33 Scattering in perturbation theory
As we’ve seen, even for extremely simple potentials the three-dimensional scattering problem can become quite complicated. For dealing with more realistic potentials, it’s very useful to have a perturbative approach instead of relying on exact solution. The framework we develop here will be a little bit different than the perturbation theory methods we’ve seen so far.
33.1 The Lippmann-Schwinger equation
Our starting point is the time-independent Schrödinger equation, (\hat{H}_0 + \hat{V}) \ket{\psi_E} = E \ket{\psi_E}. Here we’ll follow closely our treatment for time-independent Brillouin-Wigner perturbation theory (Section 20.3), except that we now have an infinite Hilbert space which will change some of the details. As in the previous finite case, we’ll begin by rewriting the equation and then inverting an operator: (E - \hat{H}_0) \ket{\psi_E} = \hat{V} \ket{\psi_E} \\ \ket{\psi_E} = \ket{\psi^{(0)}_E} + \frac{1}{E-\hat{H}_0} \hat{V} \ket{\psi_E}, where \ket{\psi^{(0)}_E} is an energy eigenstate of the unperturbed \hat{H}_0, which means it is annihilated by E - \hat{H}_0 (so you can verify that acting with E - \hat{H}_0 on the second equation gives back the first.)
Now, as we saw previously, the problem with this formal solution is that there are certain states for which the inverse operator is not defined (for example, any component of \ket{\psi^{(0)}_E} in \ket{\psi_E} gives division by zero.) In a finite system, we fix this by using linear algebra, projecting out the part of our vector space for which the operator blows up. Here, we have an infinite number of states and a continuous energy variable E. To deal with the divergence, we instead adopt the “i\epsilon prescription”: we add a small imaginary part to the denominator to regulate it.
Given a Hamiltonian of the form \hat{H} = \hat{H}_0 + \hat{V}, energy eigenstate solutions \ket{\psi_E^{\pm}} of \hat{H} satisfy the recursive equation \ket{\psi_E^{\pm}} \equiv \ket{\psi_E^{(0)}} + \frac{1}{E - \hat{H}_0 \pm i\epsilon} \hat{V} \ket{\psi_E^{\pm}} in the limit \epsilon \rightarrow 0^+, with \ket{\psi_E^{(0)}} an eigenstate of \hat{H}_0 with energy E.
So far, this is just a formal rewriting of the Schrödinger equation; it’s also still exact. We can go to position space to write the formal solution another way: \left\langle x | \psi_E^{\pm} \right\rangle = \int d^3x'\ \bra{x} \frac{1}{E-\hat{H}_0 \pm i\epsilon} \ket{x'} \bra{x'} \hat{V} \ket{\psi_E^{\pm}} + \left\langle x | \psi_E^{(0)} \right\rangle, or in terms of functions, \psi_E^\pm(x) = \psi_E^{(0)}(x) + \int d^3x' G_\pm(E,x,x') V(x') \psi_E^\pm(x'), where G_\pm is a Green’s function for the inverse operator, G_\pm(E,x,x') = \bra{x} \frac{1}{E - \hat{H}_0 \pm i\epsilon} \ket{x'}. To make progress from here, we need to specify the unperturbed Hamiltonian. Given the context of scattering, the most sensible \hat{H}_0 to use is the free Hamiltonian, \hat{H}_0 = \frac{\hat{\vec{p}}^2}{2m} = -\frac{\hbar^2}{2m} \nabla^2. (Other Hamiltonians could be useful in specific situations, but then we’d have to carefully re-think our whole scattering setup.) This lets us see explicitly that we’re solving a Green’s function equation: (E + \frac{\hbar^2}{2m} \nabla^2) G_\pm(E,x,x') = \delta^{(3)}(x-x'). This can be readily solved by going to momentum space and doing the resulting momentum integral as a contour integral; the \pm solutions correspond to contours that pick up the corresponding poles in the denominator. I’ll skip to the result: G_\pm(E,x,x') = \frac{-m}{2\pi \hbar^2} \frac{e^{\pm ik |\vec{x} - \vec{x}'|}}{|\vec{x} - \vec{x}'|} where k is the free wave number, defined as usual by \hbar^2 k^2 = 2mE. We can recognize from the functional form that the two different \pm solutions correspond exactly to incoming and outgoing spherical waves.
Solve the equation above and do the contour integration to reproduce the results for G_\pm(E,x,x').
