# Confusion about getting uncertainty by using differetiation.

1. Sep 14, 2009

### kof9595995

My question comes from my homework, but I don't think it's a homework question, so I put it here, still I will put the homework in this thread caus I think it would help.
HW problem: Show that for a free particle the uncertainty relation can also be written as
$$\delta \lambda \delta x \ge \frac{{{\lambda ^2}}}{{4\pi }}$$.
Where $$\delta \lambda$$ is the de Broglie's wave length
My solution is :
$$\delta \lambda = |\frac{{d\lambda }}{{dp}}|\delta p = \frac{h}{{{p^2}}}\delta p$$
so $$\delta \lambda \delta x = \frac{h}{{{p^2}}}\delta x\delta p \ge \frac{h}{{{p^2}}}\frac{h}{{4\pi }} = \frac{1}{{4\pi }}{(\frac{h}{p})^2} = \frac{{{\lambda ^2}}}{{4\pi }}$$

Although I got the expected result, but I really doubt if it's legal to use differentiation here. Because $$\delta p$$ is a standard deviation not increment. Using differentiation just means you map the range $$\delta p$$ to another range $$\delta \lambda$$ as if they were increments. $$\delta p$$ is the standard deviation for $$\delta p$$ distribution, but how do you know the $$\delta \lambda$$ is the standard deviation for $$\delta \lambda$$ distribution?

And I try to work out a counterexample:
Suppose W and G are two physical quantities, G follows the normal distribution
$$G = \frac{{10}}{{\sigma \sqrt {2\pi } }}\exp ( - \frac{{{x^2}}}{{2{\sigma ^2}}})$$

$$W = 100G = \frac{{1000}}{{\sigma \sqrt {2\pi } }}\exp ( - \frac{{{x^2}}}{{2{\sigma ^2}}})$$

So W and G should have the same standard deviation $$\sigma$$ (I'm not quite sure , am I correct at this?), but differentiation tells you the standard deviation should be 100 times of the first distribution.

EDIT: My counterexample is wrong, please ignore it.

Last edited: Sep 14, 2009
2. Sep 14, 2009

### kof9595995

Basicly what I'm trying to ask is:
if g=g(f), and I used $$\delta g=\frac{{dg}}{{df}}\delta f$$. Then actually I presumed
$$\sqrt{<(g-<g>)^2>}\approx\frac{{dg}}{{df}}\sqrt{<(f-<f>)^2>}$$, but I can't see how to prove this.

3. Sep 15, 2009

### Avodyne

You're correct that it's only approximately true. It's like approximating f(x) near x=a by f(a)+f'(a)(x-a).

4. Sep 15, 2009

### kof9595995

I understand in the differentiation case why f(a)+f'(a)(x-a) is the first order approximation. But I just can't see in any way a standard deviation should behave like this, even in an approximation sense.
So $$\delta g=\frac{{dg}}{{df}}\delta f$$ just seems to me more like an abuse of the notation delta here

5. Sep 16, 2009

### kof9595995

Anybody can help, or is there any really correct way of doing this problem? My head really exploded, any help is appreciated.

6. Sep 16, 2009

### Avodyne

7. Sep 17, 2009

### kof9595995

Em, so I can use first order part of Taylor expansion to prove it, that's very helpful, thanks