Hydrogen Atom
- Quantum Mechanics Unclassified
· 14 min 
from 「Modern Quantum Mechanics」: Sakurai, J. J.
A naive but quite accurate model of hydrogen.
Introduction
The Schrödinger equation allows one to calculate the stationary states and also the time evolution of quantum systems. Exact analytical answers are available for the non-relativistic hydrogen atoms and also for the hydrogen-like atoms (atom or ion consisting an only electron), such as $\ce{H}$, $\ce{He+}$, $\ce{Li^2+}$, $\ce{Be^3+}$, etc.
Schrödinger equation
The Hamiltonian of a hydrogen-like atom with the atomic number $Z$ is the radial kinetic energy operator and Coulomb attraction force between $Z$ positive protons and a negative electron. Using the time-independent Schrödinger equation, ignoring all spin-coupling interactions, and using the reduced mass $ \mu=\dfrac{m_em_N}{m_e+m_N} $, the equation is written as:
\[ \left( -\frac{\hbar^2}{2\mu}\laplacian-\frac{Ze^2}{4\pi\vpmt r} \right)\psi = E\psi\]
Expanding the Laplacian in spherical coordinates:
\[ \left[ -\frac{\hbar^2}{2\mu}\left\{ \frac{1}{r^2}\pdv{ }{r}\left(r^2\pdv{ }{r}\right) + \frac{1}{r^2\sin\theta}\pdv{ }{\theta}\left(\sin\theta\pdv{ }{\theta}\right) + \frac{1}{r^2\sin^2\theta}\pdvn{2}{ }{\varphi} \right\} - \frac{Ze^2}{4\pi\vpmt r} \right] \psi = E\psi \]
This is a separable, partial differential equation that can be solved in terms of special functions. When the wavefunction is separated as a product of functions $R(r)$, $\Theta(\theta)$ and $\Phi(\varphi)$ three independent differential functions appears:
\[ \psi(r,\theta,\varphi) = CR(r)Y(\theta,\varphi)=CR(r)\Theta(\theta)\Phi(\varphi) \]
\[ \frac{1}{R}\odv{ }{r}\left(r^2\odv{R}{r}\right) - \frac{2\mu r^2}{\hbar^2}\left(-\frac{Ze^2}{4\pi\vpmt r}-E\right) = -\frac{1}{Y}\left[ \frac{1}{\sin\theta}\pdv{ }{\theta}\left(\sin\theta\pdv{Y}{\theta}\right) + \frac{1}{\sin^2\theta}\pdvn{2}{Y}{\varphi} \right] \]
Since the left and right sides consist of only variables respectively, the whole term is constant. Let’s put it $l(l+1)$.
\[ \frac{1}{\Theta}\left[\sin\theta\odv{ }{\theta}\left(\sin\theta\odv{\Theta}{\theta}\right)\right] + l(l+1)\sin^2\theta = -\frac{1}{\Phi}\odvn{2}{\Phi}{\varphi} \]
With the same logic, let’s put the whole term $m_l^2$. Remark that $l$ and $m_l$ are complex numbers. We finally get three differential equations:
\[ \therefore \begin{cases} \dps \frac{1}{R}\odv{ }{r}\left(r^2\odv{R}{r}\right) - \frac{2\mu r^2}{\hbar^2}\left(-\frac{Ze^2}{4\pi\vpmt r}-E\right) = l(l+1) \nl\nl \dps \frac{1}{\Theta}\left[\sin\theta\odv{ }{\theta}\left(\sin\theta\odv{\Theta}{\theta}\right)\right] + l(l+1)\sin^2\theta = m_l^2 \nl\nl \dps \frac{1}{\Phi}\odvn{2}{\Phi}{\varphi} = -m_l^2 \end{cases} \]
We know the solution for $\Phi$ and $\Theta$, where $P_l^{m_l}$ is the associated Legendre function.
\[ \begin{align*} \Phi(\varphi) &\propto e^{im_l\phi} \nl \Theta(\theta) &\propto P_l^{m_l}(\cos\theta) \end{align*} \]
We also get the condition for $m_l$ and $l$.
\[ \begin{align*} &\bullet\; m_l\in\Z ,\; \abs{m_l} \le l \nl &\bullet\; l\in\Z ,\; l\ge0 \end{align*} \]
Therefore, $Y(\theta,\varphi)$ becomes the spherical harmonic function by setting appropriate coefficients.
\[ \therefore Y(\theta,\varphi) = DP_l^{m_l}(\cos\theta)e^{im_l\varphi} = Y_l^{m_l}(\theta,\varphi) \]
Let’s then take a look at $R$.
\[ \odv{ }{r}\left(r^2\odv{R}{r}\right) - \frac{2\mu r^2}{\hbar^2}\left( -\frac{Ze^2}{4\pi\vpmt r} -E\right)R = l(l+1)R \]
Let’s define a new function $u(r) \coloneqq rR(r)$:
\[ -\frac{\hbar^2}{2\mu}\odvn{2}{u}{r} + \left[ -\frac{Ze^2}{4\pi\vpmt r}+\frac{\hbar^2}{2\mu}\frac{l(l+1)}{r^2} \right]u = Eu \]
We can simplify the equation by nondimensionalization:
\[ \begin{align*} \kappa &\coloneqq \frac{\sqrt{-2\mu E}}{\hbar} \nl \lambda &\coloneqq \frac{Z\mu e^2}{2\pi\vpmt\hbar^2\kappa} \nl \rho & \coloneqq \kappa r \end{align*} \]
\[ \Rightarrow \odvn{2}{u}{\rho} = \left[ 1-\frac{\lambda}{\rho}+\frac{l(l+1)}{\rho^2} \right]u \]
It is hard to solve this differential equation directly; let’s try an asymptotic analysis. We should find converging solutions.
