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The hydrogen atom

The hydrogen atom is composed of two charges (positive and negative) in a stable orbit.

Newtonian mechanics

In classical mechanics, the hydrogen atom reduces to the case of the two-body problem. This is if we consider the truly non-relativistic limit, in which case the electromagnetic potential reduces to the electric potential, and the resolution is exactly that of the usual two-body problem. The system is

Electrodynamics

In the case of electrodynamics, the hydrogen atom is famously unstable, as can be seen by deriving its Liénard–Wiechert potential. Let's consider our proton and electron as two charge point particles, with the Nambu-Goto action

\begin{equation} S([\gamma])= m \int_\Sigma dvol(\gamma^*g) + q \int_\Sigma \gamma^* A \end{equation}

Non-relativistic quantum mechanics

For the case of a fixed classical central charge, we get

\begin{equation} -\frac{\hbar^2}{2m} \Delta \psi(\vec{x}, t) + \frac{1}{4\pi \varepsilon_0} \frac{e}{\hat{r}^2} \psi(\vec{x}, t) = i\hbar \frac{\partial}{\partial t} \psi(\vec{x}, t) \end{equation}

where $\hat{r}^{-2}$ is the obvious multiplicative operator based on the usual position operators,

In spherical coordinates, this will translate to

\begin{equation} -\frac{\hbar^2}{2m} \left[ \frac{1}{r^2} \frac{\partial}{\partial r} (r^2 \frac{\partial}{\partial r}) + \frac{1}{r^2 \sin(\theta)} \frac{\partial}{\partial \theta} \right] \psi(\vec{x}, t) + \frac{1}{4\pi \varepsilon_0} \frac{e}{r^2} \psi(\vec{x}, t) = i\hbar \frac{\partial}{\partial t} \psi(\vec{x}, t) \end{equation}