$$ J^{\mu\nu} = i(x^\mu\partial^\nu-x^\nu\partial^\mu). \tag{3.16}$$ We will soon see that these six operators generate the three boosts and three rotations of the Lorentz group.
To determine the commutation rules of the Lorentz algebra, we can now simply compute the commutators of the differential operators (3.16). The result is $$ [J^{\mu\nu},J^{\rho\sigma}]=i( g^{\nu\rho} J^{\mu\sigma} - g^{\mu\rho} J^{\nu\sigma} - g^{\nu\sigma} J^{\mu\rho} + g^{\mu\sigma} J^{\nu\rho} ). \tag{3.17}$$

This is from p.39 of Peskin&Schroeder's Quantum Field Theory book. It is written that (3.16) operators generate the Lorentz group. So, are the operators of (3.16) themselves Lorentz transformations?
Also, I cannot find a way to derive (3.17) from the definition (3.16). How does the metric $g^{\nu\rho}$ appear? Could anyone please help me?

  • $\begingroup$ How is it related to the appearance of the metric?? $\endgroup$ – Keith Mar 15 '18 at 13:56
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    $\begingroup$ It appears because $\partial^\mu x^\nu = g^{\mu\nu}$. $\endgroup$ – Prahar Jul 9 '18 at 1:07

The metric appears because \begin{equation} \partial^\mu x^\nu = \partial^\mu g^{\rho\nu} x_\rho = g^{\rho\nu} \partial^\mu x_\rho = g^{\rho\nu} \delta^\mu_\rho = g^{\mu\nu} \end{equation} (write out the meaning of $\partial^\mu$ to see why this must be so). Using this relation, it is straightforward to derive equation (3.17) from (3.16).

As to whether the $J^{\mu\nu}$ are Lorentz transformations, ZeroTheHero is entirely correct; using that \begin{equation} \exp(-i\omega_{\mu\nu}J^{\mu\nu}) := \sum_{n=0}^\infty \frac{(-i\omega_{\mu\nu}J^{\mu\nu})^n}{n!} = \lim_{n\to\infty} \left(1 - \frac{i\omega_{\mu\nu}J^{\mu\nu}}{n}\right)^n , \tag{1} \end{equation} (with some restrictions on $\omega_{\mu\nu}, J^{\mu\nu}$) it can be shown that the generators $J^{\mu\nu}$ "generates" a representation of the Lorentz group through repeated multiplication/composition [cf. the last equality in (1)].


Of course the $J^{\mu\nu}$ are not Lorentz transformations, no more than $L_x$ or $L_y$ are rotations. The $J^{\mu\nu}$ are generators of the Lorentz group in the sense that exponential $$ \exp\left(-i \omega_{\mu \nu}J^{\mu\nu}\right) \tag{1} $$ is a Lorentz transformation, just like $\exp[-i(\zeta_xL_x+\zeta_yL_y+\zeta_zL_z]$ is a rotation.

As with the rotation group, where it is possible to factorize a general rotation as a product $\exp[-i\alpha L_z]\exp[-i\beta L_y]\exp[-i \gamma L_z]$, with $\alpha$ a function of $\zeta_x,\zeta_y,\zeta_z$ it is possible to write Eq.(1) in various factored form of course with different parameters as complicated functions of the $\omega_{\mu\nu}$.

  • $\begingroup$ Thank you for your explanation. Could you explain the commutation relations too? $\endgroup$ – Keith Mar 15 '18 at 14:03
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    $\begingroup$ What is there to explain? You just power through the commutators much like you would for angular momentum. $\endgroup$ – ZeroTheHero Mar 15 '18 at 14:03

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