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Homework Help: Casimir's trick, with a stupid question

  1. Mar 23, 2012 #1
    1. The problem statement, all variables and given/known data
    There's a similar discussion on the forum at https://www.physicsforums.com/showthread.php?t=162812, but I'm not sure it's correct, and it doesn't entirely apply to my problem.

    From QED/QFT, we use Casimir's trick when calculating the S-matrix to find
    [tex] \sum_{s=1}^2u^{(s)}_{ \alpha}( \mathbf{p}) \bar u^{(s)}_{ \beta}( \mathbf{p}) = \bigg( \frac{-i \gamma \cdot p+m}{2m} \bigg)_{ \alpha \beta} [/tex]
    [tex]\begin{multline} \frac{1}{2} \sum_s \sum_{s'} \lvert \bar u' \gamma_4 u \lvert^2= \frac{1}{2} \sum_s \sum_{s'} \bar u^{(s)}_{ \delta}( \mathbf{p}) ( \gamma_4)_{ \delta \gamma} u^{(s')}_{ \gamma}( \mathbf{p}') \bar u^{(s')}_{ \beta}( \mathbf{p}') ( \gamma_4)_{ \beta \alpha} u^{(s)}_{ \alpha}( \mathbf{p})
    \\~ = \frac{1}{2}( \gamma_4)_{ \delta \gamma} \bigg( \frac{-i \gamma \cdot p+m}{2m} \bigg)_{ \gamma \beta}( \gamma_4)_{ \beta \alpha} \bigg( \frac{-i \gamma \cdot p+m}{2m} \bigg)_{ \alpha \delta}= \frac{1}{2} \text{Tr} \bigg[ \gamma_4 \bigg( \frac{-i \gamma \cdot p' +m}{2m} \bigg) \gamma_4 \bigg( \frac{-i \gamma \cdot p -m}{2m} \bigg) \bigg]\end{multline}[/tex]

    I'm using old notation, so in most modern texts you would probably do something more aligned with Griffiths, http://scienceworld.wolfram.com/physics/CasimirTrick.html.

    My problem involves a slight variation, and I'm not sure how to deal with it. Instead of
    [tex] \lvert \bar u' \gamma_4 u \lvert , [/tex]
    I've reached a point where I'm stuck with
    [tex] \lvert \bar u' u \lvert [/tex]

    3. The attempt at a solution
    Well, I still have
    [tex] \sum_{s=1}^2u^{(s)}_{ \alpha}( \mathbf{p}) \bar u^{(s)}_{ \beta}( \mathbf{p}) = \bigg( \frac{-i \gamma \cdot p+m}{2m} \bigg)_{ \alpha \beta} [/tex]
    But my sticking point is
    [tex] \frac{1}{2} \sum_s \sum_{s'} \lvert \bar u' u \lvert^2= \frac{1}{2} \sum_s \sum_{s'} \bar u^{(s)}_{ \delta}( \mathbf{p}) u^{(s')}_{ \gamma}( \mathbf{p}') \bar u^{(s')}_{ \beta}( \mathbf{p}') u^{(s)}_{ \alpha}( \mathbf{p})= \frac{1}{2} \bigg( \frac{-i \gamma \cdot p+m}{2m} \bigg)_{ \gamma \beta} \bigg( \frac{-i \gamma \cdot p+m}{2m} \bigg)_{ \alpha \delta}[/tex]

    The sad thing is that I think my situation is actually easier to deal with. Any insight would be appreciated.
    Last edited: Mar 23, 2012
  2. jcsd
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