- #1
weirdoguy
- 1,095
- 1,028
Hello everyone, my first post :shy:
I'm reading Zee's 'QFT in a Nutshell' and I came to one thing that bothers me - he's short discussion of steepest-descent approximation. I've known this thing for quite a long time now, but I've never seen the approximation of the corrections. Here is what he writes:
[tex]I = e^{-(1/\hbar)f(a)} (\frac {2\pi \hbar}{f''(a)})^{1/2} e^{-O(\hbar ^{1/2})}[/tex]
Of course we take the limit in which Planck's constant is small, and that is where problem occurs. Because in this limit [tex]e^{-O(\hbar ^{1/2})}[/tex] will approach infinity, and that is not what it should be like, right? Any thoughts about this issue? I just think that what he wrote is simply incorrect.
I tried to derive this approximation by not neglecting the cubic termis in (x-a), and I don't even see why there is a square root of Planck's constant...
Sorry for my english, it's been a long time since I wrote something in this language :shy:
I'm reading Zee's 'QFT in a Nutshell' and I came to one thing that bothers me - he's short discussion of steepest-descent approximation. I've known this thing for quite a long time now, but I've never seen the approximation of the corrections. Here is what he writes:
[tex]I = e^{-(1/\hbar)f(a)} (\frac {2\pi \hbar}{f''(a)})^{1/2} e^{-O(\hbar ^{1/2})}[/tex]
Of course we take the limit in which Planck's constant is small, and that is where problem occurs. Because in this limit [tex]e^{-O(\hbar ^{1/2})}[/tex] will approach infinity, and that is not what it should be like, right? Any thoughts about this issue? I just think that what he wrote is simply incorrect.
I tried to derive this approximation by not neglecting the cubic termis in (x-a), and I don't even see why there is a square root of Planck's constant...
Sorry for my english, it's been a long time since I wrote something in this language :shy: