Start here
A packet of matter waves (the striped band; the colours are the phase of the wave
function) is fired from the left at a target , the faint white disk, and scatters off
it. The dashed circle is the detector : the three panels on the right are measured
there as the wave crosses it: the wave sorted by momentum, how much goes to each angle
(the cross section) and the share of each partial wave. Units: ħ = m = 1.
Pick another Target ; switch Show to Scattered only to see what the
target adds to the wave; raise k₀ for a faster packet with shorter waves, or the
width for a beam more like a plane wave; point at the momentum picture to see the
momentum the target gives. Every change restarts the collision; done appears under
the picture when the packet has gone.
Look for the circular wave spreading from the target and the shadow behind it, the
measured curve (blue) settling onto the exact one (green) once the packet has gone, far
from what classical particles would do (violet), and the totals in the table agreeing. The experiments take the ideas one at a time: what
scatters and how it is counted, the Born approximation, partial waves, interference and
Bragg reflection.
1. A packet meets a disk
The packet comes in from the left, moving along $x$, nearly a plane wave across the
disk. Watch the reflected crescent and the circular wave spreading from the disk. Switch
Show to Scattered only : that is $\psi-\psi_\text{inc}$, the difference the
disk makes, computed exactly because the free packet's motion is known in closed form.
How the scattering is counted: the scattered wave's outward flux through the detector
circle, added up over the collision and divided by the beam's incident flux per unit width,
is the differential cross section $d\sigma/d\theta$ (the blue curve): how much of
the beam is sent to each angle $\theta$, counted from the forward direction. Its total
over all angles, $\sigma$ (the table's first line), is a length in two dimensions: the
width of beam the disk takes out. Here $\sigma=8.3$, well over the 3.55 that classical
particles would give (the violet dashes and the table's classical line: they
bounce off where $V=E$, at $r=1.77$, and a disk that wide blocks $2r$ of the beam). More
than half of $\sigma$ (4.4) is the forward peak within $\pm40^\circ$, which classical particles
haven't got: the wave diffracted round the edge, which is what makes the shadow
(experiment 2). Once the packet has gone, the table shows $\sigma$ four ways, and
the blue curve lies on the exact one (green).
2. The shadow is the optical theorem
Behind the disk, a shadow. In the Scattered only view the shadow is a wave
scattered straight forward, opposite in phase to the incident one, so that the two cancel
there. The interference between them carries probability away from the forward
direction: the table's shadow line is that loss, and it equals the total
scattered, line for line (8.30 and 8.30). That is the optical theorem : the
amplitude scattered straight forward, $f(0)$, fixes how much is scattered in all
directions, $\sigma=-\sqrt{8\pi/k}\,\mathrm{Re}\,[e^{i\pi/4}f(0)]$ in two dimensions (the
counterpart of $\sigma=\frac{4\pi}{k}\,\mathrm{Im}\,f(0)$ in three). The table's
optical theorem line is $\sigma$ computed from $f(0)$ alone, out of the measured
partial waves (experiment 6).
3. Twice the classical cross section
A bigger and harder disk ($a=2.5$, a wall $V_0=15$ high and $w=0.15$ soft) and a
shorter wave ($k_0=3$, $\lambda=2.1$), so that the disk is eight wavelengths round
($kr\approx8$). The violet dashes are what classical particles would do. One with
impact parameter $b$ meets the wall at the angle $\alpha$ to its normal, with
$\sin\alpha=b/r$, and bounces off through $\theta=\pi-2\alpha$; so $b=r\cos(\theta/2)$, and
$$\frac{d\sigma}{d\theta}=\Big|\frac{db}{d\theta}\Big|=\frac r2\Big|\sin\frac\theta2\Big|,\qquad
\sigma=\int_{-\pi}^{\pi}\frac{d\sigma}{d\theta}\,d\theta=2r,$$
the width the disk blocks: 5.25 here, with $r=2.63$ where $V=E$. Away from forward the
wave does what the particles do: beyond $90^\circ$ the measured and the classical curves
hold 3.71 and 3.74. Forward it does not: within $\pm30^\circ$ there is a peak of 5.7,
as much as the disk sends everywhere else, where classical particles go hardly at all
(0.18).
It is the shadow of experiment 2: behind the disk the scattered wave must cancel the
incident one, so it carries forward as much as the disk takes out of the beam. So
$\sigma=11.4$, twice the classical $2r$ and a little more: for a disk many wavelengths
wide it tends to $4r$ (the extinction paradox ), slowly (2.3 times for the disk of
experiment 1, 2.2 here).
