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Quantization of vector field in the Coulomb gauge

  1. Nov 20, 2011 #1
    I have a technical question and at the time being I can't ask it to a professor. So, I'm here:

    If I try to quantize the vector field in the Coulomb gauge (radiation gauge)

    [tex] A_0(x)=0,\quad \vec\nabla\cdot\vec A=0. [/tex]

    by imposing the equal-time commutation relation

    [tex] [A_i(x),E_j(y)]=-i\delta_{ij}\delta(\vec x-\vec y)[/tex]

    then I should find

    [tex]\partial_i[A_i,E_j]=[\vec\nabla\cdot\vec A,E_j]=0, [/tex]
    since [itex] \vec\nabla\cdot\vec A=0, [/itex] which is inconsistent with [itex] \partial_i\delta_{ij}\delta(\vec x-\vec y)\neq 0[/itex].

    My question is simply how to take this divergence

    [tex]\partial_i[A_i,E_j]=[\vec\nabla\cdot\vec A,E_j] [/tex]

    I'm getting
    [tex] \partial_i[A_i,E_j]=[\vec\nabla\cdot\vec A,E_j]+A_i\partial_i E_j-(\partial_i E_j)A_i .[/tex]
    I must be missing something in the math here. Can anyone help me?
  2. jcsd
  3. Nov 20, 2011 #2


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    I don't remember much of this, but if you can write [itex]\vec E=-\nabla\phi[/itex], then the last two terms cancel each other out.
  4. Nov 20, 2011 #3


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    [tex]\partial^{x}_i[A_i(x),E_j(y)]=[\vec\nabla\cdot\vec A,E_j(y)] [/tex]

    you are not differentiating with respect to y. If you want to avoid confussion just set y = 0.

  5. Nov 20, 2011 #4


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    Kaku's QFT p.110 seems to be addressing your question:

    "If we impose canonical commutation relations, we find a further complication.

    [Ai(x,t), Ej(y,t)] = −iδijδ(x⃗ − y⃗)

    However, this cannot be correct because we can take the divergence of both sides of the equation. The divergence of Ai is zero, so the left-hand side is zero, but the right hand side is not. As a result, we must modify the canonical commutation relations as follows:

    [Ai(x,t), Ej(y,t)] = −iδijδ(x⃗ − y⃗)

    where the right-hand side must be transverse; that is:

    δij = ∫d3k/(2π)3 exp(ik·(x-x') (δij - kikj/k2)

    [In other words, in Coulomb gauge only the transverse part is quantized, so only the transverse part appears in the commutator.]

    EDIT: In other books they make this even more explicit by putting a "transverse part" operator on both A and E on the left hand side.
    Last edited: Nov 20, 2011
  6. Nov 21, 2011 #5
    Thank you all,

    Sam, you solve my puzzle. I just puted [itex]\partial_i [/itex] and forgot that this is a differentiation only over [itex]x[/itex]. Shame on me!
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