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Gauge Fixing in Yang-Mills gauge theory

  1. Dec 22, 2013 #1
    I've come across countless sources that gauge fix SU(N) Yang-Mills fields using the typical U(1) gauges (e.g. Lorenz gauge, coulomb gauge, temporal gauge, etc). However, I can't find a single one where they prove that this gauge fixing is valid for all field configurations... I've tried to derive it myself, but the non-abelian nature of the problem makes it very difficult. I'll sketch out the basics below, if anyone can prove this to me or point me to a proof I would really appreciate it. I'm especially interested in a proof of the existence of a Lorenz gauge transformation for any field configuration in either SU(2) or the more general SU(N).

    SU(N) vector potential is written as: [itex]A_\mu \equiv A^a_\mu T_a[/itex] where [itex]T_a[/itex] are the SU(N) generators and [itex][T_a,T_b] = if_{abc}T^c[/itex] for structure constants [itex]f_{abc}[/itex]. For SU(2) the generators are the Pauli matrices (divided by 2).

    The vector potential transforms as: [itex]A_\mu \rightarrow UA_\mu U^\dagger +\dfrac{i}{g}U\partial_\mu U^\dagger[/itex] where [itex]U\equiv e^{i\theta}[/itex] is an element of SU(2) and [itex]\theta\equiv \theta^a T_a[/itex].

    What I am trying to prove is that for any [itex]A_\mu[/itex] there exists a U such that [itex]\partial^\mu A'_\mu = 0[/itex] (or another gauge condition). For the U(1) case its very straightforward and you end up with a wave equation for [itex]\theta[/itex] with a source term which is a function of A. However for the general SU(N) case I'm stuck.. Even for the specific case of SU(2) I can't prove its existence (I would be happy to just have a proof for this case)!
     
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  3. Dec 23, 2013 #2
    Can a moderator please move this thread back to classical physics and delete this post? I may not have specified this but I am working purely with classical fields and there is nothing quantum about this question
     
  4. Dec 23, 2013 #3

    Vanadium 50

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    If this gets any answer at all, it will be here, not in Classical.
     
  5. Dec 24, 2013 #4
    Ok so I've been trying to solve this problem, and I've actually managed to convince myself that for SU(N) gauge fields the Lorenz gauge is NOT possible in general.. However if you just google "non-abelian lorenz gauge" you'll find a huge number of papers which implicitly assume the Lorenz gauge is ok to use for SU(N) fields. I don't yet have a formal proof, but I'll sketch out the argument below:

    For any specific generator Ti, you can produce a unitary matrix U that makes [itex]\partial_\mu A^\mu_i = 0[/itex], by just setting [itex]U=e^{i\alpha T_i}[/itex] and [itex]\partial_\mu\partial^\mu\alpha = -g\partial_\mu A^\mu_i[/itex]. Similarly, any generator Tj that commutes with Ti will have [itex]\partial_\mu A^\mu_j = 0[/itex]. For U(N), you can have the Lorenz condition satisfied for ALL fields because all of the generators commute with the U(1) subgroup. However for SU(N) there is no generator that commutes with all the rest! So at best you can only make some subgroup of the fields satisfy the Lorenz condition.

    Now, the reason this isn't a proof is that I haven't demonstrated that my method of making a field satisfy the Lorenz condition is the only method. There may be some other one that gets around this problem, but I've been looking at SU(2) in depth and I really don't think it exists. Any unitary transformation is going to mix at least 2 of the 3 fields, so that even if you fix 1 field you can't touch the others without breaking it! You end up with the usual U(1) situation for 1 field and some incredibly complex function of the gauge parameters for the other fields if you require the Lorenz condition, and I wouldn't even know where to begin trying to prove a solution exists for all field configurations..
     
  6. Dec 24, 2013 #5
    On second thought I think I just answered my own question lol. I had been assuming U is an element of SU(N), as one of my sources specifically stated. However I just realized that ANY unitary matrix would leave the Lagrangian unchanged, since the determinant never even comes into play. Therefore although an SU(N) gauge field only has fields corresponding to SU(N), it is left invariant by U(N) and the method I showed above can be used to prove the existence of the Lorenz gauge condition.

    As this question is now much simpler, I'm hoping someone can give their opinion on this? I just want to make sure my proof is correct, especially since it contradicts a source I'm using which states explicitly that det(U)=1.
     
  7. Dec 25, 2013 #6
    Scratch that last post... The U(1) subgroup of SU(N) trivially leaves the lagrangian invariant. Because the trace of all the SU(N) generators is 0, a U(1) transformation actually leaves the field A_i unchanged. So I still have no answer to this problem, and I still suspect that no proof exists for the Lorenz condition in the general non-Abelian case
     
  8. Dec 27, 2013 #7

    DarMM

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    You can essentially derive an extremely nonlinear equation for ##\theta_i## using the condition ##\partial^{\mu}A^{'}_{\mu} = 0## and the expansion above for ##A^{'}_{\mu}##.

