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Smallest possible uncertainty in the positon of an electron

  1. Sep 11, 2009 #1
    1. The problem statement, all variables and given/known data
    Show that the smallest possible uncertainty in the position of an electron whose speed is given by [tex]\beta=v/c[/tex] is
    [tex]\Delta x_{min}=\frac{h}{4\pi m_{0}c}(1-\beta^{2})^{1/2}[/tex]


    2. Relevant equations
    [tex]\Delta x \Delta p=\frac{h}{4\pi}[/tex]

    [tex]p=mv= \frac{m_{0}}{\sqrt{1-\beta^{2}}}v[/tex]

    3. The attempt at a solution
    so from the momentum equation, i multiply in c:

    [tex]p=\frac{m_{0}c}{\sqrt{1-\beta^{2}}}\frac{v}{c}
    =\frac{m_{0}c}{\sqrt{1-\beta^{2}}}\beta[/tex]

    then i diffrenciate with respect to [tex]\beta[/tex] to get this:

    [tex]\frac{dp}{d\beta}=\frac{m_{0}c}{\sqrt{1-\beta^{2}}}(1+\frac{\beta^{2}}{1-\beta^{2}})[/tex]

    So:

    [tex]\Delta p=\frac{m_{0}c}{\sqrt{1-\beta^{2}}}(1+\frac{\beta^{2}}{1-\beta^{2}})\Delta\beta[/tex]

    So then i assumed that for minimum [tex]\Delta x[/tex] we need maximum [tex]\Delta p[/tex] and thus maximum [tex]\Delta \beta[/tex], which will give us [tex]\Delta \beta=1[/tex] (because max value of v=c)

    But then when i sub this into the uncertainty equation i still have the [tex](1+\frac{\beta^{2}}{1-\beta^{2}})[/tex] term i cant get rid off.

    And are the assumtions i am making valid?
     
  2. jcsd
  3. Sep 11, 2009 #2

    kuruman

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  4. Sep 11, 2009 #3
    ah ok thanks tahnks, thats a bit helpful.
    But i am still a bit unclear. That mean i dont have to do any rigourous maths like diffrenciations?

    and also i am not clear on the part they said, "when [tex]\Delta p[/tex] exceeds [tex]mc[/tex], the uncertainty in energy is greater than [tex]mc^{2}[/tex]" Why?
     
  5. Sep 11, 2009 #4

    gabbagabbahey

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    In SR, the relationship between energy and momentum is:

    [tex]E=\sqrt{(pc)^2+(m_0c^2)^2}[/tex]

    So, [itex]\Delta E \approx \left\vert\frac{\partial E}{\partial p}\right\vert\Delta p=[/itex]____?
     
  6. Sep 11, 2009 #5

    gabbagabbahey

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    You can simplify this further:

    [tex]1+\frac{\beta^{2}}{1-\beta^{2}}=\frac{(1-\beta^2)+\beta^{2}}{1-\beta^{2}}=\frac{1}{1-\beta^{2}})[/tex]

    So using this method, the maximum [itex]\Delta p[/itex] would be [itex]m_{0}c(1-\beta^{2})^{-3/2}[/itex], but this is greater than [itex]m_{0}c(1-\beta^{2})^{-1/2}[/itex] by a factor of [itex](1-\beta^{2})^{-1}[/itex] and the energy argument provides tighter restrictions on [itex]\Delta p[/itex].
     
  7. Sep 11, 2009 #6
    hi gabbagabbahey, sidetrack, what is your signature all about?
     
  8. Sep 11, 2009 #7

    gabbagabbahey

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    It's just a line from the (hilarious IMO) TV series "Better off Ted".
     
  9. Sep 11, 2009 #8
    oh i see, quite a new series, wasn't air in my country though. thanks.
     
  10. Sep 11, 2009 #9
    sorry to be a bit daft, but still cat really get to the end.

    Let me summarize wat i understand so far first.

    The maximum possible uncertanty in E of a partical can be [tex]m_{0}c^{2}[/tex], because anything beyond that it would have enough energy to form another partical.

    and from

    [tex]E=\sqrt{(pc)^{2}+(m_{0}c^{2})^{2}} [/tex]

    we get

    [tex]\frac{\partial E}{\partial p}=\frac{pc^{2}}{\sqrt{(pc)^{2}+(m_{0}c^{2})^{2}}}[/tex]

    So the uncertainty relation with be:

    [tex]\Delta E=\frac{pc^{2}}{\sqrt{(pc)^{2}+(m_{0}c^{2})^{2}}} \Delta p[/tex]

    so set [tex] \Delta E=m_{0}c^2 [/tex] and [tex]p=(1-\beta^{2})^{(-1/2)}m_{0}c \beta[/tex]

    But i end up getting

    [tex] \Delta p= \frac{m_{0}c}{\beta}}[/tex]

    am i doing anything wrong along the way?
     
    Last edited: Sep 12, 2009
  11. Sep 12, 2009 #10
    anyone?
     
  12. Sep 15, 2009 #11
    hi bina, I think you are close to the answear in the very first post...
    just cancel the term [tex](1+\frac{\beta^{2}}{1-\beta^{2}})[/tex], since b should be regarded as a given value. it is only dv we need to deal with. however, dv is c*db right? :)
     
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