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How do I calculate this Poisson bracket in QED?

  1. Sep 9, 2014 #1
    1. The problem statement, all variables and given/known data
    To calculate a certain Dirac bracket I need to calculate this Poisson bracket (Weinberg QTF 1 p.349 first eq.)
    where F is any functional of matter fields and their conjugates and pi is the conjugate to the vector potential. It should be zero.

    2. Relevant equations
    The Poisson bracket for two functionals is defined as
    $$[U,V]_P=\int d^3x\left[\frac{\delta U}{\delta A^i(\mathbf{x})}\frac{\delta V}{\delta \Pi_i(\mathbf{x})}-\frac{\delta V}{\delta A^i(\mathbf{x})}\frac{\delta U}{\delta \Pi_i(\mathbf{x})}\right]$$
    $$\boldsymbol{\Pi}=\dot{\mathbf{A}}+\nabla A^0$$
    $$\nabla\cdot\boldsymbol{\Pi}=-J^0=\nabla^2 A^0$$

    3. The attempt at a solution
    I tried plugging things into the definition of the Poisson bracket:
    $$[F,\Pi_i(\mathbf{z})]_P=\int d^3x\left[\frac{\delta F}{\delta A^j(\mathbf{x})}\frac{\delta \Pi_i(\mathbf{z})}{\delta \Pi_j(\mathbf{x})}-\frac{\delta \Pi_i(\mathbf{z})}{\delta A^j(\mathbf{x})}\frac{\delta F}{\delta \Pi_j(\mathbf{x})}\right]=\int d^3x\left[\frac{\delta F}{\delta A^j(\mathbf{x})}\delta_{ij}\delta^3(\mathbf{x}-\mathbf{z})-\text{something}\right]$$
    I'm not sure what the something is, but it has to be δF/δA(x) for the PB to vanish. Since these are variational derivatives and not straight partials, I reasoned that from the above definition of ∏, a variation of A causes a variation in ∏ like
    $$\delta \Pi^i(\mathbf{z})=\frac{\partial}{\partial t}\delta A^i(\mathbf{z})$$
    so the variational derivative is
    $$\frac{\delta \Pi_i(\mathbf{z})}{\delta A^j(\mathbf{x})}=\delta_{ij}\frac{\partial}{\partial t}\delta^3(\mathbf{z}-\mathbf{x})$$
    This is zero and not at all what I'm looking for. Integration by parts does not help here either.
    Last edited: Sep 9, 2014
  2. jcsd
  3. Sep 9, 2014 #2
    So I was thinking chain rule. Say I have the variational derivative δF/δA(x) and I expand in terms of ∏ derivatives:
    $$\frac{\delta F}{\delta A^i}=\frac{\delta \Pi_j}{\delta A^i}\frac{\delta F}{\delta \Pi_j}$$
    Of course the sum goes over the ∏s. But in the OP i calculated the variational derivative of ∏ w.r.t. A:
    $$\frac{\delta\Pi_j(\mathbf{x})}{\delta A^i(\mathbf{z})}=\delta_{ji}\frac{\partial}{\partial t}\delta^3(\mathbf{x}-\mathbf{z})$$
    But the delta tensor and function are symmetric w.r.t. index and argument permutation, so the following equality holds (maybe):
    $$\frac{\delta\Pi_j(\mathbf{x})}{\delta A^i(\mathbf{z})}=\frac{\delta \Pi^i(\mathbf{z})}{\delta A^j(\mathbf{x})}$$
    I think this is right.
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