# Steepest-descent approximation

1. Sep 8, 2013

### weirdoguy

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:

$$I = e^{-(1/\hbar)f(a)} (\frac {2\pi \hbar}{f''(a)})^{1/2} e^{-O(\hbar ^{1/2})}$$

Of course we take the limit in which Planck's constant is small, and that is where problem occurs. Because in this limit $$e^{-O(\hbar ^{1/2})}$$ 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:

2. Sep 8, 2013

### Staff: Mentor

Why does it approach infinity? If ℏ goes to zero, the square root does the same, and the expression goes to e^0 = 1.
Or does the O mean something different?

$e^{-(1/\hbar)}$ will go to zero.

3. Sep 8, 2013

### UltrafastPED

And the other factor goes to zero ... and since 1 x 0 = 0 the expression vanishes.

4. Sep 8, 2013

### D H

Staff Emeritus
In the limit of small ħ, exp(-O(ħ1/2)) approaches one, not infinity. Perhaps you are thinking of exp(O(ħ-1/2)). That's a very different quantity from exp(-O(ħ1/2)).

5. Sep 8, 2013

### weirdoguy

Oh, my bad... I don't know why I thought that it goes to infinity, I had graph of a wrong function in my mind.

So now, I still have a problem - how did he get this exp(-O(ħ1/2)) factor? Because I don't see why it is a square root of ħ...

6. Sep 8, 2013

### vanhees71

If you tell us, from where the book starts, we may be able to help you. Is it the application of this method to the generating functional in QFT. Then my QFT manuscript may help you too:

http://fias.uni-frankfurt.de/~hees/publ/lect.pdf

the $\hbar$ (loop) expansion is found on p. 125ff

7. Sep 8, 2013

### weirdoguy

It is probably applied later in the book, but it firstly appears in the very begining. I made a screen of everything that there is about this issue, not that much though...

And thank you vanhees71 for lecture, I'll check it later

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8. Sep 8, 2013

### D H

Staff Emeritus
For a reasonably well-behaved function f(q), the error in that Taylor expansion is going to be dominated by the last included term, $\frac 1 2 f''(q_0)(q-q_0)^2$.

Look at $\int_{-\infty}^{\infty} \exp\left(\frac{-1}{2h} f''(q_0)(q-q_0)^2\right)\, dq$. Sans some scale factors, that's just the Gaussian integral.

9. Sep 11, 2013

### weirdoguy

Well, I know, and I know how to integrate it. But still, I can't see why corrections are in the form of exponent with $-O(\sqrt{\hbar})$.
But now - I tried to see what will happen if we won't neglect higher order terms. So, we have:
$$f(q)=f(a)+\frac{1}{2}f''(a)(q-a)^2+\frac{1}{3!}f'''(a)(q-a)^3+\ldots$$
$$I=\int_\mathbb{R}dq\exp\left(\frac{-1}{\hbar}f(a)+\frac{-1}{\hbar}\frac{1}{2}f''(a)(q-a)^2+\frac{-1}{\hbar}\frac{1}{3!}f'''(a)(q-a)^3+\ldots \right)$$
And now I leave the exponent with the quadratic term, and expand the exponent with qubic and higher terms in Taylor series:
$$I=\int_\mathbb{R}dq\exp\left(\frac{-1}{\hbar}f(a)+\frac{-1}{\hbar}\frac{1}{2}f''(a)(q-a)^2\right)\left[1+\left(\frac{-1}{3!\hbar}f'''(a)(q-a)^3+\ldots\right)+\left(\frac{-1}{3!\hbar}f'''(a)(q-a)^3+\ldots\right)^2 \right]$$
Now, $I$ is a sum, first summand is of course just our basic integral, which I will denote by $I_0$:
$$I=I_0+\int_\mathbb{R}dq\exp\left(\frac{-1}{\hbar}f(a)+\frac{-1}{\hbar}\frac{1}{2}f''(a)(q-a)^2\right)\left[\left(\frac{-1}{3!\hbar}f'''(a)(q-a)^3+\ldots\right)+\left(\frac{-1}{3!\hbar}f'''(a)(q-a)^3+\ldots\right)^2 \right]$$
First non-zero integral will be the one with $(q-a)^4$:
$$-\frac{f^{(IV)}(a)}{4!\hbar}e^{\frac{-f(a)}{\hbar}}\int_\mathbb{R}dqe^{-\frac{f''(a)}{2\hbar}(q-a)^2}(q-a)^4=-\frac{3f^{(IV)}(a)}{4!\hbar}\left(\frac{\hbar}{f''(a)}\right)^2 \sqrt{\frac{2\pi\hbar}{f''(a)}} e^{\frac{-f(a)}{\hbar}}$$
Combining this with $I_0$ we get:
$$I= \sqrt{\frac{2\pi\hbar}{f''(a)}} e^{\frac{-f(a)}{\hbar}} \left(1+C\cdot\hbar+\ldots\right)$$
Where $C$ is a constant. I looked at terms with $(q-a)^6$ and it will give contribution to the terms with $\hbar$ and $\hbar^2$. Anyway, the conclusion is that still I see no way of exponent correcions with $-O(\hbar^{1/2})$ :shy:

10. Sep 11, 2013

### D H

Staff Emeritus
Don't look at those higher order terms. Look instead at the last term that was included.

In quoting just the last part of my previous post, you omitted the key point of that post. Once again,

For a reasonably well-behaved function f(q), the error in that Taylor expansion is going to be dominated by the last included term, $\frac 1 2 f''(q_0)(q-q_0)^2$.​

You don't need to look at those higher order derivatives in the expansion of f(q) if that function is "well-behaved". By well behaved I mean that there exists some finite C>0 such that for all q in the interval of interest,
$$\left|\sum_{r=n+1}^{\infty} \frac 1 {r!} f^{r}(a)(q-a)^r\right|<C\left|\frac 1 {n!} f^{n}(a)(q-a)^n\right|$$

In other words, for a well behaved function, the last included term in the Taylor expansion (in this case, the second order derivative term) bounds the error. If this is the case, that exp(-O(ħ1/2)) multiplicative factor falls right out.