Change of the order of integration including Dirac delta

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  • #1
GIM
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Can somebody help me? I am studying Faddeev-Popov trick, following the Peskin and Schroeder's QFT book, but I can't understand one thing. After they inserted the Faddeev-Popov identity,
$$I = \int {{\cal D}\alpha \left( x \right)\delta \left( {G\left( {{A^\alpha }} \right)} \right)\det \left( {\frac{{\delta G\left( {{A^\alpha }} \right)}}{{\delta \alpha }}} \right)}$$
they exchanged the order of integration, but to my knowledge, since it includes delta function, there's no guarantee to exchange the order. Where can I find the reasoning for this?

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  • #2
andrewkirk
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More detail is needed. For the order of integration to be switched, there must be at least two nested integrals, but only one is shown here. I expect the clue to validity to lie in the full statement of the nested integrals.

Also, the integration limits and ##dx##-type terms are missing.
 
  • #3
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but to my knowledge, since it includes delta function, there's no guarantee to exchange the order.

At the intuitive level when dealing with the Dirac delta function, or any generalized function for that matter, you consider it a test function that is for all practical purposes the generalized function. This is because in the weak topology of the space of test functions you can find a test function arbitrarily close to a generalized function. As a test function you can do exchange of integration etc no worries.

Unfortunately sometimes in physics liberties is taken with rigor which can be an issue for those with a math background like myself. You will even find the use of the delta function squared which is totally wrong because it doesn't exist. But physicists seem to not worry about it and it works OK in practice (it exists in a space even larger than generalized functions so there is a reason it works). I recommend the following book which will help a lot:
https://www.amazon.com/dp/0521558905/?tag=pfamazon01-20

Thanks
Bill
 
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  • #4
Demystifier
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it exists in a space even larger than generalized functions so there is a reason it works
Does this space has a name?
 
  • #6
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Yes - Colombeau Generalized Functions:
http://arxiv.org/abs/math-ph/0611069

They have the very nice feature of any two can be multiplied - but have a drawback I cant recall - its been a while since I looked into it.
So in this language, if ##\langle x|x'\rangle=\delta(x-x')##, what is the probability density ##\rho(x)## associated with the state ##|x'\rangle##? How to do the corresponding calculus and how can I see that ##\int dx\, \rho(x)=1##?
 
  • #7
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So in this language, if ##\langle x|x'\rangle=\delta(x-x')##, what is the probability density ##\rho(x)## associated with the state ##|x'\rangle##?

Its singular - its allocated infinity at 0, but that is just a heuristic because c*infinity is still infinity yet in general multiplying by c gives a different vector (in a RHS sense).

Thanks
Bill
 
  • #8
Demystifier
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Its singular - its allocated infinity at 0, but that is just a heuristic because c*infinity is still infinity yet in general multiplying by c gives a different vector (in a RHS sense).
Are you saying that Colombeau generalized functions are not useful in this case?

Or should one, perhaps, write
$$\langle x|x'\rangle = \sqrt{\delta(x-x')}\; ?$$
 
  • #9
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Are you saying that Colombeau generalized functions are not useful in this case?

Colombeau generalized functions are a super-set of Schwartz distributions so has exactly the same interpretation as it does there. The Dirac delta function is considered zero but with a singularity at zero. Of course that's just a heuristic because c times a delta function is different - yet heuristically is still zero with a singularity at zero.

The advantage of this super-set is any two distributions can be multiplied.

Thanks
Bill
 
  • #10
Demystifier
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Colombeau generalized functions are a super-set of Schwartz distributions so has exactly the same interpretation as it does there. The Dirac delta function is considered zero but with a singularity at zero. Of course that's just a heuristic because c times a delta function is different - yet heuristically is still zero with a singularity at zero.

The advantage of this super-set is any two distributions can be multiplied.
I still don't get it in a practical sense. The example I presented leads to
$$\rho(x)=\delta(x-x')\delta(x-x').$$
This implies
$$\int\, dx\rho(x)=\delta(0)$$
which is a problem because one wants
$$\int\, dx\rho(x)=1$$
Can one resolve this problem in terms of Colombeau generalized functions? How?

The more I think about it, the resolution in post #8 looks more right to me.
 
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  • #11
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Can one resolve this problem in terms of Colombeau generalized functions? How?

It exists but what it means I am not familiar with it enough to know. Intuitively you simply multiply two test function that FAPP is the same as a Dirac delta. That's the way I view it when I occasionally come across it.

Thanks
Bill
 
  • #12
Demystifier
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It exists but what it means I am not familiar with it enough to know.
I believe the correct resolution is in post #8, which is equivalent to Eq. (1) in my "quantum myths and facts" paper.
 
