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Weinberg's QFT text

  1. Feb 13, 2004 #1
    Hi. I am new to this discussion group, so forgive me if this is not the appropriate forum for this question. A friend of mine and I are going through the first volume of Weinberg's QFT trilogy. My question regards the first problem at the end of chapter two. I think I completely understand the procedure for computing a Lorentz transformed positive definite spin state, but I can't seem to compute the little group element corresponding to the Lorentz transformation given in this problem. The matrix multiplication is very lengthy and I really get caught up in the algebra. Is there an easier way to compute the little group matrix than his formula (W = L^(-1)(lambda*p)*Lambda*L(p))? I would like to hear from someone who has done this problem already and, if possible to see their solution. The little group for mass positive definite spin states is the 3-d rotation group and I know the rotation for this particular problem is about the x1 axis, but the algebra is ridiculous. I have spent a lot of time on this problem and would greatly appreciate some help. Thanks.
    Merius
     
  2. jcsd
  3. Feb 19, 2004 #2

    jeff

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    Science Advisor

    It's not so bad. Since Λ is a boost along the z-axis, it's only non-vanishing elements are

    Λ11 = Λ22
    Λ33 = Λ00
    Λ03 = Λ30.

    Since p is in the y-direction, the only non-vanishing elements of L(p) are

    L11 = L33
    L22
    L02 = L20
    L00.

    One way to start off is by writing the little group element as

    W(Λ,p) = L-1(Λp)Q(p)

    where Q(p) ≡ ΛL(p) and note that the above immediately yields

    Q1ν = Λ11L1ν
    Q2ν = Λ22L2ν
    Q0ν = Λ00L0ν + Λ03L3ν
    Q3ν = Λ30L0ν + Λ33L3ν.

    Noting that Λ11 = Λ22 = L11 = L33 = 1 simplifies things even further. Finally, L-1(Λp) = L(p') in which p'i = -Λiνpν.
     
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