Prerequisite Coulomb wave function
Photoelectron Momentum Spectrum
When calculating the photoionization of hydrogen-like atoms, when the external field disappears, the probability of each energy eigenstate (scattering state) is fixed. However, the eigenstate coefficient of momentum will still change (unless time is infinite). To obtain the three-dimensional momentum distribution of the electron at infinite time, we can directly project the wave function onto the Coulomb wave function (asymptotic plane wave). In fact, such a momentum spectrum is usually called angular resolved energy spectrum (after all, it is the eigenstate of energy). For convenience, we still call it the momentum spectrum directly.
\begin{equation}
P( \boldsymbol{\mathbf{k}} ) = \left\lvert f( \boldsymbol{\mathbf{k}} ) \right\rvert ^2 = \left\lvert \left\langle \Psi^{(-)}_{ \boldsymbol{\mathbf{k}} } \middle| \Psi( \boldsymbol{\mathbf{r}} ) \right\rangle \right\rvert ^2
\end{equation}
Normalized to
\begin{equation}
\int P( \boldsymbol{\mathbf{k}} ) \,\mathrm{d}^{3}{k} = \iint P( \boldsymbol{\mathbf{k}} ) \,\mathrm{d}{\Omega} \cdot k^2 \,\mathrm{d}{k} = 1
\end{equation}
So how to calculate the energy spectrum of the ionized photoelectron (PE or photo-electron)? First look at the distribution of momentum, normalized to
\begin{equation}
\int_0^\infty P(k) \,\mathrm{d}{k} = 1
\end{equation}
The energy distribution is normalized to
\begin{equation}
\int P(E) \,\mathrm{d}{E} = \int P(E) k \,\mathrm{d}{k} = 1
\end{equation}
The relationship between the two is $P(k) = kP(E)$.
We only need to project the total wave function onto the subspace with the absolute value of momentum $k$. Compare the above formulas
\begin{equation}
P(k) = k^2 \int P( \boldsymbol{\mathbf{k}} ) \,\mathrm{d}{\Omega}
\end{equation}
However, the wave function of hydrogen-like atoms is generally performed in spherical coordinates, and we try to directly use the (radial) Coulomb function to calculate it. Project the wave function to the normalized Coulomb spherical wave eq. 9 to get
\begin{equation}
P(l, m, k) = \left\lvert f_{l,m}(k) \right\rvert ^2
\end{equation}
among them
\begin{equation}
f_{l,m}(k) = \left\langle C_{l,m}(k) \middle| \Psi( \boldsymbol{\mathbf{r}} ) \right\rangle = \sqrt{\frac{2}{\pi}} \int F_l(k, r) \psi_{l,m}(r) \,\mathrm{d}{r}
\end{equation}
Among them, $\psi_{l,m}(r)$ is the scaled radial wave function
eq. 1 .
Normalized to1
\begin{equation}
\sum_{l,m} \int P(l, m, k) \,\mathrm{d}{k} = 0
\end{equation}
Compare
eq. 3 and
eq. 4 to get
\begin{equation}
P(k) = \sum_{l,m} P(l, m, k) = \frac{2}{\pi} \sum_{l,m} \left\lvert \int F_l(k, r) \psi_{l,m}(r) \,\mathrm{d}{r} \right\rvert ^2
\end{equation}
Although this formula seems to only include the distribution of radial kinetic energy, it actually has angular kinetic energy, which is reflected in the $l$ quantum number
2.
1. ^ Why is there no $k^2$ item for $k$ points? This depends on the normalization of the Coulomb spherical wave eq. 10 .
2. ^ Think about how $l$ in the differential equation of the Coulomb function determines the angular kinetic energy ? Note that it has nothing to do with $m$.