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Chemistry LibreTexts

424: Variation Method for the Quartic Oscillator

  • Page ID
    135915
  • Approximate Methods: The Quartic Oscillator

    For unit mass the quartic oscillator has the following energy operator in atomic units.

    \( H = - \frac{1}{2} \frac{d^{2}}{dx^{2}} \blacksquare + kx^{4} \blacksquare\) \( \int_{- \infty}^{ \infty} \blacksquare dx\)

    Suggested trial wavefunction: \( \psi (x; \beta ) := ( \frac{2 \beta}{ \pi})^{ \frac{1}{4}} exp( - \beta x^{2})\)

    Demonstrate that the wavefunction is normalized.

    \[ \int_{- \infty}^{ \infty} \psi (x; \beta )^{2} dx~assume,~ \beta > 0 \rightarrow 1\]

    Evaluate the variational energy integral.

    \[ E( \beta ) := \int_{- \infty}^{ \infty} \psi (x, \beta ) - \frac{1}{2} \frac{d^{2}}{dx^{2}} \psi (x, \beta ) dx + \int_{- \infty}^{ \infty} \psi (x, \beta ) x^{4} \psi (x, \beta ) dx |_{simplify}^{assume,~ \beta > 0} \rightarrow \frac{1}{16} \frac{8 \beta ^{3}}{\beta ^{2}}\]

    Minimize the energy with respect to the variational parameter \( \beta\) and report its optimum value and the ground-state energy.

    β := 1 β := Minimize(E, β) β = 0.90856 E(β) = 0.68142

    Plot the optimum wavefunction and the potential energy on the same graph.

    Screen Shot 2019-02-08 at 12.46.36 PM.png

    Calculate the classical turning point and the probability that tunneling is occurring.

    \[ \begin{align} x_{ctp} &= 0.68142^{ \frac{1}{4}} \\[4pt] &= 0.90856 \end{align} \]

    \[ 2 \int_{x_{ctp}}^{ \infty} \psi (x, \beta )^{2} dx \approx 0.083265\]

    Compare the variational result to energy obtained by numerically integrating Schrödinger's equation for the quartic oscillator using the numerical integration algorithm provided below.

    Numerical Solutions for Schrödinger's Equation

    Integration limit: xmax := 3 Effective mass: μ := 1 Force constant: k := 1

    Potential energy: \( V(x) := kx^{4}\)

    Numerical integration of Schrödinger's equation:

    Given

    \( \frac{-1}{2 \mu} \frac{d^{2}}{dx^{2}} \Phi (x) + V(x) \Phi (x) = energy \Phi (x)\)

    \( \Phi (-x_{max} = 0\)

    \( \Phi '(-x_{max} = 0.1\)

    \( \Phi := Odesolve (x, x_{max}\)

    Normalize wavefunction: \( \Phi (x) := \frac{ \Phi (x)}{ \sqrt{ \int_{-x_{max}}^{x_{max}} \Phi (x) ^{2} dx}}\)

    Enter energy guess: Energy = 0.6679864

    Screen Shot 2019-02-08 at 12.53.09 PM.png

    Compare the variational and numerical solutions for the quartic oscillator by putting them on the same graph.

    Screen Shot 2019-02-08 at 12.53.15 PM.png