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When dealing with a system of two energy levels $E_1$ and $E_2$ in laser theory, the following rate equations are written:

$$\frac{d N_1}{dt} = A_{21}N_2 - B_{12} \rho ( \nu ) N_1 + B_{21} \rho ( \nu ) N_2$$ $$\frac{d N_2}{dt} = -A_{21}N_2 + B_{12} \rho ( \nu ) N_1 - B_{21} \rho ( \nu ) N_2$$

where $N_1$ and $N_2$ are the number of particles with energy $E_1$ and $E_2$ (respectively) per unit volume, $A_{21}$ is «the rate of spontaneous emission from state 2 to state 1, $B_{12}$ is the rate of stimulated absorption from state 1 to state 2, and $B_{21}$ is the rate of stimulated emission from state 2 to state 1». $A,B$ are the Einstein coefficients.

This computation is used as a first description of the emission in semiconductors, for example, and optics.

Consider this document as a reference (pp. 8-9). According to it, $\rho (\nu)$ is the energy density (of the electromagnetic field inside the material) per unit volume and per unit frequency.

Why is such density used here? That is: why, in oder to obtain the probability per unit time that one photon will be absorbed by one electron, the quantity $B_{12} \rho ( \nu )$ is needed?

Shouldn't a "number of photons" be used here instead?

My question is also due to dimensions. The rates $dN_1/dt$ and $dN_2/dt$ should have in fact dimensions $1/(s \cdot cm^3)$. $A$ has dimensions $1/s$, so the terms $A_{21} N_2$ certainly are $1/(s \cdot cm^3)$.

If the $B$ coefficients are also $1/s$, considering that $\rho(\nu)$ is $J/(Hz \cdot cm^3)$, the terms with $B$ cannot have the same dimensions as the terms with $A$.

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The dimensions and meaning of the B coefficients are not the same as the A coefficient.

The probability of spontaneous emission does not depend on the radiation environment of the atom, whereas absorption and stimulated emission do.

Given that, one has a choice of how one encodes that in terms of the B coefficients, which are only a property of the atom and which can be calculated quantum mechanically.

The conventional choice (there is another that uses the mean specific intensity of the radiation field) uses the energy density at a given frequency, because what matters is how many photons of that frequency are near the atom at any time. The energy density is conveniently proportional to that.

Thus we say that the probability of a transition per unit time is the product of the B coefficients and the energy density and this defines what the B coefficient is (and its dimensions). As I said, other choices are available.

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  • $\begingroup$ Ok and thank you. Which is the proportionality factor between the number of photos at a frequency $\nu$ and the energy density $\rho ( \nu )$ at that frequency? $\endgroup$ – BowPark Jul 13 '16 at 9:47
  • $\begingroup$ @BowPark ? Each photon has an energy $h\nu$. If there are $n(\nu)$ per unit volume, then the energy density is $n(\nu)h\nu$. $\endgroup$ – Rob Jeffries Jul 13 '16 at 10:03
  • $\begingroup$ Ok, I was looking for the (quite obvious, though) relation between the energy density $\rho (\nu)$ and the photon density $n(\nu)$. Thank you. $\endgroup$ – BowPark Jul 13 '16 at 17:26
  • $\begingroup$ @RobJeffries Is A the probability or the rate of change of probability? $\endgroup$ – ado sar Dec 30 '19 at 16:51

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