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.

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.

Quantum scattering: partial wave analysis

Reference: Griffiths, David J. (2005), Introduction to Quantum Mechanics, 2nd Edition; Pearson Education – Problems 11.3-11.4.

For spherically symmetric potentials, the general solution to the Schrödinger equation for the scattering problem has the form

$\displaystyle \psi\left(r,\theta\right)=A\left(e^{ikz}+f\left(\theta\right)\frac{e^{ikr}}{r}\right) \ \ \ \ \ (1)$

where ${f}$ is the scattering amplitude. The first term on the RHS represents the incident plane wave, and the second term on the RHS represents the scattered wave. One way of finding ${f}$ is partial wave analysis, which Griffiths describes in detail in his section 11.2, so I won’t repeat the whole derivation here. It is worth, however, giving an overview of the technique to highlight the main concepts.

The main idea is to use the solution of the three-dimensional Schrödinger equation for a spherically symmetric potential that we considered earlier. The solution splits into two factors: a spherical harmonic ${Y_{l}^{m}\left(\theta,\phi\right)}$ that depends only on the angular coordinates and is independent of the potential, and a radial function ${R\left(r\right)}$ that depends only on the radial coordinate and does depend on the potential. The function ${u\left(r\right)\equiv rR\left(r\right)}$ satisfies the radial equation

$\displaystyle -\frac{\hbar^{2}}{2m}\frac{d^{2}u}{dr^{2}}+\left(V+\frac{\hbar^{2}}{2m}\frac{l(l+1)}{r^{2}}\right)u=Eu \ \ \ \ \ (2)$

In the region far from the target, where ${V}$ is very small, we can neglect the potential term and get the equation

$\displaystyle -\frac{\hbar^{2}}{2m}\frac{d^{2}u}{dr^{2}}+\frac{\hbar^{2}}{2m}\frac{l(l+1)}{r^{2}}u=Eu \ \ \ \ \ (3)$

This ODE can be solved in general using Hankel functions, and if we restrict our attention only to outgoing waves (since we’re considering only particles scattering outwards from the target), we end up with an overall wave function of

$\displaystyle \psi\left(r,\theta\right)=A\left\{ e^{ikz}+k\sum_{l=0}^{\infty}i^{l+1}\left(2l+1\right)a_{l}h_{l}^{\left(1\right)}\left(kr\right)P_{l}\left(\cos\theta\right)\right\} \ \ \ \ \ (4)$

where

$\displaystyle k\equiv\frac{\sqrt{2mE}}{\hbar} \ \ \ \ \ (5)$

the Hankel function of the first kind is defined by

$\displaystyle h_{l}^{\left(1\right)}\left(x\right)\equiv j_{l}\left(x\right)+in_{l}\left(x\right) \ \ \ \ \ (6)$

(with ${j_{l}}$ and ${n_{l}}$ being spherical Bessel functions), and the ${P_{l}}$ being Legendre polynomials. The coefficients ${a_{l}}$ are what must be calculated for the particular potential being used.

For large ${r}$ (far from the target), we can use the asymptotic forms of the ${h_{l}^{\left(1\right)}\left(kr\right)}$ which are

$\displaystyle h_{l}^{\left(1\right)}\left(kr\right)\sim\left(-i\right)^{l+1}\frac{e^{ikr}}{kr} \ \ \ \ \ (7)$

so in this region the wave function becomes

 $\displaystyle \psi\left(r,\theta\right)$ $\displaystyle \approx$ $\displaystyle A\left\{ e^{ikz}+f\left(\theta\right)\frac{e^{ikr}}{r}\right\} \ \ \ \ \ (8)$ $\displaystyle f\left(\theta\right)$ $\displaystyle \equiv$ $\displaystyle \sum_{l=0}^{\infty}\left(2l+1\right)a_{l}P_{l}\left(\cos\theta\right) \ \ \ \ \ (9)$

so we get an explicit formula for the scattering amplitude ${f}$.