4. The Born approximation
A smooth bump, $V_0=1$, half the packet's energy $E=2$: about a sixth of the packet
(18%) is scattered, nearly all of it (96%) within $60^\circ$ of forward. To first order in
$V$ (the first Born approximation ) the scattered amplitude is the Fourier transform
of the potential at the momentum transfer $\hbar\vec q=\hbar(\vec k_\text{out}-\vec k_\text{in})$,
the momentum the particle gains in turning through $\theta$, of size $q=2k\sin(\theta/2)$:
$$\frac{d\sigma}{d\theta}=\frac{m^2}{2\pi\hbar^4k}\,|\tilde V(q)|^2,\qquad
\tilde V(\vec q)=\int V(\vec r)\,e^{-i\vec q\cdot\vec r}\,d^2r$$
(dashed grey). The measured curve follows it over the forward peak, and Born's total is
6% above the exact one (the table: 2.83 against 2.67).
Born is good when the wave crossing the target is hardly changed by it: the extra phase
it picks up on the way through, roughly $V_0a/\hbar v=mV_0a/\hbar^2k$, must be small. Here
it is 0.5. Lower $V_0$ to 0.3 (0.15) and Born and the exact answer agree to 1%, but then
less than 2% of the packet scatters: that is what weak means. Raise $V_0$ to 2 (1.0): the
cross section grows less than $V_0^2$ (2.6 times, not 4) and Born is 65% too high. Make
the bump attractive ($V_0=-1$): Born cannot tell the sign, $|\tilde V|^2$ is the same; the
exact answer can (2.48 instead of 2.67). For the hard disk of experiment 1 the phase is 6,
and Born's total, 294, is 35 times the real one: a target that stops the wave is as far
from weak as can be.
The momentum picture shows it at a glance (experiment 5): outside the circle Born's
$d\sigma/d\theta$ (grey dashes) and the measured one (blue) are drawn on one scale. For
this bump they lie together over the forward lobe; at $V_0=2$ Born's lobe is twice the
other (12.6 against 5.53 straight ahead). Pick the Hard disk : Born's lobe is
fifty times the measured one (303 against 5.66), which hardly leaves the circle, and
straight back, where the ring is plainly lit, Born has almost nothing (0.14
at $180^\circ$ against 0.73).
5. Scattering puts the wave on a circle
The same bump, in the Scattered only view, so that momentum space shows
$\psi-\psi_\text{inc}$ alone. The incident packet is a spot at $(k_0,0)$; the scattering
is elastic, the energy and so $|\vec k|$ are unchanged, and it can only turn $\vec k$, not
shorten it: the scattered wave fills in the circle $|\vec k|=k_0$, brightest where
$|f(\theta)|^2$ is largest. The Born approximation says the brightness at $\vec k$ is
$|\tilde V(\vec k-\vec k_0)|^2$: the potential's Fourier transform, centred on the incident
spot and read round the circle. That is the grey dashed curve outside the circle, Born's
$d\sigma/d\theta$ toward each $\vec k$, its distance from the circle in proportion; the blue
one, on the same scale, is the measured $d\sigma/d\theta$, growing as the scattered wave
crosses the detector.
Point at the circle to move $\vec k$ round it. The pink arrow from the incident spot to
$\vec k$ is the momentum transfer $\vec q=\vec k-\vec k_0$, of length $2k_0\sin(\theta/2)$
(the readout under the picture: $q=k_0=2$ at $60^\circ$), and Born's curve there is the
transform at that $\vec q$, for this bump $\tilde V(q)=2\pi a^2V_0\,e^{-q^2a^2/2}$. Once the
packet has gone, the two agree over the forward lobe (at $30^\circ$, 1.07 measured and
1.08 Born); further round, where $q$ is larger than $1/a$, Born's falls away faster than
the measured (at $60^\circ$, 0.058 against 0.171): $\tilde V$ of a Gaussian falls as a
Gaussian, and the second-order terms that Born leaves out fall more slowly.
