# Sphere on a plane: Two-dimensional scattering from a finite curved region

James R. Anglin1 and Etienne Wamba1,2,3 1Physics Department and State Research
Center OPTIMAS, University of Kaiserslautern, Erwin-Schrödinger-Straße,
D-67663 Kaiserslautern, Germany
2Faculty of Engineering and Technology, University of Buea, P.O. Box 63, Buea,
Cameroon
3STIAS, Wallenberg Research Centre at Stellenbosch University, Stellenbosch
7600, South Africa

###### Abstract

Non-relativistic particles that are effectively confined to two dimensions can
in general move on curved surfaces, allowing dynamical phenomena beyond what
can be described with scalar potentials or even vector gauge fields. Here we
consider a simple case of piecewise uniform curvature: a particle moves on a
plane with a spherical extrusion. Depending on the latitude at which the
sphere joins the plane, the extrusion can range from an infinitesimal bump to
a nearly full sphere that just touches the plane. Free classical motion on
this surface of piecewise uniform curvature follows geodesics that are
independent of velocity, while quantum mechanical scattering depends on
energy. We compare classical, semi-classical, and fully quantum problems,
which are all exactly solvable, and show how semi-classical analysis explains
the complex quantum differential cross section in terms of interference
between two classical trajectories: the sphere on a plane acts as a kind of
double slit.

###### pacs:

02.40.Ky, 03.65.Nk, 03.65.Sq

##  I Introduction

Considerable study has been given to the quantum dynamics of a non-
relativistic particle confined to an arbitrary curved two-dimensional surface
embedded in flat three-dimensional space. While it has been necessary even
recently just to point out that in a curved space the Laplacian in the
Schrödinger equation must become the appropriate Laplace-Beltrami
operatorBernardVoon , much of the literature has concerned the addition of
potential terms involving the intrinsicdeWitt  or extrinsicJensenKoppe ;
daCosta ; Liu ; WangZong  curvature of the surface of motion, whether arising
from dynamical confinement in the full three-dimensional dynamics JensenKoppe
; daCosta ; WangZong  or from considerations of operator ordering in
constrained quantization deWitt ; Liu ; WangZong . The theory has moved beyond
these basic points: quantum motion in curved space has been given a phase
space representation GFH ; vector potentials have been incorporated Oliveira ;
and relativistic extensions have been provided, in the form of theories for
Dirac electrons on curved surfaces of topological insulators TakaneImura  or
even in wormhole geometries realized with graphene GonzalezHerrero .
Scattering of two-dimensional particles from curved regions has been analysed
perturbatively in Born approximation OMA18 , including lattices of small
surface bumps; perturbative scattering from curved surfaces with delta-
function defects has been studied in BM19 . An experiment studying electrons
on the curved inner surfaces of multi-electron bubbles in liquid helium has
even been reported Bubbles .

Explicit exact solutions for quantum particle motion on a curved surface have
only been provided in a few simple cases, however, including spheresLiu ,
ellipsoidsBernardVoon ; Mladenov , and toriWangZong . Here we provide explicit
solutions for a significantly different kind of curved surface, namely one
formed by joining a portion of a sphere to an infinite plane, to make a bump
or bubble that bulges out of the plane, which is otherwise flat. This ‘sphere-
on-a-plane’ surface thus has piecewise uniform curvature, namely zero (in the
planar portion) and negative (on the spherical portion). Three examples of the
kind of surface we mean are shown in Fig. 1. We will consider the finite
curved portion of this surface—the spherical extrusion—as a scatterer, and
compute its differential cross section: classically, semi-classically, and
quantum mechanically. Our paper can thus be considered complementary to Ref.
OMA18 , inasmuch as we consider a single class of simple geometries rather
than a generic bumpy surface, but go beyond Born approximation.

Figure 1: Surfaces of piecewise uniform curvature: a portion of a sphere
inserted into a plane. The radius of the sphere, and the latitude at which the
sphere joins the plane, are independent parameters. We express this latitude
as polar angle from the top of the sphere, denoted with $\alpha$. The examples
shown here have $\alpha=\pi/4$, $\pi/2$, and $3\pi/4$, from left to right.

We do this because we wish to consider particle deflection by surface
curvature as an analog to the specular reflection from hard walls which
defines dynamical billiardsSinai ; Heller . Billiard models are good tools for
examining the relationship between quantum and classical mechanics, because in
billiard models the classical trajectories do not depend on energy. Any
particle trajectory can be traversed at any energy, depending on how fast the
particle travels the path. Even though the classical phase space is four-
dimensional, therefore, one can describe the classical motion completely in
terms of orbits in the two-dimensional position space, and then compare this
to quantum wave functions in two-dimensional position space—and only the
quantum energy needs to be considered as a varying parameter, because the
classical paths are energy-independent.

This convenient and instructive feature of billiard models is shared by curved
surfaces, in which the classical motion is at arbitrary constant speed along
geodesics that are the same for all speeds, and thus independent of particle
energy. The ultimate interest in motion on curved surfaces in general may lie
in complex geometries that provide chaotic motion as billiards can, but our
contribution here will only be to consider the simplest case of scattering, in
an infinite plane, from an inserted spherical extrusion. The billiard analog
of our spherical extrusion would be a hard disk barrier. While the only thing
a hard disk can do is to reflect particles specularly, the spherical extrusion
can affect particle motion slightly, or greatly, depending on whether the
spherical extrusion is only a tiny bump, or a nearly full sphere.

###  I.1 Structure of the paper

In Section II we will describe our geometry precisely and introduce our
notation, and then solve the classical scattering problem, deriving a
surprisingly compact exact expression for the classical differential cross
section for spheres of arbitrary size and joining latitude.

In Section III we will then briefly discuss the issue of whether the
potentials applied to three-dimensional quantum particles, in order to
constrain their low-energy motion to follow a two-dimensional surface, must
always induce a non-constant effective potential within that surface. Although
particular effective potentials have been derived in the literature, we will
show that in general any potential whatever is possible, depending on exactly
how the three-dimensional motion is constrained to a surface, and that it is
therefore legitimate to consider the simplest case, in which the two-
dimensional potential vanishes and two-dimensional dynamics involves only the
intrinsic geometry of the surface. We will then exactly solve the quantum
scattering problem for our sphere-on-a-plane geometry with zero potential.

In Section IV we will construct the semiclassical approximation for the
sphere-on-a-plane scattering problem, and compare it to both our classical and
quantum results. The semiclassical approximation to the differential cross
section will turn out to be essentially the classical result, except with a
kind of two-slit interference pattern superimposed on it. We will find the
semiclassical approximation to be excellent whenever the quantum wavelength is
shorter than the radius of the circle on which the sphere meets the plane.

Finally in Section V we will briefly discuss our results and conclude.

##  II The Classical Problem

###  II.1 Motion on the sphere-in-a-plane

####  II.1.1 Notation and coordinates

Our geometry will be described as follows. The radius of the spherical
extrusion will be denoted $R$. The sphere is joined to the plane at polar
angle $\alpha$, so that the radius in the plane of the joining circle is
$R\sin\alpha$. See Fig. 2 for a side view of our extrusion geometry, which
looks somewhat different for $\alpha<\pi/2$ and $\alpha>\pi/2$. The two cases
of $\alpha<\pi/2$ and $\alpha>\pi/2$ will sometimes need to be considered
separately in our derivations, but all our results will come out as unified
formulas that apply in both regimes.

Figure 2: View of our surface in section from the side. The radius of the
sphere is $R$; it is joined to the plane at polar angle $\alpha$. For
$\alpha<\pi/2$, as in the sub-figure a), the spherical part of the surface is
a convex bump. For $\alpha>\pi/2$, as in sub-figure b), it is a more-than-
hemispherical bubble. In both cases the radius of the junction circle, within
the plane, is $R\sin\alpha$.

The Lagrangian for a classical particle of mass $M$ on a two-dimensional
surface is

$\mathscr{L}=\frac{M}{2}\sum_{i,j=1}^{2}g_{ij}\frac{d{r}_{i}}{dt}\frac{d{r}_{j}}{dt}$
(1)

where $r_{1,2}$ are arbitrary coordinates in the surface and $g_{ij}$ is the
two-dimensional metric tensor for the surface in those coordinates. Since our
problem has rotational symmetry around the center axis of the sphere, we will
use dimensionless polar coordinates $\rho=r/R$ and $\phi$ such that the
embedding of our surface in three-dimensional flat space is expressed in 3D
Cartesian coordinates as

$\left(\begin{array}[]{c}x\\\ y\\\
z\end{array}\right)=\left\\{\begin{array}[]{lcl}\left(\begin{array}[]{c}R\cos\phi\,\sin\rho\\\
R\sin\phi\,\sin\rho\\\ R\cos\rho\end{array}\right)&,&\rho<\alpha\\\
\left(\begin{array}[]{c}R(\rho-\alpha+\sin\alpha)\cos\phi\\\
R(\rho-\alpha+\sin\alpha)\sin\phi\\\
R\cos\alpha\end{array}\right)&,&\rho>\alpha\;.\end{array}\right.$ (2)

This implies the two-dimensional metric

$\left(\begin{array}[]{cc}g_{\rho\rho}&g_{\rho\phi}\\\
g_{\phi\rho}&g_{\phi\phi}\end{array}\right)=R^{2}\left(\begin{array}[]{cc}1&0\\\
0&g(\rho)\end{array}\right)$ (3)

where

$g(\rho)=\left\\{\begin{array}[]{lcl}\sin^{2}\rho&,&\rho<\alpha\\\
(\rho-\alpha+\sin\alpha)^{2}&,&\rho>\alpha\end{array}\right.\;.$ (4)

We will use these coordinates and this metric throughout this paper; in this
classical section we will also use the Cartesian $x,y$ coordinates for
straight-line trajectories in the plane.

####  II.1.2 Geodesics

In this classical problem the trajectories on planes and spheres are well
known: the geodesics are straight lines on the plane, and on the sphere they
are great circles. If $\alpha\geq\pi/2$ there will exist some closed great
circle orbits, such as the equator, which remain on the sphere and never enter
the plane. There are likewise many trajectories in the plane that simply pass
by the sphere without ever touching it. We are interested here, however, in
trajectories that cross from the plane onto the sphere and then back into the
plane. These must consist of half-lines and great circle arcs which connect on
the contact circle; the non-trivial question is, Which half-lines and arcs
must connect together to make up a total trajectory?

