---
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:

$$\begin{aligned}
    \boxed{
        \frac{1}{\lambda_0}
        = \frac{\omega}{2 \pi c}
        = R_\mathrm{H} \bigg( \frac{1}{n_i^2} - \frac{1}{n_f^2} \bigg)
    }
\end{aligned}$$

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:

$$\begin{aligned}
    \mathrm{Ry}
    \equiv 2 \pi \hbar c R_\mathrm{H}
    = |E_1|
    = \frac{\mu}{2 \hbar^2} \bigg( \frac{q^2}{4 \pi \varepsilon_0} \bigg)^2
    \approx 13.61 \:\mathrm{eV}
\end{aligned}$$

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.