Change of the order of integration including Dirac delta

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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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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.
 
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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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Demystifier

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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##?
 
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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).

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Bill
 

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
$$\langle x|x'\rangle = \sqrt{\delta(x-x')}\; ?$$
 
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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
 

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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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.

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Bill
 

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.
 
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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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Demystifier

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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
$$_{ren}\!<x|x'>_{ren}=\frac{\delta(x-x')}{\sqrt{\delta(0)}}=\sqrt{\delta(x-x')}$$
 

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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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
 
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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
 
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Demystifier

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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##.
 
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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
 

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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Demystifier

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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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