Daily Archives: Mon, 15 June 2015

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.