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Time dependent dispersion (Quantum Mechanics)

  1. Sep 4, 2011 #1
    1. The problem statement, all variables and given/known data
    The initial wave function [itex]\Psi (x,0) [/itex] of a free particle is a normalized gaussian with unitary probability. Let [itex]\sigma = \Delta x[/itex] be the initial variance (average of the square deviations) with respect to the position; determine the variance [itex]\sigma (t)[/itex] in a moment later.

    2. Relevant equations
    I'm not sure which equations should I start from... but, as for a gaussian distribution, I'm thinking of
    [itex]\Psi (x,t) = \int^{\infty}_{-\infty} a(k) e^{i(kx - \omega t)} dk [/itex],
    being

    [itex]a(k) = \frac{A \sigma}{\sqrt{2\pi}} \exp{\left[-\frac{(k - k_0)^2 {\sigma}^2}{2}\right]} [/itex]


    3. The attempt at a solution
    I don't know how to start it... once I describe the wave function, find A, I don't know how to construct this time dependent dispersion. I know the packet will 'spread', and its width will increase, but no ideas on how to describe this in terms of the dispersion... I know the result is [itex]\sigma (t) = \sqrt{\sigma^2(0) + (\frac{\hbar t}{2m \sigma(0)})^2}[/itex]
     
  2. jcsd
  3. Sep 7, 2011 #2

    diazona

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    Homework Helper

    Sorry you haven't gotten an answer on this yet, but if it's still relevant: I'd start by determining the dispersion relation [itex]\omega(k)[/itex] for the particle and plugging that in. The eventual goal is to take the explicitly time-dependent expression you have,
    [tex]\Psi (x,t) = \int^{\infty}_{-\infty} a(k,\sigma) e^{i(kx - \omega t)} dk[/tex]
    and rewrite it as an implicitly time-dependent expression
    [tex]\Psi (x,t) = \int^{\infty}_{-\infty} a(k,\sigma(t)) e^{ikx} dk[/tex]
    which should give you the expression you need.
     
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