Prerequisite Stationary Schrödinger equation in spherical coordinate system
, atomic unit system
This article uses the atomic unit system . hydrogen-like atom is defined as an atom/ion with $Z$ protons in the nucleus (nuclear charge $+Ze$), an atom/ion with an extranuclear electron, such as a hydrogen atom and a helium atom that has lost an electron $\mathrm{He}^+$, Lithium ion that lost two electrons $\mathrm{Li}^{++}$.
The stationary-state Schrödinger equation of a hydrogen-like atom is
\begin{equation}
-\frac{1}{2m} \boldsymbol{\nabla}^2 \psi( \boldsymbol{\mathbf{r}} ) - \frac{Z}{r} \psi( \boldsymbol{\mathbf{r}} ) = E \psi( \boldsymbol{\mathbf{r}} )
\end{equation}
The hydrogen-like atom is the only atom (ion) that has an analytical solution.
We only discuss the bound state here, that is, the solution of $E < 0$. Mathematically, we can find a solution when $E$ takes any value less than zero, but only when $E$ takes a specific discrete value can these wave functions be normalized (otherwise it has no physical meaning). Since hydrogen-like atoms have spherical symmetry, the wave function in spherical coordinates has the simplest form. The wave function is expressed as
\begin{equation}
\psi_{nlm} (r,\theta ,\phi) = R_{nl}(r) Y_{l,m}(\theta, \phi)
\end{equation}
Among them, $n$ is
principal quantum number ($n = 1, 2, \dots$), $l$ is
angular quantum number ($l = 0, 1, \dots, n - 1$), and $m$ is
magnetic quantum number ($m = -l, -l+1, \dots, l$). $R_{nl}(r)$ is the normalized
radial wave function, and $Y_{l,m}(\theta, \phi)$ is the normalized
spherical harmonic function (see "List of spherical harmonic functions
").
Fig. 1：Probability density of hydrogen atom wave function $x$-$z$ cross section of $ \left\lvert \psi_{nlm}( \boldsymbol{\mathbf{r}} ) \right\rvert ^2$, the size is proportional. The three numbers in each figure are quantum numbers $n, l, m$, and the probability of electrons appearing inside the white circle is 0.95. The corresponding value of the color bar is linear, and the value range of the color bar in each sub-graph is different. See also Matlab drawing program
.
Radial wave function $R_{nl}(r)$
If the motion of the nucleus is ignored, the following $a$ is the Bohr radius ($a=1$ in atomic units), if not ignored, $a$ is the reduced Bohr radius. Note that the functions of $Z$ and $a$ are to shrink the radial wave function about the origin by a factor of $Z/a$ (and keep the wave function normalized).
\begin{equation}
R_{nl}(r) = \sqrt{ \left(\frac{2 Z}{na} \right) ^3 \frac{(n - l - 1)!}{2n (n + l)!}} \left(\frac{2Zr}{na} \right) ^l L_{n-l-1}^{2l+1} \left(\frac{2Zr}{na} \right) \mathrm{e} ^{-Zr/(na)}
\end{equation}
Where $L_n^l(x)$ is
associated Laguerre polynomial.
The function graph of $r R_{n,l}(r)$ at $Z = 1$1 such as fig. 2
Fig. 2：Radial wave function graph (using atomic units
, $a = 1$)
The first few radial wave functions are given below. Note that the values of all radial wave functions are real numbers.
\begin{equation} n = 1 \qquad
R_{10}(r) = 2 \left(\frac{Z}{a} \right) ^{3/2} \exp\left(-Zr/a\right)
\end{equation}
\begin{equation}
n = 2 \qquad
\left\{\begin{aligned}
R_{20}(r) &= \frac{1}{\sqrt 2} \left(\frac{Z}{a} \right) ^{3/2} \left(1 - \frac12 \frac{Zr}{a} \right) \exp\left(-\frac{Zr}{2a}\right) \\
R_{21}(r) &= \frac{1}{2\sqrt{6}} \left(\frac{Z}{a} \right) ^{3/2} \frac{Zr}{a} \exp\left(-\frac{Zr}{2a}\right)
\end{aligned}\right. \end{equation}
\begin{equation}
n = 3 \qquad
\left\{\begin{aligned}
R_{30}(r) &= \frac{2}{3\sqrt {3}} \left(\frac{Z}{a} \right) ^{3/2} \left(1 - \frac23 \frac{Zr}{a} + \frac{2}{27} \frac{Z^2r^2}{a^2} \right) \exp\left(-\frac{Zr}{3a}\right) \\
R_{31}(r) &= \frac{8}{27\sqrt 6} \left(\frac{Z}{a} \right) ^{3/2} \left(1 - \frac16 \frac {Zr}{a} \right) \frac {Zr}{a} \exp\left(-\frac{Zr}{3a}\right) \\
R_{32}(r) &= \frac{4}{81\sqrt {30}} \left(\frac{Z}{a} \right) ^{3/2} \frac{Z^2r^2}{a^2} \exp\left(-\frac{Zr}{3a}\right) \end{aligned}\right.
\end{equation}
More $R_{n,l}$ available MathematicaOr
Wolfram Alpha to generate, the command is like
n = 4; l = 1;
Sqrt[(2*Z/(n*a))^3 * Factorial[n-l-1]/(2*n*Factorial[n+l])]
*(2*Z*r/(n*a))^l*LaguerreL[n-l-1, 2l+1, 2Z*r/(n*a)] * Exp[-Z*r/(n*a)]
Or use Matlab for numerical calculation
Rnl = @(Z,n,l,r)sqrt((2*Z/n)^3*factorial(n-l-1)/(2*n*factorial(n+l))) *...
