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On the use of Gauge Theories

  1. Jul 15, 2012 #1
    Once Ive asked here why you physicist use Gauge theories with such confidence and the overall answer was "because it works". This probably is true but perhaps is also a bit disappointing to me because I was looking something more fundamental. Ive recently thought of something that may be the answer I was looking but I want to share it with you because your opinions / corrections are always very helpful:

    Noether Theorem says that for every symmetry there is a global conserved charge. When the symmetry is local (the parameter varies from spacetime to spacetime) is like having an infinite number of symmetries (one for each point of spacetime) and that says that there is an infinite number of conserved charges. (one for each point of spacetime). This (with a bit of imagination or a bit of math) can be interpretated as if there is only one conserved charge but that this conservations applies not only globally but also locally in every point of spacetime.

    So if we want a Lagrangian to describe a local conservation of charge, then, the only way to do it is by using a gauge theory.

    Am I too wrong? Cant we use this idea to give a theoretical reason on using Gauge Theories and to explain why theyve been so successful?

    Thanks in advance for your answers.
  2. jcsd
  3. Jul 15, 2012 #2


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    You are right, there is an infinite number of conserved charges. You understand this quite easily in (Q)ED, in QCD it's much harder due to the algebra, but conceoptually it's rather similar. We start with Gauss law

    [tex]\nabla E - \rho = 0[/tex]

    This is a local equation of constraint, i.e. no time derivative. In addition it holds for all times, i.e.

    [tex]\partial_t(\nabla E - \rho) = 0[/tex]

    as well.

    Now you can look at the canonical formulation of QED - which is not common knowledge b/c many textbooks do path integrals today.

    The ideas is simple: The A° component is an unphysical d.o.f. b/c there is no time derivative in the Lagrangian (due to the anti-symm. of the field strength tensor). Therefore A° acts as Lagrangian multiplier. You can see this by integrationg by parts, i.e.

    [tex]f\,\partial_i A_0 \to -A_0 \partial_i f[/tex]

    where f are all terms multiplied by A° in the Lagrangian. Doing this you find that f is nothing else but the Gauss law, i.e. collection all A° terms in the Lagrangian you get

    [tex]\mathcal{L}_{A_0} \sim A_0(\nabla E - \rho)[/tex]

    That means that you will get the Gauss law from the Euler-Lagrange equations as follows:

    [tex]\frac{\delta S}{\delta (\partial_0 A_0)} = 0[/tex]

    b/c there is no such time derivative; therefore

    [tex]\frac{\delta S}{\delta A_0} = 0[/tex]

    which results in the Gauss-law.

    After having derived this equation one can chose the A°=0 gauge which eliminates A° from the theory and one can derive the canonical conjugate variables to the vector potential Ai which is nothing else but the electric field Ei. As a last step one can derive the Hamiltonian in terms of these d.o.f. Ai and Ei where the magnetic field Bi is expressed in terms of Ai.

    The details of the Hamiltonian do not matter, but one should check first that the Gauss law is conserved:

    [tex][H,\nabla E - \rho] = 0[/tex]

    So the infinitly many conserved "local charges" are just the "Gauss laws" at each spacetime point.

    One should not that in QED in canonical quantization the equation

    [tex]\nabla E - \rho = 0[/tex]

    does not hold as an operator equation; it holds as an equation acting on physical states, i.e. on the physical Hilbert space. That means we have two equations

    [tex](\nabla E - \rho)|\text{phys.}\rangle = 0[/tex]
    [tex][H,\nabla E - \rho] = 0[/tex]

    Which says that the "charges" defined by the Gauss law vanishes and that the Gauss law itself is a constant of motion. It is exactly this local operator which distinguishes a gauge theory from a theory with a global symmetry.

    In QCD the nabla is replaced by the covariant derivative and the charge density has a quark and a gluon part, therefore the math is much more complex, but you end up with a SU(3) version of this equation where you have 8 local Gauss law constraints.

    To understand the details you may have a look at

    Quantum Mechanics of Gauge Fixing
    F. Lenz, H.W.L. Naus, K. Ohta, M. Thies
    Univ Erlangen Nurnberg, Inst Theoret Phys, Staudtstr 7, D 91058 Erlangen, Germany and Univ Tokyo, Inst Phys, Tokyo 153, Japan
    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.
    Last edited: Jul 15, 2012
  4. Jul 15, 2012 #3


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    Here is the complete reference to this really very nice paper on gauge fixing and quantization for Abelian Gauge Theories (QED, Abelian Higgs model):

    F. Lenz, H.W.L. Naus, K. Ohta, M. Thies, Annals of Physics 233, 17–50 (1994)
    http://dx.doi.org/10.1006/aphy.1994.1059 [Broken]
    Last edited by a moderator: May 6, 2017
  5. Jul 15, 2012 #4


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    ... but I am afraid that all what I have explained results in nothing else but a reformulation; the question "why gauge theory?" is replaced by "why local conservation laws?" or "why local symmetries" and the answer is still "because it works".

    I don't think that we will ever be able to answer these "why-questions"; that does not mean that they are useless - they express our inquisitiveness and therefore serve as a driving force of research; but in the end always result in new "why-questions".
  6. Jul 15, 2012 #5
    Thanks Tom and Vanshees, awesome answers. Very enlightening! I will take a look around that paper and if there is some difficulty I will come here again,

    Yes, you are right, mathematically. Its obvious that no matter how beautiful some law is, if it doesnt work it doesnt work. Nevertheless, I think that they are not useless because, as you said, drive our force of research (well, "yours" more than "our"). And to me, this answers give order to my ideas and understanding of what is going on.
  7. Jul 15, 2012 #6


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    Remark: you can introduce arbitary globally conserved charges as follows: define

    [tex]G(x) = \nabla E - \rho [/tex]
    [tex]Q[\theta] = \int_{\mathbb{R}^3}d^3x\, \theta(x)\,G(x) [/tex]

    then for every (time independent) test function θ(x) this Q[θ] is a globally conserved charge which commutes with the Hamiltonian H.

    In addition this Q[θ] acts as a generator of a time independent gauge transformations of the fields A, E and ψ (the fermion fields in QED) implemented via a unitary operator

    [tex]U[\theta] = e^{-iQ[\theta]}[/tex]

    The fact that the physical states are invariant w.r.t. to gauge transformations θ(x) is then expressed as

    [tex]|\psi\rangle \to |^\theta\psi\rangle = U[\theta]\,|\psi\rangle = |\psi\rangle [/tex]

    using the fact that G(x) and therefore Q[θ] annihilates these states and that therefore U[θ] reduces to the identity on the physical subspace
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