The differential cross-section is given by a rather ugly formula:

 $\displaystyle D\left(\theta\right)$ $\displaystyle =$ $\displaystyle \frac{d\sigma}{d\Omega}=\left|f\left(\theta\right)\right|^{2}\ \ \ \ \ (10)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \sum_{l=0}^{\infty}\sum_{l'=0}^{\infty}\left(2l+1\right)\left(2l'+1\right)a_{l}a_{l'}P_{l}\left(\cos\theta\right)P_{l'}\left(\cos\theta\right) \ \ \ \ \ (11)$

Integrating this over solid angle and using the orthonormality of the ${P_{l}}$, we get the total cross-section

$\displaystyle \sigma=4\pi\sum_{l=0}^{\infty}\left(2l+1\right)\left|a_{l}\right|^{2} \ \ \ \ \ (12)$

Finally, to get a consistent formula where everything is in spherical coordinates, we need to convert the plane wave ${e^{ikz}}$ to spherical coordinates which gives the final result for the wave function:

$\displaystyle \psi\left(r,\theta\right)=A\sum_{l=0}^{\infty}i^{l}\left(2l+1\right)\left[j_{l}\left(kr\right)+ika_{l}h_{l}^{\left(1\right)}\left(kr\right)\right]P_{l}\left(\cos\theta\right) \ \ \ \ \ (13)$

To find the ${a_{l}}$s, we need to solve the Schrödinger equation for the region near the target, where ${V\ne0}$, and then match this solution up to the exterior solution 13 at the boundary between the two regions. Doing this requires a well-defined boundary, so it would seem that this method doesn’t work for potentials that fall off continuously out to infinity.

Example 1 As a simple example of how this boundary matching process works, we consider the case of hard-sphere scattering. We have a sphere that is impenetrable so that

$\displaystyle V=\begin{cases} \infty & r\le a\\ 0 & r>a \end{cases} \ \ \ \ \ (14)$

An infinite potential means that ${\psi\left(r,\theta\right)=0}$ for ${r\le a}$, so the boundary condition occurs over the sphere ${r=a}$. Matching this to 13, we get

$\displaystyle \sum_{l=0}^{\infty}i^{l}\left(2l+1\right)\left[j_{l}\left(ka\right)+ika_{l}h_{l}^{\left(1\right)}\left(ka\right)\right]P_{l}\left(\cos\theta\right)=0 \ \ \ \ \ (15)$

We can work out ${a_{l}}$ by multiplying this equation by ${P_{l'}\left(\cos\theta\right)}$ and integrating over ${\theta}$, using the orthonormality of the ${P_{l}}$:

 $\displaystyle \sum_{l=0}^{\infty}i^{l}\left(2l+1\right)\left[j_{l}\left(ka\right)+ika_{l}h_{l}^{\left(1\right)}\left(ka\right)\right]\int_{0}^{\pi}P_{l}\left(\cos\theta\right)P_{l'}\left(\cos\theta\right)$ $\displaystyle =$ $\displaystyle 0\ \ \ \ \ (16)$ $\displaystyle \sum_{l=0}^{\infty}i^{l}\left(2l+1\right)\left[j_{l}\left(ka\right)+ika_{l}h_{l}^{\left(1\right)}\left(ka\right)\right]\delta_{ll'}$ $\displaystyle =$ $\displaystyle 0\ \ \ \ \ (17)$ $\displaystyle i^{l'}\left(2l'+1\right)\left[j_{l'}\left(ka\right)+ika_{l'}h_{l'}^{\left(1\right)}\left(ka\right)\right]$ $\displaystyle =$ $\displaystyle 0 \ \ \ \ \ (18)$

Therefore, the coefficients are

$\displaystyle a_{l}=i\frac{j_{l}\left(ka\right)}{kh_{l}^{\left(1\right)}\left(ka\right)} \ \ \ \ \ (19)$

[Note that Griffiths’s eqn 11.33 is wrong: there shouldn’t be a minus sign on the RHS.]

