MIT Department of Chemstry 5.74, Sprng 005: Introductory Quantum Mechancs II Instructor: Professor Andre Tokmakoff p. 81 Rate of Absorpton and Stmulated Emsson The rate of absorpton nduced by the feld s π ( ω )= ω k ˆ µ δ ω E 0 ( ) ( k ) The rate s clearly dependent on the strength of the feld. The varable that you can most easly measure s the ntensty I (energy flux through a unt area), whch s the tme-averaged value of the Poyntng vector, S S = c (E B) 4π c I = S = c E = E 0 4π 8π Another representaton of the ampltude of the feld s the energy densty I 1 U = = E 0 (for a monochromatc feld) c 8π Usng ths we can wrte 4π = ω k µ ˆ δ ω U ( ) ( k ω) or for an sotropc feld where E 1 0 xˆ = E 0 ŷ = E 0 ẑ = E 0 4π = ω µ δ ω ω) U ( ) k ( k or more commonly = B k U (ω k ) 4π k B k = µ Ensten B coeffcent (ths s sometmes wrtten as B k = π ( ) µ k when the energy densty s n ν).
p. 8 U can also be wrtten n a quantum form, by wrtng t n terms of the number of photons N ω N ω= E 0 U = N 8π π c B s ndependent of the propertes of the feld. It can be related to the absorpton cross-secton, σ A. total energy absorbed / unt tme σ A = total ncdent ntensty (energy / unt tme / area ) ω ω Bk U (ω k ) = = I c U (ω k ) ω σ A = B k c More generally you may have a frequency dependent absorpton coeffcent σ ω B k ( ) B k g ( ω) where g(ω) s a lneshape functon. A ( ) ω = The golden rule rate for absorpton also gves the same rate for stmulated emsson. We fnd for two levels m and n : w nm = w mn B nm U ( ω nm)= B nm U ( ω nm ) snce U ( ω nm )= U ( ωmn ) B nm = B mn The absorpton probablty per unt tme equals the stmulated emsson probablty per unt tme. Also, the cross-secton for absorpton s equal to an equvalent cross-secton for stmulated emsson, (σ A ) = ( σ ). nm SE mn
p. 8 Now let s calculate the change n the ntensty of ncdent lght, due to absorpton/stmulated emsson passng through sample (length L) where the levels are thermally populated. di = N n σa dx + N m σ SE I dx di = (N N )σ dx n m a I N, N m These are populaton of the upper and lower states, but expressed as a n populaton denstes. If N s the molecule densty, N = N e βe n n Z N=N N n m s the thermal populaton dfference between states. Integratng over a pathlength L: I = e σ a I 0 e N L N σ al for hgh freq. N N N: cm σ : cm L :cm n or wrtten as Beer s Law: I A = log = C L C : mol / lter :lter / mol cm I 0 = 0 N σ A
p. 84 SPONTANEOUS EMISSION What doesn t come naturally out of sem-classcal treatments s spontaneous emsson transtons when the feld sn t present. To treat t properly requres a quantum mechancal treatment of the feld, where energy s conserved, such that annhlaton of a quantum leads to creaton of a photon wth the same energy. We need to treat the partcles and photons both as quantzed objects. You can deduce the rates for spontaneous emsson from statstcal arguments (Ensten). For a sample wth a large number of molecules, we wll consder transtons between two states m and n wth E m > E n. The Boltzmann dstrbuton gves us the number of molecules n each state. N m / N n = e ω mn / kt For the system to be at equlbrum, the tme-averaged transtons up W must equal those down mn W. In the presence of a feld, we would want to wrte for an ensemble nm? N B U (ω ) = N B U (ω mn ) m nm mn n mn but clearly ths can t hold for fnte temperature, where N m < N n, so there must be another type of emsson ndependent of the feld. So we wrte W nm = W mn N A + B U (ω mn )) = N m ( nm nm n B mn U (ω mn )
