# 424: Variation Method for the Quartic Oscillator

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