This problem can be solved with two-dimensional geometrical constructions,
using the fact that great circles on the sphere become ellipses with semi-
major axis $R$ when projected into the $x,y$ plane. It is both faster and more
easily related to less symmetrical problems, however, to use the Euler-
Lagrange equations of motion for the Lagrangian (1). For the $\phi$ coordinate
they yield

$MR^{2}g(\rho)\frac{d\phi}{dt}=J$ (5)

for some constant $J$. By differentiating our embedding (2) with respect to
$t$ we can confirm that $J$ is nothing but the angular momentum of the
particle about the central axis of the sphere:

$J\equiv M[x(t)\frac{d}{dt}y(t)-y(t)\frac{d}{dt}x(t)]\;.$ (6)

This Cartesian form for $J$ is convenient for the straight-line trajectories
in the plane.

The radial equation for $\rho$ has the energy as a first integral, allowing
the usual reduction to a first-order equation:

$\frac{d\rho}{dt}=\pm\frac{v}{R}\sqrt{1-\frac{J^{2}}{M^{2}v^{2}R^{2}g(\rho)}}$
(7)

for a constant $v>0$ that can be seen, by inserting (6) and (7) in (2), to be
the particle’s constant speed. Every incident trajectory begins on the $-$
branch of the $\pm$, with $\rho$ decreasing monotonically towards its minimum
value, where the trajectory makes its closest approach to the ‘north pole’ of
the sphere. There, where $\dot{\rho}$ is instantaneously zero, the branch
changes from $-$ to $+$, and thereafter $\rho$ increases monotonically, as the
particle exits the sphere and continues away in the plane to infinity.

####  II.1.3 Scattering trajectories

In the plane, Eqns. (5) and (7) are simply polar representations of straight
lines, which can also be represented more simply in Cartesian terms. Without
loss of generality we can take an initial straight-line trajectory which
encounters the spherical extrusion from the negative $x$ direction, moving in
the positive $x$ direction:

$\left(\begin{array}[]{c}x(t)\\\
y(t)\end{array}\right)_{\text{in}}=\left(\begin{array}[]{c}x(0)\\\
b\end{array}\right)+vt\left(\begin{array}[]{c}1\\\ 0\end{array}\right)\;,$ (8)

where $b$ is the impact parameter. Unless $|b|<R\sin\alpha$ the incident
particle will simply pass by the spherical extrusion without any scattering,
so we can restrict our attention to these cases. For later convenience we will
define the angle $\beta\in[-\pi/2,\pi/2]$ such that

$b=R\sin\alpha\,\sin\beta\;.$ (9)

By construction, therefore, this incident particle trajectory meets the
spherical extrusion at $\rho=\alpha$, $\phi=\pi-\beta$. The angular momentum
$J$ of this trajectory is determined by the velocity $v$ and impact parameter
$b$: it is easily found from (6) that

$J=-Mvb.$ (10)

Once the particle moves onto the spherical surface, it maintains its constant
speed $v$ but follows the unique great circle on the sphere which (i) has
angular momentum $J=-Mvb$ and (ii) meets the joining circle at
$(\rho,\phi)=(\alpha,\pi-\beta)$. The exactly equivalent description of this
motion in polar coordinates, less geometrically clear but more computationally
convenient, is that the particle’s radius $\rho$ will decrease from
$\rho_{1}=\alpha$ (the $-$ branch in (7)) until it reaches the turning point
$\rho_{0}$ at which $d\rho/dt=0$, then change to the $+$ branch of (7) and
increase until it again reaches $\rho=\alpha$, where the particle exits from
the spherical extrusion and re-enters the plane. The particle’s angle
coordinate at this time, $\phi=\phi_{\mathrm{exit}}$, can be computed
explicitly from

$\displaystyle\phi_{\mathrm{exit}}$ $\displaystyle=$
$\displaystyle\pi-\beta+2\int_{\alpha}^{\rho_{0}}\\!d\rho\,\frac{d\phi/dt}{d\rho/dt}$
(11) $\displaystyle=$
$\displaystyle\pi-\beta+2\int_{\alpha}^{\rho_{0}}\\!d\rho\,\frac{\sin\alpha\,\sin\beta}{\sin\rho\,\sqrt{\sin^{2}\rho-\sin^{2}\alpha\,\sin^{2}\beta}}$
$\displaystyle=$
$\displaystyle\pi-\beta+2\tan^{-1}\left(\frac{\sin\alpha\,\sin\beta\,\cos\rho}{\sqrt{\sin^{2}\rho-\sin\alpha\,\sin\beta}}\right)\Big{|}_{\rho=\rho_{0}}^{\rho=\alpha}$
$\displaystyle=$
$\displaystyle\pi-\beta+2\left(\tan^{-1}(\cos\alpha\,\tan\beta)-\frac{\pi}{2}\right)$
$\displaystyle=$ $\displaystyle 2\tan^{-1}(\cos\alpha\,\tan\beta)-\beta\;.$

The same result can be reached from great circle geometry but the derivation
is longer.

Returning now to Cartesian coordinates for the further motion in the plane,
the point of exit from the sphere and reentry to the plane is

$\left(\begin{array}[]{c}x\\\
y\end{array}\right)_{\mathrm{reentry}}=R\sin\alpha\left(\begin{array}[]{c}\cos\phi_{\mathrm{exit}}\\\
\sin\phi_{\mathrm{exit}}\end{array}\right)$ (12)

and therefore the final part of the trajectory which continues further into
the plane and has the same energy constant $\Omega$, hence the same speed $v$,
must be

$\left(\begin{array}[]{c}x(t)\\\
y(t)\end{array}\right)_{\text{out}}=R\sin\alpha\left(\begin{array}[]{c}\cos\phi_{\mathrm{exit}}\\\
\sin\phi_{\mathrm{exit}}\end{array}\right)+v(t-t_{\mathrm{exit}})\left(\begin{array}[]{c}\cos\theta\\\
\sin\theta\end{array}\right)$ (13)

for some scattering angle $\theta$.

The angular momentum (6) of this exiting trajectory must also be the same
$J=-Mvb=-MvR\sin\alpha\,\sin\beta$ of the incident trajectory. We therefore
have

$-\sin\beta=\cos\phi_{\mathrm{exit}}\,\sin\theta-\sin\phi_{\mathrm{exit}}\,\cos\theta\equiv\sin(\theta-\phi_{\mathrm{exit}})\;.$
(14)

One solution to this equation would be $\theta\to
2\tan^{-1}(\cos\alpha\,\tan\beta)\pm\pi$, but this would imply
$\dot{\rho}=-(v/R)\cos(\beta)<0$ at the moment of exit, when at this point we
must be on the $+$ branch of (7). Hence the only scattering angle which
satisfies the radial equation of motion as well as (14) is

$\theta=2[\tan^{-1}(\cos\alpha\,\tan\beta)-\beta]\qquad\mathrm{mod}(2\pi)\;.$
(15)

For our use in Section IV, below, we recall that
$\sin\beta=b/(R\sin\alpha)=-J/(MvR\sin\alpha)$, which means that
$\sin\beta\to-m/(kR\sin\alpha)$ when in the quantum problem $J\to\hbar m$ and
$Mv\to\hbar k$. This means that the classical scattering angle is a function
of $\alpha$ and $J/(MvR)\to m/(kR)=:\mu$,

$\displaystyle\theta$ $\displaystyle=\Theta\Big{(}\alpha,\frac{J}{MvR}\Big{)}$
(16) $\displaystyle\Theta(\alpha,\mu)$
$\displaystyle=2\sin^{-1}\Big{(}\frac{\mu}{\sin\alpha}\Big{)}-2\tan^{-1}\Big{(}\frac{\mu\cos\alpha}{\sqrt{\sin^{2}\alpha-\mu^{2}}}\Big{)}\;.$

The classical scattering angle $\theta$ will appear in the semi-classical
theory of Section IV, in this form as $\Theta(\alpha,\mu)$.

####  II.1.4 Scattering angle for given impact parameter

We recall that $|\beta|\leq\pi/2$ by definition, because for scattering the
impact parameter $b=R\sin\alpha\,\sin\beta$ must lie between $\pm
R\sin\alpha$. If we vary the impact parameter from $-R\sin\alpha$ to
$+R\sin\alpha$, $\tan\beta$ increases monotonically from $-\infty$ to
$+\infty$. If $\alpha>\pi/2$, then $\cos\alpha\,\tan\beta$ decreases
monotonically from $+\infty$ to $-\infty$, and hence both
$\tan^{-1}(\cos\alpha\,\tan\beta)$ and $-\beta$ decrease monotonically from
$\pi/2$ to $-\pi/2$. This means that $\theta$ decreases from $2\pi$ to
$-2\pi$, modulo $2\pi$. In other words, when the spherical extrusion is larger
than a hemisphere, the scattering angle covers the full circle, so that every
possible scattering angle, including directly backwards, occurs for some
impact parameter. In fact $\theta$ actually sweeps through a range of $4\pi$,
meaning that every scattering angle is obtained twice, for two different
values of the impact parameter $b$.

If $\alpha<\pi/2$, on the other hand, so that the spherical extrusion is less
than a hemisphere, then $\tan^{-1}(\cos\alpha\,\tan\beta)$ increases
monotonically with $b$ while $-\beta$ decreases. As a result, $\theta$ does
not change monotonically with $b$, but has an extremum when

$\displaystyle\frac{d\theta}{d\beta}$ $\displaystyle=$ $\displaystyle
2\left(\frac{\cos\alpha}{1-\sin^{2}\alpha\,\sin^{2}\beta}-1\right)\to 0$ (17)
$\displaystyle\Longrightarrow$
$\displaystyle\beta\to\pm\beta_{c}=\pm\sin^{-1}\left(\frac{1}{2\cos\frac{\alpha}{2}}\right)$
$\displaystyle\Longrightarrow$ $\displaystyle\theta\to\pm\theta_{c}=\pm
2\sin^{-1}\left(\tan^{2}\frac{\alpha}{2}\right)\;.$

So if the extrusion is smaller than a hemisphere, no particles are scattered
by more than the angle $\pm\theta_{c}$, and these scattering angles are
caustic angles at which scattered trajectories “pile up” because the
scattering angle is reversing direction as a function of impact parameter.
Those scattering angles $|\theta|<\theta_{c}$ which are realized are again
each realized twice, for different impact parameters.