6. Partial waves: slow particles scatter in all directions
A small hard disk and a slow packet, $ka\approx1$ (here 0.8), in a bigger box, since
slow waves are long. A partial wave is the part of the wave with angular momentum
$m\hbar$ about the target, $e^{im\theta}$ times a radial wave. A round target scatters
each one into itself, only shifting the phase of its outgoing part by $2\delta_m$:
$\delta_m$ is the phase shift of wave $m$, and $S_m=e^{2i\delta_m}$, the factor the
outgoing wave acquires, its S-matrix element. The bars show $\sin^2\delta_m$ for
each $m$ (the green ticks: the exact values), and
$$\sigma=\frac4k\sum_m\sin^2\delta_m,$$
so the bars are each wave's share of $\sigma$. Here the $m=0$ wave does most of the
scattering ($\sin^2\delta_0=0.998$; $m=\pm1$, at 0.16 each, add a quarter; $m=\pm2$
nothing), and the cross section is spread over every direction, forward only three times
backward. Classically a particle with angular momentum $m\hbar=\hbar kb$ passes at
distance $b=m/k$, so waves with $m$ beyond $ka$ miss the disk. Speed the packet up
($k_0=2$, $ka=2$): more partial waves join in, roughly up to $m\approx ka$, with $m=\pm1$
now the strongest ($\sin^2\delta_1=0.96$), and the scattering turns forward, seven times
the backward. The dots show $|S_m|=|e^{2i\delta_m}|$, which must be 1, as what goes in must
come out (probability conserved: unitarity ), as a check on the measurement.
7. A resonance: the packet lingers
An attractive well, with a quasi-bound state near the packet's energy: a state that
would be bound if the well were a little deeper, and that lives for a while behind the
centrifugal barrier of angular momentum $3\hbar$. Part of the wave is caught, circles
inside the well, and leaks out long after the rest has gone, in all directions. The
$m=\pm3$ phase shift runs up steeply by nearly two radians as the energy crosses the
resonance, through $\pi/2$ at $k\approx1.0$ ($E\approx0.5$), just below the packet's
$k_0=1.2$ and within its spread of energies: there $\sin^2\delta_3=1$, the unitarity
limit , as much as one partial wave can scatter, $4/k$. At $k_0$ its bar still reads
0.85. The run goes on a little longer than usual here, until the well has emptied:
stopped before then, the dots $|S_{\pm3}|$ fall short of 1 by what is still inside.
8. A screened Coulomb potential
Charge scattering: strongly peaked forward and falling off steeply with angle, as
Rutherford found (his $1/\sin^4(\theta/2)$ law is for three dimensions). Most of the
scattering is by the long, weak tail of the potential, which deflects the wave only a
little: that is the forward peak, hundreds of times the backward scattering. The screening
(length $\ell$, the distance over which the charge is cancelled by its surroundings) keeps
the total cross section finite.
One thing to notice: straight forward, the measured curve (blue) falls well short of the
partial-wave and exact curves (orange and green, 28), though the totals agree. The peak is
made by the tail of the potential, some 20 across, and a wave from so wide a source takes a
long way, $r\sim kD^2/2\pi$, hundreds here, to settle into its far-field form: on the
detector circle at $R=18$ it has not, and is still broader and lower, with the same total.
The partial waves are read off with the exact Hankel functions at $R$ and do not mind.
Shorten $\ell$ to 1 and the source is small enough for the peak to form inside the
circle.
9. Two scatterers interfere
Two disks, a distance $d$ apart across the beam. The waves scattered by each
interfere, and $d\sigma/d\theta$ has fringes spaced by about $\lambda/d$ in angle
($36^\circ$ here: peaks at $0$, $\pm37^\circ$, $\pm83^\circ$, $\pm141^\circ$ and
$180^\circ$): the double slit, done with obstacles instead of holes. The target is not
round, so no single partial wave scatters into itself and the panel below shows no phase
shifts.
10. The double slit
A wall with two openings, $a=1.2$ wide and $d=5$ apart. Behind it, two circular waves
overlap and make fringes. There is no incident wave behind a wall to subtract, so the graph
shows the flux of $\psi$ itself through the far half of the detector, from $-90^\circ$ to
$90^\circ$: every direction there is, since $\sin\theta=k_y/k$ runs from $-1$ to $1$. The
dashed curve is Fraunhofer's , the far-field pattern of plane waves leaving the
openings,
$$\frac{d\sigma}{d\theta}\propto\cos^2\theta\,\cos^2\Big(\frac{kd\sin\theta}{2}\Big)\Big(\frac{\sin(ka\sin\theta/2)}{ka\sin\theta/2}\Big)^2,$$
averaged over the packet's spread of energies: dark fringes where $d\sin\theta=(n+\frac12)\lambda$,
here $\pm18^\circ$ (measured at $\pm17^\circ$; the next, at $\pm70^\circ$, is washed out by
the packet's spread of wavelengths and the falling envelope), and bright ones between, near
$d\sin\theta=n\lambda$ ($0$ and $\pm39^\circ$; the falling envelope pulls the second in, to
$\pm37^\circ$ measured). The two are scaled to the same height: how much gets through
slits about a wavelength wide with soft edges is beyond Fraunhofer (the table has it: 3.0
per unit of incident flux, against the openings' width of 2.4), but where it goes is not.
