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Confusion about getting uncertainty by using differetiation.

  1. Sep 14, 2009 #1
    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
    [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.
     
    Last edited: Sep 14, 2009
  2. jcsd
  3. Sep 14, 2009 #2
    Basicly what I'm trying to ask is:
    if g=g(f), and I used [tex]\delta g=\frac{{dg}}{{df}}\delta f[/tex]. Then actually I presumed
    [tex]\sqrt{<(g-<g>)^2>}\approx\frac{{dg}}{{df}}\sqrt{<(f-<f>)^2>}[/tex], but I can't see how to prove this.
     
  4. Sep 15, 2009 #3

    Avodyne

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    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).
     
  5. Sep 15, 2009 #4
    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 [tex]\delta g=\frac{{dg}}{{df}}\delta f[/tex] just seems to me more like an abuse of the notation delta here
     
  6. Sep 16, 2009 #5
    Anybody can help, or is there any really correct way of doing this problem? My head really exploded, any help is appreciated.
     
  7. Sep 16, 2009 #6

    Avodyne

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  8. Sep 17, 2009 #7
    Em, so I can use first order part of Taylor expansion to prove it, that's very helpful, thanks
     
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