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[qft] Srednicki 2.3 Lorentz group generator commutator

  1. Feb 12, 2009 #1
    1. The problem statement, all variables and given/known data
    Verify that (2.16) follows from (2.14). Here [tex]\Lambda[/tex] is a Lorentz transformation matrix, [tex]U[/tex] is a unitary operator, [tex]M[/tex] is a generator of the Lorentz group.

    2. Relevant equations
    2.8: [tex]\delta\omega_{\rho\sigma}=-\delta\omega_{\sigma\rho}[/tex]

    [tex]M^{\mu\nu}=-M^{\nu\mu}[/tex]

    2.14: [tex]U(\Lambda}^{-1})M^{\mu\nu}U(\Lambda})=\Lambda^\mu_\rho\Lambda^\nu_\sigma M^{\rho\sigma}[/tex]

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


    2.12: [tex]U(1+\delta\omega)=1+{i \over 2\hbar}\delta\omega_{\mu\nu}M^{\mu\nu}[/tex]

    3. The attempt at a solution
    I assume that [tex]\Lambda[/tex] is a small transformation, as hinted by Srednicki: [tex]\Lambda=1+\delta\omega[/tex] and rewrite the 2.14:

    [tex](1-{i \over 2\hbar}\delta\omega_{\rho\sigma}M^{\rho\sigma})M^{\mu\nu}(1+{i \over 2\hbar}\delta\omega_{\rho\sigma}M^{\rho\sigma})=
    (\delta^\mu_\rho+\delta\omega^\mu_\rho)(\delta^\nu_\sigma+\delta\omega^\nu_\sigma)M^{\rho\sigma}[/tex].

    Then I cross out the [tex]M^{\mu\nu}[/tex] that come on the both sides, throw out the double-omega pieces and rewrite the omegas in the following manner

    [tex]\delta\omega^\nu_\sigma=g^{\rho\nu}\delta\omega_{\rho\sigma}[/tex]

    I come to

    [tex]
    [M^{\mu\nu},M^{\rho\sigma}]=2i\hbar (g^{\mu\sigma}M^{\rho\nu}-g^{\rho\nu}M^{\mu\sigma})
    [/tex]

    which seems to be incorrect. Where do I make the mistakes, if I do? How to derive the (2.16) without involvement of certain expression of the Lorentz generator?

    Thanks.
     
  2. jcsd
  3. Feb 12, 2009 #2
    you shouldn't just throw away the omega. Take into accout the fact that omega_ab and omega_ba are not independent but related by the relation omega_ab=-omega_ba. i.e. when you consider omega_ab, you shoud also consider omega_ba.
     
  4. Feb 12, 2009 #3
    I throw away the double omega's, that is [tex]\delta\omega^\mu_\rho\delta\omega^\nu_\sigma[/tex]. I think this is reasonable, because this kind of term is second-order (very small).

    I have also come to having [tex]\delta\omega_{\rho\sigma}[/tex] on both sides, before cancelling them out (that is, we are only dealing with asymmetric pieces).

    Or have I misunderstood your comment?
     
  5. Feb 12, 2009 #4
    I guess the expressioni "throw away" in my comment was misleading.
    I'll explain it again.

    What you actually got in the calculation must have been someting like this.
    [tex]\delta\omega_{\rho\sigma}[M^{\mu\nu},M^{\rho\sigma}]=2i\hbar \delta\omega_{\rho\sigma} (g^{\mu\sigma}M^{\rho\nu}-g^{\rho\nu}M^{\mu\sigma})[/tex]

    Now, you want to take away [tex]\delta\omega_{\rho\sigma}[/tex] from the above expression, as the above equation holds for arbitrary [tex]\delta\omega[/tex]

    If each component [tex]\delta\omega_{\rho\sigma}[/tex] is truely arbitrary, you can do that. Yet, in your case, [tex]\delta\omega[/tex] is antisymmetric so that some of its components are related. You should take this fact into account.
     
  6. Feb 13, 2009 #5
    Yes, I have got the equation

    [tex]\delta\omega_{\rho\sigma}[M^{\mu\nu},M^{\rho\sigma}]=2i\hbar \delta\omega_{\rho\sigma} (g^{\mu\sigma}M^{\rho\nu}-g^{\rho\nu}M^{\mu\sigma})[/tex]

    but [tex]M^{\rho\sigma}[/tex] is antisymmetric, too, and I thought this gives us the right to cancel the [tex]\delta\omega_{\rho\sigma}[/tex] 's.

    Thank you for pointing out the mistake, however, I still do not know how I take into account that some of the components are related. Should I try to write a new equation with interchanged indices [tex]\rho\leftrightarrow\sigma[/tex] and try to combine the two? Is there some resource in the net that would give me this kind of elementary understanding about equations with antisymmetric coefficients?
     
  7. Feb 13, 2009 #6
    In the above equation, [tex]\rho [/tex] and [tex]\sigma [/tex] are summed over. Lets expand this summation and look into the 12 and 21 components, as an example.

    [tex]\delta\omega_{\rho\sigma}[M^{\mu\nu},M^{\rho\sigma}]=2i\hbar \delta\omega_{\rho\sigma} (g^{\mu\sigma}M^{\rho\nu}-g^{\rho\nu}M^{\mu\sigma}) [/tex]

    ->

    [tex]... + \delta\omega_{12}[M^{\mu\nu},M^{12}] + \delta\omega_{21}[M^{\mu\nu},M^{21}] +...= ... + 2i\hbar \delta\omega_{12} (g^{\mu 2}M^{1\nu}-g^{1\nu}M^{\mu 2}) + 2i\hbar \delta\omega_{21} (g^{\mu 1}M^{2\nu}-g^{2\nu}M^{\mu 1})+...[/tex]

    First, lets assume every component of [tex]\delta\omega[/tex] is independent. Then, for the above equation to hold for an arbitrary [tex]\delta\omega[/tex], the following equations should be satisfied.

    ...
    [tex][M^{\mu\nu},M^{12}] = 2i\hbar (g^{\mu 2}M^{1\nu}-g^{1\nu}M^{\mu 2})[/tex]
    [tex][M^{\mu\nu},M^{21}] = 2i\hbar (g^{\mu 1}M^{2\nu}-g^{2\nu}M^{\mu 1})[/tex]
    ...

    What if there is a constraint that [tex]\delta\omega_{12} = -\delta\omega_{21} [/tex]?
     
    Last edited: Feb 13, 2009
  8. Feb 13, 2009 #7
    [tex]\delta\omega_{12}[M^{\mu\nu}, M^{12}]-\delta\omega_{12}[M^{\mu\nu}, M^{21}]=2i\hbar\delta\omega_{12}(g^{\mu2}M^{1\nu}-g^{1\nu}M^{\mu2}-g^{\mu 1}M^{2\nu}+g^{2\nu}M^{\mu 1}[/tex]

    that, by crossing out [tex]2\delta\omega_{12}[/tex] and writing rho instead of 1 and sigma instead of 2 is

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

    and this is the same as (2.16). Thank you very much, weejee, that was a really great help :)
     
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