Subsections


A.30 Three-dimensional scattering

This note introduces some of the general concepts of three-di­men­sion­al scattering, in case you run into them. For more details and actual examples, a quantum mechanics text for physicists will need to be consulted; it is a big thing for them.

Figure A.21: Scattering of a beam off a target.
\begin{figure}
\centering
\setlength{\unitlength}{1pt}
\begin{picture}(...
...a,\phi)\frac{e^{{\rm i}p_\infty r/\hbar}}{r}$}}
\end{picture}
\end{figure}

The basic idea is as sketched in figure A.21. A beam of particles is sent in from the far left towards a three-di­men­sion­al target. Part of the beam hits the target and is scattered, to be picked up by surrounding detection equipment.

It will be assumed that the collision with the target is elastic, and that the particles in the beam are sufficiently light that they scatter off the target without transferring kinetic energy to it. In that case, the target can be modeled as a steady potential energy field. And if the target and/or incoming particles are electrically neutral, it can also be assumed that the potential energy decays fairly quickly to zero away from the target. (In fact, a lot of results in this note turn out not apply to a slowly decaying potential like the Coulomb one.)

It is convenient to use a spherical coordinate system $(r,\theta,\phi)$ with its origin at the scattering object and with its axis aligned with the direction of the incoming beam. Since the axis of a spherical coordinate system is usually called the $z$-​axis, the horizontal coordinate will now be indicated as $z$, not $x$ like in the one-di­men­sion­al analysis done earlier.

In the energy eigenfunctions, the incoming particle beam can be represented as a one-di­men­sion­al wave. However, unlike for the one-di­men­sion­al scattering of figure 7.22, now the wave is not just scattered to the left and right, but in all directions, in other words to all angles $\theta$ and $\phi$. The far-field behavior of the energy eigenfunctions is

\begin{displaymath}
\fbox{$\displaystyle
\psi_E \sim C^{\rm{l}}_{\rm{f}} e^{...
...r}\quad r\to\infty
\qquad p_\infty \equiv \sqrt{2mE}
$} %
\end{displaymath} (A.216)

Here $E$ is the kinetic energy of the incoming particles and $m$ the mass. Therefore $p_\infty$ is what classical physics would take to be the momentum of the particles at infinity. The first term in the far field behavior allows the incoming particles to be described, as well as the same particles going out again unperturbed. If some joker removes the target, that is all there is.

The second term describes the outgoing scattered particles. The constant $C_{\rm {f}}(\theta,\phi)$ is called the “scattering amplitude.” The second term also contains a factor $e^{{{\rm i}}p_{\infty}r/\hbar}$ consistent with wave packets that move radially away from the target in the far field.

Finally, the second term contains a factor 1/$r$. Therefore the magnitude of the second term decreases with the distance $r$ from the target. This happens because the probability of finding a particle in a given detection area should decrease with distance. Indeed, the total detection area is $4{\pi}r^2$, where $r$ is the distance at which the detectors are located. That increases proportional to $r^2$, and the total number of particles to detect per unit time is the same regardless of where the detectors are located. Therefore the probability of finding a particle per unit area should decrease proportional to 1/$r^2$. Since the probability of finding a particle is proportional to the square of the wave function, the wave function itself must be proportional to 1/$r$. The second term above makes it so.

Consider now the number of particles that is detected in a given small detection area ${\rm d}{A}$. The scattered stream of particles moving towards the detection area has a velocity $v$ $\vphantom0\raisebox{1.5pt}{$=$}$ $p_\infty$$\raisebox{.5pt}{$/$}$$m$. Therefore in a time interval ${\rm d}{t}$, the detection area samples a volume of the scattered particle stream equal to ${\rm d}{A}$ $\times$ ${v}{\rm d}{t}$. The chances of finding particles are proportional to the square magnitude of the wave function times that volume. Using the asymptotic wave function above, the number of particles detected will be

\begin{displaymath}
{\rm d}I = \mbox{[constant] }
\vert C_{\rm {f}}(\theta,\phi)\vert^2 \frac{{\rm d}A}{r^2} {\rm d}t
\end{displaymath}

The constant includes the factor $p_\infty$$\raisebox{.5pt}{$/$}$$m$. The constant must also account for the fact that the wave function is not normalized, and that there is a continuous stream of particles to be found, rather than just one particle.

