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

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[\tfrac12\omega_{\rho\sigma}J^{\rho\sigma} -\epsilon_\rho P^\rho\right] U^{-1}(\Lambda,a) & = \tfrac12(\Lambda\omega\Lambda^{-1})_{\mu\nu} J^{\mu\nu} \\ & \quad - (\Lambda\epsilon - \Lambda\omega\Lambda^{-1}a)_\mu P^\mu \tag{2.4.7} \end{align} Equating coefficients of $$\omega_{\rho\sigma}$$ and $$\epsilon_\rho$$ on both sides of this equation (and using (2.3.10)) we find \begin{align} U(\Lambda,a) J^{\rho\sigma} U^{-1}(\Lambda,a) & = \Lambda_\mu^{\ \rho}\Lambda_\nu^{\ \sigma} (J^{\mu\nu} - 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

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

• Welcome, by the way! What are the coefficients of ω and ϵ on the LHS of equation (2.4.7)? – Deschele Schilder Apr 20 '19 at 11:10
• In equation (2.3.10),$(\Lambda^{-1})^{\rho}_{\nu} = \Lambda_{\nu}^{\rho} = \eta_{\mu\nu}\eta_{\rho\sigma}\Lambda_{\sigma}^{\nu}$ , shouldn't the first $\Lambda$ and the second ${\Lambda}^{-1}$ have a $\rho$ and $\eta$ as indices instead of a $\rho$ and $\nu$? – Deschele Schilder Apr 20 '19 at 11:36

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 $$(\frac 1 2 \Lambda {\Lambda}^{-1}J-\Lambda {\Lambda}^{-1} a P)$$.