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I How to calculate self energy Feynman diagram like that

  1. Oct 4, 2016 #1
    Hi all ,

    I try to calculate the squared amplitude of the following self energy digram :

    B-l.png


    where se is massless Dirac fermion , and vrm is massless right handed neutrino. x is a scalar with mass m ..

    I wrote the nominator of this process as:

    N = ## \bar{u}(p) (1-\gamma_5) (p\!\!\!/ +k\!\!\!/ ) (1+\gamma_5) u(p) \to p\!\!\!/ (1-\gamma_5) (p\!\!\!/ +k\!\!\!/ ) (1+\gamma_5) \to p\!\!\!/ (1-\gamma_5) (p\!\!\!/ +k\!\!\!/ ) \to p_\alpha \gamma^\alpha (1-\gamma_5) (p_\alpha \gamma^\alpha + k_\beta \gamma^\beta ) \to p_\alpha p_\alpha \gamma^\alpha \gamma^\alpha +p_\alpha k_\beta\gamma^\alpha \gamma^\beta - p_\alpha \gamma^\alpha \gamma_5 p_\alpha \gamma^\alpha - p_\alpha \gamma^\alpha \gamma_5 k_\beta \gamma^\beta ##

    I don't know how the first term as ## p_\alpha p_\alpha \gamma^\alpha \gamma^\alpha ## can be calculated or the third term .. also I don't know how ## P.K## which comes from the second term can be calculated from the process kinematics ? while ## P^2 =0 ## and ## K^2 = m_\chi^2##

    Note that: the momentums of the particles in the Feynman diagram are se(p), vrm(p+k) and x(k)

    Also the "x se vrm " vertices come from these Lagrangian terms:

    ## y (\bar{\nu_R^c} \chi s + \bar{s} \chi \nu_R^c ) ## (where s-> se, ##\chi ## -> x and \nu -> vrm)

    Thanks ..
     
  2. jcsd
  3. Oct 5, 2016 #2
    For the terms you asked for: Wikipedia has a useful list of identities: https://en.wikipedia.org/wiki/Gamma_matrices#Feynman_slash_notation. I am not sure if I follow your steps though. Are there traces implied over your terms? Also, ##k^2 \neq m_\chi^2##, as ##k^\mu## is the loop momentum you are integrating over.

    Edit: Also, for a self energy you would not necessarily write it between spinors, probably you could expand more on what you actually want to calculate...
     
    Last edited: Oct 5, 2016
  4. Oct 5, 2016 #3
    Yes, sure .. I forgot Tr [] :(

    Now it getting worse .. the nominator becomes:

    ## N = Tr [ (1-\gamma_5) (p\!\!\!/ +k\!\!\!/ ) (1+\gamma_5) \to (1-\gamma_5) (p\!\!\!/ +k\!\!\!/ ) ] \to 0 ##

    since:

    Trace of any product of an odd number of ## \gamma^\mu ## is zero and trace of ## \gamma^5 ## times a product of an odd number of ## \gamma^\mu ## is also zero ..
     
  5. Oct 5, 2016 #4

    ChrisVer

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    Gold Member

    Why do you write same indices more than once?
     
  6. Oct 5, 2016 #5
    Which indices ? I think the veritecs and momentums are all clear in the question
     
  7. Oct 6, 2016 #6
    But you only have a trace if there are spinors. In this case all your terms just need traces (and I think you are missing a factor of two from ##(1-\gamma_5)(1-\gamma_5) = 2(1-\gamma_5)##).

    For example

    would be more readable as ##p_\alpha p_\beta\gamma^\alpha\gamma^\beta##.
     
  8. Oct 6, 2016 #7

    vanhees71

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    2016 Award

    It's not only "more readable" but it's a meaningful expression. Each index can occur maximally twice, and in this case one must be an upper and one a lower index, and this implies by convention that you sum over this index from 0 to 3 (Einstein summation convention). If the same index occurs more than twice or two same upper or lower indices, than the expression is wrong.
     
  9. Oct 6, 2016 #8
    So some thing is not clear now, do I write the spinors ? that is the right ?

    Yap .. but forget about numbers now, actually it will be a factor of 1/2, because ##P_L = (1-\gamma^5)/2## and ##P_R = (1+\gamma^5)/2##

    I will say why .. please look the following Feynman digram of the process

    loop.png

    I have in N ## p\!\!\!/ ## comes from the spinors and ## p\!\!\!/ ## appear in ## (p\!\!\!/+k\!\!\!/) ## from the fermion propagtor, so it's the same momentum or p .. I wonder if it will be meaningfull if i used different indecis for like ## p_\alpha \gamma^\alpha (1-\gamma_5) (p_\delta \gamma^\delta + k_\beta \gamma^\beta ) ## ?
     

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  10. Oct 6, 2016 #9
    The self energy ##\Sigma(p)## is usually defined as just the 1PI blob, for which at lowest order your diagram would contribute. There are no spinors involved. You can of course write down the expression for ##\bar{u}(p)\Sigma(p)u(p)##, but I am not aware what this expression would mean. What observable do you want to calculate?


    That is why you call it ##p## both times. But the contracted indices mean summation e.g. ##p_\alpha\gamma^\alpha = \sum_{\alpha=0}^3 p_\alpha\gamma^\alpha##. The sums in the two ##\slash{p}## expressions have their own indices.

    That would be the expression.
     
  11. Oct 6, 2016 #10
    So now "P.K" or ##p_\mu k^\mu## what does it equal ? 'd we use CMF kinematics here ? but I wonder how ..
     
  12. Oct 6, 2016 #11
    Well , ##k^\mu## is the loop momentum you are integrating over, so you need to calculate something like ##p_\mu \int \frac{d^4k}{(2\pi)^4} \frac{k^\mu}{D}##, where ##D## means the denominator corresponding to your diagram.
     
  13. Oct 6, 2016 #12
    I want to calculate the mass of a particle ## s ## on the loop level ..
     
  14. Oct 7, 2016 #13
    so you indeed just need the self energy
     
  15. Oct 7, 2016 #14

    vanhees71

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    What's you interaction? I can only guess from your a bit strange posting #1 that it might be the coupling
    $$\mathcal{L}_{\text{int}}=g \overline{\psi} \mathrm{i} \gamma^5 \psi \phi,$$
    where ##\psi## is a Dirac spinor and ##\phi## a (pseudo-)scalar field. The corresponding Feynman rule is that the three-point vertex is ##-g \gamma^5##. The propagator lines are
    $$G(p)=\frac{\gamma_{\mu} p^{\mu}+m}{p^2-m^2+\mathrm{i} 0^+}, \quad D(p)=\frac{1}{p^2-M^2+\mathrm{i} 0^+},$$
    for the fermion and boson respectively.

    Now apply the Feynman rules. The self-energy is truncated, i.e., the external legs are not considered. They just tell you that you have a self-energy. Now you have to read the diagram against the direction of the spinor legs. Then taking into account the symmetry factor you get
    $$\sigma(p)=-g^2 \int \mathrm{d}^4 l \frac{1}{(2 \pi)^4} \gamma^5 \mathrm{i} G(l+p) \gamma^5 \mathrm{i} D(l).$$
    Now you can use standard formulae (e.g., regularizing with dimensional regularization) and evaluate the diagram, which is linearly divergent, i.e., you need a wave-function and a mass counter term to make the diagram finite, as expected.
     
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