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Freedom to pick a gauge

  1. Sep 5, 2011 #1
    I'm trying to understand how much freedom one has to pick a gauge to use for gauge fixing a free EM field and if so, on what conditions that freedom depends.
    Is there a particular reason (besides Lorentz covariance) to pick the Lorenz gauge? Are the alternatives (R[itex]\xi[/itex] for instance) still meaningful?
  2. jcsd
  3. Sep 5, 2011 #2


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    In general you are free to pick any gauge; Lorentz gauge is convenient in many cases, but other's are as well. You could chose an axial gauge for example - and even in QCD sometimes Coulomb gauge is used to study IR phenomena (not scattering, of course).

    afaik the only restriction in chosing a gauge is due to the requirement that the gauge condition g(A)=0 specifies a 'global section' of the fibre bundel defined by the 'gauge group * space-time'. b/c the U(1) bundle is trivial this is trivial in QED, but for non-abelian gauge groups like SU(N) afaik there are many gauges (including Lorentz gauge, Coulomb gauge, axial gauge) which fail to define such a global section and which are therefore plagued with so-called Gribov ambiguities, i.e. non-unique solutions of g(A)=0. In addition there may be 'gauges' not defining a gauge at all, i.e. not intersecting all fibers of the bundle.

    afaik - but I am not sure about that - it has been shown that there are no global sections w/o ambiguities for SU(N).

    For perturbation theory this doesn't matter b/c the Gribov ambiguities are well-separated from A=0 and are therefore non-perturbative.
  4. Sep 5, 2011 #3
    So I can pick any gauge and the choice is just because of convenience.
    This makes me think that different gauges don't lead to different physics, for instance unphysical states in one gauge are unphysical in any other gauge, is that right?
    If it's correct then as the Coulomb gauge is not Lorentz covariant, how can a boost transform a physical (according to the Coulomb gauge) state into a nonphysical state, but still physical according to the Lorenz gauge?
  5. Sep 6, 2011 #4


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    Yes, different gauges result in the same physics; that's the meaning of gauge invariance.

    Unfortunately not! There are "physical" and "unphysical gauges".

    In the Lorentz gauge in QCD (due to its non-abelian character) it's NOT possibe to formulate the theory entirely in terms of physical degrees of freedom (two gluon polarizations). Instead one has four gluons (!) one gauge condition (the Lorentz gauge condition) plus so-called Fadeev-Popov ghost fields (which decouple in an abelian U(1) therefore you don't need them in QED). The gauge condition plus the unphysical ghosts "cancel" two unphysical gluons resulting in a physical and gauge invariant theory (BRST symmetry, Slavnov-Taylor identities).

    In the axial gauge in QCD no ghosts arise and one can formulate the theory entirely in terms of physical degrees of freedom, i.e. two gluon (two polarization states). A left-over from the complete gauge-fixing is a "non-abelian Coulomb-potential" which cannot written down in closed form, unfortunately.

    Fixing a gauge means restricting the theory to a certain "physical" subsector of a large an unphysical Hilbert space. Different gauges lead to different gauge fixing and to different degrees of freedom in general. What is invariant is the extracted physics, i.e. matrix elements, scattering cross sections etc.

    After gauge fixing there is no gauge symmetry anymore, i.e. a gauge transformation is just the identity. In addition a physical boost operator does not violate the gauge condidition, simply b/c the quantum mechanical generators of gauge transformations (something like generalized Gauss law constraints) are constants of motion, i.e. they commute with the physical Hamiltonian. Therefore any gauge transformation is reduced to the identity in the physical Hilbert space (the physical states are gauge singuletts) ... and the identity commutes with all other physical operators like boosts. So a boost of a physical state in Coulomb gauge transforms this state into a new physical state respecting Coulomb gauge as well. After gauge fixing there is no way do transform betweeen different gauges.

    The big technical problem is to show explicitly that the theory is still Lorentz covariant (even so it has been formulated in a non-Lorentz covariant gauge). In order to do that one has to calcluate all Poincare generators in Coulomb gauge and show that their algebra still closes - even after regularization, i.e. that no anomalies are introduced. In additon every boost has to be calculated explicitly (many many sheets of paper) whereas in Lorentz gauge everything is manifestly covariant.

    That's the reason why whenever one needs Lorentz invariance (e.g. for scattering cross sections where you want to transform to lab frame, c.o.m. frame, ... quite frequently) one uses covariant gauges, whereas in low-energy physics when you want to calculate hadron spectra or something like that you may use physical gauges w/o manifest Lorentz covariance.
  6. Sep 6, 2011 #5
    When you say "the theory" are you referring to, say, all the currents that I can calculate from the Lagrangian? Or the Lagrangian itself?

