# Second order Born approximation in scattering theory

Reference: Griffiths, David J. (2005), Introduction to Quantum Mechanics, 2nd Edition; Pearson Education – Problem 11.15.

The first order Born approximation in scattering theory is derived from the integral form of the Schrödinger equation

$\displaystyle \psi\left(\mathbf{r}\right)=\psi_{0}\left(\mathbf{r}\right)+\int g\left(\mathbf{r}-\mathbf{r}_{0}\right)V\left(\mathbf{r}_{0}\right)\psi\left(\mathbf{r}_{0}\right)d^{3}\mathbf{r}_{0} \ \ \ \ \ (1)$

where

$\displaystyle g\left(\mathbf{r}-\mathbf{r}_{0}\right)\equiv-\frac{m}{2\pi\hbar^{2}}\frac{e^{ik\left|\mathbf{r}-\mathbf{r}_{0}\right|}}{\left|\mathbf{r}-\mathbf{r}_{0}\right|} \ \ \ \ \ (2)$

is the Green’s function and ${\psi_{0}}$ is a free particle wave function. The first order Born approximation assumes that the wave function isn’t changed much by the scattering process, so we can approximate the integral equation by replacing the full wave function ${\psi}$ in the integrand by the incident wave function:

$\displaystyle \psi\left(\mathbf{r}\right)=\psi_{0}\left(\mathbf{r}\right)+\int g\left(\mathbf{r}-\mathbf{r}_{0}\right)V\left(\mathbf{r}_{0}\right)\psi_{0}\left(\mathbf{r}_{0}\right)d^{3}\mathbf{r}_{0} \ \ \ \ \ (3)$

We can generate a second order approximation by inserting 3 into the integral in 1. When we do this, it’s important to keep track of the position vectors that apply to each integration. We’ll relabel ${\mathbf{r}_{0}}$ as ${\mathbf{r}_{1}}$ in 1, and then insert 3 into it:

 $\displaystyle \psi\left(\mathbf{r}\right)$ $\displaystyle =$ $\displaystyle \psi_{0}\left(\mathbf{r}\right)+\int g\left(\mathbf{r}-\mathbf{r}_{1}\right)V\left(\mathbf{r}_{1}\right)\psi\left(\mathbf{r}_{1}\right)d^{3}\mathbf{r}_{1}\ \ \ \ \ (4)$ $\displaystyle$ $\displaystyle \approx$ $\displaystyle \psi_{0}\left(\mathbf{r}\right)+\int g\left(\mathbf{r}-\mathbf{r}_{1}\right)V\left(\mathbf{r}_{1}\right)\psi_{0}\left(\mathbf{r}_{1}\right)d^{3}\mathbf{r}_{1}+\ \ \ \ \ (5)$ $\displaystyle$ $\displaystyle$ $\displaystyle \int g\left(\mathbf{r}-\mathbf{r}_{1}\right)V\left(\mathbf{r}_{1}\right)\int g\left(\mathbf{r}_{1}-\mathbf{r}_{0}\right)V\left(\mathbf{r}_{0}\right)\psi_{0}\left(\mathbf{r}_{0}\right)d^{3}\mathbf{r}_{0}d^{3}\mathbf{r}_{1}\nonumber$

The first two terms just repeat the first order Born approximation, so to calculate the second order approximation we need to work out the third term involving the double integral. Using 2 we have for this term (which we’ll call ${I_{2}}$)

$\displaystyle I_{2}=\left(\frac{m}{2\pi\hbar^{2}}\right)^{2}\int\int\frac{e^{ik\left|\mathbf{r}-\mathbf{r}_{1}\right|}}{\left|\mathbf{r}-\mathbf{r}_{1}\right|}\frac{e^{ik\left|\mathbf{r}_{1}-\mathbf{r}_{0}\right|}}{\left|\mathbf{r}_{1}-\mathbf{r}_{0}\right|}V\left(\mathbf{r}_{1}\right)V\left(\mathbf{r}_{0}\right)\psi_{0}\left(\mathbf{r}_{0}\right)d^{3}\mathbf{r}_{0}d^{3}\mathbf{r}_{1} \ \ \ \ \ (6)$

In the original derivation, we simplified things by taking ${\mathbf{r}}$ to point to the detector, which is assumed to be very far from the scattering region. This allows us to approximate the Green’s function:

$\displaystyle g\left(\mathbf{r}-\mathbf{r}_{0}\right)\approx-\frac{m}{2\pi\hbar^{2}}\frac{e^{ikr}}{r}e^{-i\mathbf{k}\cdot\mathbf{r}_{0}} \ \ \ \ \ (7)$

This approximation is still valid for the first factor in the integrand in 6, but not for the second factor, since ${\mathbf{r}_{0}}$ and ${\mathbf{r}_{1}}$ are position vectors that both refer to locations within the scattering region (that is, where ${V\ne0}$). Therefore we can write

$\displaystyle I_{2}\approx\left(\frac{m}{2\pi\hbar^{2}}\right)^{2}\frac{e^{ikr}}{r}\int\int e^{-i\mathbf{k}\cdot\mathbf{r}_{1}}\frac{e^{ik\left|\mathbf{r}_{1}-\mathbf{r}_{0}\right|}}{\left|\mathbf{r}_{1}-\mathbf{r}_{0}\right|}V\left(\mathbf{r}_{1}\right)V\left(\mathbf{r}_{0}\right)\psi_{0}\left(\mathbf{r}_{0}\right)d^{3}\mathbf{r}_{0}d^{3}\mathbf{r}_{1} \ \ \ \ \ (8)$

Example Soft-sphere scattering. A soft sphere is defined by the potential

$\displaystyle V\left(\mathbf{r}\right)=\begin{cases} V_{0} & r\le a\\ 0 & r>a \end{cases} \ \ \ \ \ (9)$

where ${V_{0}>0}$ is a constant. Griffiths works out the first order Born approximation for the scattering amplitude in his example 11.4, so we’ll consider the second order term here. We’ll also consider only the case of low energy scattering, which is defined by the condition ${ka\ll1}$. We start by writing out 8 for this case.

$\displaystyle I_{2}\approx\left(\frac{m}{2\pi\hbar^{2}}\right)^{2}\frac{e^{ikr}}{r}V_{0}^{2}\int\int e^{-i\mathbf{k}\cdot\mathbf{r}_{1}}\frac{e^{ik\left|\mathbf{r}_{1}-\mathbf{r}_{0}\right|}}{\left|\mathbf{r}_{1}-\mathbf{r}_{0}\right|}\psi_{0}\left(\mathbf{r}_{0}\right)d^{3}\mathbf{r}_{0}d^{3}\mathbf{r}_{1} \ \ \ \ \ (10)$

To do the integral, we use an incident plane wave for ${\psi_{0}}$:

 $\displaystyle \psi_{0}\left(\mathbf{r}_{0}\right)$ $\displaystyle =$ $\displaystyle Ae^{i\mathbf{k}'\cdot\mathbf{r}_{0}}\ \ \ \ \ (11)$ $\displaystyle \mathbf{k}'$ $\displaystyle \equiv$ $\displaystyle k\hat{\mathbf{z}} \ \ \ \ \ (12)$

If we do the ${\mathbf{r}_{0}}$ integration first, we can take the polar axis to be along ${\mathbf{r}_{1}}$. Then

