1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Evaluation of Feynman propagator in position space

  1. Oct 1, 2016 #1
    1. The problem statement, all variables and given/known data

    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.

    2. Relevant equations

    3. 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?
     
  2. jcsd
  3. Oct 1, 2016 #2
    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.
     
  4. Oct 1, 2016 #3
    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?
     
  5. Oct 2, 2016 #4
    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}##
     
  6. Oct 2, 2016 #5
    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)##?
     
  7. Oct 2, 2016 #6
    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}##.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Evaluation of Feynman propagator in position space
Loading...