The linewidths come out very naturally from Maxwell's Equations by treating the atom as a tiny classical antenna. I do the calculations for the 2p-1s transition in hydrogen on my blogsite here: The Semi-Classical Calculation

The idea is that from the Schroedinger equation, the superposition of the s and p states gives you get a tiny oscillating dipole about 1 Angstrom in length, a frequency of 2.5 x 10^15 Hz, and a wavelength of 1200 Angstroms.

From the ratio of the antenna size to the wavelength, you can easily calculate the radiation resistance: it is on the order of 100 micro-ohms.

The "current" in the antenna can be estimated by taking the charge of the electron times the frequency. This is not quite right but it's close enough for what we're doing here. Oddly enough, for the hydrogen atom it comes out to the seemingly macroscopic value of around 1 milliamp.

The power in the antenna is just I-squared-R, which gives us 100 pico-watts. But what about the energy in the excited state? It's 3/4 Rydberg, which works out to around .000 001 pico-joules. So clearly the "lifetime" of the excited state is around $\ 10^{-8}$ seconds.

You can continue with the classical analysis to get your linewidth by taking the ratio of the "lifetime" to the single-cycle time. It's called the Q factor in antenna theory. Or your can use the language of quantum mechanics, as other posters have done here, and express the linewidth broadening in terms of the uncertainty principle. It's exactly the same thing.

But you can't apply the uncertainty principle until you calculate the decay constant, or "lifetime". And that comes straight out of classical antenna theory.

The linewidths come out very naturally from Maxwell's Equations by treating the atom as a tiny classical antenna. I do the calculations for the 2p-1s transition in hydrogen on my blogsite here: The Semi-Classical Calculation

The idea is that from the Schroedinger equation, the superposition of the s and p states gives you get a tiny oscillating dipole about 1 Angstrom in length, a frequency of 2.5 x 10^15 Hz, and a wavelength of 1200 Angstroms.

From the ratio of the antenna size to the wavelength, you can easily calculate the radiation resistance: it is on the order of 100 micro-ohms.

The "current" in the antenna can be estimated by taking the charge of the electron times the frequency. This is not quite right but it's close enough for what we're doing here. Oddly enough, for the hydrogen atom it comes out to the seemingly macroscopic value of around 1 milliamp.

The power in the antenna is just I-squared-R, which gives us 100 pico-watts. But what about the energy in the excited state? It's 3/4 Rydberg, which works out to around .000 001 pico-joules. So clearly the "lifetime" of the excited state is around 10-8 seconds.

You can continue with the classical analysis to get your linewidth by taking the ratio of the "lifetime" to the single-cycle time. It's called the Q factor in antenna theory. Or your can use the language of quantum mechanics, as other posters have done here, and express the linewidth broadening in terms of the uncertainty principle. It's exactly the same thing.

But you can't apply the uncertainty principle until you calculate the decay constant, or "lifetime". And that comes straight out of classical antenna theory.

EDIT: Just noticed that the accepted answer claims you calculate the linewidth by simply applying the Heisenberg Uncertainty principle. Surely this is quite incorrect. In the example given, the "lifetime" is taken as given...but it is precisely the lifetime (which is inverse to the linewidth) that we want to calculate. You don't get the linewidth by applying Heisenberg to the lifetime...you get it by dividing the lifetime by the speed of light. And that begs the question...how did you get the lifetime?

As I've explained already, you get the lifetime by applying the classical antenna equations to the vibrating atom. Its a very classical calculation.

The linewidths come out very naturally from Maxwell's Equations by treating the atom as a tiny classical antenna. I do the calculations for the 2p-1s transition in hydrogen on my blogsite here: The Semi-Classical Calculation

The idea is that from the Schroedinger equation, the superposition of the s and p states gives you get a tiny oscillating dipole about 1 Angstrom in length, a frequency of 2.5 x 10^15 Hz, and a wavelength of 1200 Angstroms.

From the ratio of the antenna size to the wavelength, you can easily calculate the radiation resistance: it is on the order of 100 micro-ohms.

The "current" in the antenna can be estimated by taking the charge of the electron times the frequency. This is not quite right but it's close enough for what we're doing here. Oddly enough, for the hydrogen atom it comes out to the seemingly macroscopic value of around 1 milliamp.

The power in the antenna is just I-squared-R, which gives us 100 pico-watts. But what about the energy in the excited state? It's 3/4 Rydberg, which works out to around .000 001 pico-joules. So clearly the "lifetime" of the excited state is around 10-8 seconds.

You can continue with the classical analysis to get your linewidth by taking the ratio of the "lifetime" to the single-cycle time. It's called the Q factor in antenna theory. Or your can use the language of quantum mechanics, as other posters have done here, and express the linewidth broadening in terms of the uncertainty principle. It's exactly the same thing.

But you can't apply the uncertainty principle until you calculate the decay constant, or "lifetime". And that comes straight out of classical antenna theory.

The linewidths come out very naturally from Maxwell's Equations by treating the atom as a tiny classical antenna. I do the calculations for the 2p-1s transition in hydrogen on my blogsite here: The Semi-Classical Calculation

The idea is that from the Schroedinger equation, the superposition of the s and p states gives you get a tiny oscillating dipole about 1 Angstrom in length, a frequency of 2.5 x 10^15 Hz, and a wavelength of 1200 Angstroms.

From the ratio of the antenna size to the wavelength, you can easily calculate the radiation resistance: it is on the order of 100 micro-ohms.

The "current" in the antenna can be estimated by taking the charge of the electron times the frequency. This is not quite right but it's close enough for what we're doing here. Oddly enough, for the hydrogen atom it comes out to the seemingly macroscopic value of around 1 milliamp.

The power in the antenna is just I-squared-R, which gives us 100 pico-watts. But what about the energy in the excited state? It's 3/4 Rydberg, which works out to around .000 001 pico-joules. So clearly the "lifetime" of the excited state is around $\ 10^{-8}$ seconds.

You can continue with the classical analysis to get your linewidth by taking the ratio of the "lifetime" to the single-cycle time. It's called the Q factor in antenna theory. Or your can use the language of quantum mechanics, as other posters have done here, and express the linewidth broadening in terms of the uncertainty principle. It's exactly the same thing.

But you can't apply the uncertainty principle until you calculate the decay constant, or "lifetime". And that comes straight out of classical antenna theory.