Scattering by a Coulomb potential #

1. Abstract #

Consider a scenario in which an electron with charge $q_B$ collides with an atomic nucleus with charge $q_A$ from a distance. For instance, we can visualize the process of an electron ($q_A=-e$, where e represents the elementary charge) colliding from afar with a proton, which is the nucleus of a hydrogen atom ($q_B=+e$).

The time-independent Schrödinger equation in relative coordinate can be written as,

\begin{align} \label{tise0} \left[ -\frac{\hbar^2}{2\mu}\nabla^2+\frac{q_Aq_B}{4\pi\varepsilon_0 r} \right]\psi_c(\mathbf{r})=E\psi_c(\mathbf{r}) \end{align}

Here, $\hbar$ stands for Planck’s constant, $\varepsilon_0$ for the permittivity of vacuum, $\mu$ for the reduced mass of A and B, and $E$ represents the energy of collision.

To simplify, we collect coefficients and rewrite the equation as, \begin{align} \left[ \nabla^2-\frac{U_0}{r}+k^2\right]\psi_c(\mathbf{r})=0 \end{align} where we define, \begin{align} U_0\equiv \frac{2\mu}{\hbar^2}\frac{q_Aq_B}{4\pi\varepsilon_0},~~ k\equiv \sqrt{\frac{2\mu E}{\hbar^2}}. \end{align} Now, we want to determine the responce when a incident-plane wave defined by momentum $k$ will collide to a proton. To do so, we solve equation \eqref{tise0} under the boundary condition, \begin{align} \label{bc0} \psi_c(\mathbf{r})|_{z\to -\infty}=e^{ikz}. \end{align}

The result appears as, \begin{align} &\psi_c(\mathbf{r})\Bigr|_{|r-z|\to\infty} \nonumber\\ &=\exp{\biggl(ikz-i\gamma \ln\Bigl[kr(1-\cos\theta)\Bigr]\biggr)} \cdot\left(1+\frac{\gamma^2}{ikr(1-\cos\theta)}+\cdots\right) \nonumber\\ &\hspace{13em}+f(k,\theta)\frac{\exp \Bigl[ikr -i\gamma \ln(2kr)\Bigr] }{r} \label{asymcoulomb} \end{align} Note that this expansion is not applicable on or near the z-axis where $|r-z|\to 0$. The original boundary condition \eqref{bc0} is only determined at $z\to -\infty$, with no conditions on x and y, but the conclusion derived is strictly valid only for regions distant from the z-axis.

We have \begin{align} \gamma=\frac{U_0}{2k}=\frac{q_Aq_B}{4\pi\varepsilon_0}\frac{1}{\hbar\nu}=\frac{q_Aq_B}{4\pi\varepsilon_0}\frac{\mu}{\hbar^2 k}, \end{align} where $\gamma$ is known as the Sommerfeld parameter, and $\nu=\frac{\hbar k}{\mu}$ represents the relative velocity between the two particles. Additionally, \begin{align} f(k,\theta)=-\frac{\gamma}{2k\sin^2\frac{\theta}{2}} \cdot\exp\left[-i\gamma\ln\left(\sin^2\frac{\theta}{2}\right)\right] \cdot\frac{\Gamma(1+i\gamma)}{\Gamma(1-i\gamma)} \end{align} represents the Coulomb scattering amplitude, a measure of how much the spherical wave is distorted in the direction of angle $\theta$ relative to the incident direction of the plane wave with wave number $k$.

$\Gamma(x)$ denotes the gamma function, and it can sometimes be written as, \begin{align} \frac{\Gamma(1+i\gamma)}{\Gamma(1-i\gamma)}=\exp(2i\sigma_0) \end{align} $\sigma_0\equiv\arg(\Gamma(1+i\gamma))$ is referred to as the phase shift.

The first term on the right-hand side of equation \eqref{asymcoulomb} represents the incident wave, while the second term denotes the outgoing spherical wave. The term including $\ln$ inside the exponential function is due to logarithmic divergence, which disturb the plane wave form of $e^{ikz}$. This is a consequence of the Coulomb potential being a long-range potential, implying that the influence of scattering is felt even at great distances. In other words, it signifies that the scattering radius of the Coulomb potential is infinite.