(2*Z*r/n).^l .* laguerreL(n-l-1, 2*l+1, 2*Z*r/n) .* exp(-Z*r/n);
Nature
We require that each bound state of the hydrogen atom meets the normalization condition
\begin{equation}
\int \left\lvert \Psi_{n,l,m}( \boldsymbol{\mathbf{r}} ) \right\rvert \,\mathrm{d}^{3}{r} = \int \int_0^\infty \left\lvert R_{n,l}(r) Y_{l,m}( \hat{\boldsymbol{\mathbf{r}}} ) \right\rvert ^2 r^2 \,\mathrm{d}{r} \,\mathrm{d}{\Omega} = 1
\end{equation}
First do the angular integration, the spherical harmonic function has satisfied the normalization condition
\begin{equation}
\int \left\lvert Y_{l,m}( \hat{\boldsymbol{\mathbf{r}}} ) \right\rvert ^2 \,\mathrm{d}{\Omega} = 1
\end{equation}
The normalization condition to obtain the radial wave function is
\begin{equation}
\int [rR_{n,l}(r)]^2 \,\mathrm{d}{r} = 1
\end{equation}
Looking at the orthogonality again, we know that the eigenstates of the Hamiltonian are orthogonal to each other (where at least one of $n',l',m'$ is different from $n, l, m$)
\begin{equation}
\int \Psi_{n,l,m}^*( \boldsymbol{\mathbf{r}} ) \Psi_{n',l',m'}( \boldsymbol{\mathbf{r}} ) \,\mathrm{d}^{3}{r}
= \int \int_0^\infty R_{n,l}(r) R_{n',l'}(r) Y_{l,m}^*( \hat{\boldsymbol{\mathbf{r}}} ) Y_{l',m'}( \hat{\boldsymbol{\mathbf{r}}} ) r^2 \,\mathrm{d}{r} \,\mathrm{d}{\Omega} = 0
\end{equation}
The same can be done in the diagonal direction first, if at least one of $l',m'$ is different from $l, m$, then the integral is directly zero, and the radial wave function does not require any orthogonality conditions (and indeed does not satisfy). But if the two spherical harmonic functions are the same, that is, $l' = l$, $m' = m$, $n' \ne n$, then the angular integral is equal to 1, and the radial wave function satisfies
\begin{equation}
\int \int_0^\infty R_{n,l}(r) R_{n',l}(r) r^2 \,\mathrm{d}{r} = 0
\end{equation}
Radial probability distribution
Let's find the radial probability distribution $P(r)$. The definition of $P(r)$ is: The probability of finding a particle in $r \in [a, b]$ (thick spherical shell) is $\int_a^b P(r) \,\mathrm{d}{r} $. Since the square of the modulus length of the wave function is the three-dimensional probability density, there is
\begin{equation}
\int_a^b P(r) \,\mathrm{d}{r} = \int_a^b \int \left\lvert \frac1r \psi_{l,m}(r) Y_{l,m}( \hat{\boldsymbol{\mathbf{r}}} ) \right\rvert ^2 \,\mathrm{d}{\Omega} r^2 \,\mathrm{d}{r}
= \int_a^b \left\lvert \psi_{l,m}(r) \right\rvert ^2 \,\mathrm{d}{r}
\end{equation}
Is true for any $a, b > 0$, so there is
\begin{equation}
P(r) = \left\lvert \psi_{l,m}(r) \right\rvert ^2
\end{equation}
Arbitrary wave function can be expressed as the superposition of all eigenwave functions
\begin{equation}
\Psi( \boldsymbol{\mathbf{r}} ) = \frac1r \sum_{l,m} \psi_{l,m}(r) Y_{l,m}( \hat{\boldsymbol{\mathbf{r}}} )
\end{equation}
Its radial probability distribution is
\begin{equation}
\int_a^b P(r) \,\mathrm{d}{r} = \int_a^b \int \left\lvert \frac1r \sum_{l,m}\psi_{l,m}(r) Y_{l,m}( \hat{\boldsymbol{\mathbf{r}}} ) \right\rvert ^2 \,\mathrm{d}{\Omega} r^2 \,\mathrm{d}{r}
= \sum_{l,m} \int_a^b \left\lvert \psi_{l,m}(r) \right\rvert ^2 \,\mathrm{d}{r}
\end{equation}
Holds for any $a, b > 0$, so there is
\begin{equation}
P(r) = \sum_{l,m} \left\lvert \psi_{l,m}(r) \right\rvert ^2
\end{equation}
Momentum Representation Momentum Distribution
To require the wave function under the momentum representation, we need to project the wave function of the position representation onto the eigenvector of the normalized momentum, that is, the three-dimensional Fourier transform
\begin{equation}
\psi_{nlm}( \boldsymbol{\mathbf{p}} ) = \left\langle \boldsymbol{\mathbf{p}} \middle| \psi \right\rangle = \frac{1}{\sqrt{2\pi}} \int \exp\left(- \mathrm{i} \boldsymbol{\mathbf{p}} \boldsymbol\cdot \boldsymbol{\mathbf{r}} \right) \psi( \boldsymbol{\mathbf{r}} ) \,\mathrm{d}^{3}{r}
\end{equation}
This integral is the most convenient to complete in spherical coordinates. We will illustrate the specific method with an example (see
ex. 1 ).
Just as the distribution function of position under the position representation is $ \left\lvert \psi( \boldsymbol{\mathbf{r}} ) \right\rvert ^2$, the distribution function of momentum under the momentum representation is $ \left\lvert \psi( \boldsymbol{\mathbf{p}} ) \right\rvert ^2$ (also in accordance with the measurement theory).
1. ^ We will see that $r R_{l,m}(r)$ is more commonly used than $R_{l,m}(r)$ in the future
