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Heisenberg Uncertainty in wavelength and position

  1. Jul 20, 2015 #1
    1. The problem statement, all variables and given/known data

    Show that ## \Delta\lambda\Delta\ x>lamdba^2/4*pi##




    3. The attempt at a solution

    When I substitute de Broglie's p=h/lambda I get the equation of

    ##\frac {\Delta\x}{\Delta\lambda} > 1/(4*pi )##
     
  2. jcsd
  3. Jul 20, 2015 #2

    Orodruin

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    You need to show more of your work. How are we supposed to tell where you went wrong if you only provide us with your final result? (I anyway have a pretty good idea of where you have gone wrong, but I want to see exactly what you did first.)
     
  4. Jul 20, 2015 #3
    Here is my working
     

    Attached Files:

  5. Jul 20, 2015 #4

    Orodruin

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    The first row is not correct. ##p = h/\lambda## does not imply ##\Delta p = h/\Delta \lambda##. What is ##d\lambda/dp##?
     
  6. Jul 20, 2015 #5
    I have attached a worked solution in which I came to the right answer so I believe it must be right. However, I still don't understand intuitively why ##p = h/\lambda## does not imply ##\Delta p = h/\Delta \lambda##
     

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  7. Jul 20, 2015 #6

    Orodruin

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    Because it is a matter of how a small change in ##\lambda## changes ##p##. If you used your formula, a small change in ##\lambda## would give a huge change in ##p##.
     
  8. Jul 20, 2015 #7

    BvU

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    If ##\ \ y = {1\over x}\ \ ## then surely ##\ \ {dy\over dx} = -{1\over x^2}\ \ \Rightarrow \ \ dy = -{dx\over x^2}\ ## . Change d to ##\Delta## and voila !

    (Sorry for barging in, Oro...)
     
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