    The only problem is that in the ##U(1)## case this reduces to:
    ##\partial^2 \theta = -\partial^{\mu}A_{\mu}##

    which, since ##\partial^2## has an inverse, poses the unique solution:
    ##\theta = -(\partial^2)^{-1}\partial^{\mu}A_{\mu}##

    However in the non-Abelian case you can't obtain a result like this as there will be ##\theta, A_{\mu}## cross terms. In fact you can prove the solution isn't unique, there will be several fields along one gauge orbit which obey the Lorenz condition. These being known as Gribov copies.
     
  9. Jan 1, 2014 #8
    Yes, this is the basic problem I'm facing
    I have seen references to this, that the Lorenz condition is not a completely fixed gauged in non-abelian theories. However, I still haven't seen a proof that there must be at least one solution to the Lorenz condition! The question of non-unique gauge solutions is interesting and I'm curious how you would go about completely eliminating ALL gauge degrees of freedom for a non-abelian theory, but my original question of whether any solution exists is far more important to me
     
  10. Jan 3, 2014 #9

    tom.stoer

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    You can't! At least discrete gauge copies will always survive in non-abelian gauge theories, regardless which gauge family (Lorentz, Coulomb, axial, ...) you chose.
     
  11. Jan 14, 2014 #10
    How can that be possible? Surely you can ALWAYS just fix a specific gauge, so why wouldn't there be some analytic expression for which that fixed gauge is the only solution? For example, in the Abelian case take the Lorenz gauge. If you make certain assumptions about the fields then this gauge is unique, but in general it admits the addition of any function [itex]\phi[/itex] where [itex]\partial_\mu\partial^\mu\phi = 0[/itex]. However, setting boundary conditions on the fields (as is typically done) immediately fixes [itex]\phi[/itex] up to some constant term. I guess you could call this constant term a gauge degree of freedom, but it is completely unphysical since it doesn't even effect the 4-potential...

    Anyway, this is off topic. My original question is why does a solution HAVE to exist, not how many are there or how you can fix it.
     
  12. Jan 14, 2014 #11

    tom.stoer

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    You should find some papers on so-called Gribov copies explaining the obstacles to unique gauge fixing.

    So your expectation that there has to be a solution. OK, fine. But the question is a but more subtle. The mathematical structure you have to deal with is a vector bundle which locally (!) looks like a direct product of the base space (pseudo-Riemann 4-manifold) with the gauge algebra SU(N); but globally this not a direct product, unfortunately. So gauge fixing means to specify a condition which fixes a submanifold in this vector bundle such that for every fiber there is one and only one solution.

    If there is no solution that means that the condition eliminates a physical part of the vector bundle, i.e. it is too strong. If there are a couple of solutions that means that a physical field configuration has more than one equivalent representations in the vector bundle, i.e. they are counted multiple times, e.g. in the path integral.

    So the physical expectation is that a gauge condition - together with a constraint equation of the fields - eliminates too unphysical d.o.f. globally. However the mathematical problem is that there is absolutely no reason why this should be possible by one unique gauge condition. And as far as I know there are rigorous proofs that there is no globally specified gauge condition which does not have these Gribov copies.

    Afaik the Lorentz gauge is always used in the "perturbative domain", i.e. for "small fields", and b/c the Gribov copies are specified by "large fields" they are not "close to the small fields", so for perturbative treatments nobody cares about these effects.

    However for IR physics where perturbative approaches relying on "small fields" are not sufficient, one has to take the gauge artifacts into account. Afaik there the Lorentz gauge is not very popular. But anyway, the gauge artifacts are always present and one has to take them into account.

    I'll try to find some references for you.
     
  13. Jan 15, 2014 #12
    If I remember correctly then Weinberg seem to have said somewhere in his vol. 2 that axial gauges are not plagued by gribov copies,but all he has shown is that Fadeev popov determinant is independent of gauge field and there are no ghost loop.Actually it is possible to remove gribov copies in spacelike planar gauge as done here.one can also see page 42 here
     
    Last edited: Jan 15, 2014
  14. Jan 15, 2014 #13

    tom.stoer

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    Yes, the Fadeev-Popov determinant evaluated for small gauge fields does not force you to introduce ghosts. But that does not provide a proof for large non-trivial gauge fields. I remember a paper which says that no such gauge is possible, but I remember similar statements like yours that there are special gauges w/o Gribov copies. So perhaps I should better say that I am not sure about this and that I have to double-check carefully ...
     
  15. Jan 15, 2014 #14
    This is all interesting, and I appreciate the comments, but none of this addresses my original question!

