\displaystyle{\theta(x^{0}-y^{0})\int\ \frac{d^{3}p}{(2\pi)^{3}}}\ \bigg[\frac{1}{2E_{\vec{p}}}e^{-iE_{\vec{p}}(x^{0}-y^{0})}e^{i\vec{p}(\vec{x}-\vec{y})} \bigg]+\theta(y^{0}-x^{0})\int\ \frac{d^{3}p}{(2\pi)^{3}}\ \bigg[\frac{1}{2E_{\vec{p}}}e^{iE_{\vec{p}}(x^{0}-y^{0})} e^{i\vec{p}(\vec{x}-\vec{y})}\bigg]##

The first term gives ##=\displaystyle{\theta(x^{0}-y^{0})\int\ \frac{d^{3}p}{(2\pi)^{3}}}\ \bigg[\frac{1}{2E_{\vec{p}}}e^{-iE_{\vec{p}}(x^{0}-y^{0})}e^{i\vec{p}(\vec{x}-\vec{y})} \bigg]=\displaystyle{\theta(x^{0}-y^{0})\int\ \frac{d^{3}p}{(2\pi)^{3}}}\ \bigg[\frac{1}{2E_{\vec{p}}}e^{-ip\cdot{(x-y)}} \bigg]\bigg|_{p^{0}=E_{\vec{p}}}=D(x-y)##.

How do I convert the second term to ##-D(y-x)##?

 

[Fightfish]

I think it should be just ##D(y-x)## and not ##-D(y-x)##. The trick is to flip the sign of ##\vec{p}##, which is a valid operation because we are integrating over the entire ##p##-volume, and all other quantities only depend on the magnitude of ##\vec{p}##.