  • #13
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I believe the correct resolution is in post #8, which is equivalent to Eq. (1) in my "quantum myths and facts" paper.

I think <x|x'> is a Dirac Delta - not it's square root.

Added Later:
After thinking about it I am not that sure. I need to think about it a bit more.

Thanks
Bill
 
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  • #14
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Hmmmmm. What would happen if you cube it?
There is no any physical reason to do it.

I think <x|x'> is a Dirac Delta - not it's square root.
Well, one can always change a normalization of the state. If ##<x|x'>=\delta(x-x')##, one can introduce
$$|x>_{ren}=\frac{|x>}{\delta^{1/4}(0)}$$
Then we have
$$_{ren}\!<x|x'>_{ren}=\frac{\delta(x-x')}{\sqrt{\delta(0)}}=\sqrt{\delta(x-x')}$$
 
  • #16
GIM
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I am sorry, @andrewkirk.
After insertion of the identity, it becomes
\[\int {{\cal D}A\,{e^{iS\left[ A \right]}}} = \det \left( {\frac{{\delta G\left( {{A^\alpha }} \right)}}{{\delta \alpha }}} \right)\int {{\cal D}\alpha } \int {{\cal D}A\,{e^{iS\left[ A \right]}}\delta \left( {G\left( {{A^\alpha }} \right)} \right)} \]
And thank you very much@bhobba! But still I have a question. Following you, I might relax delta function by the limit of some enough continuous function. However, this functional integral is divergent from the beginning. Then, I can't use the Fubini's theorem for changing the order of integration. What am I wrong?
 
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  • #17
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What about the representation of state |f> which would be f(x) = <x|f>, and f(x') = <x'|f> = ∫ <x'|x><x|f> which would imply<x'|x> is the Dirac Delta function.

Thanks
Bill
 
  • #18
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this functional integral is divergent from the beginning.

I don't really know your context, but I suspect its related to some kind of path integral. There are definite issues of convergence that require some quite advanced math to resolve:
http://www.mathnet.or.kr/mathnet/kms_tex/99937.pdf

If it helps with your problem I don't really know. I was responding to your query about interchange of integration.

Thanks
Bill
 
  • #19
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What about the representation of state |f> which would be f(x) = <x|f>, and f(x') = <x'|f> = ∫ <x'|x><x|f> which would imply<x'|x> is the Dirac Delta function.
Yes, but instead of working with ##|f>##, I can work with ##|f>_{ren}=c|f>##. It may happen that ##<f|f>\neq 1##, and the appropriate choice of ##c## may fix it by leading to ##_{ren}\!<f|f>_{ren}=1##.
 
  • #20
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Yes, but instead of working with ##|f>##, I can work with ##|f>_{ren}=c|f>##. It may happen that ##<f|f>\neq 1##, and the appropriate choice of ##c## may fix it by leading to ##_{ren}\!<f|f>_{ren}=1##.

I am not getting it. Can you elaborate?

Thanks
Bill
 
  • #21
GIM
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I am sorry for my ignorance of latex. This was my equation.
\begin{equation}
\int {{\cal D}A\,{e^{iS\left[ A \right]}}} = \det \left( {\frac{{\delta G\left( {{A^\alpha }} \right)}}{{\delta \alpha }}} \right)\int {{\cal D}\alpha } \int {{\cal D}A\,{e^{iS\left[ A \right]}}\delta \left( {G\left( {{A^\alpha }} \right)} \right)}
\end{equation}
 
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  • #23
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I am not getting it. Can you elaborate?
OK, let me be more systematic and use standard conventions as much as possible. So let's start from the standard relations
$$<x|x'>=\delta(x-x')$$
$$\int dx\, |x><x|=1$$
Then we have
$$\psi(x)=<x|\psi>=\int dx'\, <x|x'><x'|\psi>$$
$$\int dx\, |\psi(x)|^2 = \int dx\, <\psi|x><x|\psi>=<\psi|1|\psi>=<\psi|\psi>$$
That all looks fine when ##<\psi|\psi>=1##. However, for some states ##<\psi|\psi>\neq 1##. For instance, if we take ##|\psi>=|x'>##, we have
$$<x'|x'>=\delta(0)=\infty$$
and this is the problem. So how to avoid that problem?