 $\displaystyle \left|\mathbf{r}_{1}-\mathbf{r}_{0}\right|$ $\displaystyle =$ $\displaystyle \sqrt{r_{1}^{2}+r_{0}^{2}-2r_{0}r_{1}\cos\theta}\ \ \ \ \ (13)$ $\displaystyle I_{2}$ $\displaystyle \approx$ $\displaystyle \left(\frac{m}{2\pi\hbar^{2}}\right)^{2}\frac{Ae^{ikr}}{r}V_{0}^{2}\int\int e^{-i\mathbf{k}\cdot\mathbf{r}_{1}}\frac{e^{ik\sqrt{r_{1}^{2}+r_{0}^{2}-2r_{0}r_{1}\cos\theta}}}{\sqrt{r_{1}^{2}+r_{0}^{2}-2r_{0}r_{1}\cos\theta}}e^{i\mathbf{k}'\cdot\mathbf{r}_{0}}d^{3}\mathbf{r}_{0}d^{3}\mathbf{r}_{1} \ \ \ \ \ (14)$

We can’t really make much headway with this integral without invoking the low energy assumption. In that case, all the exponentials are approximately 1, so we get

$\displaystyle I_{2}\approx\left(\frac{m}{2\pi\hbar^{2}}\right)^{2}\frac{Ae^{ikr}}{r}V_{0}^{2}\int\int\frac{d^{3}\mathbf{r}_{0}d^{3}\mathbf{r}_{1}}{\sqrt{r_{1}^{2}+r_{0}^{2}-2r_{0}r_{1}\cos\theta}} \ \ \ \ \ (15)$

Doing the ${\mathbf{r}_{0}}$ first, we have

 $\displaystyle \int\frac{d^{3}\mathbf{r}_{0}}{\sqrt{r_{1}^{2}+r_{0}^{2}-2r_{0}r_{1}\cos\theta}}$ $\displaystyle =$ $\displaystyle \int_{0}^{a}\int_{0}^{\pi}\int_{0}^{2\pi}\frac{r_{0}^{2}\sin\theta\; d\phi d\theta dr_{0}}{\sqrt{r_{1}^{2}+r_{0}^{2}-2r_{0}r_{1}\cos\theta}}\ \ \ \ \ (16)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 2\pi\int_{0}^{a}\int_{0}^{\pi}\frac{r_{0}^{2}\sin\theta\; d\theta dr_{0}}{\sqrt{r_{1}^{2}+r_{0}^{2}-2r_{0}r_{1}\cos\theta}}\ \ \ \ \ (17)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{2\pi}{r_{1}}\int_{0}^{a}r_{0}\left[r_{0}+r_{1}-\left|r_{0}-r_{1}\right|\right]dr_{0}\ \ \ \ \ (18)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{2\pi}{r_{1}}\int_{0}^{r_{1}}r_{0}\left[r_{0}+r_{1}-\left(r_{1}-r_{0}\right)\right]dr_{0}+\ \ \ \ \ (19)$ $\displaystyle$ $\displaystyle$ $\displaystyle \frac{2\pi}{r_{1}}\int_{r_{1}}^{a}r_{0}\left[r_{0}+r_{1}-\left(r_{0}-r_{1}\right)\right]dr_{0}\nonumber$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{4}{3}\pi r_{1}^{2}+2\pi a^{2}-2\pi r_{1}^{2}\ \ \ \ \ (20)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 2\pi a^{2}-\frac{2}{3}\pi r_{1}^{2} \ \ \ \ \ (21)$

The integral over ${\mathbf{r}_{1}}$ is now easy, and we get

 $\displaystyle I_{2}$ $\displaystyle \approx$ $\displaystyle \left(\frac{m}{2\pi\hbar^{2}}\right)^{2}\frac{Ae^{ikr}}{r}V_{0}^{2}\int_{0}^{a}\int_{0}^{\pi}\int_{0}^{2\pi}\left(2\pi a^{2}-\frac{2}{3}\pi r_{1}^{2}\right)r_{1}^{2}\sin\theta\; d\phi d\theta dr_{1}\ \ \ \ \ (22)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{Ae^{ikr}}{r}\frac{8m^{2}V_{0}^{2}a^{5}}{15\hbar^{4}} \ \ \ \ \ (23)$

The second order correction to the scattering amplitude is the coefficient of ${\frac{Ae^{ikr}}{r}}$ so the correction is

$\displaystyle f_{2}=\frac{8m^{2}V_{0}^{2}a^{5}}{15\hbar^{4}} \ \ \ \ \ (24)$

Combining this with the first order correction taken from Griffiths’s Example 11.4 we get

 $\displaystyle f\left(\theta,\phi\right)$ $\displaystyle \approx$ $\displaystyle -\frac{2ma^{3}V_{0}}{3\hbar^{2}}+\frac{8m^{2}V_{0}^{2}a^{5}}{15\hbar^{4}}\ \ \ \ \ (25)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{2ma^{3}V_{0}}{3\hbar^{2}}\left(1-\frac{4mV_{0}a^{2}}{5\hbar^{2}}\right) \ \ \ \ \ (26)$

# Optical theorem

Reference: Griffiths, David J. (2005), Introduction to Quantum Mechanics, 2nd Edition; Pearson Education – Problem 11.19.

There is a simple relationship between the total cross section and the scattering amplitude in three dimensional scattering. The formulas are (in terms of phase shifts):

 $\displaystyle f\left(\theta\right)$ $\displaystyle =$ $\displaystyle \frac{1}{k}\sum_{l=0}^{\infty}\left(2l+1\right)e^{i\delta_{l}}\sin\delta_{l}P_{l}\left(\cos\theta\right)\ \ \ \ \ (1)$ $\displaystyle \sigma$ $\displaystyle =$ $\displaystyle \frac{4\pi}{k^{2}}\sum_{l=0}^{\infty}\left(2l+1\right)\sin^{2}\delta_{l} \ \ \ \ \ (2)$

If ${\theta=0}$, we can use the fact that ${P_{l}\left(1\right)=1}$ for all ${l}$ (from the definition) so taking the imaginary part of ${f\left(0\right)}$ we get

$\displaystyle \Im f\left(0\right)=\frac{1}{k}\sum_{l=0}^{\infty}\left(2l+1\right)\sin^{2}\delta_{l} \ \ \ \ \ (3)$

from which we get the optical theorem:

$\displaystyle \sigma=\frac{4\pi}{k}\Im f\left(0\right) \ \ \ \ \ (4)$

# Born approximation in one dimension

Reference: Griffiths, David J. (2005), Introduction to Quantum Mechanics, 2nd Edition; Pearson Education – Problem 11.17.

Using Griffiths’s Green’s function for the one-dimensional Schrödinger equation:

$\displaystyle \psi\left(x\right)=\psi_{0}\left(x\right)-\frac{im}{\hbar^{2}k}\int_{-\infty}^{\infty}e^{ik\left|x-x_{0}\right|}V\left(x_{0}\right)\psi\left(x_{0}\right)dx_{0} \ \ \ \ \ (1)$

we can work out the Born approximation in the one-dimensional case. The idea is we replace ${\psi\left(x_{0}\right)}$ inside the integral by the incident plane wave form ${\psi_{0}\left(x_{0}\right)}$. Assuming that the potential is zero outside a finite distance from the origin, we want the wave function in two regions: ${x\ll0}$ and ${x\gg0}$. The former will give the reflected wave and the latter the transmitted wave.

