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:


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$$\gamma(\nu)=-\left[\frac{3n}{\pi}\right]^{\frac{2}{3}}\frac{h}{8m}+\nu$$


$$\frac{A_{12}}{A_{21}}=\alpha$$ 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.


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


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.

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    $\begingroup$ 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? $\endgroup$
    – Rococo
    Oct 18, 2016 at 23:58
  • $\begingroup$ @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 $\endgroup$ Oct 19, 2016 at 0:45

1 Answer 1


Your states 1 and 2 are states of an atom, i.e. the nucleus and all bound electrons taken together, and consequently your labelling should already be taking the Fermi statistics of electrons on a single atom. In other words you do not say "atom a is in state 1 twice" you would say "I excited one electron to put the atom in state 2 and then excited a second to put it in some other state 3" and this labelling scheme simply does not bother to include state forbidden by the exclusion principle.

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 $\mathbf{y}$.

  • $\begingroup$ by the 12 and 21 labelling I mean that the electron has gone from state one to state two and the electron has gone from state two to state one respectively. I have done this because the Einstein's coefficients mention the directionality of electrons as in spontaneous and stimulated emissions, the electron moves from state two to state one. I have edited and added my reflections, please do take a look and comment your thoughts. And thanks! $\endgroup$ Oct 19, 2016 at 5:15
  • $\begingroup$ I'm not sure what you mean here. Fermi-dirac statistics describes the statistics of state occupation numbers, not the transitions between them. The only way it effects transitions is by forbidding certain final states and altering the equilibrium populations. $\endgroup$ Oct 19, 2016 at 8:59
  • $\begingroup$ I mean exactly what your are saying because I'm talking about the change in equilibrium populations and their distribution in the levels(as due to the exclusion principle, the distribution would alter). Hence, shouldn't we use FD distribution function rather than the MB distribution function when we are talking about atomic excitations or rather when if we introduce spin. Here I have assumed a simple two level system and done my calculations. $\endgroup$ Oct 19, 2016 at 9:07
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    $\begingroup$ OK my point was that states on different atoms are distinguishable, so MB statistics is correct in this case and when considering a single atom, if I excite the atom twice I do not increase the population of the higher energy state by 2, I put the atom into some totally different state not included in the two level picture, and Fermi statistics has already been taken account of in writing a list of the allowed atomic states. $\endgroup$ Oct 19, 2016 at 9:19
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    $\begingroup$ That would be the whole of atomic physics $\endgroup$ Oct 19, 2016 at 10:12

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