# 405: Numerical Solutions for the Quartic Oscillator

- Page ID
- 137729

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

Set parameters:

Increments: n = 100

Integration limits: xmin = -3

xmax = 3

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

Effective mass: \( \mu\) = 1

Force constant: k = 1

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 x_{i} = xmin + (i - 1) \( \Delta\)

\[ V_{i,~j} = if \bigg[ i =j,~k (x_i)^4 ,~0 \bigg] \]

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

Hamiltonian matrix: H = T + V

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

Display three eigenvalues: m = 1 .. 3

E_{m} =

\( \begin{array}{|r|}

\hline \\

0.6680 \\

\hline \\

2.3936 \\

\hline \\

4.6968 \\

\hline

\end{array} \)

Calculate associated eigenfunctions:

k = 1 .. 3

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

Plot the potential energy and selected eigenfunctions:

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 the virial theorem is satisfied for the harmonic oscillator for which n = 4.

\( \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} \\

\psi (3)^{T} T \psi(3) & \psi (3)^{T} V \psi(3) & E_{3}

\end{pmatrix} = \begin{pmatrix}

"Kinetic~Energy" & "Potential~Energy" & "Total~Energy" \\

0.4453 & 0.2227 & 0.6680 \\

1.5958 & 0.7979 & 2.3936 \\

3.1312 & 1.5656 & 4.6968

\end{pmatrix} \)