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Chemistry LibreTexts

416: Numerical Solutions for a Modified Harmonic Potential

  • Page ID
    137745
  • This tutorial deals with the following potential function:

    \[ V(x, d) = \bigg| _{ \infty ~otherwise}^{ \frac{1}{2} k(x-d)^2~if~x \geq 0 + d \leq 0}\]

    If d = 0 we have the harmonic oscillator on the half-line with eigenvalues 1.5, 3.5, 5.5, ... for k = \( \mu\) = 1. For large values of d we have the full harmonic oscillator problem displaced in the x-direction by d with eigenvalues 0.5, 1.5, 2.5, ... for k = \( \mu\) = 1. For small to intermediate values of d the potential can be used to model the interaction of an atom or molecule with a surface.

    Integration limit: xmax = 10

    Effective mass: \( \mu\) = 1

    Force constant: k = 1

    Potential energy minimum: d = 5

    Potential energy:

    \[ V(x,d) = \frac{k}{2} (x-d)^2\]

    Integration algorithm:

    Given

    \[ \]

    Normalize wavefunction:

    \[ \psi (x) = \frac{ \psi (x)}{ \sqrt{ \int_{0}^{x_{max}} \psi (x)^2 dx}}\]

    Energy guess: E = 0.5

    Screen Shot 2019-02-19 at 12.16.32 PM.png

    Calculate average position:

    \[ X_{avg} = \int_{0}^{x_{max}} \psi (x) x \psi (x) dx = 5\]

    Calculate potential and kinetic energy:

    \[ V_{avg} = \int_{0}^{x_{max}} \psi (x) V(x,d) \psi (x) dx = 0.25\]

    \[ T_{avg} = E - V_{avg} = 0.25\]

    Exercises:

    • For d = 0, k = \( \mu\) = 1 confirm that the first three energy eigenvalues are 1.5, 3.5 and 5.5 Eh. Start with xmax = 5, but be prepared to adjust to larger values if necessary. xmax is effectively infinity.
    • For d = 5, k = \( \mu\) = 1 confirm that the first three energy eigenvalues are 0.5, 1.5 and 2.5 Eh. Start with xmax = 10, but be prepared to adjust to larger values if necessary.
    • Determine and compare the virial theorem for the exercises above.
    • Calculate the probability that tunneling is occurring for the ground state for the first two exercises. (Answers: 0.112, 0.157)