For ${x\ll0}$ (and at a distance where ${V\left(x\right)=0}$) we have

 $\displaystyle e^{ik\left|x-x_{0}\right|}$ $\displaystyle =$ $\displaystyle e^{-ikx}e^{ikx_{0}}\ \ \ \ \ (2)$ $\displaystyle \psi_{0}\left(x_{0}\right)$ $\displaystyle =$ $\displaystyle Ae^{ikx_{0}} \ \ \ \ \ (3)$

where ${A}$ is the normalization constant. The Born approximation for this region is

 $\displaystyle \psi\left(x\right)$ $\displaystyle =$ $\displaystyle Ae^{ikx}-\frac{im}{\hbar^{2}k}e^{-ikx}\int_{-\infty}^{\infty}e^{ikx_{0}}V\left(x_{0}\right)Ae^{ikx_{0}}dx_{0}\ \ \ \ \ (4)$ $\displaystyle$ $\displaystyle =$ $\displaystyle A\left[e^{ikx}-\frac{im}{\hbar^{2}k}e^{-ikx}\int_{-\infty}^{\infty}e^{2ikx_{0}}V\left(x_{0}\right)dx_{0}\right] \ \ \ \ \ (5)$

The scattering amplitude and reflection coefficient for the reflected particle are therefore

 $\displaystyle f_{R}$ $\displaystyle =$ $\displaystyle -\frac{im}{\hbar^{2}k}\int_{-\infty}^{\infty}e^{2ikx_{0}}V\left(x_{0}\right)dx_{0}\ \ \ \ \ (6)$ $\displaystyle R$ $\displaystyle =$ $\displaystyle \left|f_{R}\right|^{2}\ \ \ \ \ (7)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left(\frac{m}{\hbar^{2}k}\right)^{2}\left|\int_{-\infty}^{\infty}e^{2ikx_{0}}V\left(x_{0}\right)dx_{0}\right|^{2} \ \ \ \ \ (8)$

[Note that we’re assuming that ${x so although the limits of the integral are infinite, we’re implicitly assuming that ${V=0}$ for all ${x_{0}>x}$ so the integral isn’t really over an infinite range.]

For ${x\gg0}$ we have

 $\displaystyle e^{ik\left|x-x_{0}\right|}$ $\displaystyle =$ $\displaystyle e^{ikx}e^{-ikx_{0}}\ \ \ \ \ (9)$ $\displaystyle \psi_{0}\left(x_{0}\right)$ $\displaystyle =$ $\displaystyle Ae^{ikx_{0}} \ \ \ \ \ (10)$

The Born approximation here is

 $\displaystyle \psi\left(x\right)$ $\displaystyle =$ $\displaystyle Ae^{ikx}-\frac{im}{\hbar^{2}k}e^{ikx}\int_{-\infty}^{\infty}e^{-ikx_{0}}V\left(x_{0}\right)Ae^{ikx_{0}}dx_{0}\ \ \ \ \ (11)$ $\displaystyle$ $\displaystyle =$ $\displaystyle Ae^{ikx}\left[1-\frac{im}{\hbar^{2}k}\int_{-\infty}^{\infty}V\left(x_{0}\right)dx_{0}\right] \ \ \ \ \ (12)$

The scattering amplitude and transmission coefficient are therefore

 $\displaystyle f_{T}$ $\displaystyle =$ $\displaystyle 1-\frac{im}{\hbar^{2}k}\int_{-\infty}^{\infty}V\left(x_{0}\right)dx_{0}\ \ \ \ \ (13)$ $\displaystyle T$ $\displaystyle =$ $\displaystyle \left|f_{T}\right|^{2}\ \ \ \ \ (14)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 1+\left(\frac{m}{\hbar^{2}k}\right)^{2}\left|\int_{-\infty}^{\infty}V\left(x_{0}\right)dx_{0}\right|^{2} \ \ \ \ \ (15)$

The Born approximation fails for transmission in this case, since ${T>1}$ which is impossible. We can still get an estimate of ${T}$ from ${T=1-R}$.

# Impulse approximation in scattering theory

Reference: Griffiths, David J. (2005), Introduction to Quantum Mechanics, 2nd Edition; Pearson Education – Problem 11.14.

In classical scattering theory, the simplest approximation is the impulse approximation, in which a particle’s path is assumed to be a straight line right through the scattering region, and the total impulse resulting from the component of force perpendicular to the particle’s trajectory is calculated. This impulse ${I}$ is assumed to be a small fraction of the particle’s incoming horizontal momentum ${p}$, so the scattering angle should be small, and given approximately by

$\displaystyle \theta\approx\arctan\frac{I}{p} \ \ \ \ \ (1)$

As an example, we’ll apply the impulse approximation to Rutherford scattering of a charge ${q_{1}}$ travelling with kinetic energy ${E}$ from another charge ${q_{2}}$ at rest. We assume ${q_{1}}$ has an impact parameter ${b}$ (that is, if the particle didn’t interact with the target, it would pass by with a closest approach distance of ${b}$).

The impulse is the change in momentum that a force produces over a given time, so the required impulse from the perpendicular component of a force is

$\displaystyle I=\int F_{\perp}dt \ \ \ \ \ (2)$

We’ll take ${q_{1}}$‘s straight line trajectory to be along the ${x}$ axis and place ${q_{2}}$ at ${y=b}$ on the ${y}$ axis. Then the Coulomb force between ${q_{1}}$ and ${q_{2}}$ is

$\displaystyle F=\frac{q_{1}q_{2}}{4\pi\epsilon_{0}r^{2}} \ \ \ \ \ (3)$

with a perpendicular component of

$\displaystyle F_{\perp}=F\frac{b}{r}=\frac{q_{1}q_{2}b}{4\pi\epsilon_{0}r^{3}} \ \ \ \ \ (4)$

The impulse is

$\displaystyle I=\frac{q_{1}q_{2}b}{4\pi\epsilon_{0}}\int_{-\infty}^{\infty}\frac{dt}{r^{3}} \ \ \ \ \ (5)$

To convert this to an integral over ${r}$ we can use the fact that ${q_{1}}$‘s speed is constant at

$\displaystyle v=\frac{dx}{dt}=\sqrt{\frac{2E}{m}} \ \ \ \ \ (6)$

So we have

 $\displaystyle x$ $\displaystyle =$ $\displaystyle \sqrt{r^{2}-b^{2}}\ \ \ \ \ (7)$ $\displaystyle dx$ $\displaystyle =$ $\displaystyle \frac{r\; dr}{\sqrt{r^{2}-b^{2}}}=\sqrt{\frac{2E}{m}}dt\ \ \ \ \ (8)$ $\displaystyle I$ $\displaystyle =$ $\displaystyle \frac{q_{1}q_{2}b}{4\pi\epsilon_{0}\sqrt{2E/m}}\times2\int_{b}^{\infty}\frac{dr}{r^{2}\sqrt{r^{2}-b^{2}}} \ \ \ \ \ (9)$

where the ${r}$ integral is over the range of ${r}$ from its closest approach when ${r=b}$ out to infinity. We’ve doubled the integral to account for the incoming (${-\infty) and outgoing (${0) legs of the journey.

