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Faddeev-Popov Trick

  1. Jan 16, 2013 #1
    I'm reading Peskin and Schroeder and when I got to the section on Quantization of the Electromagnetic field in Chapter 9, I encountered the Faddeev-Popov Trick.

    Conceptually, I got what's going on - badly divergent integrals from redundantly integrating over physically equivalent field configurations and a gauge-fixing trick to isolate the interesting part of the path integral counting each physical configuration once and only once, yada, yada.

    However, I didn't quite get the math. I absorbed enough to move on, but it's coming back to bite me now in Chapter 16 on Quantization of Non-Abelian Gauge Fields.

    So, here is the initial equation that threw me off. It's eq. 9.53 and I don't quite get why it is equal to 1.

    1 = ∫ D [itex]\alpha[/itex] (x) [itex]\delta[/itex] (G(A[itex]^{\alpha}[/itex])) det([itex]\delta[/itex]G(A[itex]^{\alpha}[/itex])/[itex]\delta\alpha[/itex])

    I've been assuming the determinant is the Jacobian of the transformation. The delta is the gauge-fixing condition G(A) = 0, but can somebody fill in any extra steps that shows why this is equal to 1? Any extra insight on the what's going on even conceptually would be nice too.

    Also, I suck at using Latex, but an equation two down from this one is a slightly simpler form if that helps elucidate things.

    Anyhow, if I can clear that up, I'm sure the rest will fall into place.

    Thanks!
     
  2. jcsd
  3. Jan 17, 2013 #2
    try this
    δ(f(x))=δ(x)/|f'(x)|
     
  4. Jan 17, 2013 #3
    ∫ D α (x) δ (G(Aα)) |(δG(Aα)/δα|
    =∫ D α (x) (δ (Aα))/|(δG(Aα)/δ(Aα)|) |(δG(Aα)/δα|
    =∫ D α (x) {δ (Aα)}(|δ(Aα)/δα|)
    =∫ D α (x) {δα/|δ(Aα)/δα|}(|δ(Aα)/δα|)
    =∫ D α (x) δα
    =1
    edit:just made mistakes with subscript and superscript ,but it is understandable.
     
  5. Jan 17, 2013 #4
    Hi andrien,

    Thanks for the reply!

    I think the delta function rule you put up in post #2 will be key. I always forget about that one!

    I'm still a little uneasy with a couple of the steps in post #3, but don't have time right now to look into it more.

    I'll try and think about it more later tonight (or this weekend) and respond back.

    dm4b
     
  6. Jan 17, 2013 #5
    To be more explicit, the formula you are looking at is the functional-integral version of

    ##\int dx \delta(f(x))|f'(x)| = 1##

    As an intermediate step you might prove the following formulas in N dimensions:

    ##\delta^{(N)}(A\vec{x}) = \delta^{(N)}(\vec{x})/(\det A)##

    and thus

    ##\int d^N x \delta^{(N)}(A\vec{x}) \det A = 1##

    (here A is an N by N matrix).

    Then generalize to infinite dimensions (i.e., functional integrals) by waving your hands in the right way.
     
    Last edited: Jan 17, 2013
  7. Jan 17, 2013 #6
    Alright, thanks guys, this has cleared up a lot.

    There is only one step I am a little uneasy with still.

    A[itex]^{\alpha}_{\mu}[/itex](x) = A[itex]_{\mu}[/itex](x) + (1/e)[itex]\partial_{\mu}[/itex][itex]\alpha[/itex](x)

    So, why wouldn't this be the case since A is a function of x, as well.

    [itex]\delta[/itex](A[itex]^{\alpha}[/itex]) = [itex]\delta[/itex](x) / | [itex]\frac{\partial A^{\alpha}}{\partial x}[/itex] |

    Instead of

    [itex]\delta[/itex](A[itex]^{\alpha}[/itex]) = {δα/|δ(Aα)/δα|}

    from line 4 of andrien's post (post #3)

    Once I make myself feel okay about that, everything else is falling into place!
     
  8. Jan 18, 2013 #7
    Aαμ(x) = Aμ(x) + (1/e)∂μα(x)
    this has nothing to do with delta function,it seems like it is related to local gauge invariance.But this is completely off the line.
     
  9. Jan 18, 2013 #8
    well, that A is the very one that is in the delta function in question. See page 295 of Peskin and Schroeder.
     
  10. Jan 18, 2013 #9
    it is still off the line.The integral is over α which is already a function of x.No explicit representation necessary.that relation has nothing to do with the delta function formula.It is just showing the local gauge invariance.
     
  11. Jan 18, 2013 #10
    I'm not sure what you mean by "off the line"

    Sorry, I am also confused by how that relation can have nothing to do with the delta function. It seems to me that the delta function written out explicitly is:

    [itex]\delta[/itex] [ A[itex]^{\alpha}_{\mu}[/itex](x) ]

    = [itex]\delta[/itex] [ Aμ(x) + (1/e)∂μα(x) ]

    This is what they have on page 295 and it's why I am still confused between the 2nd and 3rd equation in post #6.

    The best reason I can come up with going with the 3rd one as you did is because the integral is a functional integral, therefore the delta should be of a function (i.e. the 3rd equation in post #6, or the one you used) and should not be of a number, x. Although, I am not completely sold on my reasoning there.
     
    Last edited: Jan 18, 2013
  12. Jan 18, 2013 #11
    I just realized you were saying the same thing in your last post, I think

    Well, maybe I got this wrapped up then.

    Anyhow, thanks again for the help!
     
  13. Jan 18, 2013 #12

    dextercioby

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    My advice would be to learn BRST quantization.
     
  14. Jan 18, 2013 #13
    I'm as far as section 16.2. Looks like that may be section 16.4, so I just might very soon ;-)
     
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