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Quantum Mechanics Background

A standard problem in quantum mechanics is the solution of the harmonic oscillator problem with the Hamiltonian

\begin{displaymath}
H = -\frac{\hbar^2}{2m} \frac{d^2}{dx^2} + \frac{1}{2} m \omega^2 x^2
\end{displaymath}

The eigenvalue problem is

\begin{displaymath}
H \psi_n(x) = E_n \psi_n(x)
\end{displaymath}

The notation is simplified if we change variables to

\begin{displaymath}
\rho = \sqrt{\frac{m\omega}{\hbar}} x
\end{displaymath}

so that the Hamiltonian is now

\begin{displaymath}
H = \hbar\omega H_0
\end{displaymath}

with

\begin{displaymath}
H_0 = \left[-\frac{1}{2}\frac{d^2}{d\rho^2} + \frac{1}{2}\rho^2\right]
\end{displaymath}

If energy is measured in units of $\hbar\omega$, then the eigenvalue problem reads

\begin{displaymath}
H_0 \psi_n(\rho) = E_n \psi_n(\rho)
\end{displaymath}

The eigenenergies are $E_n = n + 1/2$ for $n = 0,1,\ldots{}$ and the eigenfunctions are

\begin{displaymath}
\psi_n(\rho) = \frac{1}{\sqrt{2^n n! \sqrt{\pi}}} e^{-\rho^2/2}H_n(\rho).
\end{displaymath}

The $H_n(\rho)$ are Hermite polynomials.

Often the harmonic oscillator form is just the first term in a Taylor's series approximation to a more complicated potential. In some cases it is important to keep higher terms. Thus we consider the symmetric anharmonic oscillator potential, obtained by adding a quartic term:

\begin{displaymath}
H^{\rm anh} = H_0 + \frac{1}{2}\epsilon \rho^4
\end{displaymath}

where $\epsilon$ may be small or large.

This problem cannot be solved in closed form, but can be solved numerically in the basis of the eigenstates of $H_0$. In that case the eigenvalue problem becomes a matrix eigenvalue problem. It reads

\begin{displaymath}
h^{\rm anh} {\vec u} = E {\vec u}
\end{displaymath}

where ${\vec u}$ is a column vector containing the expansion coefficients of the anharmonic oscillator wave function in terms of harmonic oscillator wave functions

\begin{displaymath}
\psi_{\rm anh}(\rho) = \sum_{i=0}^{\infty} u_i \psi_i(\rho)
\end{displaymath}

and the hamiltonian matrix is


\begin{displaymath}
h^{\rm anh}_{ij} = \int_{-\infty}^{\infty} d\rho    
\psi_i(\rho) H^{\rm anh}\psi_j(\rho)
\end{displaymath}

The integral can be done exactly with the result that
\begin{displaymath}
h^{\rm anh}_{ij} = (i+1/2)\delta_{ij} + \epsilon g_{ij}
\end{displaymath} (1)

where
\begin{displaymath}
g_{ij} = \left\{\begin{array}{ll}
\frac{3}{2}(m^2 + m + \f...
...or }   i=j\pm 4\\
0 & {\rm otherwise}
\end{array}\right.
\end{displaymath} (2)

with $m = \min(i,j)$.


next up previous
Next: Numerical Analysis Problem Up: anharm_osc Previous: anharm_osc
Carleton DeTar 2002-10-18