This contents are based on the Physics of Atoms and Molecules (2nd Edition) Sec. 12.5, “The Coulomb potential”, written by B. H. Bransden and C. J. Joachain¹.

Physics of Atoms and Molecules

B. H. Bransden and C. J. Joachain, Benjamin Cummings

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2. Scattering Problem #

We solve the time-independent Schrödinger equation, transformed into \begin{align} \label{tise} \left[ \nabla^2-\frac{U_0}{r}+k^2\right]\psi_c(\mathbf{r})=0, \end{align} under the boundary conditions¹² \begin{align} \label{bc} \psi_c(\mathbf{r})|_{z\to -\infty}=e^{ikz}. \end{align} The boundary conditions assume a process where a plane wave with wave number $k$ is incident from $z\to-\infty$, scattered by an atomic nucleus existing at the origin, and becomes an outgoing spherical wave.

2.1. Reason to Consider a Plane Wave as an Incident Wave #

Why do we consider a solution that enters as a plane wave?

One answer to this question is the Fourier transformation. For example, a wave packet that can be prepared in an experiment will always be a wave packet within a finite range. Therefore, an ideal plane wave does not exist, and we have to either treat it approximately as a plane wave or treat it as a finite range wave packet. If the former, there is no problem with the discussion of plane waves. However, in the latter case, it cannot be expressed as a plane wave. Here comes the Fourier transformation. When an arbitrary initial state is prepared, it can be expressed as a superposition of plane waves, so if you know the response to a plane wave, you know the response to an arbitrary initial state. That’s the aim.

This is one reason to consider a solution that enters as a plane wave.

3. Scattering State #

Now, the problem of the Coulomb potential, like an atom, can be solved by variable separation in spherical or parabolic coordinates. The problem we want to consider now is how a plane wave is scattered by the Coulomb potential. In problems with a specific direction, such as a plane wave, there is a spherical and a specific linear direction. In that case, it is convenient to separate variables using the parabolic coordinate system.

3.1. Parabolic Coordinate #

The parabolic coordinate is defined by⁴³ \begin{align} \left\{ \begin{array}{l} \xi= r+z,\hspace{4em} (0\leq \xi < \infty)\\ \eta=r-z,\hspace{4em} (0\leq \eta < \infty)\\ \phi=\tan^{-1}\left(\frac{y}{x}\right),\hspace{1.5em} (0\leq \phi < 2\pi)\ \end{array} \right. \end{align} In the parabolic coordinate system, the Laplacian is derived as follows. \begin{align} \label{e3} \nabla^2=\frac{4}{\xi+\eta}\left(\frac{\partial}{\partial \xi}\xi\frac{\partial}{\partial \xi}+\frac{\partial}{\partial \eta}\eta\frac{\partial}{\partial \eta}\right) +\frac{1}{\xi\eta}\frac{\partial^2}{\partial \phi^2} \end{align}

3.2. Solution in $\phi$ #

Looking at the Laplacian in the parabolic coordinate, we notice that it seems to be solvable only in the $\phi$ direction. Therefore, we assume that the scattering solution can be separated by variables. Namely \begin{align} \psi_c(\mathbf{r})=\psi_c(\xi, \eta)\psi(\phi). \end{align}

We omit the specific steps, but if you substitute $\psi(\phi)$ into \eqref{tise} and solve it, you can write \begin{align} \psi(\phi)\propto e^{im\phi}. \end{align}

Since we are considering a problem of scattering solutions from an incident plane wave and a spherically symmetric potential, we can expect that there will be no dependence on $\phi$ from the scattering solutions obtained⁵. In other words, we know that $m=0$. Therefore, we adopt $m=0$ and get⁶¹ \begin{align} \label{psicr} \psi_c(\mathbf{r})=\psi_c(\xi, \eta). \end{align}

