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 \...
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...
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...
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...
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...
I have got an answer to my question (from A.KASSLER lab. in Paris) : there is no activation energy for the spontaneous emission. There is an exponential decay as usual when a level is coupled to a continuum. Vacuum fluctuations are of course involved in the process but standard radiation damping...
Hello,
using the words of chemical thermodynamics could we say that stimulated emission of light is "catalysed" by incident photons, as well as absorption ?
In that way, does it mean that there is a kind of "activation energy" in the emission process ?
Without incident photon...