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Euclidean Feynman rules for QED

  1. Jul 10, 2015 #1
    Hi, i have some trouble with feynman rules after wick's rotation. I don't understand how the propagators transform. In particular if i take the photon's propagator in minkowskian coordinates i dont understand where the factor "-i" goes after the transformation.

    ## \frac{-i\eta_{\mu\nu}}{p^2} \rightarrow \frac{\delta_{\mu\nu}}{p_E^2} ##
    Last edited: Jul 10, 2015
  2. jcsd
  3. Jul 11, 2015 #2
    I think you get an extra i from the ##{p^0}## integral

    & {p^0} = ip_E^0 \cr
    & \downarrow \cr
    & dp_{}^0 = idp_E^0 \cr} ##.
  4. Jul 11, 2015 #3
    So if i start from ## \frac{i}{p_\mu \gamma^\mu - m} ## using euclidean clifford algebra i should find ## \frac{1}{p_\mu \gamma^\mu + m} ##? Because in many textbooks i find ## \frac{-i}{p_\mu \gamma^\mu + m} ##. Maybe they start from another definition of fermion propagator?
  5. Jul 11, 2015 #4
    Watch out that the i is just conventional. Other metric choices give no i for example (-+++)
    i must vanish as its a Euclidian space and indeed as physichu said the integral measure will give one i and togather
    with -i it will give +1.

    One more thing to remember is that propagators always
    carry an intergal over the internal momentum and as long as one keep that in mind then
    one can jiggle whatever with Feynman diagrams and in the end put all things to their palces.
  6. Jul 11, 2015 #5
    when you will change the metric to say (-+++) then the dirac equation will change and the lagrangian will change and action will change etc etc.
    that is why one should keep in mind which convention is being used. You may note the difference in Peskin and Srednicki due to different metric.
    On top of that different ppl prefer different reps of dirac matrices which also make difference in many identities.

    Wick rotation=euclidean space, among others, is used to get the same sign in the denominator of the propagator i.e the p^2 term gets same sign for the time and space component and which is then easily integrated in polar cordinates. Note that p^2 sign changes if you use (-+++) vs. (+---) metric.

    later i will post how to do it in case of fermion propagator e.g Peskin 4.106
  7. Jul 11, 2015 #6
    for the fermion propagator starting from 4.106 and neglecting the 2pi and exp,


    S(x-y) & = & \int d^{4}p\frac{i(p_{\mu}\gamma^{\mu}+m)}{p^{2}-m^{2}}\\

    S(x-y)_{E} & = & \int idp_{0}d^{3}p\frac{i(p_{\mu}\gamma^{\mu}+m)}{i^{2}p_{0}^{2}-\vec{p}^{2}-m^{2}}\\

    & = & i\int d^{4}p_{E}\frac{i((p_{\mu}\gamma^{\mu})_{E}+m))}{-p_{0}^{2}-\vec{p}^{2}-m^{2}}\\

    & = & -\int d^{4}p_{E}\frac{((p_{\mu}\gamma^{\mu})_{E}+m))}{-p_{E}^{2}-m^{2}}\\

    & = & \int d^{4}p_{E}\frac{((p_{\mu}\gamma^{\mu})_{E}+m))}{p_{E}^{2}+m^{2}}


    where in the second line I wick rotated i.e p_0 ---->ip_0 everywhere. I hope this helps.
  8. Jul 12, 2015 #7
    When you go to euclidean space ## (p_\mu \gamma^\mu)_M \rightarrow -(p_\mu \gamma^\mu)_E ## so the equation become


    S(x-y)_{E} & = & \int d^{4}p_{E}\frac{(-(p_{\mu}\gamma^{\mu})_{E}+m))}{p_{E}^{2}+m^{2}} \\

    & = & \int d^{4}p_{E}\frac{(-(p_{\mu}\gamma^{\mu})_{E}+m))}{(-(p_{\mu}\gamma^{\mu})_{E}+m)((p_{\mu}\gamma^{\mu})_{E}+m)} \\

    & = & \int d^{4}p_{E}\frac{1}{(p_{\mu}\gamma^{\mu})_{E}+m}


    Or not?
  9. Jul 12, 2015 #8
    I think you miss an ##i## in the ##{p_0}## term.
  10. Jul 12, 2015 #9
    your starting assumption is wrong. p-slah dont get negative in wick rotation, its only the 0th term of p that gets and extra i.

    I dont see that the whole p-slash gets negative according to Peskin rep. of Dirac matrices. Only the p_0 term of p-slash gets an i but you cant take it out as its a sum, other terms of p-slash dont have an i so how can you take it out and what is the justification of the - sign ?

    I think that what I have posted above is correct, only the p_0 gets an extra i becasue we Wick-rotate in p_0 plane to avoide the poles
    so gamma-matrices have nothing to do with this rotation.
  11. Jul 12, 2015 #10
    Wick rotation is a diffeomorphism so the metric of your space change ## \eta_{\mu\nu} \rightarrow -\delta_{\mu\nu} ##, so for the clifford algebra gamma matrices must satisfy

    \{\gamma_\mu,\gamma_\nu \}=-2\delta_{\mu\nu}
  12. Jul 12, 2015 #11
    I really dont know where you got this negative metric prescription (maybe I overlooked?? dosent makes sense though). can you provide a ref. please?

