Hot answers tagged cavity-qed
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 ...
Half life variations have been suspected for decades, and almost all (maybe all…but I do not pretend to have a comprehensive knowledge) have been shown to be caused by limits in experimental design. This latest set from this (now expanded) group (they did another paper on this topic a couple of years ago ...
Using a two-state atom dropped or conveyed through the cavity with a precisely controlled speed such that the Rabi oscillation (atom in $|g\rangle$, cavity mode in $|1\rangle$ $\leftrightarrow$ atom in $|e\rangle$, cavity mode in $|0\rangle$) makes just one $\pi$-rotation. These transactions are well described by the Jaynes-Cummings model.
A quite detailed article (in german though) discussing this "effect" can be found here: Radioaktivität: Wer hat an der Uhr gedreht? At the end of the article there is a list of interesting related papers and literature. According to this article first experiments trying to use artificially produced neutrinos to influence the decay rates of radioactive ...
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