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QED spin matrix

  1. Sep 21, 2014 #1
    Hello everyone, I have a problem with deriving following equality:
    u^{s}(p) \bar{u}^{s}(p) = 1/2 ((\slashed{p} + m)(1+\gamma^5 \slashed{s}))

    where s is spin 4-vector. I know how to calculate this tensor product when there is spin sum in front of it, but without the sum, I am clueless. Can someone help me, please? Thank you.
  2. jcsd
  3. Sep 22, 2014 #2


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    This identity cannot be true if there is no sum over the "s".
  4. Sep 22, 2014 #3
    No, it is true only without the sum over s. Now I see there might have appeared small misconception: the "s" on the left side is not the same "s" as on the right side of the equality. On the right side, the "s" is four vector.
    For example, in particle's rest frame,

    s = (0, \vec{s})

    where the 3-vector is a unit vector in the direction of the spin of a particle.
    With the sum, we would have the completeness relation

    \sum_{s} u^{s}(p) \bar{s}^s (p) = \slashed p + m
  5. Sep 22, 2014 #4


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    Where did you find it? How can it be true if you donot sum over spin degrees of freedom? Calling it "s" or "r", it does not matter. As well as the completeness relation
    [tex]\sum_{s} u^{ s }( p ) \bar{ u }^{s} ( p ) = p_{ \mu } \gamma^{ \mu } + m ,[/tex]
    you also have the followings
    [tex]\frac{1}{2} ( 1 + \gamma_{5} \gamma^{ \mu } s_{ \mu } ) u^{ s } ( p ) = u^{ s } ( p ) ,[/tex]
    [tex]\frac{1}{2} \bar{ u }^{ s } ( p ) ( 1 - s_{ \mu } \gamma^{ \mu } \gamma_{ 5 } ) = \bar{ u }^{ s } ( p ) .[/tex]
  6. Sep 22, 2014 #5
    I think you can find the answer in Bjorken&Drell's famous book "Relativistic Quantum Mechanics", I remenber it's in the first few chapters.
  7. Sep 23, 2014 #6
    Thank you, there is not exactly what I needed, but it helped me to understand at least part of it.
    I still don't see how to get from those projection operators from Bjorken&Drell to my expression.
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