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Finding tensor in QED

  1. Mar 8, 2015 #1
    In the process $$e^+e^- \rightarrow \gamma \gamma$$ Capture.PNG
    for which the amplitude can be written as: $$M= \epsilon^*_{1\nu}\epsilon^*_{2\mu}(A^{\mu\nu}+\tilde{A}^{\mu\nu})$$, where $$\epsilon_i$$ is the polarization vector of a photon.

    How can one find the tensors $$A^{\mu\nu}$$ and $$\tilde{A}^{\mu\nu}$$?
     
  2. jcsd
  3. Mar 8, 2015 #2

    Orodruin

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    Are you familiar with Feynman rules?
     
  4. Mar 8, 2015 #3
    Hey @Orodruin I am to a certain extent but I can't figure out what are the tensors here if you may help?
     
  5. Mar 8, 2015 #4

    Orodruin

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    They are simply the amplitudes you would get by applying the Feynman rules to the corresponding diagram.
     
  6. Mar 8, 2015 #5
    But I already posted the amplitude $M$ in the question above. I do not think you are saying that $M$ equals to tensors $A's$, right? If so, may you please elaborate?
     
  7. Mar 8, 2015 #6

    Orodruin

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    M is an invariant. The amplitudes A are given by applying the Feynman rules to the diagrams and factoring out the photon polarizations.
     
  8. Mar 8, 2015 #7
    Oh I see, then I comprehended this the wrong way. So I use Feynman rules for QED, but I am not familiar with how to factor out photon polarization? If you may guide me with one of the calculations and I could do the other.
     
  9. Mar 8, 2015 #8
    @Orodruin Do you the name of this process? I am trying to self-learn QFT and I am not doing that systematically. If you could guide me or link me to appropriate rules so I can try this?
     
  10. Mar 8, 2015 #9

    Orodruin

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    It is just basic electron-positron annihilation. It should be available in any QED textbook.
     
  11. Mar 8, 2015 #10
    Aha, I searched and found it to be for the first let's say $$(ie)(-ie) \bar{\nu}(q) \gamma^{\rho} \epsilon_{\rho} \frac{i}{\gamma^{\rho} p_{\rho} - \gamma^{\rho}k_{1\rho} - m} \gamma^{\rho}\epsilon_{\rho} u(p)$$ If this is correct, how come this is a tensor with both indices as indicated above?
     
    Last edited: Mar 8, 2015
  12. Mar 9, 2015 #11
    I am afraid A^{\mu\nu} represent photon propagators and not amplitude. @Orodruin
     
  13. Mar 9, 2015 #12

    Orodruin

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    Your expression in #10 looks on the way to becoming fine. You have not factorised out the ##\epsilon##s and you are using the same name for several different summation indices.

    Edit: Also, I may not reply for a while, I am really on vacation ... :rolleyes:
     
  14. Mar 9, 2015 #13
    What do you precisely mean by factorizing out the epsilons? You mean to take them outside only? And concerning my propagator worry. They could not be propagators, no? Please reply when you can. Enjoy your vacation.
     
  15. Mar 9, 2015 #14

    Hepth

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    He means if you have a scalar product of epi
    silon and something, you rewrite it as eps^mu p_mu, and pull the epsilon out.

    If its epsilon to a different index, you make eps_a = eps_mu g^{mu a}

    Sorry, on mobile phone....
     
  16. Mar 9, 2015 #15
    Sorry I didn't quite understand what you wrote
     
  17. Mar 9, 2015 #16

    Hepth

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    Sorry, Im back on a PC. Im saying if you have something such as :

    ## A = (\epsilon_1 \cdot p_2) (\epsilon_2 \cdot p_1) - (\epsilon_1 \cdot \epsilon_2) (p_1 \cdot p_2) ##

    that you can factor out the epsilons by :

    ## A = \epsilon_1^{\mu} \epsilon_2^{\nu}(p_{2 \mu} p_{1 \nu} - g_{\mu \nu} (p_1 \cdot p_2)) ##
    and just say
    ## A = \epsilon_1^{\mu} \epsilon_2^{\nu} M_{\mu \nu} ## where ## M_{\mu \nu} = p_{2 \mu} p_{1 \nu} - g_{\mu \nu} (p_1 \cdot p_2)##

    that is what he meant by "factor" out the polarizations. Its common practice.

    If you do both of the diagrams, and factor out the polarizations, and add the diagrams together, you should be able to get it into a form matching your original equation. The A's in your case are just generic tensors that you have to solve for, and the way you solve them is by doing the diagram.


    Also, there are no photon propagators in those diagrams.
     
  18. Mar 9, 2015 #17
    But in my case they already have a Lorentz index. I'm afraid I'm missing something here. So yes I understand what is meant by factoring out epsilons but how does this come about in my case, now that's what is confusing me. Anyhow if you seem M up in the question, the epsilons have a star *. How is this related to this scenario of factoring things out?
     
  19. Mar 10, 2015 #18

    ChrisVer

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    That's easy to see if you understand that they are telling you that you use the Feynman rules to write [itex]M[/itex], and then from that expression you get, you factor out the epsilons and get the form of [itex]M[/itex] you have posted and the [itex]A[/itex]'s.
     
  20. Mar 10, 2015 #19

    Hepth

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    This is your problem. Remember to have control over your own indices, and NEVERrepeat their use. I can rewrite this as :

    $$(ie)(-ie) \bar{\nu}(q) \gamma^{\mu} \epsilon_{2 \mu} \frac{i}{\gamma^{\rho} p_{\rho} - \gamma^{\rho}k_{1\rho} - m} \gamma^{\nu}\epsilon_{1 \nu} u(p)$$

    because the choice in labeling your indices when converting from a dot product or slashed-object to explicitly writing out the gamma matrices is arbitrary, but must never be repetitive.

    Also, in the written diagrams it has the "1" photon associated with the ##\nu## vertex. So when you write out the diagram, it tells you how to write it.

    As for the complex conjuagates on the polarizations, this is to designate whether the photon is incoming our outgoing. It will have come from which state (initial or final) the A operator acted on.

    Does that help some?
     
    Last edited: Mar 10, 2015
  21. Mar 10, 2015 #20
    Great, it helps yes. So you meant to write $$\epsilon_{1\nu}$$ instead of $$\epsilon_{2\nu}$$ at the end of your amplitude if I am not mistaken. Thank you, you have been a great help!
     
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