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Linear Harmonic Oscillator

  1. Jul 9, 2010 #1
    Hello there,

    Can anyone help me, I am struggling with solving LHO in two dimension,but in the polar coordinates.
    I transfer laplacian into polar from decart coordinates, write Ψ=ΦR, and do fourier separation method for solving differential equation. But I do not know how to solve differential equation on R.

    I will write to where I came tomorrow, or in a few hours.

    Thanks!!!
     
  2. jcsd
  3. Jul 9, 2010 #2
    Why trouble yourself by going into polar co-ordinates? Just look for solution of the form Ψ = X(x)Y(y), for the Schrodinger equation in cartesian co-ordinates and you'll get two equations for a Linear Harmonic Oscillator. Surely, for the case of two one dimensional equation, you know the solution, multiply them, and add there energy levels to get the answer. I think the lowest energy state would have the energy of (hbar)*w, where w is the natural frequency of oscillation and the wavefunction is a Gaussian in radial co-ordinate r.
     
  4. Jul 9, 2010 #3
    In this problem, both [itex]\phi[/itex] and [itex]z[/itex] are cyclic coordinates (they do not enter explicitly in the Schroedinger equation), so the corresponding conjugate momenta [itex]p_{\phi} \equiv l_{z}[/itex] and [itex]p_{z}[/itex] commute with the Hamiltonian. Therefore, the stationary states of the system are of the form:

    [tex]
    \psi_{m, p_{z}}(\rho, \phi, z) = \frac{1}{(2 \pi \hbar)^{1/2}} \, e^{\frac{i}{\hbar} \, p_{z} \, z} \, \frac{1}{\sqrt{2 \, \pi}} \, e^{i \, m \, \phi} \, R_{m}(\rho), \ m = \ldots, -1, 0, 1, \ldots
    [/tex]

    Substitute this into the Schroedinger equation and you will get the following oridnary differential equation:

    [tex]
    -\frac{\hbar^{2}}{2 \, \mu} \, \left[ \frac{1}{\rho} \, \frac{d}{d \rho}\left(\rho \, \frac{dR_{m}(\rho)}{d \rho}\right) - \frac{m^{2}}{\rho^{2}} \, R_{m}(\rho) - \frac{p^{2}_{z}}{\hbar^{2}} \, R_{m}(\rho) \right] + \frac{\mu \, \omega^{2} \, \rho^{2}}{2} \, R_{m}(\rho) = E \, R_{m}(\rho)
    [/tex]

    The normalization condition is:

    [tex]
    \int_{0}^{\infy}{\rho \, R^{2}_{m}(\rho) \, d\rho} = 1
    [/tex]

    Introduce a dimensionless argument:

    [tex]
    \rho = a x, R_{m}(\rho) = \frac{1}{a} \, y_{m}(x)
    [/tex]

    with:

    [tex]
    a = \left(\frac{\hbar}{\mu \, \omega}\right)^{\frac{1}{2}}
    [/tex]

    and:

    [tex]
    \epsilon = \frac{2}{\hbar \, \omega} \, \left( E - \frac{p^{2}_{z}}{2 \, \mu} \right)
    [/tex]

    you will get the following equation:

    [tex]
    y''_{m} + \frac{1}{x} \, y'_{m}(x) + \left(\epsilon - x^{2} - \frac{m^{2}}{x^{2}}\right) \, y_{m}(x) = 0
    [/tex]

    with the normalization condition:

    [tex]
    \int_{0}^{\infty}{x \, y^{2}_{m}(x) \, dx} = 1
    [/tex]

    Usually, we want to get rid of the first derivative in the differential equation. We can achieve this by the substitution:

    [tex]
    y_{m}(x) = \frac{z_{m}(x)}{\sqrt{x}}
    [/tex]

    A direct substitution should convince you that the equation satisfied by [itex]z_{m}(x)[/itex] is:

    [tex]
    z''_{m} + \left(\epsilon - x^{2} - \frac{m^{2} - 1/4}{x^{2}}\right) \, z_{m} = 0
    [/tex]

    and the normalization condition reads:

    [tex]
    \int_{0}^{\infty}{z^{2}_{m}(x) \, dx} = 1
    [/tex]

