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Solving two coupled PDE from a quantum problem

  1. Dec 1, 2013 #1
    Hello Guys.
    I have to solve two coupled PDE coming from a quantum physical problem, which possess a cylindrical symmetry. They look like this
    [itex]\left\{-\frac{i\hbar}{2M}\left( D^2_\rho + \frac{1}{\rho}D_\rho + D^2_z - \frac{L_z^2}{\hbar^2 \rho^2} \right) + \frac{M\omega^2_\rho \rho^2}{2} + \frac{M\omega^2_z z^2}{2} + g\psi^2_0 - \mu \right\} f^+_\nu = \epsilon_\nu f^-_\nu[/itex]
    [itex]\left\{-\frac{i\hbar}{2M}\left( D^2_\rho + \frac{1}{\rho}D_\rho + D^2_z - \frac{L_z^2}{\hbar^2 \rho^2} \right) + \frac{M\omega^2_\rho \rho^2}{2} + \frac{M\omega^2_z z^2}{2} + 3g\psi^2_0 - \mu \right\} f^-_\nu = \epsilon_\nu f^+_\nu[/itex]

    where
    [itex]f^{+,-}_\nu = f^{+,-}_\nu(\rho, z, \phi) [/itex]

    [itex] D_x \equiv \frac{\partial}{\partial x} [/itex]

    [itex] \psi_0^2 = \frac{\mu}{g} \left(1 - \frac{\rho^2}{R^2_\rho} - \frac{z^2}{R^2_z}\right)[/itex]

    [itex] L_z [/itex] is the projection of the angular momentum operator, which operates like this on his eigenfunctions:
    [itex] L_z (\frac{\exp{(im\phi)}}{\sqrt(2\pi)}) = \hbar m \frac{\exp{(im\phi)}}{\sqrt(2\pi)}[/itex]

    and [itex]\hbar, M, \omega_\rho, \omega_z, R_\rho, R_z, \epsilon_\nu [/itex] and [itex]\mu[/itex] are constants.

    Actually, I have to find the energies [itex]\epsilon[/itex].
    I would like to use the fact that I know how the operator Lz acts, but also, I know how the operator on the z coordinate acts. I would like to end up with only a radial equation on ρ.
    To be clearer, I rewrite the equations like this (grouping operators acting on different variables, and plugging the expression for [itex]\psi_0[/itex]):

    [itex]\left\{-\frac{i\hbar}{2M}\left( D^2_\rho + \frac{1}{\rho}D_\rho \right) + \frac{M\omega^2_\rho \rho^2}{2} -\frac{\rho^2}{R_\rho} \right\} f^+_\nu +\left\{ -\frac{i\hbar}{2M} D^2_z + \frac{M\omega^2_z z^2}{2} -\frac{z^2}{R_z}\right\}f^+_\nu -\frac{i\hbar}{2M} \left\{-\frac{L_z^2}{\hbar^2 \rho^2} \right\}f^+_\nu= \epsilon_\nu f^-_\nu[/itex]

    And here comes my question. I know how Lz acts on his eigenfunctions, but I also know how the operator on z acts on his eigenfunctions ( because it is the 1D quantum oscillator).
    Then, am I allowed to write the solution in the following way?
    [itex]f^{+,-}_\nu = \sum_{m} \sum_{n} f^{+,-}_{n,m,\nu}(\rho) \psi_n(z) \frac{\exp{(im\phi)}}{\sqrt(2\pi)}[/itex]

    where
    [itex] \psi_n(z) [/itex] are the eigenfunctions of the harmonic oscillator?

    If this is possible, the equation will be simplified in a 1D equation in ρ, replacing the harmonic operator in z, and the operator Lz with their eigenvalues.
    Thanks for the attention.
     
  2. jcsd
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