4. INTERACTION OF LIGHT WITH MATTER

Transcription

1 Andre Tokmakoff, MIT Department of Chemstry, /8/ INTERACTION OF LIGHT WITH MATTER One of the most mportant topcs n tme-dependent quantum mechancs for chemsts s the descrpton of spectroscopy, whch refers to the study of matter through ts nteracton wth lght felds (electromagnetc radaton). Classcally, lght-matter nteractons are a result of an oscllatng electromagnetc feld resonantly nteractng wth charged partcles. Quantum mechancally, lght felds wll act to couple quantum states of the matter, as we have dscussed earler. Lke every other problem, our startng pont s to derve a Hamltonan for the lght-matter nteracton, whch n the most general sense would be of the form H = H M + H L + H LM. (4.1) The Hamltonan for the matter H M s generally (although not necessarly) tme ndependent, whereas the electromagnetc feld H L and ts nteracton wth the matter H LM are tme-dependent. A quantum mechancal treatment of the lght would descrbe the lght n terms of photons for dfferent modes of electromagnetc radaton, whch we wll descrbe later. However, we wll start wth a common semclasscal treatment of the problem. For ths approach we treat the matter quantum mechancally, and treat the feld classcally. For the feld we assume that he lght only presents a tme-dependent nteracton potental that acts on the matter, but the matter doesn t nfluence the lght. (Quantum mechancal energy conservaton says that we expect that the change n the matter to rase the quantum state of the system and annhlate a photon from the feld. We won t deal wth ths rght now). In that case, we can really largely gnore H L, and we have a Hamltonan that can be solved n the nteracton pcture representaton: H H M + H LM = H + V t Here, we ll derve the Hamltonan for the lght-matter nteracton, the Electrc Dpole Hamltonan. It s obtaned by developng a classcal Hamltonan for the nteracton of charged partcles wth an electromagnetc feld, and then substtutng quantum operators for the matter. ( t) (4.)

2 p ˆ# x ˆx 4- (4.) In order to get the classcal Hamltonan, we need to work through two steps: (1) We need to descrbe electromagnetc felds, specfcally n terms of a vector potental, and () we need to descrbe how the electromagnetc feld nteracts wth charged partcles. Bref summary of electrodynamcs Let s summarze the descrpton of electromagnetc felds that we wll use. A dervaton of the plane wave solutons to the electrc and magnetc felds and vector potental s descrbed n the appendx. Also, t s helpful to revew ths materal n Jackson 1 or Cohen-Tannoud, et al.. > Maxwell s Equatons descrbe electrc and magnetc felds E, B > To construct a Hamltonan, we must descrbe the tme-dependent nteracton potental (rather than a feld). > To construct a potental representaton of E and B, you need a vector potental A r,t and a scalar potental ( r,t). For electrostatcs we normally thnk of the feld beng related to the electrostatc potental through E = #, but for a feld that vared n tme and n space, the electrodynamc potental must be expressed n terms of both A and. > In general an electromagnetc wave wrtten n terms of the electrc and magnetc felds requres 6 varables (the x,y, and z components of E and B). Ths s an overdetermned problem; Maxwell s equatons constran these. The potental representaton has four varables ( A x, A y, A z and ), but these are stll not unquely determned. We choose a constrant a representaton or guage that allows us to unquely descrbe the wave. Choosng a gauge such that = (Coulomb gauge) leads to a plane-wave descrpton of E and B : A r,t + 1 # A( r,t) c #t = (4.4) A = (4.5)

3 4- Ths wave equaton allows the vector potental to be wrtten as a set of plane waves: k r #$t A( r,t) = A ˆ e # k r #$t + A* ˆ e. (4.6) Ths descrbes the wave oscllatng n tme at an angular frequency ω and propagatng n space n the drecton along the wavevector k, wth a spatal perod = k. The wave has an ampltude A whch s drected along the polarzaton unt vector ˆ. Snce A =, we see that k ˆ = or k ˆ. From the vector potental we can obtan E and B E = A t = # A ˆ$ & e ' B = A A ' e = k ˆ# ( k %r #t e k %r #t k $r %&t ( ) % k $r %&t % e ) * (4.7) (4.8) If we defne a unt vector along the magnetc feld polarzaton as ˆb = ( k ˆ ) k = ˆk ˆ, we see that the wavevector, the electrc feld polarzaton and magnetc feld polarzaton are mutually orthogonal ˆk ˆ ˆb. Also, by comparng eq. (4.6) and (4.7) we see that the vector potental oscllates as cos ωt, whereas the feld oscllates as sn ωt. If we defne 1 E = A (4.9) then, 1 B = k A (4.1) E( r,t) = E ˆ sn( k r # $t) (4.11)

4 4-4 B( r,t) = B ˆbsn ( k r #t ). (4.1) Note, E B = k = c.

