Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Uncertainty in the wavepacket

  1. Aug 30, 2007 #1

    T-7

    User Avatar

    Hi,

    I'm puzzled by a couple of formulae in the answer sheet to a problem set I'm working on.

    To calc. the new uncertainty in the position of a group of electrons, initially localised to [tex]\pm1\mum[/tex], after time t, it uses the factor:

    [tex]\left(1+\frac{\hbar^{2}(\Delta k^{4}t^{2})}{4m^{2}}\right)^{1/2}[/tex]

    The equation I would use for the probability density of this wave is

    [tex]
    \left|\Psi(z,t)\right|^{2} = \frac{\pi.(\Delta k^{2})}{\left(1+\frac{\hbar^{2}(\Delta k^{4}t^{2})}{4m^{2}}\right)^{1/2}}.exp\left(\frac{(\Delta k^{2})(z-vt)^{2}}{2\left(1+\frac{\hbar^{2}(\Delta k^{4}t^{2})}{4m^{2}}\right)}\right)
    [/tex]

    The denominator of the exp. component describes the increasing width of the wave packet (whereas [tex]\left(1+\frac{\hbar^{2}(\Delta k^{4}t^{2})}{4m^{2}}\right)^{1/2}[/tex] describes the decreasing peak height). I am puzzled by the square root sign they've introduced. Shouldn't they have simply applied it without?

    Clearly, one needs to calc. [tex]\Delta k[/tex] to determine the new uncertainty in position, but I'm also puzzled by the claim that [tex]\Delta p\Delta z \approx \sqrt{2}[/tex] (nowhere justified). How do they figure that one?

    Cheers!
     
    Last edited: Aug 30, 2007
  2. jcsd
  3. Aug 30, 2007 #2
    The uncertainty of a gaussian is usually taken to be its std. dev. The denominator in the exponent is usually [tex]2\sigma^2[/tex] where sigma is the S.D.
     
  4. Aug 30, 2007 #3

    T-7

    User Avatar

    Thank you. :smile:
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook