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+---
+title: "Hydrogen atom"
+sort_title: "Hydrogen atom"
+date: 2023-10-15
+categories:
+- Physics
+- Quantum mechanics
+layout: "concept"
+---
+
+The quantum-mechanical calculation of the **hydrogen atom** is,
+in my opinion, the single most important model in all of physics:
+miraculously, it is possible to find closed-form solutions
+for the wave function of an electron in a proton's potential well.
+The results are highly educational, and also qualitatively
+tell us a lot about all other chemical elements.
+
+We start from the time-independent Schrödinger equation,
+where $$\mu$$ is the [reduced mass](/know/concept/reduced-mass/)
+of the electron-proton system,
+and $$V$$ is the proton's Coulomb potential:
+
+$$\begin{aligned}
+    E \psi
+    = - \frac{\hbar^2}{2 \mu} \nabla^2 \psi + V \psi
+\end{aligned}$$
+
+In [spherical coordinates](/know/concept/spherical-coordinates/)
+$$(r, \theta, \varphi)$$ it becomes as follows,
+where $$V$$ only depends on $$r$$:
+
+$$\begin{aligned}
+    E \psi
+    = - \frac{\hbar^2}{2 \mu}
+    \bigg( \pdvn{2}{\psi}{r} + \frac{2}{r} \pdv{\psi}{r}
+    + \frac{1}{r^2} \pdvn{2}{\psi}{\theta} + \frac{1}{r^2 \tan{\theta}} \pdv{\psi}{\theta}
+    + \frac{1}{r^2 \sin^2{\theta}} \pdvn{2}{\psi}{\varphi} \bigg)
+    + V \psi
+\end{aligned}$$
+
+We will use the method of *separation of variables*
+by making the following ansatz,
+such that the Schrödinger equation takes the form below:
+
+$$\begin{aligned}
+    \psi(r, \theta, \varphi)
+    = R(r) \: Y(\theta, \varphi)
+\end{aligned}$$
+
+$$\begin{aligned}
+    E R Y
+    &= - \frac{\hbar^2}{2 \mu}
+    \bigg( R'' Y + \frac{2 R' Y}{r} +
+    + \frac{R Y_{\theta\theta}}{r^2} + \frac{R Y_\theta}{r^2 \tan{\theta}}
+    + \frac{R Y_{\varphi\varphi}}{r^2 \sin^2{\theta}} \bigg)
+    + V R Y
+\end{aligned}$$
+
+After multiplying by $$- 2 \mu r^2 / (\hbar^2 R Y)$$,
+each term depends on $$r$$ or $$(\theta, \varphi)$$, but not both:
+
+$$\begin{aligned}
+    0
+    &= \bigg( r^2 \frac{R''}{R} + 2 r \frac{R'}{R} - \frac{2 \mu}{\hbar^2} r^2 V + \frac{2 \mu}{\hbar^2} r^2 E \bigg)
+    + \bigg( \frac{Y_{\theta\theta}}{Y} + \frac{1}{\tan{\theta}} \frac{Y_\theta}{Y}
+    + \frac{1}{\sin^2{\theta}} \frac{Y_{\varphi\varphi}}{Y} \bigg)
+\end{aligned}$$
+
+Since these two groups are independent,
+this equation can only hold if there exists
+a *separation constant* $$C$$ such that:
+
+$$\begin{aligned}
+    C
+    &= r^2 \frac{R''}{R} + 2 r \frac{R'}{R}
+    - \frac{2 \mu}{\hbar^2} r^2 ( V - E )
+    \\
+    &=
+    - \frac{Y_{\theta\theta}}{Y}
+    - \frac{1}{\tan{\theta}} \frac{Y_\theta}{Y}
+    - \frac{1}{\sin^2{\theta}} \frac{Y_{\varphi\varphi}}{Y}
+\end{aligned}$$
+
+Now we have two simpler equations than the one we started with.
+We multiply them by $$R$$ and $$Y$$ respectively,
+and define $$C \equiv \ell (\ell + 1)$$ to help us later
+($$\ell$$ is unknown for now).
+The results are the **radial equation**
+and the **angular equation**:
+
+$$\begin{aligned}
+    \boxed{
+        \begin{aligned}
+            \ell (\ell + 1) R
+            &= r^2 R'' + 2 r R' - \frac{2 \mu}{\hbar^2} r^2 (V - E) R
+            \\
+            - \ell (\ell + 1) Y
+            &= Y_{\theta\theta} + \frac{1}{\tan{\theta}} Y_\theta + \frac{1}{\sin^2\theta} Y_{\varphi\varphi}
+        \end{aligned}
+    }
+\end{aligned}$$
+
+Note that this calculation has not really been specific to hydrogen so far:
+it is applicable to all spherically symmetric quantum systems.
