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A Maxwell field commutation relations

  1. Nov 2, 2016 #1
    Maxwell field commutation relations

    I'm looking at Aitchison and Hey's QFT book. I see in Chapter 7, (pp. 191-192), they write down the canonical momentum for the Maxwell field [itex]A^\mu(x)[/itex]:

    [tex]
    \pi^0=\partial_\mu A^\mu \\
    \pi^i=-\dot{A}^i+\partial^i A^0
    [/tex]

    and then write down the commutation relations

    [tex]
    [\hat{A}_\mu(\boldsymbol{x},t),\hat{\pi}_\nu(\boldsymbol{y},t)]=ig_{\mu\nu}\delta^3(\boldsymbol{x}-\boldsymbol{y})
    [/tex]

    and state that if you assume the commutation relations

    [tex]
    [\hat{A}_\mu(\boldsymbol{x},t),\hat{A}_\nu(\boldsymbol{y},t)]=[\hat{\pi}_\mu(\boldsymbol{x},t),\hat{\pi}_\nu(\boldsymbol{y},t)]=0
    [/tex]

    we see that the spatial derivatives of the [itex]\hat{A}[/itex]'s commute with the [itex]\hat{A}[/itex]'s, and with each other, at equal times.

    They state it as if it's obvious, so maybe I'm missing something, but I don't see why the spatial derivatives of the [itex]\hat{A}[/itex]'s commute with the [itex]\hat{A}[/itex]'s, and with each other.
     
  2. jcsd
  3. Nov 3, 2016 #2

    vanhees71

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    You just take the derivatives with respect to spatial components ##\vec{x}## or ##\vec{y}##, and you see that the equatl-time commutators stay 0. It's different for the time derivative, because here you have the same time argument in both entries of the commutator. It's the specialty of time vs. space coordinates in the Hamiltonian formalism, which is never manifestly Lorentz covariant!
     
  4. Nov 5, 2016 #3
    Ah, I see. The spatial arguments of the [itex]A_\mu[/itex] and [itex]A_\nu[/itex] are different, of course... I seem to have overlooked that. Thanks!
     
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