The integral evaluates to

 $\displaystyle \int_{b}^{\infty}\frac{dr}{r^{2}\sqrt{r^{2}-b^{2}}}$ $\displaystyle =$ $\displaystyle \left.\frac{\sqrt{r^{2}-b^{2}}}{b^{2}r}\right|_{b}^{\infty}\ \ \ \ \ (10)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{b^{2}}\ \ \ \ \ (11)$ $\displaystyle I$ $\displaystyle =$ $\displaystyle \frac{q_{1}q_{2}}{2\pi\epsilon_{0}b\sqrt{2E/m}} \ \ \ \ \ (12)$

The incoming momentum is

$\displaystyle p=m\frac{dx}{dt}=\sqrt{2mE} \ \ \ \ \ (13)$

so

 $\displaystyle \theta$ $\displaystyle \approx$ $\displaystyle \arctan\frac{I}{p}\ \ \ \ \ (14)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \arctan\left[\frac{q_{1}q_{2}}{4\pi\epsilon_{0}bE}\right]\ \ \ \ \ (15)$ $\displaystyle b$ $\displaystyle \approx$ $\displaystyle \frac{q_{1}q_{2}}{4\pi\epsilon_{0}E}\cot\theta\ \ \ \ \ (16)$ $\displaystyle \tan\theta$ $\displaystyle \approx$ $\displaystyle \frac{q_{1}q_{2}}{4\pi\epsilon_{0}bE} \ \ \ \ \ (17)$

 $\displaystyle b$ $\displaystyle =$ $\displaystyle \frac{q_{1}q_{2}}{8\pi\epsilon_{0}E}\cot\frac{\theta}{2}\ \ \ \ \ (18)$ $\displaystyle \tan\frac{\theta}{2}$ $\displaystyle =$ $\displaystyle \frac{q_{1}q_{2}}{8\pi\epsilon_{0}bE} \ \ \ \ \ (19)$

For small ${\theta}$, ${\tan\frac{\theta}{2}\approx\frac{\theta}{2}}$ so

 $\displaystyle \frac{\theta}{2}$ $\displaystyle \approx$ $\displaystyle \frac{q_{1}q_{2}}{8\pi\epsilon_{0}bE}\ \ \ \ \ (20)$ $\displaystyle \theta$ $\displaystyle \approx$ $\displaystyle \frac{q_{1}q_{2}}{4\pi\epsilon_{0}bE} \ \ \ \ \ (21)$

which is consistent with 17.

# Born approximation for a spherical delta function shell

Reference: Griffiths, David J. (2005), Introduction to Quantum Mechanics, 2nd Edition; Pearson Education – Problem 11.13.

Earlier we looked at scattering from a delta function spherical shell for a low energy incident particle, using partial wave analysis. This was a fairly complex task, as it involved matching interior and exterior wave functions at the delta function boundary.

Here, we’ll calculate the scattering amplitude using the first Born approximation. For a spherically symmetric potential, the approximation is

$\displaystyle f\left(\theta\right)\approx-\frac{2m}{\hbar^{2}\kappa}\int_{0}^{\infty}V\left(r\right)r\sin\left(\kappa r\right)dr \ \ \ \ \ (1)$

where

$\displaystyle \kappa\equiv2k\sin\frac{\theta}{2} \ \ \ \ \ (2)$

For a delta function potential

$\displaystyle V\left(r\right)=\alpha\delta\left(r-a\right) \ \ \ \ \ (3)$

where ${\alpha}$ is a constant representing the strength of the delta function, so we get

$\displaystyle f\left(\theta\right)\approx-\frac{2m\alpha a}{\hbar^{2}\kappa}\sin\left(\kappa a\right) \ \ \ \ \ (4)$

For low energy, ${ka\ll1}$ so ${\kappa a\ll1}$ as well, so ${\sin\left(\kappa a\right)\approx\kappa a}$, and we get

$\displaystyle f\left(\theta\right)\approx-\frac{2m\alpha a^{2}}{\hbar^{2}} \ \ \ \ \ (5)$

Our earlier result using partial wave analysis is

$\displaystyle f\left(\theta\right)\approx-\frac{\beta a}{1+\beta} \ \ \ \ \ (6)$

where

$\displaystyle \beta\equiv\frac{2m\alpha a}{\hbar^{2}} \ \ \ \ \ (7)$

This gives a differential cross section and total cross section of

 $\displaystyle \frac{d\sigma}{d\Omega}$ $\displaystyle =$ $\displaystyle \left|f\left(\theta\right)\right|^{2}=\beta^{2}a^{2}\ \ \ \ \ (8)$ $\displaystyle \sigma$ $\displaystyle =$ $\displaystyle 4\pi\beta^{2}a^{2} \ \ \ \ \ (9)$

The low energy result 5 from the Born approximation is, in terms of ${\beta}$:

$\displaystyle f\left(\theta\right)\approx-\beta a \ \ \ \ \ (10)$

so it agrees with the partial wave result if ${\beta\ll1}$. This is equivalent to the condition that ${\alpha\ll\hbar^{2}/2ma}$, in other words, that the potential is weak. This was the main assumption in deriving the Born approximation, so in this limit, the results are consistent.

# Scattering from the Yukawa potential

Reference: Griffiths, David J. (2005), Introduction to Quantum Mechanics, 2nd Edition; Pearson Education – Problems 11.11-11.12.

We’ve looked at the Yukawa potential as an example of the variational principle, so here we’ll look at scattering by a Yukawa potential, using the first Born approximation. The Yukawa potential in its general form is

$\displaystyle V\left(r\right)=\beta\frac{e^{-\mu r}}{r} \ \ \ \ \ (1)$

where ${\beta}$ and ${\mu}$ are constants. Since the potential is spherically symmetric, we can use the Born approximation in the form

$\displaystyle f\left(\theta\right)\approx-\frac{2m}{\hbar^{2}\kappa}\int_{0}^{\infty}V\left(r_{0}\right)r\sin\left(\kappa r\right)dr \ \ \ \ \ (2)$

where

 $\displaystyle \kappa$ $\displaystyle =$ $\displaystyle 2k\sin\frac{\theta}{2}\ \ \ \ \ (3)$ $\displaystyle k$ $\displaystyle =$ $\displaystyle \frac{\sqrt{2mE}}{\hbar} \ \ \ \ \ (4)$

We get

 $\displaystyle f\left(\theta\right)$ $\displaystyle \approx$ $\displaystyle -\frac{2m\beta}{\hbar^{2}\kappa}\int_{0}^{\infty}e^{-\mu r}\sin\left(\kappa r\right)dr\ \ \ \ \ (5)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{2m\beta}{\hbar^{2}\left(\kappa^{2}+\mu^{2}\right)} \ \ \ \ \ (6)$

We did the integral using Maple, but if you want to do it by hand, you can do it with two integrations by parts:

