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The quantum mechanics of the structure of atoms as determined by the electromagnetic forces inside them correctly describes the location and coupling of the different energy levels in essentially all atoms and molecules. However, "classical" quantum mechanics predicts that atoms in excited states will remain there for all time, and it does not provide a mechanism for atoms in excited states to radiate a photon and move to a lower level.

How does QED justify this process?

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Of course quantum mechanics does provide an answer to the question of atomic transitions. It would be very impractical to use QED to compute all of these transitions. QED could be used to compute corrections if necessary. – Raskolnikov Jun 15 '11 at 20:42
up vote 7 down vote accepted

If your atom were alone, it would be described by its Hamiltonian $H_{a}$. The eigenstate of its Hamiltonian correspond to the energy level of the atoms and would be, by definition, steady states.

However, as stated by @Vladimir Kalitvianski, the atom is not alone.

Suppose you are performing a CQED (cavity QED) experiment with your atom. You put your atome in a single-mode optical cavity. The field of the cavity is described by its Hamiltonian $H_c=\sum_{n\ge0} n\hbar\omega|n\rangle\langle n|$, where $|n\rangle$ is tha n-photons Fock state. If the atom and the cavity interact, you need to describe this interaction by another term $H_i$ such that the global Hamiltonian is now $$ H=H_a+H_c+H_i.$$ Because of the $H_i$ term, neither the eigenstates of $H_a$ (a.k.a the atomic energy levels) nor the eigenstates of $H_c$ (a.k.a the Fock states) are eigenstate of $H$. They are not steady state anymore. For example, for a two-level atom described by its ground state $|g\rangle$ and its excited state $|e\rangle$, two of these steady states are quantum superpositions of $|e\rangle_a|0\rangle_c$ and $|g\rangle_a|1\rangle_c$. One of the consequences is that we can have vacuum Rabi oscillation: If you put an excited atom in a cavity with no light inside ($|e\rangle_a|0\rangle_c$), the state of the atom+cavity ensemble oscillates between $|e\rangle_a|0\rangle_c$ and $|g\rangle_a|1\rangle_c$. In other words, the atom releases the photon and recaptures it periodically.

If you want more details, look for Jaynes-Cummings model on the internet, on Wikipedia or in your favourite quantum mechanics textbook.

If you have several cavities, and/if your cavity is big enough, the atom is coupled to several modes, and the recapture of the photon takes longer. You can then make your cavity grow such that your atom "sees" many modes, which you can describe by a continuum, with a density of mode. From the atom's point of view, there is no difference between sitting of the middle of many infinitely large cavities (one per direction of the space) and sitting in free space. In that limit, the oscillation becomes an exponential decay, the recapture of the photon being postponed to an infinitely distant time. To compute that decay, you should use Fermi Golden Rule, invented by Dirac [!] to get the decay rate.

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An electron is never alone. It is always coupled to the quantum field oscillators. Preparing an atom in an excited state means the entire excitation energy is concentrated in the electron motion. Due to permanent interaction with quantum (photon) oscillators, the occupation numbers of some oscillators start to grow from zero to one. At the same time the probability amplitude for the electron to stay on the same level decays and that to appear on a lower level grows too. So, in the end you have an electron on a lower energy level and one or more photons created (oscillators excited). This process of the energy redistribution in the system with many coupled degrees of freedom takes some time (life time of the level).

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