According to the above expression, the number of particles detected in a given area ${\rm d}{A}$ is proportional to its three-di­men­sion­al angular extent

\begin{displaymath}
{\rm d}\Omega \equiv \frac{{\rm d}A}{r^2}
\end{displaymath}

This is the so-called “solid angle” occupied by the detection area element. It is the three-di­men­sion­al generalization of two-di­men­sion­al angles. In two dimensions, an element of a circle with arc length ${\rm d}{s}$ occupies an angle ${\rm d}{s}$$\raisebox{.5pt}{$/$}$$r$ when expressed in radians. Similarly, in three dimensions, an element of a sphere with area ${\rm d}{A}$ occupies a solid angle ${\rm d}{A}$$\raisebox{.5pt}{$/$}$$r^2$ when expressed in “steradians.”

In those terms, the number ${\rm d}{I}$ of particles detected in an infinitesimal solid angle ${\rm d}\Omega$ is

\begin{displaymath}
{\rm d}I = \mbox{[constant] } \vert C_{\rm {f}}(\theta,\phi)\vert^2 {\rm d}\Omega {\rm d}t
\end{displaymath}

As noted, the constant of proportionality depends on the rate at which particles are sent at the target. The more particles are sent at the target, the more will be deflected. The number of particles in the incoming beam per unit beam cross-sectional area and per unit time is called the “luminosity” of the beam. It is related to the square of the wave function of the incoming beam through the relation

\begin{displaymath}
{\rm d}I_{\rm {b}} = \mbox{[constant] }
\vert C^{\rm {l}}_{\rm {f}}\vert^2 {\rm d}A_{\rm {b}} {\rm d}t
\end{displaymath}

Here ${\rm d}{A}_{\rm {b}}$ is a cross sectional area element of the incoming particle beam and ${\rm d}{I}_{\rm {b}}$ the number of particles passing through that area.

Physicist like to relate the scattered particle flow in a given infinitesimal solid angle ${\rm d}{\Omega}$ to an equivalent incoming beam area ${\rm d}{A_b}$ through which the same number of particles flow. Therefore they define the so-called “differential cross-section” as

\begin{displaymath}
\fbox{$\displaystyle
\frac{{\rm d}\sigma}{{\rm d}\omega} \equiv \frac{{\rm d}A_{\rm{b,equiv}}}{{\rm d}\Omega}
$}
\end{displaymath} (A.217)

The quantity ${\rm d}{A}_{\rm {b,equiv}}$ can be thought of as the infinitesimal area of the incoming beam that ends up in the infinitesimal solid angle ${\rm d}\Omega$. So the differential cross-section is a scattered particle density expressed in suitable terms.

Note how well chosen the term “differential cross-section” really is. If physicists had called it something like scattered cross-section density, or even simply scattered cross-section, nonexperts would probably have a pretty good guess what physicists were talking about. But cross section by itself can mean anything. There is nothing in the term to indicate that it is a measure for how many particles are scattered. And preceding it by differential is a stroke of genius because it is not a differential, it is a differential quotient. This will confuse mathematically literate nonexperts even more.

The differential cross section does not depend on how many particles are sent at the target, nor on wave function normalization. Following the expressions for the particle flows given above, the differential cross section is simply

\begin{displaymath}
\fbox{$\displaystyle
\frac{{\rm d}\sigma}{{\rm d}\omega}...
...\theta,\phi)\vert^2}{\vert C^{\rm{l}}_{\rm{f}}\vert^2}
$} %
\end{displaymath} (A.218)

Moreover, the particle flow in an incoming beam area ${\rm d}{A}_{\rm {b}}$ may be measured using the same experimental techniques as are used to measure the deflected particle flow. Various systematic errors in the experimental method will then cancel in the ratio, giving more accurate values.

The total area of the incoming beam that gets scattered is called the “total cross-section” $\sigma$:

\begin{displaymath}
\fbox{$\displaystyle
\sigma \equiv A_{\rm b,total}
$}
\end{displaymath} (A.219)

Of course, the name is quite below the normal standards of physicists, since it really is a total cross-section. Fortunately, physicist are clever enough not to say what cross section it is, and cross-section can mean many things. Also, by using the symbol $\sigma$ instead of something logical like $A_{\rm {b}}$ for the differential cross-section, physicists do their best to reduce the damage as well as possible.

If you remain disappointed in physicists, take some comfort in the following term for scattering that can be described using classical mechanics: the “impact parameter.” If you guess that it describes the local physics of the particle impact process, it is really hilarious to physicists. Instead, think centerline offset; it describes the location relative to the centerline of the incoming beam at which the particles come in; it has no direct relation whatsoever to what sort of impact (if any) these particles end up experiencing.

