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How is this derivation of a field transformation, in Weinberg's QFT book, performed?

Physics Asked by Aleksandrochka on December 21, 2020

I am reading Weinberg’s book Quantum theory of fields.

Could you explain to me the following things? Vol.1, page 60 (transcribed from this image):

To first order in $omega$ and $epsilon$, we have then
begin{align}
U(Lambda,a) left[tfrac12omega_{rhosigma}J^{rhosigma} -epsilon_rho P^rhoright] U^{-1}(Lambda,a)
& =
tfrac12(LambdaomegaLambda^{-1})_{munu} J^{munu}
& quad – (Lambdaepsilon – LambdaomegaLambda^{-1}a)_mu P^mu
tag{2.4.7}
end{align}

Equating coefficients of $omega_{rhosigma}$ and $epsilon_rho$ on both sides of this equation (and using (2.3.10)) we find
begin{align}
U(Lambda,a) J^{rhosigma} U^{-1}(Lambda,a) & = Lambda_mu^{ rho}Lambda_nu^{ sigma} (J^{munu} – a^mu P^nu + a^nu P^mu)
tag{2.4.8}
U(Lambda,a) P^rho U^{-1}(Lambda,a) & = Lambda_mu^{ rho} P^mu
tag{2.4.9}
end{align}

How are we equating the coefficients? How we find the formula (2.4.8)?

The previous formula that the chapter refers to is

$$ (Lambda^{-1})^{rho}_{nu} = Lambda_{nu}^{rho} = eta_{munu}eta_{rhosigma}Lambda_{sigma}^{nu} tag{2.3.10}$$

One Answer

Equation (2.4.9) follows easy:

The coefficient of $eta$ on the LHS of (2.4.7) is $-U P U^{-1}$. And on the RHS it is $-Lambda P$ (here we don't have to worry about the indices). Set these two equal and (2.4.9) follows.

The coefficient of $omega$ on the LHS of (2.4.7) is $U J U^{-1}$. The coefficient on the RHS is $(frac 1 2 Lambda {Lambda}^{-1}J-Lambda {Lambda}^{-1} a P)$.

I'll leave it to you to put in the indices after which you can use (2.4.10) and obtain (2.4.8), though I have the feeling that this is where you are stuck. Try harder! I'm not supposed to give you the full answer.

Answered by Deschele Schilder on December 21, 2020

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