Answer:
To be filled in! But see e.g. David Tong’s scattering notes.Now we can substitute back in to the Lippmann-Schwinger equation to find a rewritten form in the presence of a free \hat{H}_0:
\psi_E^\pm(x) = \psi_E^{(0)}(x) - \frac{m}{2\pi \hbar^2} \int d^3x' \frac{e^{\pm ik|\vec{x} - \vec{x}'|}}{|\vec{x} - \vec{x}'|} V(x') \psi_E^\pm(x').
Now let’s specialize just a little bit more to the scattering problem. Since we’re interested in scattering solutions, we don’t really need the full wavefunction solution, we just care about it at large distances. If we have a localized scattering potential, then |\vec{x}| \sim r, and also |\vec{x}| \gg |\vec{x}'|, since \vec{x}' only integrates over where the potential is nonzero.
Based on this assumption, we can simplify the difference between vectors: |\vec{x} - \vec{x}'| \rightarrow \sqrt{|x|^2 + |x'|^2 - 2\vec{x} \cdot \vec{x}'} \approx r \sqrt{1 - \frac{2}{r} \hat{x} \cdot \vec{x}' + ...} \approx r - \hat{x} \cdot \vec{x}'. Then e^{ik|\vec{x} - \vec{x}'|} \approx e^{ikr} e^{-ik\hat{x} \cdot \vec{x}'} = e^{ikr} e^{-i\vec{k}' \cdot \vec{x}'} defining the momentum vector \vec{k'} \equiv k\hat{x}. The number of vectors we’re defining here is proliferating a bit, so it’s worth stopping for a moment to think about the geometry and take inventory of the vectors we’ve defined, in the context of scattering:
- \vec{x}: encodes the coordinates of our full solution for the wavefunction (plane wave + scattering wave.)
- \vec{x}': integration variable, runs only over the region with non-zero potential V(\vec{x}').
- \vec{k}: points in the direction of the incoming plane wave.
- \vec{k}': points in the same direction as \vec{x}.
For the scattering problem \vec{x} is at large r and pointing in the outgoing scattering direction, which means that we can think of \vec{k}' as the scattered, outgoing momentum vector. It has the same magnitude as the incoming \vec{k} because energy/momentum is conserved.
Plugging this expansion back in and choosing the outgoing spherical wave for scattering, we have a key result:
In the context of scattering with |\vec{x}| = r \gg |\vec{x}'|, the Lippmann-Schwinger equation takes the form \psi(\vec{x}) = \mathcal{N} e^{i\vec{k} \cdot \vec{x}} - \frac{m}{2\pi \hbar^2} \frac{e^{ikr}}{r} \int d^3x' e^{-i\vec{k}' \cdot \vec{x}'} V(\vec{x}') \psi(\vec{x}'), from which we identify the scattering amplitude, f(\theta, \phi) = f(\vec{k}, \vec{k}') = -\frac{m}{2\pi \hbar^2} \int d^3x' e^{-i\vec{k}' \cdot \vec{x}'} V(\vec{x}') \psi(\vec{x}').
This gives us an alternative (and much more direct) way to obtain the scattering amplitude and thus the differential cross section d\sigma / d\Omega. However, this is still just an implicit integral equation, since the solution we want \psi(\vec{x}') appears on both sides. However, we already know how to deal with that in a perturbative setup: we’ll series expand the solution in a small parameter and then match, order by order.
33.2 Born approximation
To proceed, we write the full solution \psi(\vec{x}) as an infinite series, \psi(\vec{x}) = \sum_{n=0}^\infty \psi_n(\vec{x}), with the implicit assumption that each consecutive term is suppressed by a small parameter contained in V(\vec{x}). Then we identify the zeroth-order term with the unperturbed wavefunction - in the case of scattering, just our incoming plane wave, \psi_0(\vec{x}) = \mathcal{N} e^{i\vec{k} \cdot \vec{x}}. If we plug the whole series in and compare term by term, we obtain the recursive relation \psi_n(\vec{x}) = -\frac{m}{2\pi \hbar^2} \int d^3x' e^{-i\vec{k}' \cdot \vec{x}'} V(\vec{x}') \psi_{n-1}(\vec{x}'). We see that as assumed, this results in a series where each term has another power of V(\vec{x}) contained in it. Taking this series setup and truncating at n-th order is known as the nth Born approximation. The most commonly used version is just to keep the first term (which is sometimes just called the “Born approximation” on its own):
The scattering amplitude to first Born approximation (first order in an expansion in V(x)) is given by
f(\theta, \phi) = -\frac{m}{2\pi \hbar^2} \int d^3x' e^{i\vec{q} \cdot \vec{x}'/\hbar} V(\vec{x}') \\ = -\frac{m}{2\pi \hbar^2} \tilde{V}(\vec{q}) where \tilde{V}(q) is the Fourier transform of the potential V(\vec{x}), and \vec{q} = \hbar(\vec{k} - \vec{k}') is the momentum transfer.
