Prerequisite Quantum Scattering
1This article uses the atomic unit system . Coulomb wave function is a hydrogen-like atomThe wave function of the scattering state. It is the solution of the stationary Schrödinger equation to the energy $E > 0$ under the Coulomb potential energy $V(r) = Z/r$ ($Z < 0$).
There are two commonly used boundary conditions for Coulomb wave functions. The Coulomb wave function of plane wave incidence (spherical wave exit) with wave number $ \boldsymbol{\mathbf{k}} $ is recorded as $\psi_{ \boldsymbol{\mathbf{k}} }^{(+)}( \boldsymbol{\mathbf{r}} )$, and the scattering state of plane wave exit (spherical wave incidence) is recorded as $\psi_{ \boldsymbol{\mathbf{k}} }^{(-)}( \boldsymbol{\mathbf{r}} )$. The first type of boundary strip is used to calculate the scattering
\begin{equation}
\psi_{ \boldsymbol{\mathbf{k}} }^{(\pm)}( \boldsymbol{\mathbf{r}} ) \to \frac{1}{(2\pi)^{3/2}} \left\{ \mathrm{e} ^{ \mathrm{i} \boldsymbol{\mathbf{k}} \boldsymbol\cdot \boldsymbol{\mathbf{r}} + \mathrm{i} \eta \ln\left(kr - \boldsymbol{\mathbf{k}} \boldsymbol\cdot \boldsymbol{\mathbf{r}} \right) }
+ f( \hat{\boldsymbol{\mathbf{k}}} ) \frac{ \exp\left[ \mathrm{i} kr - \mathrm{i} \eta \ln\left(2kr\right) \right] }{r} \right\}
\qquad
( \boldsymbol{\mathbf{r}} \boldsymbol\cdot \boldsymbol{\mathbf{k}} \to \mp\infty)
\end{equation}
In the parabolic coordinate system
can be solved
2 The analytical formula of the Coulomb wave function that satisfies the boundary is
\begin{equation}
\psi_{ \boldsymbol{\mathbf{k}} }^{(\pm)}( \boldsymbol{\mathbf{r}} ) = \frac{1}{(2\pi)^{3/2}} \Gamma(1\pm \mathrm{i} \eta) \mathrm{e} ^{-\pi\eta/2} \mathrm{e} ^{ \mathrm{i} \boldsymbol{\mathbf{k}} \boldsymbol\cdot \boldsymbol{\mathbf{r}} } {_1F_1}(\mp \mathrm{i} \eta; 1; \pm \mathrm{i} kr - \mathrm{i} \boldsymbol{\mathbf{k}} \boldsymbol\cdot \boldsymbol{\mathbf{r}} )
\end{equation}
Among them, $\eta = mZ/k$ is called
Sommerfeld parameter, $\rho = kr$, $k$ is the wave number of a plane wave with energy $E = k^2/(2m)$, and $Z$ is the product of the charge of the nucleus and electrons (pay special attention to $Z < 0$). The above formula is satisfied
\begin{equation}
\psi_{ \boldsymbol{\mathbf{k}} }^{(+)}( \boldsymbol{\mathbf{r}} ) = \psi_{- \boldsymbol{\mathbf{k}} }^{(-)}(r)^*
\end{equation}
\begin{equation}
\langle{\psi_{ \boldsymbol{\mathbf{k}} }^{(\pm)}}|{\psi_{ \boldsymbol{\mathbf{k}} '}^{(\pm)}}\rangle = \delta( \boldsymbol{\mathbf{k}} - \boldsymbol{\mathbf{k}} ')
\end{equation}
Fig. 1：$\psi_{ \boldsymbol{\mathbf{k}} }^{(+)}( \boldsymbol{\mathbf{r}} )$, $Z = -5$, $ \boldsymbol{\mathbf{k}} = \boldsymbol{\mathbf{y}} $, $\eta = -5$. When $y \to -\infty$ is close to a plane wave. From the perspective of classical mechanics, the closer the particle is to the nucleus, the more the incident deflection. Note that the plane waves on the left and right sides will intersect and interfere after being deflected.
Another boundary condition is to use separate variables to solve the Schrodinger equation in spherical coordinates. The eigenstate is
\begin{equation}
\psi_{l,m}( \boldsymbol{\mathbf{r}} ) = \frac{1}{r} \sqrt{\frac{2}{\pi}} F_l(\eta, kr) Y_{l,m}( \hat{\boldsymbol{\mathbf{r}}} )
\end{equation}
Orthogonal normalization conditions are
\begin{equation}
\left\langle \psi_{k', l',m'} \middle| \psi_{k, l,m} \right\rangle = \delta_{l,l'}\delta_{m,m'} \delta(k'-k)
\end{equation}
Coulomb wave function in spherical coordinates
Prerequisite Schrodinger radial equation
The relationship between the Coulomb wave function in spherical coordinates and the "Coulomb plane wave" above can be compared to the relationship between plane waves and spherical waves . That is, the space formed by the plane wave base in all directions with the same $k$ is the same as the space formed by the spherical wave base, which is an infinite-dimensional subspace with energy $E = k^2/(2m)$.
