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Energy-Time Uncertainty of Gaussian

  1. Mar 22, 2012 #1
    1. The problem statement, all variables and given/known data
    Check the energy-time uncertainty principle for:
    [tex]\Psi(x,0)=Ae^{-ax^2}e^{ilx}[/tex]
    using the observable x.

    2. Relevant equations
    [tex]
    \Delta{E}\Delta{t}\geq\hbar/2\\
    \Delta{E}=\sigma_H=\sqrt{<H^2>-<H>^2}\\
    \Delta{t}=\frac{\sigma_x}{\frac{d<x>}{dt}}
    [/tex]


    3. The attempt at a solution
    First I find the wavefunction with time dependence
    [tex]
    \Psi(x,t)=\frac{1}{\sqrt{2\pi}}\int{\phi(k)e^{ikx-i\frac{\hbar k^{2}t}{2m}}dk}\\
    \phi(k)=\frac{1}{\sqrt{2\pi}}\int{\Psi(x,0)e^{-ikx}dx}
    [/tex]
    This gives me a fairly complicated expression
    [tex]
    \Psi(x,t)=\frac{Aexp(\frac{-ax^{2}+lix-\frac{tl^{2}i\hbar}{2m}}{w^{2}})}{w}\\
    A^{2}=\sqrt{\frac{2a}{\pi}},w^{2}=1+\frac{\hbar2ita}{m}
    [/tex]
    Just finding <H> (not to mention <H^2> is giving me a lot of trouble. Is grinding out the derivatives and integrals the only way to do this problem?

    The only other options I can think of:

    1: <H> and <H^2> should not change in time (?), so I can just use the wavefunction at t=0 to find them? (But, still have to use the wavefunction with time to get the std dev of x)

    2: Considering [tex] \frac{e^{ikx}}{\sqrt{2\pi}} [/tex] are like orthonormal eigenfunctions for H, could I say:

    [tex]
    <H>=\int{\phi(k)^{2}E(k)dk}
    [/tex]
     
  2. jcsd
  3. Mar 24, 2012 #2
    remember that

    [itex] \Psi (x, t) = e^{(-i/ \hbar) H t} \Psi (x, 0) [/itex]

    H commutes past the time evolution operator

    x does not, use a series expansion to determine the correct commutation relation

    Furthermore complete the square in the gaussian
     
  4. Mar 24, 2012 #3
    You are definitely on the right track. Your expression is going to be ugly. Now, set
    [tex] z\equiv 2 \hbar at/m[/tex]
    and use
    [tex] \lvert \psi \lvert^2= \sqrt{ \frac{2a}{ \pi}} \frac{1}{ \sqrt{1+ z^2}} e^{-l^2/2a}e^{a\big\{ \frac{(ix+l/2a)^2}{(l+iz)}+ \frac{(-ix +l/2a)^2}{(1-iz)}\big\}}[/tex]
    Expand the term in the brackets {} of the right-hand exponent, and plug it back into
    [tex] \lvert \psi(x,t) \lvert^2[/tex]
    where your psi should be
    [tex] \psi(x,t)= \bigg( \frac{2a}{ \pi}\bigg)^{1/4} \frac{1}{ \sqrt{1+2i \hbar at/m}} e^{-l^2/4a}e^{a(ix+l/2a)^2/(1+2ia \hbar t/m)} [/tex]
     
    Last edited: Mar 24, 2012
  5. Mar 24, 2012 #4
    Thanks for the help. I think I got it.

    As a check on comprehension, I'd be interested in knowing if #2 in the original post is right or wrong and why. It seems to me that phi(k) above is for energy eigenfunctions what the momentum space wavefunction is for momentum eigenfunctions. That is, roughly, both show the amount of each eigenfunction in psi.
     
  6. Mar 25, 2012 #5
    Code (Text):
     
    Sorry, I misunderstood your last question, so I'm editing my post. I'm going to have to come back to your #2 point later, since it's getting late where I am. The following was written when I misunderstood your question, so I'm just going to post it rather than waste it. If we're being honest, this problem is designed just to make you do some math. It doesn't really help you with any physical intuition. By convention, the time evolution operator is
    [tex] \hat U (t,t_0) = e^{-i(t-t_0) \, \hat H/ \hbar}[/tex]
    or in application,
    [tex] \big\lvert \, \psi(t) \big>= e^{-i(t-t_0) \, \hat H/ \hbar} \big\lvert \, \psi(t_0) \big> [/tex]
    So, for example, if you've got an infinite square well, at t_0 =0, in its ground state,
    [tex] \big\lvert \, \psi_{n=1} (t=0) \big>= \big\lvert \, \psi_1 (0) \big>= \sqrt{ \frac{2}{a}} \sin\frac{ \pi x}{a} [/tex]
    You know the energy is
    [tex]E_1= \frac{ \hbar^2 \pi^2}{2ma^2} [/tex]
    So at a future time t,
    [tex] \big\lvert \, \psi_1 (t) \big>= e^{-it \, \hat H/ \hbar} \bigg(\sqrt{ \frac{2}{a}} \sin\frac{ \pi x}{a}\bigg) = \bigg(\sqrt{ \frac{2}{a}} \sin\frac{ \pi x}{a}\bigg)e^{-it \, E_1/ \hbar} =\sqrt{ \frac{2}{a}} \sin\frac{ \pi x}{a}e^{-it \, \hbar \pi^2/2m a^2}[/tex]
    Time evolution is really pretty simple. It gets convoluted with Green's functions and stuff like that, but really for all intents and purposes, you should just think of it as in the example above.
     
    Last edited: Mar 25, 2012
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