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Is it true that a spontaneously light-emitting atom changes its lifetime if it is put between two parallel plates that are so near that they attract each other through the Casimir effect?

Thus: does the Casimir effect, and the changes in the vacuum state it induces, influence spontanenous emission?

If so, who measured this effect for the first time? Where can I read about it?

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2 Answers 2

This is true. The simple explanation is this: For calculating the decay rate of an excited state, you use Fermi's Golden Rule, which involves the matrix element $$|\langle f | V | i \rangle|^2$$ where $f$ and $i$ denote the final and initial state, respectively.

Since the final state contains the electron in its groundstate together with a photon created by this decay, the nature of your cavity determines what the matrix element will be: For example, if your cavity forbids standing waves of the emission frequency, decay is suppressed. The study of these effects goes under the name of Cavity QED

I would not say that this is due to the Casimir effect. Rather, this effect and the Casimir effect are both due to the boundary conditions created by your plates.

I don't exactly know who first studied this. I suggest consulting a review article such as this one.

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Presence of something else next to an excited atom influences the lifetime of the excited state. Any such presence is described with some additional interaction energy.

In case of a cavity QED, you can get suppression of radiation rate: $e^{-\gamma_1 t}$ with smaller $\gamma_1$ due to suppression of the corresponding part of the electromagnetic spectrum of the whole system.

In case of a neighboring similar atom in the ground state, you can obtain a decrease of the life-time due to additional independent channel of the excitation deactivation - a resonance transition, for example. So the resulting decay rate will be $e^{-(\gamma_1+\gamma_2) t}$.

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