###  II.2 Differential cross sections

While $\theta(\beta)$ is the natural solution to the particle motion as an
initial value problem, the differential cross section $|db/d\theta|$ defines
what infinitesimal range of impact parameters $b$ contributes to scattering
within an infinitesimal range of scattering angles $\theta$. In this sense it
is a final value problem instead of an initial value problem, but it is
normally used to give probabilistic answers to a probabilistically posed
initial value question: if an ensemble of incident particles encounters the
scatterer with an evenly distributed range of impact parameters, then
$|db/d\theta|$ as a function of $\theta$ gives the probability density for
scattering at the angle $\theta$.

####  II.2.1 Hard disk example

Since differential cross sections may be more familiar in quantum mechanics
than they are classically, we briefly review this classical scattering theory
for the simple example of a hard disk scatterer of radius $a$. This geometry
is shown in Fig. 3, and from the figure we can immediately see that for impact
parameter $b=a\sin\beta>0$ the exit angle is $\theta=\pi-2\beta$. For negative
$b$, it is clear from the problem’s vertical reflection symmetry that we must
simply reverse the sign of $\theta$. It follows that in general

$\theta=\left\\{\begin{matrix}\pi\,\mathrm{sgn}(\beta)-2\beta\;,&\beta\neq
0\\\ \pi\;,&\beta=0.\end{matrix}\right.$ (18)

Figure 3: Classical scattering from a hard disk of radius $a$. Left: a
particle with impact parameter $b=a\sin\beta$ is reflected. Right: Polar plot
of the cardioid $r=(1/4)\sin(|\theta|/2)$, which is the differential cross
section divided by the total cross section.

To obtain the differential cross section, however, we must invert this
relation and find $b=a\sin\beta$ as a function of $\theta$. For the hard disk
this can be easily done: $\beta=(1/2)(\pi\,\mathrm{sgn}(\theta)-\theta)$, so
that for $b=a\sin\beta$ we obtain

$b(\theta)=a\,\mathrm{sgn}(\theta)\cos\frac{\theta}{2}\;.$ (19)

The differential cross section is then

$\Big{|}\frac{db}{d\theta}\Big{|}=\frac{a}{2}\sin\frac{|\theta|}{2}\;,$ (20)

which is the cardioid curve plotted in the right panel Fig. 3. The shape of
the cardioid means that when an ensemble of particles with evenly distributed
impact parameters strike the disk from the left, many are scattered nearly
backward, while only few strike near enough to the edges of the disk to
receive only slight deflections. The total cross section is identically equal
to the disk diameter $2a$, because the integral of the differential cross
section over the full range of scattering angles is by definition equal to the
integral over all $b$ that produce any deflection, which is the range
$-a<b<a$.

####  II.2.2 Spherical extrusions

To compute the classical differential cross section for our spherical
extrusions, therefore, we need to invert our expression (15) for
$\theta(\beta)$ and obtain $b=R\sin\alpha\,\sin\beta$ as a function of
$\theta$ instead. Surprisingly, this can be done quite compactly, if we begin
by using (15) to compute

$\displaystyle\sin\frac{\theta}{2}$ $\displaystyle=$
$\displaystyle\frac{\cos\alpha\,\sin\beta-\sin\beta}{\sqrt{1+\cos^{2}\alpha\tan^{2}\beta}}$
(21) $\displaystyle=$
$\displaystyle\sin\beta\,\cos\beta\frac{\cos\alpha-1}{\sqrt{1-\sin^{2}\alpha\sin^{2}\beta}}\;.$

Squaring both sides of (21) produces a quadratic equation for $\sin^{2}\beta$.
Solving this and then applying trigonometric identities, and checking the
results against (15) itself to discard spurious roots produced by the squaring
of (21), eventually yields two solutions for $b=R\sin\alpha\,\sin\beta$ for
each $\theta$, which we denote as $b(\theta)\to b_{\pm}(\theta)$:

$\displaystyle b_{+}$ $\displaystyle=$
$\displaystyle-R\,\mathrm{sgn}(\cos\alpha)\,\mathrm{sgn}(\theta)\cos^{2}\frac{\alpha}{2}$
$\displaystyle\times\left(\sqrt{\left(1+\sin\frac{|\theta|}{2}\right)\left(\tan^{2}\frac{\alpha}{2}+\sin\frac{|\theta|}{2}\right)}\right.$
$\displaystyle\qquad+\left.\sqrt{\left(1-\sin\frac{|\theta|}{2}\right)\left(\tan^{2}\frac{\alpha}{2}-\sin\frac{|\theta|}{2}\right)}\right)$
$\displaystyle b_{-}$ $\displaystyle=$
$\displaystyle-R\,\mathrm{sgn}(\theta)\cos^{2}\frac{\alpha}{2}$
$\displaystyle\times\left(\sqrt{\left(1+\sin\frac{|\theta|}{2}\right)\left(\tan^{2}\frac{\alpha}{2}+\sin\frac{|\theta|}{2}\right)}\right.$
$\displaystyle\qquad-\left.\sqrt{\left(1-\sin\frac{|\theta|}{2}\right)\left(\tan^{2}\frac{\alpha}{2}-\sin\frac{|\theta|}{2}\right)}\right)\;.$

For $\alpha<\pi/2$, Eqn. (II.2.2) is only valid for
$|\theta|<\theta_{c}=2\sin^{-1}(\tan^{2}\frac{\alpha}{2})$, because no
scattering occurs with angles $|\theta|>\theta_{c}$. At the largest possible
scattering angles, which are either $\pm\theta_{c}$ for $\alpha<\pi/2$ or
$\pm\pi$ for $\alpha>\pi/2$, the two branches $b_{\pm}$ coincide. For
$\theta\to 0^{\pm}$, $b_{-}\to 0$ while $b_{+}\to\mp
R\,\mathrm{sgn}(\cos\alpha)\sin\alpha$.

Figure 4: Examples of pairs of trajectories with the same scattering angle
$\theta$, for $\alpha=\pi/4$ and $\theta=0.25$ (left panel) and
$\alpha=3\pi/4$, $\theta=-5\pi/8$ (right panel). The trajectories are shown
projected onto the plane, so that great circle arcs on the sphere appear as
elliptical arcs. In each case the two trajectories enter the spherical
extrusion from the left with two different impact parameters
$b_{\pm}(\alpha,\theta)$, and exit at two different exit angles
$\phi_{\mathrm{exit}}$, but emerge from the extrusion on parallel paths. One
of the two trajectories in the right panel is shown dashed, to aid in
distinguishing the two trajectories where their planar projections cross.
Portions of the trajectories between the entry and exit tangents (dotted
lines) are shown thicker, because the lengths of these thicker portions will
appear in the semi-classical theory of Section IV.

Since both branches of $b_{\pm}$ contribute scattering in the $\theta$
direction, the differential cross section at $\theta$ has contributions from
both: $|db/d\theta|=\sum_{\pm}|db_{\pm}/d\theta|$. It can be shown using
trigonometric identities, or by plotting, that for $\alpha>\pi/2$, $b_{\pm}$
are both monotonically decreasing functions of $\theta$ over their full ranges
of support in $\theta$, while for $\alpha<\pi/2$, $b_{-}$ is decreasing and
$b_{+}$ is increasing. In the sum of the two branches, therefore, the first
square root terms in each $b_{\pm}$ always cancel each other in the
differential cross section, while the second square root terms add together.

The final result for the classical differential cross section of the spherical
extrusion is thus this surprisingly compact expression full of half-angles:

$\displaystyle\Big{|}\frac{db}{d\theta}\Big{|}$ $\displaystyle=$
$\displaystyle 2R\sin\alpha\,D(\alpha,\theta)$ (23) $\displaystyle
D(\alpha,\theta)$ $\displaystyle=$
$\displaystyle\frac{\sqrt{1+\sin\frac{|\theta|}{2}}\left(1-2\cos^{2}\frac{\alpha}{2}\sin\frac{|\theta|}{2}\right)}{4\sin\alpha\sqrt{\tan^{2}\frac{\alpha}{2}-\sin\frac{|\theta|}{2}}}\;.$

The integral of this differential cross section over all scattering angles
(i.e. from $-\theta_{c}$ to $+\theta_{c}$ for $\alpha<\pi/2$ and from $-\pi$
to $+\pi$ for $\alpha>\pi/2$) is exactly $2R\sin\alpha$, as indeed it must be,
because this is simply the full range of $b$ for which the particle touches
the sphere and can be deflected. The dimensionless function $D(\alpha,\theta)$
thus gives the angular distribution of scattered particles, and is independent
of $R$; it is plotted for a selection of different values of $\alpha$ in Fig.
5. The area inside the closed curve defined as having radius
$r(\theta)=D(\alpha,\theta)$ is identically 1 for all $\alpha$, if we count
the caustic limits $\pm\theta_{c}$ as closing the curve for $\alpha<\pi/2$.
The divergences of $D(\alpha,\theta)$ at the caustics are only inverse square
roots $\sim(|\theta|-\theta_{c})^{-1/2}$, which are integrable.

Figure 5: Normalized differential cross section $D(\alpha,\theta)$ as polar
plots in $\theta$ for the different values of $\alpha$ shown by the plot
labels. To better show the shapes of the curves, the different plots are not
all drawn to the same scale; the true size of the contours is determined by
the fact that they all enclose unit area.

We can also note a few simple exact results for special cases of $\alpha$. For
$\alpha\to 0$ the maximum scattering angle $\theta_{c}\to 0$ as well—a slight
bump deflects only slightly—and so for all the very small angles into which
scattering occurs we can write

$\lim_{\alpha\to\epsilon^{+}}D(\alpha,\theta)=\frac{1/2}{\epsilon\sqrt{\epsilon^{2}-2|\theta|}}\;.$
(24)

Here $\epsilon$ is a small (positive) value of the parameter $\alpha$. If we
let $\alpha\to 0$ while increasing $R\sim 1/\alpha$ so that $a=R\sin\alpha$
remains fixed, then our spherical extrusion becomes a disk of finite radius
$a$ with vanishing curvature. It nonetheless retains its finite total cross
section $2a$—and this seems paradoxical, because in this limit of vanishing
curvature the spherical extrusion should have no effect whatever on particle
motion, since it is nothing but an arbitrarily designated circular region in
the plane. The resolution of this paradox is in the definition of cross
section as the measure of a set of trajectories that are affected by the
spherical region. This definition takes no account of how greatly the
trajectories are affected; any deflection at all is accounted as scattering,
no matter how slight the deflection may be. When we take the limit of
$R\to\infty$ and $\alpha\to 0$ so that $R\sin\alpha$ stays constant, we
approach the case where the sphere has no effect by keeping the same set of
trajectories affected, but reducing the amount by which they are deflected.

