Appendix K: Transmission cross section in the narrow slit limit

In this appendix we derive the transmission cross-section of the idealised slit from Section 7.3, in the tunneling limit $ka \ll 1$. We first consider a simple scaling argument which gives the dependence on $k$ and $a$, and then find the correct prefactor.

The slit system possesses a scaling property (shared by all hard-walled quantum systems): it is invariant under a rescaling of space $V{({\mathbf r})}\mapsto V(\alpha {\mathbf r})$ provided there is a corresponding rescaling of wavevector $k \mapsto \alpha k$. Our slit potential is defined by a single parameter $a$, the slit half-width. The particle wavelength $\lambda \equiv 2\pi/k$ is the only other length scale in the problem. This means that for a given incident wave, the form of the full scattering wavefunction, and hence the ratio of the quantum to classical transmission cross-sections, can only depend on $ka$. (In other words, if the wavelength changes in proportion to the slit size, we are viewing the original problem with a new unit of length.) The classical cross-section is proportional to $a$. Therefore the quantum cross-section must have the exact form

$$\sigma_{{\mbox{\tiny T}}}(k,\phi) = a \cdot f_\phi(ka),$$ (K.1)

where $f_\phi$ is a function dependent on the incident angle $\phi$.

We seek $f_\phi$ in the $ka \ll 1$ limit. For a hard-walled system solutions of Schrodinger's equation become those of the Helmholtz equation with Dirichlet boundary conditions. Close to the slit (a distance $r \ll \lambda$) they become well approximated by the $k\rightarrow0$ limit, namely solutions of Laplace's equation. Thus we can use `electrostatics' at these small distances, and then match to the small-$ka$ tails of the Helmholtz solutions to find the transmitted flux. The solutions obtained in this way have been known for over a century, and were first found by Lord Rayleigh ([168] in which the scattering solutions are derived for $ka \ll 1$ Dirichlet and Neumann apertures and their inverses, in both 2D and 3D. See p. 270-1 for our particular case). The transmission is dominated by scattering of incoming dipole ($l{=}1$) radiation to outgoing dipole radiation; we will justify this shortly.

The dipole parts of the 2D Laplace solution can be written generally as

$$\psi \; = \; \left\{ \begin{array}{ll} c_1 r\cos\theta + (... ...d_{-1}/r)\cos\theta' & \mbox{right side} . \end{array}\right.$$ (K.2)

The solution of Laplace's equation in the slit problem with this asymptotic ($r \gg a$) form is well known (see [168,148]), and can be found for instance using a conformal map $w(z) = (z-i a)^{1/2}(z+i a)^{1/2}$, where $z \equiv x + i y$. ($\psi$ is then a simple linear function of $w$.) There results a linear relation between the coefficients

$$c_{-1} \; = \; d_{-1} \; = \; A (c_1 + d_1) ,$$ (K.3)

where $A$ is a constant dependent on the slit size $a$ alone. The term in brackets is the difference in wavefunction $x$-gradients between the left and right sides. It is determined by the `unscattered' wave $\psi_{{\mbox{\tiny0}}}$, to be precise by the $l{=}1$ component of $\psi_{{\mbox{\tiny0}}}$. This gradient difference `drives' the dipole terms $c_{-1}$ and $d_{-1}$. The transmitted flux is in turn proportional to $\vert d_{-1}\vert^2$.

The dimensional scaling of (K.3) implies $A \sim a^2$, so at constant $k$ the transmitted flux, and hence $\sigma_{{\mbox{\tiny T}}}$, grows like $a^4$. Comparison with (K.1) immediately gives the scaling $\sigma_{{\mbox{\tiny T}}}\sim k^3 a^4$. This $\lambda^{-3}$ dependence of cross section for a given aperture size is the 2D equivalent of Rayleigh's famous $\lambda^{-4}$ law for dipole light scattering in 3D explaining why the sky appears blue [167] (see also [105] p.418).

