1. PF Contest - Win "Conquering the Physics GRE" book! Click Here to Enter
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

  1. Feb 22, 2010 #1

    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?

  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).
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook