# 457: Variation Method Using the Wigner Function - The Quartic Oscillator

Define potential energy:

$V(x) = x^4$

Display potential energy:

Choose trial wave function:

$\psi (x, \beta = \left( \frac{2 \beta}{ \pi} \right)^{ \frac{1}{4}} exp( - \beta x^2)$

Calculate the Wigner distribution function:

$W(x, p, \beta ) = \frac{1}{2 \pi} \int_{- \infty}^{ \infty} \psi \left(x + \frac{s}{2}, \beta \right) exp(isp) \psi (\left( x- \frac{s}{2}, \beta \right) ds~ \bigg|_{assume,~ \beta > 0}^{simplify} \rightarrow \frac{1}{ \pi} e^{ \frac{-1}{2} \frac{4 \beta ^2 x^2 + p^2}{ \beta}}$

Evaluate the variational integral:

$E( \beta ) = \int_{- \infty}^{ \infty} \int_{- \infty}^{ \infty} W(x, p, \beta ) \left( \frac{p^2}{2} + V(x) \right)dx~dp$

Minimize the energy integral with respect to the variational parameter, $$\beta$$.

$$\beta$$ = 1 $$\beta$$ = Minimize (E, $$\beta$$) $$\beta$$ = 0.90856 E( $$\beta$$) = 0.68142

Calculate and display the coordinate distribution function:

$Px(x, \beta ) = \int_{- \infty}^{ \infty} W(x, p, \beta ) dp$

Classical turning points: $$x_{cl} = 0.681^{ \frac{1}{4}}~~~ x_{cl} = 0.90842$$

Probability that tunneling is occurring:

$2 \int_{0.908}^{ \infty} Px (x, \beta ) dx = 0.08345$

Calculate and display the momentum distribution function:

$Pp(p, \beta ) = \int_{- \infty}^{ \infty} W(x, p, \beta ) dx$

Display the Wigner distribution function:

N = 60 i = 0 .. N xi = $$-3 + \frac{6i}{N}$$ j = 0 .. N pj = $$-5 + \frac{10j}{N}$$ Wigneri, j = W( xi, pj, $$\beta$$)