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 . We first consider a simple scaling argument which gives the dependence on and , 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
provided there is a corresponding rescaling of wavevector
.
Our slit potential is defined by a single parameter , the slit
half-width.
The particle wavelength
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 .
(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 .
Therefore the quantum cross-section must have the exact form

We seek in the 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 ) they become well approximated by the limit, namely solutions of Laplace's equation. Thus we can use `electrostatics' at these small distances, and then match to the small- 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 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 () radiation to outgoing dipole radiation; we will justify this shortly.

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

(K.2) |

where is a constant dependent on the slit size alone. The term in brackets is the difference in wavefunction -gradients between the left and right sides. It is determined by the `unscattered' wave , to be precise by the component of . This gradient difference `drives' the dipole terms and . The transmitted flux is in turn proportional to .

The dimensional scaling of (K.3) implies , so at constant the transmitted flux, and hence , grows like . Comparison with (K.1) immediately gives the scaling . This dependence of cross section for a given aperture size is the 2D equivalent of Rayleigh's famous 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 and with to all those with . However because of the associated and factors, by dimensional arguments all the elements of this matrix (other than the already considered above) will scale like powers of higher than 2. It is important to realise that the Laplace coefficients () directly connect to the respective coefficients of Bessel functions of type for and type for on the left (right) sides, which in turn determine the incoming and outgoing waves of angular momentum on those same sides. Therefore the resulting contribution to due to excitation by the channel on the left and re-radiation into the channel on the right will scale as . This contribution will therefore be smaller than the dipole-dipole (, ) component by a factor . In fact, because of symmetry about the -axis, the system only couples even to even and odd to odd ; this guarantees that the most significant correction is in fact a factor down. Thus in our limit we are justified in using only p-wave scattering: dipole `absorption' from , and dipole re-radiation into . (There will also be equal dipole re-radiation into , which interferes with but does not affect the conductance).

We now can find the prefactor.
The connection between the Laplace dipole solution ()
and the outgoing
dipole coefficient (see Section 7.2)
is found by matching to the small- form
[7] of the Neumann part of the Hankel function.
This gives
.
Combining this with (K.3) and the correct Laplace slit result
[168,148]
gives

(K.5) |

where in the second equality we used (K.4) with substituted. Applying Eq.(7.6) gives the desired transmission cross section as stated in Eq.(7.15).

Note that
is equal to
, 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 .
This follows from substitution of Eq.(K.1)
into Eq.(7.12).