 $\displaystyle \int_{0}^{\infty}e^{-\mu r}\sin\left(\kappa r\right)dr$ $\displaystyle =$ $\displaystyle -\left.\frac{\cos\left(\kappa r\right)e^{-\mu r}}{\kappa}\right|_{0}^{\infty}-\frac{\mu}{\kappa}\int_{0}^{\infty}e^{-\mu r}\cos\left(\kappa r\right)dr\ \ \ \ \ (7)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{\kappa}-\left.\frac{\sin\left(\kappa r\right)e^{-\mu r}}{\kappa}\right|_{0}^{\infty}-\frac{\mu^{2}}{\kappa^{2}}\int_{0}^{\infty}e^{-\mu r}\sin\left(\kappa r\right)dr\ \ \ \ \ (8)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{\kappa}-\frac{\mu^{2}}{\kappa^{2}}\int_{0}^{\infty}e^{-\mu r}\sin\left(\kappa r\right)dr\ \ \ \ \ (9)$ $\displaystyle \left(1+\frac{\mu^{2}}{\kappa^{2}}\right)\int_{0}^{\infty}e^{-\mu r}\sin\left(\kappa r\right)dr$ $\displaystyle =$ $\displaystyle \frac{1}{\kappa}\ \ \ \ \ (10)$ $\displaystyle \int_{0}^{\infty}e^{-\mu r}\sin\left(\kappa r\right)dr$ $\displaystyle =$ $\displaystyle \frac{\kappa}{\mu^{2}+\kappa^{2}} \ \ \ \ \ (11)$

We can find the total cross section by integrating the differential cross section over solid angle:

 $\displaystyle \frac{d\sigma}{d\Omega}$ $\displaystyle =$ $\displaystyle \left|f\left(\theta\right)\right|^{2}\ \ \ \ \ (12)$ $\displaystyle \sigma$ $\displaystyle =$ $\displaystyle \frac{4m^{2}\beta^{2}}{\hbar^{4}}\int_{0}^{\pi}d\theta\int_{0}^{2\pi}d\phi\;\frac{\sin\theta}{\left(\kappa^{2}+\mu^{2}\right)^{2}}\ \ \ \ \ (13)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{8\pi m^{2}\beta^{2}}{\hbar^{4}}\int_{0}^{\pi}d\theta\frac{\sin\theta}{\left(4k^{2}\sin^{2}\frac{\theta}{2}+\mu^{2}\right)^{2}}\ \ \ \ \ (14)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{16\pi m^{2}\beta^{2}}{\mu^{2}\hbar^{4}\left(\mu^{2}+4k^{2}\right)}\ \ \ \ \ (15)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{16\pi m^{2}\beta^{2}}{\mu^{2}\hbar^{2}\left(\mu^{2}\hbar^{2}+8mE\right)} \ \ \ \ \ (16)$

Again, we did the integral using Maple. To do it by hand, we use the trig identity

$\displaystyle \sin\theta=2\sin\frac{\theta}{2}\cos\frac{\theta}{2} \ \ \ \ \ (17)$

followed by the substitution

 $\displaystyle u$ $\displaystyle =$ $\displaystyle \sin\frac{\theta}{2}\ \ \ \ \ (18)$ $\displaystyle du$ $\displaystyle =$ $\displaystyle \frac{1}{2}\cos\frac{\theta}{2}\; d\theta \ \ \ \ \ (19)$

This gives

 $\displaystyle \int_{0}^{\pi}d\theta\frac{\sin\theta}{\left(4k^{2}\sin^{2}\frac{\theta}{2}+\mu^{2}\right)^{2}}$ $\displaystyle =$ $\displaystyle 4\int_{0}^{1}\frac{u\; du}{\left(4k^{2}u^{2}+\mu^{2}\right)^{2}}\ \ \ \ \ (20)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left.-\frac{4}{8k^{2}\left(4k^{2}u^{2}+\mu^{2}\right)}\right|_{0}^{1}\ \ \ \ \ (21)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{1}{2k^{2}\left(4k^{2}+\mu^{2}\right)}+\frac{1}{2k^{2}\mu^{2}}\ \ \ \ \ (22)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{2}{\left(4k^{2}+\mu^{2}\right)\mu^{2}} \ \ \ \ \ (23)$

# First Born approximation: soft-sphere scattering

Reference: Griffiths, David J. (2005), Introduction to Quantum Mechanics, 2nd Edition; Pearson Education – Problem 11.10.

The first Born approximation for the scattering amplitude comes from the integral form of the Schrödinger equation:

$\displaystyle \psi\left(\mathbf{r}\right)=\psi_{0}\left(\mathbf{r}\right)-\frac{m}{2\pi\hbar^{2}}\int\frac{e^{ik\left|\mathbf{r}-\mathbf{r}_{0}\right|}}{\left|\mathbf{r}-\mathbf{r}_{0}\right|}V\left(\mathbf{r}_{0}\right)\psi\left(\mathbf{r}_{0}\right)d^{3}\mathbf{r}_{0} \ \ \ \ \ (1)$

This equation is valid for all ${\mathbf{r}}$, even for positions near to the origin where the scattering potential ${V}$ could be significantly different from zero. In a scattering problem, the detector is usually situated far from the scattering region, so for all ${\mathbf{r}}$ of interest, ${r\gg r_{0}}$ and we’re well outside the region where ${V\ne0}$. In such cases, we can approximate (see Griffiths, section 11.4.2 for details) the integral equation by

 $\displaystyle \psi\left(\mathbf{r}\right)$ $\displaystyle \cong$ $\displaystyle Ae^{ikz}-\frac{m}{2\pi\hbar^{2}}\frac{e^{ikr}}{r}\int e^{-i\mathbf{k}\cdot\mathbf{r}_{0}}V\left(\mathbf{r}_{0}\right)\psi\left(\mathbf{r}_{0}\right)d^{3}\mathbf{r}_{0}\ \ \ \ \ (2)$ $\displaystyle$ $\displaystyle =$ $\displaystyle A\left[e^{ikz}-\frac{m}{2\pi\hbar^{2}A}\frac{e^{ikr}}{r}\int e^{-i\mathbf{k}\cdot\mathbf{r}_{0}}V\left(\mathbf{r}_{0}\right)\psi\left(\mathbf{r}_{0}\right)d^{3}\mathbf{r}_{0}\right] \ \ \ \ \ (3)$

where ${\mathbf{k}\equiv k\hat{\mathbf{r}}}$ is a vector pointing from the origin to the detector (that is, parallel to ${\mathbf{r}}$). The first term on the RHS represents the incoming plane wave, as usual.