How does the contribution of higher multipole radiation scale? The exact Laplace solution with general asymptotic form will involve generalizing Eq.(K.3) to a linear matrix relation connecting all the $\{ c_m \}$ and $\{ d_m \}$ with $m = 1 \cdots \infty$ to all those with $m = -1 \cdots -\infty$. However because of the associated $r^m$ and $r^{-m}$ factors, by dimensional arguments all the elements of this matrix (other than the $A$ already considered above) will scale like powers of $a$ higher than 2. It is important to realise that the Laplace coefficients $\{ c_m \}$ ($\{ d_m \}$) directly connect to the respective coefficients of Bessel functions of $J_{m}$ type for $m>0$ and $Y_{-m}$ type for $m<0$ on the left (right) sides, which in turn determine the incoming and outgoing waves of angular momentum $l = \vert m\vert$ on those same sides. Therefore the resulting contribution to $\sigma_{{\mbox{\tiny T}}}$ due to excitation by the $l$ channel on the left and re-radiation into the $l'$ channel on the right will scale as $\sim a^{l+l'}$. This contribution will therefore be smaller than the dipole-dipole ($l{=}1$, $l'{=}1$) component by a factor $\sim (ka)^{l+l'-2}$. In fact, because of symmetry about the $x$-axis, the system only couples even $l$ to even $l'$ and odd $l$ to odd $l'$; this guarantees that the most significant correction is in fact a factor $(ka)^2$ down. Thus in our $ka \ll 1$ limit we are justified in using only p-wave scattering: dipole `absorption' from $\psi_{{\mbox{\tiny0}}}$, and dipole re-radiation into $\psi_{{\mbox{\tiny T}}}$. (There will also be equal dipole re-radiation into $\psi_{{\mbox{\tiny R}}}$, which interferes with $\psi_{{\mbox{\tiny0}}}$ but does not affect the conductance).

We now can find the prefactor. The connection between the Laplace dipole solution ($m=-1$) and the outgoing dipole coefficient $q^+_1$ (see Section 7.2) is found by matching to the small-$kr$ form [7] of the Neumann part $Y_1(kr)$ of the Hankel function. This gives $q^+_1 = (i\pi k/2) d_{-1}$. Combining this with (K.3) and the correct Laplace slit result $A = a^2/4$ [168,148] gives

$$q^+_1 \; = \; \frac{i\pi k a^2}{8} \, c_1 .$$ (K.4)

Here we assumed that $d_1 = 0$ which is strictly true only when the slit is closed, since the unscattered wave is zero on the right side. However at $ka \ll 1$ it is a good approximation because the transmitted wave is much weaker than the incident. With a unit incident plane wave at angle $\phi$, the unscattered wave is Eq.(7.1) with $\gamma_{\bf k}= 0$, giving $c_1 = 2i k \cos \phi$. Now taking the large-$kr$ form of the Hankel function and comparing to Eq.(7.3) gives the scattering amplitude

$$f_{{\mbox{\tiny T}}}(\theta') \; = \; \sqrt{\frac{2}{\pi k}} \, q^+_1 \cos \theta'$$

$$\; = \; - \sqrt{\frac{\pi k^3}{8}} \, a^2 \cos \phi \, \cos \theta' ,$$ (K.5)

where in the second equality we used (K.4) with $c_1$ substituted. Applying Eq.(7.6) gives the desired transmission cross section $\sigma_{{\mbox{\tiny T}}}$ as stated in Eq.(7.15).

Note that $\sigma_{{\mbox{\tiny R}}}$ is equal to $\sigma_{{\mbox{\tiny T}}}$, because the reflected dipole strength is equal to the transmitted. (The dipole is also `two-headed', radiating in phase on both sides).

It is interesting to note that in any hard-walled scattering system, the conductance can only be a function of the product $ka$. This follows from substitution of Eq.(K.1) into Eq.(7.12).