The total cross section can be found by integrating the differential cross section over all deflection angles:

\begin{displaymath}
\sigma = \int_{\rm all} \frac{{\rm d}A_{\rm {b,equiv}}}{{\rm d}\Omega} {\,\rm d}\Omega
\end{displaymath}

In spherical coordinates this can be written out explicitly as
\begin{displaymath}
\fbox{$\displaystyle
\sigma = \int\limits_{\theta=0^+}^{...
...rm{f}}\vert^2}
\sin\theta {\,\rm d}\theta{\rm d}\phi
$} %
\end{displaymath} (A.220)


A.30.1 Partial wave analysis

Jim Napolitano from RPI and Cornell notes:

The term Partial Wave Analysis is poorly defined and overused.
Gee, what a surprise! For one, they are component waves, not partial waves. But you already componently assumed that they might be.

This discussion will restrict itself to spherically symmetric scattering potentials. In that case, the analysis of the energy eigenfunctions can be done much like the analysis of the hydrogen atom of chapter 4.3. However, the boundary conditions at infinity will be quite different; the objective is not to describe bound particles, but particles that come in from infinity with positive kinetic energy and are scattered back to infinity. Also, the potential will of course not normally be a Coulomb one.

But just like for the hydrogen atom, the energy eigenfunctions can be taken to be radial functions times spherical harmonics $Y_l^m$:

\begin{displaymath}
\psi_{Elm}(r,\theta,\phi) = R_{El}(r) Y_l^m(\theta,\phi)
\end{displaymath} (A.221)

These energy eigenfunctions have definite angular momentum in the $z$-​direction $m\hbar$, as well definite square angular momentum $l(l+1)\hbar^2$. The radial functions $R_{El}$ will not be the same as the hydrogen $R_{nl}$ ones.

The incoming plane wave $e^{{{\rm i}}p_{\infty}z/\hbar}$ has zero angular momentum in the $z$-​direction. Unfortunately, it does not have definite square angular momentum. Instead, it can be written as a linear combination of free-space energy eigenfunctions with different values of $l$, hence with different square angular momentum:

\begin{displaymath}
e^{{\rm i}p_\infty z/\hbar}
= \sum_{l=0}^\infty c_{{\rm ...
...eta)
\qquad
c_{{\rm {w}},l} = {\rm i}^l \sqrt{4\pi(2l+1)}
\end{displaymath} (A.222)

See {A.6} for a derivation and the precise form of the spherical Bessel functions $j_l$.

Now finding the complete energy eigenfunction corresponding to the incoming wave directly is typically awkward, especially analytically. Often it is easier to solve the problem for each term in the above sum separately and then add these solutions all together. That is where the name partial wave analysis comes from. Each term in the sum corresponds to a partial wave, if you use sufficiently lousy terminology.

The partial wave analysis requires that for each term in the sum, an energy eigenfunction is found of the form $\psi_{El}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $R_{El}Y_l^0$. The required behavior of this eigenfunction in the far field is

\begin{displaymath}
\fbox{$\displaystyle
\psi_{El} \sim
\left[c_{{\rm{w}}...
...r)\right]
Y_l^0(\theta) \qquad\mbox{for } r\to\infty
$} %
\end{displaymath} (A.223)

Here the first term is the component of the incoming plane wave corresponding to spherical harmonic $Y_l^0$. The second term represents the outgoing deflected particles. The value of the coefficient $c_{{\rm {f}},l}$ is determined in the solution process.

Note that the above far field behavior is quite similar to that of the complete energy eigenfunction as given earlier in (A.216). However, here the coefficient $C^{\rm {l}}_{\rm {f}}$ was set to 1 for simplicity. Also, the radial part of the reflected wave function was written using a Hankel function of the first kind $h_l^{(1)}$. This Hankel function produces the same $e^{{{\rm i}}p_{\infty}r/\hbar}$$\raisebox{.5pt}{$/$}$$r$ radial behavior as the second term in (A.216), {A.6} (A.25). However, the Hankel function has the advantage that it becomes exact as soon as the scattering potential becomes zero. It is not just valid at very large $r$ like the bare exponential.