Since the scattering amplitude ${f}$ is the coefficient of ${e^{ikr}/r}$ inside the square brackets, we have

$\displaystyle f\left(\theta,\phi\right)=-\frac{m}{2\pi\hbar^{2}A}\int e^{-i\mathbf{k}\cdot\mathbf{r}_{0}}V\left(\mathbf{r}_{0}\right)\psi\left(\mathbf{r}_{0}\right)d^{3}\mathbf{r}_{0} \ \ \ \ \ (4)$

This formula still doesn’t help us much, since we still need to know the wave function ${\psi}$ inside the scattering region where ${V\ne0}$. The Born approximation assumes that the potential is weak, so that the incoming plane wave ${Ae^{ikz}}$ doesn’t change much after it scatters. That is, we assume that, for all points where the integrand is non-zero:

$\displaystyle \psi\left(\mathbf{r}_{0}\right)\approx\psi_{0}\left(\mathbf{r}_{0}\right) \ \ \ \ \ (5)$

The incident plane wave has a wave vector of magnitude ${k}$ that is parallel to ${\hat{\mathbf{z}}}$, which we can write as

$\displaystyle \mathbf{k}'\equiv k\hat{\mathbf{z}} \ \ \ \ \ (6)$

For some position ${\mathbf{r}_{0}}$ with ${z}$ component ${z_{0}}$:

$\displaystyle kz_{0}=\mathbf{k}'\cdot\mathbf{r}_{0} \ \ \ \ \ (7)$

so the assumption above amounts to saying that

$\displaystyle \psi\left(\mathbf{r}_{0}\right)\approx Ae^{i\mathbf{k}'\cdot\mathbf{r}_{0}} \ \ \ \ \ (8)$

This assumption gives us an approximation for ${f}$:

$\displaystyle f\left(\theta,\phi\right)\approx-\frac{m}{2\pi\hbar^{2}}\int e^{i\left(\mathbf{k}'-\mathbf{k}\right)\cdot\mathbf{r}_{0}}V\left(\mathbf{r}_{0}\right)d^{3}\mathbf{r}_{0} \ \ \ \ \ (9)$

It’s important to remember that, from the point of view of the integral, ${\mathbf{k}}$ and ${\mathbf{k}'}$ are constant, with the direction of ${\mathbf{k}=k\hat{\mathbf{r}}}$ being how the polar angles ${\theta}$ and ${\phi}$ are specified. The vector ${\mathbf{k}'=k\hat{\mathbf{z}}}$ is always the same as it specifies the direction of the incident particle.

For a spherically symmetric potential, the integral can be simplified a bit by defining the vector

$\displaystyle \boldsymbol{\kappa}\equiv\mathbf{k}'-\mathbf{k} \ \ \ \ \ (10)$

The vector is the base of an isosceles triangle with sides ${\mathbf{k}'}$ and ${\mathbf{k}}$, so since the angle between ${\mathbf{k}'}$ and ${\mathbf{k}}$ is ${\theta}$ (${\mathbf{k}'}$ is the direction to the detector and ${\mathbf{k}}$ is the direction of the incident particle, so ${\theta}$ is the scattering angle), we can divide the isosceles triangle into two symmetric right angled triangles by drawing a line from the origin to the the midpoint of ${\boldsymbol{\kappa}}$. The length of the base is ${\kappa/2}$ which is also ${k\sin\frac{\theta}{2}}$, so

$\displaystyle \kappa=2k\sin\frac{\theta}{2} \ \ \ \ \ (11)$

Letting the polar axis in the integral 9 lie along ${\boldsymbol{\kappa}}$ we get ${\left(\mathbf{k}'-\mathbf{k}\right)\cdot\mathbf{r}_{0}=\kappa r_{0}\cos\theta_{0}}$ [where ${\theta_{0}}$ is the polar angle of integration, not ${\theta}$!] and

 $\displaystyle f\left(\theta\right)$ $\displaystyle \approx$ $\displaystyle -\frac{m}{2\pi\hbar^{2}}\int_{0}^{\infty}\int_{0}^{\pi}\int_{0}^{2\pi}e^{i\kappa r_{0}\cos\theta_{0}}V\left(r_{0}\right)r_{0}^{2}\sin\theta_{0}d\phi_{0}d\theta_{0}dr_{0}\ \ \ \ \ (12)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{2m}{\hbar^{2}\kappa}\int_{0}^{\infty}V\left(r_{0}\right)r_{0}\sin\left(\kappa r_{0}\right)dr_{0} \ \ \ \ \ (13)$

Example Soft-sphere scattering. A soft sphere is defined by the potential

$\displaystyle V\left(\mathbf{r}\right)=\begin{cases} V_{0} & r\le a\\ 0 & r>a \end{cases} \ \ \ \ \ (14)$

where ${V_{0}>0}$ is a constant. [The hard sphere takes ${V_{0}=\infty}$.] From 13, we can get the Born approximation for the scattering amplitude:

 $\displaystyle f\left(\theta\right)$ $\displaystyle \approx$ $\displaystyle -\frac{2mV_{0}}{\hbar^{2}\kappa}\int_{0}^{a}r\sin\left(\kappa r\right)dr\ \ \ \ \ (15)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{2mV_{0}}{\hbar^{2}\kappa^{3}}\left[\sin\left(\kappa a\right)-a\kappa\cos\left(\kappa a\right)\right] \ \ \ \ \ (16)$

with the ${\theta}$ dependence given by the definition of ${\kappa}$ in 11.

For low energy scattering ${\kappa a\ll1}$ and we can expand the sin and cos.

 $\displaystyle \sin\left(\kappa a\right)-a\kappa\cos\left(\kappa a\right)$ $\displaystyle =$ $\displaystyle \kappa a-\frac{\left(\kappa a\right)^{3}}{3!}+\ldots-\kappa a\left(1-\frac{\left(\kappa a\right)^{2}}{2!}+\ldots\right)\ \ \ \ \ (17)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\left(\kappa a\right)^{3}}{3}+\ldots \ \ \ \ \ (18)$

To this order, the scattering amplitude is

$\displaystyle f\left(\theta\right)\approx-\frac{2mV_{0}a^{3}}{3\hbar^{3}} \ \ \ \ \ (19)$

which agrees with equation 11.82 in Griffiths.

# Phase shift in the spherical delta function shell

Reference: Griffiths, David J. (2005), Introduction to Quantum Mechanics, 2nd Edition; Pearson Education – Problem 11.7.

We can apply 3-d partial wave analysis using phase shifts to the problem of the spherical delta function shell. Restricting our attention to the ${l=0}$ term, we found earlier that the wave function for points outside the sphere is

$\displaystyle \psi_{ext}=\frac{A}{kr}\left[\sin kr+ka_{0}e^{ikr}\right] \ \ \ \ \ (1)$

where

 $\displaystyle a_{0}$ $\displaystyle =$ $\displaystyle -\frac{\beta e^{-ika}\sin^{2}ka}{\left(\beta\sin ka+ka\cos ka-iak\sin ka\right)k}\ \ \ \ \ (2)$ $\displaystyle k$ $\displaystyle \equiv$ $\displaystyle \frac{\sqrt{2mE}}{\hbar}\ \ \ \ \ (3)$ $\displaystyle \beta$ $\displaystyle \equiv$ $\displaystyle \frac{2ma\alpha}{\hbar^{2}} \ \ \ \ \ (4)$

and ${a}$ is the radius of the sphere, and ${\alpha}$ is the strength of the delta function in the potential: ${V\left(r\right)=\alpha\delta\left(r-a\right)}$.

