Why does the rate of transitions depend on the population of the ground state?

[marco1235]

Hi all. I need a help about a stupid thing which is puzzling me! I'm studying a paper where it's described a classical excitation of a fluorophore from a ground state to an excited state.

See the attached image!

There are two laser beams which act on the sample, the first one which is an excitation laser (it runs the $L_{0}$ to $L_{1}$ transition) and a second laser beam called STED (yes the same of the "new" fancy way of getting super resolute images, I'm actually studying the theory of this super resolution technique).

Now the eqs are the classical Einstein like, but what I don't understand is (let's take for example the first one) why there's a factor like $h_{exc}\sigma_{01} ( n_1- n_0)$. I would had expected something like "minus" a quantity which told me that the temporal variation of the ground state is diminished because of absorption. In fact the second term of the right hand side is of the form $1/\tau_{vibr} * n_3$ which is telling me that atoms are coming from the level $L_3$.

Same reasoning applies for the other equations, but it would be great to have an understanding of just the first one, the other will be the same thing.
Thanks in advance and forgive me if my English is not so correct. I did my best.

Have a wonderful day everybody! Ciao

 

[Ygggdrasil]

Note that in the first equation you have n_1 - n_0, not n_0 - n_1. It is a negative number, just written by flipping the n terms instead of with the negative sign explicitly out in front of the term.

 

[marco1235]

Yeah I noticed it, but the question is, why that term with a difference in population number and not a simple $- h_{exc} \sigma_{01} n_1$ !? I don't understand the part with $n_0$.
It has no sense to me.

 

[marco1235]

I would had written the equation like this: $dn_{0}/dt = -h_{exc}\sigma_{01} * n_1 + 1/\tau_{vibr} * n_3$

Not as in the paper. Maybe now my objection is more clear, I hope so.

 

[Ygggdrasil]

The +h_excσ₀₁∗n₁ term accounts for stimulated emission.

 

[marco1235]

Did you mean the $+h_{ecx}\sigma n_0$ ? $n_0$ not $n_1$ as you wrote in the last message

 

[Ygggdrasil]

No, what I wrote is correct. Can you explain why you think $dn_{0}/dt = -h_{exc}\sigma_{01} * n_1 + 1/\tau_{vibr} * n_3$ is the correct form? Which transition does the $-h_{exc}\sigma_{01}$ term account for and what transition does the $1/\tau_{vibr} * n_3$ account for?

 

[marco1235]

Solved! I went a bit fast and after reviewing Einstein coefficients everything is right!

 

[Ygggdrasil]

The basic thing to remember is that the rate of transitions from state A to state B will almost always depend on the population of state A. So, for example, if you're looking at the rate of excitation of electrons from L0 to L1, the rate will be $h_{exc}\sigma_{01}*n_o$ not $h_{exc}\sigma_{01}*n_1$. Similarly, because stimulated emission is a transition from L1 to L0, the rate will be $h_{exc}\sigma_{01}*n_1$. Thus, when you look at the contributions of excitation from L0 to L1 and stimulated emission from L1 to L0 on the population of n0, the $h_{exc}\sigma_{01}*n_o$ term will have a negative sign (because it involves electrons transitioning from L0 to L1, and the $h_{exc}\sigma_{01}*n_1$ will have a positive sign (because it involves electrons transitioning from L1 to L0).