To be sure, for a slowly decaying potential like the Coulomb one, the Hankel function is no better than the exponential. However, the Hankel function is very closely related to the Bessel function $j_l$, {A.6}, allowing various helpful results to be found in table books. If the potential energy is piecewise constant, it is even possible to solve the complete problem using Bessel and Hankel functions. These functions can be tied together at the jumps in potential in a way similar to addendum {A.27}.

In terms of the asymptotic behavior above, the differential cross section is

\begin{displaymath}
\fbox{$\displaystyle
\frac{{\rm d}\sigma}{{\rm d}\omega}...
...,{\underline l}} Y_l^0(\theta)Y_{\underline l}^0(\theta)
$}
\end{displaymath} (A.224)

This can be verified using {A.6} (A.25), (A.216), and (A.218). The Bessel functions form the incoming wave and do not contribute. For the total cross-section, note that the spherical harmonics are orthonormal, so

\begin{displaymath}
\fbox{$\displaystyle
\sigma = \frac{\hbar^2}{p_\infty^2} \sum_{l=0}^\infty \vert c_{{\rm{f}},l}\vert^2
$}
\end{displaymath}

One special case is worth mentioning. Consider particles of such low momentum that their typical quantum wave length, $2\pi\hbar$$\raisebox{.5pt}{$/$}$$p_\infty$, is gigantic compared to the radial size of the scattering potential. Particles of such large wave lengths do not notice the fine details of the scattering potential at all. Conversely, normally the scattering potential only notices the incoming partial wave with $l$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0. That is because the Bessel functions are proportional to

\begin{displaymath}
(p_\infty r/\hbar)^l
\end{displaymath}

for small arguments. If the wave length is large compared to the typical radial size $r$ of the scattering potential, this is negligible unless $l$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0. Now $l$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0 corresponds to a wave function that is the same in all directions; it is proportional to the constant spherical harmonic $Y_0^0$. If only the partial wave that is the same in all directions gets scattered, then the particles get scattered equally in all directions (if they get scattered at all.)

Coincidently, equal scattering in all directions also happens in another case: scattering of classical point particles from a hard elastic sphere. That is very much the opposite case, because negligible uncertainty in position requires high, not low, energy of the particles. In any case, the similarity between the two cases is is superficial. If a beam of classical particles is directed at a hard sphere, only an area of the beam equal to the frontal area of the sphere gets scattered. But if you work out the scattering of low-energy quantum particles from a hard sphere, you get a total scattering cross section that is 4 times bigger.


A.30.2 Partial wave amplitude

This subsection gives some further odds and ends on partial wave analysis, for the incurably curious.

Recall that a partial wave has an asymptotic behavior

\begin{displaymath}
\psi_{El} \sim
\left[c_{{\rm {w}},l}j_l(p_\infty r/\hba...
... r/\hbar)\right]
Y_l^0(\theta) \qquad\mbox{for } r\to\infty
\end{displaymath}

The first term corresponds to the wave function of the incoming particles. The second term is the effect of the scattering potential.

Physicists like to write the coefficient of the scattered wave as

\begin{displaymath}
\fbox{$\displaystyle
c_{{\rm{f}},l} = {\rm i}k c_{{\rm{w}},l} a_l
$}
\end{displaymath} (A.225)

They call the so-defined constant $a_l$ the “partial wave amplitude” because obviously it is not a partial wave amplitude. Confusing people is always funny.

Now every partial wave by itself is a solution to the Hamiltonian eigenvalue problem. That means that every partial wave must ensure that particles cannot just simply disappear. That restricts what the partial wave amplitude can be. It turns out that it can be written in terms of a real number $\delta_l$:

\begin{displaymath}
\fbox{$\displaystyle
a_l = \frac{1}{k} e^{{\rm i}\delta_l} \sin\delta_l
$}
\end{displaymath} (A.226)

The real number $\delta_l$ is called the “phase shift.”

Some physicist must have got confused here, because it really is a phase shift. To see that, consider the derivation of the above result. First the asymptotic behavior of the partial wave is rewritten in terms of exponentials using {A.6} (A.24) and (A.25). That gives

\begin{displaymath}
\psi_{El} \sim \ldots
\left[e^{-{\rm i}p_\infty r/\hbar}...
...)^{l+1}e^{{\rm i}p_\infty r/\hbar}(1+2{\rm i}k a_l)
\right]
\end{displaymath}

The dots stand for common factors that are not important for the discussion. Physically, the first term above describes spherical wave packets that move radially inwards toward the target. The second term describes wave packets that move radially outwards away from the target.

