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Scattering matrix to traces/dirac spinors

  1. May 20, 2009 #1

    Hepth

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    Gold Member

    I've been searching online and in my qft books (im an early phd student) and I can't find a clear explanation. If you have one, or can simply direct me to a page that does please do so.

    For a generic scattering/decay matrix :

    [tex]
    \sum _{polarization}
    \left|M|^2\right.=\sum _{polarization} \bar{u}_i\left(F^*\left[\gamma ^5\right]\right) u_f \bar{u}_f \left(F\left[\gamma ^5\right]\right)u_i
    [/tex]
    Where F[gamma5] is just a linear function of gamma5.

    You simplify by "knowing" that since they're Dirac spinors you can rewrite to a trace of the sums over the u's.

    [tex]\text{Tr}\left[\left(F^*\left[\gamma ^5\right]\right)\sum _{\epsilon } u_f \bar{u}_f \left(F\left[\gamma ^5\right]\right){\sum _{\epsilon } \bar{u}_i}u_i\right]
    [/tex]

    Then using completeness
    [tex]
    {\sum _{\epsilon } \bar{u}_f}u_f\right = p+m
    [/tex]
    [tex]
    {\sum _{\epsilon } \bar{u}_i}u_i\right = p-m
    [/tex]

    or something like that.

    My questions is how to get the first step.
    How are we rearranging the operators and distributing the sums and using the trace.
    Is it that the sum over polarization is a sum over both polarizations:
    [tex]
    \sum _{\epsilon } =\sum _{\epsilon 1} \sum _{\epsilon 2}
    [/tex]

    so they only operate on their spinors. So then the trace must come in when we need to get the ui's next to eachother.

    Anyone know offhand?
     
  2. jcsd
  3. May 20, 2009 #2
    Your remark on the polarizations is correct, as far as I can tell.

    As for the trace, remember that it has a cyclic property: Tr[ABCD] = Tr[BCDA] = Tr[CDAB] etc, hence you can get the ui's next to each other (which is also what you want).
     
  4. May 20, 2009 #3

    Hepth

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    Hmm maybe its just the index technique I'm not familiar with.
    In Peskin and Schroeder, just below equaiton 5.3 they state :
    "Working with the frst half of 5.2 and writing in spinor indices so we can
    freely move the v next to the vbar we have"
    Then they have :

    [tex]
    \sum _{s,s'} \bar{v}_a^{s'} \gamma _{\text{ab}}^{\mu } u_b^s \bar{u}_c^s \gamma _{\text{cd}}^{\nu } v_d^{s'} = (p-m)_{\text{da}}\text{ }\gamma _{\text{ab}}^{\mu } (p+m)_{\text{bc}}\text{ }\gamma _{\text{cd}}^{\mu } = \text{trace}\left[(p-m) \gamma ^{\mu } (p+m) \gamma ^{\mu }\right]
    [/tex]
     
  5. May 20, 2009 #4

    Hepth

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    Right I could see that, but where do we get the trace from anyway? How do we just throw it in? You cant say:
    M*M = Sum[ABCD] = Tr[BCAD]

    I'm starting to reason out that it has to do with suming over the indices.
     
  6. May 20, 2009 #5
    Well you're equation a scalar quantity ( M^2 ) to some expression containing spinors. But such an equation can only make sense if the expression on the right hand side is a scalar quantity as well. So the expression implicitly already contains the trace - it wouldn't make sense otherwise.

    Note furthermore that a trace is indeed nothing more but summing over indices. If you have a multiplication of matrices:
    [tex]M^a_{~b}N^b_{~c}[/tex]
    then the trace is simply summing over the remaining two indices:
    [tex]Tr[MN] = \sum_{a} M^a_{~b}N^b_{~a}[/tex]
    It's a little trickier with the spinors, but it comes down to the same thing.

    The trace is a little neater to work with, notationwise. It automatically ensures that you're working with a scalar quantity.

    (forgive me if I'm stating too many obvious things too you, but it's kinda difficult figuring where the conceptual problem lies)
     
  7. May 20, 2009 #6
    [tex]u\cdot v = \mathrm{Tr}(u^\dagger v) = \mathrm{Tr}(v u^\dagger)[/tex]
     
  8. May 24, 2009 #7
    This is right. The reason it's a trace is because the definition of the trace is the sum of all the diagonal elements, Tr A = Sum A_{ii} (Sorry, I don't know how to get Latex stuff on here!). So, you have the same index at the start and the end of your T-matrix element, that's why it's a trace. I was completely confused about this too. If you've got a copy of Griffiths' Introduction to Elementary Particles to hand, it goes through it really well there (page 245 onwards). I spent ages looking in Peskin and Schroeder, but this is much more simple. :-)

    Samantha
     
  9. May 26, 2009 #8

    nrqed

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    the first index of the first quantity is the same as the last index of the last quantity, and by definition this is the trace of the whole thing.
    For example, for matrices A, B and C,

    [tex] A_{ij} B_{jk} C_{ki} [/tex]

    is simply Tr(ABC)
     
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