Andrei Tokmakoff, MIT Department of Chemistry, 5/19/
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1 drei Tokmakoff, MT Departmet of Chemistry, 5/9/5 4-9 Rate of bsorptio ad Stimulated Emissio The rate of absorptio iduced by the field is E k " (" (" $% ˆ µ # (" &" k k (4. The rate is clearly depedet o the stregth of the field. The variable that you ca most easily measure is the itesity (eergy flux through a uit area, hich is the time-averaged value of the Poytig vector, S c S E 4" c c S E E 4 8 (4. (4. other represetatio of the amplitude of the field is the eergy desity U E c 8 (for a moochromatic field (4.4 Usig this e ca rite 4 k U " k #$µ % " k & " " ˆ (4.5 or for a isotropic field here E xˆ E yˆ E zˆ E 4 U " µ # " $ " " k k k (4.6 or more commoly k k U k (4.7 4 µ " k k Eistei coefficiet (4.8 (this is sometimes ritte as k ( k " µ he the eergy desity is i ν.
2 drei Tokmakoff, MT Departmet of Chemistry, 5/9/5 4-4 U ca also be ritte i a quatum form, by ritig it i terms of the umber of photos E U (4.9 8" is idepedet of the properties of the field. t ca be related to the absorptio cross-sectio, σ. total eergy absorbed / uit time total icidet itesity eergy / uit time / area "" # k U( k (" "" # " k c U k (4. "" c k More geerally you may have a frequecy depedet absorptio coefficiet ("# k (" k g( " here g(ω is a lieshape fuctio. The golde rule rate for absorptio also gives the same rate for stimulated emissio. We fid for to levels m ad : U ( U ( sice U ( U ( (4. The absorptio probability per uit time equals the stimulated emissio probability per uit time. lso, the cross-sectio for absorptio is equal to a equivalet cross-sectio for stimulated. emissio, SE
3 drei Tokmakoff, MT Departmet of Chemistry, 5/9/5 4-4 o let s calculate the chage i the itesity of icidet light, due to absorptio/stimulated emissio passig through sample (legth L here the levels are thermally populated. d " dx + " dx (4. m SE d m a dx " (4., m These are populatio of the upper ad loer states, but expressed as a populatio desities. f is the molecule desity, E e " # $ % & ' Z ( (4.4 " is the thermal populatio differece betee states. m tegratig over a pathlegth L: al e " # for high freq. " (4.5 # al e " : cm " : cm L :cm or ritte as eer s La: log C " L C : mol / liter :liter / mol cm (4.6 "
4 drei Tokmakoff, MT Departmet of Chemistry, 5/9/5 4-4 SPOTEOUS EMSSO What does t come aturally out of semi-classical treatmets is spotaeous emissio trasitios he the field is t preset. To treat it properly requires a quatum mechaical treatmet of the field, here eergy is coserved, such that aihilatio of a quatum leads to creatio of a photo ith the same eergy. We eed to treat the particles ad photos both as quatized objects. You ca deduce the rates for spotaeous emissio from statistical argumets (Eistei. For a sample ith a large umber of molecules, e ill cosider trasitios betee to states m ad ith E m > E. The oltzma distributio gives us the umber of molecules i each state. m / e " / kt (4.7 For the system to be at equilibrium, the time-averaged trasitios up must equal those do. the presece of a field, e ould at to rite for a esemble W ( m W? U U (4.8 but clearly this ca t hold for fiite temperature, here m <, so there must be aother type of emissio idepedet of the field. So e rite W W ( + ( m U U (4.9
5 drei Tokmakoff, MT Departmet of Chemistry, 5/9/5 4-4 f e substitute the oltzma equatio ito this ad use, e ca solve for : U ( e /kt ( " (4. For the eergy desity e ill use Plack s blackbody radiatio distributio: " U / kt " " c e # #$%$& U (4. U is the eergy desity per photo of frequecy ω. is the mea umber of photos at a frequecy ω. " # c Eistei coefficiet (4. The total rate of emissio from the excited state is + U + usig U " C (4. (4.4 otice, eve he the field vaishes (, e still have emissio. Remember, for the semiclassical treatmet, the total rate of stimulated emissio as (4.5 f e use the statistical aalysis to calculate rates of absorptio e have (4.6 The coefficiet gives the rate of emissio i the absece of a field, ad thus is the iverse of the radiative lifetime: rad (4.7
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