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Quantum Mechanics (position uncertainty)

  1. Sep 16, 2012 #1
    1. The problem statement, all variables and given/known data

    I need some help with the following problem:

    http://img827.imageshack.us/img827/4061/prob1y.jpg [Broken]

    2. Relevant equations

    For some particle in state Ψ the expectation value of x is given by:

    [itex]\left\langle x \right\rangle = \int^{+\infty}_{-\infty} x |\Psi(x, t)|^2 \ dx[/itex]

    The wave function in ground state:

    [itex]\psi_0 (x)= A e^{\frac{-m \omega}{\hbar}x^2}[/itex]

    3. The attempt at a solution

    Let α = mω/2ħ, so

    [itex]\left\langle x \right\rangle = \int^{+\infty}_{-\infty} x A^2 (e^{-\alpha x^2})^2 \ dx[/itex]

    [itex]= A^2 \int^{+\infty}_{-\infty} x (e^{-2 \alpha x^2}) \ dx[/itex]

    [itex]=A^2 \left[ \frac{-e^{-2 \alpha x^2}}{4 \alpha} \right]^{+\infty}_{-\infty} \ dx[/itex]

    But since e=∞ and e-∞=0 (i.e. the limits as x approaches ±∞) we get:

    <x>=A2 ∞ = ∞

    So, what have I done wrong here? :confused:

    Also for <x2> we have the following

    [itex]\left\langle x^2 \right\rangle = \int^{+\infty}_{-\infty} x^2 A^2 (e^{-\alpha x^2})^2 \ dx[/itex]

    But similarly here I will encounter the same problem. So how can I evaluate the <x> and <x2> without getting infinity?

    Clearly they can't equal to infinity since we must obtain:

    [itex]\Delta x = \sqrt{\left\langle x^2 \right\rangle -\left\langle x \right\rangle^2} = \sqrt{\frac{\hbar}{2m \omega}}[/itex]

    Any help is greatly appreciated.
     
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Sep 16, 2012 #2
    Check the evaluation of your upper and lower limits on your integrals again. ;)
     
  4. Sep 16, 2012 #3
    The upper limit is +∞ and the lowe limit is -∞. Here's what I did:

    [itex]A^2 \left( \frac{e^{-2 \alpha \infty^2}}{4 \alpha} - \frac{e^{2 \alpha \infty^2}}{4 \alpha} \right)[/itex]

    [itex]= A^2 \left( \frac{e^{-\infty^2}}{4 \alpha} - \frac{e^{\infty^2}}{4 \alpha} \right)[/itex]

    [itex]= A^2 (0- 0) = 0[/itex]

    Edit: this time I get 0. So <x>=0?
     
    Last edited: Sep 16, 2012
  5. Sep 16, 2012 #4
    Yes! Not infinity as you were getting before. It makes sense to, a simple ground state harmonic oscillator spends most of the time close to origin.
     
  6. Sep 16, 2012 #5
    Okay, but what about <x2>? The integral [itex]\int^{+\infty}_{-\infty} x^2 A^2 (e^{-\alpha x^2})^2 \ dx[/itex] turns out to be extremely messy. I'm just wondering if I am on the correct path? :confused:

    If I integrate that I get this:

    [itex]\left[ \frac{\sqrt{\pi/2} \ erf(\sqrt{2} \sqrt{\alpha} x)}{8 \alpha^{3/2}} - \frac{x e^{-2\alpha x^2}}{4 \alpha} \right]^{+\infty}_{-\infty}[/itex]

    But I must get <x2>=ħ/2mω so that the position uncertainty would be the square root of that.
     
  7. Sep 16, 2012 #6
    You are the right path, if you evaluate the integrals you should get the right value.
     
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