Now particles cannot just disappear. Wave packets that go in towards the target must come out again with the same amplitude. And that means that the two terms in the asymptotic behavior above must have the same magnitude. (This may also be shown mathematically using procedures like in {A.32}.)

Obviously the two terms do have the same magnitude in the absence of scattering, where $a_l$ is zero. But in the presence of scattering, the final parenthetical factor will have to stay of magnitude one. And that means that it can be written in the form

\begin{displaymath}
1+2{\rm i}k a_l = e^{{\rm i}2 \delta_l}
\end{displaymath} (A.227)

for some real number $\delta_l$. (The factor 2 in the exponential is put in because physicists like to think of the wave being phase shifted twice, once on the way in to the target and once on the way out.) Cleaning up the above expression using the Euler formula (2.5) gives the stated result.

If you add in the time dependent factor $e^{{{\rm i}}Et/\hbar}$ of the complete unsteady wave function, you can see that indeed the waves are shifted by a phase angle $2\delta_l$ compared to the unperturbed wave function. Without any doubt, the name of the physicist responsible for calling the phase angle a phase angle has been ostracized from physics. She will never be heard of again.


A.30.3 The Born approximation

The Born approximation assumes that the scattering potential is weak to derive approximate expressions for the scattering.

Consider first the case that the scattering potential is zero. In that case, the wave function is just that of the incoming particles:

\begin{displaymath}
\psi_E = e^{{\rm i}p_\infty z/\hbar} \qquad p_\infty=\sqrt{2mE}
\end{displaymath}

where $E$ is the energy of the particle and $m$ its mass.

Born considered the case that the scattering potential $V$ is not zero, but small. Then the wave function $\psi_E$ will still be close to the incoming wave function, but no longer exactly the same. In that case an approximation to the wave function can be obtained from the so-called integral Schrö­din­ger equation, {A.13} (A.42):

\begin{displaymath}
\psi_E({\skew0\vec r}) = e^{{\rm i}p_\infty z/\hbar}
- \...
...skew0\vec r}^{\,\prime}) {\,\rm d}^3{\skew0\vec r}^{\,\prime}
\end{displaymath}

In particular, inside the integral the true wave function $\psi_E$ can be replaced by the incoming wave function:
\begin{displaymath}
\psi_E({\skew0\vec r}) \approx e^{{\rm i}p_\infty z/\hbar}...
...m i}p_\infty z'/\hbar} {\,\rm d}^3{\skew0\vec r}^{\,\prime} %
\end{displaymath} (A.228)

It is not exact, but it is much better than just setting the integral to zero. The latter would make the wave function equal to the incoming wave. With the approximate integral, you get a valid approximation to the particle deflections.

To get the differential cross section, examine the behavior of (A.228) at given scattering angles $\theta$ and $\phi$ for large $r$. That produces, {D.47}:

\begin{displaymath}
\fbox{$\displaystyle
\frac{{\rm d}\sigma}{{\rm d}\omega}...
...\,\prime}\;
\right\vert^2 \qquad \mbox{($V$\ small)}
$} %
\end{displaymath} (A.229)

Here
\begin{displaymath}
{\skew0\vec p}_\infty^{\,\rm {l}} = p_\infty {\hat k}
\q...
...y {\hat\imath}_r
\qquad \mbox{(with $p_\infty=\sqrt{2mE}$)}
\end{displaymath} (A.230)

are the classical momentum vectors of the incoming and scattered particles. Note that the direction of ${\skew0\vec p}_\infty$ depends on the considered scattering angles $\theta$ and $\phi$. And that apparently the momentum change of the particles is a key factor affecting the amount of scattering.

One additional approximation is worth mentioning. Consider particles of such low momentum that their quantum wave length, $2\pi\hbar$$\raisebox{.5pt}{$/$}$$p_\infty$, is gigantic compared to the radial size of the scattering potential. Particles of such wave lengths do not notice the fine details of the scattering potential at all. Mathematically, $p_\infty$ is so small that the argument of the exponential in the differential cross section above can be assumed zero. Then:

\begin{displaymath}
\fbox{$\displaystyle
\frac{{\rm d}\sigma}{{\rm d}\omega}...
...ight\vert^2 \qquad \mbox{($V$\ and $p_\infty$\ small)}
$} %
\end{displaymath} (A.231)

The differential cross section no longer depends on the angular position. If particles get scattered at all, they get scattered equally in all directions.

Note that the integral is infinite for a Coulomb potential.