3.3. Solutions in the $\xi, \eta$ direction #

Let’s consider the equations for the $\xi$ and $\eta$. By substituting the Laplacian of the parabolic coordinate \eqref{e3} into the time-independent Schrödinger equation \eqref{tise}, we get:

\begin{gather} \left[\nabla^2-\frac{U_0}{r}+k^2\right]\psi_c(\mathbf{r})=0 \\ \left[\frac{4}{\xi+\eta}\left(\frac{\partial}{\partial \xi}\xi\frac{\partial}{\partial \xi}+\frac{\partial}{\partial \eta}\eta\frac{\partial}{\partial \eta}\right) +\frac{1}{\xi\eta}\frac{\partial^2}{\partial \phi^2}+k^2-\frac{2}{\xi+\eta}U_0\right]\psi_c(\xi, \eta)=0 \end{gather} By multiplying both sides by $\frac{\xi+\eta}{4}$, we obtain \begin{align}\label{e4} \left[\frac{\partial}{\partial \xi}\xi\frac{\partial}{\partial \xi}+\frac{1}{4} k^2\xi+\frac{\partial}{\partial \eta}\eta\frac{\partial}{\partial \eta}+\frac{1}{4} k^2\eta-\frac{U_0}{2}\right]\psi_c(\xi, \eta)=0 \end{align} We introduce separation constants $\tilde{\nu}_1, \tilde{\nu}_2$. The separation constants are determined so as to satisfy: \begin{align} \frac{U_0}{2}=\tilde{\nu}_1+\tilde{\nu}_2. \end{align}

Currently, the operators inside the brackets of equation \eqref{e4} are independent in the $\xi$ and $\eta$ parts and are written as the sum of each. In this case, we consider the solution as separable variables and write it as: \begin{align} \label{psicx} \psi_c(\xi, \eta)=f(\xi)g(\eta) \end{align} Substitute and simplify, we get: \begin{align} \left\{ \begin{aligned} & \left[\frac{d}{d \xi}\xi\frac{d}{d \xi}+\frac{1}{4} k^2\xi-\tilde{\nu}_1\right]f(\xi)=0\\ & \left[\frac{d}{d \eta}\eta\frac{d}{d \eta}+\frac{1}{4} k^2\eta-\tilde{\nu}_2\right]g(\eta)=0 \\ & \tilde{\nu}_1+\tilde{\nu}_2=\gamma k \end{aligned} \right. \label{e7} \end{align} Here we set: \begin{align} \gamma\equiv\frac{U_0}{2k}=\frac{q_Aq_B}{4\pi\varepsilon_0}\frac{1}{\hbar\nu}=\frac{q_Aq_B}{4\pi\varepsilon_0}\frac{\mu}{\hbar^2 k} \end{align} $\gamma$ is a dimensionless quantity called the Sommerfeld parameter, which characterizes whether the system is a low energy scatterer. Also, $\nu=\frac{\hbar k}{\mu}$ represents the relative velocity between two particles.

Now, let’s solve equation \eqref{e7}. To solve it, we consider the boundary condition \eqref{bc}. We are considering that the incident plane wave, with wave number $k$, comes from $z\to-\infty$ as a plane wave $e^{ikz}$. Considering this condition in the parabolic coordinate, we can write: \begin{align} \psi_c(\mathbf{r})|_{z\to -\infty}&=e^{ikz}\nonumber\\ &=\exp\left(\frac{1}{2}ik(\xi-\eta)\right) \label{bcp} \end{align} Since $\xi, \eta \geq 0$, the only region to represent $z\to -\infty$ is $\eta\to\infty$. Therefore, $z\to-\infty$ corresponds to $\eta\to\infty$. However, there is no restriction about $\xi$.