For $\alpha\to\pi$ we instead have

$D(\pi,\theta)=\frac{\sqrt{1+\sin\frac{|\theta|}{2}}}{8}\;.$ (25)

So if we let $R\to\infty$ while keeping $a=R\sin\alpha$ fixed by letting
$\alpha$ approach $\pi$, then the finite contact circle of radius $a$ becomes
the junction between the plane and a large spherical world into which
particles can disappear from the plane for a long time, emerging eventually at
some different angle from the one at which they entered, with a particular
average distribution of exit angles given by the expression above, independent
of $R$ (for fixed $a=R\sin\alpha$).

Finally, for the hemisphere we have

$D(\frac{\pi}{2},\theta)=\frac{1}{4}\cos\frac{\theta}{2}\;,$ (26)

which is exactly the same cardioid distribution of scattered particles that
would be emitted by a hard disk, if the particles were incident from the
opposite direction.

##  III The Quantum Problem

###  III.1 From three dimensions to two

Any two-dimensional problem may be considered theoretically, but to restrict
the three-dimensional motion of a physical particle to a two-dimensional
surface, one must apply some confining force in the perpendicular direction.
In Ref. daCosta  it was concluded that the effective two-dimensional theory in
quantum mechanics must therefore include a certain effective potential
determined by the extrinsic curvature of the surface. In fact this particular
result for the effective potential is not a universal formula at all, however,
but only the special case that applies if the strength of the perpendicular
confinement is kept uniform over the two-dimensional surface. It is easy to
see that even a small proportional variation of confinement strength can
produce an arbitrary effective potential within the surface.

To confirm this it suffices to take the trivial case in which the two-
dimensional surface to which three-dimensional motion should be confined is
simply the plane $z=0$ in Cartesian coordinates. We let the confinement be
enforced by a three-dimensional potential which is very slightly non-uniform
over the plane:

$V_{3D}(x,y,z)=\frac{\hbar^{2}z^{2}}{2M\lambda^{4}}\left[1+\epsilon^{2}\Phi\Bigl{(}\epsilon\frac{x}{\lambda},\epsilon\frac{y}{\lambda}\Bigr{)}\right]^{2}\;,$
(27)

where $M$ is the particle mass, $\lambda$ is the transverse confinement
length, and $\epsilon\ll 1$. The effectively two-dimensional limit applies
when we consider particle energies that are low in comparison to the
transverse confinement scale,

$\displaystyle
E=[1+2\epsilon^{2}\mathcal{E}]\,\frac{\hbar^{2}}{2M\lambda^{2}}$ (28)

for dimensionless $\mathcal{E}=\mathcal{O}(\epsilon^{0})$.

We can then introduce the scaled planar coordinates
$(\tilde{x},\tilde{y})=(\epsilon/\lambda)(x,y)$ and the Ansatz

$\Psi(x,yz)=e^{-\frac{z^{2}}{2\lambda^{2}}\left[1+\epsilon^{2}\Phi(\tilde{x},\tilde{y})\right]}\sum_{n=0}^{\infty}\epsilon^{4n}\psi_{n}\Bigl{(}\epsilon\frac{x}{\lambda},(\epsilon\frac{y}{\lambda}\Bigr{)}H_{2n}\Bigl{(}\frac{z\sqrt{1+\epsilon^{2}\Phi}}{\lambda}\Bigr{)}$
(29)

for the three-dimensional wave functions of energy eigenstates, with $H_{n}$
being the Hermite polynomials. The orthogonality of the Hermite polynomials
then lets us read back out

$\displaystyle\psi_{n}(\tilde{x},\tilde{y})=\epsilon^{-4n}\int\\!dz\,e^{-\frac{z^{2}}{2\lambda^{2}}[1+\epsilon^{2}\Phi(\tilde{x},\tilde{y})]}H_{2n}\Bigl{(}\frac{z\sqrt{1+\epsilon^{2}\Phi}}{\lambda}\Bigr{)}\Psi(x,y,z)\;.$
(30)

Inserting (30) into the three-dimensional time-independent Schrödinger
equation then reveals

$\displaystyle\mathcal{E}\psi_{0}$ $\displaystyle=$
$\displaystyle-\frac{1}{2}\left[\frac{\partial^{2}}{\partial\tilde{x}^{2}}+\frac{\partial^{2}}{\partial\tilde{y}^{2}}\right]\psi_{0}+\frac{1}{2}\Phi(\tilde{x},\tilde{y})\psi_{0}+\mathcal{O}(\epsilon^{2})\;.$

Even a proportionally very small non-uniformity $\epsilon^{2}\Phi$ in strong
perpendicular confinement will thus produce a significant effective potential
in the two-dimensional low-energy dynamics.

For confinement to surfaces more complicated than a plane, Ref. daCosta  has
shown that the there is a non-trivial effective potential in two dimensions
even when the perpendicular confinement strength is uniform. We now point out,
however, that it may be a difficult matter in practice to achieve such uniform
confinement that the result of daCosta  is valid—even when the surface to
which motion should be confined is as simple as a plane or a cylinder. If the
three-dimensional potential is complicated enough to confine the particle to a
more complicated surface, then it will surely be more difficult still to make
the confinement strength uniform. The particular potential derived in daCosta
must therefore be considered as a somewhat academic result. Quasi-two-
dimensional motion that is realized in an experiment is likely to be subject
to an effective potential that depends arbitrarily on the details of exactly
how the confinement to two dimensions is achieved.

Conversely, however, it is in principle possible for practically any two-
dimensional potential to be achieved, if the perpendicular confinement is
implemented appropriately. The effects of local potentials on quantum motion
in two dimensions have long been familiar, while the kinetic effects of
spatial curvature, through a non-trivial Laplace-Beltrami operator, are less
well understood. It is therefore a worthwhile theoretical contribution to
analyze an informative model that involves only the Laplace-Beltrami operator,
since this simple scenario is in principle no less realistic than the uniform
confinement scenario assumed in Ref. daCosta . In this paper we will provide
such a contribution, by assuming that the effective confinement from three
dimensions to two has somehow been achieved in such a way as to leave no
potential term in the effectively two-dimensional Schrödinger equation for
quantum motion in the low-energy limit.

One result is easy to obtain immediately: introducing a curved region within
an asymptotically flat plane that extends to infinity can never create any
bound states. If there were such a bound state, then in the flat plane at
infinity it would have to decay exponentially. It would therefore have to have
negative energy, as usual for a bound state. There would therefore exist a
wave function, namely the wave function of this bound eigenstate, for which
the expectation value of the Hamiltonian was negative. The expectation value
of the energy of a particle on a curved surface without any potential,
however, is

$\langle\hat{H}\rangle=\frac{\hbar^{2}}{2M}\sum_{i,j=1}^{2}\int\\!d^{2}r\,\sqrt{g}g^{ij}\partial_{i}\psi^{*}\partial_{j}\psi\;,$
(32)

where $g^{ij}$ is the contravariant metric tensor on the two-dimensional
surface and $g$ is the determinant of its inverse matrix (the covariant
metric). The integral is invariant under arbitrary coordinate changes, so we
can show that the integrand is everywhere positive definite, for all
$\psi(\mathbf{r}$, by transforming at any point to coordinates in which the
metric tensor is $g^{ij}=\delta_{ij}$. No negative energy expectation values
can exist, therefore—and therefore no negative eigenvalues and no bound
states. Local curved regions can hold long-lived quasi-bound resonances,
however, as we will see.

For potential-free cases in which an asymptotically flat plane contains a
curved region, the quantum problem is thus essentially a scattering problem:
how do incident plane waves from infinity propagate through the curved region?
For the simple case of the sphere on a plane the quantum scattering problem
can, like its classical counterpart, be solved exactly, albeit in the form of
a Fourier series for the scattering amplitude in which the coefficients
involve Legendre and Bessel functions.

###  III.2 Time-independent Schrödinger equation for the sphere on a plane

We consider a quantum particle of mass $M$ that moves in the geometry defined
by the metric (3), with sphere radius $R$ and contact angle $\alpha$ as
before. In a quantum energy eigenstate of energy eigenvalue
$E=\hbar^{2}k^{2}/(2M)$, the wave function $\psi(\mathbf{r})$ of the particle
obeys the time-independent Schrödinger equation

$\frac{\hbar^{2}k^{2}}{2M}\Psi(\mathbf{r})=-\frac{\hbar^{2}}{2MR^{2}}\left[\frac{\partial^{2}}{\partial\rho^{2}}+\frac{1}{2}\frac{g^{\prime}(\rho)}{g(\rho)}\frac{\partial}{\partial\rho}+\frac{1}{g(\rho)}\frac{\partial^{2}}{\partial\theta^{2}}\right]\Psi,$
(33)

where $\mathbf{r}=(R\rho,\theta)$ in polar coordinates, with $g(\rho)$ again
given by (4) and $g^{\prime}(\rho)\equiv dg/d\rho$. We can then further
decompose a generic energy eigenstate into simultaneous eigenstates of energy
and angular momentum,

$\Psi(\mathbf{r})=\sum_{m}C_{m}e^{im\phi}\psi_{m}(\rho)\;,$ (34)

satisfying

$(kR)^{2}\psi_{m}(\rho)=-\left[\frac{\partial^{2}}{\partial\rho^{2}}+\frac{1}{2}\frac{g^{\prime}(\rho)}{g(\rho)}\frac{\partial}{\partial\rho}-\frac{m^{2}}{g(\rho)}\right]\psi_{m}.$
(35)

We note that $\hbar m$ is the quantum analog of the classical angular momentum
$J$, and that the wave function depends on $R$ and $k$ only through their
product $kR$.