By comparing this form of the wave function with the phase shift form, we found that

$\displaystyle a_{l}=\frac{1}{k}e^{i\delta_{l}}\sin\delta_{l} \ \ \ \ \ (5)$

To find the phase shift from 2, we need to put ${ka_{0}}$ in modulus-argument form. We can grind through the calculations by multiplying 2 top and bottom by the complex conjugate of the denominator and then finding the real and imaginary parts. This is just rather tedious algebra, so I got Maple to do it for me, with the results:

 $\displaystyle \Re\left(ka_{0}\right)$ $\displaystyle =$ $\displaystyle -\frac{\beta\sin^{2}\left(ka\right)\left(ka+\beta\sin\left(ka\right)\cos\left(ka\right)\right)}{\left(ka\right)^{2}+\beta^{2}+2ka\beta\sin\left(ka\right)\cos\left(ka\right)-\beta^{2}\cos^{2}\left(ka\right)}\ \ \ \ \ (6)$ $\displaystyle \Im\left(ka_{0}\right)$ $\displaystyle =$ $\displaystyle \frac{\beta^{2}\sin^{4}\left(ka\right)}{\left(ka\right)^{2}+\beta^{2}+2ka\beta\sin\left(ka\right)\cos\left(ka\right)-\beta^{2}\cos^{2}\left(ka\right)} \ \ \ \ \ (7)$

From this, we get

 $\displaystyle \delta_{0}$ $\displaystyle =$ $\displaystyle \arctan\left(\frac{\Im ka_{0}}{\Re ka_{0}}\right)\ \ \ \ \ (8)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \arctan\left(-\frac{\beta\sin^{2}\left(ka\right)}{ka+\beta\sin\left(ka\right)\cos\left(ka\right)}\right)\ \ \ \ \ (9)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\arctan\left(\frac{\beta\sin^{2}\left(ka\right)}{ka+\beta\sin\left(ka\right)\cos\left(ka\right)}\right) \ \ \ \ \ (10)$

For some reason, Griffiths wants to express the answer using cotangents, so using ${\arctan x=\mbox{arccot}\frac{1}{x}}$, we have

 $\displaystyle \delta_{0}$ $\displaystyle =$ $\displaystyle -\mbox{arccot}\left(\frac{ka+\beta\sin\left(ka\right)\cos\left(ka\right)}{\beta\sin^{2}\left(ka\right)}\right)\ \ \ \ \ (11)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\mbox{arccot}\left(\cot\left(ka\right)+\frac{ka}{\beta\sin^{2}\left(ka\right)}\right) \ \ \ \ \ (12)$

# Partial waves in three dimensions: hard sphere scattering

Reference: Griffiths, David J. (2005), Introduction to Quantum Mechanics, 2nd Edition; Pearson Education – Problem 11.6.

In our earlier look at partial wave analysis in quantum scattering, we found that for scattering of an incident plane wave from a spherically symmetric potential, the overall wave function for the region outside the influence of the potential is

$\displaystyle \psi\left(r,\theta\right)=A\sum_{l=0}^{\infty}i^{l}\left(2l+1\right)\left[j_{l}\left(kr\right)+ika_{l}h_{l}^{\left(1\right)}\left(kr\right)\right]P_{l}\left(\cos\theta\right) \ \ \ \ \ (1)$

where ${k=\sqrt{2mE}/\hbar}$, ${j_{l}}$ is a spherical Bessel function, ${h_{l}^{\left(1\right)}}$ is a Hankel function of the first kind and ${P_{l}}$ is a Legendre polynomial. The partial wave coefficients ${a_{l}}$ must be determined by solving the Schrödinger equation for the scattering region (where ${V\ne0}$) and matching that solution to the above wave function using boundary conditions.

It turns out that there is another way of writing the wave function in the outer region, using phase shifts. For a spherically symmetric potential ${V\left(r\right)}$, the force is the negative gradient of the potential, and is always parallel to ${\hat{\mathbf{r}}}$. This means that the torque ${\mathbf{N}=\mathbf{r}\times\mathbf{F}=0}$ for such a potential, so total angular momentum is conserved: ${\dot{\mathbf{L}}=I\mathbf{N}=0}$, where ${I}$ is the moment of inertia. This means that the amplitude of the ${l}$th component of the incident wave must be equal to the amplitude of the ${l}$th component of the scattered wave in 1, although the scattering may introduce a phase shift, just as in the one-dimensional case.

Griffiths goes through the details of the derivation of the phase shift in his section 11.3. The basic idea is to write the wave function for the incoming plane wave on its own (when ${V=0}$ so there is no scattering):

$\displaystyle \psi_{0}=A\sum_{l=0}^{\infty}i^{l}\left(2l+1\right)j_{l}\left(kr\right)P_{l}\left(\cos\theta\right) \ \ \ \ \ (2)$

and then take its asymptotic form as ${kr\gg1}$:

$\displaystyle \psi_{0}\approx\frac{A}{2ikr}\sum_{l=0}^{\infty}i^{l}\left(2l+1\right)\left(e^{ikr}-\left(-1\right)^{l}e^{-ikr}\right)P_{l}\left(\cos\theta\right) \ \ \ \ \ (3)$

We now write a wave function for the case where there is some scattering, so that ${V\ne0}$, by introducing a phase shift ${2\delta_{l}}$ in the outgoing wave term for each partial wave:

$\displaystyle \psi\approx\frac{A}{2ikr}\sum_{l=0}^{\infty}i^{l}\left(2l+1\right)\left(e^{i\left(kr+2\delta_{l}\right)}-\left(-1\right)^{l}e^{-ikr}\right)P_{l}\left(\cos\theta\right) \ \ \ \ \ (4)$

Each term in the sum is one partial wave with angular momentum number ${l}$, and we know that the amplitudes of each incoming and scattered partial wave for each individual value of ${l}$ must be the same, due to conservation of angular momentum.

Of course, we still need to find the phase shifts ${\delta_{l}}$ and to do that, we need to solve the Schrödinger equation in the region where ${V\ne0}$ and match it to 4 using boundary conditions, so there isn’t really any saving in the amount of work we need to do to find the scattering amplitudes. However, the phase shift is real, while the ${a_{l}}$s in 1 are often complex, and phase shifts have a physical interpretation that is easier to understand than the abstract ${a_{l}}$ coefficients.

To derive the relation between ${\delta_{l}}$ and ${a_{l}}$, we compare the asymptotic form of 1 with 4 (details in Griffiths), and we find that

$\displaystyle a_{l}=\frac{e^{2i\delta_{l}}-1}{2ik}=\frac{1}{k}e^{i\delta_{l}}\sin\delta_{l} \ \ \ \ \ (5)$

We can invert this to get

$\displaystyle \delta_{l}=\frac{1}{2i}\ln\left(2ika_{l}+1\right) \ \ \ \ \ (6)$

The overall scattering amplitude and cross section are

 $\displaystyle f\left(\theta\right)$ $\displaystyle =$ $\displaystyle \frac{1}{k}\sum_{l=0}^{\infty}\left(2l+1\right)e^{i\delta_{l}}\sin\delta_{l}P_{l}\left(\cos\theta\right)\ \ \ \ \ (7)$ $\displaystyle \sigma$ $\displaystyle =$ $\displaystyle \frac{4\pi}{k^{2}}\sum_{l=0}^{\infty}\left(2l+1\right)\sin^{2}\delta_{l} \ \ \ \ \ (8)$

Example Hard sphere scattering phase shifts. We’ve already found the ${a_{l}}$s for the hard sphere case:

$\displaystyle a_{l}=i\frac{j_{l}\left(ka\right)}{kh_{l}^{\left(1\right)}\left(ka\right)} \ \ \ \ \ (9)$

Using 6 to find ${\delta_{l}}$ is a bit tricky, since the complex logarithm is multi-valued. It’s easier to work from 5. From the definition of ${h_{l}^{\left(1\right)}}$ we have

