Note on Weinberg Section 2.3

This is a note I made while working my way through Section 2.3 of Weinberg's QFT book. He skips some steps (which are of course, trivial).

We want to show that equating coefficients of $\omega_{\rho\sigma}$ and $\epsilon_{\rho}$ on both sides of

(1)
\begin{align} U(\Lambda, a)\left[\frac{1}{2}\omega_{\rho\sigma}J^{\rho\sigma}-\epsilon_{\rho}P^{\rho}\right]U^{-1}(\Lambda, a) = \frac{1}{2}(\Lambda \omega\Lambda^{-1})_{\mu\nu}J^{\mu\nu} - (\Lambda\epsilon - \Lambda\omega\Lambda^{-1}a)_{\mu}P^{\mu} \end{align}

gives

(2)
\begin{align} U(\Lambda, a)J^{\rho\sigma}U^{-1}(\Lambda, a) = \Lambda_{\mu}^{\rho}\Lambda_{\nu}^{\rho}(J^{\mu\nu} - a^{\mu}P^{\nu} + a^{\nu}P^{\mu}) \end{align}
(3)
\begin{align} U(\Lambda, a)P^{\rho}U^{-1}(\Lambda, a) = \Lambda_{\mu}^{\rho}P^{\mu} \end{align}

Proving (2) is rather easy. The key is to split $(\Lambda\omega\Lambda^{-1})_{\mu\nu}$ cleverly as

(4)
\begin{align} (\Lambda\omega\Lambda^{-1})_{\mu\nu} = {(\Lambda\omega)_{\mu}}^{\rho}(\Lambda^{-1})_{\rho\nu} \end{align}

Now,

(5)
\begin{split} (\Lambda\omega\Lambda^{-1})_{\mu\nu} &= {(\Lambda\omega)_{\mu}}^{\rho}(\Lambda^{-1})_{\rho\nu}\\ &= \eta_{\rho\sigma}{(\Lambda\omega)_{\mu}}^{\rho}({\Lambda^{-1})^{\sigma}}_{\nu}\\ &= \eta_{\rho\sigma}{(\Lambda\omega)_{\mu}}^{\rho}{\Lambda_\nu}^{\sigma}\\ &= \eta_{\rho\sigma}{\Lambda_{\mu}}^{\alpha}{\omega_{\alpha}}^{\rho}{\Lambda_\nu}^{\sigma}\\ &= \eta_{\rho\sigma}\eta^{\rho\beta}{\Lambda_{\mu}}^{\alpha}\omega_{\alpha\beta}{\Lambda_\nu}^{\sigma}\\ &= {\delta_{\sigma}}^{\beta}{\Lambda_{\mu}}^{\alpha}{\Lambda_{\nu}}^{\sigma}\omega_{\alpha\beta}\\ &= \omega_{\rho\sigma}{\Lambda_{\mu}}^{\rho}{\Lambda_{\nu}}^{\sigma}\\ \end{split}

So, using (1), one gets

(6)
\begin{align} \begin{split} U(\Lambda, a)\frac{1}{2}\omega_{\rho\sigma}J^{\rho\sigma}U^{-1}(\Lambda, a) &= \omega_{\rho\sigma}\frac{1}{2}{\Lambda_\mu}^{\rho}{\Lambda_{\nu}}^{\sigma}J^{\mu\nu} + \omega_{\rho\sigma}{\Lambda_\mu}^{\rho}{\Lambda_\nu}^{\sigma}a^{\nu}P^{\mu}\\ &= \omega_{\rho\sigma}\frac{1}{2}{\Lambda_\mu}^{\rho}{\Lambda_{\nu}}^{\sigma}J^{\mu\nu} + \frac{1}{2}(\omega_{\rho\sigma}-\omega_{\sigma\rho}){\Lambda_\mu}^{\rho}{\Lambda_\nu}^{\sigma}a^{\nu}P^{\mu}\\ &= \omega_{\rho\sigma}\frac{1}{2}{\Lambda_\mu}^{\rho}{\Lambda_{\nu}}^{\sigma}J^{\mu\nu} + \frac{1}{2}\omega_{\rho\sigma}{\Lambda_\mu}^{\rho}{\Lambda_\nu}^{\sigma}(-a^{\mu}P^{\nu} + a^{\nu}P^{\mu}) \end{eplit} \end{align}

where we have used antisymmetry of $\omega_{\rho\sigma}$.

