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.(adsbygoogle = window.adsbygoogle || []).push({});

HW problem: Show that for a free particle the uncertainty relation can also be written as

[tex]\delta \lambda \delta x \ge \frac{{{\lambda ^2}}}{{4\pi }}[/tex].

Where [tex]\delta \lambda [/tex] is the de Broglie's wave length

My solution is :

[tex]\delta \lambda = |\frac{{d\lambda }}{{dp}}|\delta p = \frac{h}{{{p^2}}}\delta p[/tex]

so [tex]\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 }}[/tex]

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

And I try to work out a counterexample:

Suppose W and G are two physical quantities, G follows the normal distribution

[tex]G = \frac{{10}}{{\sigma \sqrt {2\pi } }}\exp ( - \frac{{{x^2}}}{{2{\sigma ^2}}})[/tex]

[tex]W = 100G = \frac{{1000}}{{\sigma \sqrt {2\pi } }}\exp ( - \frac{{{x^2}}}{{2{\sigma ^2}}})[/tex]

So W and G should have the same standard deviation [tex]\sigma [/tex] (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.

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# Confusion about getting uncertainty by using differetiation.

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