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3D quantum harmonic oscillator: linear combination of states

  1. May 28, 2017 #1
    1. The problem statement, all variables and given/known data

    Hi everybody! In my quantum mechanics introductory course we were given an exercise about the 3D quantum harmonic oscillator. We are supposed to write the state ##l=2##, ##m=2## with energy ##E=\frac{7}{2}\hbar \omega## as a linear combination of Cartesian states ##\psi_{n_x, n_y, n_z}##.

    2. Relevant equations

    ##Y_{nlm} (r,\theta,\varphi)=R_{nl}(r) Y_l^m(\theta,\varphi)##

    3. The attempt at a solution

    This comes at the end of a series of exercises about the QHO, so I already know that for this energy I must have ##n=2## and ##n_r=0## since ##n=2n_r+l## and ##E_n = (2n_r+l+\frac{3}{2})\hbar\omega##. I've also seen ##k## being used instead of ##n_r##. Just to make sure what I mean, this is the degree of the polynomial in the radial equation for which any coefficient ##a_{n_r+1}=0## (it terminates the series). Please note that ##n=0## is my ground state energy (and not ##n=1## as for the hydrogen atom for example).

    I've worked hard on the derivation of the radial equation, it was very long and tedious so I spare you that part. I found the following:

    ##R_{n l} (r) = \sum_{k=0}^{\infty} a_k \left( \frac{m\omega}{\hbar} r \right)^{l+2k} e^{-\frac{m\omega}{2\hbar} r^2}##

    I believe this is correct (according to this source: http://quantummechanics.ucsd.edu/ph130a/130_notes/node244.html), and after normalization (not asked, but well) my whole wave function in spherical coordinates looks like:

    ##\psi_{220} (r,\theta) = \sqrt{\frac{16}{15}} \frac{1}{\sqrt{\pi}} \left( \frac{m\omega}{\hbar} \right)^{7/4} r^2 e^{-\frac{m\omega}{2\hbar} r^2} Y_2^0 (\theta)##

    where ##Y_2^0(\theta)=\sqrt{\frac{5}{16\pi}} (3 \cos^2 \theta - 1)##.

    Now the part ##r^2 (3 \cos^2 \theta - 1) = 2z^2-x^2-y^2## and I rearranged the expressions until I got terms looking like

    ##\left(\frac{m\omega}{\pi \hbar} \right)^{1/4} \frac{1}{\sqrt{2^2 \cdot 2!}} H_2 \left( \sqrt{\frac{m\omega}{\hbar}} x \right) e^{-\frac{m\omega}{2\hbar} r^2}##

    since that is the wave function for each 1D oscillator (of course you would have to replace ##x## with ##y## and ##z## for the other states). And by the way I chose the Hermite polynomial ##H_2## because my cartesian coordinates are squared and that's the only way I could fit them in the cartesian wave function.

    So at the end I get a linear combination of cartesian states

    ##\psi(x,y,z)= \sqrt{\frac{m\omega}{\hbar}} \sqrt{\frac{2}{3}} \pi^{-1/4} \left[ \psi_2(z) - \frac{1}{2} \left( \psi_2(x)+\psi_2(y) \right) \right]##

    I don't expect anyone to correct this, my problem is just that I don't understand at all the point of doing that. I don't understand why we would want a linear combination of states and not a product. Maybe I really misunderstood the question and we were really suppose to find a linear combination of 3D states (and not 1D...)? If so, I wouldn't know what to do with my ##2z^2-x^2-y^2##...

    And one last question: why am I not just retrieving one of the 6 possible ground states associated with energy ##\frac{7}{2}\hbar \omega##? I guess that's similar to my previous question: I was expecting a product of states when transforming from spherical to cartesian coordinates, not a linear combination.

    Thank you very much in advance for your answers and suggestions.


    Julien.
     
  2. jcsd
  3. May 28, 2017 #2

    TSny

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    Hello.

    The problem states that ##m = 2##, but the spherical harmonic ##Y^0_2## implies ##m = 0##.

    As you note, ##r^2Y^m_l## yields a polynomial in ##x, y, z##. But why do you say that this implies that the wavefunction ##\psi_{n_x, n_y, n_z}(x,y,z)## should be a sum of ##\psi_{n_x}(x)##, ##\psi_{n_y}(y)##, ##\psi_{n_z}(z)## rather than a linear combination of products of ##\psi_{n_x}(x)##, ##\psi_{n_y}(y)##, and ##\psi_{n_z}(z)##?
     
    Last edited: May 28, 2017
  4. May 28, 2017 #3

    vela

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    How did you manage to break up the exponential term since ##e^{-\alpha r^2} = e^{-\alpha(x^2+y^2+z^2)} = e^{-\alpha x^2}e^{-\alpha y^2}e^{-\alpha z^2}##?
     
