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Quantum Harmonic Oscillator

The quantum mechanical version of this harmonic oscillator problem may be written as
$\displaystyle \left( -\frac {\hbar^2}{2 \mu} \frac{d^2}{dx^2}
+ \frac{1}{2} k x^2 \right) \Psi(x)$ $\textstyle =$ $\displaystyle E \Psi(x).$ (14)

By considering the limiting behavior as $x \rightarrow \infty$ and as $x \rightarrow 0$, one finds that only certain energies $E_n$ yield reasonable solutions $\Psi(x)$. The eigenvalues and associated eigenvectors are
$\displaystyle E_n$ $\textstyle =$ $\displaystyle \hbar \omega \left( n + \frac{1}{2} \right) =
h \nu \left( n + \frac{1}{2} \right),$ (15)
$\displaystyle \Psi_n(x)$ $\textstyle =$ $\displaystyle \left( \frac{\mu \omega}{\pi \hbar 2^{2n} (n!)^2} \right)^{1/4}
e...
...r} \right)
H_n \left[ \left( \frac{ \mu \omega}{\hbar} \right)^{1/2}
x \right],$ (16)

where $H_n$ are the Hermite polynomials of order $n$. The Hermite polynomials have a number of useful properties such as
$\displaystyle \frac{d}{dx} H_n(x)$ $\textstyle =$ $\displaystyle 2n H_{n-1}(x)$ (17)
$\displaystyle H_{n+1}(x)$ $\textstyle =$ $\displaystyle 2x H_n(x) - 2n H_{n-1}(x)$ (18)
$\displaystyle \int_{-\infty}^{\infty} H_n(x) H_{m}(x) e^{-x^2} dx$ $\textstyle =$ $\displaystyle \delta_{nm} \sqrt{\pi} 2^n n!.$ (19)



Subsections
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Next: Units Up: ho Previous: Classical Harmonic Oscillator
David Sherrill 2002-10-16