(Note that we’re assuming there isn’t an extra normalizing factor when we define the Fourier transform; if there is, it would typically also appear in the first equation from the normalization of the initial plane-wave state.)
So at lowest order in perturbation theory, we find that the scattering amplitude is just the Fourier transform of the scattering target potential. The angular dependence is fully encoded in the momentum transfer vector \vec{q}, which is the difference between the incoming momentum \vec{k} and outgoing momentum \vec{k}'.
In the particular case of a spherically symmetric potential, we can simplify a bit further. Choosing \vec{q} to be pointing along the z'-axis just to do the integral, we have \vec{q} \cdot \vec{x}' = qr' \cos \theta' so that the integration becomes f(\theta) = -\frac{m}{2\pi \hbar^2} \int d^3x' e^{iqr' \cos \theta' /\hbar} V(r') \\ = \frac{-m}{\hbar^2} \int_0^\infty dr'\ \frac{\hbar}{iqr'} (e^{iqr'/\hbar} - e^{-iqr'/\hbar}) V(r') r'{}^2 \\ = -\frac{2m}{q\hbar} \int_0^\infty dr'\ r' V(r') \sin \left( \frac{qr'}{\hbar} \right) Note again that the \theta dependence is buried in q. If we assume elastic scattering (energy is conserved, so k = k'), then we can simplify the formula further: rotating to the plane containing \vec{k} and \vec{k}', we have q^2 = 2k^2 - 2\vec{k} \cdot \vec{k}' = 2k^2 (1 - \cos^2 \theta) = 4k^2 \sin^2 (\theta/2) or q = 2k \sin (\theta/2).
33.3 Example: Rutherford scattering
As an application, let’s have a look at the classical example of Rutherford scattering, which involves alpha particles (i.e. helium-4 nuclei) scattering off of an atomic nucleus. One way to treat this problem is in terms of only the Coulomb potential of the nucleus, but the Coulomb potential dies off slowly enough that we would want to be more careful about how we treat our scattering states; they would not be “close enough” to free even at very large distances.
Fortunately for us, the extension of the Coulomb potential to infinite distance doesn’t really match the real experimental situation; Rutherford’s actual experiments were with atoms, not nuclei, and atoms are neutral. We just need to adopt a more careful treatment of the nuclear Coulomb potential to include the presence of the electrons that neutralize it. This seems like a significant complication, but there is a really nice trick that we can use to make it very simple to deal with. We start with the spherically symmetric formula for the scattering amplitude from above, and then using the fact that \nabla^2 e^{i\vec{q} \cdot \vec{x}'/\hbar} = \frac{-q^2}{\hbar^2} e^{i\vec{q} \cdot \vec{x}'/\hbar} we can rewrite it as f(\theta) = \frac{m}{2\pi q^2} \int d^3x' \nabla^2 e^{i\vec{q} \cdot \vec{x}'/\hbar} V(\vec{x}') and then integrate by parts to obtain f(\theta) = -\frac{m}{2\pi q^2} \int d^3x' e^{i\vec{q} \cdot \vec{x}'/\hbar} \nabla^2 V(\vec{x}'). So far, this is just a rearrangement of the original equation. But now, we have a rewritten form which is especially nice for dealing with the Coulomb potential, which we know satisfies the Poisson equation: if our probe has charge +Ze, then \nabla^2 V(\vec{x}) = -4\pi Ze \rho(\vec{x}), where \rho(\vec{x}) is the charge density of the target. Keeping our specific application in mind, our scattering probe is now significantly massive compared to our target; we haven’t discussed this case before, but it just requires a very simple modification, which is working in terms of the reduced mass \mu instead of the probe mass m. Plugging back in with this change, the scattering amplitude is f(\theta) = \frac{2Ze\mu}{q^2} \int d^3x' e^{i\vec{q} \cdot \vec{x}'/\hbar} \rho(\vec{x}) = \frac{2Ze\mu}{q^2} F(q), defining the form factor F(q) as the Fourier transform of the charge density \rho(\vec{x}). We obtain for the differential cross section \frac{d\sigma}{d\Omega} = |f(\theta)|^2 = \frac{4Z^2 e^2 \mu^2}{q^4} |F(q)|^2.
