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Problem with Gaussian Integral

  1. May 7, 2014 #1
    I'm reading a book on Path Integral and found this formula
    [itex]\int_{-\infty}^{\infty }e^{-ax^2+bx}dx=\sqrt{\frac{\pi}{a}}e^{\frac{b^2}{4a}} [/itex]
    I Know this formula to be correct for a and b real numbers, however, the author applies this formula for a and b pure imaginary and I do not understand why this is correct. This is like saying
    [itex]\int_{-\infty}^{\infty}e^{-ix^2 }dx=\int_{-\infty}^{\infty}cos(x^2) dx +i\int_{-\infty}^{\infty}sen(x^2) dx=\sqrt{\pi}[/itex]
     
  2. jcsd
  3. May 7, 2014 #2

    stevendaryl

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    You're right, if [itex]a[/itex] is pure imaginary, then the integral is undefined. However, like lots of things in physics, we can get a value in a hand-wavy way. There are multiple ways of doing it. One way is just to get the answer for [itex]a[/itex] being real, and analytically continuing it to the case where [itex]a[/itex] is imaginary. Another way is to put in a damping factor of the form [itex]e^{-\lambda x^2}[/itex] and take the limit as [itex]\lambda \Rightarrow 0[/itex].

    I don't think that there is a consensus about how to make path integrals rigorous in all cases.
     
  4. May 7, 2014 #3
    I can understand that in physics there is no room for mathematical rigor and many things must be done heuristically, but this really does not make sense and I'm shocked.
    I've seen things like the delta function and its Fourier transform and I learned to accept it.
    Tanks for your answer
     
  5. May 7, 2014 #4

    vanhees71

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    Sometimes, it's good to think about these questions. Here, it's of course simple to make [itex]a[/itex] complex with a positive real part and evaluate the then well-defined integral and afterwards letting the real part go to [itex]0[/itex]. The tricky issue here is the square root [itex]\sqrt{a}[/itex], which has an essential singularity at [itex]a=0[/itex], and the final result depends on which branch of the square root you evalute it finally for [itex]a[/itex] purely imaginary. Here it's clear that one should take the branch, which is positive for positive real [itex]a[/itex], and then the analytica continuation to purely imaginary [itex]a[/itex] is unique (putting the branch cut along the negative real axis as is standard).

    It's, however, indeed true that path integrals are not rigorously defined in general in the strict sense of pure mathematics.
     
  6. May 7, 2014 #5

    strangerep

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    Rubbish.

    You should be shocked, and you should not accept such nonsense. I used to really hate it when my Physics lecturers used sloppy arguments -- on things which my Applied Maths lecturers were able to present easily with reasonable rigor. Really, there is (usually) no excuse for such sloppiness.

    Both those things are defined rigorously -- within the Schwarz theory of Distributions (Generalized Functions). Possibly you think they're non-rigorous because Physics lecturers just give the rules for using those techniques but without the full underlying mathematics?

    In the case of your simple Gaussian integral, one can potentially make sense of it as a distribution instead of an ordinary function. But that depends on the context in which it's being used. The various other ways of making sense of it using convergence factors and limit arguments can be regarded as a manifestation of the fact that a Schwarz distribution is generally a limit of a sequence of ordinary functions satisfying certain conditions (which I'll skip for now).

    Path integrals are a harder case: there is no satisfactory translation-invariant Lebesgue measure on an infinite dimensional space (which is what a path space usually is). Therefore, one attempts to other measures, such as Gaussian measure. That's basically what's going on when the (quadratic) part of the free Hamiltonian is absorbed into the measure -- with a Wick rotation and various analyticity theorems, the path integral may become better defined. For more info, try this.
     
  7. May 8, 2014 #6
    No, I know there is rigorous math for those subjects, what I meant is that a physicist does not need to be a mathematician(Remenber Fymman's definition of mathematics:"waste time showing the obvious"). I also know that some physicits are more "mathematical" than others.
    I content myself to understand things like the delta function intuitively, however the result that motivated me to put this thread really puzzled me maybe because of my poor math.
    The explanations you(and the others) gave me seem very sofisticated for me, I was hoping there were some easier or more conventional explanations like "Residue Theory" for instance.
     
  8. May 8, 2014 #7

    strangerep

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    A lot of distribution theory stuff can sound sophisticated to the uninitiated, but the truth is that acquiring a basic understanding of distribution theory is not so difficult if one has a textbook at the right level. Physicists need at least a basic grasp of that stuff. If you're capable of understanding complex variables and residue theory, then you're certainly capable of acquiring a basic understanding of distribution theory.

    For this purpose, Bhobba often recommends https://www.amazon.com/The-Theory-D...&keywords=introduction+to+distribution+theory (but I haven't read it).

    There's also a briefer (but perhaps more demanding) short course in Appendix A of Nussenzveig's https://www.amazon.com/Causality-Di...F8&qid=1399607773&sr=1-1&keywords=nussenzveig.
     
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