  5. May 28, 2017 #4

    TSny

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    Yes, you want linear combinations of 3D states ##\psi_{n_x, n_y, n_z}(x, y, z) = \psi_{n_x}(x) \psi_{n_y}(y) \psi_{n_z}(z)##
    You will get a different polynomial if you use ##Y^2_2## rather than ##Y^0_2##. But, you should be able to express the polynomial as combinations of Hermite polynomials. This should help in finding the coefficients in the expansion ##\psi(x, y, z) = \sum c_{n_x, n_y, n_z} \psi_{n_x, n_y, n_z}(x, y, z)##
     
  6. May 28, 2017 #5
    @TSny Oops sorry I made a mistake. ##m=0## in the problem. My bad.

    @vela Good point, I didn't realise that...

    Okay I guess I go back to the drawing board, I will post my progress as soon as I have something.

    Thank you very much for your help.

    Julien.
     
  7. May 28, 2017 #6
    Okay here is my new attempt:

    ##\psi_{220}(r,\theta)=\sqrt{\frac{16}{15}} \frac{1}{\sqrt{\pi}} \left( \frac{\hbar}{m\omega} \right)^{1/4} \left( \frac{m\omega}{\hbar} \right)^2 r^2 e^{-\frac{m\omega}{2\hbar} r^2} \sqrt{\frac{5}{16\pi}} (3 \cos^2\theta-1)##
    ##=\frac{1}{\sqrt{3}\pi} \left( \frac{m\omega}{\hbar} \right)^{7/4} r^2 (3\cos^2\theta - 1) e^{-\frac{m\omega}{2\hbar} r^2}##
    ##= \frac{1}{\sqrt{3} \pi} \left( \frac{m\omega}{\hbar} \right)^{7/4} (2z^2-x^2-y^2) e^{-\frac{m\omega}{2\hbar} (x^2+y^2+z^2)}##
    ##=\frac{1}{\sqrt{3} \pi}\left( \frac{m\omega}{\hbar} \right)^{7/4} \left[\frac{1}{2} H_2 \left(\sqrt{\frac{m\omega}{\hbar} z}\right) + 1 - \frac{1}{4} H_2\left(\sqrt{\frac{m\omega}{\hbar} x}\right) - \frac{1}{2} - \frac{1}{4} H_2\left(\sqrt{\frac{m\omega}{\hbar} y}\right) - \frac{1}{2} \right] e^{-\frac{m\omega}{2\hbar} (x^2+y^2+z^2)}##
    ##=\left( \frac{m\omega}{\pi \hbar} \right)^{3/4} \left( \frac{1}{\sqrt{2^2 \cdot 2!}} \right)^3 e^{-\frac{m\omega}{2\hbar} (x^2+y^2+z^2)} \left( \frac{m\omega}{\pi \hbar} \right) 8 \sqrt{\frac{2}{3}} \left( \frac{1}{\pi} \right)^{1/4} \left[ H_2\left(\sqrt{\frac{m\omega}{\hbar} z}\right) - \frac{1}{2} H_2\left(\sqrt{\frac{m\omega}{\hbar} x}\right) - \frac{1}{2} H_2\left(\sqrt{\frac{m\omega}{\hbar} y}\right) \right]##
    ##= \frac{8}{\pi^{1/4}} \sqrt{\frac{2}{3}} \frac{m\omega}{\pi \hbar} \left[ \psi_{002}(z) - \frac{1}{2} \left( \psi_{200}(x) + \psi_{020}(y) \right) \right]##

    I wrote all my steps this time so that the train of thought is easier to follow. I took your remarks in consideration and tried to bring the spherical wave function in a linear combination of cartesian wave functions. Not quite sure I labeled the wave functions correctly, and I still don't quite get what this whole thing means.

    Thank you in advance.


    Julien.
     
  8. May 28, 2017 #7

    TSny

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    OK, that's starting to look good. But note that , for example, ##\psi_{002}(z)## is actually a function of ##x##, ##y##, and ##z##:
    ##\psi_{002} = \psi_0(x) \psi_0(y)\psi_2(z)##,
    where each of the ##\psi##'s in the expression on the right is a 1D Harmonic oscillator function.

    Also, I don't think your final expression is normalized (if that matters).

    As far as what the whole thing means, you can think of the 3D Cartesian functions as forming a basis. You have expanded the given wavefunction (in spherical coordinates) in terms of the Cartesian basis.
     
  9. May 29, 2017 #8
    @TSny Ah yes, I oversaw the x,y,z dependance indeed. Your explanation makes sense about the basis. I think I can close this topic now, thank you very much for your input.

    Julien.
     
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