p. 85 If we substtute the Boltzmann equaton nto ths and use B mn = B nm, we can solve for A nm : A nm = B nm U (ω mn )(e ω mn /kt 1 ) For the energy densty we wll use Planck s blackbody radaton dstrbuton: U ( ) ω 1 ω = e ω mn π c /kt 1 U ω N ω U ω s the energy densty per photon of frequency ω. Nω s the mean number of photons at a frequency ω. The total rate of emsson from the excted state s ω A nm = B Ensten A coeffcent π nm c w = B U (ω nm ) + A usng U (ω ) = N π nm nm nm nm ω = B π nm (N +1) c ω C Notce, even when the feld vanshes (N 0), we stll have emsson. Remember, for the semclasscal treatment, the total rate of stmulated emsson was ω w nm = B N π nm ( ) c If we use the statstcal analyss to calculate rates of absorpton we have ω w mn = B N π mn c The A coeffcent gves the rate of emsson n the absence of a feld, and thus s the nverse of the radatve lfetme: 1 τ rad = A
p. 86 Quantzed Radaton Feld Background Our treatment of the vector potental has drawn on the monochromatc plane-wave soluton to the wave-equaton for A. The quantum treatment of lght as a partcle descrbes the energy of the lght source as proportonal to the frequency ω, and the photon of ths frequency s assocated wth a cavty mode wth wavevector k =ω/ c that descrbes the number of oscllatons that the wave can make n a cube wth length L. For a very large cavty you have a contnuous range of allowed k. The cavty s mportant for consderng the energy densty of a lght feld, snce the electromagnetc feld energy per unt volume wll clearly depend on the wavelength λ = π/ k of the lght. Boltzmann used a descrpton of the lght radated from a blackbody source of fnte volume at constant temperature n terms of a superposton of cavty modes to come up wth the statstcs for photons. The classcal treatment of ths problem says that the energy densty (modes per unt volume) ncreases rapdly wth ncreasng wavelength. For an equlbrum body, the energy absorbed has to equal the energy radated, but clearly as frequency ncreases, the energy of the radated lght should dverge. Boltzmann used the detaled balance condton to show that the partcles that made up lght must obey Bose-Ensten statstcs. That s the equlbrum probablty of fndng a photon n a partcular cavty mode s gven by 1 f ( ω ) = e ω/kt 1 From our perspectve (n retrospect), ths should be expected, because the quantum treatment of any partcle has to follow ether Bose-Ensten statstcs or Ferm-Drac statstcs, and clearly lght energy s somethng that we want to be able to ncrease arbtrarly. That s, we want to be able to add mode and more photons nto a gven cavty mode. By summng over the number of cavty modes n a cubcal box (usng perodc boundary condtons) we can determne that the densty of cavty modes (a photon densty of states), g ( ω ) = ω π c Usng the energy of a photon, the energy densty per mode s ω ω = π c g ( ) ω and so the probablty dstrbuton that descrbes the quantum frequency dependent energy densty s ) g ( ) ( ) ω/kt u (ω = ω ω f ω = ω π c e 1 1