+
+
+
+## Angular equation
+
+Let us keep this generality, by keeping $$V$$ unspecified for now,
+In that case, the radial equation cannot be solved yet,
+but the angular one can. We separate the variables again:
+
+$$\begin{aligned}
+    Y(\theta, \varphi) = \Theta(\theta) \: \Phi(\varphi)
+\end{aligned}$$
+
+Insert this into the equation and multiply by
+$$\sin^2\theta / (\Theta \Phi)$$ to get a clean separation:
+
+$$\begin{aligned}
+    \sin^2{\theta} \frac{\Theta''}{\Theta} + \sin{\theta} \cos{\theta} \frac{\Theta'}{\Theta}
+    + \ell (\ell + 1) \sin^2{\theta}
+    = - \frac{\Phi''}{\Phi}
+\end{aligned}$$
+
+Each term depends on $$\theta$$ or $$\varphi$$ but not both,
+so there exists a constant $$m^2$$ such that:
+
+$$\begin{aligned}
+    m^2
+    &= \sin^2{\theta} \frac{\Theta''}{\Theta} + \sin{\theta} \cos{\theta} \frac{\Theta'}{\Theta} + \ell (\ell + 1) \sin^2{\theta}
+    \\
+    &= - \frac{\Phi''}{\Phi}
+\end{aligned}$$
+
+These are two distinct equations;
+multiplying by $$\Theta$$ and $$\Phi$$ respectively yields:
+
+$$\begin{aligned}
+    \boxed{
+        \begin{aligned}
+            m^2 \Theta
+            &= \sin^2{\theta} \:\Theta'' + \sin{\theta} \cos{\theta} \:\Theta' + \ell (\ell + 1) \sin^2{\theta} \:\Theta
+            \\
+            - m^2 \Phi
+            &= \Phi''
+        \end{aligned}
+    }
+\end{aligned}$$
+
+Clearly the latter is the simplest, so we start there.
+It is an eigenvalue problem for $$m^2$$,
+but it looks like a harmonic oscillator equation,
+so the solutions are easily found to be:
+
+$$\begin{aligned}
+    \boxed{
+        \Phi(\varphi) = e^{i m \varphi}
+    }
+\end{aligned}$$
+
+Because the coordinate $$\varphi$$ is only defined in the interval $$[0, 2\pi]$$,
+we demand periodic boundary conditions $$\Phi(0) = \Phi(2 m \pi)$$,
+which tells us that $$m$$ is an integer.
+
+The other equation, for $$\Theta$$, needs a bit more work.
+We write it out like so:
+
+$$\begin{aligned}
+    0
+    &= \dvn{2}{\Theta}{\theta} + \frac{\cos{\theta}}{\sin{\theta}} \dv{\Theta}{\theta}
+    + \Big( \ell (\ell + 1) - \frac{m^2}{\sin^2{\theta}} \Big) \Theta
+\end{aligned}$$
+
+And then perform a change of variables $$\theta \to \xi$$
+where $$\xi \equiv \cos{\theta}$$, leading to:
+
+$$\begin{aligned}
+    0
+    &= - \dv{}{\theta} \bigg( \sin{\theta} \dv{\Theta}{\xi} \bigg)
+    - \cos{\theta} \dv{\Theta}{\xi}
+    + \bigg( \ell (\ell + 1) - \frac{m^2}{\sin^2{\theta}} \bigg) \Theta
+    \\
+    &= \sin^2{\theta} \dvn{2}{\Theta}{\xi}
+    - 2 \cos{\theta} \dv{\Theta}{\xi}
+    + \bigg( \ell (\ell + 1) - \frac{m^2}{\sin^2{\theta}} \bigg) \Theta
+    \\
+    &= (1 - \xi^2) \dvn{2}{\Theta}{\xi}
+    - 2 \xi \dv{\Theta}{\xi}
+    + \bigg( \ell (\ell + 1) - \frac{m^2}{1 - \xi^2} \bigg) \Theta
+\end{aligned}$$
+
+This result can be recognized as
+[Legendre's generalized equation](/know/concept/legendre-polynomials/),
+a known eigenvalue problem for $$\ell (\ell + 1)$$,
+which has solutions when $$\ell$$ is a non-negative integer.