For $\rho>\alpha$, (35) is just the usual Bessel equation for a radial
eigenfunction in the plane, with the ordinary radial coordinate
$r=R(\rho-\alpha+\sin\alpha)$ in the plane outside the sphere. So the
normalized real solution for $\rho>\alpha$ can be written

$\displaystyle\psi_{m}(\rho)=$
$\displaystyle\cos(\delta_{m})J_{m}\big{(}kR(\rho-\alpha+\sin\alpha)\big{)}$
$\displaystyle-\sin(\delta_{m})Y_{m}\big{(}kR(\rho-\alpha+\sin\alpha)\big{)}$
$\displaystyle\equiv$
$\displaystyle\mathrm{Re}\left(e^{i\delta_{m}}H_{m}\big{(}kR(\rho-\alpha+\sin\alpha)\big{)}\right)\;,$
(36)

for some phase shift $\delta_{m}(kR)$, where $J_{m}$ and $Y_{m}$ are the
Bessel and Neumann functions, respectively, and $H_{m}=J_{m}+iY_{m}$ is the
Hankel function. For $\rho<\alpha$, however, (35) is the associated Legendre
equation of order $\lambda$ such that $\lambda(\lambda+1)=(kR)^{2}$,

$\lambda(kR)=\sqrt{(kR)^{2}+\frac{1}{4}}-\frac{1}{2}\;.$ (37)

This equation has only one solution that is regular at $\rho=0$, the
associated Legendre function $P_{\lambda}^{m}(\cos\rho)$, and so we must have
$\psi_{m}(\rho)=C_{m}P_{\lambda}^{m}(\cos\rho)$ for some real $C_{m}(kR)$.

The coefficients $C_{m}$ and $\delta_{m}$ are determined by imposing
continuity of $\psi_{m}(\rho)$ and $\psi^{\prime}_{m}(\rho)$ across the
junction at $\rho=\alpha$ between the sphere and the plane. The result for the
phase shift $\delta_{m}$ is

$\tan\delta_{m}=\frac{kRP^{m}_{\lambda}(\cos\alpha)J^{\prime}_{m}(kR\sin\alpha)+\sin\alpha
J_{m}(kR\sin\alpha)P^{{}^{\prime}m}_{\lambda}(\cos\alpha)}{kRP^{m}_{\lambda}(\cos\alpha)Y^{\prime}_{m}(kR\sin\alpha)+\sin\alpha
Y_{m}(kR\sin\alpha)P^{{}^{\prime}m}_{\lambda}(\cos\alpha)}\;,$ (38)

where the primes denote differentiation of each function with respect to its
argument. In the two-dimensional version of standard partial wave scattering
theory, the $\delta_{m}$ determine all scattering features, in the sense that
an energy eigenstate of the scattering form

$\displaystyle\Psi_{k}(\mathbf{r})$
$\displaystyle=e^{ikx}+\psi_{\mathrm{sc}}(\mathbf{r})$
$\displaystyle\lim_{r\to\infty}\psi_{\mathrm{sc}}$
$\displaystyle=\sqrt{\frac{2}{\pi kr}}e^{i\frac{\pi}{4}}f(\theta)e^{ikr}$ (39)

can be composed out of many $\psi_{m}$ as

$\Psi_{k}(\mathbf{r})=\sum_{m=-\infty}^{\infty}e^{i\delta_{m}}i^{m}e^{im\theta}\psi_{m}(\rho)\;.$
(40)

This follows from the asymptotic forms of the Bessel and Neumann functions and
from the expansion of the two-dimensional plane wave in polar coordinates. The
result by elementary algebra is just as for partial waves in three dimensions,

$f(\theta)=\sum_{m=-\infty}^{\infty}\sin\delta_{m}e^{i\delta_{m}}e^{im\theta}\;,$
(41)

which implies the two-dimensional optical theorem

$\displaystyle\sigma:=$
$\displaystyle\lim_{r\to\infty}r\oint\\!d\theta\,|\psi_{\mathrm{sc}}|^{2}=\frac{2}{\pi
k}\oint\\!d\theta\,|f|^{2}$ $\displaystyle=$
$\displaystyle\frac{4}{k}\sum_{m=-\infty}^{\infty}\sin^{2}\delta_{m}=\frac{4}{k}\mathrm{Im}[f(0)]\;.$
(42)

We can therefore obtain exact differential cross-sections
$d\sigma/d\theta=2|f(\theta)|^{2}/(\pi k)$ for arbitrary $kR$ and $\alpha$,
from the Fourier series (41) for $f(\theta)$, with $\delta_{m}$ given in terms
of Bessel, Neumann, and associated Legendre functions by (38). Using the
asymptotic behaviors of the Bessel functions at small argument, and of the
associated Legendre functions at small order $\lambda$, it is straightforward
to show that for $kR\ll 1$, scattering becomes approximately isotropic
($f(\theta)=\sqrt{k\sigma/4}\times[1+\mathcal{O}(kR)^{2}]$), with

$\lim_{kR\to 0}\sigma=4\pi^{2}k^{3}R^{4}\sin^{8}\frac{\alpha}{2}\;.$ (43)

For $\alpha\ll 1$ as well as $kR\ll 1$, this agrees with the first Born
approximation result of $\sigma=(\pi^{2}/64)k^{3}R^{4}\sin^{8}\alpha$ which is
obtained from Eqn. (23) of Ref. OMA18 , when we express our locally curved
surface in the notation of Ref. OMA18 . (The quantity $G(r)$ of Ref. OMA18
becomes $G(r)\to(r/R)\theta(R\sin\alpha-r)$ in our case, $\theta(x)$ being the
step function, and the parameters $\lambda_{1,2}$ of Ref. OMA18  are both zero
for us.) The first Born approximation of Ref. OMA18  is only valid in our case
when both $\alpha$ and $kR$ are small, so that the total effect of the curved
region can be a small perturbation; the results of OMA18 , on the other hand,
are valid for small local curvature deformations that are much more general in
form than our sphere on a plane.

Although at long wavelengths the isotropic quantum scattering is simpler than
the caustics and cardioids of the classical differential cross section, at
shorter wavelengths the angular dependence of quantum scattering becomes more
complicated. See Fig. 6 for some examples, which illustrate most of the
typical features for all $kR$ and $\alpha$; the classical differential cross
sections in each case are shown in dots, for comparison. Once $kR$ is larger
we can recognize that the quantum differential cross section mainly oscillates
around the classical value, but beyond this there are significant differences.

Figure 6: Quantum differential cross sections divided by $R$, for $kR=20$.
Although Cartesian axes are shown because the plots extend so far to the
right, these are actually polar plots, with deflection angle $\theta=0$
corresponding to the horizontal-right direction, and radius from the origin at
(0,0) giving $R^{-1}d\sigma/d\theta=2|f(\theta)|^{2}/(\pi kR)$, for scattering
into the corresponding deflection angle. The corresponding classical
differential cross sections divided by $R$, $R^{-1}db/d\theta$ as given by
(23), are superimposed as dashed contours. Upper panel: a slightly-less-than-
hemisphere ($\alpha=0.47\pi$). Lower panel: a three-quarter sphere
($\alpha=3\pi/4$).

At small deflection angles $\theta$ there is a dramatic spike of quantum
scattering probability. In the limit of large $kR$, where one might expect
from classical correspondence that the total quantum cross section would
approach the classical value $2R\sin\alpha$, the narrow forward spike actually
contributes a further $2R\sin\alpha$ in cross section all by itself, bringing
the total quantum cross section to $4R\sin\alpha$ as $kR\to\infty$. As we will
see in the next section, however, this large forward lobe in the differential
cross-section is actually such a basic wave-mechanical scattering feature that
it is not really due to the peculiar geometry of the sphere on a plane. Any
round obstacle of the same size—for example, a hard disk—will produce a very
similar spike of forward scattering for wavelengths much shorter than the
obstacle size. The effect is a matter wave analogue of the optical Poisson
spot, and while it is in principle a real phenomenon it is in practice
difficult to observe because its existence depends on the incident particle
beam having a coherence width broader than the target width (in this case
$2R\sin\alpha$). At large $kR$, furthermore, where the forward spike is most
dramatic, the spike becomes so narrowly concentrated around deflection angle
zero that it could be difficult to distinguish it in experiments from the
background of unscattered particles in the incident beam. There will usually
be little reason for experimentalists to work hard to resolve the forward
spike, moreover, since scattering experiments are usually intended to probe
the internal structure of the scattering target, and the forward spike does
not depend on this internal structure. In effect the forward spike is a
dramatic wave-mechanical effect that is not actually very important.

The oscillations of $|f(\theta)|^{2}$ around the classical $db/d\theta$
persist at high $kR$; as we will see, they are a basic interference effect due
to the presence of two classical trajectories for each scattering angle
$\theta$. The classical caustics which occur for $\alpha<\pi/2$ when these two
trajectories merge are, as usual, softened by wave mechanics. The quantum
scattering probability extends smoothly beyond these classical caustics; the
appearance of quantum particles in a classically forbidden zone can be
considered a form of tunnelling even though we do not have a potential barrier
of the usual kind.

Figure 7: Quantum total cross sections $\sigma$ divided by sphere radius $R$,
as functions of $kR$, for a quarter-sphere ($\alpha=\pi/4$, upper panel) and a
slightly-more-than-hemisphere ($\alpha=5\pi/8$, lower panel). The high-$kR$
limit $\sigma/R=4\sin\alpha$ is shown as a horizontal dashed line; the semi-
classical approximation with tunnelling neglected, defined below in Section
IV, is shown as a dotted curve. For sub-hemispheres like the case in the upper
panel, the cross section oscillates as it approaches the asymptotic limit,
following the semi-classical curve closely for $kR\gg 1$ but with a small
systematic deficit. As $\alpha$ approaches $\pi/2$ from below the oscillations
at lower $kR$ become more complicated, but settle down to sinusoidal
oscillation around $4\sin\alpha$ at higher $kR$. For $\alpha>\pi/2$ sharp
peaks appear at all $kR=\sqrt{l(l+1)}$ for whole $l$ (vertical grid lines in
the right plot). As $\alpha$ approaches $\pi$ and $kR$ increases, these peaks
become narrower.