5 4-5 Hamltonan for radaton feld nteractng wth charged partcle Now, let s fnd a classcal Hamltonan that descrbes charged partcles n feld n terms of the vector potental. Start wth Lorentz force on a partcle wth charge q (1) F = q( E + v B). (4.1) Here v s the velocty of the partcle. Wrtng ths for one drecton (x) n terms of the Cartesan components of E, v and B, we have: () F x = q( E x + v y B z v z B y ). (4.14) In Lagrangan mechancs, ths force can be expressed n terms of the total potental energy U as () F x = U x + d dt # U & % $ v ( x ' (4.15) Usng the relatonshps that descrbe E and B n terms of A and, nsertng nto eq. (4.14) and workng t nto the form of eq. (4.15), we can show that: (4) U = q qv # A (4.16) Ths s derved n CTDL, and you can confrm by pluggng t nto eq. (4.15). Now we can wrte a Lagrangan n terms of the knetc and potental energy of the partcle L = T U (4.17) (5) L = 1 mv + qv A q# (4.18) The classcal Hamltonan s related to the Lagrangan as H = pv L (6) = pv 1 mv qv A q# (4.19) Recognzng p = L v = mv + qa (4.) we wrte v = m 1 ( p qa ). (4.1) Now substtutng (4.1) nto (4.19), we have:

6 4-6 H = 1 m p ( p qa) 1 m ( p qa ) q m ( p qa ) A + q# (4.) H = 1 p qa( r,t) $ m # % + q& r,t (4.) Ths s the classcal Hamltonan for a partcle n an electromagnetc feld. In the Coulomb, the last term s dropped. gauge = We can wrte a Hamltonan for a collecton of partcles n the absence of a external feld and n the presence of the EM feld: p H = $ '# + V m ( r )&. (4.4) % + V r 1 % H = ( $ p m q A( r ) '. (4.5) # & q + Expandng: H = H # p m A+ A p # A (4.6) m Generally the last term s consdered small. That s the energy of partcles hgh relatve to ampltude of potental. (Ths term should be consdered for extremely hgh feld strength, whch sgnfcantly dstorts the potental bndng molecules together). For weak felds we have: V ( t) = q H = H + V ( t) (4.7) q p m A + A p (4.8) Now we are n a poston to substtute the quantum mechancal momentum for the classcal. Here the vector potental remans classcal, and only modulates the nteracton strength. V ( t) = p = (4.9) # q m A+ A (4.)

7 4-7 We can show that A = A. Notce A = A operatng on a wavefuncton A # Coulomb gauge A =. Now we have: + A (chan rule). For nstance f we are # + A # = A, but we are workng n the V ( t) = # = $ q # m q A m A p (4.1) For a sngle charge partcle our nteracton Hamltonan s V ( t) = q m A p + c.c. = q % m A k r $t ˆ# p e &' ( )* (4.) Electrc Dpole Approxmaton Under most crcumstances, we can neglect the wavevector dependence of the nteracton potental. If the wavelength of the feld s much larger than the molecular dmenson #, k then e k r 1. We do retan the spatal dependence for certan types of lght matter nteractons. In that case we defne r as the center of mass of a molecule and expand e k r = e k r ek ( r r ) + = e k r # $ 1 + k r r % & (4.) For nteractons, wth UV, vsble, and nfrared (but not X-ray) radaton, k r r <<1, and settng r = means that e k r 1. We retan the second term for quadrupole transtons: charge dstrbuton nteractng wth gradent of electrc feld and magnetc dpole.

8 Electrc Dpole Hamltonan Now, usng A = E, V ( t) = qe % m & ˆ# $ p e t ˆ# $ p e +t ' ( (4.4) V ( t) = qe ( m ˆ# $ p )snt = q m E(t)$ p or for a collecton of charged partcles (molecules): V t = % ' & $ q m ( ( ˆ # p )* ) Ths s known as the electrc dpole Hamltonan (EDH). (4.5) E sn+t (4.6) + Harmonc Perturbaton: Matrx Elements To better nterpret ths, let s evaluate the transton rates nduced by V(t). For a perturbaton V t = V sn t the rate of transtons nduced by feld s w k = V k + ( E k # E + $ ) % & E k # E # $ Let s look at the matrx elements for the EDH. ' ( (4.7) We can evaluate the matrx element k p usng V k = k V = qe m k ˆ# $ p (4.8) whch gves r, H # $ = p m (4.9)

9 4-9 k p = m k r H H r = m ( k r E E k r k ) = m k k r (4.4) So we have or for a collecton of partcles V k = qe k k ˆ# $r (4.41) V k = E k k & ) ˆ# $ % q r ' ( * + = E k k ˆ# $µ (4.4) = E k µ kl µ s the dpole operator. The dpole moment can be expressed more generally as the dstrbuton of charge n the molecule: So µ = dr r ( r ) (4.4) V ( t) = µ E(t) (4.44) Then the rate of transtons between quantum states nduced by the electrc feld s w k = E k µ kl = E µkl % & # ( E k $ E $ )+( E k $ E + )' ( % & # ( k $ )+# ( k + )' ( (4.45) Ths s an expresson for the absorpton spectrum snce the rate of transtons can be related to the power absorbed from the feld. More generally we would express the absorpton spectrum n terms of a sum over all ntal and fnal states, the egenstates of H : w f = E µ & f # ( $ f % $ )+# ( $ f + $ ' )( ), f (4.46)

10 4-1 The strength of nteracton between lght and matter s gven by the matrx element µ f f µ ˆ#. The scalar part f µ says that you need a change of charge dstrbuton between f and to get effectve absorpton. Ths matrx element s the bass of selecton rules based on the symmetry of the states. The vector part says that the lght feld must proect onto the dpole moment. Ths allows nformaton to be obtaned on the orentaton of molecules, and forms the bass of rotatonal transtons.