+Those solutions are called the *associated Legendre polynomials*
+$$P_\ell^m(x)$$ of degree $$\ell$$ and order $$m$$.
+For a given $$\ell$$, there exist $$2 \ell + 1$$
+such "polynomials" (they actually contain square roots too)
+indexed by the integer $$m$$ in the range $$[-\ell, \ell]$$,
+so e.g. for $$\ell = 2$$ there is $$m = -2, -1, 0, 1, 2$$.
+We now have:
+
+$$\begin{aligned}
+    Y_\ell^m(\theta, \varphi)
+    \propto P_\ell^m(\cos{\theta}) \: e^{i m \varphi}
+\end{aligned}$$
+
+We are still missing a constant factor,
+found by imposing the normalization condition:
+
+$$\begin{aligned}
+    \int_0^{2 \pi} \int_0^\pi |Y_\ell^m|^2 \sin{\theta} \dd{\theta} \dd{\varphi}
+    = 1
+\end{aligned}$$
+
+Calculating the normalization constant (not shown here) leads to the
+following definition of the so-called **spherical harmonics**
+$$Y_\ell^m$$ of degree $$\ell$$ and order $$m$$:
+
+$$\begin{aligned}
+    \boxed{
+        Y_\ell^m(\theta, \varphi)
+        = (-1)^m \sqrt{\frac{(2 \ell + 1) (\ell - m)!}{4 \pi (\ell + m)!}} \: P_\ell^m(\cos{\theta}) \: e^{i m \varphi}
+    }
+\end{aligned}$$
+
+These are important functions:
+the wave function of any spherically symmetric quantum system
+is a superposition of $$Y_\ell^m$$ with $$r$$-dependent coefficients.
+And, as befits a (component of a) wave function,
+they form an orthonormal basis, specifically:
+
+$$\begin{aligned}
+    \int_0^{2 \pi} \int_0^\pi Y_\ell^m \:Y_{\ell'}^{m'} \:\sin\theta \:d\theta \:d\varphi = \delta_{\ell\ell'} \delta_{mm'}
+\end{aligned}$$
+
+
+
+## Radial equation
+
+With the angular part solved, we now turn to the radial part.
+Introducing $$u(r) = r R(r)$$, such that the derivatives of $$R(r)$$ become:
+
+$$\begin{aligned}
+    R'
+    = \frac{r u' - u}{r^2}
+    \qquad\qquad
+    R''
+    = \frac{r^2 u'' - 2 r u' + 2 u}{r^3}
+\end{aligned}$$
+
+Inserting this into the radial equation
+and cancelling some of the terms:
+
+$$\begin{aligned}
+    \ell (\ell + 1) \frac{u}{r}
+    &= \frac{r^2 u'' - 2 r u' + 2 u}{r} + \frac{2 r u' - 2 u}{r} - \frac{2 \mu}{\hbar^2} (V - E) r u
+    \\
+    &= r u'' - \frac{2 \mu}{\hbar^2} (V - E) r u
+\end{aligned}$$
+
+After multiplying by $$\hbar^2 / (2 \mu r)$$ and rearranging, this turns into:
+
+$$\begin{aligned}
+    E u
+    = - \frac{\hbar^2}{2 \mu} u'' + \bigg( V + \frac{\hbar^2}{2 \mu} \frac{\ell (\ell + 1)}{r^2} \bigg) u
+\end{aligned}$$
+
+Here it is useful to define an **effective potential** $$V_{\mathrm{eff}}(r)$$ as below.
+Keep in mind that $$\ell$$ is known after solving the angular equation:
+
+$$\begin{aligned}
+    V_{\mathrm{eff}}
+    \equiv V + \frac{\hbar^2}{2 \mu} \frac{\ell (\ell + 1)}{r^2}
+\end{aligned}$$
+
+This yields a relation of the same form as the
+time-independent Schrödinger equation, just with $$V$$ replaced by
+$$V_{\mathrm{eff}}$$. This is the "true" **radial equation**,
+an eigenvalue problem for $$E$$:
+
+$$\begin{aligned}
+    \boxed{
+        E u
+        = - \frac{\hbar^2}{2 \mu} u'' + V_{\mathrm{eff}} u
+    }
+\end{aligned}$$
+
+Now, finally, we specialize for the hydrogen atom.
