- #1
Mesmerized
- 54
- 0
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 \int_{-\infty}^\infty dx e^{-1/2ax^ 2} =(\frac {2\pi}{a})^ {1/2}
To compute the integral I=\int_{-\infty}^\infty dq e^{-1/h f(q)} 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, f(q)=f(a)+1/2f''(a)(q-a)^{2}+O[(q-a)^{3}] we obtain
I = e^{-(1/h)f(a)} (\frac {2\pi h}{f''(a)})^{1/2} e^{-O(h ^{1/2})}
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 e^{-O(h ^{1/2})}?? 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.
The only prior information required is the Gaussian integral \int_{-\infty}^\infty dx e^{-1/2ax^ 2} =(\frac {2\pi}{a})^ {1/2}
To compute the integral I=\int_{-\infty}^\infty dq e^{-1/h f(q)} 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, f(q)=f(a)+1/2f''(a)(q-a)^{2}+O[(q-a)^{3}] we obtain
I = e^{-(1/h)f(a)} (\frac {2\pi h}{f''(a)})^{1/2} e^{-O(h ^{1/2})}
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 e^{-O(h ^{1/2})}?? 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.