The Schrödinger radial equation of the Coulomb potential (eq. 10 ) is
\begin{equation}
-\frac{1}{2m} \frac{\mathrm{d}^{2}{u}}{\mathrm{d}{r}^{2}} + \left[\frac{Z}{r} + \frac{l(l+1)}{2mr^2} \right] u = \frac{k^2}{2m}u
\end{equation}
Among them, $u(r)$ is the Scaled wave function, $l$ is the angular quantum number, and $Z$ is the product of the nuclear charge and the electronic charge (note that it is less than zero). The equivalent potential energy in square brackets is like
fig. 2 .
Fig. 2：Radial equivalent potential energy in spherical coordinates
Use $\eta = mZ/k$ and $\rho = kr$ in exchange for yuan, get
\begin{equation}
\frac{\mathrm{d}^{2}{u}}{\mathrm{d}{\rho}^{2}} + \left[1 - \frac{2\eta}{\rho} - \frac{l(l+1)}{\rho^2} \right] u = 0
\end{equation}
Two linearly independent solutions are
Coulomb function of the first kind $F_l(\eta, \rho)$ and
Coulomb function of the second kind $G_l(\eta, \rho)$
. In the scattering state, we assume $k > 0$ ($E > 0$). But theoretically we can also take $k$ as a positive imaginary number ($E < 0$), so that when $E$ is the bound state energy, we can get the bound state radial wave function (
eq. 3 ).
It can be proved that Coulomb spherical wave of complete orthogonal normalization is
\begin{equation}
\left\lvert C_{l,m}(k) \right\rangle = \frac{1}{r} \sqrt{\frac{2}{\pi}} F_l(\eta, kr) Y_{l,m}( \hat{\boldsymbol{\mathbf{r}}} )
\end{equation}
Satisfy
\begin{equation}
\left\langle C_{l',m'}(k') \middle| C_{l,m}(k) \right\rangle = \delta_{l,l'}\delta_{m,m'}\delta(k-k')
\end{equation}
Now the Coulomb wave function of the first boundary condition (eq. 1 ) can be expanded as
\begin{equation}
\psi_{ \boldsymbol{\mathbf{k}} }^{(\pm)}( \boldsymbol{\mathbf{r}} ) = \sum_{l,m} a_{l,m}^{(\pm)}( \boldsymbol{\mathbf{k}} ) \left\lvert C_{l,m}(k) \right\rangle
\end{equation}
among them
\begin{equation}a_{l,m}^{(\pm)}( \boldsymbol{\mathbf{k}} ) = \frac{ \mathrm{i} ^l}{k} \exp\left[\pm \mathrm{i} \sigma_l(\eta)\right] Y_{l,m}^* ( \hat{\boldsymbol{\mathbf{k}}} )
\end{equation}
$\sigma_l$ is the Coulomb phase shift (
eq. 8 ). Note that $a_{l,m}$ is not $ \langle{C_{l,m}(k)}|{\psi_{ \boldsymbol{\mathbf{k}} }^{(\pm)}}\rangle $, the latter's modulus length is infinite, which can also be recorded as $a_{l,m}\delta(0)$.
projection
Similar to the case of plane waves (Fourier transform), to project a spherical harmonic expansion wave function $ \left\lvert f \right\rangle $ onto the Coulomb wave function, first project onto the Coulomb spherical wave, and then perform unitary transformation
\begin{equation}
\langle{\psi_{ \boldsymbol{\mathbf{k}} }^{(\pm)}}|{f}\rangle = \sum_{l,m} a_{l,m}^{(\pm)}( \boldsymbol{\mathbf{k}} )^* \left\langle C_{l,m}(k) \middle| f \right\rangle
\end{equation}
If the scaled radial wave function of $ \left\lvert f \right\rangle $ is $u_{l,m}(r)$, then
\begin{equation}
\langle{\psi_{ \boldsymbol{\mathbf{k}} }^{(\pm)}}|{f}\rangle = \frac{1}{k} \sum_{l,m} g_{l,m}^\pm(k) Y_{l,m}( \hat{\boldsymbol{\mathbf{k}}} )
\end{equation}
among them
\begin{equation}
g_{l,m}^\pm(k) = \sqrt{\frac{2}{\pi}} \mathrm{i} ^{-l} \mathrm{e} ^{\mp \mathrm{i} \sigma_l} \int_0^\infty F_l(\eta, kr) u_{l,m}(r) \,\mathrm{d}{r}
\end{equation}
What is the difference between the Fourier transform of the wave function and the projection to the Coulomb wave function? For example, TDSE for the ionization of hydrogen atoms, if you want to find the momentum distribution at $t = +\infty$, in theory, you only need to do Fourier transform when $t$ is large enough, but if $t$ is not large enough (the electric field has disappeared), the ionization wave packet The Coulomb force experienced cannot be ignored, so although the instantaneous momentum distribution is obtained, it is different from that of $t = +\infty$. At this time, because the momentum operator is not compatible with the Hamiltonian, momentum is not conserved. But if it is projected to $\psi_{ \boldsymbol{\mathbf{k}} }^{(-)}$ when the electric field disappears, since it is the Hamiltonian eigenfunction, the modulus (probability) of the projection will not change with time.
1. ^ One of the reference materials is Wikipedia, and the other is the appendix of "F Morales et al 2016 J. Phys. B: At. Mol. Opt. Phys. 49 245001". It is said that [17] also has it.
2. ^ derived in [19] .
