30 Jul 2025
Quantum Scattering Theory
Cross sections, wave packets, Green functions, partial waves, Coulomb and Born scattering, the optical theorem, phase shifts, scattering length, and effective range.
For a localized potential $V(\mathbf r)$ and energy $E=\hbar^2k^2/(2\mu)$, choose the incident wave along $z$. The stationary scattering state has the large-distance form
\[\boxed{ \psi(\mathbf r)\underset{r\to\infty}{\sim} e^{ikz}+f(\theta,\phi)\frac{e^{ikr}}r. }\]The first term is the incident plane wave; the second is an outgoing spherical wave. This boundary condition excludes incoming waves from infinity.
Differential and total cross sections
The probability current is
\[\mathbf j=\frac{\hbar}{2\mu i} (\psi^*\nabla\psi-\psi\nabla\psi^*).\]For $e^{ikz}$, $j_{\rm in}=\hbar k/\mu$. For the outgoing term, keeping the leading power at large $r$,
\[j_{\mathrm{sc},r} =\frac{\hbar k}{\mu r^2}\lvert f(\theta,\phi)\rvert^2.\]The rate crossing $r^2d\Omega$ divided by incident flux is therefore
\[\boxed{\frac{d\sigma}{d\Omega}=\lvert f(\theta,\phi)\rvert^2,} \qquad \boxed{\sigma_{\rm tot}=\int\lvert f\rvert^2d\Omega.}\]$f$ has dimensions of length and a cross section has dimensions of area.
Scattering of a wave packet
A physical incident state is a packet,
\[\Psi(\mathbf r,t)=\frac{1}{(2\pi)^{3/2}}\int a(\mathbf k) \psi_{\mathbf k}^{(+)}(\mathbf r) e^{-i\hbar k^2t/(2\mu)}d^3k,\]where the scattering states use the plane-wave convention $\langle\psi_{\mathbf k}^{(+)}\mid\psi_{\mathbf k’}^{(+)}\rangle =(2\pi)^3\delta^3(\mathbf k-\mathbf k’)$. Thus $a(\mathbf k)$ is concentrated near $\mathbf k_0$ and $\int\lvert a(\mathbf k)\rvert^2d^3k=1$. Before collision, the incoming part is localized; after collision, the outgoing packet separates spatially. The stationary cross section is recovered when the packet is narrow enough that $f$ and the incident speed vary negligibly across its momentum width.
Green function and scattering amplitude
The stationary Schrödinger equation is
\[(E-H_0)\psi=V\psi, \qquad H_0=-\frac{\hbar^2}{2\mu}\nabla^2.\]Define the outgoing Green function by
\[(E-H_0)G^{(+)}(\mathbf r-\mathbf r') =\delta^3(\mathbf r-\mathbf r').\]Since
\[(\nabla^2+k^2)\frac{e^{ikR}}R=-4\pi\delta^3(\mathbf R),\]its normalization is
\[\boxed{ G^{(+)}(\mathbf R) =-\frac{\mu}{2\pi\hbar^2}\frac{e^{ikR}}R. }\]Adding a free incident solution gives the Lippmann-Schwinger equation
\[\psi(\mathbf r)=e^{i\mathbf k\cdot\mathbf r} -\frac{\mu}{2\pi\hbar^2} \int\frac{e^{ik\lvert\mathbf r-\mathbf r'\rvert}} {\lvert\mathbf r-\mathbf r'\rvert} V(\mathbf r')\psi(\mathbf r')d^3r'.\]For $r$ much larger than the range of the potential,
\[\lvert\mathbf r-\mathbf r'\rvert\simeq r-\hat{\mathbf r}\cdot\mathbf r', \qquad \frac1{\lvert\mathbf r-\mathbf r'\rvert}\simeq\frac1r.\]Comparison with the asymptotic definition identifies $\mathbf k’=k\hat{\mathbf r}$ and
\[\boxed{ f(\mathbf k',\mathbf k) =-\frac{\mu}{2\pi\hbar^2} \int e^{-i\mathbf k'\cdot\mathbf r'} V(\mathbf r')\psi(\mathbf r')d^3r'. }\]This formula is exact; its unknown scattering state remains inside the integral.
Born approximation
When the scattered wave is small inside the interaction region, replace $\psi(\mathbf r’)$ by the incident plane wave. The first Born amplitude is
\[\boxed{ f_B(\mathbf q) =-\frac{\mu}{2\pi\hbar^2} \int e^{-i\mathbf q\cdot\mathbf r}V(\mathbf r)d^3r, \qquad \mathbf q=\mathbf k'-\mathbf k. }\]Elastic scattering has $\lvert\mathbf k’\rvert=\lvert\mathbf k\rvert=k$, so
\[q^2=\lvert\mathbf k'-\mathbf k\rvert^2 =2k^2(1-\cos\theta)=4k^2\sin^2\frac\theta2.\]The approximation requires the potential-induced wave to remain small; it is generally reliable for weak potentials or sufficiently large incident energy.