Example 2 The delta-function spherical shell. Given the potential

$\displaystyle V\left(r\right)=\alpha\delta\left(r-a\right) \ \ \ \ \ (20)$

we want to find ${f\left(\theta\right)}$. We’ll take the incident particle to have very low energy, so that ${ka\ll1}$. Since ${k=2\pi/\lambda}$ this amounts to saying that ${\lambda\gg a}$ so that the wavelength of the particle is much greater than the size of the target. From Planck’s formula ${E=h\nu=h/\lambda}$, this is equivalent to the particle having a low energy. In the text, Griffiths shows that the cross section can be expanded in powers of ${\left(ka\right)^{l}}$, so for low energy, only the ${l=0}$ term is significant, so we’ll restrict our analysis to that case.

The exterior solution is given by the ${l=0}$ term from 13:

$\displaystyle \psi_{ext}=A\left[j_{0}\left(kr\right)+ika_{0}h_{0}^{\left(1\right)}\left(kr\right)P_{0}\left(\cos\theta\right)\right] \ \ \ \ \ (21)$

We can use the small argument forms of the Bessel and Hankel functions:

 $\displaystyle j_{0}\left(kr\right)$ $\displaystyle \approx$ $\displaystyle \frac{\sin kr}{kr}\ \ \ \ \ (22)$ $\displaystyle h_{0}^{\left(1\right)}\left(kr\right)$ $\displaystyle \approx$ $\displaystyle -i\frac{e^{ikr}}{kr} \ \ \ \ \ (23)$

Also, ${P_{0}=1}$ so we get, for ${r\ge a}$:

$\displaystyle \psi_{ext}=\frac{A}{kr}\left[\sin kr+ka_{0}e^{ikr}\right] \ \ \ \ \ (24)$

For the internal function for ${r, the potential is ${V=0}$, so the general solution of the radial equation 2 is

 $\displaystyle u\left(r\right)$ $\displaystyle =$ $\displaystyle B\sin kr+D\cos kr\ \ \ \ \ (25)$ $\displaystyle R\left(r\right)$ $\displaystyle =$ $\displaystyle \frac{u\left(r\right)}{r}\ \ \ \ \ (26)$ $\displaystyle$ $\displaystyle =$ $\displaystyle B\frac{\sin kr}{r}+D\frac{\cos kr}{r} \ \ \ \ \ (27)$

This solution must be valid at ${r=0}$, so only the ${\frac{\sin kr}{r}}$ term is finite there, so we must have ${D=0}$, giving

$\displaystyle \psi_{in}\left(r\right)=B\frac{\sin kr}{r} \ \ \ \ \ (28)$

The wave function must be continuous at the boundary ${r=a}$, so

$\displaystyle \frac{A}{ka}\left[\sin ka+ka_{0}e^{ika}\right]=B\frac{\sin ka}{a} \ \ \ \ \ (29)$

Since the delta function is infinite at ${r=a}$, we can’t assume that the derivative of the wave function is continuous there, but we can use the same method that we used in analyzing the delta function well to get another boundary condition. The radial equation 2 for ${u}$ is the same as the one dimensional Schrödinger equation that we solved for the delta function well, except that the delta function here is a barrier rather than a well, so we replace ${-\alpha}$ by ${+\alpha}$ to get the condition on the derivative. The radial equation for ${l=0}$ is

$\displaystyle -\frac{\hbar^{2}}{2m}\frac{d^{2}u}{dx^{2}}+\alpha\delta(x)u=Eu \ \ \ \ \ (30)$

Now if we integrate this equation term by term across the boundary, we get, for some value of ${\epsilon}$:

 $\displaystyle -\frac{\hbar^{2}}{2m}\int_{-\epsilon}^{\epsilon}\frac{d^{2}u}{dr^{2}}dr+\alpha\int_{a-\epsilon}^{a+\epsilon}\delta(r-a)u\; dr$ $\displaystyle =$ $\displaystyle E\int_{a-\epsilon}^{a+\epsilon}u\; dr\ \ \ \ \ (31)$ $\displaystyle -\frac{\hbar^{2}}{2m}\frac{du}{dr}\Big|_{a-\epsilon}^{a+\epsilon}+\alpha u(a)$ $\displaystyle =$ $\displaystyle E\int_{a-\epsilon}^{a+\epsilon}u\; dr \ \ \ \ \ (32)$

