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3D harmonic oscillator- expected value of distance

  1. Jun 9, 2013 #1
    1. The problem statement, all variables and given/known data
    Hey! I got this problem about 3D harmonic oscillator, here it goes:
    A particle can move in three dimensions in a harmonic oscillator potential
    ##V(x,y,z)=\frac{1}{2}m\omega^2(x^2+y^2+z^2)##. Determine the ground state wave function. Check by explicitly counting that it is properly normalized.
    Calculate the expectation value ##<r>##, where ##r=\sqrt{x^2+y^2+z^2}## is a distance to the origin.

    2. Relevant equations
    ##\alpha=\sqrt{\frac{m\omega}{\hbar}}##
    ##\omega=\sqrt{\frac{k}{m}}##
    ##V(x,y,z)=\frac{1}{2}m\omega^2(x^2+y^2+z^2)##
    ##\psi(x,y,z)==X(x)Y(y)Z(z)##
    ##-\frac{\hbar^2}{2m}[\frac{\xi^2\psi}{\xi x^2}+\frac{\xi^2\psi}{\xi y^2}+\frac{\xi^2\psi}{\xi z^2}]+\frac{k}{2}(x^2+y^2+z^2)\psi=E\psi##
    ##E_x+E_y+E_z=E##

    3. The attempt at a solution
    I pretty much calculated the wave function for the ground state, and it came out to be equal to ##\psi_0(x,y,z)=(\frac{\alpha}{\pi})^\frac{3}{4} exp(-\alpha (x^2+y^2+z^2))##
    with energy equal to ##\frac{3}{2}\hbar\omega##. I don't understand what do they mean by explicitly calculating proper normalization, and I have trouble with calculating the expectation value for the distance from the point of origin. My first guess was to do this:
    ##\int_{-\infty}^{+\infty}\int_{-\infty}^{+\infty}\int_{-\infty}^{+\infty} \psi^{*}(x,y,z) r \psi(x,y,z)\,dx\,dy\,dz.## But it was quite a while since I've done multpile integrals, so I used wolframalpha for this:
    http://www.wolframalpha.com/input/?...)^(3/4))*exp(-a(x^2+y^2+z^2)(1/2)))^2)+dxdydz
    I am pretty shure that notation is wrong, but you probably get my point. I don't know if my way of thinking is correct. Help?
     
    Last edited: Jun 9, 2013
  2. jcsd
  3. Jun 9, 2013 #2
    In general: [itex] < ψ_{i} | ψ_{j} > = δ_{ij}. [/itex]
    Use the normalization condition [itex] ∫^{\infty}_{-\infty}dxdydz|ψ(x,y,z)|^{2}=1 [/itex] to check the ground state is properly normalized.

    To calculate [itex] < r > [/itex] use spherical coordinates.
     
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