The excited states of Hydrogen, like all atoms, have a non-zero lifetime due to interactions with the vacuum electromagnetic field. Rigorously speaking, these excited orbital states are not the true eigenstates of the combined atom-light system, as those eigenstates should not decay. For comparison, the $1s$ state, combined with zero photons in any mode, is truly the ground state of the combined atom-light system.

Formally, $\vert 1s;0\rangle$ is an eigenstate of the total Hamiltonian $$H_T=H_{atom}+H_{EM},$$ whereas $\vert 2p;0\rangle$ is not. The true excited eigenstates (I suppose) should be some superposition state of the excited orbital and excited photon number states.

$$\vert \psi_{2p} \rangle ?= \sum_i a_i\vert 1s;n_i\rangle + \sum_j b_j \vert 2p;n_j\rangle$$

My question is simple, can we write down the actual eigenstates of an excited level of the Hydrogen atom in the hydrogen orbital/photon number basis?

If these eigenstates are too complicated, can the nature of the true excited state be at least described in more detail? And, to be clear, I want to ignore fine, hyperfine, and spin structure unless it's necessary. Hopefully, the full machinery of QFT is not needed and some appropriately large volume $V$ can be used, but I am unsure about that.

  • $\begingroup$ As you noticed, the hydrogen atom does not exist independently of the electromagnetic field. More precisely... the hydrogen atom is bound by the electromagnetic field, so in the end it is a state of the vacuum field, rather than a system that exists independently of the field and that only couples to the field. One photon in an infinite size vacuum is, as far as I can tell, a meaningless concept. In quantum field theory bound states like atoms seem to appear as poles in the S-matrix instead, if I remember correctly. $\endgroup$ Jun 19 at 10:58

1 Answer 1


There is a simplified model of the atom-photon interaction in a cavity (so the photon modes are discrete) which is exactly solvable, and the solution agrees with experiment. The solution is too long for an SE answer though.

  • $\begingroup$ Interesting, I normally thought of Jaynes Cummings as behavior specific to an atomic in a high finesse cavity. What is a bit confusing there (to me) is that the coupling of an atom to the cavity is an inverse function of the cavity volume. Which would suggest that as $V$ becomes large, there should be no hybridization with photon modes, and which is why the JC in my mind wouldn't apply in free space. Is there some density of states argument that cancels it out? $\endgroup$
    – KF Gauss
    Jun 19 at 13:27
  • $\begingroup$ In a large system each photon mode has a $1/\sqrt{V}$ in its amplitude, so the coupling goes to zero, but, as you say, there is a large density of photon states in the neighbourhood of $\Delta E$. This means that Fermi's Golden Rule applies and that excited states decay exponentially to the atomic ground state with no nearby photons. $\endgroup$
    – mike stone
    Jun 19 at 14:08
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    $\begingroup$ There is also the Lee model in QFT that has an infinity of modes. A recent paper is arxiv.org/pdf/2001.07781.pdf $\endgroup$
    – mike stone
    Jun 19 at 14:49
  • $\begingroup$ Thanks so much. In a truly isolated hydrogen atom in vacuum, the excited state should be infinitely long lived, but that definitely feels unphysical. What's the right way to resolve that question? $\endgroup$
    – KF Gauss
    Jun 20 at 5:31

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