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Static electric field in Quantum Field Theory?

  1. Apr 22, 2012 #1
    In quantum field theory do we "describe" static electric fields with proper combinations of transvers propagating photons? Is that a basis? Is there a basis using the longitudinal and timelike photons to describe static fields?

    Thanks for any help!
     
  2. jcsd
  3. Apr 22, 2012 #2

    tom.stoer

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    The reduction of the degrees of freedom from four to two is related to gauge fixing and looks different based on the gauge condition you use. The most transparent approach (which unfortunately is rarely discussed in lectures and textbooks) is the A°=0 gauge.

    A° is not a dynamical field but a Lagrangian multiplier. This is due to the fact that there is no conjugate momentum for A°, i.e. no time derivative ∂0A0 b/c F°°=0 due to anti-symmetry of Fαβ. Therefore fixing A°=0 is reasonable from the very beginning.

    This leaves us with a constraint generated by the Lagrangian multiplier A°, the so-called Gauss law G(x) ~ 0. This constraint is time-independent, i.e. commutes with the Hamiltonian [H,G(x)]=0 and can be solved for physical states, i.e. not as an operator equation G(x)=0 which would contradict commutation relations, but

    G(x)|phys> = 0.

    The presence of the Gauss law is related to a residual gauge symmetry, i.e. time-independent gauge transformations χ(x), ∂0χ(x) = 0 respecting

    A'°(x) = A°(x) - ∂0χ(x) = 0.

    Solving the Gauss law constraint is equivalent to solving the Poisson equation

    ΔA°(x) = ρ(x)

    in classical electrodynamics resulting

    A°(x) = Δ-1 ρ(x)

    which generates the 'static' Coulomb interaction term for the charge density.

    [tex]V_\text{Coulomb} = e^2 \int_{\mathbb{R}^3 \times \mathbb{R}^3}d^3x\;d^3y\;\frac{\rho(x)\,\rho(y)}{4\pi\,|x-y|}[/tex]

    My recommendatio is always

    Quantum Mechanics of Gauge Fixing
    F. Lenz, H.W.L. Naus, K. Ohta, M. Thies
    Annals of Physics, Volume 233, Issue 1, p. 17-50.
    Abstract: In the framework of the canonical Weyl gauge formulation of QED, the quantum mechanics of gauge fixing is discussed. Redundant quantum mechanical variables are eliminated by means of unitary transformations and Gauss′s law. This results in representations of the Weyl-gauge Hamiltonian which contain only unconstrained variables. As a remnant of the original local gauge invariance global residual symmetries may persist. In order to identify these and to handle infrared problems and related "Gribov ambiguities," it is essential to compactify the configuration space. Coulomb, axial, and light-cone representation of QED are derived. The naive light-cone approach is put into perspective. Finally, the Abelian Higgs model is studied; the unitary gauge representation of this model is derived and implications concerning the symmetry of the Higgs phase are discussed.
     
  4. Apr 22, 2012 #3
    Thank you Tom! So A° is useful in classical electrodynamics but in QED we can just discard A°, ouch %^). I want A° to still be important %^(.

    Didn't Feynman use all four polarizations of the electromagnetic field in his version of QED but that two of the polarizations tended to cancel in some sense?

    Thanks again!
     
    Last edited: Apr 22, 2012
  5. Apr 22, 2012 #4
    In Feynman's book Quantum Electrodynamics, page 126, "The Sum over Four Polarizations"?
     
  6. Apr 22, 2012 #5

    tom.stoer

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    It has not so much to do with classical electrodynamics or quantum electrodynamics but with the Hamiltonian analysis. A° is non-dynamical and if you want to express your theory in terms of dynamical d.o.f. = physical polarizations you must eliminate it.

    It is - but not as a dynamical field.

    There are several ways to quantize gauge theories. In QED the Lorentz gauge condition is sometimes preferred due to its explicit covariance. In this gauge A° is not eliminated and the unphysical d.o.f. freedom cancel in physical amplitudes. In QCD additional d.o.f., so called Fadeev-Popov ghosts are required in the Lorentz gauge to cancel the unphysical d.o.f.

    But afaik in all these gauges there is no low-energy potential term. If you want to calculate scattering cross sections and use perturbation theory this is fine, but for the low-energy regime, bound states etc. physical gauges seem to be more appropriate.
     
  7. Apr 23, 2012 #6
    Tom, thank you for your expertise!
     
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