As for (3), well that is truly trivial since

(7)
\begin{align} (\Lambda\epsilon)_{\mu} = {\Lambda_{\mu}}^{\rho}\epsilon_{\rho} \end{align}

and a straight comparison yields the desired result.

## The Lie Algebra of the Poincare Group

In this section, we will see that equations (2) and (3) are quite potent, since they help us recover (or if you will, uncover) the Lie Algebra of the Poincare group. By Lie Algebra here, I mean all the commutator rules connecting the set of $J^{\mu\nu}$'s to the set of $P^{\mu}$'s, which completely characterize how these operators act on group elements.

Specifically, for the infinitesimal transformation $U(1+\omega,\epsilon)$ we have the Taylor expansion,

(8)
\begin{align} U(1+\omega,\epsilon) = 1 + \frac{i}{2}\omega_{\mu\nu}J^{\mu\nu} - i\epsilon_{\mu}P^{\mu} \end{align}

Now, letting $\Lambda$ and $a$ in (2) be infinitesimal, i.e. ${\Lambda^{\mu}}_{\nu} = {\delta^{\mu}}_{\nu} + {\omega^{\mu}}_{\nu}$ and $a^\mu = \epsilon^\mu$ where ${\omega^{\mu}}_{\nu}$ and $\epsilon^\mu$ are infinitesimal quantities, the LHS of (2) becomes

(9)
\begin{split} U(\Lambda,a)J^{\rho\sigma}U^{-1}(\Lambda, a) &= \left(1 + \frac{i}{2}\omega_{\rho\sigma}J^{\rho\sigma} - i\epsilon_{\rho}P^{\rho}\right)J^{\rho\sigma}\left(1-\frac{i}{2}\omega_{\rho\sigma}J^{\rho\sigma} + i\epsilon_{\rho}P^{\rho}\right)\\ &= \left(J^{\rho\sigma} + \frac{i}{2}\omega_{\mu\nu}J^{\mu\nu}J^{\rho\sigma} - i\epsilon_{\alpha}P^{\alpha}J^{\rho\alpha}\right)\left(1-\frac{i}{2}\omega_{\mu'\nu'}J^{\mu'\nu'} + i\epsilon_{\alpha'}P^{\alpha'}\right)\\ &= J^{\rho\sigma} + \frac{i}{2}\omega_{\mu\nu}J^{\mu\nu}J^{\rho\sigma} -\frac{i}{2}\omega_{\mu\nu}J^{\rho\sigma}J^{\mu\nu} + i\epsilon_{\alpha}J^{\rho\sigma}P^{\alpha} - i\epsilon_{\alpha}P^{\alpha}J^{\rho\sigma} \end{split}

The RHS of (2) becomes

(10)
\begin{split} {\Lambda_{\mu}}^{\rho}{\Lambda_{\nu}}^{\sigma}(J^{\mu\nu}-a^{\mu}P^{\nu}+a^{\nu}P^{\mu}) &= ({\delta_{\mu}}^{\rho} + {\omega_{\mu}}^{\rho})({\delta_{\nu}}^{\sigma} + {\omega_{\nu}}^{\sigma}(J^{\mu\nu}-\epsilon^{\mu}P^{\nu}+\epsilon^{\nu}P^{\mu})\\ &= ({\delta_{\mu}}^{\rho}{\delta_{\nu}}^{\sigma} + {\delta_{\mu}}^{\rho}{\omega_{\nu}}^{\sigma} + {\delta_{\nu}}^{\sigma}{\omega_{\mu}}^{\rho})(J^{\mu\nu}-\epsilon^{\mu}P^{\nu}+\epsilon^{\nu}P^{\mu})\\ &= J^{\rho\sigma} + {\omega_{\mu}}^{\rho}J^{\mu\sigma} + {\omega_{\nu}}^{\sigma}J^{\rho\nu} - \epsilon^{\rho}P^{\sigma} + \epsilon^{\sigma}P^{\rho} \end{split}