Scattering in a Coulomb field
Let $V(r)=\kappa_C/r$. Because the Coulomb field is long-ranged, introduce $e^{-\eta r}$ and take $\eta\to0^+$ after transforming:
\[\begin{aligned} \int e^{-i\mathbf q\cdot\mathbf r} \frac{e^{-\eta r}}r\,d^3r &=\frac{4\pi}{q}\int_0^\infty e^{-\eta r}\sin(qr)dr\\ &=\frac{4\pi}{q^2+\eta^2} \longrightarrow\frac{4\pi}{q^2}. \end{aligned}\]Thus the screened Born limit gives
\[f_C(\theta) =-\frac{2\mu\kappa_C}{\hbar^2q^2} =-\frac{\mu\kappa_C} {2\hbar^2k^2\sin^2(\theta/2)}.\]Using $E=\hbar^2k^2/(2\mu)$,
\[\boxed{ \frac{d\sigma_C}{d\Omega} =\frac{\kappa_C^2}{16E^2\sin^4(\theta/2)}. }\]This is the Rutherford angular dependence. The unscreened Coulomb potential also produces a long-range logarithmic phase, so the short-range asymptotic phase conventions used below must not be applied to it without modification.
Partial-wave analysis and phase shifts
For a central short-range potential, expand the incident wave as
\[e^{ikz}=\sum_{l=0}^{\infty} i^l(2l+1)j_l(kr)P_l(\cos\theta).\]Since
\[j_l(kr)\sim\frac1{kr} \sin\left(kr-\frac{l\pi}{2}\right),\]the incident plane wave contains, in each $l$ channel, equal incoming and outgoing radial flux. Outside the potential range, the radial solution obeys $u_l’’+[k^2-l(l+1)/r^2]u_l=0$ and may be written
\[\boxed{ u_l(r)\sim A_l \sin\left(kr-\frac{l\pi}{2}+\delta_l\right). }\]$\delta_l$ is the phase shift. Writing the sine as two exponentials shows that the outgoing-to-incoming ratio changes from its free value by
\[S_l=e^{2i\delta_l}.\]Elastic flux conservation gives $\lvert S_l\rvert=1$, so $\delta_l$ is real. Keep the incoming coefficient equal to the plane-wave value and replace only the outgoing coefficient by $S_l$. Subtracting the free outgoing part gives
\[f(\theta)=\frac1{2ik}\sum_{l=0}^{\infty} (2l+1)(S_l-1)P_l(\cos\theta).\]Since $(e^{2i\delta_l}-1)/(2i)=e^{i\delta_l}\sin\delta_l$,
\[\boxed{ f(\theta)=\frac1k\sum_{l=0}^{\infty} (2l+1)e^{i\delta_l}\sin\delta_lP_l(\cos\theta). }\]Legendre orthogonality,
\[\int P_l(\cos\theta)P_{l'}(\cos\theta)d\Omega =\frac{4\pi}{2l+1}\delta_{ll'},\]then gives
\[\boxed{ \sigma_{\rm tot}=\frac{4\pi}{k^2} \sum_{l=0}^{\infty}(2l+1)\sin^2\delta_l. }\]
Optical theorem
At $\theta=0$, $P_l(1)=1$, so
\[\operatorname{Im}f(0) =\frac1k\sum_l(2l+1) \operatorname{Im}[e^{i\delta_l}\sin\delta_l] =\frac1k\sum_l(2l+1)\sin^2\delta_l.\]Comparison with the total cross section gives
\[\boxed{ \sigma_{\rm tot}=\frac{4\pi}{k}\operatorname{Im}f(0). }\]The relation follows from probability conservation, even though it connects the total cross section to the strictly forward amplitude.
Scattering length and effective range
At sufficiently small $k$, a finite-range potential is dominated by $l=0$. Its amplitude is
\[f_0(k)=\frac{e^{2i\delta_0}-1}{2ik} =\boxed{\frac1{k\cot\delta_0-ik}}.\]At zero energy and outside the potential range, the radial equation is $u_0’‘=0$. Normalize its solution as
\[u_0(r)\longrightarrow1-\frac ra.\]The intercept $a$ is the scattering length. Matching this limit to $\sin(kr+\delta_0)$ gives
\[\lim_{k\to0}k\cot\delta_0=-\frac1a, \qquad f_0(0)=-a, \qquad \boxed{\sigma_{k\to0}=4\pi a^2.}\]For a short-range potential, the next even power defines the effective range:
\[\boxed{ k\cot\delta_0=-\frac1a+\frac12r_ek^2+O(k^4). }\]If the zero-energy solution is normalized as above, comparison of the finite- and zero-energy radial equations gives Bethe’s integral form
\[\boxed{ r_e=2\int_0^\infty \left[\left(1-\frac ra\right)^2-u_0^2(r)\right]dr. }\]The integral terminates effectively outside the interaction range because the two terms then coincide. Substitution into $f_0(k)$ describes the leading finite-energy correction to scattering-length behavior.
The outgoing Green equation for $r>0$, screened-Coulomb transform, Rutherford form, partial-wave optical theorem, and scattering-length limit are checked in the Maxima worksheet; every printed residual is zero.
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