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Kinetic Energy and the Heisenberg Uncertainty Principle

  1. Jun 22, 2012 #1
    1. The problem statement, all variables and given/known data

    This is not a problem as such. Just a derivation for which I've been given a solution which I cannot seem to find.


    2. Relevant equations
    Ke = 1/2 mv2 = ρ2/2m
    hbar << 2ΔxΔp

    Δp≈p as the average magnitude of p is small.

    3. The attempt at a solution

    p >> hbar/2Δx
    p2 = hbar2/4(Δx)2
    so p2/2m=ke=hbar2/8(Δx2)m

    whereas I've been given the result that T=h2/2m(Δx2)
     
  2. jcsd
  3. Jun 26, 2012 #2

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    Hello ConorB,

    One of the reasons that might make this confusing is the various incarnations that Heisenberg's uncertainty principle takes form.

    In its most generic form, it is often expressed by

    [tex] \Delta x \Delta p \approx h [/tex]

    Note that the [itex] h [/itex] above is not even [itex] \hbar [/itex]. It's just the normal Planck's constant.

    If you really want the formal version of Heisenberg's uncertainty principle, you'll first need to know the shape of the wave-function in question. If you know the shape of the wave-function, you can calculate the standard deviation of position, [itex] \sigma_x [/itex],and also the standard deviation of momentum, [itex] \sigma_p [/itex]. Once you have those, the ultimate version of the uncertainty principle is

    [tex] \sigma_x \sigma_p \geq \frac{\hbar}{2} [/tex]

    where [itex] \hbar = \frac{h}{2 \pi} [/itex]. The special case where they are equal is if the shape of the wave-function is the Gaussian shape (i.e. bell curve).

    But I don't think the solution you were given used the formal version of the uncertainty principle. I'm guessing it is using the more approximate version of [itex] \Delta x \Delta p \approx h [/itex]. Starting with that, plug [itex] \Delta p [/itex] into your original equation for kinetic energy and see what happens. :smile:
     
    Last edited: Jun 26, 2012
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