 $\displaystyle h_{l}^{\left(1\right)}$ $\displaystyle \equiv$ $\displaystyle j_{l}+in_{l}\ \ \ \ \ (10)$ $\displaystyle a_{l}$ $\displaystyle =$ $\displaystyle i\frac{j_{l}}{k\left(j_{l}+in_{l}\right)}\ \ \ \ \ (11)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{i}{k}\frac{j_{l}\left(j_{l}-in_{l}\right)}{j_{l}^{2}+n_{l}^{2}}\ \ \ \ \ (12)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{j_{l}}{k\left(j_{l}^{2}+n_{l}^{2}\right)}\left(n_{l}+ij_{l}\right) \ \ \ \ \ (13)$

We can write this in modulus-argument form:

 $\displaystyle \left|a_{l}\right|$ $\displaystyle =$ $\displaystyle \frac{\left|j_{l}\right|\sqrt{j_{l}^{2}+n_{l}^{2}}}{k\left(j_{l}^{2}+n_{l}^{2}\right)}\ \ \ \ \ (14)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\left|j_{l}\right|}{k\sqrt{j_{l}^{2}+n_{l}^{2}}}\ \ \ \ \ (15)$ $\displaystyle \delta_{l}=\arg a_{l}$ $\displaystyle =$ $\displaystyle \arctan\frac{j_{l}}{n_{l}} \ \ \ \ \ (16)$

When comparing this with 5 we have to be careful to get the right quadrant for ${\delta_{l}}$. We do this in the usual way by looking at the signs of ${j_{l}}$ and ${n_{l}}$. If they are both positive, then ${\delta_{l}}$ is in the first quadrant, if ${j_{l}>0}$ and ${n_{l}<0}$, we’re in the second quadrant, and so on.

# Phase shift in one-dimensional scattering

Reference: Griffiths, David J. (2005), Introduction to Quantum Mechanics, 2nd Edition; Pearson Education – Problem 11.5.

In a one-dimensional scattering problem, if we have an incident wave coming in from the left and scattering off a potential that is non-zero in the region ${-a\le x\le0}$ and infinite for ${x>0}$, then the wave effectively hits a brick wall at ${x=0}$ and is fully reflected back in the ${-x}$ direction. Since the total probability of finding the particle can’t change (it won’t partially or fully disappear), the amplitude of the incoming wave function must equal the amplitude of the reflected wave function heading in the ${-x}$ direction. However, interaction with the potential in the region ${-a\le x\le0}$ could change the phase of the wave function, since the phase doesn’t affect the amplitude. That is, the wave function in the region ${x<-a}$ has the form

$\displaystyle \psi\left(x\right)=A\left(e^{ikx}-e^{i\left(2\delta-kx\right)}\right) \ \ \ \ \ (1)$

where ${2\delta}$ is the total phase change due to the reflection through the potential (the 2 is there since the wave function undergoes one phase shift of ${\delta}$ as it travels towards the origin and another phase shift of ${\delta}$ as it travels back towards ${-x}$, so ${2\delta}$ is the total phase shift). The quantity ${k=\sqrt{2mE}/\hbar}$ as usual.

To find ${\delta}$ we must, as usual, solve the Schrödinger equation in the region ${-a\le x\le0}$ and apply boundary conditions.

Example Suppose the potential is a half-finite square well of form

$\displaystyle V=\begin{cases} 0 & x<-a\\ -V_{0} & -a\le x\le0\\ \infty & x>0 \end{cases} \ \ \ \ \ (2)$

where ${V_{0}}$ is a positive constant. Then the solution of the Schrödinger equation inside the well is

 $\displaystyle \psi\left(x\right)$ $\displaystyle =$ $\displaystyle Be^{ik'x}+Ce^{-ik'x}\ \ \ \ \ (3)$ $\displaystyle k'$ $\displaystyle \equiv$ $\displaystyle \frac{\sqrt{2m\left(E+V_{0}\right)}}{\hbar} \ \ \ \ \ (4)$

Because of the infinite barrier at ${x=0}$, we must have ${\psi\left(0\right)=0}$ so ${C=-B}$ and

 $\displaystyle \psi\left(x\right)$ $\displaystyle =$ $\displaystyle B\left(e^{ik'x}-e^{-ik'x}\right)\ \ \ \ \ (5)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 2iB\sin k'x \ \ \ \ \ (6)$

Since the potential is finite at ${x=-a}$, both ${\psi}$ and ${\psi'}$ must be continuous there. This gives, from 1 and 6

 $\displaystyle A\left(e^{-ika}-e^{i\left(2\delta+ka\right)}\right)$ $\displaystyle =$ $\displaystyle -2iB\sin k'a\ \ \ \ \ (7)$ $\displaystyle ikA\left(e^{-ika}+e^{i\left(2\delta+ka\right)}\right)$ $\displaystyle =$ $\displaystyle 2ik'B\cos k'a \ \ \ \ \ (8)$

Dividing the second equation by the first gives

$\displaystyle ik\frac{e^{-ika}+e^{i\left(2\delta+ka\right)}}{e^{-ika}-e^{i\left(2\delta+ka\right)}}=-k'\cot k'a \ \ \ \ \ (9)$

We can solve for ${e^{2i\delta}}$ by multiplying the LHS by ${e^{-ika}/e^{-ika}}$, so we get

 $\displaystyle e^{2i\delta}$ $\displaystyle =$ $\displaystyle e^{-2ika}\frac{ik+k'\cot k'a}{k'\cot k'a-ik}\ \ \ \ \ (10)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -e^{-2ika}\frac{k-ik'\cot k'a}{k+ik'\cot k'a} \ \ \ \ \ (11)$

Plugging this into 1 we see that the reflected wave is

 $\displaystyle \psi_{ref}\left(x\right)$ $\displaystyle =$ $\displaystyle -Ae^{i\left(2\delta-kx\right)}\ \ \ \ \ (12)$ $\displaystyle$ $\displaystyle =$ $\displaystyle Ae^{-2ika}\frac{k-ik'\cot k'a}{k+ik'\cot k'a}e^{-ikx} \ \ \ \ \ (13)$

The amplitude of the reflected wave is

 $\displaystyle \left|\psi_{ref}\right|^{2}$ $\displaystyle =$ $\displaystyle \left|A\right|^{2}\left|\frac{k-ik'\cot k'a}{k+ik'\cot k'a}\right|^{2}\ \ \ \ \ (14)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left|A\right|^{2}\frac{k^{2}+\left(-k'\cot k'a\right)^{2}}{k^{2}+\left(k'\cot k'a\right)^{2}}\ \ \ \ \ (15)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left|A\right|^{2} \ \ \ \ \ (16)$

So the reflected wave has the same amplitude as the incident wave, as required.

For a very deep well, so that ${V_{0}\gg E}$, ${k'\rightarrow\infty}$ so from 11

 $\displaystyle e^{2i\delta}$ $\displaystyle \rightarrow$ $\displaystyle -e^{-2ika}\left[\frac{-ik'\cot k'a}{ik'\cot k'a}\right]=e^{-2ika}\ \ \ \ \ (17)$ $\displaystyle \delta$ $\displaystyle \rightarrow$ $\displaystyle -ka \ \ \ \ \ (18)$