    So if I had an operator O which mixes physical with unphysical degrees of freedom, which would turn a physical state into a nonphysical one (and perhaps viceversa) there's no chance that there is a gauge according to which both the original and the transformed state are physical?
    Can this (or the converse) be shown by calculating a general commutator between O and the Q of BRST? I mean, given that O, would it be correct to try to show that there's no possible Q that doesn't commute with O?
  7. Sep 6, 2011 #6


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    With "the theory" I mean all physical observables, i.e. matrix elements, scattering cross sections, spectra, form factors, ...
    A current is in general not gauge invariant (the electromagnetic current is, but the color-current in QCD isn't).

    Let's see if I understand what you have in mind.

    Let's assume we are working in QCD, A°=0 gauge, with one residual gauge symmetry, i.e. time-independent gauge transformations which do not violate thie s gauge. Then we have the Gauss law constraint saying that Ga(x)|phys> = 0. And in addition Gauss law generates gauge transformations. Now what do you have in mind?
    1) Change the gauge. i.e. do no longer consider A°=0 but something else?
    2= Or stay within the A°=0 gauge but find an operator that sends |phys> to |non-phys'>?

    Regarding the first idea I think it's not possible simply b/c these are different theories (at least in the canonical formalism, perhaps not in generalized BRST). Regarding the second idea I would say that it is inded possible.

    Consider Ga(x)|phys> = 0

    Introduce a new operator Fb(y) with

    Fb(y)|phys> = |non-phys>

    Then calculate

    Ga(x) Fb(y)|phys> = [Ga(x), Fb(y)]|phys> + Fb(y)Ga(x)|phys>

    The second term vanishes whereas the first term reduces to something like CabcEc(x) times a delta function.

    Using some operator Fa for which the new operator Ea is linear in the gauge fields one immediately sees that it may contain an unphysical (i.e. longitudinal) gluon; that means that indeed Fa sends the physical state to a non-physical one ...

    But what does that mean? In the canonical formalism a residual gauge transformation with gauge function f, generated by the Gauss law constraint respecting the A°=0 gauge, is implemented as unitary operator

    U[f] = exp i ∫d3x fa(x) Ga(x)

    But it is not allowed to transform only the states of the Hilbert space according to U[f] |.>; at the same time one has to transform all operators O according U[f] O U[f].

    Now you have two 'gauges', the transformation between these two gauges is implemented as U[f] with some gauge function f. The states before and after the transformation reside in the same kinematical Hilbert space, two states are not identical in general, but two physical states are identical b/c Ga(x) annihilates all physical states, therefore U[f] reduces to the identity

    U[f] |phys> = exp i ∫d3x fa(x) Ga(x) |phys> = (1 + i ∫d3x fa(x) Ga(x) + ...)|phys> = (1 + 0 + ...)|phys>

    That means that not only does U map physical states to physical states; on a physical state U is the identity! That means that a gauge transformation respecting the A°=0 gauge does not change anything within the physical Hilbert space.

    As said in the very beginning I do not know how to enlarge this to a framwork in which one can change the A°=0 gauge to something else, simply b/c I think that this cannot be formulated in one single Hilbert space. Perhaps BRST will help.
  8. Sep 6, 2011 #7
    My idea can be put in simpler terms, suppose we only have a noninteracting EM field. No currents, just fields.
    For instance, I can say that a component S[itex]\perp[/itex] of the spin operator of the photon which is perpendicular to the k vector is unobservable, because it generates a transformation that mixes a transverse polarization with the longitudinal one, i.e. if I see this from a BRST point of view, that operator maps states from the cohomology of physical states to 'somewhere' outside.
    Could there be a different gauge such that the transformation generated by S[itex]\perp[/itex] is an endomorphism of the new cohomology?
    Or are different cohomologies generated by different gauges simply isomorphic in some way and contain the same objects? (in which case the question would be trivially answered)
  9. Sep 6, 2011 #8


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    There should be some isomorphism, otherwise different gauges wouldn't be equivalent. But I am not sure, I do not have enough experience rearding these BRST topics.
  10. Sep 6, 2011 #9
    I see. Well you've certainly given me good ideas, and good examples that I can keep and think about!

    PS can I restart/continue the topic somewhere else or would it be considered double posting?
  11. Sep 6, 2011 #10


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    Usually it is sufficient to refer to this thead just to ensure that others don't have to write the same things twice.
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