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Imagine you have an ensemble of 3-level systems whose dynamics is a Markov chain Further, suppose that transition from level 1 to level 3 requires the absorption of a "yellow" photon (high energy), while transitions between the other levels involve the emission or absorption of "infrared" photons (low energy). (the systems are in contact with a thermal reservoir that exchanges energy with the ensemble, mostly infrared but sometimes yellow, and there's also a source of "yellow" light shining on them). My question is this: is there a useful way to calculate the rate at which this ensemble is converting yellow light to infrared light in terms of the transition probabilities and the occupation numbers (how many systems are in a given state at that time)? Even if the ensemble of systems is not in a stationary state?

Thank you.

Edit: Following dan_fulea's suggested notation, say that the current distribution is given by the $ 1 \times 3$ row vector $q$, and that the transition matrix for a particle's state is the $3 \times 3$ matrix $A$. Each system (particle) updates its state every unit of time.

The best I have been able to come up with for the "instantaneous" rate of conversion of yellow light into its energy equivalent in infrared light is something like $r = q_1a_{13}a_{32}a_{21}/3$. The reasoning is that for a particle to contribute to the conversion of energy starting right now, it has to be in state 1, and then get knocked into state 3, and then successively fall back down to state 2 and then state 1. Really, this is a per-particle rate, and would need to be multiplied by $N$ to get the overall rate. The division by 3 is just because the cyclic path considered, 1->3->2->1, takes 3 time steps.

But that's a little cheesy, since the particle could bounce around between states 2 and 3 before it finally got back to state 1. So maybe I need to consider all such paths, and average the time spent in them. (I also forgot that the diagonal elements of $A$ are typically not zero, they may even be dominant. So the particle may remain in the same state for stretches of time).

There are other problems. I don't really like the fact that it depends only on the state 1 occupation number, $q_1$, and not the others. That would seem to indicate that the flow of energy through the system is increased by having all the particles in state 1. But if energy is to be absorbed and then dissipated, they have to spend time in the other two states.

To fix this, I could add two more similar terms proportional to $q_2$ and $q_3$ that account for particles that are at different stages in the cycle at the moment. But then I would get an answer independent of the occupation numbers, depending only on the transition matrix. This also seems dumb. Surely the distribution of the particles among the 3 states affects their overall ability to absorb and dissipate energy.

So that's where my thinking is right now.

Willie Betmore
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    Which are the "states"? Only three, the levels 1,2,3? Or levels with energy information? Do we have "steps" of equal length? What is "the rate of converting"? Please take time and translate the situation in a mathematical language, we cannot do the modeling. – dan_fulea Aug 16 '18 at 21:16
  • Okay, good point. I will try to come back and translate it into math later. For the time being...there are three states. I am considering that each system has the same 3x3 transition matrix. That way, the number of systems in a given state is nearly proportional to the probability of being in a given state. What is the overall rate at which 1->3 transitions are resulting in a 3->2 transition followed by a 2->1 transition? – Willie Betmore Aug 16 '18 at 21:27
  • In this case, the modeling could be the following one. We use an initial distribution of particles in states. If there are $10^7$ particles (or systems), $2\cdot 10^6$ of them in state one, $5\cdot 10^6$ of them in state two, $3\cdot 10^6$ of them in state three, then the initial distribution is given by the row vector $q=\left[\frac 2{10}\ \frac 5{10}\ \frac 3{10}\ \right]$. Then the transition matrix $A$ of the system from one time unit to the next one, assuming each particle / system is acted by one transition in each step, brings $q$ to $qA$, then $qAA$, etc Spectral properties of $A$... – dan_fulea Aug 16 '18 at 21:34
  • Exactly. That's the type of situation I have in mind. I would be happy to have a way to define the rate I'm talking about even in the stationary state (left eigenvector of A). So maybe I should start there first. But what I'm really hoping for is a way to define the "instantaneous" rate of conversion, even when the distribution is not stationary. – Willie Betmore Aug 16 '18 at 21:51
  • I edited my post to include a little more detail on how I have been thinking about the problem. – Willie Betmore Aug 17 '18 at 20:21
  • If we must implement a transition matrix $A(q)$ depending on $q=(q_1,q_2,q_3)$, $q_1+q_2+q_3=1$, then the modeling is no longer discrete. Instead, we can first use only two parameters, $$q_1,\ q_2$$ inside the (solid) triangle $\Bbb T$ given by $0\le q_1,q_2\le q_1+q_2\le 1$, and now we need either a discrete, or a continuous transition process "from $\Bbb T$ to $\Bbb T$", i.e. having the states $\Bbb T$. The transition has to be given by the physical need... This is the best i can do in the given framework... – dan_fulea Aug 18 '18 at 12:37
  • The transition matrix A is constant for a particle (and it is the same for all particles). Of course the discrete time is unphysical, but I'm not really concerned about that. So I'm treating the trajectory of each particle through the 3-level state space as a finite Markov chain. Even if it's not very realistic. It is the whole ensemble of N particles whose absorption and dissipation behavior should be a function of $q_i$. – Willie Betmore Aug 19 '18 at 03:12
  • I should probably also have mentioned that this transition matrix A is as nice as they come. All entries are positive, so it is regular, irreducible, ergodic, etc. This is because of microscopic reversibility. Anything that happens microscopically can also go the other way, although it may be less probable. – Willie Betmore Aug 19 '18 at 03:54

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