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406: Numerical Solutions for Morse Oscillator

  • Page ID
    137730
  • Schrödinger's equation is integrated numerically for the first three energy states for the Morse oscillator. The integration algorithm is taken from J. C. Hansen, J. Chem. Educ. Software, 8C2, 1996.

    Set parameters:

    n = 300

    xmin = -2

    xmax = 12

    \[ \Delta = \frac{xmax - xmin}{n-1}\]

    \( \mu\) = 1

    D = 2

    \( \beta\) = 2

    xe = 0

    Calculate position vector, the potential energy matrix, and the kinetic energy matrix. Then combine them into a total energy matrix.

    i = 1 .. n j = 1 .. n xi = xmin + (i - 1) \( \Delta\)

    \[ V_{i,~j} = if \bigg[ i =j,~D [ 1 - exp [ \beta ( x_i - x_e )]]^2 ,~0 \bigg] \]

    \[ T_{i,~j} = if \left[ i=j, \frac{ \pi ^{2}}{6 \mu \Delta ^{2}}, \frac{ (-1)^{i-j}}{ (i-j)^{2} \mu \Delta^{2}} \right] \]

    Hamiltonian matrix: H = T + V

    Find eigenvalues: E = sort(eigenvals(H))

    Display three eigenvalues: m = 1 .. 3

    Em =

    \( \begin{array}{|r|}
    \hline \\
    0.8750 \\
    \hline \\
    1.8750 \\
    \hline \\
    2.0596 \\
    \hline
    \end{array} \)

    Calculate associated eigenfunctions:

    k = 1 .. 3

    \[ \psi (k) = eigenvec (H, E_k)\]

    Plot the potential energy and selected eigenfunctions:

    Screen Shot 2019-02-06 at 7.32.57 PM.png

    For \(V = ax^n\), the virial theorem requires the following relationship between the expectation values for kinetic and potential energy:

    \[<T> = 0.5n<V>. \]

    The calculations below show that virial theorem is not satisfied for the Morse oscillator. The reason is revealed in the following series expansion in \(x\). The expansion contains cubic, quartic and higher order terms in \(x\), so the virial theorem does not apply to the quartic oscillator.

    \( D (1 - exp( - \beta x))^2\) converts to the series \( D \beta ^2 x^2 + (-D) \beta ^3 x^3 + \frac{7}{12} D \beta ^4 x^4 + O(x^5)\)

    \( \begin{pmatrix}
    "Kinetic~Energy" & "Potential~Energy" & "Total~Energy" \\
    \psi (1)^{T} T \psi(1) & \psi (1)^{T} V \psi(1) & E_{1} \\
    \psi (2)^{T} T \psi(2) & \psi (2)^{T} V \psi(2) & E_{2}
    \end{pmatrix} = \begin{pmatrix}
    "Kinetic~Energy" & "Potential~Energy" & "Total~Energy" \\
    0.3750 & 0.5000 & 0.8750 \\
    0.3754 & 1.4996 & 1.8750
    \end{pmatrix} \)