    The asymptotic behavior for [itex]z_{m}(x)[/itex] for both large and small values of [itex]x[/itex] is obtained by keeping the dominant terms in the equation for the relevant region:

    [tex]
    z''_{m 0} - \frac{m^{2} - 1/4}{x^{2}} \, z_{m 0} = 0, \ x \rightarrow 0
    [/tex]

    This is an Euler equation and has solutions of the form [itex]z_{m 0} \tilde x^{k}[/itex]. Substituting, we get an algebraic equation for [itex]k[/itex]:

    [tex]
    k^{2} - k - \left(m^{2} - \frac{1}{4}\right) = 0
    [/tex]

    the solutions of which are:

    [tex]
    k_{1/2} = \frac{1 \pm 2 \, |m|}{2}
    [/tex]

    We take:

    [tex]
    z_{m 0} \sim x^{|m| + \frac{1}{2}}, \ x \rightarrow 0
    [/tex]

    For large x, we have:

    [tex]
    z''_{m \infty} - x^{2} \, z_{m \infty} = 0
    [/tex]

    which is of the same form as in the one dimensional case, so we have:

    [tex]
    z_{m \infty} \sim e^{-\frac{x^{2}}{2}}, \ x \rightarrow \infty
    [/tex]

    Finally, we capture the asymptotic behavior by writing:

    [tex]
    z_{m}(x) = x^{|m| + 1/2} \, e^{-x^{2}/2} \, v_{m}(x)
    [/tex]

    After some differentiation and algebraic manipulation, you should get the following equation for [itex]v_{m}(x)[/itex]:

    [tex]
    v''_{m} + \left(\frac{2 |m| + 1}{x} - 2 x \right) \, v'_{m} + \lambda \, v_{m} = 0
    [/tex]

    where

    [tex]
    \lambda = \epsilon - 2 |m| - 1
    [/tex]

    Using a change of variables in the argument:

    [tex]
    x = a \, t^{\alpha}
    [/tex]

    you will get:

    [tex]
    t \, \ddot{v}_{m} + \left[ 1 + 2 \, \alpha \, |m| - 2 a^{2} \, \alpha \, t^{2 \, \alpha} \right] \, \dot{v}_{m} + \lambda \, (a \alpha)^{2} t^{2 \, \alpha - 1} = 0[/tex]

    Compare this with the differential equation for the Kummer confluent hypergeometric equation [itex]_{1}F_{1}(a; c; t)[/itex]:

    [tex]
    t \ddot{y} + (c - t) \, \dot{y} - a \, y = 0
    [/tex]

    we see that we should have:

    [tex]
    2 \, \alpha = 1, \; 2 \, \alpha \, a^{2} = 1 \Rightarrow \alpha = \frac{1}{2}, \; a = 1 \Rightarrow x = t^{\frac{1}{2}} \Leftrightarrow t = x^{2}
    [/tex]

    So, we can write:

    [tex]
    v_{m}(x) = _{1}F_{1}(-\frac{\lambda}{4}, |m| + 1, x^{2})
    [/tex]

    Collecting everything back, the radial wavefunction is:

    [tex]
    R_{n m}(a \, x) = C_{n, m} \, x^{|m|} \, e^{-\frac{x^{2}}{2}} \, _{1}F_{1}\left( -n, |m| + 1, x^{2} \right), n \in \mathbb{N}_{0}, m \in \mathbb{Z}
    [/tex]

    where [itex]C_{n, m}[/itex] is a normalization constant. The energy eigenvalues associated with motion in the plane of the oscillator are:

    [tex]
    E_{n, m, p_{z}} - \frac{p^{2}_{z}}{2 \, \mu} = E'_{n, m} = \hbar \, \omega \, \left( |m| + 2 \, n + \frac{1}{2} \right)
    [/tex]
     
  5. Jul 10, 2010 #4
    I have attached a pdf file into this post.

    Thank you for responding!

    You asked me why I do it in polar coordinates, when it`s much easier in decarte. Well I simply want to practice on my own, I want to solve one problem in a few ways. I have just finished basic cousre of quantum mechanics, but we haven`t done LHO in polar coordinates.
     

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