A anharmonic oscillator is characterized by potential
$V\left(x\right)=\frac{m{\omega}^{2}}{2}{x}^{2}+\lambda {x}^{4}$
where
$\lambda $
is a positive constant.
With
$\lambda =0$
the potential of the harmonic oscillator is obtained.
It is not possible to analytically solve the Schrödinger equation.
Unless otherwise specified we have set the following values:
$\hslash \text{}=1$;
$m=1$;
$\omega =1$.
Each graph displays the harmonic potential
$\frac{m{\omega}^{2}}{2}{x}^{2}$
(in blue color)
and the quadratic potential
$\lambda {x}^{4}$
(in red color).


The Mathematica notebook, given a potential V(x),
allows to numerically solve the equation, finding the energies (eigenvalues) and
the relative functions (eigenfunctions).

EIGENVALUES



$\lambda $=0
versus
$\lambda $=0.01
The total potential is almost harmonic (the harmonic component is the blue line).
The comparison shows a small difference in the eigenvalues.



$\lambda $=0
versus
$\lambda $=0.1
The value of the total potential depends on both the harmonic
component (blue color) and the quadratic component (red color).
The deviation of the eigenvalues is still small.



$\lambda $=0
versus
$\lambda $=1
The potential is nearly quadratic (red color).
The deviation of the eigenvalues is significant.

EIGENVALUES and EIGENFUNCTIONS



$\lambda $=0
versus
$\lambda $=0.01
The total potential is almost harmonic (the harmonic component is the blue line).
The comparison shows a small difference in the eigenvalues.



$\lambda $=0
versus
$\lambda $=0.1
The value of the total potential depends on both the harmonic
component (blue color) and the quadratic component (red color).
The deviation of the eigenvalues is still small.



$\lambda $=0
versus
$\lambda $=1
The potential is nearly quadratic (red color).
The deviation of the eigenvalues is significant.