Therefore, the boundary condition \eqref{bcp} can be interpreted as follows:

• The function $g(\eta)$ in the $\eta$ direction must take the form $e^{-ik\eta/2}$ as $\eta\to\infty$
• The function $f(\xi)$ in the $\xi$ direction must take the form $e^{ik\xi/2}$ everywhere in $\xi$

Therefore, considering the boundary conditions, the solution in the $\xi$ direction is thought to be: \begin{align}\label{e8} f(\xi)=e^{ik\xi/2}. \end{align}

Assuming that the solution is given by equation \eqref{e8}, let’s substitute it into the first equation of \eqref{e7}. In the first equation of \eqref{e7}, only the separation constants are undetermined, so let’s derive them. Then, from: \begin{align} \left[\frac{d}{d \xi}\xi\frac{d}{d \xi}+\frac{1}{4} k^2\xi-\tilde{\nu}_1\right]e^{ik\xi/2}=0 \end{align} We get: \begin{align} \left(\frac{1}{2}ik-\tilde{\nu}_1\right)e^{ik\xi/2}=0 \label{e9} \end{align} Since equation \eqref{e9} must always hold regardless of $\xi$, the bracket on the left side must always be zero. Therefore, the separation constant $\tilde{\nu}_1$ must have the following value: \begin{align} \tilde{\nu}_1=\frac{1}{2}ik \end{align}

Now that we know $\tilde{\nu}_1$, from the third equation of \eqref{e7}, we find that the other separation constant $\tilde{\nu}_2$ is: \begin{align} \tilde{\nu}_2&=\gamma k -\tilde{\nu}_1 \nonumber \\ &=\gamma k -\frac{1}{2}ik. \end{align}

Finally, let’s solve the second differential equation of \eqref{e7} in the $\eta$ direction. The $\eta$ direction must take the form $e^{-ik\eta/2}$ at $\eta\to-\infty$, so we assume that the known part is in the following form and solve for $h(\eta)$. \begin{align} \label{gh} g(\eta)=h(\eta)e^{-ik\eta/2}. \end{align}

Substituting into the second equation of \eqref{e7} and considering the differential equation for $h(\eta)$, we get: \begin{align} \left[\frac{d}{d \eta}\eta\frac{d}{d \eta}+\frac{1}{4} k^2\eta-\tilde{\nu}_2\right]\cdot \bigl(h(\eta)e^{-ik\eta/2}\bigr)=0 \end{align} which gives us: \begin{align}\label{ghsub} e^{-ik\eta/2}\left[\eta\frac{d^2h(\eta)}{d \eta^2}+(1-ik\eta)\frac{d h(\eta)}{d\eta}-\gamma k h(\eta)\right]=0. \end{align} However, since equation \eqref{ghsub} must hold for any $\eta$, it must satisfy: \begin{align} \label{hetadiff} \left[\eta\frac{d^2}{d \eta^2}+(1-ik\eta)\frac{d}{d\eta}-\gamma k\right]h(\eta)=0. \end{align} If this is solved, we can substitute into equation \eqref{gh} to solve for $g(\eta)$.

The form of differential equation \eqref{hetadiff} is known as the Kummer-Laplace differential equation, and its solutions are known as confluent hypergeometric functions. Since equation \eqref{hetadiff} is a second-order linear partial differential equation, the general solution can be written as a linear combination of two independent solutions. These two solutions chosen are the regular solution and irregular solution at the origin. In physics, we are interested in the regular solution, where the function does not diverge at the origin. The regular solution of the confluent hypergeometric functions is written as $M(a,b,z)$, so $h(\eta)$ is written as follows: \begin{align} \label{heta} h(\eta)=cM(-i\gamma, 1,ik\eta) \end{align} Here, $M(a,b,0)=1$ is satisfied⁷, and $c$ is a normalization constant. This gives us $h(\eta)$ completely.

From the above, the scattering solution \eqref{psicr} is, from equations \eqref{psicx}, \eqref{gh} and \eqref{heta}, \begin{align} \psi_c(\mathbf{r})&=\psi_c(\xi, \eta) \nonumber\\ &=ce^{ik\xi/2}e^{-ik\eta/2}M(-i\gamma, 1,ik\eta) \nonumber\\ &=c e^{ikz}M(-i\gamma, 1,ik\eta). \label{psisol} \end{align}

4. Asymptotic form of the scattering state #

To illustrate the current problem, it is represented like this.

