4.4. Spontaneous Emission

4.4. Spontaneous Emission

4-19 4.4. SPONTANEOUS EMISSION What doesn’t come naturally out of semi-classical treatments is spontaneous emission—transitions when the field isn’t present. To treat it properly requires a quantum mechanical treatment of the field, where energy is conserved, such that annihilation of a quantum leads to creation of a photon with the same energy. We need to treat the particles and photons both as quantized objects. You can deduce the rates for spontaneous emission from statistical arguments (Einstein). For a sample with a large number of molecules, we will consider transitions between two states m and n with EEmn> . The Boltzmann distribution gives us the number of molecules in each state. −ωmn /kT NNmn/ = e (4.71) For the system to be at equilibrium, the time-averaged transitions up Wmn must equal those down Wnm . In the presence of a field, we would want to write for an ensemble ? NBUmnm()ωω mn= NBU nmn () mn (4.72) but clearly this can’t hold for finite temperature, where NNmn< , so there must be another type of emission independent of the field. So we write WWnm= mn (4.73) NAm() nm+= BU nm()ωω mn NBU n mn() mn Andrei Tokmakoff, MIT Department of Chemistry, 5/19/2005 4-20 If we substitute the Boltzmann equation into this and use Bmn= B nm , we can solve for Anm : ωmn /kT ABUnm=− nm(ω mn )( e 1) (4.74) For the energy density we will use Planck’s blackbody radiation distribution: ω3 1 U ()ω = (4.75) π 23c ωmn /kT −1 e U ω Nω Uω is the energy density per photon of frequency ω. Nω is the mean number of photons at a frequency ω. ω3 ∴ AB= Einstein A coefficient (4.76) nmπ 23c nm The total rate of emission from the excited state is ω3 wBU=+()ω A using UN()ω = (4.77) nm nm nm nm nm π 23C ω3 =+BN()1 (4.78) π 23c nm Notice, even when the field vanishes ( N → 0) , we still have emission. Remember, for the semiclassical treatment, the total rate of stimulated emission was ω3 wBN= () (4.79) nmπ 23c nm If we use the statistical analysis to calculate rates of absorption we have ω3 wBN= (4.80) mnπ 23c mn The A coefficient gives the rate of emission in the absence of a field, and thus is the inverse of the radiative lifetime: 1 τ = (4.81) rad A.

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