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Help with understanding the uncertainty principle

  1. Feb 25, 2008 #1
    [SOLVED] Help with understanding the uncertainty principle

    I may have posted this in the wrong forum. If so I am sure a moderator could move it to the correct one.

    Problem statement:
    I have an assignment in which I need to calculate the lowest possible mean of an electron's KE , <KE>, based on the uncertainty of the electron's moment.

    Given data:
    The radius in which the electron is allowed to move, d=2.81*10^-10

    Attempt at solution:
    The uncertainty of the electrons position corresponds to d, deltax=d

    Using the uncertainty principle I got, deltap = hbar/(2*deltax), where deltap is the lowest possible deviation.

    I got deltaKE from, deltaKE = deltap^2/(2m), (m is the electron mass)

    A fellow student explained to me that because my problem involves a stationary system (an electron bound to an atom), <KE> should be zero, which gives me:

    (deltaKE)^2 = <KE^2> - <KE>^2 => (deltaKE)^2 = <KE^2>

    But how do I proceed from here?
    Last edited: Feb 25, 2008
  2. jcsd
  3. Feb 25, 2008 #2


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    Ah... the good ol' "fellow student". Unfortunately, the fellow student is often not correct. For a stationary state [itex]\delta <H>[/itex] is zero, not <KE> (where H=KE + PE). The way you were proceeding previously by considering delta p as related to the given delta x looks fine to me.
  4. Feb 25, 2008 #3


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    for example, if the electron is confined to an atom of size 'a' then the KE is roughly
    \frac{\hbar^2}{m a^2}
    (c.f. particle in a box energy levels (\frac{\hbar^2 \pi^2}{2 m a^2}), etc).
  5. Feb 25, 2008 #4
    I don't think I quite understand. Did you, as I did, derive the expression for the KE from KE = p^2/(2m) ?
    I simply substituted p with deltap to get the min deviation in KE.

    I'm not sure how I am supposed use the expression of H.
  6. Feb 25, 2008 #5


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    yeah. that's right.

    dont use it for anything; I was just pointing out that in a stationary state it is the *total* energy which has no variance, not the *kinetic* energy.
  7. Feb 25, 2008 #6
    Ok, but is the expression (deltaKE)^2 = <KE^2> valid?

    If so, can I just take the square root to get the mean KE?
  8. Feb 25, 2008 #7


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    no... but we are just getting rough estimates here. That's all that is required. That's why we are using the "equation"

    KE "=" (delta p)^2/2m

    where (delta p) is given by \hbar divided by (delta x).
  9. Feb 25, 2008 #8
    I see. Thanks! You've made it a lot clearer.

    I'll post again if I need more help.
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