+Coulomb's law tells us the attractive force $$F(r)$$
+between the electron and the proton,
+which we integrate to find the potential energy $$V(r)$$:
+
+$$\begin{aligned}
+    F(r)
+    = \frac{q^2}{4 \pi \varepsilon_0 r^2}
+    \qquad \implies \qquad
+    V(r)
+    = - \frac{q^2}{4 \pi \varepsilon_0 r}
+\end{aligned}$$
+
+Where $$q < 0$$ is the electron's charge,
+and $$\varepsilon_0$$ is the permittivity of free space.
+Note that $$V < 0$$, so there is a natural distinction
+between **bound states** $$E < 0$$
+(where the electron is trapped in the proton's well),
+and **scattering states** $$E > 0$$
+(where the electron is free).
+The true radial equation, after dividing by $$E$$, is now given by: 
+
+$$\begin{aligned}
+    u
+    = - \frac{\hbar^2}{2 \mu E} u'' + \bigg( \frac{\hbar^2}{2 \mu E} \frac{\ell (\ell + 1)}{r^2}
+    - \frac{\hbar^2 \mu}{\hbar^2 \mu} \frac{q^2}{4 \pi \varepsilon_0 r E} \bigg) u
+\end{aligned}$$
+
+For brevity, let us introduce new constants
+$$\kappa$$ and $$\rho_0$$, defined as follows:
+
+$$\begin{aligned}
+    \kappa
+    \equiv \frac{\sqrt{-2 \mu E}}{\hbar}
+    \qquad\qquad
+    \rho_0
+    \equiv \frac{\mu q^2}{2 \pi \varepsilon_0 \hbar^2 \kappa}
+\end{aligned}$$
+
+Where $$E < 0$$, as we are interested in bound states.
+Now the radial equation has become:
+
+$$\begin{aligned}
+    0
+    = \frac{1}{\kappa^2} u'' + \Big( \frac{\rho_0}{\kappa r} - \frac{\ell (\ell + 1)}{\kappa^2 r^2} - 1 \Big) u
+\end{aligned}$$
+
+To clean this up further,
+we switch to the dimensionless variable $$\rho \equiv \kappa r$$, yielding:
+
+$$\begin{aligned}
+    0
+    = u'' + \Big( \frac{\rho_0}{\rho} - \frac{\ell (\ell + 1)}{\rho^2} - 1 \Big) u
+\end{aligned}$$
+
+We then choose the following ansatz for $$u(\rho)$$, where $$v(2 \rho)$$ is unknown:
+
+$$\begin{aligned}
+    u(\rho)
+    &= w(2 \rho) \: \rho^{\ell + 1} \: e^{- \rho}
+\end{aligned}$$
+
+For reference, we also calculate its first and second derivatives:
+
+$$\begin{aligned}
+    u'(\rho)
+    &= \Big( 2 \rho w' + (\ell + 1 - \rho) w \Big) \rho^\ell \: e^{- \rho}
+    \\
+    u''(\rho)
+    &= \bigg( 4 \rho^2 w'' + 4 (\ell + 1 - \rho) \rho w'
+    + \big( \rho^2 - 2 \rho (\ell + 1) + \ell (\ell + 1) \big) w \bigg) \rho^{\ell-1} \: e^{- \rho}
+\end{aligned}$$
+
+Inserting this into the radial equation and dividing out all common factors gives:
+
+$$\begin{aligned}
+    0
+    &= 4 \rho w'' + 4 (\ell + 1 - \rho) w' + \big( \rho_0 - 2 (\ell + 1) \big) w
+\end{aligned}$$
+
+Let us rearrange this to put it in a more suggestive form.
+Keep in mind that $$w = w(2 \rho)$$:
+
+$$\begin{aligned}
+    0
+    &= (2 \rho) w'' + \Big( (2 \ell + 1) + 1 - (2 \rho) \Big) w' + \Big( \frac{\rho_0}{2} - \ell - 1 \Big) w
+\end{aligned}$$
+
+This can be recognized as [Laguerre's generalized equation](/know/concept/laguerre-polynomials/),
+a well-known eigenvalue problem for $$\lambda \equiv (\rho_0 / 2 \!-\! \ell \!-\! 1)$$.
+It has solutions when $$\lambda$$ is a non-negative integer,
+in other words for $$\rho_0 = 2n$$ with $$n = 1, 2, 3,...$$,
+which also tells us that $$\ell$$ cannot be larger than $$n - 1$$.
