I've been confused about the sign conventions used in Weinberg's QFT book for a long time.

Here's my question: The generators $J^{\mu\nu}$ are defined in this book as $$U(1+\omega)=1+\frac{i}{2}\omega_{\mu\nu}J^{\mu\nu}\tag{2.4.3}$$ (on page 59), and in Maggiore's book as $$U(1+\omega)=1-\frac{i}{2}\omega_{\mu\nu}J^{\mu\nu}.$$ The metric conventions are the opposite and the former uses passive transformation while the latter uses active transformation.

If the parameters ${\omega^\mu}_\nu$ are the same in the two cases, then the left hand sides of the two equations should represent opposite transformations. On the other hand, the right hand sides show they are the same. What accounts for the discrepancy?

  • 1
  • $\begingroup$ @AccidentalFourierTransform The metric convention is (-,+,+,+), so the signs do differ. $\endgroup$ – Xavier Mar 22 '16 at 16:10
  • $\begingroup$ @AccidentalFourierTransform Thanks a lot! I've got one last question, see the edits. $\endgroup$ – Xavier Mar 22 '16 at 18:19
  • $\begingroup$ the parameters $\omega$ are not the same in both cases: one is the negative of the other (actively rotating $\pi$ radians is the same as passively rotating $-\pi$ radians). Therefore, both $U,J$ are the same in both conventions, but one $\omega$ is the opposite of the other. $\endgroup$ – AccidentalFourierTransform Mar 22 '16 at 18:38
  • $\begingroup$ @AccidentalFourierTransform Yes, letting $U$s be the same would make the two $\omega$s differ by a minus sign, which, however, is the same as saying that identicle $\omega$s represent opposite $U$s, and both views lead to contradiction. $\endgroup$ – Xavier Mar 23 '16 at 3:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.