If we take the limit as ${\epsilon\rightarrow0}$, the integral on the right tends to zero, since it is the integral of a continuous function over an infinitesimally small interval. The first term on the left, however, will not be zero, since derivative of the wave function is not continuous when the potential is infinite. Thus we get

$\displaystyle \lim_{\epsilon\rightarrow0}\frac{du}{dr}\Big|_{a-\epsilon}^{a+\epsilon}=\frac{2m\alpha}{\hbar^{2}}u\left(a\right) \ \ \ \ \ (33)$

To simplify the notation in what follows I’ll use the following shorthand:

 $\displaystyle s$ $\displaystyle \equiv$ $\displaystyle \sin ka\ \ \ \ \ (34)$ $\displaystyle c$ $\displaystyle \equiv$ $\displaystyle \cos ka\ \ \ \ \ (35)$ $\displaystyle e$ $\displaystyle \equiv$ $\displaystyle e^{ika} \ \ \ \ \ (36)$

From 24 and 28

 $\displaystyle u_{ext}\left(r\right)$ $\displaystyle =$ $\displaystyle r\psi_{ext}\left(r\right)\ \ \ \ \ (37)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{A}{k}\left(s+ka_{0}e\right)\ \ \ \ \ (38)$ $\displaystyle \left.\frac{du_{ext}}{dr}\right|_{r=a}$ $\displaystyle =$ $\displaystyle A\left(c+ika_{0}e\right)\ \ \ \ \ (39)$ $\displaystyle u_{in}\left(r\right)$ $\displaystyle =$ $\displaystyle r\psi_{in}\left(r\right)\ \ \ \ \ (40)$ $\displaystyle$ $\displaystyle =$ $\displaystyle Bs\ \ \ \ \ (41)$ $\displaystyle \left.\frac{du_{in}}{dr}\right|_{r=a}$ $\displaystyle =$ $\displaystyle Bkc \ \ \ \ \ (42)$

From 33 we get

 $\displaystyle \lim_{\epsilon\rightarrow0}\frac{du}{dr}\Big|_{a-\epsilon}^{a+\epsilon}$ $\displaystyle =$ $\displaystyle \left.\frac{du_{ext}}{dr}\right|_{r=a}-\left.\frac{du_{in}}{dr}\right|_{r=a}\ \ \ \ \ (43)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left(A-Bk\right)c+Aika_{0}e\ \ \ \ \ (44)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{2m\alpha}{\hbar^{2}}u\left(a\right)\ \ \ \ \ (45)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{2m\alpha}{\hbar^{2}}Bs \ \ \ \ \ (46)$

To get rid of the constants ${A}$ and ${B}$ we use 29:

$\displaystyle B=\frac{s+ka_{0}e}{ks}A \ \ \ \ \ (47)$

so we get from 44 and 46

$\displaystyle \left(1-\frac{s+ka_{0}e}{s}\right)c+ika_{0}e=\frac{s^{2}+ka_{0}es}{sk}\frac{\beta}{a} \ \ \ \ \ (48)$

where

$\displaystyle \beta\equiv\frac{2m\alpha a}{\hbar^{2}} \ \ \ \ \ (49)$

Solving for ${a_{0}}$ gives (restoring the full notation):

$\displaystyle a_{0}=-\frac{\beta e^{-ika}\sin^{2}ka}{\left(\beta\sin ka+ka\cos ka-iak\sin ka\right)k} \ \ \ \ \ (50)$

For ${ka\ll1}$ we can approximate ${\sin ka\approx ka}$, ${\cos ka\approx1}$ and ${e^{-ika}\approx1}$, so

 $\displaystyle a_{0}$ $\displaystyle \approx$ $\displaystyle -\frac{\beta\left(ka\right)^{2}}{\left(\beta ka+ka-i\left(ak\right)^{2}\right)k}\ \ \ \ \ (51)$ $\displaystyle$ $\displaystyle \approx$ $\displaystyle -\frac{\beta a}{1+\beta} \ \ \ \ \ (52)$

where we dropped the imaginary term in the denominator since it is of second order in ${ak}$.

