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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

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$$|e,0\rangle \rightarrow a(t)|e,0\rangle + \sum_k b_k(t) |g,1_k\rangle$$ 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.
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  • $\begingroup$ Thank you very much. So in summary: the spectrum of emission is fundamentally a Lorentzian and not a delta dirac. The delta diract spectrum representation is a simplification of what is fundamentally going on (it is a fictive, simplified and wrong representation of things, not an idealized/perfect one). $\endgroup$
    – StarBucK
    Jan 11 '20 at 17:20
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    $\begingroup$ @StarBucK Yes, correct. You would get a dirac-like spectrum only if $\Gamma=0$. $\endgroup$ Jan 11 '20 at 17:23
  • $\begingroup$ Thank you for your clear answer. $\endgroup$
    – StarBucK
    Jan 11 '20 at 17:23

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