4.1. Asymptotic form of the solution #

Let’s see what kind of solutions the scattering state has in the region far away from the atomic nucleus. The reasons for considering the asymptotic form are as follows:

• It’s hard to understand solution \eqref{psisol} since it’s abstract.
• For application, there are problems where we can consider only the Coulomb potential far away from the atomic nucleus, and we want to connect this to future applications (see section 4.3.2 for details).
• To find the normalization constant $c$.
  • Specifically, by finding the asymptotic series of solution \eqref{psisol} as $z \to -\infty$, we expect to find a term including $e^{ikz}$ in the asymptotic series, and we determine $c$ so that the coefficient of this term is 1.

To investigate the asymptotic form of solution \eqref{psisol}, we need to know the asymptotic form of the confluent hypergeometric function $M(a,b,z)$. The asymptotic form of $M(a,b,z)$ can be written as

from Eq. (13.5.1) of reference⁸. Here, $(a)_n$ denotes the Pochhammer symbol, which is defined as \begin{align} (a)_n =a(a+1)(a+2)\cdots(a+n-1) =\frac{\Gamma(a+n)}{\Gamma(a)}, ~~(a)_0=1 \end{align} according to Eq. (6.1.22) of reference⁸.

In the case where the arguments of the confluent hypergeometric function are $M(-i\gamma, 1,ik\eta)$, since $\arg(ik\eta)=\pi/2$, we can use Eq. \eqref{Masy}. That is, \begin{align} &\hspace{-2em}M(-i\gamma, 1,ik\eta)\Bigr|_{\eta\to\infty} \nonumber\\ &\hspace{-3em}= \frac{\Gamma(1)}{\Gamma(1+i\gamma)}e^{\pi \gamma}(ik\eta)^{i\gamma}\sum_{n=0}^{\infty}\frac{(-i\gamma)_n(-i\gamma)_n}{n!}(-ik\eta)^{-n} +\frac{\Gamma(1)}{\Gamma(-i\gamma)}e^{ik\eta}(ik\eta)^{-i\gamma-1}\sum_{n=0}^{\infty}\frac{(1+i\gamma)_n(1+i\gamma)_n}{n!}(ik\eta)^{-n} \\ &\hspace{-3em}= \frac{e^{\pi \gamma}}{\Gamma(1+i\gamma)}(ik\eta)^{i\gamma }\sum_{n=0}^{\infty}\frac{{(-i\gamma)_n}^2}{n!}(-ik\eta)^{-n} +\frac{e^{ik\eta}}{\Gamma(-i\gamma)}(ik\eta)^{-i\gamma}\sum_{n=0}^{\infty}\frac{{(1+i\gamma)_n}^2}{n!}(ik\eta)^{-n-1} \end{align} By using the complex number relation \begin{align} z^c=\exp(c\ln z)=\exp\left[c(\ln|z|+i\arg z)\right] \end{align} we can make the following transformation for the multiplier. \begin{align} (ik\eta)^{i\gamma}&=\exp\left[i\gamma\left\{\ln(k\eta)+i\frac{\pi}{2}\right\}\right] \\ (ik\eta)^{-i\gamma}&=\exp\left[-i\gamma\left\{\ln(k\eta)+i\frac{\pi}{2}\right\}\right] \end{align} Here, we use the fact that $k>0$ and $\eta>0$. Continuing the transformation gives \begin{align} &M(-i\gamma, 1,ik\eta)\Bigr|_{\eta\to\infty} \nonumber \\ &= \frac{e^{\pi\gamma/2}}{\Gamma(1+i\gamma)}\exp\bigl[i\gamma\ln(k\eta)\bigr] \sum_{n=0}^{\infty}\frac{{(-i\gamma)_n}^2}{n!}(-ik\eta)^{-n} -\frac{e^{\pi\gamma/2}e^{ik\eta}}{\Gamma(-i\gamma)}\exp\bigl[-i\gamma\ln(k\eta)\bigr] \sum_{n=0}^{\infty}\frac{{(1+i\gamma)_n}^2}{n!}(ik\eta)^{-n-1} \\ &= \frac{e^{\pi\gamma/2}}{\Gamma(1+i\gamma)}\exp\bigl[i\gamma\ln(k\eta)\bigr] \left[1+\frac{\gamma^2}{ik\eta}+O((k\eta)^{-2})\right] -\frac{e^{\pi\gamma/2}}{\Gamma(-i\gamma)}\exp\bigl[ik\eta-i\gamma\ln(k\eta)\bigr] \frac{1}{ik\eta}\left[1+\frac{(1+i\gamma)^2}{ik\eta}+O((k\eta)^{-2})\right] \label{Masy2} \end{align} By inserting this expansion into solution \eqref{psisol} and considering $\eta\to\infty$, we get \begin{align} \psi_c(\mathbf{r})|_{\eta\to\infty}&=c \frac{e^{\pi\gamma/2}}{\Gamma(1+i\gamma)}\exp\bigl[ikz+i\gamma\ln(k\eta)\bigr] \left[1+\frac{\gamma^2}{ik\eta}+O((k\eta)^{-2})\right] \nonumber \\ &+ c\frac{e^{\pi\gamma/2}}{\Gamma(-i\gamma)}\exp\bigl[ik(z+\eta)-i\gamma\ln(k\eta)\bigr] \frac{1}{ik\eta}\left[1+\frac{(1+i\gamma)^2}{ik\eta}+O((k\eta)^{-2})\right] \end{align} Usually, the order in the curly braces of the second term is up to $O((k\eta)^{-1})$ when considering the $\eta^{-1}$ coefficient of the second term, but I’ll write it for now.

