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Appendix L: Explicit relation of cross section to Landauer formula

In this appendix we derive the relation (7.23) formally in a self-contained fashion. No reference to the semiclassical derivation of Section 7.1 is made. Therefore this provides an independent derivation of our main result (7.1).

We first present the special case $\gamma_{\bf k}= 0$ corresponding to a QPC system embedded in a hard wall (at $x=0$). The reflection off the unbroken wall is diagonal in the half-plane partial-wave basis of Section 7.2. We can expand the `unscattered wave' in Bessel functions [continuing the expansion of Eq.(7.20)], to give

\psi_{{\mbox{\tiny0}}}{({\mathbf r})}\; = \;
4 \sum_{l=1}^...
...{2}$}- \theta)]
\sin [l(\mbox{\small$\frac{\pi}{2}$}- \phi)].
\end{displaymath} (L.1)

Recognising this as being composed entirely of a sum of the half-plane partial-wave bases (7.13) on the left side, we can read off the coefficients $p^-_l = 2 (-i)^l \sin [l(\mbox{\small$\frac{\pi}{2}$}- \phi)]$. (Note that the outgoing coefficients $p^+_l$ are also equal to this). With the QPC system now placed in the wall, the outgoing amplitudes on the right side are $q^+_n = \sum_l t_{nl} p^-_l$. This gives a flux
$\displaystyle \Gamma_{{\mbox{\tiny T}}}(\phi) \;$ $\textstyle =\;$ $\displaystyle \frac{\hbar}{m} \sum_n \vert q^+_n\vert^2 \; = \;
\frac{\hbar}{m} \sum_{ll'} (p^-_l)^* (t^\dag t)_{ll'} p^-_{l'}$  
  $\textstyle =$ $\displaystyle 4\frac{\hbar}{m} \sum_{l l'} (i)^{l-l'} \sin[l(\mbox{\small$\frac{\pi}{2}$}- \phi)]
\; (t^\dag t)_{ll'}$  
    $\displaystyle \hspace{.5in} \sin[l'(\mbox{\small$\frac{\pi}{2}$}- \phi)] .$ (L.2)

Using the definition of cross section (7.4) we can write
$\displaystyle \int_{-\pi/2}^{\pi/2} \!\!\!\!d\phi \, \sigma_{{\mbox{\tiny T}}}(k,\phi) \;$ $\textstyle =\;$ $\displaystyle \frac{1}{v} \int_{-\pi/2}^{\pi/2} \!\!\!\! d\phi \, \Gamma_{{\mbox{\tiny T}}}(\phi)$  
  $\textstyle =$ $\displaystyle \frac{4}{k} \sum_{l l'} (t^\dag t)_{ll'}
\cdot \int_{-\pi/2}^{\pi/2} \!\!\!\! d\phi \sin[l(\mbox{\small$\frac{\pi}{2}$}- \phi)]$  
    $\displaystyle \hspace{.5in} \sin[l'(\mbox{\small$\frac{\pi}{2}$}- \phi)] ,$ (L.3)

from where the orthogonality of the radial functions allows us to replace the final integral by $(\pi/2) \delta_{ll'}$, whereupon the sum becomes a trace, giving
\int_{-\pi/2}^{\pi/2} \!\! d\phi \, \sigma_{{\mbox{\tiny T}}}(k,\phi) \; = \;
\lambda \mbox{Tr} (t^\dag t),
\end{displaymath} (L.4)

completing the proof.

Now considering a general wall profile with a general $\gamma_{\bf k}$, the situation becomes more subtle. An attempt to expand $\psi_{{\mbox{\tiny0}}}$ in Bessel functions as in (L.1) gives additional terms which are not part of the half-plane basis (7.13), namely an $l{=}0$ (s-wave) term and terms with angular dependence $\cos [l(\mbox{\small$\frac{\pi}{2}$}- \theta)]$ for all $l\geq 1$. This would suggest that the basis (7.13) is incomplete. However, all these new terms can be expressed as sums of the $\sin [l(\mbox{\small$\frac{\pi}{2}$}- \theta)]$ functions already present, if we are careful to consider the large radius limit $kr \rightarrow \infty$ before the maximum angular momentum limit $l_{{\mbox{\tiny max}}} \rightarrow \infty$. In this limit $l_{{\mbox{\tiny max}}} \ll kr$, a given basis state will have negligible $x$-wavevector near the wall, compared to the wall profile width, so the `softness' of the wall profile will not be apparent. Rather, the wall will appear as a hard reflector at $x \approx 0$. We write the entire left side wavefunction