A further quantum effect which is not obvious in any individual plot of
$|f(\theta)|^{2}$ shows up when we plot the total cross-section $\sigma$ as a
function of $kR$, as in Fig. 7. For super-hemispherical cases $\alpha>\pi/2$
the cross section shows sharp resonance peaks whenever $kR=\sqrt{l(l+1)}$ for
whole number $l$; these precise values correspond to the energy levels
$\hbar^{2}l(l+1)/(2MR^{2})$ of a particle on an isolated full sphere of radius
$R$, of which the corresponding eigenfunctions are the spherical harmonics
$Y_{lm}(\rho,\theta)$. At higher $l$ and for $\alpha$ closer to $\pi$ the
resonances become narrower and narrower, but their location does not change
and they are present even for $\alpha$ only slightly greater than $\pi/2$. For
sub-hemispherical cases $\alpha<\pi/2$, however, these peaks are completely
absent. A clue to what causes these sharp peaks can be seen when we examine
the partial wave scattering probabilities $\sin^{2}(\delta_{m})$, as shown in
Fig. 8. For larger $kR$ these probabilities generally fall abruptly to
(essentially) zero for $|m|>kR\sin\alpha$; this is because for $|m|$ beyond
this limit the centrifugal barrier prevents particles with energy
$\hbar^{2}k^{2}/(2M)$ from even reaching the contact circle at
$r=R\sin\alpha$. When $k$ is close to $\sqrt{l(l+1)}$ for whole-number $l$,
however, finite scattering probability extends somewhat past the
$kR\sin\alpha$ limit, in an approximately half-Gaussian profile. The total
cross section is simply the sum of all the partial wave probabilities, and so
these additional non-negligible $\sin^{2}\delta_{m}$ contributions from
$|m|>kR\sin\alpha$ are responsible for the sharp peaks in the cross section as
a function of $kR$. As we will see in the next section, these contributions to
scattering from angular momenta $kR\sin\alpha<|\hbar m|<kR$ are due to
tunnelling into resonances related to the closed classical orbits within the
sphere, when it overhangs the plane for $\alpha>\pi/2$.

Figure 8: Partial wave scattering probabilities $\sin^{2}\delta_{m}$ (vertical
axes) as functions of angular quantum number $m$ (horizontal axes), for a
nearly-full sphere ($\alpha=7\pi/8$), at $kR=100$ (top), $kR=101$ (bottom),
and $kR=\sqrt{100\cdot 101}$ (middle). Vertical lines are at $m=kR\sin\alpha$.
Similar plots for almost all $kR$ resemble the top and bottom panels
qualitatively, with complicated jumbles of points that all fall to zero for
$|m|>kR\sin\alpha$. Whenever $kR$ is close to $\sqrt{l(l+1)}$ for any whole
number $l$, however, a few non-zero points extend past the usual
$kR\sin\alpha$ limit. These anomalous points can in general be fitted quite
well to a Gaussian. Similar patterns are seen for all $\alpha>\pi/2$.

All of these features can be understood analytically by using the Wentzel-
Kramers-Brillouin-Jeffreys semi-classical approximation to the quantum
problem, at least qualitatively. Close quantitative agreement between the
exact quantum treatment and the semi-classical approximation may require quite
high $kR$, since the proportional errors in the semi-classical results are of
order $(kR\sin\alpha)^{-1/2}$ in general, and (particularly for small
$\alpha$) there can be unfortunately large $\alpha$-dependent pre-factors. At
large enough $kR\sin\alpha$ to make these errors small, computing the exact
differential cross section becomes numerically challenging; it also becomes a
visual challenge to to compare polar plots with very many radial fringes. In
the next section we will therefore develop the semi-classical theory which
becomes exact in the limit $kR\to\infty$, and show with numerical plots that
for modest $kR\lesssim 100$ the agreement with the exact differential cross
sections is already good enough to confirm the large-$kR$ trend.

##  IV The Semi-Classical Problem

For the conventional Schrödinger problem of motion in a potential, the semi-
classical regime consists of cases in which the quantum wavelength is short
compared to the scale on which the potential varies spatially. In this regime
we apply the Wentzel-Kramers-Brillouin-Jeffreys (WKBJ) approximation to solve
the time-independent Schrödinger equation for the quantum wave function using
the method of characteristics. The wave function propagates simply along the
characteristic curves in space, and these curves are a set of classical
trajectories. Because of the need to find these classical trajectories as the
first step towards solving the quantum problem, in general the WKBJ method is
significantly more complicated in two or more dimensions than it is in the
one-dimensional case that is more commonly presented in textbooks. Since our
problem has exact rotational symmetry, however, we can reduce it to a one-
dimensional radial problem before applying the WKBJ approximation. We will
then see that for the sphere on a plane the WKBJ method provides a good
approximation to exact quantum scattering whenever the wavelength is short
compared to the contact radius $R\sin\alpha$ between the sphere and the plane.

###  IV.1 The WKBJ approximation

The WKBJ approximation requires, first of all, that $kR$ be large. One then
expands

$\psi_{m}(\rho)=\exp\left[i\sum_{n=0}^{\infty}(kR)^{1-n}\int_{\rho_{0}}^{\rho}\\!d\rho^{\prime}\,K_{mn}(\rho^{\prime})\right]$
(44)

and solves the dimensionless radial Schrödinger equation (35) order-by-order
in $(kR)^{-1}$. The result is

$\displaystyle\psi_{m}(\rho)=$
$\displaystyle\left(g(\rho)-\frac{m^{2}}{(kR)^{2}}\right)^{-\frac{1}{4}}\sum_{\pm}A_{\pm}e^{\pm
ikR\int_{\rho_{0}}^{\rho}\\!d\rho^{\prime}\,K_{m0}(\rho^{\prime})}$
$\displaystyle+\mathcal{O}(kR)^{-1}$ (45)

for

$K_{m0}(\rho)=\sqrt{1-\frac{m^{2}}{(kR)^{2}g(\rho)}}\;,$ (46)

where as we recall from (4), $g(\rho)=\sin^{2}\rho$ for $\rho<\alpha$ (on the
sphere), while for $\rho>\alpha$ we instead have
$g(\rho)=(\rho-\alpha+\sin\alpha)^{2}$, which corresponds to $1/r^{2}$ for the
usual radial coordinate $r=\rho-\alpha+\sin\alpha$ in the plane.

In (IV.1) we have taken the lower limit of $\rho$ integration in the phase to
be the turning point $\rho_{0}(kR/m)$ such that $g(\rho_{0})=m^{2}/(kR)^{2}$.
Choosing any other lower limit would simply shift the coefficients $A_{\pm}$,
but choosing the turning point is convenient because it is at this point that
the fourth-root prefactor in (IV.1) diverges and the WKBJ approximation breaks
down. As usual we can interpolate through this breakdown region by
approximating the radial Schrödinger equation there as an Airy equation.

If we assume that for $\rho<\rho_{0}$, where (IV.1) is again valid, the wave
function $\psi_{m}(\rho)$ consists only of the $\pm$ branch in (IV.1) that
decays as $\rho$ decreases (evanescent wave in the classically forbidden
region), then we obtain the condition $A_{-}=iA_{+}$. This assumption about
the behavior of $\psi_{m}$ for $\rho<\rho_{0}$ will turn out to be valid
except for a few special cases (the tunnelling resonances) that we will
mention below. This means that for large $\rho\to\infty$ we will have

$\displaystyle\lim_{\rho\to\infty}\psi_{m}(\rho)$
$\displaystyle=2\frac{A_{+}e^{i\pi/4}}{\sqrt{r}}\mathrm{Re}\left(e^{-i\pi/4}e^{ikR\int_{\rho_{0}}^{\rho}\\!d\rho^{\prime}\,K_{m0}(\rho^{\prime})}\right)$
$\displaystyle\equiv$ $\displaystyle
2\frac{A_{+}e^{i\pi/4}}{\sqrt{r}}\mathrm{Re}\left(e^{i(\delta_{m}+\frac{\pi}{4})}e^{ik\int_{\frac{|m|}{k}}^{r}\\!dr^{\prime}\,\sqrt{1-\frac{m^{2}}{k^{2}r^{2}}}}\right)$
(47)

for

$\delta_{m}=kR\int_{\rho_{0}}^{\alpha}\\!d\rho\,\sqrt{1-\frac{m^{2}}{(kR)^{2}g(\rho)}}-k\int_{\frac{|m|}{k}}^{r}\\!dr^{\prime}\,\sqrt{1-\frac{m^{2}}{k^{2}r^{2}}}\;.$
(48)

Since

$\lim_{r\to\infty}H_{m}(kr)\propto\frac{1}{\sqrt{r}}e^{-i\pi/4}e^{ik\int_{\frac{|m|}{k}}^{r}\\!dr^{\prime}\,\sqrt{1-\frac{m^{2}}{k^{2}r^{2}}}}\;,$
(49)

we can recognize by comparison with (III.2) above that $\delta_{m}$ as given
by (48) is indeed the semiclassical result for the scattering phase shift. The
integrals in (48) can be performed explicitly, leading to our semiclassical
expression for the scattering amplitude
$f(\theta)=\sum_{m}e^{i\delta_{m}}e^{im\theta}\sin\delta_{m}$.

For $|m|>kR\sin\alpha$, (48) implies immediately that $\delta_{m}=0$. We can
see this by considering $\rho_{0}$ as a function of $m/(kR)$, as shown in Fig.
9. What Fig. 9 shows is that for $|m|>kR\sin\alpha$ we have $\rho_{0}>\alpha$,
so that the turning point in $\rho$ lies in the plane, fully outside the
spherical extrusion; the centrifugal barrier turns the particle away before it
reaches the contact circle at $r=R\sin\alpha$. For $|m|<kR\sin\alpha$, in
contrast, the turning point $\rho_{0}<\alpha$ is on the spherical extrusion,
where $g(\rho)$ represents the curved spherical surface. For
$|m|>kR\sin\alpha$ and hence $\rho_{0}>\alpha$, therefore, we have
$g(\rho)=(\rho-\alpha+\sin\alpha)^{2}=r^{2}/R^{2}$ and the two integrals in
(48) cancel identically, leaving $\delta_{m}=0$ according to (48).

This conclusion conflicts, however, with the $\delta_{m}\not=0$ for
$|m|>kR\sin\alpha$ shown in the middle panel of Fig. 8, even though the
$kR\doteq 100$ of Fig. 8 should be large enough for semiclassical methods to
work. To explain this discrepancy, we must now note the exceptions to our
assumption of purely decaying $\psi_{m}(\rho)$ for $\rho<\rho_{0}$. As Fig. 9
shows, for $\alpha>\pi/2$ and $kR\sin\alpha<|m|<kR$ there are two additional
turning points inside the outermost turning point $\rho_{0}$. For large $kR$
the tunnelling transmission amplitude through the centrifugal barrier peak at
$\rho=\alpha$ is still exponentially small, and so for most $k$ these
additional turning points will have negligible effect on scattering. For
$(kR)^{2}=l(l+1)$ for integer $l$, however, or very close, tunnelling through
the centrifugal peak at $\rho=\alpha$ _does_ significantly change the phase
relationship between $A_{\pm}$ in the $\rho>\rho_{0}$ region. This is clearly
what is responsible for the anomalous $\delta_{m}\not=0$ for
$|m|>kR\sin\alpha$ that were noted, for these special values of $k$ and only
for $\alpha>\pi/2$, in Fig. 8 above. We can therefore identify the sharp peaks
in the cross section at these same values of $k$, which were noted in Fig. 7
(again only for $\alpha>\pi/2$), as due to tunnelling through the centrifugal
barrier peak into quasi-bound resonances with angular momentum
$|m|>kR\sin\alpha$. These resonances correspond classically to closed great
circle orbits on the spherical extrusion when it overhangs the plane for
$\alpha>\pi/2$.

