Download Intensities and Distributions - Class Notes | PHYS 552 and more Study notes Optics in PDF only on Docsity! 1 Intensities and Distributions We want to calculate what is expected as far as the excitation of vibrational states and spontaneous vs. stimulated emission. Before we leave the Einstein coefficients, let’s see what we can do with them. Go back to Planck’s Law of radiation: Remember that the density of modes (light) with their frequency between and dω ω ω+ is: ( ) ( ) 2 2 3/ cd dρ ω ω ω ω π= At thermal equilibrium the probability that the mode oscillator is thermally excited to its nth state is: [ ] [ ] n n exp -E / kT P( ) exp -E / kT n n = ∑ n 1E n 2 ω⎛ ⎞= +⎜ ⎟ ⎝ ⎠ (zero point energy cancels) Define [ ]u exp - / kTω= ( ) n n n n 0 uP( ) u 1 u u n ∞ = ⎡ ⎤ ⎢ ⎥ ⎢ ⎥= = − ⎢ ⎥ ⎢ ⎥⎣ ⎦ ∑ The mean number of photons excited in the field mode at temperature T is: ( ) nn nP( ) 1 u nu n n n< >= = −∑ ∑ ( ) n un 1 u u u 1 unu ∂ < >= − = ∂ −∑ [ ] u 1n 1 u exp / kT 1ω < >= = − − This is the “Planck thermal excitation function.” 2 The mean energy density of the radiation in the modes of frequency ω and temperature T is: ( )( ) nTW d dω ω ωρ ω ω< > =< > [ ] 3 2 3( ) c exp / kT 1T dW d ω ωω ω π ω < > = − ( )TW dω ω< > = energy / unit volume / unit angular frequency range Planck’s Law of Blackbody Radiation We can rewrite the expression mean # of photons un 1 u < > = − to get nu 1 n < > = + < > Remember ( ) nP( ) 1 u un = − So we have: ( ) n 1 n nP( ) 1 n n + < > = + < > n = 0 has the largest probability. Characterize a probability distribution by its factorial moments. The rth factorial moment is: ( )( )( ) ( )( ) ( )n n 1 n 2 n r 1 n n 1 n 2 . . . n r 1 P( ) n n< − − − + >= − − − +∑ rr! n= < > Use this to calculate the fluctuations: ( ) ( ) ( )22 2 2 2n n n P( ) n n ; but n n 1 2 n n nΔ = − < > =< > − < > < − >= < >∑ So ( ) ( )2 2n n nΔ = < > + < > Fluctuation: ( )1/ 22n n nΔ = < > + < > <n> 1>> n <n>Δ ∼ We can only measure this with perfect detectors and instantaneous measurements. 5 Light Beams with External Sources We have in general (not just at thermal equilibrium) ( )3 2 3 21 21/ c B Aω π = ; but at steady state. Define a “saturation radiative energy density”. ( )3 2 3sW / cω π= then s 21 21W B A= So, at 15 -13 10 sω ≈ × -14 -3sW 10 Jmd dω ω≅ The intensity of the saturating light source: 3 21 s 2 2 21 AI cW c B cs ω π = = = -6 -2I 3 10 Wmsd dω ω≅ × If the conventional light source has 10 -16 10 sωΔ ×∼ 5 -2I 1.8 10 Wms strongest dω ≈ ×∫ Total Beam Intensity! All the strongest conventional sources are < 1/10 of this intensity. So all molecules decay by 21A for conventional sources!! Directions of the 3 Einstein processes: So, the spontaneous emission, 21A , is what leads to the absorption of the incoming intensity. 6 How long does it take to get to the saturation value? Assume a gas – no other ways of deactivation (only fluorescence). Let’s look at the optical excitation of a two-state atom (with the total intensity W . ( )1 2 2 2 1N / - N / N A N N Bd dt d dt W= = + − < > At steady-state ( )2 2 1N A N N B 0ω+ − < >= Remembering that 21 21W B As = , we have ( )2 2 1N W N N 0s W+ − < >= , and 1 2N N N= + 2 BN N N A 2B W 2s W W W W < > < > = = + < > + < > Steady state atomic population: 1 WA BN N N A 2B W 2 s s WW W W + < >+ < > = = + < > + < > 7 Mean rates of the Einstein transitions for steady state: ( ) ( )1 N B NA 2 s s s W W WW W W W + < > < > < >= + < > absorption ( ) ( )2 N B NA 2 s s s W W WW W W W + < > < > < >= + < > stimulated emission ( )2 N A NA 2s W W W < > = + < > spontaneous emission So, 2 2 NN A A⎛ ⎞→ ⎜ ⎟ ⎝ ⎠ at full saturation. Spontaneous emission cannot happen faster than a particular limit. Stimulated emission can increase without limit. • Let’s look at rates: ( )1 2 2 2 1 N N- N A N N B W t t ∂ ∂ = = + − < > ∂ ∂ The general time dependent solution of the equation above: 1 2N N N= + ( ) ( )1 1N ( ) N 0 N exp - A 2B N2 2 s s s s W W W Wt W t W W W W ⎧ ⎫+ < > + < > = − + < > +⎡ ⎤⎨ ⎬ ⎣ ⎦+ < > + < >⎩ ⎭ 2 2 1N ( ) from N N Nt = −