In the classical Rabi oscillation, where you solve the Bloch equations to find the dynamics of your spin-1/2 moment, you ALWAYS use a two-level system. The system is best suited for that, independent of your detuning.

Why? Because the a spin 1/2 system, which has two levels, is representable in the Bloch-sphere, which shows up as the equivalence between the SU(2) group and the SO(3) groups.

The case, in which the classical Bloch equations won't be sufficient, is the case when you have spins more than 1/2 (so more than 2 levels), and your Hamiltonian would make higher levels interactions to each other (which is rare). So if you have spin-1 system, for example, then the Bloch sphere would not be enough to represent the dynamics of your system, because the Bloch-sphere looks only at the *orientation* of your system, which is a vector quantity that you classically represent by the Bloch vector. If you have a higher spin, then a contribution would show up from higher multipole moments, which is *alignment* rather than orientation, that are represented with higher order tensors $T^{(K)}$.

Luckily, even though higher multipole moments could exist in many atoms, their effect isn't really pronounced. The reason is that we usually use light to interact with atoms, and light as polarizations which could be either linear $(\pi)$ or circularly polarized $(\sigma^+,\sigma^-)$, or a superposition of those. Therefore the interaction of higher orders would only exist in a pronounced way if you push your system to have not only dipole transitions, but also quardrupole and octupole and higher order transitions.

The most complicated case with light is when you have linearly polarized light for pumping. In that case you'll induce alignment rather than orientation. A paper that discusses that is this:

http://arxiv.org/abs/physics/0605234

For more details, look at the irreducible representations of Multipole moments. It's a huge topic, into which I wouldn't recommend going, unless you have something to do with research. The article would give you an introduction into that.

Good luck, hope that helps.

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 (spike in values) around $n = n_{av} = |\alpha|^2$.
You can get it from the Stirling formula for large $n$ ($n>10$) by approximating

$n! \approx \sqrt{2 \pi n}\left(\frac{n}{e}\right)^n$

and than trying to find the maximum of the function

$\frac{e^{-n_{av}}|n_{av}|^{n}}{{n!}} \approx 1 /\sqrt{2 \pi n}\left(\frac{e}{n}\right)^n e^{-n_{av}}|n_{av}|^{n}$

with respect to $n$ as it was the continuous variable.

Calculating and putting the derivative to zero you get

$e^{-n_{av}} /(2 \sqrt{2 \pi})e^n n_{av}^n n^{-3/2-n}(1 + 2 n_{av} \log n - 2 n \log n) = 0 $

and the approximate solution, neglecting the small $1$ in the bracket, is

$n=n_{av}$.

Than you can Taylor-expand $g(n+1)^{1/2}$ up to the first order around $n_{av}$ i.e.
$g(n+1)^{1/2} \approx g (n_{av}+1)^{1/2} + g/[2 (n_{av}+1)^{1/2}](n-n_{av})$ to care only about the most contributing terms in the sum.
The energies get linear in $n$ now like for the harmonic oscillator.
You get both the revival time and the short time decay using this approximation
after some page of further math (which is not in the paper as it was the Letter).
Revivals are much simpler since now all cosines oscillate with the multiple
of the same frequency $g/[2 (n_{av}+1)^{1/2}]$ and therefore they are seen directly to rephase. The revival time is therefore
$2 \pi (n_{av}+1)^{1/2}/g$ while you get the decay time $1/g$ using small time expansion
of the sum terms for short times to sum them up to the real dropping exponent.
Paradoxically revivals were not noticed or ignored by Cummings in his original paper deriving $1/g$ collapse rate even if the approximate
formula he derived has them in it and they were discovered later by Eberly
in Phys. Rev. lett. 44, 1980, page 1323:
http://journals.aps.org/prl/abstract/10.1103/PhysRevLett.44.1323 .
He derives a complicated compact formula for the arbitrary case of the nonzero detuning that is long time periodic behavior of the full revival after the time
$2 \pi (n_{av})^{1/2}/g \approx 2 \pi (n_{av}+1)^{1/2}/g$ that resembles slightly
the oscillatory motion of the weakly spreading free particle wave packet if one calculated the overlap of it with itself from the time $0$ (the autocorrelation function) and with the revival period while the fast oscillations are some phase.

