Calculating Excited State Populations

If one or more electrons in an atom occupies a state higher in energy than an unoccupied state, we consider the atom to be in an excited state.  An atom that is not in an excited state is in the ground state.  Consider a dilute gas composed of a single atomic species.  It is of interest to calculate the number of atoms which are in an excited state, because only atoms in an excited state can emit light via spontaneous or stimulated emission.

Clearly there will be more than one excited state for an atom, but for the moment let’s zoom in on just two states.  $latex N_2$ is defined as the number of atoms in the upper energy state per unit volume, and $latex N_1$ is the number of atoms in the lower energy state per unit volume, which may or may not be the ground state.  Thermodynamics tells us that in equilibrium, the ratio

$latex \displaystyle \frac{N_2}{N_1} = e^{-\frac{E_2 – E_1}{k_B T}} $

where $latex k_B$ is Boltzmann’s constant and $latex T$ is the temperature.  As the energy difference between state 2 and state 1 increases, the relative population of state 2 decreases exponentially.  If we want to make a laser, which is based on stimulated light emission, this is problematic, because there will always be many more atoms in lower energy states in equilibrium, and stimulated emission requires atoms in the upper energy states.  A laser or optical amplifier must not operate in equilibrium.

If there is more than one state at energy $latex E_2$, we say that they are degenerate, and the degeneracy $latex g_2$ equals the number of states at that energy.  Incorporating degeneracy, the ratio above becomes

$latex \displaystyle \frac{N_2}{N_1} = \frac{g_2}{g_1} e^{-\frac{E_2 – E_1}{k_B T}} $

But now let’s move away from equilibrium.  We want to write rate equations that allow us to track how the population of atoms in each state evolve with time.  Our rate equations should incorporate the three types of optical transitions we’ve mentioned: absorption, spontaneous and stimulated emission, along with other possible processes.  Einstein (yes, that Einstein) proposed so-called A and B coefficients to describe these rates.  In his formulation, we say that the rate of spontaneous emission is given by

$latex \displaystyle \frac{dN_2}{dt} \biggr\rvert_{spon} = -A_{21} N_2$

where $latex A_{21}$ is the Einstein A coefficient, whose value depends on the type of atom and transition.  Einstein was just using ‘common sense’ here, to say that the rate of spontaneous emission should be proportional to the number of excited atoms.  Each spontaneous emission event ‘destroys’ one excited atom, which is why the time derivative of $latex N_2$ is negative for this process.  Assuming the atom transitions to state 1 after the spontaneous emission event, we can immediately write

$latex \displaystyle \frac{dN_1}{dt}\biggr\rvert_{spon} = A_{21} N_2$

which is positive.  We can write down similar ‘common sense’ rate equations for the other optical processes, as well.  Considering optical absorption, what do we expect the rate to depend upon?  For starters, if we’re talking about optical transitions between states 1 and 2, absorption requires an atom in state 1.  We may therefore expect that the absorption rate should depend on the density of atoms in state 1.  In addition, absorption requires an incoming photon to be absorbed.  Therefore, we might expect the absorption rate to depend on the spectral density of photons at the proper transition frequency; or equivalently, the spectral density of optical energy.  We write the absorption rate as:

$latex \displaystyle \frac{dN_2}{dt}\biggr\rvert_{abs} = B_{12} N_1 \rho(\nu) $

where $latex B_{12}$ is the Einstein B coefficient, and $latex \rho(\nu)$ is the spectral density of optical energy at frequency $latex \nu = (E_2 – E_1) / h$.  As usual, the rate of atoms entering state 1 due to this process is just the negative of the rate of atoms leaving state 2:

$latex \displaystyle \frac{dN_1}{dt}\biggr\rvert_{abs} = – \frac{dN_2}{dt}\biggr\rvert_{abs} $

Finally we come to stimulated emission, which requires an excited atom and an incoming photon, so we write:

$latex \displaystyle \frac{dN_2}{dt}\biggr\rvert_{stim} = -B_{21} N_2 \rho(\nu) $

$latex \displaystyle = – \frac{dN_1}{dt}\biggr\rvert_{stim} $

We can put together all of these optical processes into a single rate equation for $latex N_2$, which would be:

$latex \displaystyle \frac{dN_2}{dt} \biggr\rvert_{optical} = -A_{21} N_2 + B_{12} N_1 \rho(\nu) – B_{21} N_2 \rho(\nu) $

$latex \displaystyle = -\frac{dN_1}{dt} \biggr\rvert_{optical} $

There may also be ‘dark’ processes that transfer atoms into and out of these states, which we will write down later.  Solving these rate equations will allow us to find the dynamic and nonequilibrium population of carriers in the two energy states.