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Photon degreees of freedom

  1. Jan 16, 2012 #1
    I just learned about the quantization of the em field. Then EM field 4-potential has components A^mu, giving you 4 degrees of freedom. That gets reduced to 2 degrees of freedom (2 polarization states), that gets done to impose the Lorenz gauge onto free photons. SO what the hell happens to the so called "scalar" A^0 and "longitudinal" A^3 polarization states. How is it that the EM field has 4 degrees of freedom in one case and 2 in an other.
     
  2. jcsd
  3. Jan 16, 2012 #2
    Only the radiative fields are photons with two degrees of freedom.

    The static fields have the third (radial) degree of freedom. These have to be quantized too but you can't use ordinary photons. So you call them "virtual" photons.
     
  4. Jan 17, 2012 #3
    what about real photons that couple to fermion fields?
     
  5. Jan 17, 2012 #4

    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 in the 'static' Coulomb interaction term for the charge density.

    So imposing A°(x)=0 and G(x)~0 explicitly reduces the number of d.o.f. from 4 to 2, i.e. leaves us with two transversal photons. This approach can even be generalized to QCD and provides a non-perturbative, physical quantization (where by physical I mean that no unphysical degrees of freedom like 4 gluons and Fadeev-Popov ghosts have to be used). The drawback especially in scattering calculations is that the A°=0 gauge is not manifest Lorentz covariant, i.e. covariance is not visible explicitly.

    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.
     
  6. Jan 17, 2012 #5
    oh, so that's how it is.
    Thanks a lot!!!
     
  7. Jan 17, 2012 #6

    tom.stoer

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    ... in A°=0 gauge ;-)

    You should definitly try to understand different approaches to gauge fixing. I'll bet that in more advanced lectures e.g. on QCD you will hardly find this A°=0 approach but Lorentz gauge plus Fadeev-Popov / BRST.
     
  8. Jan 17, 2012 #7
    i haven't take a QFT class yet, but I'm learning ahead. any good books that cover gauge fixing or is it only published in journal form?
     
  9. Jan 17, 2012 #8

    tom.stoer

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    I would have a look at Weinberg, but I do not know which method he is teaching in his textbooks. Ryder discussses Fadeev-Popov in Lorentz gauge, as far as I remember.
     
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