Figure 9: The function $\lambda^{2}/g(\rho)$ for a sub-hemispherical and a
super-hemispherical case (left panel, $\alpha=0.45\pi$; right panel,
$\alpha=0.75\pi$), and for selected examples of $\lambda=m/(kR)$ (namely: 0.8
and 1.2 in the left panel; 0.5, 0.8, and 1.1 in the right panel). The
intersection $\lambda^{2}/g(\rho)=1$ occurs at the turning point
$\rho=\rho_{0}$. The vertical grid line marks $\rho=\alpha$ in both cases,
where the form of $g(\rho)$ changes; for $\alpha>\pi/2$ the change in
$g(\rho)$ gives the centrifugal barrier a maximum at $\rho=\alpha$, with an
inner well at smaller $\rho$. In the sub-hemispherical left panel we see that
for $|m|>kR\sin\alpha$ the turning point is outside the junction circle,
$\rho_{0}>\alpha$, while for $|m|<kR\sin\alpha$ we instead have
$\rho_{0}<\alpha$. The same is true for the outermost turning points in the
super-hemispherical right panel, but within the range $kR\sin\alpha<|m|<kR$
there are classical orbits at energy $\hbar^{2}k^{2}/(2M)$ and angular
momentum $\hbar m$ that remain inside the sphere ($\rho<\alpha$). Within this
$m$ range the quantum particle can potentially tunnel through the centrifugal
barrier, modifying the scattering behavior.

Although we are confident in this tunnelling explanation of the sharp
scattering resonances at $k=\sqrt{l(l+1)}$ for $\alpha>\pi/2$, we will now
ignore these sharp resonances and leave detailed study of these cases of
tunnelling to future work. The project may be of some interest as an exercise
in semiclassical technique, though for large $kR$ the resonances would likely
be too narrow to observe experimentally, even if the rest of our model could
be realized. We assume from now on that (48) can be applied for all $kR$.

###  IV.2 The forward scattering spike

This means that we take (48) as correctly implying that $\delta_{m}=0$ for
$|m|>kR\sin\alpha$. This lets us simplify the semiclassical scattering
amplitude $f_{SC}(\theta)$ by truncating the infinite sum over $m$ into a sum
from $-\lfloor kR\sin\alpha\rfloor$ to $\lfloor kR\sin\alpha\rfloor$, where
$\lfloor kR\sin\alpha\rfloor$ is the largest whole number not greater than
$kR\sin\alpha$:

$\displaystyle f_{SC}(\theta)$ $\displaystyle=\sum_{m=-\lfloor
kR\sin\alpha\rfloor}^{\lfloor
kR\sin\alpha\rfloor}e^{im\theta}e^{i\delta_{m}}\sin\delta_{m}$
$\displaystyle=\frac{i}{2}\sum_{m=-\lfloor kR\sin\alpha\rfloor}^{\lfloor
kR\sin\alpha\rfloor}e^{im\theta}\left(1-e^{2i\delta_{m}}\right)$
$\displaystyle\equiv\frac{i}{2}\frac{\sin\Big{(}\big{(}\lfloor
kR\sin\alpha\rfloor+\frac{1}{2}\big{)}\theta\Big{)}}{\sin\frac{\theta}{2}}+\tilde{f}(\theta)$
$\displaystyle\tilde{f}(\theta)$ $\displaystyle:=-\frac{i}{2}\sum_{m=-\lfloor
kR\sin\alpha\rfloor}^{\lfloor
kR\sin\alpha\rfloor}e^{i(2\delta_{m}+m\theta)}\;.$ (50)

For large $kR\sin\alpha$, the first term in the third line of (IV.2) is a
sharp peak around $\theta=0$: it is responsible for the large forward spikes
of scattering probability that we found in our exact quantum calculations,
above. Because of the general form of partial wave scattering amplitudes
$e^{i\delta_{m}}\sin\delta_{m}=(i/2)(1-e^{2i\delta_{m}})$, which is itself
required by unitarity, this forward-spike term only depends on one feature of
$\delta_{m}$—namely, that it vanishes for $|m|>kR\sin\alpha$. This means, as
we mentioned above, that the same forward-scattering spike will appear for any
scattering target for which $\delta_{m}$ has the same finite range of support
in $m$. In the semi-classical limit this simply means that the scattering
target has to have the same classical cross section $2R\sin\alpha$.

###  IV.3 Classical paths

The remaining $\tilde{f}(\theta)$ part of the scattering amplitude does depend
on the particular form of $\delta_{m}$, as given semi-classically by (48). The
integrals in (48) can be evaluated in closed form:

$\displaystyle\delta_{m}$
$\displaystyle=\frac{kR}{2}\Delta(\alpha,\frac{m}{kR})-\frac{m}{2}\Theta(\alpha,\frac{m}{kR})$
$\displaystyle\Delta(\alpha,\mu)$
$\displaystyle:=2\cos^{-1}\left(\frac{\cos\alpha}{\sqrt{1-\mu^{2}}}\right)-2\sqrt{\sin^{2}\alpha-\mu^{2}}\;,$
(51)

and where $\Theta(\alpha,\mu)$ from (16) is the classical scattering angle, as
a function of sphere contact angle $\alpha$ and scaled angular momentum $\mu$.
The path length difference $R\Delta(\alpha,\mu)$ is likewise the length of the
portion of the classical path for $\alpha$ and $\mu$ which is shown thick in
Fig. 5, (i.e. the length of the classical path between the tangents to the
contact circle perpendicular to the incident and scattered directions), minus
$2R\sin\alpha$ (which would be the distance between those two tangents, if
there were no spherical extrusion and the particle just continued straight).
We have stated (IV.3) simply as the result of performing the integrals in
(48), which it is, but of course this is not a mere coincidence of
integration. It can be derived from the classical mechanics for our system’s
Lagrangian, and in appropriately generalized form it holds for any
rotationally symmetric scattering target.

Eqn. (IV.2) gives $\tilde{f}(\theta)$ as a sum of
$\exp{\big{(}ikR\,S\big{(}m/(kR))\big{)}}$ over $m$, for

$\displaystyle S(\mu)$ $\displaystyle=2\frac{\delta_{\mu kR}}{kR}+\mu\theta$
$\displaystyle=2\cos^{-1}\left(\frac{\cos\alpha}{\sqrt{1-\mu^{2}}}\right)$
$\displaystyle\quad-2\sqrt{\sin^{2}\alpha-\mu^{2}}+\mu\big{(}\theta-\Theta(\alpha,\mu)\big{)}\;.$
(52)

In the semi-classical limit of large $kR\sin\alpha$ it is accurate to
approximate this sum as an integral _and_ also approximate the integral using
the method of stationary phase. This approximation is accurate even though,
for most of the terms in the sum, the change in the summand between one term
and the next is _not_ small. In fact the approximation is accurate precisely
because, for most terms in the sum, the summand rotates in phase between one
term and the next rapidly and almost randomly: this means that most of the sum
simply cancels. Around certain values of $m$, however, the phase of the
summands changes only slowly with $m$, allowing the sum to be treated as a
Riemann sum. Which values of $m$ it is, for which $kR\,S\big{(}m/(kR)\big{)}$
changes slowly, depends on $\theta$ and $\alpha$; the special values of $m$
are found by treating $\mu$ as if were a continuous variable, and finding the
points at which $S(\mu)$ is stationary, $dS/d\mu=0$.

As may again seem like a miraculous coincidence, differentiation of $S(\mu)$
with respect to $\mu$, including differentiating both $\Theta$ and $\Delta$,
yields exactly the result that one would obtain if one forgot to differentiate
$\Theta$ and $\Delta$:

$\displaystyle\frac{\partial}{\partial\mu}S(\mu)$
$\displaystyle=\theta-\Theta(\alpha,\mu)\quad\hbox{and so}$
$\displaystyle\Theta\left(\alpha,\frac{m}{kR}\right)$
$\displaystyle\stackrel{{\scriptstyle!}}{{=}}\theta\;\mathrm{mod}(2\pi)$ (53)

determines the saddlepoints $m_{\pm}$. This result is again no coincidence,
however, but can be derived from the classical dynamics for any rotationally
symmetric scattering target. Even the mod$(2\pi)$ here appears inevitably,
because in $\tilde{f}(\theta)$ we are evaluating a discrete sum of phases,
after all, and a phase which changes by nearly $2\pi$ between successive
summand terms is equivalent to a slowly changing phase. The result is that the
$\tilde{f}(\theta)$ sum (48) is dominated by $m$ around precisely the
classical angular momenta $J/\hbar$ which contribute to scattering into the
angle $\theta$, as given by (15). In particular we have two stationary points
$\mu\to m_{\pm}kR$ for $m_{\pm}(\alpha,\theta)$:

$m_{\pm}=-kb_{\pm}(\alpha,\theta)$ (54)

as given by (II.2.2) in Section II, above.

###  IV.4 Classical differential cross section

Let us now examine the higher derivatives of $S(\mu)$ with respect to $\mu$,
at $\mu\to m_{\pm}kR$. Since the classical impact parameters
$b_{\pm}(\alpha,\theta)$ as given in (II.2.2) are proportional to the sphere
radius $R$, but independent of momentum, we can further see that the Taylor
series of $S(mkR)$ around $m_{\mp}$ will be of the form

$S\big{(}(m_{\pm}+\Delta
m)kR\big{)}=S(m_{\pm}kR)+\sum_{n=2}^{\infty}\frac{1}{n!}X_{n}(\alpha,\theta)\left(\frac{\Delta
m}{kR}\right)^{n}$ (55)

for coefficients $X_{n}$ that do not depend on $kR$. This means that in the
limit $kR\to\infty$, $\tilde{f}(\theta)$ in (48) is indeed given by two
Riemann sums for integrals with respect to $\mu=m/(kR)$, each of which
describes the integral of a complex exponential, of which the phase has a
large prefactor. We can therefore indeed apply the method of stationary phase
to these integrals, approximating them as Gaussians, up to corrections that
are smaller by factors of $1/(kR)$.

