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The commutator of generators of Lorentz transformation and translation is as follow: $$[M^{\mu\nu},P^\sigma]=i(P^\mu\eta^{\nu\sigma}-P^\nu\eta^{\mu\sigma} ).$$ Then from this we usually say that the generators of translations transform as vectors under the Lorentz group.

I don't quit understand this statement, could anyone explain it?

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Notation: I will write a Poincaré transformation as ${x'}^\mu = {\Lambda^\mu}_\nu x^\nu + a^\mu$, the operator representing this transformation on the Hilbert space is $U(\Lambda, a)$. An infinitesimal transformation with ${\Lambda^\mu}_\nu = \delta^\mu_\nu + {\omega^\mu}_\nu$ and $a^\mu = \epsilon^\mu$ can be expanded as $$ U(\delta + \omega, \epsilon) = 1 + \frac{i}{2} \omega_{\rho\sigma} M^{\rho\sigma} - i \epsilon_\rho P^\rho + \cdots \tag{1} $$

Definition: Operators $V^\mu$ transform as a vector under Poincaré transformations iff $$ U(\Lambda, a) V^\rho U^{-1}(\Lambda, a) = {\Lambda_\mu}^\rho V^\mu \tag{2} $$ which I would say makes sense, intuitively.

Theorem: The condition (2) is equivalent to

  • $[M^{\mu\nu}, V^\sigma] = i \left(V^\mu \eta^{\nu\sigma} - V^\nu \eta^{\mu\sigma} \right)$ (what you wrote)
  • and $[P^\mu, V^\sigma] = 0$.

That should answer your question. One direction of the equivalence is easy to show by inserting the expansion (1) into (2) and equation coefficients of $\omega$ and $\epsilon$.
(For the other direction, one could probably take the derivative of (2) with respect to $\omega$ and $\epsilon$ and show that both sides satisfy the same differential equation, but I don't think I've ever seen anyone actually do this proof.)

Notes:

  • I've taken this more or less from Weinberg QFT 1, Chapters 2.3 - 2.4.
  • The situation is pretty similar in quantum mechanics, just with SO(3) instead of the Poincaré group. Link
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  • $\begingroup$ For equation (2), isn't that a vector should transform as $\Lambda^\rho_{\ \mu}V^{\mu}$, instead of $\Lambda^{\ \rho}_{\mu}V^{\mu}$? I mean the positions of the indexs on $\Lambda$ seem not right for me, could you explain again? Thank you! $\endgroup$ – Nahc Jul 6 '15 at 15:34
  • $\begingroup$ You could equivalently write $U^{-1} V^\rho U = {\Lambda^\rho}_\mu V^\mu$. I agree that this would be a bit nicer (I was following Weinberg). $\endgroup$ – Noiralef Jul 6 '15 at 17:04
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A representation of the Lorentz group implies a set of matrices $(S_{\mu\nu})_a{}^b$ (one for each $\mu,\nu$) that satisfy the Lorentz algebra, i.e. satisfies a relation of the form (I am not keeping track of signs) $$ [S_{\mu\nu} , S_{\rho\sigma} ] = i ( \eta_{\mu\rho} S_{\nu\sigma} + \cdots ) $$ What this means is that there is a vector space, with vectors $\psi^a$ so that $$ [ \psi_a , M_{\mu\nu} ] = (S_{\mu\nu})_a{}^b \psi_b $$ A particular representation of the Lorentz algebra is the vector representation, for which $a=\mu$ and $$ (S_{\mu\nu})_\rho{}^\sigma = i (\eta_{\mu\rho} \delta_\nu^\sigma - \delta^\sigma_\mu \eta_{\nu\rho} ) $$ The vectors in this representation are therefore of the form $A_\mu$. Then $$ [ A_\rho , M_{\mu\nu} ] = (S_{\mu\nu})_\rho{}^\sigma A_\sigma = - i (\eta_{\mu\rho} \delta_\nu^\sigma - \delta^\sigma_\mu \eta_{\nu\rho} ) A_\sigma = - i ( \eta_{\mu\rho} A_\nu - \eta_{\nu\rho} A_\mu ) $$ This is precisely the algebra that $M_{\mu\nu}$ satisfies with $P_\mu$. This therefore indicates that $P_\mu$ transforms in the vector representation of the Lorentz algebra (as it should)

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