Using $\eta=r-z=r(1-\cos\theta)=2r\sin^2\frac{\theta}{2}$ to represent the incident wave and spherical wave, we get \begin{align} \psi_c(\mathbf{r})\Bigr|_{r-z\to\infty} &=c \frac{e^{\pi\gamma/2}}{\Gamma(1+i\gamma)}\exp\bigl[ikz+i\gamma\ln{kr(1-\cos\theta)}\bigr] \left[1+\frac{\gamma^2}{ikr(1-\cos\theta)}+\cdots\right] \nonumber \\ &+ c\frac{e^{\pi\gamma/2}}{\Gamma(-i\gamma)}\exp\left(ikr-i\gamma\left[\ln(2kr)+\ln\sin^2\frac{\theta}{2}\right]\right) \frac{1}{2ikr\sin^2\frac{\theta}{2}}\left[1+\frac{(1+i\gamma)^2}{ikr(1-\cos\theta)}+\cdots\right]\hspace{3em} \label{psiasy2} \end{align} It can be seen that the first term of \eqref{psiasy2} represents a plane wave and the second term represents a spherical wave (this becomes clear when you remove the logarithmic part of the exponential function).

4.2. Determination of the normalization constant #

For normalization, as decided in \eqref{bc}, we make the coefficient of the exponential function of the incident wave as 1. Therefore, it is understood that we should decide \begin{align} &c \frac{e^{\pi\gamma/2}}{\Gamma(1+i\gamma)}=1,\nonumber \\ &\to c=e^{-\pi\gamma/2}\Gamma(1+i\gamma) \end{align} By setting $c$ in this way, we can transform it as follows. \begin{align} \hspace{-2em}\psi_c(\mathbf{r})\Bigr|_{r-z\to\infty} &=\exp\bigl[ikz+i\gamma\ln{kr(1-\cos\theta)}\bigr] \left[1+\frac{\gamma^2}{ikr(1-\cos\theta)}+\cdots\right] \nonumber \\ &+ \frac{\Gamma(1+i\gamma)}{\Gamma(-i\gamma)}\frac{1}{2ik\sin^2\frac{\theta}{2}} \exp\left(-i\gamma\ln\sin^2\frac{\theta}{2}\right) \frac{\exp\left[ikr-i\gamma\ln(2kr)\right]}{r} \left[1+\frac{(1+i\gamma)^2}{ikr(1-\cos\theta)}+\cdots\right]\hspace{3em} \\ &\hspace{-6em}\mbox{Using $\Gamma(1+z)=z\Gamma(z)$ and $\Gamma(-i\gamma)=\frac{\Gamma(1-i\gamma)}{-i\gamma}$,} \nonumber \\ &=\exp\bigl[ikz+i\gamma\ln{kr(1-\cos\theta)}\bigr] \left[1+\frac{\gamma^2}{ikr(1-\cos\theta)}+\cdots\right] \nonumber \\ &+ \frac{\Gamma(1+i\gamma)}{\Gamma(1-i\gamma)}\frac{-\gamma}{2k\sin^2\frac{\theta}{2}} \exp\left(-i\gamma\ln\sin^2\frac{\theta}{2}\right) \frac{\exp\left[ikr-i\gamma\ln(2kr)\right]}{r} \left[1+\frac{(1+i\gamma)^2}{ikr(1-\cos\theta)}+\cdots\right]\hspace{3em} \end{align} Here, we define the Coulomb scattering amplitude \begin{align} \label{csa} f(k,\theta)\equiv \frac{-\gamma}{2k\sin^2\frac{\theta}{2}} \exp\left[-i\gamma\ln\left(\sin^2\frac{\theta}{2}\right)\right]\frac{\Gamma(1+i\gamma)}{\Gamma(1-i\gamma)} \end{align} And get \begin{align} \psi_c(\mathbf{r})\Bigr|_{r-z\to\infty} &=\exp\bigl[ikz+i\gamma\ln{kr(1-\cos\theta)}\bigr] \left[1+\frac{\gamma^2}{ikr(1-\cos\theta)}+\cdots\right]\nonumber \\ &\hspace{3em}+ f(k,\theta) \frac{\exp\left[ikr-i\gamma\ln(2kr)\right]}{r} \left[1+\frac{(1+i\gamma)^2}{ikr(1-\cos\theta)}+\cdots\right] \hspace{1em}\label{end} \end{align} If we align the orders at the asymptotics, it would be \begin{align} \psi_c(\mathbf{r})\Bigr|_{r-z\to\infty} &=\exp\bigl[ikz+i\gamma\ln{kr(1-\cos\theta)}\bigr] \left[1+\frac{\gamma^2}{ikr(1-\cos\theta)}+\cdots\right]\nonumber \\ &\hspace{3em}+ f(k,\theta)\frac{\exp\left[ikr-i\gamma\ln(2kr)\right]}{r} \end{align}

That’s the end of the derivation up to this point.

4.3. Notes #

Let’s add a few notes here for further clarity.

4.3.1. Coulomb Scattering Amplitude #

The Coulomb scattering amplitude, which appears in Equation \eqref{csa}, represents the degree of scattering when a particle is incident at a momentum $k$ and is reflected in the direction $\theta$. A characteristic feature of this amplitude is its divergence as $\theta \to 0, \pi$.

If we temporarily ignore the logarithm inside the exponential function and everything except 1 inside the parentheses in the first and second lines of \eqref{end}, the first and second lines become $e^{ikz}$ and $\frac{e^{ikr}}{r}$, respectively. This suggests that they might represent a plane wave advancing in the positive z-direction and an outgoing spherical wave. However, because the content of the exponential function is a divergent logarithm, it cannot be described as a perfect plane wave, no matter how far apart.

Even if it cannot be described as a plane wave, for very large $\eta=r-z$, the effect of the Coulomb potential affects only the phase of the wave function (because $\gamma\ln(kr(1-\cos\theta))$ is entirely a real function). In other words, even though the wave pattern of the first term is distorted, it disappears if we take the absolute value.

If you calculate the flux (the flow of probability) $\mathbf{j}(\mathbf{r})$, the result will be the same as for a simple plane wave.

4.3.2. Why the Asymptotic Form is Important #

Now, the asymptotic form of the scattering state $\psi_c(\mathbf{r})$ is of interest in physics. Let’s explain why the asymptotic form is important.

This problem is one of the few that can be solved in all regions as the solution \eqref{psisol} under the Coulomb potential. Generally, the region near the atomic nucleus is not under a Coulomb potential, and only the region very far away can be considered as such. To observe the result of scattering experimentally, you basically look at a distance far enough away from the size of the atomic nucleus, so in that region, it becomes a linear combination of the differential equations under the Coulomb potential.

That’s why the asymptotic form is important.

5. References and Footnotes #