# Fermi-Dirac statistic vs. Maxwell Boltzmann statistics in atomic excitation of lasers systems

In lasers, while relating Einstein's coefficients to the energy density(which depends upon the frequency) we get, $$U(\nu)=\frac{\frac{B}{A}}{e^\frac{h\nu}{k_{B}T}-1}$$ Where $B$ is the coefficient of spontaneous emission and $A$ is the coefficient of stimulated emission.

We further relate this energy density to the planck's formula of energy density of black body radiation, which is $$U(\nu)=\frac{\frac{8(\pi)h\nu^{3}}{c^{3}}}{e^\frac{h\nu}{k_{B}T}-1}$$ While doing all of this we assume population inversion( for a two state system) and assume the atoms of the gas to behave in a Maxwell-Boltzmannian manner. This would hold true because excited atoms follow Maxwell-Boltzmann statistics. Let the ground state have a energy $E_{1}$ and a population $N_{1}$ and the first excited state have a energy $E_{2}$ and a population $N_{2}$, we then relate them by $$\frac{N_{2}}{N_{1}}=e^{\frac{-h\nu}{k_{B}T}}$$ My question is this: Excitation can be of different forms, atomic, thermal, etc; but when we talk of atomic excitation, we need to address the excited electrons within the atoms and henceforth introduce the concept of spin because electrons are fermions which obey Pauli's exclusion principle, so now how can we relate the population of electrons in ground and excited states in a MB-ian manner. Don't we have to use Fermi-Dirac statistics here? If we do use FD statistics, then to what energy density would we relate the energy density of the the coefficients to (because we cannot relate it to the Planck's black body radiation energy density)?

I have below my reflections:

$$\frac{N_{2}}{N_{1}}=\frac{1}{e^{\frac{-E_{f}+E}{k_{B}T}}+1}$$

Where $$\frac{-E_{f}+E}{k_{B}T}=\frac{-\left[\frac{3n}{\pi}\right]^{\frac{2}{3}}\frac{h^{2}}{8m}+h\nu}{k_{B}T}$$

Where I define a function $\gamma$ which varies with frequency as$$\gamma(\nu)=-\left[\frac{3n}{\pi}\right]^{\frac{2}{3}}\frac{h}{8m}+\nu$$

$$U(\nu)=\frac{B_{21}}{A_{21}}\frac{1}{\frac{A_{12}}{A_{21}}}\frac{1}{e^{\frac{\gamma(\nu)h}{k_{B}T}}+1}$$

$$\frac{A_{12}}{A_{21}}=\alpha$$ Now multiplying the numerator and denominator by $\alpha$ I obtained an equation that I used to compare to the Planck's BB radiation energy density.

$$U(\nu)=\frac{B_{21}}{A_{21}}\alpha\frac{1}{\alpha^{2}e^{\frac{\gamma(\nu)h}{k_{B}T}}-{[-\alpha^{2}+\alpha}]}$$

Now by comparison of the above formula to BB radiation energy density, I obtained $\frac{B_{21}}{A_{12}}\alpha=\frac{8(\pi)h\nu^{3}}{c^{3}}$ and $-\alpha^{2}+\alpha=1$ This quadratic equation yields two real roots and by comparing $\alpha^{2}=1$ we obtain in total three possible values of alpha,I.e.

Case one:$\alpha= 1.618$

Case two:$\alpha=-0.618$

Case three:$\alpha=1$

Now using this in the expression of energy density I obtained Let $\frac{B_{21}}{A_{21}}=\beta$

Case one : $$U(\nu)\approx\beta\frac{1}{e^{\frac{\gamma(\nu)h}{k_{B}T}+0.5}+e^{-0.5}}$$

Case two : $$U(\nu)\approx\beta\frac{1}{e^{\frac{\gamma(\nu)h}{k_{B}T}-0.5}+e^{0.5}}$$

Case three : $$U(\nu)=\beta\frac{1}{e^{\frac{\gamma(\nu)h}{k_{B}T}}}$$ Introducing another term $\epsilon$, which is the chemical potential of the system we observe, $$\gamma(\nu)=-E_{f}+\nu-\epsilon$$ Case three changes as $$U(\nu)=\beta\frac{1}{e^{\frac{(-E_{f}+\nu-\epsilon)h}{k_{B}T}}}$$ Now the Taylor's series approximation of e^x is $$e^{x}=1+\frac{x}{1!}+\frac{x^2}{2!}+...\approx1+x$$ At low frequencies $$\frac{\gamma(\nu)h}{k_{B}T}<<1$$ $$e^{\frac{(-\left[\frac{3n}{\pi}\right]^{\frac{2}{3}}\frac{h}{8m}+\nu- \epsilon)h}{k_{B}T}}\approx\frac{(-E_{f}+\nu-\epsilon)h}{k_{B}T }+1$$ Thus the energy density becomes

$$U(\nu)\approx\beta\frac{k_{B}T}{(-E_{f}+\nu-\epsilon)h+k_{B}T}$$

Comparing this with the planck's BB radiation energy density, $$\frac{h\nu}{k_{B}T}+1=\frac{(-E_{f}+\nu-\epsilon)h}{k_{B}T }+1$$ Hence I say that $$E_{f}\alpha\frac{-∂U}{∂N}$$ where the latter term is $\epsilon$, this is true because the Millikan potential(chemical potential of an electron), there is a similar dependence between Fermi energy and chemical potential.

• I am slightly confused by some of the things you have said here. But I think the issue is this: the usual derivation of Einstein coefficients idealizes the atom as a two-state system, and in such a system Fermi statistics play no role. A real atom is not a two-state system, and in some cases Pauli exclusion might affect the transitions which are possible. This would indeed modify the theory, but it should be in a straightforward way (basically just changing the degeneracy of a given transition). Am I at least getting at your question? – Rococo Oct 18 '16 at 23:58
• @Rococo Yeah, it's just that when we consider atomic excitation then we are to consider spin of the electrons which would bring in FD statistics rather than the MB statistics used to describe the population of the electrons. Hence when we do so, how can we then reduce the formula obtained for energy density(when we equate rate of absorption to rate of emission) to the planck's energy density for BB radiation – Naveen Balaji Oct 19 '16 at 0:45

If the electrons in states relevant to the lasing transition are strongly localised (which in a gas they will be) then the question of two electrons from different atoms trying to occupy the same state never arises and in this case Fermi-Dirac statistics reduces to Maxwell-Boltzmann statistics. In other words I can treat the electrons as distinguishable because I can say one is on the atom at position $\mathbf{x}$ while that one is on the atom at position $\mathbf{y}$.