\psi{({\mathbf r})}\; = \; f^-(\theta) \frac{e^{-ikr}}{\sqrt{r}} +
f^+(\theta) \frac{e^{ikr}}{\sqrt{r}} ,
\end{displaymath} (L.5)

in a general form where the incoming and outgoing angular distributions are apparent. Therefore, even for general $\gamma_{\bf k}$, the apparent hardness of the wall forces $f^-(\theta)$ and $f^+(\theta)$ to go smoothly to zero as $\vert\theta\vert
\rightarrow \pi/2$. This ensures that the original basis set is sufficient to represent all asymptotic wavefunctions, since $\sin [l(\mbox{\small$\frac{\pi}{2}$}- \theta)]$, $l=1\cdots\infty$ form a complete set in the interval $[-\mbox{\small$\frac{\pi}{2}$},\mbox{\small$\frac{\pi}{2}$}]$. The set $\cos [l(\mbox{\small$\frac{\pi}{2}$}- \theta)]$, $l=0\cdots\infty$ form another complete set in the same interval; both sets are not required and we are able to choose just one [7]. Because of the hard wall boundary condition in $\theta$-space, the former set is more appropriate.

We are still left with the issue of finding the incoming $p^-_l$ coefficients given an unscattered wave $\psi_{{\mbox{\tiny0}}}$ for general $\gamma_{\bf k}$. The problem is subtle, but can be understood when we consider the order of limits above: restricting $l \leq l_{{\mbox{\tiny max}}}$ allows the incoming representation $f^-(\theta)$ of a single plane wave $e^{i{\bf k}\cdot{\bf r}}$ to become a well-defined, narrowly-peaked delta-like function about the incident angle. This single plane wave will also cause a similar delta peak in the opposite direction, which it turns out is irrelevant because it contributes instead only to outgoing $f^+(\theta)$. Thus we have the important result that we can ignore the reflected wave (making its phase shift $\gamma_{\bf k}$ irrelevant) in calculating $p^-_l$, because this wave can only contribute to $f^+(\theta)$ in the interval $[-\mbox{\small$\frac{\pi}{2}$},\mbox{\small$\frac{\pi}{2}$}]$. We can use stationary-phase integration (method of steepest descents) [7] applied to a single plane wave to show this, and find the delta strength, as follows. Taking care with the definitions of angle (see Fig. 7.1a) we have

e^{i{\bf k}\cdot{\bf r}} \; = \; e^{-ikr\cos(\theta-\phi)} \; = \;
f^-(\theta) \frac{e^{-ikr}}{\sqrt{r}} ,
\end{displaymath} (L.6)

where $f^-(\theta) = e^{ikr[1 - \cos(\theta-\phi)]}$. Expanding the cos as a quadratic about the stationary point $\theta\approx\phi$, and making the usual stationary-phase replacement
e^{i g(x)} \; \rightarrow \; \sqrt{\frac{2\pi}{-i g''(x_0)}} \,
e^{i g(x_0)} \, \delta(x - x_0)
\end{displaymath} (L.7)

gives $f^-(\theta) = e^{i\pi/4} (2\pi/k)^{1/2} \, \delta(\theta-\phi)$. The other stationary point $\theta\approx\phi + \pi$ results in the same delta strength but an additional factor $e^{2ikr}$, which turns the incoming $e^{-ikr}$ to an outgoing $e^{ikr}$, thus contributing only to $f^+(\theta)$. Generally, we can write the expression for incoming partial-wave expansion
p^-_l \; = \; \sqrt{\frac{2k}{\pi}} \, e^{-i\pi/4} \, (i)^l...
...-(\theta) \,
\sin [l(\mbox{\small$\frac{\pi}{2}$}- \theta)] ,
\end{displaymath} (L.8)

which is easily derived using orthogonality of the angular functions and the asymptotic form of $H^{(2)}_l(kr)$. Substituting the narrowly-peaked $f^-(\theta)$ resulting from the single incident plane wave gives
p^-_l \; = \; 2 (-i)^l \sin[l(\mbox{\small$\frac{\pi}{2}$}- \phi)].
\end{displaymath} (L.9)

These are the same incoming partial wave coefficients as derived above for the case $\gamma_{\bf k}= 0$, which is as expected since the reflected wave phase has not entered into our considerations. Thus our proof (L.4) holds for general $\gamma_{\bf k}$.

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Next: Bibliography Up: Dissipation in Deforming Chaotic Previous: Appendix K: Transmission cross
Alex Barnett 2001-10-03