Only within a degree or so of $\pm90^\circ$, where the detector runs into the wall
(shaded), is nothing measured. In momentum space the transverse momenta $k_y$ are the
diffraction orders. Move the slits apart ($d$) and the fringes close up; make them
narrower ($a$) and the envelope spreads.
11. Bragg reflection from a crystal
A square crystal of $7\times7$ atoms, $c=2$ apart, turned by $\alpha=30^\circ$, in a beam
of wavelength $\lambda=2$. Each row of atoms reflects a little of the wave, like a
half-silvered mirror, and lets the rest through; the waves reflected by successive rows add
up only if their paths differ by whole wavelengths, $2c\sin\alpha=n\lambda$ (Bragg). Here
$2c\sin30^\circ=\lambda$: a strong beam leaves at $2\alpha=60^\circ$, about a quarter of the
packet (26%, the table). In momentum space the circles are $\vec k_0+\vec G$: the
reciprocal lattice , the wavevectors $\vec G=\frac{2\pi}{c}(n_1,n_2)$ (turned with the
crystal) of the plane waves that repeat with the lattice, hung on the incident spot. A
scattered wave $\vec k_0+\vec G$ must also be on the energy circle $|\vec k|=k_0$; where a
lattice point sits on it (filled), there is a Bragg beam: Ewald's construction .
The pink arrow, at $60^\circ$, ends on it: the momentum transfer to a Bragg beam is a
vector of the reciprocal lattice, $q=2k_0\sin\alpha=2\pi/c$, which is Bragg's law again.
Born's curve is the crystal's Fourier transform, the kinematic theory of diffraction.
12. Off the Bragg angle
The same crystal turned to $\alpha=20^\circ$: $2c\sin\alpha/\lambda=0.68$, no lattice
point on the energy circle, and the beam mostly goes through; a sixth as much goes to
$2\alpha$ (4.5%). The pink arrow, at $2\alpha=40^\circ$, points along a lattice vector
$\vec G$ but reaches only 0.68 of it ($q=2k_0\sin\alpha=2.15$, $2\pi/c=3.14$): the
crystal gives momentum in lattice vectors, and none ends on the circle. Turn the crystal
slowly back to $30^\circ$ and the reflected beam comes back, strong only within a few
degrees of the Bragg angle (the crystal is small, so its lattice points are not sharp).
Or keep $\alpha=20^\circ$ and shorten the wavelength until $2c\sin\alpha=\lambda$: $k_0=4.6$,
and the beam reflects to $40^\circ$ (18%). At $\alpha=60^\circ$ it is the columns that
reflect, $2c\cos\alpha=\lambda$, to $-60^\circ$.
13. The impact parameter
A narrow packet aimed off centre, with impact parameter $b$: how far from the
centre it would pass if nothing deflected it. Large $b$: it passes by, barely deflected.
Small $b$: it bounces off. A classical particle's deflection depends on $b$ alone; a
wave's spreads, the more so the narrower the packet.
14. Your own target
Type any $V(x,y)$, using $x$, $y$, $r$ and $\theta$. Sharp edges are softened slightly
(an edge sharper than the grid can't be followed in time).
How it works
A wave packet, nearly a plane wave across the target, runs into a potential $V(x,y)$ and
scatters, with $\hbar=m=1$. The picture is $\psi$ coloured by phase (the colour wheel of the
other demos) and lit by $|\psi|$; the potential is drawn faintly, and the dashed circle is
the detector. Far away the scattered part is an outgoing circular wave,
$$\psi\to e^{ikx}+f(\theta)\,\frac{e^{ikr}}{\sqrt r},\qquad \frac{d\sigma}{d\theta}=|f(\theta)|^2,$$
and in two dimensions a cross section is a length. Everything on the right is measured on
the detector circle while the packet goes by. A target that isn't round can be turned
(turn ): the same as sending the packet in from another direction, with $\theta$
still the scattering angle.
