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Fadeev-Popov ghosts and renormalization

  1. Jul 22, 2011 #1
    I have a question about Fadeev-Popov (FP) ghosts.
    FP ghosts brake the gauge symmetry of the lagrangian and solve the problem of the divergence of the physical amplitudes, due to the gauge equivalence of the states. My question is: is it a renormalization process? I mean, the renormalization group is based on a rescaling of the parameters of the theory (fields, coupling constants, etc..) but not (as far as I know) on the introduction of new fields (physical or not, like FP).. So how can I see the FP ghosts in the contest of renormalization? Am I wrong on something?
    Sorry for my english.. Thanks in advice!
     
  2. jcsd
  3. Jul 24, 2011 #2
    Nobody?
     
  4. Jul 24, 2011 #3

    tom.stoer

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    The Fadeev Popov ghost do not brake gauge invariance, they fix gauge invariance. Look at the solar system and the orbits of the planet. The original formulation (a Lagrangian for the Kepler problem) is symmetric w.r.t. to SO(3) rotations. Now you fix the z-Direction as the axis for the orbital angular momentum and solve the Kepler problem.

    After havig done that you could still rotate the orbital angular momentum and describe a different system with a different orientation. So the SO(3) symmetry is still there in the space of solutions, you only pick one specific direction, one representative.

    This is what you are doing with fixing a gauge slice in the path integral.

    The infinity due to the gauge degrees of freedom which you are eliminating by this trick has nothing to do with renormalization. Renormalization (or the infinities of the physical amplitudes) arise after the Fadeev-Popov trick. They do exist in theories w/o gauge symmetry as well.
     
  5. Jul 25, 2011 #4
    Tom.stoer,

    aren't you confusing physical symmetry (like SO(3) symmetry in the Kepler problem which is, global and physical) and gauge invariance (which is mostly local and always a reduncancy in the description of the problem). I think the Kepler example is a bad example of gauge fixing.

    Otherwise I agree. Gauge-equivalent configurations are physically the same, it's just having many different labels for the same physical state. Fadeev-Popov construction fixes the gauge redundancy, such that each physical state is represented once. This has nothing to do with renormalization.
     
  6. Jul 25, 2011 #5

    tom.stoer

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    No confusion from my side, only the problem to find a clear example w/o too much complexity. I'll try to find a better example.
     
  7. Jul 25, 2011 #6
    Thank you very much! Now it's clearer to me what happens. It was a stupid question, hope Mr. Fadeev will forgive me :P
     
  8. Jul 25, 2011 #7
    Maybe I'm the confused one, but I don't think there are any gauge symmetry (redundancy) in the Kepler problem and therefore no gauge-fixing is required. All configurations related by a SO(3) transformation are physically distinct and not "gauge-equivalent".

    Anyhow, samuelr85 got the point so no reason to go too much into this.

    Samuel, Mr. Fadeev will forgive you, don't you worry. It's Mr. Popov you have to worry about! ;)
     
  9. Jul 25, 2011 #8

    tom.stoer

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    I never said that there is a gauge symmetry in the Kepler problem; I only want to use this as a simple (but unfortunately confusing) example.
     
  10. Jul 25, 2011 #9
    Oh sorry, I misunderstood your point.
     
  11. Jul 26, 2011 #10

    tom.stoer

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    OK, here's a rather simple example:

    Think about an integral

    [itex]F = \int dr\, f(r)[/itex]

    Now we rewrite this as

    [itex]F = (2\pi)^{-1} \int d\phi \int dr\,f(r) = (2\pi)^{-1} \int d^2r\,f(r,\phi) [/itex]

    and interpret it as a two-dim. integral with a "gauge symmetry", namely the rotation in the angle \phi. Instead of working backwards (= throw \phi out and arrive at the physical function f(r)) we introduce a delta function

    [itex]F = \int d^2r\,\delta(\phi-\phi_0)\,f(r,\phi) [/itex]

    The problem in gauge theory is that in general one cannot seperate physical (r) from unphysical (\phi) coordinates, which means that the delta function is not simply a function "in the unphysical fields" but determined by an expression where physical and unphysical fields are "intertwined". Ther most general case is therefore a gauge-fixing function G(r,\phi). So we try

    [itex]F = \int d^2r\,\delta(G(r,\phi)\,f(r,\phi) [/itex]

    For G(r,\phi) to be a "good gauge choice" it has to have zeros such that G(r,\phi) = 0 determines (implicitly) exactly one \phi for each r. If this is the case we can change variables from \phi to G:

    [itex]\int d^2r\,\delta(G(r,\phi)\,f(r,\phi) = \int dG\,dr\,\|\partial G / \partial\phi\|^{-1} \delta(G)\,f(r) = \int dG\,\|\partial G / \partial\phi\|^{-1}\,\int dr\,f(r) = \int dG\,\|\partial G / \partial\phi\|^{-1}\,F[/itex]

    So unfortunately we pick up a contribution from this gauge fixing function G which we interprete as the "volume of the gauge group", the 2\pi in the very beginning, wich we have to compensate:

    [itex]F = \int d^2r\,\|\partial G / \partial\phi\|\,\delta(G(r,\phi)\,f(r,\phi) = \int d^2r\,\|G^\prime\|\, \delta(G)\,f[/itex]

    This last expression contains the Jacobian and remains valid even if the origial function f and the gauge fixing function G are not specified in terms of r and \phi. The only condition (which is not valid in gauge theory globally!!!) is that this G defines a "global cut".
     
  12. Jul 28, 2011 #11

    blechman

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    I DON'T WANT TO HIJACK THE THREAD! But I thought I would just chime in with a clarification on this point. If you're not interested, then please ignore!

    I just wanted to say that your Kepler problem example is actually a good example of SPONTANEOUS SYMMETRY BREAKING: there is an SO(3) symmetry of the Hamiltonian, namely the rotational symmetry of the 1/r potential. But when speaking of the solar system, the universe had to make a choice for a plane for the solar system to be in, and once that is done, the symmetry is spontaneously broken (nonlinearly realized, there is the "Goldstone mode" that tips the solar system on its side, etc...).

    Anyway, that's not what FP gauge fixing is (there is no symmetry breaking there, just throwing out gauge degrees of freedom). tom.stoer's future example is more appropriate.

    I'm sorry if this is seen as beating a dead horse, but I just thought I'd throw that in. We now return you to your regularly scheduled thread....
     
  13. Jul 28, 2011 #12

    tom.stoer

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    thanks bechman, you comment is helpful; I hope samuel will come back ...
     
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