Here is how. Define the renormalized state
$$|x'>_{ren}=c|x'>$$
where ##c## is a positive constant. Then we have
$$_{ren}\!<x'|x'>_{ren}=c^2\delta(0)$$
Thus we see that taking
$$c=\frac{1}{\sqrt{\delta(0)}}$$
leads to the desired result
$$_{ren}\!<x'|x'>_{ren}=1$$
Consequently
$$\int dx\, |\psi_{ren}(x)|^2 =1$$
as is required by the probabilistic interpretation, where
$$\psi_{ren}(x)=<x|\psi>_{ren}$$
$$|\psi>_{ren}=c|\psi>=c|x'>=|x'>_{ren}$$
This means that
$$\psi_{ren}(x)=c\delta(x-x')=\frac{\delta(x-x')}{\sqrt{\delta(0)}}=\sqrt{\delta(x-x')}$$
Note also that
$$_{ren}\!<x|x'>_{ren}=\delta_{xx'}$$
with the Kronecker ##\delta##, rather than the Dirac one.
 
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  • #25
strangerep
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$$c=\frac{1}{\sqrt{\delta(0)}}$$[...]
Oh... my... Gawd! :nb) :eek: :-p

I'd love to see you try and make that rigorous. :wink: :biggrin:
 
  • #26
ddd123
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I'd be content just having this passage explained to me:

$$\frac{\delta(x-x')}{\sqrt{\delta(0)}}=\sqrt{\delta(x-x')}$$

Thanks :oldtongue:
 
  • #27
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Oh... my... Gawd! :nb) :eek: :-p

I'd love to see you try and make that rigorous. :wink: :biggrin:
It's very simple indeed to make it more rigorous. One can always replace the ##\delta## "function" with a true function, such as a very narrow Gaussian parameterized with a small width ##\epsilon##. Or instead of a Gaussian, an even better choice is the narrow "wall" function ##\delta_{\epsilon}(x)## defined by
$$\delta_{\epsilon}(x)=1/\epsilon \;\;{\rm for}\;\; |x|<\epsilon/2$$
$$\delta_{\epsilon}(x)=0 \;\;{\rm for}\;\; |x|>\epsilon/2$$
$$\delta_{\epsilon}(x)=1/2\epsilon \;\;{\rm for}\;\; |x|=\epsilon/2$$
It satisfies
$$\int_{-\infty}^{\infty} dx \, \delta_{\epsilon}(x)=1$$
By Taylor expansion ##f(x)=f(0)+xf'(0)+x^2f''(0)/2+...## one obtains
$$\int_{-\infty}^{\infty} dx \, f(x)\delta_{\epsilon}(x)=f(0) +\epsilon^2\frac{f''(0)}{24}+...=f(0) +{\cal O}(\epsilon^2)$$
Repeating my previous renormalization procedure by a replacement ##\delta\rightarrow\delta_{\epsilon}## at the right places, one obtains
$$\psi_{ren}(x)=\frac{\delta_{\epsilon}(x-x')}{\sqrt{\delta_{\epsilon}(0)}}
=\sqrt{\delta_{\epsilon}(x-x')}$$
which is well defined for an arbitrarily small positive ##\epsilon##. Putting also the limit ##\epsilon\rightarrow 0## in the right places, one covers also the ##\delta## "function" case. But such pedantry makes all equations more cumbersome, so for the sake of practical calculus I would prefer not to take care of all these details. For me, it's sufficient to know that I can do it rigorously as sketched above, if I really want to.
 
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  • #28
Demystifier
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I'd be content just having this passage explained to me:
Here is an explanation at the level of practical calculus. When ##x\neq x'##, both sides are zero, so there is a match. When ##x=x'##, the left-hand side is
$$\frac{\delta(0)}{\sqrt{\delta(0)}}=\sqrt{\delta(0)}$$
which equals the right-hand side.

If you want a more rigorous argument, see the hint in the post above.
 
  • #29
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Note also that physics textbooks often do some similar "illegitimate" manipulations. For example, for the Dirac ##\delta## in the momentum space ##\delta^4(k)## physicists often write
$$\delta^4(0)=\frac{TV}{(2\pi)^4}$$
where ##V## is the volume of the "laboratory" and ##T## is the time duration of the experiment. Without carrying about rigor, in this way they obtain results which agree with experiments.
 
  • #30
vanhees71
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Come on, it doesnt' make sense to take a square or a square root of a ##\delta## distribution (NOT FUNCTION). In QT you must have
$$\langle x | x' \rangle=\delta(x-x'),$$
because otherwise the entire Dirac formalism of bras and kets breaks down, and it's well defined in the sense of distributions. You can take various weak limits to define this properly.
 