What is measured. The split-operator method on a $256\times256$ grid, with an
absorbing border. On the detector circle (radius $R$) the outward flux of the scattered wave
is added up over the collision and divided by the incident fluence (the beam's probability
per unit width): that is the measured $d\sigma/d\theta$, averaged over the packet's spread of
energies. Behind a wall, the flux of $\psi$ itself is measured instead of the scattered
flux.
What is computed from it. At the packet's central energy $E_0=k_0^2/2$, the time
Fourier transform of $\psi_\text{sc}$ on the circle, $\int\psi_\text{sc}(R,\theta,t)\,e^{iE_0t}dt$,
picks out the stationary scattered wave at that one energy. For a round target it is a sum
of partial waves, $\sum_m\beta_m\,\frac{S_m-1}{2}\,H_m(kR)\,e^{im\theta}$ with
$H_m=J_m+iY_m$ the outgoing Hankel function and $\beta_m$ the incident wave's own partial-wave
amplitudes ($i^m$ for a plane wave; a beam of finite width carries a little less of the
higher $m$, and its $\beta_m$ are taken from its momentum amplitudes on the energy circle).
Expanding in $e^{im\theta}$ and dividing by $\beta_mH_m(kR)$ gives $S_m=e^{2i\delta_m}$
exactly, without assuming $kR\gg1$. A target that isn't round mixes the partial waves and
has no phase shifts. From the $S_m$,
$$f(\theta)=\frac{e^{-i\pi/4}}{\sqrt{2\pi k}}\sum_m(S_m-1)\,e^{im\theta},\qquad
\sigma=\frac1k\sum_m|S_m-1|^2=\frac4k\sum_m\sin^2\delta_m,$$
which are the orange curve and the table's partial waves line; the optical
theorem line is $-\sqrt{8\pi/k}\,\mathrm{Re}\,[e^{i\pi/4}f(0)]$ from the same $S_m$. The
two agree when $|S_m|=1$. For a potential of $r$ alone, $\delta_{-m}=\delta_m$.
Why they should agree. The measured curve is the flux of the very same scattered
wave, but summed over all the energies in the packet, while the partial waves are taken at
$k_0$ alone: a longer packet (length ) has a narrower spread of energies and brings
the two closer. The exact curve and line are what time-independent scattering
theory gives for the same $V(r)$ at $k_0$: the radial equation integrated outwards for each
$m$ and matched to $J_m(kr)$ and $Y_m(kr)$ beyond the potential. For a hard disk of radius
$a$ that gives $\tan\delta_m=J_m(ka)/Y_m(ka)$. The first Born approximation, with this sign
convention (from the two-dimensional Green's function $-\frac i4H_0(kr)$), and its phase
shifts, small and negative for a repulsive potential, are
$$f(\theta)=-\frac{m}{\hbar^2}\,\frac{e^{i\pi/4}}{\sqrt{2\pi k}}\,\tilde V(\vec q),\quad
\vec q=\vec k_\text{out}-\vec k_\text{in},\qquad
\delta_m\approx-\pi\frac{m}{\hbar^2}\int_0^\infty V(r)\,J_m^2(kr)\,r\,dr.$$
Round the momentum circle, and classically. After the collision the scattered
wave moves freely, and the part of it leaving along $\theta$ has its momentum along
$\theta$: the probability on the circle $|\vec k|=k_0$ toward $\theta$ is the incident
fluence times $d\sigma/d\theta$. So both curves are drawn round it, on one scale: the
measured one, the flux through the detector, and Born's,
$\frac{m^2}{2\pi\hbar^4k}|\tilde V(\vec k-\vec k_0)|^2$ for $\vec k$ on the circle, the
potential's transform at the momentum transfer $q=2k\sin(\theta/2)$. The classical curve
is a hard disk's, of the radius where the wall's $V$ equals $E$ and a particle coming head
on is turned back, $r=a+w\ln(V_0/E-1)$: $\frac r2|\sin\frac\theta2|$, and $2r$ in all
(experiment 3). A wave's $\sigma$ tends to $4r$ as $kr\to\infty$, as much again diffracted
forward as the disk reflects. The wall is soft, over $w$: a particle that meets it at the
angle $\alpha$ to its normal turns back where $V=E\cos^2\alpha$, further out than $r$, and
is bent on its way past too. Rays through the disk of experiment 1 itself ($w=0.25$) send
less straight back than the hard disk (0.73 against 0.89), and so does the wave (0.73).
New in 2026 for BrainFlux. Source:
github.com/gallicchio/brainflux , MIT licence.