# Weinberg's QFT text

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

jeff
Originally posted by merius
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))?

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

&Lambda;11 = &Lambda;22
&Lambda;33 = &Lambda;00
&Lambda;03 = &Lambda;30.

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(&Lambda;,p) = L-1(&Lambda;p)Q(p)

where Q(p) &equiv; &Lambda;L(p) and note that the above immediately yields

Q1&nu; = &Lambda;11L1&nu;
Q2&nu; = &Lambda;22L2&nu;
Q0&nu; = &Lambda;00L0&nu; + &Lambda;03L3&nu;
Q3&nu; = &Lambda;30L0&nu; + &Lambda;33L3&nu;.

Noting that &Lambda;11 = &Lambda;22 = L11 = L33 = 1 simplifies things even further. Finally, L-1(&Lambda;p) = L(p') in which p'i = -&Lambda;i&nu;p&nu;.