p. 87 The Quantum Vector Potental So, for a quantzed feld, the feld wll be descrbed by a photon number N k j, whch represents the number of photons n a partcular mode (k, j ) wth frequency ω = ck n a cavty of volume v. For lght of a partcular frequency, the energy of the lght wll be N kj ω. So, the state of the electromagnetc feld can be wrtten: ϕ EM = N k,j, N k,,,j N 1 1 k,j If my matter absorbs a photon from mode k, then the state of my system would be ϕ EM = N, 1, N k,j k k,j, 1,j What I want to do s to wrte a quantum mechancal Hamltonan that ncludes both the matter and the feld, and then use frst order perturbaton theory to tell me about the rates of absorpton and stmulated emsson. So, I am gong to partton my Hamltonan as a sum of a contrbuton from the matter and the feld: H 0 = H EM + H M If the matter s descrbed by expressed as product states: ϕ M, then the total state of the E.M. feld and matter can be ϕ =ϕ EM ϕ M And we have egenenerges E= E EM + E M
p. 88 Now, f I am watchng transtons from an ntal state to a fnal state k, then I can express the ntal and fnal states as: ϕ = ;N,N,N,N, I 1 ϕ = k; N, N, N,, N ±1, F 1 + : emsson : absorpton matter k feld N N 1 Where I have abbrevated N N k, j, the energes of these two states are: E I = E + N j ( ω j ) j E F = E k + N ( ω j )± ω j j ω j = ck j k So lookng at absorpton, we can wrte the Golden Rule Rate for transtons between states as: π = δ(e k E ω) ϕ F V ( t ) ϕ I Now, let s compare ths to the absorpton rate n terms of the classcal vector potental: π δ ω ω) q m A k ˆ p = ( k j v kj k, j If these are to be the same, then clearly V( t)must have part that looks lke ( p ˆ )that acts on the matter, but t wll also need another part that acts to lower and rase the photons n the feld. Based on analogy wth our electrc dpole Hamltonan, we wrte: V()= t q 1 ˆ ˆ ˆ * ˆ (p j A + p k j A k, j k k, j ) m v k, j
p. 89 ˆ ˆ where A k,j and A k,j are lowerng/rasng operators for photons n mode k. These are operators n the feld states, whereas p k remans only an operator n the matter states. So, we can wrte out the matrx elements of V as q 1 ϕ V t = k p ˆ, N 1, ˆ F () ϕ I k A, N, m v 1 ( ) = ω k v k ˆ µ A Comparng wth our Golden Rule expresson for absorpton, We see that the matrx element = π δω ( k ω) ω k E 0 µ k ω A ( ) = E 0 E but 0 4vω = N ω 8π So we can wrte π = N vω ˆ A k, j = π a v ω k, j ˆ π = a v ω A k,j k j, where a, a are lowerng, rasng operators. So A = π k ( r ωt ) ( ωt ˆ ) + a k r ˆ j ( a e e kj kj ) v ω k,j
p. 90 So what we have here s a system where the lght feld looks lke an nfnte number of harmonc oscllators, one per mode, and the feld rases and lowers the number of quanta n the feld whle the momentum operator lowers and rases the matter: H= H EM + H M + V ()= t H 0 + V ( t ) 1 H EM = ω k (a kj a kj + ) k, j H M = p m + V ( r, t ) V t ()= q m A p π ( p a ( ωt ) k ( r ωt ) = q ˆ j ) k r + a e k,je k,j k,j m v ω k = V ( ) + V (+) Let s look at the matrx elements for absorpton (ω k > 0) V ( ) ˆ k, N 1, N = q π k, N 1 ( p)a, N m v ω q π = N k p ˆ m v ω π ω = N µ ˆ k v and for stmulated emsson (ω k < 0)
p. 91 + k, N +1 V ( ), N = q π k, N +1 ( p)a, N m v ω ˆ q π = N 1 k ˆ + ˆ p m v ω = π ω v N + 1 µ ˆ k We have spontaneous emsson! Even f there are no photons n the mode (N k = 0), you can stll have transtons downward n the matter whch creates a photon. Let s play ths back nto the summaton-over-modes expresson for the rates of absorpton/emsson by sotropc feld. π ω + w dω V ( ) k = δ ω ω) dω k, N +1, N (π ) ( k j = π ω 8π (π ω)(n +1) µ k (π c) number densty per mode average over polarzaton ( 1) 4 N + ω = µ k c ω = B k (N +1) π c energy densty per mode So we have the result we deduced before.
