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I understand that the idea of emission or absorption cannot be incorporated in the framework of non-relativistic QM. We need QFT or relativistic QM to explain the creation and annihilation of particles, which indeed is the central phenomena in an emission/absorption process.

I have 2 questions here :

  1. How exactly is spontaneous emission of photons because of atomic transitions of electrons modelled in QFT?

  2. Why don't we have any phenomena like spontaneous absorption? Is such a transition rate zero?

Emilio Pisanty
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Tushar Gopalka
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2 Answers2

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If by "spontaneous absorption" you mean a process where you have an atom sitting in vacuum, with no photons around, and it magically gets excited, then that process is forbidden by the conservation of energy. We also wouldn't call it "absorption", since nothing is getting absorbed.

If, instead, you mean the inverse process to spontaneous emission ─ i.e., the atom is somewhere with a nonzero photon flux, and it spontaneously absorbs one of those photons to go into an excited state ─ then yes, this process is perfectly possible, though normally we call it "absorption". If you have any materials that are not transparent at hand, then this is the process that turns them from transparent to not-transparent.

Emilio Pisanty
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$\newcommand{\Ket}[1]{\left|#1\right>}$ $\newcommand{\BKet}[1]{\left<#1\right|}$

To answer your question one only needs to second quantize the photon field and use normal quantum mechanics. For reference you can check up 'Advanced Quantum Mechanics' Sakurai Chapter 2 for a semi Field theory treatment. But the essentials of the process becomes quite clear.

Spontanious emission is the emission process that happens even when there are no available photons for an atom or some QM system to interact with. That is this process seemingly happens on its own. The key for spontaneous emission s that the Photon field and thus the interaction exists even when the number of available particles is zero.

Let us make ourselves a toy physical world where you have two hydrogen atoms and they interact is some way to exchange energy. The rule in our world is that the photon emitted by one atom will be absorbed by the other.

Scenario 1: atom 1 is in ground state and atom 2 is in the exited state. Now what possibilities do we have for available processes?

Say atom 2 emits a photon and moves to the lower state and atom 1 absorbs the photon and moves to an exited state. $$\Ket{\Psi^2_E}\Ket{\Psi^1_{GroundState}} \rightarrow \Ket{\Psi^2_{E-\epsilon}}\Ket{\Psi^1_\epsilon}$$

Scenario 2: Both atoms are in the ground state. In this case neither atom has a lower state to move to so that the other atom can make an excitation. So there is no excitation in this scenario.

$$\Ket{\Psi^2_{GroundState}}\Ket{\Psi^1_{GroundState}} \rightarrow ??$$

Scenario 1 is similar to spontaneous emission where an atom interacts with the photon vacuum state, dumps a photon to the photon field and moves to an available lower exited state. The Photon field now has one photon in the corresponding energy level.

$$\Ket{\Psi_E}\Ket{0} \rightarrow \Ket{\Psi_{E-\hbar \omega}}\Ket{1\hbar \omega}$$

Like in Scenario 2 as both atoms where in their ground states neither of them can exchange energy. Similarly in case of spontaneous absorption, if the atom to absorb a photon from the photon field there should be an available lower energy state for the field to fall off to. As there is no such available energy level as the Photon field is already at its ground state, spontaneous absorption doesn't happen.

$$\Ket{\Psi_{GroundState}}\Ket{0} \rightarrow ?? $$

Of course all this is hand waving. You can check up the reference for more rigorous details.

Kolandiolaka
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