# How can hydrogen spectrum be monomode (focusing on the lowest emission band)?

Consider you have an hydrogen atom. This atom can emit light under spontaneous emission for example, but the light it will emit will only be at some very specific frequencies: https://en.wikipedia.org/wiki/Emission_spectrum

I take one atom of hydrogen perfectly at rest and I look at the light it emits. I excite the electron on the level just after the ground state so that I can focus on the lowest light ray hydrogen can emit, where I call the energy gap $$\hbar \omega_0$$.

The dynamic of spontaneous emission can be described with the following hamiltonian (Wigner weisskopf model https://www.mpi-hd.mpg.de/personalhomes/palffy/Files/Spontaneous.pdf):

$$H=\frac{\hbar \omega_0}{2} \sigma_z + \sum_k \hbar \omega_k a_k^{\dagger}a_k + \sum_k g_k \left(a_k \sigma_+ + a_k^{\dagger} \sigma_- \right)$$

Solving the dynamic, you find the evolution for a time $$t$$:

$$|e,0\rangle \rightarrow a(t)|e,0\rangle + \sum_k b_k(t) |g,1_k\rangle$$

The spontaneous emission process is thus understood as an entanglement between the atom and the many modes of the field. Tracing out the atom we would have a mixed state involving many frequency on the field and not only the frequency $$\omega_0$$.

Thus: why do we say the hydrogen would emit a photon at frequency $$\omega_0$$ only ? From the spontaneous emission model we see that the light state after emission is not $$|1_{\omega_0}\rangle$$ but actually involves many different modes.

My question is on the conceptual level, I don't want to take in account possible doppler effect that would spread frequencies and give a continuum in emission. I want to understand why "in theory", in a perfect world, the hydrogen atom would emit at a single frequency

$$|e,0\rangle \rightarrow a(t)|e,0\rangle + \sum_k b_k(t) |g,1_k\rangle$$ The spontaneous emission process is thus understood as an entanglement between the atom and the many modes of the field. Tracing out the atom we would have a mixed state involving many frequency on the field and not only the frequency $$\omega_0$$.
You are correct so far. But the story continues. The article about the Wigner-Weisskopf model linked by you actually gives the solution for the amplitudes $$a(t)$$ and $$b_k(t)$$. The results given on page 3 are
• $$|a(t)|^2 = e^{-\Gamma t}$$, where $$\Gamma$$ is a bunch of constants.
This means that the probability of the atom being in the excited state decays with the life-time $$\tau=1/\Gamma$$.
• for $$t \to\infty$$:$$\quad$$ $$|b_k(t)|^2=\frac{|g_k|^2}{\Gamma^2/4+(\omega_k-\omega_0)^2}$$.
This means that most photons are in the frequency range between $$\omega_0-\Gamma/2$$ and $$\omega_0+\Gamma/2$$. This is the well-known Lorentzian spectral line shape. It is a necessary consequence of the life-time $$\tau=1/\Gamma$$ above. It has nothing to do with a broadening of spectral lines by the Doppler effect due to different speeds of the atoms.
• @StarBucK Yes, correct. You would get a dirac-like spectrum only if $\Gamma=0$. Jan 11 '20 at 17:23