    What I see in " Qauntum Field and Strings; A course for Mathematicians " deligne & kazhdan on pg 221 eq 7.4 and 7.5 that is other than what you mentioned above. Both metric are given as g_E and g_M. the whole idea to move to Euclidean space is to be able to use the positive definitive metric but its not just the negative of Minkowskian metric.

    One only wants to change the sign of one component of the integration variable to integrate. i.e (-+++)--->(++++) OR (+---)--->(----) in doing so only the relative sign in time and space components changes.
  13. Jul 12, 2015 #12
    I don't have a ref for this...... Anyway wick rotation is a change of coordinates, so you get a new metric tensor. So you have to take "euclidean" gamma matricies define as above
  14. Jul 12, 2015 #13
    i don't understand when you say " but its not just the negative of Minkowskian metric", because i got the negative of euclidean metric tensor
  15. Jul 12, 2015 #14
    okay no problem but again can you get the consistent result with your -ive metric?
    try to match the workout relation anywhere either in Peskin or Srednicki or somewhere else.

    I guess that results wont match as we only change the 0th component and not the whole metric.

    The whole pt for the wick roatation here is not to go to Eucledian space but to remove the sign difference in
    time and space components in the denominator.

    I suggest please take a look into Peskin or Srednicki or Ryder or any other text.
  16. Jul 12, 2015 #15
  17. Jul 12, 2015 #16
    then I guess there is a glitch in your computation. I suggest try again and stick to only changing the 0th component of the integration variable.

    If you like then you can start from 3.120 of Peskin's book.
  18. Jul 12, 2015 #17
    I have just scanned the 4th reply and its okay so far and see, he is not saying to take the negative of the Minskowskian metric but he is giving you
    a +ive metric for Eucledian space.
    A metric tensor has 4 components and in Minkowskian metric tensor the time and space component have a sign difference, if you just take the negative of the Minkowskain tensor then that sign difference remain and we dont want that.
  19. Jul 12, 2015 #18
    I don't take the negative of the Minkowskain tensor, when you change your coordinate over a manifold the metric tensor transform ( as a rank 2 tensor) and you get the negative of euclidean metric (----) or ##-\delta_{\mu\nu}## in this particular case
  20. Jul 12, 2015 #19
    say you are right then can you match results of any text? how would you remove the sign difference in the time and space components of the integration variable if you follow your method?
  21. Jul 12, 2015 #20
    Oh I see....., you mean that Minkowskian metric is mapped to (----) metric and not that Minkowskian metric = (----). In the concerned relation just change the 0th component and whatever relation results, you will notice that its Eucledian space and that the metric is positive definitive i.e the square of the metric is +ive like q0^2 +q1^2 +........qn^2.

    As long as its positive definitive then we have no problem but again I dont see how you get the extra negative sign in the propagator. If you are confused then post me the exact equation then I will try to help.

    Further the anti-comm. relation of gamma matrices ( above in your post) in Eucledian space is not the same gamma matrices as you see in Peskin e.g.
  22. Jul 12, 2015 #21
    Above i use two fact :

    - ## \not p \not p = \gamma_\mu \gamma_\nu p^\mu p^\nu = \frac{1}{2} \{ \gamma_\mu ,\gamma_\nu \} p^\mu p^\nu = \frac{1}{2} (-2\delta_{\mu\nu}) p^\mu p^\nu = -p_E^2##
    - ## \not p_M = p_0\gamma^0 - p_i\gamma^i \rightarrow ip_0 i\gamma^0 - p_i\gamma^i=- \not p_E ## because ## i\gamma_M^0=\gamma_E^0 ##
  23. Jul 12, 2015 #22
    in the 1st eq; Just substitute the gamma matrices in the 3rd term (e.g use Peskin 3.25) and see if you get that minus.

    In the second eq. you dont get the i with gamma-0 but just an i with p0. Gamma matrices are still in the original Minkowskian reps ( whatever that is ). Remember that we only rotated the p0 hence it affects only the p0 everywhere in the integral.

    I really dont see the need to change the rep. of gamma matrices for whatever the reason.

    If we move systematically then all we wanted is to integrate but due to the denominator sign difference in the p^2 term we had to rotate the axis and what we found out is that after rotation we are in Euclidean space where the metric is positive definitive. This is the end ! .

    Now to see if this is the right approach I would test it out and compute some relation given already in Text.

    maybe these notes are of any help http://hitoshi.berkeley.edu/230A/clifford.pdf
  24. Jul 13, 2015 #23
  25. Jul 13, 2015 #24
    Dear andrex904 there is no need to change the reps of gamma matrices.

    Best I can comment on the -ive metric is that the -ive sign is per convention, if the metric is (+---) then negative sign and if the metric is (-+++) then the +ive sign in your above equation.

    Remember that the Euclidean space does not have time or temporal component so in your equation above the greek indices are actually only the spatial indices and the negative sign is according to the convention of the Minkowskian metric used (+---) or (-+++). But never there is an i in front of gamma0.
    Last edited: Jul 13, 2015
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