1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
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: 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.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook