# relation between line width and life time

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?

• 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? – Sanya Jul 12 '16 at 19:24
• @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 – WetSavannaAnimal Jul 18 '16 at 0:47

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 $\omega$ in the equation is not defined. I doubt there should be an oscillatory term in the decay probability. – DanielSank Jul 18 '16 at 13:38
• @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. – WetSavannaAnimal Jul 18 '16 at 13:54