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Steepest descent approximation

  1. Apr 6, 2012 #1
    Hi all, I am reading now Zee's book "Quantum Field Theory in a Nutshell", there in Apendix 2 of Chapter I.2 the method of steepest descent is briefly described. The part where I have a question is almost self contained and half a page long, so I attached the screen shot of it (formula 19). Anyway, the question is the following:
    The only prior information required is the Gaussian integral [tex]\int_{-\infty}^\infty dx e^{-1/2ax^ 2} =(\frac {2\pi}{a})^ {1/2}[/tex]
    To compute the integral [tex]I=\int_{-\infty}^\infty dq e^{-1/h f(q)}[/tex] in the limit of very small h, we say that the integral is dominated by the minimum of f, and expanding f near that point up to quadratic terms, [tex]f(q)=f(a)+1/2f''(a)(q-a)^{2}+O[(q-a)^{3}][/tex] we obtain
    [tex]I = e^{-(1/h)f(a)} (\frac {2\pi h}{f''(a)})^{1/2} e^{-O(h ^{1/2})}[/tex]
    So the first part of the right hand side is ok, it's just the Gaussian integral, but how he knows that the corrections are of the form [tex]e^{-O(h ^{1/2})}[/tex]?? Please, at least tell the general idea how it can be shown. Simply keeping also the terms of the next (third) order, i.e. (q-a)^3, gives a very hard integral, at least for me, to evaluate.

    Attached Files:

  2. jcsd
  3. May 1, 2012 #2
    It does not seem obvious : see for exemple,

    methode_de_laplace by NoƩ Cuneo

    where the demonstration is very strict !
  4. May 2, 2012 #3
    Keeping the third order terms results in the incomplete Airy function as a solution. This would be used to compute approximations where the quadratic term doesn't dominate. When this happens the f'' may be heading for zero. This is not the origin of the O(1/h^1/2) term. That comes from doing an expansion in powers of 1/h.
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