To obtain the decay time: Now we have:

$P_e(t) \approx \sum_{n=0}^{\infty}{\frac{e^{-n_{av}}n_{av}^{n}}{n!}} \times$
$\cos^2{[g (n_{av}+1)^{1/2} + g/[2 (n_{av}+1)^{1/2}](n-n_{av})]}$

or

$P_e(t) \approx \sum_{n=0}^{\infty}{\frac{e^{-n_{av}}n_{av}^{n}}{n!}} \times$
$[\cos{[2 g (n_{av}+1)^{1/2} t + g/[ (n_{av}+1)^{1/2}](n-n_{av}) t]}/2+1/2]$

Now we use to formula for the $\cos (a+b)=\cos a \cos b - \sin a \sin b$ twice and drop all terms that
have $\sin (...t) \approx 0$ in it since they they are small and terms with $\cos (...t)$ except the most oscillatory with $2g$
not having the running $n$ we put to $1$ for the same reason.
The result is:

$P_e(t) \approx \sum_{n=0}^{\infty}{\frac{e^{-n_{av}}n_{av}^{n}}{n!}} \times$
$[\cos [2 g (n_{av}+1)^{1/2} t] \cos{[g/[ (n_{av}+1)^{1/2}]n t]}/2+1/2]$

Now using the formula for $\cos(x)= (e^{ix}+e^{-ix})/2$

the series can be readily summed up from the series exponent expansion

$e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!}$

(note the
terms containing the exponent of the exponent):

$P_e(t) \approx (1/4)\cos [2 g (n_{av}+1)^{1/2} t] e^{-n_{av}} [e^{n_{av}e^{i g/[ (n_{av}+1)^{1/2}]t}}$
$+e^{n_{av}e^{-i g/[ (n_{av}+1)^{1/2}t]}}]$ +1/2$

The one more step is to expand again for small $t$ (we care only about the real
parts while the imaginary have the small $\sin (...t)$ again):

$e^{-i g/[ (n_{av}+1)^{1/2}t} \approx 1 - t^2 g^2 /2/ (n_{av}+1)$

and so

$e^{i g/[ (n_{av}+1)^{1/2}t} \approx 1 - t^2 g^2 /2/ (n_{av}+1)$

approximating

$n_{av}+1 \approx n_{av}$
we obtain
$P_e(t) \approx (1/2) \cos [2 g (n_{av}+1)^{1/2} t] e^{-g^2 t^2/2}+ 1/2$
so the decay time is
$1/g$

## Best Answer

The Rabi oscillation is a simplified picture of the atom-light interaction. The atom is considered to be a pure 2-level system and the energy difference of the two atomic states is given by $E = \hbar \omega_0$.

Now, if we shine light onto the atom, the electric dipole moment of the light couples these two states. If the light is resonant with the atomic transition it drives the atom from its ground state into its excited state. However, since the frequency of the atom and the light does not change, the light is also resonant with the inverse process. Therefore, in the next sequence, the light "takes" the excited atom and brings it back to its ground state. The mathematical description is actually very simple: The probability amplitude $c_g(t)$ of the atom to be in ground state at time $t$ behaves like a harmonic oscillators with a

complexdamping factor, $$ \frac{d^2 c_g}{dt^2} - i \Delta \frac{dc_g}{dt} + \frac{\omega^2_0}{4}c_g = 0$$ The same is true for the probability amplitude of the excited state, $$ \frac{d^2 c_e}{dt^2} + i \Delta \frac{dc_e}{dt} + \frac{\omega^2_0}{4}c_e = 0$$ The physical picture of the "damping term" for the ground state is, that the atom leaves the ground state and "goes" into the excited state.My picture for these equations is that we have two oscillations in

eachequation:If $\Delta=0$ we obtain pure harmonic oscillators and the atomic state oscillates between its extreme: fully in the ground state and fully in the excited state.

If however $\Delta \ne 0$ the picture becomes more interesting, because we have to choose some initial conditions to solve the equations. Usually we take $c_g(t=0)=1$ and $c_e(t=0)=0$, which breaks the symmetry of the two equations. Thus the atom periodically reaches this initial state, however, it never reaches the "fully excited state". Using the two oscillations from above, the atomic and the laser, we can come up with the following simple picture: If these frequencies are not equal, the two oscillations are not synchronised. Thus, if we start with the same phase at $t=0$ the electric dipole moment of the light excited the atom. However, with time passing the two oscillations build up a phase difference, which grows in time. At a certain point in time the two oscillations are in "opposite direction". Thus, instead of exciting the atom, the light bringst it back into the ground state.

However, the most interesting feature for the $\Delta \ne 0$ case is that the Rabi frequency increases with increasing detuning, $\Omega = \sqrt{w^2_0 + \Delta^2}$. Using the phase shift picture, one can argue that the detuning can only accelerate the atomic oscillation.