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I don't see why the following relation should be correct.

The scattering rate $R_s$ of atoms in a laser beam is

$R_s=\frac{1}{T_s}$.

$T_s$ is the time interval between the first and second excitation of the electron. Now, the probability to find the electron in the excited state $\rho_{22}$ equals

$\rho_{22}=\frac{\tau}{T_s}$, $\tau$ being the life time.

Now, I dont understand the following. If one plugs the realtion for $\rho_{22}$ in the first equation, one obtains

$R_s=\frac{\rho_{22}}{\tau}\stackrel{?}{=}\Gamma \rho_{22}$.

Why is this correct? I mean, why is $\tau=\frac{1}{\Gamma}$.

The life time equals one over the line width?

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  • $\begingroup$ I have often heard the argument $\Delta f \Delta t \approx 1$, thus life time is 1/linewidth. That's not really a good answer to me, but maybe that's really why this is used? $\endgroup$ – Sanya Jul 12 '16 at 19:24
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    $\begingroup$ @Sanya There are definite theoretical grounds for that relationship: it's a general property of Fourier transform pairs. See my treatment of this question here $\endgroup$ – WetSavannaAnimal Jul 18 '16 at 0:47
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It's simply a Fourier transform relationship. If an atom is in a metastable state, with the latter coupled to the electromagnetic field, then it is fairly straightforward to show that the probability amplitude for decay to happen in the time interval $[t,\,t+\mathrm{d}t]$ is:

$$\psi(t) = \left\{\begin{array}{ll}\frac{1}{\sqrt{\tau}}\,\exp\left(i\,\omega_0\,t - \frac{t}{2\,\tau}\right);& t\geq0\\0&\text{otherwise}\end{array}\right.$$

where $\hbar\,\omega$ is the transition energy. I show how this calculation is done in my answer here. Witness that the probability density $|\psi|^2$ is the memoryless exponential distribution with lifetime $\tau$. If you work out the Fourier transform of $\psi$ then the power spectral density to get the lineshape, you get a Lorentzian lineshape whose linewidth is given by the relationship you ask about.

The Heisenberg-like inequalities that people pull out of thin air without explanation are all manifestations of the relationships between second moments of functions and their power spectral densities. See my treatment of this question here. If you don't want to spend your life feeling like killing the next person who shoves slapdash Heisenberg like relationships down your throat without explanation as though it were some kind of Physicist Masonic handshake, then you need to study these kinds of behaviors of the Fourier transforms carefully. Even the Heisenberg uncertainty principle itself ultimately comes down to this (or this is one way to derive it): quantum states in conjugate co-ordinates are mapped to one-another by the Fourier transform.

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  • $\begingroup$ The $\omega$ in the equation is not defined. I doubt there should be an oscillatory term in the decay probability. $\endgroup$ – DanielSank Jul 18 '16 at 13:38
  • $\begingroup$ @DanielSank It's a probability amplitude, the probability is not oscillatory. If you check out the answer I link, you'll see what it means. It is the amplitude to find a two state system coupled to the EM field in its excited state in the time interval $[t,\,t+d t)$. It sits together with a continuous function $\psi(\omega, t)$ that defines the amplitude to find the EM field mode in frequency interval $[\omega, \omega+d\omega)$ and together they define the quantum state of the whole coupled system. $\endgroup$ – WetSavannaAnimal Jul 18 '16 at 13:54

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