Spectral line width calculations

[Mjolnir]

So I'm in an intro quantum physics course, and while I'm sure this is a really simple problem, I'm just not getting the math to work out here.

Say you excite the atoms of some gas such that they emit light at a wavelength of 5500 angstroms as they fall back to the ground state. Now if the light intensity falls off with I(t) = I*e^{-r*t} (with r = 5*10^(-7) Hz), we can get the time dependence of the wave function to be e^{-r*t/2}*e^{-i*\omega_{0}*t}, correct? Now given this, how would you go about finding the spread of wavelengths of the spectral line? Now the idea I would think is to find <E^2> - <E>^2, and then just set the standard deviation of lamda equal to h*c over this? Anyway, as I said the math just isn't working out. So as much as I hate my first post here to be a question, if someone could maybe layout a general process for doing this (without actually plugging in the values, of course; the idea is to understand this on my own), it would be much appreciated.

 

[Meir Achuz]

Fourier transform the wave function using exp(-iEt/h).
This gives you the WF in energy space.
Its absolute value squared gives you the line shape.

 

[country boy]

... we can get the time dependence of the wave function to be e^{-r*t/2}*e^{-i*\omega_{0}*t}, correct? Now given this, how would you go about finding the spread of wavelengths of the spectral line? ...

You can get an approximation of the width without having to perform the Fourier transform. Let Df be the difference in frequency between two waves at the upper and lower bounds of the distribution. As time t progresses the phase difference between the two extreme waves will increase as 2pi*Df*t. The waves will strongly interfere when the phase difference reaches pi at time T, so that 2*Df*T = 1. Now T can be equated with the time-width of your waveform, approximately 4/r (2 times 2/r), and the spread in your frequencies is approximately Df = r/8.

This doesn't give the exact answer, which depends on details in the shape of the distribution. But it is close, and it helps demonstrate what is going on.

 

[Meir Achuz]

I gave you the method of doing it.
If you just want the answer, the width of the line is gamma=2r.
gamma is the "full width at half maximum".