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6

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 ...


5

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 ...


3

It is really quite difficult. This gets you to the original paper by Cummings in Phys. Rev. Lett. 140 in 1965, page A1051: http://journals.aps.org/pr/abstract/10.1103/PhysRev.140.A1051 . You must notice first that if $P_e(t)$ is given as you wrote then the coefficients $\frac{e^{-\alpha^2}|\alpha|^{2n}}{{n!}}$ multiplying $\cos()^2$ terms are localized ...


2

If $kT \gt \hbar \omega$, the thermal energy of the system is able to populate noticeably the states of the system you are trying to study. The quantum dynamical effects will be masked by thermal noise as mentioned by DanielSank in the comments. The surrounding degrees of freedom he mentions can be in the system itself, just not the degrees of freedom being ...


2

it is not exact and is impossible to compute exactly i recommend to use the Euler method to approximate your series by an integral plus some extra corrections , this Euler-Maclaurin summation converges fast to the exact solution with only a few terms.


2

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.


2

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 ...


1

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 ...


1

A single photon mode is described by a creation operator. The problem is not complicated and really generic in bosonic problems. How to include the driving is just a matter of taste. Suppose your circuit is described by the degrees of freedom $x$ and $p$ (say position and momentum). A force $f$ would be added as a term $x\cdot f$ in the Hamiltonian for ...


1

A short, mathematical answer to the question is found in the properties of Fourier transforms. The temporal response of the environment to a perturbation is given by the Fourier transform of its frequency response to the same perturbation. Therefore, if a broad range of frequencies in the bath are perturbed, the response occurs over a narrow range of times. ...


1

Assuming the question is about creating superpositions of eigenstates of the atomic Hamiltonian, one can do this by shining coherent electromagnetic radiation at the frequency $\omega$, such that $E = \hbar \omega$ is the energy difference between the two atomic states. This causes Rabi oscillations, so that the quantum state oscillates between the ground ...


1

The OP's characterization of these regimes is not quite universal, and there is some variation. This is partly because there are five important timescales for a Rabi-like system: the resonance and driving frequencies, $\omega_0$ and $\omega$, and the coupling strength $g$, but also the cavity decay rate $\kappa$ and the atomic decay rate $\gamma$. The ...


1

since this expansion follows directly from maxwell's equations in Lorentz gauge, it should still hold identically for the cavity because the former also holds. However what changes are the boundary conditions so that the sum over momentum goes from being over $\mathbb{R}$ to being over $\mathbb{N}$ for $p = 2 \pi n\hbar/L$



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