+Then the solutions are the so-called *associated Laguerre polynomials*
+$$L_{n - \ell - 1}^{2 \ell + 1}(2 \rho)$$, therefore:
+
+$$\begin{aligned}
+    u(\rho)
+    \propto \rho^{\ell + 1} \: e^{-\rho} \: L_{n - \ell - 1}^{2 \ell + 1}(2 \rho)
+\end{aligned}$$
+
+We are still missing a constant factor,
+found by imposing the normalization condition:
+
+$$\begin{aligned}
+    \int_0^\infty R^2 \: r^2 \dd{r}
+    = 1
+\end{aligned}$$
+
+Calculating the normalization constant (not shown here)
+leads to this radial solution $$R_{n\ell}(r)$$:
+
+$$\begin{aligned}
+    R_{n \ell}(r)
+    = \sqrt{\frac{(n - \ell - 1)!}{2 n (n + \ell)!}} \: (2 \kappa)^{3/2}
+    \: (2 \kappa r)^\ell \: e^{-\kappa r} \: L_{n - \ell - 1}^{2 \ell + 1}(2 \kappa r)
+\end{aligned}$$
+
+Meanwhile, by isolating the definitions of $$\kappa$$ and $$\rho_0$$ for $$E$$,
+we find the eigenenergies to be:
+
+$$\begin{aligned}
+    E
+    = - \frac{\hbar^2}{2 \mu} \kappa^2
+    = - \frac{\hbar^2}{2 \mu} \bigg( \frac{\mu q^2}{2 \pi \varepsilon_0 \hbar^2 \rho_0} \bigg)^2
+\end{aligned}$$
+
+Since $$\rho_0 = 2 n$$, these allowed **Bohr energies** $$E_n$$
+of the electron are as follows:
+
+$$\begin{aligned}
+    \boxed{
+        E_n
+        = - \frac{1}{n^2} \frac{\mu}{2 \hbar^2} \bigg( \frac{q^2}{4 \pi \varepsilon_0} \bigg)^2
+    }
+\end{aligned}$$
+
+At this point, it is customary to also define
+the **reduced Bohr radius** $$a_0^*$$, given by:
+
+$$\begin{aligned}
+    \boxed{
+        a_0^*
+        \equiv \frac{1}{n \kappa}
+        = \frac{4 \pi \varepsilon_0 \hbar^2}{\mu q^2}
+    }
+    \approx 5.295 \times 10^{-11} \:\mathrm{m}
+    = 0.5295 \:\mathrm{\AA}
+\end{aligned}$$
+
+The non-reduced **Bohr radius** $$a_0$$ simply uses
+the electron's raw mass $$m_e$$ instead of $$\mu$$.
+Roughly speaking, $$a_0^*$$ is the most probable electron-proton distance
+after a measurement of the electron's position while it is in its ground state.
+This is often to used to write $$R_{n \ell}(r)$$ as:
+
+$$\begin{aligned}
+    \boxed{
+        R_{n \ell}(r)
+        = \sqrt{\frac{(n - \ell - 1)!}{2 n (n + \ell)!}} \bigg( \frac{2}{n a_0^*} \bigg)^{3/2}
+        \bigg( \frac{2 r}{n a_0^*} \bigg)^\ell e^{- r / n a_0^*} \: L_{n - \ell - 1}^{2 \ell + 1}\Big(\frac{2 r}{n a_0^*}\Big)
+    }
+\end{aligned}$$
+
+
+
+## Quantum numbers
+
+Multiplying the angular and radial parts together,
+we thus arrive at the following expression
+for the full wave function $$\psi_{n \ell m}$$:
+
+$$\begin{aligned}
+    \boxed{
+        \psi_{n \ell m}(r, \theta, \varphi)
+        = R_{n \ell}(r) \: Y_\ell^m(\theta, \varphi)
+    }
+\end{aligned}$$
+
+The indices $$n$$, $$\ell$$, and $$m$$ are the **quantum numbers**,
+which describe the state of the electron.
+There is also a fourth not shown here, the **spin quantum number**,
+which is $$+1/2$$ or $$-1/2$$ for spin-up or spin-down electrons respectively.
+
+The **principal quantum number** $$n$$, often called the **shell number**,
+gives the energy level (shell) of the electron,
+because the other numbers do not appear in $$E_n$$'s formula.
+Since $$E_n = E_1 / n^2$$,
+the energy differences decrease with increasing $$n$$,
+so electrons in higher shells can be excited more easily
+(i.e. they need less energy to get excited).
+
+The **azimuthal quantum number** $$\ell$$ gives the **subshell**
+of shell $$n$$ in which the electron is located.