In particular we note the second derivative of $S(\mu)$ at $\mu=b_{\pm}/R$,

$\displaystyle\frac{\partial^{2}}{\partial\mu^{2}}S(\mu)$
$\displaystyle\equiv\frac{\partial}{\partial\mu}\big{(}\theta-\Theta(\alpha,\mu)\big{)}$
$\displaystyle=R\frac{\partial\Theta}{\partial
b_{\pm}}\equiv\left(\frac{1}{R}\frac{\partial
b_{\pm}}{\partial\theta}\right)^{-1}\;.$ (56)

Our stationary phase approximation to $\tilde{f}(\theta)$ is therefore

$\displaystyle\tilde{f}(\theta)$
$\displaystyle\doteq-\frac{i}{2}\sum_{\pm}e^{ik\Delta(\alpha,\frac{m_{\pm}}{kR})}\sqrt{kR}\int_{-\infty}^{\infty}\\!d\xi\,e^{i\frac{R}{2\partial_{\theta}b_{\pm}}\xi^{2}}$
(57) $\displaystyle=-i\sqrt{\frac{\pi
k}{2}}\sum_{\pm}\sqrt{|\partial_{\theta}b_{\pm}|}\,e^{ikR\Delta\big{(}\alpha,-\frac{b_{\pm}(\alpha,\theta)}{R}\big{)}}e^{i\frac{\pi}{4}\mathrm{sgn}\,\partial_{\theta}b_{\pm}}\;.$

Finally inserting (57) into (IV.2), and using $f(\theta)\to
f_{\mathrm{SC}}(\theta)$ in the quantum differential cross section
$d\sigma/d\theta=2|f(\theta)|^{2}/(\pi k)$, yields the semi-classical
differential cross section as composed of three interfering terms:

$\frac{d\sigma}{d\theta}=\left|\frac{\sin\Big{(}\big{\lfloor}kR\sin\alpha\rfloor+\frac{1}{2}\big{)}\theta\Big{)}}{\sqrt{2\pi
k}\sin\frac{\theta}{2}}-\sum_{\pm}\sqrt{|\partial_{\theta}b_{\pm}|}\,e^{ikR\Delta(\alpha,-\frac{b_{\pm}}{R})}e^{i\frac{\pi}{4}\mathrm{sgn}\,\partial_{\theta}b_{\pm}}\right|^{2}\;.$
(58)

The incoherently adding “diagonal” terms in this semi-classical differential
cross section consist of a sharp spike which approaches
$2R\sin\alpha\delta(\theta)$ for $kR\to\infty$, plus exactly the classical
differential cross section from Section II. The qualitatively non-classical
interference terms, however, do not vanish for large $kR$; they simply
oscillate more and more rapidly with $\theta$, providing many fine fringes.
For large $kR$, the two terms with $\sqrt{|b_{\pm}|}$ factors dominate the
differential cross section everywhere except at $\theta=0$, and these can be
recognized as a kind of double-slit interference pattern, due to the two
classical paths which exist for every scattering angle $\theta$ over the
surface of the sphere on a plane.

###  IV.5 Comparison of exact and semi-classical differential cross sections

Figure 10 compares the exact and sem-classical differential cross sections for
two cases, both with $kR=40$, one with a (slightly) less than hemispherical
extrusion ($\alpha=0.49\pi$), and one with an almost full sphere
($\alpha=0.85\pi$). It can be seen that the semi-classical approximation
represents the angular positioning of the interference fringes quite well,
even for this modestly large $kR$, though it does not always fully capture
their amplitudes.

It can further be noted that the semi-classical approximation works better for
the super-hemispherical case than for the sub-hemispherical one. There are two
reasons for this. First of all the simple semi-classical method breaks down at
the caustic at $\theta=\theta_{c}$, where the two $b_{\pm}$, and therefore the
two $m_{\pm}$, coincide. Approximating the sum over $m$ in (IV.2) as two
separate Gaussian integrals is no longer a good approximation as we approach
$\theta_{c}$. And for $|\theta|>\theta_{c}$ there are no stationary points
$m_{\pm}$, because there are no classical trajectories, so the simple semi-
classical method does not work at all. Improved semi-classical methods exist
to resolve such problemsBerry1 , but like the tunnelling resonances above, we
leave them for future work.

Secondly, however, the semi-classical approximation works less well for
$\alpha<\pi/2$ even for $|\theta|<\theta_{c}$, because although the extrema of
$S(\mu)$ at $m_{\pm}kR$ are both nice maxima for all $\theta$, for
$\alpha<\pi/2$ the extremum at $\mu=-b_{+}/R$ is, for most
$|\theta|<\theta_{c}$, only a shallow dip within a steep slope. For
sufficiently large $kR$ this still implies a deep minimum in the phase
$kR\,S(\mu)$, but $kR$ has to be considerably larger to make the stationary
phase approximation good around $m_{+}$ for $\alpha<\pi/2$ than it does for
$\alpha>\pi/2$. Quantum corrections may therefore be needed for accurate
differential cross sections at moderate $kR$ for sub-hemispherical extrusions.

Figure 10: Exact (solid) and semi-classical (dashed) differential cross
sections (divided by $R$) for $kR=40$ and $\alpha=0.49\pi$ (upper plot),
$\alpha=0.85\pi$ (lower plot). The classical differential cross sections are
also shown (dotted) for comparison. The forward scattering spikes, suppressed
to show detail for other angles, extend in the exact differential cross
sections to 25.1 ($\alpha=0.49\pi)$ and to 3.66 ($\alpha=0.85\pi$), while the
corresponding semi-classical values of $|f(0)|^{2}$ are 26.6 and 4.31. The
semi-classical approximation already captures the angular spacing of the
radial fringes well even at $kR\lesssim 20$, but higher $kR$ is required for
the lengths of all the radial fringes to converge on the exact values,
especially for $\alpha<\pi/2$. In particular the semi-classical approximation
breaks down for $\alpha<\pi/2$ near and beyond the caustic
$|\theta|=\theta_{c}$.

###  IV.6 Exact and semi-classical total cross sections

Using the optical theorem $\sigma(k)=(4/k)\mathrm{Im}f(0)$, we can conclude
from (IV.2) and (57), with our results from Section 2 that
$b^{\prime}_{+}(\theta=0)=0$ and $b^{\prime}_{-}(\theta=0)<0$, that in the
semi-classical limit $kR\sin\alpha\gg 1$ we have

$\displaystyle\sigma$ $\displaystyle\to
R\Big{(}\tilde{\sigma}_{\mathrm{SC}}(\alpha,kR)+\mathcal{O}(kR)^{-1}\Big{)}$
(59) $\displaystyle\tilde{\sigma}_{\mathrm{SC}}$
$\displaystyle=4\sin\alpha-\frac{2\sqrt{\pi\cot\frac{\alpha}{2}}}{\sqrt{kR}}\cos\big{[}2kR(\alpha-\sin\alpha)-\frac{\pi}{4}\big{]}\;.$

The dotted curves in Fig. 7 are this $\tilde{\sigma}_{\mathrm{SC}}$. The
agreement with the full quantum cross section is indeed good for
$kR\sin\alpha\gg 1$, except for the tunnelling resonance spikes for
$\alpha>\pi/2$ that we have already discussed, and for a small but systematic
discrepancy for $\alpha<\pi/2$, which we presume is due to the semi-classical
failures at the classical caustics $\theta=\pm\theta_{c}$.

Apart from the problems that we have postponed to future work, concerning
caustics for $\alpha<\pi/2$ and tunnelling resonances for $\alpha>\pi/2$, the
semi-classical method clearly describes quantum scattering from the sphere on
a plane quite well qualitatively for moderately large $kR$, with quantitative
convergence as $kR\to 0$. The striking non-classical features of quantum
scattering are well explained semi-classically by the Poisson-spot-like
forward spike and interference between the two classical trajectories that
exist for each scattering angle $\theta$.

##  V Conclusions

Our paper has been a technical exercise in theoretical physics, addressing a
problem which is not only idealized but frankly exotic. We can hardly say that
we have answered an urgent question; we must rather admit that our question is
mainly interesting because of how thoroughly it can be answered. The same
admission must be made for many solvable model problems, which nevertheless
have their place in theoretical physics. Geodesic motion in curved spaces
really is physically important, in the four-dimensional context of General
Relativity; our two-dimensional example of piecewise uniform curvature may at
least be of pedagogical value in providing some intuition about how curvature
can affect motion.

More complicated two-dimensional cases than ours may also be of interest for
understanding subtleties in the relationships between quantum and classical
mechanics, possibly including subtleties of quantum chaos, because of the
convenient billiard-like feature that the classical trajectories do not depend
on energy, and so the entirety of classical dynamics can be visualized in two
dimensions without loss of information, even though the phase space is
actually four-dimensional. In our simple and symmetrical case, quantum-
classical correspondence has turned out to be relatively straightforward, but
interference and caustics and tunnelling have all appeared, even here. Further
models of free particles on curved surfaces may be useful for theoretical
studies of the quantum-classical frontier.

Experimental or even practical realizations of that general problem may not
even be so far-fetched, moreover. Curved surfaces can be realized in graphene,
for instance, and in such cases curvature will certainly influence band
structure. The $V(x,y)$ potential effects which have assumed to be absent will
surely be important in any such real two-dimensional problems; our results may
nevertheless indicate an important feature of the interaction between
curvature and potential energy, even though we have neglected the latter. We
have observed that curvature alone cannot induce bound states, but as the
right panel of our Figure 9 shows, curvature can have dramatic effects on
_effective_ potentials: it can significantly distort centrifugal barriers. If
there is a potential well within a region of negative curvature, like our
spherical extrusion, the softening of the centrifugal barrier due to curvature
may allow more bound states to exist than one would expect without accounting
for curvature. And we can conjecture, conversely, that if regions of positive
curvature could somehow be realized, potential wells in these regions might
possess fewer bound states.

JRA thanks Ali Mostafazadeh for drawing the nicely complementary paper OMA18
to his attention.

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