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Elementary Commutators

We need $[p_x,\cos \frac{\pi x}{L}]$ and $[p_x,\sin\frac{\pi x}{L}]$:

\begin{displaymath}[p_x,\cos \frac{\pi x}{L}]= \imath \hbar \frac{\pi}{L} \sin \...
...\right )
= p_x \cos \frac{\pi x}{L} - \cos \frac{\pi x}{L} p_x
\end{displaymath}


\begin{displaymath}[p_x,\sin \left ( \frac{\pi x}{L} \right )]= -\imath \hbar \f...
...\pi x}{L} \right ) - \sin \left ( \frac{\pi x}{L} \right ) p_x
\end{displaymath}

and $[p_x^2,\cos\left ( \frac{\pi x}{L} \right )]$ (and its sine equivalent):

\begin{displaymath}[p_x^2,\cos \left ( \frac{\pi x}{L} \right )]
= 2 \imath \hba...
...\frac{\pi}{L} \right )^2
\cos \left ( \frac{\pi x}{L} \right )
\end{displaymath}


\begin{displaymath}[p_x^2,\sin \left ( \frac{\pi x}{L} \right )]
= - 2 \imath \h...
...(\frac{\pi}{L}\right )^2 \sin \left ( \frac{\pi x}{L} \right )
\end{displaymath}




1998-05-13