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A Lorentz transformation of field with components

  1. Nov 17, 2016 #1


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    I am not looking for a solution, just a "starting point"/guidance for calculating the expression:
    [itex][M^{\mu \nu} , \phi_a][/itex]
    with [itex]M^{\mu \nu}[/itex] being the angular-momentum operators and [itex]\phi_a[/itex] being the field's component, which happens to transform under Lorentz Transformations:
    [itex]x'^\mu = x^\mu + \delta \omega^{\rho \sigma} X ^\mu_{\rho \sigma}~~ X^{\mu}_{\rho \sigma} = \eta^{\mu}_\rho x_\sigma - \eta^\mu_\sigma x_\rho[/itex]
    [itex]\phi_a'(x') = \phi_a(x) + \frac{1}{2} \delta \omega^{\mu \nu} (\Sigma_{\mu \nu})_a^b \phi_b(x)[/itex]

    I have a tendency of writing [itex]M^{\mu \nu} =i( x^\mu \partial^\nu - x^\nu \partial^\mu )[/itex] but I am not sure that this can help.
    Last edited: Nov 17, 2016
  2. jcsd
  3. Nov 17, 2016 #2


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    In case you want a solution, I believe there's one in Atkinson's textbook, you should check it out.
  4. Nov 17, 2016 #3


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    1) You have a factor of half missing in the Lorentz transformation of the coordinates: [tex]\bar{x}^{\tau} = x^{\tau} + \delta x^{\tau} ,[/tex] [tex]\delta x^{\tau} = \frac{1}{2}\omega_{\mu\nu} (\eta^{\tau \mu} x^{\nu} - \eta^{\tau \nu} x^{\mu}) . \ \ \ \ (1)[/tex]
    2) The commutator [itex][M , \varphi ][/itex] means that you are subjecting the field operator to an infinitesimal Lorentz transformation.
    3) As an operator, [itex]\varphi[/itex] transforms by (infinite dimensional) unitary representation of the Lorentz group [tex]\bar{\varphi}_{a}(x) = e^{-\frac{i}{2}\omega_{\mu\nu}M^{\mu\nu}} \ \varphi_{a}(x) \ e^{\frac{i}{2}\omega_{\mu\nu}M^{\mu\nu}} .[/tex] Expanding this to first order in [itex]\omega[/itex], you obtain [tex]\delta \varphi_{a}(x) = - \frac{i}{2}\omega_{\mu\nu}[M^{\mu\nu}, \varphi_{a}(x)] , \ \ \ \ \ \ \ \ \ (2)[/tex] [tex]\delta \varphi_{a}(x) \equiv \bar{\varphi}_{a}(x) - \varphi_{a}(x) .[/tex]
    4) Since [itex]\varphi_{a}(x)[/itex] is a finite-component field on Minkowski space, Lorentz group mixes its components by (finite-dimensional representation) matrix [itex]D(\omega)[/itex], and transform its argument by [itex]\Lambda^{-1}[/itex] [tex]\bar{\varphi}_{a}(x) = D_{a}{}^{b} (\omega) \varphi_{b}(\Lambda^{-1}x) . \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (3)[/tex] Infinitesimally, we may write [tex]D_{a}{}^{b} (\omega) = \delta_{a}{}^{b} - \frac{i}{2} \omega_{\mu\nu} \left( \Sigma^{\mu\nu}\right)_{a}{}^{b} ,[/tex] [tex]\Lambda^{-1}x = x - \delta x .[/tex] Using these expressions and keeping only the terms linear in [itex]\omega[/itex], Eq(3) becomes [tex]\delta \varphi_{a}(x) = - \delta x^{\tau} \partial_{\tau}\varphi_{a} - \frac{i}{2}\omega_{\mu\nu} \left(\Sigma^{\mu\nu}\right)_{a}{}^{b} \varphi_{b} .[/tex]
    Now, in this expression, substitute (1) and (2) to obtain [tex][M^{\mu\nu} , \varphi_{a}(x)] = i \left(x^{\mu}\partial^{\nu} - x^{\nu}\partial^{\mu}\right) \varphi_{a} + \left(\Sigma^{\mu\nu}\right)_{a}{}^{b} \varphi_{b} .[/tex]
    This expression has no meaning in field theory: on the left hand side, you have [itex]M^{\mu\nu}[/itex] which is an operator constructed out of [itex]\varphi_{a}[/itex] and its conjugate [itex]\pi^{a}[/itex], but the differential "operator" on the RHS, [itex]x^{[\mu} \partial^{\nu]}[/itex] is not an operator in field theory. If the generator [itex]M^{\mu\nu}[/itex] acts on the field [itex]\varphi[/itex] according to [tex][M^{\mu\nu} , \varphi (x)] = i (x^{\mu}\partial^{\nu} - x^{\nu}\partial^{\mu}) \varphi (x) ,[/tex] this means that [itex]\varphi[/itex] is a scalar field, i.e., the spin matrix [itex]\Sigma = 0[/itex].
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