Do spontaneous and stimulated emission obey to a Boltzmann's statistics ?

[fred_run]

Are spontaneous and stimulated emission selected by a Boltzmann's statistics ?

Consider 2 levels(m,n) oscillators in thermal equilibrium with Einstein's coefficients Amn (spontaneous emission), Bmn (stimulated emission), Bnm (absorption) and r(f) the energy density at the frequency f (black body). The ratio of probality for spontaneous emission upon all emission is :

Amn/(Amn+r.Bmn)=1-exp(-hf/kT)=(sum (0 to hf) exp(-E/kT)dE)/(sum (0 to infinity) exp(-E/kT)dE)

That's the statistical part in a Boltzmann's statistics corresponding to photons with energy less than hf. This is actually the condition which defines the spontaneous emission.

In the same way, the ratio of probality for stimulated emission upon all emission is :

rBmn/(Amn+r.Bmn)=exp(-hf/kT)=(sum (hf to infinity) exp(-E/kT)dE)/(sum (0 to infinity) exp(-E/kT)dE)

That's the statistical part in a Boltzmann's statistics corresponding to photons with energy more than hf. This is actually the condition which defines the stimulated emission.

Does it mean that the statistical distribution of spontaneous and stimulated emission in thermal equilibrium obeys to a Boltzmann's statistics ?

Is it really self consistent with the previous theory of black body proposed by Einstein with its coefficients ?

 

[f95toli]

Generally speaking, Boltzmann's statistics is just the classical limit of the "correct" Fermi or Bose statistics, i.e it is valid when you can consider you system to be made up on "classical particles" and quantum effects can be neglected.
Boltzmann's statistics is usually a very good approximation as long as you are looking at large ensemble of particles and kBT is of the same order order of magnitude (or higher) than the energies involved.

Also, note that the Einstein model is really semi-classical meaning it is by no means a fulll and correct description of stimulated emission (for that you need QED); but it does work very well in most practical situations.

 

[fred_run]

Thanks for these general remarks. Boltzmann's statistics is a limit in classical or semiclassical matters.

However, my understanding is that Einstien introduced the coefficients Amn, Bmn, Bnm and QED is able to evaluate their magnitude by operators and wave functions, with different models. But in QED I haven't met such a simple scheme :
(1) the properties of Amn imply that the energy of the incident radiation E<hf,
(2) the properties of Bmn imply that the energy of the incident radiation E>hf,
by means of Boltzmann's statistics. This statistics was employed by Einstein when considering the number of atoms in the level m and in the level n. But it seems that the proportion of spontaneous process in the black body case, considering only emissions, is given by this Boltzmann's statistics. That's was in fact the main object of my question.

Is there such an explicit logical link in QED ?

 

[fred_run]

Using Latex formula,

the probability of spontaneous emission over all emissions is computed according a Boltzmann's law :
$$\frac{A^{n}_{m}}{{A^{n}_{m}}+{\rho}{B^{n}_{m}}}=1-exp(-h\upsilon/kt)=\frac{\int^{h\upsilon}_{0}exp(-E/kT)dE}{\int^{\infty}_{0}exp(-E/kT)dE}$$

so $$E<h\upsilon$$.

And the probability of stimulated emission over all emissions is computed according a Boltzmann's law :
$$\frac{{\rho}{B^{n}_{m}}}{{A^{n}_{m}}+{\rho}{B^{n}_{m}}}=exp(-h\upsilon/kt)=\frac{\int^{\infty}_{h\upsilon}exp(-E/kT)dE}{\int^{\infty}_{0}exp(-E/kT)dE}$$

so $$E>h\upsilon$$.

Does it make a sense in QED ?

 

[Cthugha]

I am a bit puzzled. How do you know the probability of spontaneous emission compared to stimulated emission if you do not include the number of photons present(which is the reason for the occurence of stimulated emission).

 

[fred_run]

The coefficients are related to the density of radiation $$\rho$$ in thermal equilibrium (black body). Probabilities are then implicit.

$${A^{n}_{m}}dt$$ is according the author related to the probability of a spontaneous emission, P(Spontaneous emission).

$${{\rho}B^{n}_{m}}dt$$ is related to the probability of a stimulated emission, P(Stimulated emission).

$${({A^{n}_{m}}+{\rho}B^{n}_{m}})dt$$ is related to the probability of emission, P(Emission).

So applying the theorem of Bayes, noting that the probability P(Emission|Spontaneous emission)=1 and P(Emission|Stimulated emission)=1. For an example :
P(Spontaneous emission|Emission) = P(Emission|Spontaneous emission).P(Spontaneous emission)/P(Emission) .

So,

$${P(Spontaneous \:emission|Emission)=\frac{{A^{n}_{m}}}{{A^{n}_{m}}+{\rho}B^{n}_{m}}}$$

$${P(Stimulated \:emission|Emission)=\frac{{\rho}B^{n}_{m}}{{A^{n}_{m}}+{\rho}B^{n}_{m}}}$$

Is that correct ?

 

[fred_run]

Please find following a modified aspect of the statistics presented. We consider only one frequency $$\upsilon$$ in an infinitesimal interval $$d\upsilon$$ and the incident radiation is quantified. Then if r is the number of incident photons :

$$P(Spontaneous \ emission|Emission)=\frac{A^{n}_{m}}{{A^{n}_{m}}+{\rho}{B^{n}_{m}}}=1-exp(-h\upsilon/kt)=\frac{exp(-0h\upsilon/kT)}{\sum^{\infty}_{r=0}exp(-rh\upsilon/kT)}$$]

so it implies that the number of incident photons is r<1.

$$P(Stimulated \ emission|Emission)=\frac{{\rho}{B^{n}_{m}}}{{A^{n}_{m}}+{\rho}{B^{n}_{m}}}=exp(-h\upsilon/kt)=\frac{\sum^{\infty}_{r=1}exp(-rh\upsilon/kT)}{\sum^{\infty}_{r=0}exp(-rh\upsilon/kT)}$$]

so it implies that the number of incident photons is r>=1.

The probability of a spontaneous emission over all emissions should be the same as the Boltzmann's probability of having less than one photon in the external radiation.

The probability of a stimulated emission over all emissions should be the same as the Boltzmann's probability of having more than one photon in the external radiation.