All my experience with textbook problems of quantum mechanics shows that the energy levels associated with the bound states of a confined quantum system are discrete and sharp. For example, the energy levels of the hydrogen atom. Why is it then said that excited states of an atom have an energy width? Where does the width come from? This fact doesn't match with the examples I know in quantum mechanics. If this question is asked before can some one give the links?
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Leaving aside Doppler broadening and the other main practical reasons for line broadening, the fundamental reason is that the excited atom is coupled to all modes of the EM field equally, or at least there is an extremely wide frequency band of modes that are coupled.
So the atom "tries" to couple its excess energy into all of the modes. As it does so, destructive interference hinders the process for modes that have large frequency separation from the center frequency defined by the energy gap. If you model this broadband coupling mathematically and assume truly equal coupling to all modes at once, you get a Lorentzian lineshape whose breadth is proportional to the coupling strength. This Lorentzian linewidth tends to be narrow compared to the other, more "practical" reasons I mentioned above. I show how to do this calculation in my answer here. This is essential Wigner-Weisskopf Theory. The transition rate, which is proportional to the frequency linewidth, when reckoned by a Fermi Golden Rule calculation, is given by:
$$\Gamma_{rad}(\omega) = \frac{4\, \alpha\, \omega^3\,| \langle 1|\mathbf{r}|2\rangle |^2}{3 \,c^2}$$
with $\alpha$ being the fine structure constant, $\omega$ the center frequency and $\langle 1|\mathbf{r}|2\rangle$ being the overlap between the two electronic states on either side of the transition.
Selene Routley
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