Commutators of Lorentz generators

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  • #1
nrqed
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Main Question or Discussion Point

Equation 2.14 of Srednicki is:

[tex] U(\Lambda)^{-1} M^{\mu \nu} U(\Lambda) = \Lambda^{\mu}_{\,\,\rho} \Lambda^{\nu}_{\, \, \sigma} M^{\rho \sigma} [/tex]

He says that writing [itex] \Lambda = 1 + \delta \omega [/itex], one obtains the usual commutation relation of the [itex] M_{\mu \nu} [/itex]:

[tex] [ M^{\mu \nu},M^{\rho \sigma} ] = i ( g^{\mu \rho} M^{\nu \sigma} - (\mu \leftrightarrow \nu}) - (\rho \leftrightarrow \sigma) [/tex]

I get, starting from the first equation

[tex] [ M^{\mu \nu},M^{\rho \sigma} ] \delta \omega_{\rho \sigma} = i (\eta^\mu_\rho \delta \omega^\nu_\sigma M^{\rho \sigma} + \eta^\nu_\sigma \delta \omega^\mu_\rho M^{\rho \sigma} ) [/tex]

I don't see how to extract the commutator because the omega on the right side contain one of the indices that is not summed over. Its' not clear to me how to proceed. Any suggestion?
 

Answers and Replies

  • #2
StatusX
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I think it should work if you stick in a kronecker delta.
 
  • #3
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Check the right hand side of the equation, you can write


[tex]

\delta\omega^\nu_\sigma = \delta\omega_{\rho\sigma} \eta^{\rho \nu}

[/tex]

Edit:

Also M is anti-symmetric, which means you can write it as

[tex]

M^{uv} = \frac{1}{2} (M^{uv} - M^{vu})

[/tex]
 
Last edited:
  • #4
what level of quantum physics is this? grad?
 
  • #5
nrqed
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Check the right hand side of the equation, you can write


[tex]

\delta\omega^\nu_\sigma = \delta\omega_{\rho\sigma} \eta^{\rho \nu}

[/tex]
Of course! (Slapping myself right now...I should have seen it). Thanks!!
 
  • #6
Fredrik
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what level of quantum physics is this? grad?
This is from a book on quantum field theory. You would normally study two books on quantum mechanics before this.
 

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