The scattering amplitude, differential cross section and total cross section are, from 9, 11 and 12:

 $\displaystyle f\left(\theta\right)$ $\displaystyle =$ $\displaystyle a_{0}=-\frac{\beta a}{1+\beta}\ \ \ \ \ (53)$ $\displaystyle \frac{d\sigma}{d\Omega}$ $\displaystyle =$ $\displaystyle \left|f\left(\theta\right)\right|^{2}=\frac{\beta^{2}a^{2}}{\left(1+\beta\right)^{2}}\ \ \ \ \ (54)$ $\displaystyle \sigma$ $\displaystyle =$ $\displaystyle 4\pi\left|a_{0}\right|^{2}=4\pi\frac{\beta^{2}a^{2}}{\left(1+\beta\right)^{2}} \ \ \ \ \ (55)$

As the strength ${\alpha}$ of the delta function gets higher, ${\beta\rightarrow\infty}$ and the cross section tends to ${4\pi a^{2}}$, which is the cross section for a hard sphere. Thus even though the delta function presents an infinite barrier, if it is of finite strength, the total cross section is less than that of a hard sphere.

Quantum scattering: scattering amplitude and differential cross section

Reference: Griffiths, David J. (2005), Introduction to Quantum Mechanics, 2nd Edition; Pearson Education – Problem 11.2.

In quantum mechanics, the study of scattering is conceptually similar to the classical study of the scattering of waves. In a typical scattering experiment, a stream of particles travels in from (essentially) infinity and encounters a target (represented by a potential). As a result, particles can scatter away from the target in some radial direction. The probability of a given particle scattering in a particular direction can be measured by placing detectors or particle counters at various positions around the target.

In three dimensions, we can represent the incoming particle by a plane wave and the scattered particle by a spherical wave. For the plane wave, the surface probability density of the wave is constant as the wave travels (since the wave doesn’t spread out), so we can represent this by the free particle wave function ${Ae^{ikz}}$, where ${k=\sqrt{2mE}/\hbar}$. We’re using only the ${e^{ikz}}$ term (and not the ${e^{-ikz}}$ term), since we’re considering a particle travelling in the ${+z}$ direction only.

For the spherical wave, the surface probability density decreases according to the inverse square law (just as the radiation from a star falls off according to the same law), so the wave function itself must decrease as ${1/r}$. The probability of scattering can, in general, depend on both the polar and azimuthal angles ${\theta}$ and ${\phi}$ (where the polar ${z}$ axis is the direction of the incoming particle), so the outgoing wave function has the form ${Af\left(\theta,\phi\right)e^{ikr}/r}$ where ${f}$ is the scattering amplitude. The complete wave function is then

$\displaystyle \psi\left(r,\theta,\phi\right)=A\left(e^{ikz}+f\left(\theta,\phi\right)\frac{e^{ikr}}{r}\right) \ \ \ \ \ (1)$

If the potential of the scattering target is independent of ${\phi}$, as would be the case for a potential that depends only on ${r}$, then the scattering probability cannot depend on ${\phi}$ either. It can still depend on ${\theta}$, of course, since the direction of the incoming particle breaks the symmetry of the polar angle, even if the target is spherically symmetric. [Think of throwing stones at a large spherical boulder. The probability of the stone scattering in a particular direction depends only on ${\theta}$.] In this case, we get

$\displaystyle \psi\left(r,\theta\right)=A\left(e^{ikz}+f\left(\theta\right)\frac{e^{ikr}}{r}\right) \ \ \ \ \ (2)$

These wave functions are valid for large distances from the target. The actual behaviour of the system within the region where the potential has an appreciable influence can be quite complicated and depends on the precise form of the potential function.

