# Understand Faddeev-Popov Trick: Equation 9.53 Explained

• dm4b
In summary, the author is trying to figure out why the delta function is equal to 1, and comes across a relation between ∂μα(x) and A^{\alpha}_{\mu}(x). After understanding this relation, he moves on to other issues.
dm4b
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 $\alpha$ (x) $\delta$ (G(A$^{\alpha}$)) det($\delta$G(A$^{\alpha}$)/$\delta\alpha$)

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!

try this
δ(f(x))=δ(x)/|f'(x)|

∫ 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.

Hi andrien,

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

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:
Alright, thanks guys, this has cleared up a lot.

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

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

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

$\delta$(A$^{\alpha}$) = $\delta$(x) / | $\frac{\partial A^{\alpha}}{\partial x}$ |

$\delta$(A$^{\alpha}$) = {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!

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.

andrien said:
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.

well, that A is the very one that is in the delta function in question. See page 295 of Peskin and Schroeder.

dm4b said:
well, that A is the very one that is in the delta function in question. See page 295 of Peskin and Schroeder.
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.

andrien said:
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.

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:

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

= $\delta$ [ 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:
dm4b said:
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.

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!

My advice would be to learn BRST quantization.

dextercioby said:
My advice would be to learn BRST quantization.

I'm as far as section 16.2. Looks like that may be section 16.4, so I just might very soon ;-)

## 1. What is the Faddeev-Popov trick?

The Faddeev-Popov trick is a mathematical technique used in quantum field theory to deal with the issue of overcounting in calculations involving gauge theories. It involves introducing a term into the equations that cancels out the redundant contributions from physically equivalent states, allowing for more accurate and efficient calculations.

## 2. How does the Faddeev-Popov trick work?

The Faddeev-Popov trick works by introducing a term called the "ghost field" into the equations, which cancels out the redundant contributions from physically equivalent states. This allows for a more accurate and efficient calculation of quantum field theory equations involving gauge theories.

## 3. What is Equation 9.53 in the Faddeev-Popov trick?

Equation 9.53 is a mathematical representation of the Faddeev-Popov trick, which involves introducing a term known as the "ghost field" into the equations. This equation is used to cancel out the redundant contributions from physically equivalent states in calculations involving gauge theories.

## 4. Why is the Faddeev-Popov trick important?

The Faddeev-Popov trick is important because it allows for more accurate and efficient calculations in quantum field theory involving gauge theories. By canceling out the redundant contributions from physically equivalent states, the trick helps to avoid overcounting and produces more accurate results.

## 5. What are some applications of the Faddeev-Popov trick?

The Faddeev-Popov trick has various applications in theoretical physics, including quantum chromodynamics, quantum electrodynamics, and the theory of weak interactions. It is also used in calculations involving the Standard Model of particle physics and in the development of new theories beyond the Standard Model.

Replies
36
Views
4K
Replies
1
Views
906
Replies
1
Views
2K
Replies
16
Views
2K
Replies
4
Views
2K
Replies
5
Views
3K
Replies
2
Views
2K
Replies
4
Views
1K
Replies
4
Views
1K
Replies
6
Views
1K