+It takes integer values from $$0$$ to $$n - 1$$ inclusive,
+with $$0$$, $$1$$, $$2$$, and $$3$$ respectively
+also called the $$s$$, $$p$$, $$d$$, and $$f$$ subshells.
+The electron's total angular momentum is given by $$\hbar \sqrt{\ell (\ell + 1)}$$.
+
+The **magnetic quantum number** $$m$$ splits the electrons in each subshell
+into **orbitals**, and takes integer values from $$-\ell$$ to $$\ell$$.
+The $$z$$-component of the electron's angular momentum is $$\hbar m$$.
+
+The total degeneracy of each energy level $$n$$
+can be calculated as the sum of an arithmetic series,
+and is found to be $$n^2$$ excluding spin (or $$2 n^2$$ with spin).
+
+Unsurprisingly, all these wave functions form an orthonormal basis
+(although not a *complete* one unless the scattering states with $$E > 0$$ are included):
+
+$$\begin{aligned}
+    \int_0^{2 \pi} \int_0^\pi \int_0^\infty
+    \psi_{n \ell m}^* \: \psi_{n' \ell' m'}
+    \: r^2 \sin{\theta} \dd{r} \dd{\theta} \dd{\varphi}
+    = \delta_{nn'} \delta_{\ell\ell'} \delta_{mm'}
+\end{aligned}$$
+
+When an excited electron drops from a state with energy $$E_i$$
+to a lower level $$E_f$$, it emits a photon with energy $$\hbar \omega$$,
+where $$\omega$$ is the angular frequency of the resulting
+[electromagnetic wave](/know/concept/electromagnetic-wave-equation/):
+
+$$\begin{aligned}
+    \hbar \omega
+    = E_i - E_f
+    = E_1 \bigg( \frac{1}{n_i^2} - \frac{1}{n_f^2} \bigg)
+\end{aligned}$$
+
+The corresponding vacuum wavelength is $$\lambda_0 = 2 \pi c / \omega$$,
+leading to the **Rydberg formula**,
+which was discovered empirically before the hydrogen atom had been solved:
+
+$$\begin{aligned}
+    \boxed{
+        \frac{1}{\lambda_0}
+        = \frac{\omega}{2 \pi c}
+        = \mathcal{R}\Big( \frac{1}{n_i^2} - \frac{1}{n_f^2} \Big)
+    }
+\end{aligned}$$
+
+Quantum mechanics then successfully gave a theoretical value
+to the experimentally determined **Rydberg constant**
+$$R_\mathrm{H}$$ (or $$R_\infty$$ if the raw electron mass $$m_e$$ is used):
+
+$$\begin{aligned}
+    \boxed{
+        R_\mathrm{H}
+        = \frac{|E_1|}{2 \pi \hbar c}
+        = \frac{\mu}{4 \pi \hbar^3 c} \bigg( \frac{q^2}{4 \pi \varepsilon_0} \bigg)^2
+    }
+    \approx 1.097 \times 10^7 \:\mathrm{m}^{-1}
+\end{aligned}$$
+
+The transitions from excited states to the ground state $$n_f = 1$$
+correspond to ultraviolet spectral lines known as the **Lyman series**.
+Similarly, transitions to $$n_f = 2$$ give visible lines known as the **Balmer series**,
+and transitions to $$n_f = 3$$ explain the **Paschen series** of infrared lines.
+
+The Rydberg constant is not to be confused
+with the **Rydberg energy** $$\mathrm{Ry}$$,
+which is the ionization energy of ground-state hydrogen,
+and is sometimes used as a unit in calculations:
+
+$$\begin{aligned}
+    \mathrm{Ry}
+    = 2 \pi \hbar c R_\mathrm{H}
+    = |E_1|
+    = \frac{\mu}{2 \hbar^2} \bigg( \frac{q^2}{4 \pi \varepsilon_0} \bigg)^2
+    \approx 13.61 \:\mathrm{eV}
+\end{aligned}$$
+
+The point is that the hydrogen atom's solution gave clear
+explanations for known experimental data,
+and settled the mystery of what an atom *actually looks like*.
+While other elements' atoms generally do not have such closed-form solutions
+(because they have more than one electron),
+their orbitals are qualitatively very similar.
+In short, this model is the foundation of our modern understanding of atoms.
+
+
+
+## References
+1.  D.J. Griffiths, D.F. Schroeter,
+    *Introduction to quantum mechanics*, 3rd edition,
+    Cambridge.
+2.  R. Shankar,
+    *Principles of quantum mechanics*, 2nd edition,
+    Springer.