    *edit* if you can answer my question I would be happy to continue to discuss this related one
     
  16. Jan 15, 2014 #15

    tom.stoer

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    Can you please repeat your question?
     
  17. Jan 15, 2014 #16
    Can the Lorenz gauge condition be satisfied for any SU(N) field configuration? Can you prove it?
     
  18. Jan 15, 2014 #17

    tom.stoer

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    I don't know whether there is a rigorous proof that it can be satisfied for any SU(N) field. But I know that if it can be satisfied, then the solution for A is not unique.
     
    Last edited: Jan 15, 2014
  19. Jan 16, 2014 #18

    tom.stoer

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    There is a way to see that under rather general assumptions it should be possible.

    Given a general su(N) valued connection A(x) you want to construct a SU(N) valued transformation U(x) satisfying

    [tex]A_\mu^\prime = U^\dagger(A_\mu - i\partial_\mu)U;\;\;\partial^\mu A_\mu^\prime = 0[/tex]

    Now one can proceed in two two steps.
    1) fix the temporal gauge
    2) fix the Coulomb gauge

    [tex]A_0^\prime = 0;\;\;\partial^i A_i^\prime = 0[/tex]

    For the 1st step the proof starts with

    [tex]A_0^\prime = U^\dagger(A_0 - i\partial_0)U \stackrel{!}{=} 0[/tex]

    This equation is equivalent to

    [tex]A_0 + iU\partial_0 U^\dagger \stackrel{!}{=} 0[/tex]

    which can be solved formally using the path-ordered (here: time-ordered) exponential

    [tex]U[A] = \text{T}\,\exp{\left[ -i \int_{t_0}^t ds\,A_0(s)\right]}[/tex]

    For the 2nd step I have to find a general proof for the existence of a a time-independent, unitary transformation V[A] which implements the Coulomb gauge. I know this is possible, but I have to find the reference and the assumptions (smoothness of A, boundary conditions, topology of the 4-manifold, ...).

    Note that this was not your original question, but it shows that an even stronger gauge condition is possible.

    However there are problems making these proofs useless in many contexts (canonical quantization, non-perturbative approaches):
    The 1st construction requires a time-like path and therefore a solution of the equations of motion.
    The solution is rather formal; only in rare cases an explicit construction is possible.
    The solution is not unique b/c every such construction of U is path-dependent; this can be seen by inserting closed loops

    [tex]W[C] = \text{T}\,\exp{\left[ -i \oint_C dx^\mu\,A_\mu(s)\right]} \neq 1[/tex]

    which affect both U and A' but which do not violate the gauge condition for A'. Of course this is related to the existence of Gribov ambiguities.
     
  20. Jan 16, 2014 #19
    In general these two gauge conditions can't be simultaneously satisfied. Take U(1) theory:
    [itex]A_0^\prime = A_0 + \partial_0\Theta = 0[/itex]
    This fixes [itex]\Theta[/itex] up to some time-independent component. So the requirement:
    [itex]\partial^i A_i^\prime = 0[/itex] can not be satisfied for a general time-dependent vector potential!
    This is false! [itex]A_0[/itex] is a matrix [itex]A_0^a T_a[/itex], and in general does not commute with U

    Also, there is nothing special about the Lorenz gauge. I would be satisfied with a proof of existence for ANY of the common gauges! However, even for the temporal gauge you haven't demonstrated a solution must exist.

    *EDIT* Nevermind! I see what you did, I think that is a solid proof of the temporal gauge :)
     
    Last edited: Jan 16, 2014
  21. Jan 16, 2014 #20

    tom.stoer

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    The second gauge trf. V must be independent in order not to destroy the temporal gauge.

    One can show that it's possible due to the fact that given the temporal gauge A0 = 0 on a space-like 3-manifold the time evolution generated by the physical Hamiltonian respects the temporal gauge, i.e. does never generate a gauge field configuration violating the temporal gauge. This is a result of the fact that A0 is not a dynamical variable but a Lagrange multiplier, and that the physical Hamiltonian does not contain a conjugate momentum ∏ = ∂0A0 ~ 0.

    So the proof requires that you fix the temporal gauge and evolve the spatial components using the physical Hamiltonian. Therefore the two conditions are valid in the physical subspace if the full phase space.

    Yes, they do not commute, but the derivation does not use something like that

    You start with

    ##U^\dagger(A_0 - i\partial_0)U = 0##

    ##U^\dagger A_0 U - i U^\dagger \partial_0 U = 0##

    Now multiple with U from the left and U-dagger from the right

    ## A_0 - i U U^\dagger (\partial_0 U) U^\dagger = 0##

    ## A_0 - i (\partial_0 U) U^\dagger = 0##

    Now use

    ##\partial (U U^\dagger) = 0##

    to rewrite the second term

    ## A_0 + i U \partial_0 U^\dagger = 0##
     
    Last edited: Jan 16, 2014
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