1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Commutation Relations and Unitary Operators

  1. Aug 25, 2007 #1
    I have a problem with deriving another result. Sorry I am new to this field. Please see the attached PDF - everything is there.

    Attached Files:

  2. jcsd
  3. Aug 26, 2007 #2


    User Avatar
    Science Advisor
    Homework Helper

    What you're trying to prove is wrong. Srednicki is wrong. If you make the unitary transformation of the 4-momentum operator the way it should be done, namely

    [tex] \bar{P}^{\mu}(x')=U(\Lambda)P^{\mu}(x)U^{\dagger} (\Lambda) [/tex]

    then you get the correct RHS

    [tex] \bar{P}^{\mu}(x')=U(\Lambda)P^{\mu}(x)PU^{\dagger} (\Lambda)= \Lambda^{\mu}{}_{\nu} P^{\nu}(x) [/tex]

    which leads to the commutation relation

    [tex] \left[P^{\mu}(x),M^{\nu\lambda}(x)\right]_{-}= i\hbar \ g^{\mu[\nu}P^{\lambda]}(x) [/tex]
    Last edited: Aug 26, 2007
  4. Sep 1, 2007 #3


    User Avatar
    Science Advisor

    abode_x: you got
    [tex] {i\over2\hbar}\delta\omega_{\rho\sigma}[P^\mu,M^{\rho\sigma}]
    =\delta\omega^\mu{}_\nu P^\nu [/tex]
    [tex]\delta\omega^\mu{}_\nu P^\nu
    =\delta\omega_{\rho\nu}g^{\rho\mu} P^\nu
    =\delta\omega_{\rho\sigma}g^{\rho\mu} P^\sigma[/tex]
    Next, use that [tex]\delta\omega_{\rho\sigma}[/tex] is antisymmetric, and so we can replace [tex] g^{\rho\mu} P^\sigma [/tex] with its antisymmetric part,
    [tex]{1\over2}(g^{\rho\mu} P^\sigma-g^{\sigma\mu}P^\rho)[/tex]
    This gives
    [tex]{i\over2\hbar}\delta\omega_{\rho\sigma}[P^\mu,M^{\rho\sigma}]={1\over2}\delta\omega_{\rho\sigma}(g^{\rho\mu} P^\sigma-g^{\sigma\mu}P^\rho)[/tex]
    For this to be true for any [tex]\delta\omega_{\rho\sigma}[/tex], it must be true of each coefficient. So,
    ={1\over2}(g^{\rho\mu} P^\sigma-g^{\sigma\mu}P^\rho)[/tex]
    Multiply by [tex]-2i\hbar[/tex] and switch the two terms on the right to get
    [tex] [P^\mu,M^{\rho\sigma}] =i\hbar(g^{\sigma\mu}P^\rho-g^{\rho\mu} P^\sigma)[/tex]
    Using [tex]g^{\sigma\mu}=g^{\mu\sigma}[/tex], this matches Srednicki's eq.(2.18).
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?