  • #31
vanhees71
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Note also that physics textbooks often do some similar "illegitimate" manipulations. For example, for the Dirac ##\delta## in the momentum space ##\delta^4(k)## physicists often write
$$\delta^4(0)=\frac{TV}{(2\pi)^4}$$
where ##V## is the volume of the "laboratory" and ##T## is the time duration of the experiment. Without carrying about rigor, in this way they obtain results which agree with experiments.
That's the usual short cut often used to define the square of S-matrix elements, but it's only a hand-waving shortcut. The correct way is to use true states rather than distributions (plane waves are distributions, namely momementum-eigenfunctions in position representation) for the asymptotic free states. Then there's no problem in squaring the S-matrix element and taking the weak limit to generalized momentum-eigenfunctions for the asymptotic free states. See Peskin/Schroeder for the details.
 
  • #32
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Come on, it doesnt' make sense to take a square or a square root of a ##\delta## distribution (NOT FUNCTION). In QT you must have
$$\langle x | x' \rangle=\delta(x-x'),$$
because otherwise the entire Dirac formalism of bras and kets breaks down, and it's well defined in the sense of distributions. You can take various weak limits to define this properly.
The square root in #27 certainly does make sense.
 
  • #33
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That's the usual short cut often used to define the square of S-matrix elements, but it's only a hand-waving shortcut. The correct way is to use true states rather than distributions (plane waves are distributions, namely momementum-eigenfunctions in position representation) for the asymptotic free states. Then there's no problem in squaring the S-matrix element and taking the weak limit to generalized momentum-eigenfunctions for the asymptotic free states. See Peskin/Schroeder for the details.
Yes, just like #23 is a hand-waving shortcut which can be justified by more rigorous procedure sketched in #27.
 
  • #34
samalkhaiat
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Can somebody help me? I am studying Faddeev-Popov trick, following the Peskin and Schroeder's QFT book, but I can't understand one thing. After they inserted the Faddeev-Popov identity,
$$I = \int {{\cal D}\alpha \left( x \right)\delta \left( {G\left( {{A^\alpha }} \right)} \right)\det \left( {\frac{{\delta G\left( {{A^\alpha }} \right)}}{{\delta \alpha }}} \right)}$$
they exchanged the order of integration, but to my knowledge, since it includes delta function, there's no guarantee to exchange the order. Where can I find the reasoning for this?

[Mentor's note: This text had been edited to fix the Latex formatting. everyone is reminded that there's section explaining how to make Latex work with the Physics Forums software on our help page: https://www.physicsforums.com/help/] [Broken]

There is no problem with such formal manipulations at all. The whole purpose of the Faddeev-Popov procedure is to introduce the correct integration measure,
[tex]\Delta[A] = \det \left| \frac{\delta G}{\delta \theta}\right|_{G = 0} ,[/tex]
into the path integral so that we can factor out the infinite group volume [itex]\int \mathcal{D} g(\theta)[/itex] which causes the infinite over-counting, i.e. summing over equivalent gauge field configurations. Plus, in actual calculations, you never need to interchange the order of integrations because once you choose the gauge-fixing surface [itex]G^{a}[A] = 0[/itex] so that it intersects every group-orbit exactly once, all pieces of the integrand will become independent of the group coordinates [itex]\theta^{a}(x)[/itex], and you can safely pull out the group volume [itex]\int \mathcal{D} g(\theta)[/itex]. For example in the covariant gauge [itex]G^{a} = \partial^{\mu}A^{a}_{\mu}[/itex], the path integral becomes
[tex]\int \mathcal{D}A_{\mu} \ e^{i S[A]} = \int \mathcal{D}A_{\mu} \int \mathcal{D}g(\theta) \ \prod_{x} \delta [\partial^{\mu}A_{\mu}] \det \left| \partial^{\nu}D^{ab}_{\nu}\right| \ e^{i S[A]} .[/tex] Clearly, nothing is there to prevent you from factoring out the infinite volume of the gauge group [itex]\mbox{Vol}(\mathcal{G}) = \int \mathcal{D} g[/itex]. The only important point to observe is this, your choice of gauge fixing [itex]G^{a}[A][/itex] which defines the local cross section on the principal fibre bundle (gauge slice transverse to every [itex]\mathcal{G}[/itex]-orbit) must not come with vanishing Jacobian, i.e., the matrix [tex]\Delta[A] = \frac{\delta G^{a}}{\delta \theta_{b}}|_{\vec{\theta} = 0} ,[/tex] must have an inverse.
 
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  • #35
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Thank you, @samalkhaiat! I think this would be the best correct answer, even though I didn't understand it well. I will try to understand it along this direction. Thank you again.
 

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