To understand this further, let’s start with the example of how a naive treatment of Rutherford scattering would proceed, only including the nucleus as a point charge: \rho_{\rm pt}(\vec{r}) = Z'e \delta^{(3)}(\vec{r}). Then the Fourier transform just gives a constant form factor, F_{\rm pt}(q) = Z'e. Substituting in, \frac{d\sigma_{\rm pt}}{d\Omega} = \frac{4Z^2 Z'{}^2 e^4 \mu^2}{q^4}. = \frac{Z^2 Z'{}^2 e^4 \mu^2}{k^4 \sin^4 (\theta/2)}. This is the classic Rutherford result for the angular dependence, matching the experimental observations of Geiger and Marsden for alpha scattering. However, if we think about this result carefully, there are a couple of features that don’t quite match with the story for Rutherford scattering:
In simple terms, the “surprise” of Rutherford scattering is typically stated as the presence of large backscattering, which comes from alpha particles that pass very close to the nucleus and scatter back at large angles. But the Rutherford formula, apparently, shows the opposite of this effect; the cross section is minimized at \theta = \pi in the backwards direction. How does the idea of “large backscattering” make sense?
Although the angular dependence matches well on to experiment, the differential cross section is badly divergent as q \rightarrow 0, which includes \theta \rightarrow 0 at fixed energy. If we try to integrate over d\Omega to recover the total cross section, we find that it is infinity! This doesn’t give confidence that we’re treating things correctly here. (And indeed, if we’re using a delta-function charge distribution, we should worry that our scattering states are not really being dealt with properly at infinity since we have a 1/r potential everywhere in space.)
The form factor treatment lets us understand both of these problems and put together a more complete picture of the physics. Let’s deal with the second issue first. For neutral atoms there is of course no 1/r potential extending to infinity; the nuclear electric charge is balanced by the charge of the electrons. This means that on very general grounds, we expect that as q \rightarrow 0 so we’re looking at the very long-distance part of our charge distribution, we should find F(q) \rightarrow 0. This will regulate the divergence in q and give us a finite total cross section.
To be concrete, let’s compute the correct Rutherford cross-section for hydrogen scattering (just so we can compute explicitly.) The charge density of the electron is just the probability density of its wavefunction, times its charge. So for hydrogen in the ground state, we have \rho(\vec{x}) = e \delta^{(3)}(\vec{x}) - e |\psi_{100}(r)|^2 \\ = e \left[ \delta^{(3)}(\vec{x}) - \frac{1}{\pi a_0^3} e^{-2r/a_0} \right] substituting in (Y_0^0)^2 = 1/(4\pi) in the second term. Now we take the Fourier transform: F(q) = e \int d^3x \left[ \delta^{(3)}(\vec{x}) - \frac{1}{\pi a_0^3} e^{-2Zr/a_0} \right] e^{i\vec{q} \cdot \vec{x}/\hbar}
and a few lines of algebra leads us to the result: F(q) = e \left[ 1 - \left( 1 + \frac{q^2 a_0^2}{4\hbar^2} \right)^{-2} \right]
Work out the Fourier transform set up above to show the given result for the form factor F(q) for ground-state hydrogen.
Answer:
Continuing from the first line above: F(q) = e - \frac{2e}{a_0^3} \int_0^\infty dr\ r^2 \int_{-1}^1 d(\cos \theta) e^{iqr \cos \theta/\hbar} e^{-2r/a_0} \\ = e - \frac{2e}{a_0^3} \int_0^\infty dr\ r^2 \frac{\hbar}{iqr} \left[ e^{iqr/\hbar} - e^{-iqr/\hbar} \right] e^{-2r/a_0} \\ = e - \frac{2e\hbar}{iqa_0^3} \int_0^\infty dr\ r \left[ e^{r(iq/\hbar-2/a_0)} - e^{-r(iq/\hbar+2/a_0)} \right] To simplify the remaining integrals, we integrate by parts, \int_0^\infty dr\ r e^{-cr} = -\frac{1}{c} \int_0^\infty e^{-cr} = \frac{1}{c^2}, (valid as long as the real part of c > 0 so the boundary term vanishes), giving F(q) = e - \frac{2e\hbar}{iqa_0^3} \left[ \frac{1}{(2/a_0 - iq/\hbar)^2} - \frac{1}{(2/a_0 + iq/\hbar)^2}\right] \\ = e - \frac{2e\hbar}{iqa_0} \left[ \frac{1}{(2 - iqa_0/\hbar)^2} - \frac{1}{(2 + iqa_0/\hbar)^2} \right] \\ = e - \frac{2e\hbar}{iqa_0} \left[ \frac{(2 + iqa_0/\hbar)^2 - (2 - iqa_0/\hbar)^2}{(2-iqa_0/\hbar)^2(2+iqa_0/\hbar)^2}\right] \\ = e - \frac{2e\hbar}{iqa_0} \left[ \frac{8iqa_0/\hbar}{(q^2 a_0^2 / \hbar^2 + 4)^2} \right] \\ = e - 16e \frac{1}{(q^2 a_0^2 / \hbar^2 + 4)^2} \\ = e \left[ 1 - \left( 1 + \frac{q^2 a_0^2}{4\hbar^2} \right)^{-2} \right] recovering the result above.
