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Expansion of the dilation operator

A scaling wavefunction $\psi$ (see Eq.(6.1)) whose unscaled wavenumber is $k_\mu$ will be rescaled to a new wavenumber $k = k_\mu + \delta$, where $\delta$ is the wavenumber shift. Its value at location ${\mathbf r}$ is

\begin{displaymath}
\psi(k_\mu + \delta,{\mathbf r}) \ = \ \psi(\frac{k_\mu + \...
...k_\mu}
{\mathbf r} ) \ = \
\psi({\mathbf r} + {\mathbf a}) ,
\end{displaymath} (I.1)

where the translation vector is ${\mathbf a} = (\delta/k_\mu){\mathbf r}$. Formally this translation can be performed by exponentiation of the generator of translation [174,7]:
\begin{displaymath}
\psi(k_\mu + \delta,{\mathbf r}) \ = \ e^{{\mathbf a}\cdot\...
...\cdot\nabla)
({\mathbf a}\cdot\nabla) + \cdots \right] \psi .
\end{displaymath} (I.2)

Thus we have written the scaled value in terms of local derivatives of the unscaled wavefunction. This becomes a definition of dilation only when the ${\mathbf r}$-dependence of ${\mathbf a}$ is finally substituted. Note that ${\mathbf a}$ is held constant as far as $\nabla$ is concerned. (This differs from other possible definitions of dilation [14], where the commutator $[\partial_i,r_j] = \delta_{ij}$ plays a role; however my form will be simpler to expand to high order). For instance in Cartesian coordinates, $({\mathbf a}\cdot\nabla)({\mathbf a}\cdot\nabla)$ is to be interpreted as $a_i a_j \partial_i \partial_j$ (Einstein summation assumed), because in Cartesian coordinates $\partial_i a_j = 0$.

Figure I.1: Curvilinear coordinates $(z,s)$ used to represent points ${\mathbf r}$ near the billiard boundary $\Gamma $. The orthogonal unit vectors are the vector fields ${\mathbf n}$ and ${\mathbf t}$.
\begin{figure}\centerline{\epsfig{figure=fig_ap/curvi.eps,width=0.7\hsize}}\end{figure}


next up previous
Next: Curvilinear boundary coordinates Up: Appendix I: Scaling expansion Previous: Appendix I: Scaling expansion
Alex Barnett 2001-10-03