What Determines the Lineshape in a Two-Level System with Spontaneous Emission?

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Take two level system as example, if I know the population (density matrix element) of excited and ground state, how do I get the lineshape of the spontaenous emission? Can I take the Fourier transformation on the diagonal density matrix element?
 
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Lineshape is FT of
Tr[d(0).d(t) rho] (if you do not include the omega**4 power which is typically
a constant over the line profile)
 
Thanks. What is d is your explanation? dipole moment? In my calculation, I have dipole moment to be constant and rho is time dependent
 
d is dipole moment. rho is the density matrix. Why is d constant? You need matrix elements of d to do the trace. You can get a constant reduced matrix element out of d, but you still have the Clebsch-Gordan to sum
e.g. Ly-a:(sum over m)
<100|d|21m><21m|d(t)|100>
 
Thanks. I am thinking the following question. For two-level system, if you take an average on dipole moment (trace), because of the parity, the diagonal terms vanish. So only the off-diagonal terms contribute to Tr[d(0).d(t) rho], namely, the line shape is from the off-diagonal terms? But as I read in other books, the fluorescence is same as the spontaneous spectrum (i.e. line shape obtained above), right? The fluorescence should be proportional to the diagonal terms (i.e. population), how come does this contradiction occurs?
 
Look in my example for Ly-a: Assuming a diagonal density matrix,
the Trace is
sum_m <21m|rho|21m><21m|d|100><100|d(t)|21m>

no parity issue here.
The relative contribution to any particular line is proportional to the upper level population,
which (for a diagonal density matrix) is proportional to the diagonal matrix element of rho.
Even this is not that simple, because what counts are ALL states that contribute to the line in question, e.g. 210,211,21-1, but NOT 310 for example. When you talk about diagonal matrix elements, you are referring to the operator :rho d.d(t), not rho
 
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