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 $\vec{p}$.