Equating the LHS and RHS and rearranging terms, we get

(11)
\begin{align} i\left[\frac{1}{2}\omega_{\mu\nu}J^{\mu\nu}-\epsilon_{\mu}P^{\mu}, J^{\rho\sigma}\right] = {\omega_{\mu}}^{\rho}J^{\mu\sigma} + {\omega_{\nu}}^{\sigma}J^{\rho\nu} - \epsilon^{\rho}P^{\sigma} + \epsilon^{\sigma}P^{\rho} \end{align}

Now, let us do the same thing on both sides of (3). The LHS becomes

(12)
\begin{split} U(\Lambda,a)P^{\rho}U^{-1}(\Lambda, a) &= \left(1 + \frac{i}{2}\omega_{\mu\nu}J^{\mu\nu} - i\epsilon_{\alpha}P^{\alpha}\right)P^{\rho}\left(1 - \frac{i}{2}\omega_{\mu'\nu'}J^{\mu'\nu'} + i\epsilon_{\alpha'}P^{\alpha'}\right)\\ &= \left(P^{\rho} + \frac{i}{2}\omega_{\mu\nu}J^{\mu\nu}P^{\rho} - i\epsilon_{\alpha}P^{\alpha}P^{\rho}\right)\left(1-\frac{i}{2}\omega_{\mu'\nu'}J^{\mu'\nu'} + i\epsilon_{\alpha'}P^{\alpha'}\right)\\ &= P^{\rho} + \frac{i}{2}\omega_{\mu\nu}J^{\mu\nu}P^{\rho} - i\epsilon_{\alpha}P^{\alpha}P^{\rho} - \frac{i}{2}\omega_{\mu'\nu'}P^{\rho}J^{\mu'\nu'} + i\epsilon_{\alpha'}P^{\rho}P^{\alpha'} \end{split}

whereas the RHS becomes

(13)
\begin{split} {\Lambda_{\mu}}^{\rho}P^{\mu} &= ({\delta_{\mu}}^{\rho} + {\omega_{\mu}}^{\rho})P^{\mu}\\ &= P^{\rho} + {\omega_{\mu}}^{\rho}P^{\mu} \end{split}

Equating the LHS and RHS gives

(14)
\begin{align} i\left[\frac{1}{2}\omega_{\mu\nu}J^{\mu\nu}-\epsilon_{\mu}P^{\mu},P^{\rho}\right] = {\omega_{\mu}}^{\rho}P^{\mu} \end{align}

### March 08, 2010; Under construction

Equating the coefficients of $\omega_{\mu\nu}$ on both sides of equations (11) and (12) is supposed to give us the desired commutator relations. (EXPAND)

Rewriting the LHS of (11), we have

(15)
\begin{split} i\left\{\left(\frac{1}{2}\omega_{\mu\nu}J^{\mu\nu}-\epsilon_{\mu}P^{\mu}\right)J^{\rho\sigma}-J^{\rho\sigma}\left(\frac{1}{2}\omega_{\mu\nu}J^{\mu\nu}-\epsilon_{\mu}P^{\mu}\right)\right\} \\= {\omega_{\mu}}^{\rho}J^{\mu\sigma} + {\omega_{\nu}}^{\sigma}J^{\rho\nu} - \epsilon^{\rho}P^{\sigma} + \epsilon^{\sigma}P^{\rho}\\ = \eta^{\rho\nu}\omega_{\mu\nu}J^{\mu\sigma}+\eta^{\nu\sigma}\omega_{\nu\mu}J^{\rho\nu}-\epsilon^{\rho}P^{\sigma}+\epsilon^{\sigma}P^{\rho}\\ = \eta^{\nu\rho}\omega_{\mu\nu}J^{\mu\sigma}-\eta^{\nu\sigma}\omega_{\mu\nu}J^{\rho\nu}-\epsilon^{\rho}P^{\sigma}+\epsilon^{\sigma}P^{\rho} \end{split}

Equating coefficients should give (EXPAND and CORRECT as necessary - Mar 08),

(16)
\begin{align} i\frac{1}{2}[J^{\mu\nu}, J^{\rho\sigma}] = \eta^{\nu\rho}J^{\mu\sigma}-\eta^{\nu\sigma}J^{\rho\nu} \end{align}

What about the other two terms? Does the factor of 2 produce them? (ED: CHECK!)

page revision: 21, last edited: 08 Mar 2010 03:32