If q a_0 \gg 1, then this just becomes the point-charge form factor of e, and we recover the Rutherford result from above. But at low energies qa_0 \ll 1, the leading behavior is different. It’s helpful to expand: F(q) \approx e \left[ 1 - \left( 1 - \frac{q^2 a_0^2}{2\hbar^2} + ... \right) \right] = -\frac{ea_0^2}{2\hbar^2} q^2 = -eR_H^2 (q/\hbar)^2, defining the hydrogen charge radius as R_H = \frac{a_0}{\sqrt{2}}. When expanding the charge form factor at low q^2, the charge radius is the leading deviation from simple point-charge behavior, and it provides one way of defining the size of a charged object (in this case, more or less just the Bohr radius again, as we might have guessed). But now if we plug back in to the differential cross section, we get \frac{d\sigma}{d\Omega} \approx \frac{4Z^2 e^4 \mu^2}{q^4} R_H^4 (q/\hbar)^4 = \frac{Z^2 e^4 \mu^2}{\hbar^4} a_0^4, which is no longer divergent, but is simply a constant instead. Taking the form factor into account regulates our calculation of the total cross-section, and it will now be finite at any q.
Of course, if we go to very large q we’ll find another modification needs to be done, which is that the nucleus of the atom itself is really a charge distribution with a finite size. For hydrogen, the proton itself has a charge radius of about 1 fm \sim 10^{-5} a_0. This corresponds to an energy scale of around 1 GeV (10^9 = 1 billion eV!), which we would need to get close to in order to probe the finite size of the proton. (In practice, there is another important effect that we’re neglecting, which is that alpha particle can also interact with another nucleus by a strong nuclear interaction. This requires much less energy to see, and leads to deviations from the Rutherford formula for alpha particles with energies of around 30 MeV, much lower than 1 GeV.)
Finally, we can go back to the first question: why is the Rutherford result phrased in terms of “large backscattering” when that is where the cross section is smallest? The answer is in what we are comparing the Rutherford result to, which historically would be Thomson’s “plum pudding” model in which the positive charge is distributed evenly over the whole atom. As a simple model for this, let’s just take a constant radial charge density out to a_0, \rho_{T,+} = \begin{cases} \frac{3Z}{4\pi a_0^3}, & r \leq a_0; \\ 0, & r > a_0. \end{cases}
The resulting form factor for the positive charge would be F_{T,+}(q) = \frac{4\pi \hbar}{q} \int_0^{a_0} dr\ r \left( \frac{3Z}{4\pi a_0^3} \right) \sin \left( \frac{qr}{\hbar} \right) \\ = \frac{3Z \hbar^2}{(qa_0)^3} \left[ \sin \left(\frac{qa_0}{\hbar}\right) - qa_0 \cos \left(\frac{qa_0}{\hbar}\right) \right] If we plug back in at low q^2, we see that |F_{T,+}(q)|^2 \sim 1/q^4, so the overall cross section goes as 1/q^8, dying off much more rapidly with large momentum. Moreover, what we really care about is the difference at high momentum. For large q (which occurs at large \theta \sim \pi since q = 2k \sin (\theta/2)), this form factor will be rapidly oscillating as a function of \theta; the amplitude will average to zero, and the cross section will vanish. We see that the Thomson model cross-section for a blob of positive charge predicts a much smaller amount of back-scattered high-q events compared to the Rutherford prediction with a point-charge nucleus present.