Excited state at room temperature and near 0 Kelvin

Question

What does the excited state of the electron of a hydrogen atom look like at room temperature? What does the almost ground state look like slightly above 0 Kelvin?

Answer 1

If I understand your question correctly, you want to find the state of a hydrogen atom at some arbitrary temperature $T \ne 0 $.

This is simply a mixed state described by the density operator $\rho = C \exp(- H/kT)$, where $C$ is a normalization factor such that ${\rm Tr} \, \rho =1$ and $k$ is the Boltzmann constant. Separating off the (trivial) motion of the center of mass of the system and concentrating on the relative motion of the electron and the proton, the spectrum of $H$ is given by the well known discrete values

$E_n= -m c^2 \alpha^2/2 n^2 \simeq - 13.6 \,{\rm eV}/n^2 \quad( n=1,2, \ldots$)

plus a continuous spectrum starting at $E=0$. Including the spin, the energy eigenvalue $E_n$ is $2 n^2$-fold degenerate. Denoting the probability of finding the hydrogen atom with energy $E_n$ by $p_n$, the ratio $p_n/p_1$ is thus given by

$p_n/p_1= n^2 e^{-13.6 \, {\rm eV}(1-1/n^2)/kT}$.

As $T = 300 \, {\rm K}$ corresponds to $k T \simeq \frac{1}{39} \, {\rm eV}$, already $p_2/p_1$ results in a tiny-tiny value at this temperature, being beyond good and evil. In other words, at room temperature (and even more closer to $T=0$), the admixture of excited states in the density operator can safely be neglected. Substantial contributions of excited states can only be expected if the temperature approaches a value where $k T \simeq E_2-E_1 $.