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A question about trace technology

  1. May 13, 2008 #1
    Hi everybody!

    While reading Peskin-Schroeder, i stuck in the in equation 5.4 about the unpolarized cross section of the [tex]e^- e^+ \rightarrow \mu^- \mu^+[/tex] annihilation. i didn't understand how this relationof the electron trace came to be, and where the indices came from?

    [tex] \operatorname{tr}[(p\!\!/^\prime-m_e) \gamma^\mu (p\!\!/ + m_e) \gamma^\nu] = 4[p^{\prime \mu} p^\nu +p'^\nu p^\mu - g^{\mu \nu} (p \cdot p' + m^2_e)] [/tex]


    thanx in advanced!
     
  2. jcsd
  3. May 13, 2008 #2

    malawi_glenn

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    what have you tried? Do you know how traces with gamma matrices works at all?

    if not, then I can give some relations that are useful
     
    Last edited: May 13, 2008
  4. May 13, 2008 #3
    Hi,

    I remember when I first encountered with those... very confused I was. First time, I did it "brute force". Calculated all the components on one side, check they were the same on the other. Of course it is stupid, and not the right way to do it. But by doing so, I understood how to work it out neatly.

    Everybody has his own methods...
     
  5. May 13, 2008 #4

    malawi_glenn

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    This example can be calculated on 1/2 page, if one does all calculations. But one have to know all trace identies and so on.

    [tex] \text{Tr}(\gamma ^{\alpha}\gamma ^{\beta}) = g^{\alpha \beta} [/tex]
     
  6. May 13, 2008 #5

    nrqed

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    Are you asking about how to get from the trace (on the left) to the final result or are you asking how one gets from a Feynman amplitude to a trace expression (in other words how the left hand side comes from)?

    To get from the left side to the right side is a simple application of trace formula. Look up the equations 5.5, 5.8, 5.9 etc. Or are you asking how where those trace formula come from? (using the cyclicity of the trace operation and using identities for the product of gamma matrices does the trick)
     
  7. May 13, 2008 #6
    yes, i just want to know how to get from the left hand side to the right hand side. the problem mabye is thas i am not familiar with traces of gamma matrices. i'll try to learn it then come with a more concrete question..
    but if anybody can give some hints that helps i'll be grateful.
     
  8. May 13, 2008 #7

    nrqed

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    But then it's just a matter of applying the trace formula given in equation 5.5. Just expand the expression you gave in your first post (keeping in mind that that masses simplys multiply identity matrices) and use directly the formula of 5.5. That's all there is to it.
     
  9. May 14, 2008 #8

    malawi_glenn

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    Then if you want a lot of rules and properties of gamma matrices and dirac-slash notation, you can find it on many places on the internet. And their derivation. Do you know how the minkowski metric tensor works?

    I am actually trying to make my own collection of formulas, and a collection of proofs. Just to have it all collected in a nice reference and TeX - practicing. If you are interessted you can send me a PM.
     
  10. May 14, 2008 #9
    ok tell me please how to trace this thing [tex]\operatorname{tr}(\gamma^\nu\gamma^\mu p'_\mu\gamma^\mu\gamma^\nu p_\nu)[/tex]? why it can't show [tex]\operatorname{tr}(\gamma^\nu\gamma^\mu p'_\mu\gamma^\mu\gamma^\nu p_\nu)[/tex]? the last is what i want to ask
     
    Last edited: May 14, 2008
  11. May 15, 2008 #10

    malawi_glenn

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    how did you get the same indicies on two of them? you must as general as possible, i.e:

    [tex] \operatorname{tr}[(p\!\!/^\prime-m_e) \gamma^\mu (p\!\!/ + m_e) \gamma^\nu] = \operatorname{tr}[(p'_{\alpha}\gamma^{\alpha}-m_e) \gamma^\mu (p_{\beta}\gamma^{\beta}+ m_e) \gamma^\nu] [/tex]

    Then use the trace formula for 4 gamma matrices which I mailed you a link to.
     
  12. May 18, 2008 #11
    [tex] \operatorname{tr}(8p'^\nu p^\mu) - \operatorname{tr}(2mp'^\mu \gamma^\nu) + \operatorname{tr}(2m\gamma^\mu p^\nu) + \operatorname{tr}(2m^2g^{\mu\nu}) [/tex]
    am i on the right way?
     
  13. May 18, 2008 #12

    malawi_glenn

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    no, lets take this one for instance:

    [tex] \text{Tr}(p'_{\alpha}\gamma^{\alpha}\gamma^\mu p_{\beta}\gamma^{\beta} \gamma^\nu) = \text{Tr}(p'_{\alpha}p_{\beta}\gamma^{\alpha}\gamma^\mu \gamma^{\beta} \gamma^\nu) = 4p'_{\alpha}p_{\beta}(g^{\alpha\mu}g^{\beta \nu} - g^{\alpha\beta}g^{\mu\nu}+g^{\alpha\nu}g^{\mu\beta}) = ... [/tex]
     
  14. May 18, 2008 #13
    i didn't know that we can commute p with gamma and put it out of the trace!! would you tell me why we can do this?
     
  15. May 18, 2008 #14

    malawi_glenn

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    p_a is just a number, an element in a 4-vector.

    [tex] p_a\gamma ^a = p_0\gamma ^0 + p_1\gamma ^1 + p_2\gamma ^2 +p_3\gamma ^3 [/tex]

    [tex] p_0 = E [/tex]

    [tex] p_1 = p_x [/tex] etc.
     
    Last edited: May 18, 2008
  16. May 18, 2008 #15
    thanx for help Glenn! i'll give it another shot
     
  17. May 18, 2008 #16

    malawi_glenn

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    np, just knock yourself out hehe, cheers
     
  18. May 20, 2008 #17
    ok, i have another question that should have came earlier
    [tex](p'\!\!\!\!\!/ \ -m)_{da}\gamma^\mu_{ab}(p\!\!\!\!\!/ \ +m)_{bc}\gamma^\nu_{cd}
    =\operatorname{Tr}[(p'\!\!\!\!\!/ \ -m)\gamma^\mu(p\!\!\!\!\!/ \
    +m)\gamma^\nu] [/tex]
    where the trace came from here?
     
    Last edited: May 20, 2008
  19. May 20, 2008 #18

    nrqed

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    Let's say you have three matrices A, B and C. If you are calculating the following sum:

    [tex] A_{cd} B_{de} C_{ec} [/tex]

    this is the same as

    [tex] tr(ABC) [/tex]

    The fact that the last index of the third matrix is the same as the first index of the first matrix means that if you work with the actual matrices, you have to multiply tham and trace the product. That's all there is to it.
     
  20. May 21, 2008 #19
    i'll try to prove it to make sure i got it..
    thanks nrqed!
     
  21. May 22, 2008 #20

    malawi_glenn

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