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Homework Help: How to derive Eq. (3.110) in Peskin's book?

  1. Feb 22, 2010 #1
    Hello,

    I'm having trouble to derive Eq. (3.110) in Peskin's QFT book. In the book, it's said to use
    [tex]u(\Lambda^{-1} \tilde{p})=\Lambda^{-1}_{1/2} u(\tilde{p})[/tex]
    But I ran into trouble to derive this. Here is my try:

    The equation is equivalent to
    [tex]u(p)=\Lambda^{-1}_{1/2} u(\Lambda p)[/tex]
    To make things simpler, I consider only a boost in axis-1 and a rotation in axis-3. In this way, the Lorentz transform on the momentum is given by Eq. 3.20 and 3.21 in the book, i.e.,
    [tex] \tilde{p}^0 = p^0+\beta p^1 [/tex]
    [tex]\tilde{p}^1 = p^1+\beta p^0 -\theta p^2 [/tex]
    [tex]\tilde{p}^2 = p^2+\theta p^1 [/tex]
    [tex]\tilde{p}^3 = p^3 [/tex]
    Using Eq.3.37 in the book, and working on the left-hand portion of the spinnor
    [tex] \left[ \Lambda^{-1}_{1/2} u(\tilde{p}) \right]_L =(1+i\theta\cdot \sigma/2+\beta\cdot \sigma/2) \sqrt{\tilde{p}\cdot \sigma} \xi [/tex]
    [tex]=\sqrt{(1+i\theta \sigma^3+\beta \sigma^1)\left[ (p^0+\beta p^1)-(p^1+\beta p^0 -\theta p^2)\sigma^1-(p^2+\theta p^1)\sigma^2 - p^3\sigma^3 \right]} \xi [/tex]
    Keeping the 1st order terms, I have
    [tex] \left[ \Lambda^{-1}_{1/2} u(\tilde{p}) \right]_L = \sqrt{p\cdot \sigma +i\theta (p^0\sigma^3-p^3) -i\beta(p^2\sigma^3-p^3\sigma^2)} \xi [/tex]
    This is a little different from the left-hand portion of spinnor [tex] u(p) [/tex].

    Could someone help me find out what's wrong with the above derivation?

    Thanks.
     
  2. jcsd
  3. Jun 27, 2012 #2
    Not sure I can help you, but the same law was used before while building solutions of Dirac equation in order to get u(P) from u(P-rest frame).
     
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