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Lorentz Transformation of Scalar Fields

  1. Oct 24, 2008 #1
    1. The problem statement, all variables and given/known data

    Working on an exercise from Srednicki's QFT and something is not clear.

    Show that

    [tex] [\varphi(x), M^{uv}] = \mathcal{L}^{uv} \varphi(x) [/tex]


    [tex] \mathcal{L}^{uv} = \frac{\hbar}{i} (x^u \partial^v - x^v \partial^u )[/tex]

    2. Relevant equations

    (1) [tex] U(\Lambda)^{-1} \varphi(x) U(\Lambda) = \varphi(\Lambda^{-1}x) [/tex]

    (2) [tex] \Lambda = 1 + \delta\omega[/tex]

    where [tex] \delta\omega [/itex] is an infinitesimal, and

    (3) [tex] U(\Lambda) = I + \frac{i}{2\hbar} \delta\omega_{uv} M^{uv} [/tex]

    3. The attempt at a solution

    Got the left side of (1) equal to

    [tex] \varphi(x) + \frac{i}{2\hbar}\delta\omega_{uv}[\varphi(x), M^{uv}] [/tex]

    but not sure what to do with the right side and how to get the desired derivatives. I suspect
    it has something to do with the transformation (1) of its derivative, but so far no luck.

    [tex] U(\Lambda)^{-1} \partial^u \varphi(x) U(\Lambda) = \Lambda^{u}_{ p} \bar{\partial}^p \varphi(\Lambda^{-1}x) [/tex]
    Last edited: Oct 24, 2008
  2. jcsd
  3. Oct 24, 2008 #2
    1) What does Eq. (2) give you for the inverse of \Lambda?

    2) use that expression to Taylor expand the right-hand side of your Eq. (1): that should also give you \phi(x) plus something....
  4. Oct 24, 2008 #3
    Would it be

    [tex]\Lambda^{-1} = 1 - \delta\omega [/tex]

    and taking the second order [itex] O(\delta\omega^2) [/itex] to zero.

    If I were to Taylor that then would get something like [itex] \varphi(0) + \varphi^{,}(0) [/itex]

    but that doesn't seem right
  5. Oct 24, 2008 #4
    The first part is correct, the second isn't (but getting close): check your Taylor expansion: what's \phi(x+\epsilon)?
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