\[ \begin{cases} \rho\to\infty &: u\rq\rq \approx u &\Rightarrow u \approx k_1e^{-\rho} \nl \rho\to0 &: u\rq\rq \approx l(l+1)\rho^{-2}u &\Rightarrow u \approx k_2\rho^{l+1} \end{cases} \]
Therefore, it is possible to reset $u$ using a new function $L(\rho)$.
\[ u \equiv \rho^{l+1}e^{-\rho}L(\rho) \]
We then get a differential equation of $L$.
\[ \rho\odvn{2}{L}{\rho} + 2(l+1-\rho)\odv{L}{\rho} + \left\{\lambda-2(l+1)\right\}L = 0 \]
Let’s use the Frobenius method to solve this equation.
\[ L=\sum_{j=0}^\infty c_j\rho^j \]
\[ \sum_{j=0}^\infty \Big[ j(j+1)c_{j+1} + 2(l+1)(j+1)c_j - 2jc_j + \left\{\lambda-2(l+1)\right\}c_j \Big]\rho^j = 0 \]
\[ \Rightarrow c_{j+1} = \frac{2(j+l+1)-\lambda}{(j+1)(j+2l+2)}c_j \]
It seems to be good to use an asymptotic analysis again.
\[ \begin{align*} j\to\infty \;:\; c_{j+1} \approx \frac{2}{j}c_j &\Rightarrow c_j \approx \frac{2^j}{j!}c_0 \nl &\Rightarrow L \approx c_0\sum_{j=0}^\infty \frac{2^j}{j!}\rho^j = c_0e^{2\rho} \nl &\Rightarrow c_0\rho^{l+1}e^\rho \end{align*} \]
The solution diverges, so it cannot be the right function describing the system. Therefore, we could guess that the recurrence relation has to be cut off, after a particular term. This implies the condition for $\lambda$.
\[ \begin{align*} \lambda &= 2(j+l+1) \nl &= 2n \;(n\in\N) \end{align*} \]
Let’s get back to the differential equation. By changing the variable as $x \coloneqq 2\rho$:
\[ x\odvn{2}{L}{x} + (2l+2-x)\odv{L}{x} + (n-l-1)L = 0 \]
This is the equation for generalized Laguerre function.
\[ L \propto L_{n-l-1}^{2l+1}(x) \]
We finally get the entire solution of $R$.
\[ R \propto \frac{1}{r}\rho^{l+1}e^{-\rho}L_{n-l-1}^{2l+1}(2\rho) \]
Remember that $\lambda$ was a nondimensionalizing variable, we can write $R$ as a function of $r$ where $a_0$ is Bohr radius.
\[ \begin{align*} \rho &= \frac{r}{\kappa} \nl &= \frac{Z\mu e^2}{2\pi\vpmt\hbar^2}\cdot\frac{r}{\lambda} = \frac{Z\mu e^2}{4\pi\vpmt\hbar^2}\cdot\frac{r}{n} \nl &= \frac{Zr}{na_0} \;\left( a_0 \coloneqq \frac{4\pi\vpmt\hbar^2}{\mu e^2} \right) \end{align*} \]
\[ \therefore R(r) = \left(\frac{2Zr}{na_0}\right)^l \exp\left(-\frac{Zr}{na_0}\right) L_{n-l-1}^{2l+1}\left(\frac{2Zr}{na_0}\right) \]
It’s almost done.
\[ \psi = C\left(\frac{2Zr}{na_0}\right)^l \exp\left(-\frac{Zr}{na_0}\right) L_{n-l-1}^{2l+1}\left(\frac{2Zr}{na_0}\right) Y_l^{m_l}(\theta,\varphi) \]
We should normalize the wavefunction. By some complex calculations, we get:
\[ \oint_{\Omega}\int_0^\infty \abs{\psi}^2 r^2 \,\d r\d\Omega = 1 \]
\[ \begin{align*} \therefore \psi_{nlm_l}(r,\theta,\varphi) &= \sqrt{\left(\frac{2Z}{na_0}\right)^3 \frac{(n-l-1)!}{2n \cdot (n+l)!}} \left(\frac{2Zr}{na_0}\right)^l \exp\left(-\frac{Zr}{na_0}\right) L_{n-l-1}^{2l+1}\left(\frac{2Zr}{na_0}\right) Y_l^{m_l}(\theta,\varphi) \nl &= \ket{n,l,m_l} \end{align*} \]
Now we should calculate energy, the eigenvalue of the Hamiltonian.
Remember that $\kappa$ contained $E$:
\[ E=-\frac{\hbar^2\kappa^2}{2\mu},\; \kappa = \frac{2Z}{\lambda a_0} \]
We finally get the energy, where $\RydE$ is the Rydberg unit of energy.
\[ \begin{align*} \therefore E_n &= -\frac{\hbar^2}{2\mu}\frac{Z^2}{n^2a_0^2} \nl &= -\frac{Z^2}{n^2}\RydE \;\left( \RydE \coloneqq \frac{e^2}{8\pi\vpmt a_0} \right) \end{align*} \]
Let’s summarize.
\[ \psi_{nlm_l}(r,\theta,\varphi) = \sqrt{\left(\frac{2Z}{na_0}\right)^3 \frac{(n-l-1)!}{2n \cdot (n+l)!}} \left(\frac{2Zr}{na_0}\right)^l \exp\left(-\frac{Zr}{na_0}\right) L_{n-l-1}^{2l+1}\left(\frac{2Zr}{na_0}\right) Y_l^{m_l}(\theta,\varphi) \]
\[ E_n = -\frac{Z^2}{n^2}\RydE \]
