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Change of the order of integration including Dirac delta

  1. May 29, 2016 #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?

    [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]
     
    Last edited by a moderator: May 7, 2017
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  3. May 29, 2016 #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.
     
  4. May 29, 2016 #3

    bhobba

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    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/Theory-Distributions-Nontechnical-Introduction/dp/0521558905

    Thanks
    Bill
     
    Last edited by a moderator: May 7, 2017
  5. May 30, 2016 #4

    Demystifier

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    Does this space has a name?
     
  6. May 30, 2016 #5

    bhobba

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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.

    Thanks
    Bill
     
  7. May 30, 2016 #6

    Demystifier

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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##? How to do the corresponding calculus and how can I see that ##\int dx\, \rho(x)=1##?
     
  8. May 30, 2016 #7

    bhobba

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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).

    Thanks
    Bill
     
  9. May 30, 2016 #8

    Demystifier

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    Are you saying that Colombeau generalized functions are not useful in this case?

    Or should one, perhaps, write
    $$\langle x|x'\rangle = \sqrt{\delta(x-x')}\; ?$$
     
  10. May 30, 2016 #9

    bhobba

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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.

    Thanks
    Bill
     
  11. May 30, 2016 #10

    Demystifier

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    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.
     
    Last edited: May 30, 2016
  12. May 30, 2016 #11

    bhobba

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    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
     
  13. May 30, 2016 #12

    Demystifier

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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.
     
  14. May 30, 2016 #13

    bhobba

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    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
     
    Last edited: May 30, 2016
  15. May 30, 2016 #14

    Demystifier

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    There is no any physical reason to do it.

    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. May 30, 2016 #15

    bhobba

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    Yes - after thinking about it I removed it.

    Still thinking on it.

    Thanks
    Bill
     
  17. May 30, 2016 #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?
     
    Last edited: May 30, 2016
  18. May 30, 2016 #17

    bhobba

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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
     
  19. May 30, 2016 #18

    bhobba

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    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
     
  20. May 30, 2016 #19

    Demystifier

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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##.
     
  21. May 30, 2016 #20

    bhobba

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    I am not getting it. Can you elaborate?

    Thanks
    Bill
     
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