p. 9 Appendx: Rates of Absorpton and Stmulated Emsson Here are a couple of more detaled dervatons: Verson 1: Let s look a lttle more carefully at the rate of absorpton nduced by an sotropc, broadband lght source = ( ω)ρ E ( ω) d ω where, for a monochromatc lght source ( π )= ω ω k ˆ µ δ (ω ) E 0 ( ) k For a broadband sotropc lght source ρ(ω )dω represents a number densty of electromagnetc modes n a frequency range dω ths s the number of standng electromagnetc waves n a unt volume. For one frequency we wrote: but more generally: A = A 0 e (k r ˆ ωt) + c.c. ˆ k r A = A 0k j e ( ωt) + c.c. k, j where the sum s over the k modes and j s the polarzaton component. By summng over wave vectors for a box of fxed volume, the number densty of modes n a frequency range dω radated nto a sold angle dω s 1 ω dn = dω d Ω (π) c and we get ρ E by ntegratng over all Ω 1 ω ω E ( )d d ρ ω ω= (π) c π c 4π dω Ω= dω number densty at ω
p. 9 We can now wrte the total transton rate between two dscrete levels summed over all frequences, drecton, polarzatons π = dω ω δ ω ω) 1 ω E 0 ( ) ( k (π) ˆ dω k j µ c j 8π µ k ω E 0 (ω k ) = µ k 6π c We can wrte an energy densty whch s the number densty n a range dω # of polarzaton components energy densty per mode. ω U (ω k ) = π c 8π = B k U (ω k ) E 0 rate of energy flow/c 4π k B k = µ s the Ensten B coeffcent for the rate of absorpton U s the energy densty and can also be wrtten n a quantum form, by wrtng t n terms of the number of photons N N ω= E 0 8π ω U (ω k ) = N π c The golden rule rate for absorpton also gves the same rate for stmulated emsson. We fnd for two levels : m and n w nm = w mn B nm U ( ω nm)= B nm U ( ω nm ) snce U ( ω nm )= U ( ωmn ) B nm = B mn The absorpton probablty per unt tme equals the stmulated emsson probablty per unt tme.
p. 94 Verson : Let s calculate the rate of transtons nduced by an sotropc broadband source we ll do t a bt dfferently ths tme. The unts are cgs. The power transported through a surface s gven by the Poyntng vector and depends on k. S = c E B = c ω A 0 kˆ = ω E 0 4π 8π π and the energy densty for ths sngle mode wave s the tme average of S/ c. The vector potental for a sngle mode s ˆ ( ωt) A = A 0 e kr + c.c. wth ω = ck. More generally any wave can be expressed as a sum over Fourer components of the wave vector: ˆ j k ( r ω t ) A = A k j e + c.c. V k,j The factor of V normalzes for the energy densty of the wave whch depends on k. The nteracton Hamltonan for a sngle partcle s: Vt ()= q A ρ m or for a collecton of partcles Vt ()= q A p m Now, the momentum depends on the poston of partcles, and we can express p n terms of an ntegral over the dstrbuton of partcles: p = d ( ) p ) = p δ(r r ) r p r ( r So f we assume that all partcles have the same mass and charge say electrons: ()= q d r A r,t () Vt ( ) p r m
p. 95 The rate of transtons nduced by a sngle mode s: = π δ ω ω) q A k, j k r ( ) k,j V ( k ˆ j m p ( ) And the total transton rate for an sotropc broadband source s: = ( ) k,j k,j We can replace the sum over modes for a fxed volume wth an ntegral over k : 1 d k dk k dω dω ω dω ( ) π ( π C) V k π ( ) So for the rate we have: dω =snθ dθ dø = dω π ω δ ω π ω) q ˆ r dω k p ( ) ( c) m j ( k k, j can be wrtten as k The matrx element can be evaluated n a manner smlar to before: q q k p ( r) = ˆ p m m ˆ j k δ( r r ) A = q q ˆ k [ r, H ]δ( r r ) 1 0 = ω k q ˆ k r 1 For the feld E kj = ω k µ ˆ where µ= q r k E 0 A k = = 4ω j ω k j k j
p. 96 π ω W k = dω 4 (πc) δ (ω ω) ω k E ˆ 0 dω k j µ k ω j / k for sotropc 8π µ ω E 0 = µ k 6π c For a broadband source, the energy densty of the lght I ω E U = = 0 c 8π c 4π W k = B k U (ω k ) B k = µ k We can also wrte the ncdent energy densty n terms of the quantum energy per photon. For N photons n a sngle mode: ω N ω = B k N π c where B k has molecular quanttes and no dependence or feld. Note B k = B k rato of S.E. = absorpton. The rato of absorpton can be related to the absorpton cross-secton, δ A P total energy absorbed/unt tme σ A = = I total ntensty (energy/unt tme/area) P = ω W k = ω B k U (ω k ) I= cu (ω k ) ω σ a = B k c or more generally, when you have a frequency-dependent absorpton coeffcent descrbed by a lneshape functon g ω ( ) σ ω B k g ( ω ) unts of cm a ( )= ω c