Although all real scattering experiments are done in three dimensions, we can generalize the above to lower dimensions. We’ve already looked at one-dimensional scattering, in which case the scattered particle can either continue on in the same direction or scatter directly backwards, so we get

$\displaystyle \psi\left(z\right)=\begin{cases} Ae^{ikz}+Be^{-ikz} & z\ll0\\ Ce^{ikz} & z\gg0 \end{cases} \ \ \ \ \ (3)$

where we’re assuming that the potential’s region of influence is localized within some finite distance of ${z=0}$.

For two dimensions, the outgoing wave function is a circular wave and the probability density falls off as ${1/r}$ so the outgoing wave function must fall off as ${1/\sqrt{r}}$, giving

$\displaystyle \psi\left(z,\theta\right)=A\left(e^{ikz}+f\left(\theta\right)\frac{e^{ikr}}{\sqrt{r}}\right) \ \ \ \ \ (4)$

Finally, we can relate ${f}$ to physically measureable quantities by considering the differential cross section. In classical scattering, the idea is that if we take a cross section of the incoming particle beam and divide this cross section up into little regions of area ${d\sigma}$, then a particle that crosses the area element ${d\sigma}$ will (always, classically) scatter into an element of solid angle ${d\Omega}$. Depending on the scattering potential and the angle of scattering ${\theta}$, there is a relation between the size of ${d\sigma}$ and the size of ${d\Omega}$. Basically, if we consider a larger element of cross sectional area, then the element of solid angle into which these particles scatter is also larger. The proportionality factor ${d\sigma/d\Omega}$ can depend on where we choose the element ${d\sigma}$ or, conversely, which element of ${d\Omega}$ we choose to detect so, for a spherically symmetric potential, ${d\sigma/d\Omega}$ will in general depend on ${\theta}$, the scattering angle. That is, using the (non-standard) symbol ${D\left(\theta\right)}$ for the scattering cross-section

$\displaystyle D\left(\theta\right)\equiv\frac{d\sigma}{d\Omega} \ \ \ \ \ (5)$

[Griffiths is quite right that this is a terrible name; I can remember being confused myself when I first encountered it. It’s actually the ratio of a differential cross-section ${d\sigma}$ to a differential solid angle ${d\Omega}$.]

In the quantum case, we still assume that a particle incident in ${d\sigma}$ will scatter into ${d\Omega}$. However, here we can only deal in the probability that a particle will cross the area ${d\sigma}$ as it comes in, and the probability that it will scatter into ${d\Omega}$ as it leaves. The scattering assumption is that these two probabilities are equal. If the incident particle is travelling at speed ${v}$ in the ${+z}$ direction, then the probability that it is found in a thin slice of space of area ${d\sigma}$ in time interval ${dt}$ is

$\displaystyle dP_{i}=\left|\psi_{incident}\right|^{2}dV=\left|A\right|^{2}v\; dt\; d\sigma \ \ \ \ \ (6)$

[You might wonder if it’s correct to say that the incident particle’s velocity is precisely ${v}$; what about the uncertainty principle? However, since we’re dealing with a plane wave, the particle’s position is totally unknown, so its momentum and hence velocity can be precisely known.]

The outgoing particle, also travelling at speed ${v}$, has a probability of being found in a thin radial slice of volume ${v\; dt\times r^{2}d\Omega}$. Here, ${v\; dt}$ is the thickness of the slice and ${r^{2}d\Omega}$ is the surface area of the slice. This time, the probability is

$\displaystyle dP_{o}=\left|\psi_{outgoing}\right|^{2}dV=\frac{\left|Af\right|^{2}}{r^{2}}v\; dt\; r^{2}d\Omega \ \ \ \ \ (7)$

Equating ${dP_{i}=dP_{o}}$ gives

$\displaystyle D\left(\theta\right)\equiv\frac{d\sigma}{d\Omega}=\left|f\left(\theta\right)\right|^{2} \ \ \ \ \ (8)$

Thus the differential cross section is equivalent to the square modulus of the scattering amplitude. The scattering amplitude ${f}$ is what can, in principle, be calculated from the theory, so it provides the link between theory and experiment.