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3d Harmonic oscillator

  1. Feb 21, 2016 #1
    1. The problem statement, all variables and given/known data
    The wave function for the three dimensional oscillator can be written
    ##\Psi(\mathbf r) = Ce^{-\frac{1}{2}(r/r_0)^2}##
    where ##C## and ##r_0## are constants and ##r## the distance from the origen.
    Calculate
    a) The most probably value for ##r##
    b) The expected value of ##r##
    c) The expected value of ##1/r##.

    2. Relevant equations
    Expected value of function with a normed wave function
    ##<f(r)> = \int dr \Psi(r)^*f(r) \Psi ##.

    3. The attempt at a solution
    a) It is my understanding that the most probably value is the maximum value of ##\Psi^* \Psi## or equivalently in this case the maximum of ##\Psi## which is at ##r=0##.
    The answer to the exercise however disagrees and says it's at ##r_0##.
    If the oscillator is centered at ##r_0## this makes physical sense with the most probably value of course being the centre but it doesn't seem to agree with maximizing the function.

    b) Norming the wave function
    ##1 = C^2 \int_0^\infty e^{-r^2/r_0^2}dr = C^2 \int_0^\infty r_0 e^{-s^2}ds = C^2\frac{r_0 \sqrt{\pi}}{2}## and hence that ##C^2 = \frac{2}{r_0 \sqrt{\pi}}##.
    The expected value of ##r## is then
    ##<r> = \frac{2}{r_0\sqrt{\pi}} \int_0^\infty re^{-r^2/r_0^2} dr = \frac{r_0}{\sqrt{\pi}}##.
    The answer however says ##\frac{2r_0}{\sqrt{\pi}}##.

    c) For this one, I get a divergent integral ##<1/r> = \frac{2}{r_0\sqrt{\pi}} \int_0^\infty \frac{e^{-r^2/r_0^2}}{r} dr##.
     
  2. jcsd
  3. Feb 21, 2016 #2

    PeroK

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    You're very modest to post QM under "introductory" homework. In any case, your problems stem from not thinking enough about integrating in spherical coordinates.
     
  4. Feb 21, 2016 #3
    Well it's a first course in QM for second year university students (at an European university) so I felt it being more introductory. The advanced forum I thought were mainly for graduate students.

    Thanks! I forgot I have to integrate over the volume in this case.
    Norming the function
    ##1 = C^2 \int_0^\infty \int_0^\pi \int_0^{2\pi} r^2\sin \theta e^{-r^2/r_0^2}d\phi d\theta dr = C^2 4\pi \frac{\sqrt{\pi}r_0^3}{4} = C^2 \pi \sqrt{\pi} r_0^3 \Longrightarrow C^2 = \frac{1}{\pi \sqrt{\pi }r_0^3}##.

    The expected value is then
    ##<r> = \frac{1}{\pi \sqrt{\pi} r_0^3} 4\pi \int_0^\infty r^3e^{-r^2/r_0^2}dr = \frac{1}{\pi \sqrt{\pi} r_0^3} 4\pi \frac{r_0^4}{2} = \frac{2r_0}{\sqrt{\pi}}## as expected!

    And for ##1/r##
    ##<1/r> = \frac{4}{\sqrt{\pi} r_0^3} \int_0^\infty re^{-r^2/r_0^2}dr = \frac{2}{\sqrt{\pi} r_0}##.

    However as for the a) question I'm still not sure? that doesn't have anything with integrating to do.
     
  5. Feb 21, 2016 #4

    PeroK

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    Why not calculate the probability that the particle is found near a given distance from the origin: between ##R## and ##R+ \epsilon## say?
     
  6. Feb 21, 2016 #5
    After thinking about this for a while it seems my error here just as before is that I need the volume element.

    As far as I understand it, the probability distribution is
    ##P(\mathbf r) = |\Psi(\mathbf r)|^2 = \Psi^*(\mathbf r) \Psi(\mathbf r).##
    And the probability of finding the particle within a certain volume is
    ##\int_V P(\mathbf r) dV##.
    In our case that is
    ##I = \frac{4}{\sqrt{\pi}r_0^3} \int_R^{R+\epsilon} r^2e^{-r^2/r_0^2}dr##
    So we want to find the ##R## that maximizes ##\int_R^{R+\epsilon} r^2e^{-r^2/r_0^2}dr## when ##\epsilon \to 0##.
    However since this seem complicated we can instead maximize the integrand ##r^2e^{-r^2/r_0^2}## which has a maximum for ##r=r_0##.
    So I need to account for the scale factors when I'm not working in Cartesian coordinates. Thanks for helping!
     
  7. Feb 21, 2016 #6

    PeroK

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    Yes. The thing you're missing is that the limit of an integral such as the one you have is simply the function value at the point:

    ##\lim_{\epsilon \rightarrow 0} \int_{R}^{R+\epsilon} f(r) dr = f(R)##

    So, maximising that integral is equivalent to maximising ##f##
     
  8. Feb 21, 2016 #7
    Thanks for pointing that out. I had forgotten about the mean value theorem!
     
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