11 4-11 Relaxaton Leads to Lne-broadenng Let s combne the results from the last two lectures, and descrbe absorpton to a state that s coupled to a contnuum. What happens to the probablty of absorpton f the excted state decays exponentally? k relaxes exponentally... for nstance by couplng to contnuum P k exp #$ w nk t% & We can start wth the frst-order expresson: b k = t # d k V t (4.47) t or equvalently t b = kt k e# V k ( t) (4.48) We can add rreversble relaxaton to the descrpton of b k, followng our early approach: Or usng V ( t) = E µ k snt t b = kt k e# V k ( t) w nk b (4.49) k t b = kt k e# sn#t V k w nk b t k = E # k $ ( # e # k +# ) # e ( k # )t ' %& () µ w nk k b t k (4.5) The soluton to the dfferental equaton y + ay = be t (4.51) s y( t) = Ae at + bet a +. (4.5)

12 4-1 b k ( t) = Ae w nk t / + E µ k # e ( k + )t % $ % w nk / + k + Let s look at absorpton only, n the long tme lmt: For whch the probablty of transton to k s b k ( t) = E µ # k e ( k )t & % ( $ % k w nk / '( E P k = b k = µ k 1 4 k t e k w nk / + ( k ) + w nk / 4 The frequency dependence of the transton probablty has a Lorentzan form: & ( '( (4.5) (4.54) (4.55) The lnewdth s related to the relaxaton rate from k nto the contnuum n. Also the lnewdth s related to the system rather than the manner n whch we ntroduced the perturbaton. Andre Tokmakoff, MIT Department of Chemstry, Appendx: Revew of Free Electromagnetc Feld Maxwell s Equatons (SI): (1) B = () # E = / $

13 4-1 () B # E = $ t µ µ E t (4) # B = J +$ E : electrc feld; B : magnetc feld; J : current densty; : charge densty; : electrcal permttvty; µ : magnetc permttvty We are nterested n descrbng E and B n terms of a scalar and vector potental. Ths s requred for our nteracton Hamltonan. Generally: A vector feld F assgns a vector to each pont n space. The dvergence of the feld (5) F F Fz x y z x y # F = + + s a scalar. For a scalar feld, the gradent (6) # = xˆ + yˆ + zˆ x y z s a vector for the rate of change at on pont n space. Here vectors. xˆ + yˆ + zˆ = rˆ are unt Also, the curl (7) # F = xˆ yˆ zˆ x y z F F F x y z s a vector whose x, y, and z components are the crculaton of the feld about that component. Some useful denttes from vector calculus are:

14 4-14 (8) (# F) = (9) #( ) = (1) ( F) = (# F) $ F We now ntroduce a vector potental A( r, t ) and a scalar potental ( r, t) relate to E and B, whch we wll = and ( A) Snce B (11) B = A Usng (), we have: A # E = $ # t or # A (1) $% & E + = # t ' = : From (9), we see that a scalar product exsts wth: A t (1) E + = # $ ( r, t) or conventon (14) A E = # $ t So we see that the potentals A and determne the felds B and E : (15) B( r, t) = A( r, t) = # $ # t (16) E ( r, t) ( r, t) A( r, t)

15 4-15 We are nterested n determnng the wave equaton for A and. Usng (15) and dfferentatng (16) and substtutng nto (4): (17) ( A) $ A $ # µ ( µ ) %& %& + ' +% = * $ t $ t + Usng (1): J (18) $ A # % $ & '( + ) µ + ( (* + ) µ = µ - A + A, J t. 1 $ / $ t From (14), we have: # $ A #$ E = % % # t and usng (): (19) #$ V % A $ t # & = / ' Notce from (15) and (16) that we only need to specfy four feld components ( Ax, Ay, A z, ) to determne all sx E and B components. But E and B do not unquely determne A and. So, we can construct A and n any number of ways wthout changng E and B. Notce that f we change A by addng where s any functon of r and t, ths won t change B ( ( B) ) can change to # $, but we # =. It wll change E by ( t ) $ = % #. Then E and B wll both be unchanged. Ths property of # t changng representaton (gauge) wthout changng E and B s gauge nvarance. We can transform between gauges wth: () A ( r, t) = A( r, t) +#$ ( r, t) # $ = % # t (1) ( r, t) ( r, t) ( r, t) gauge transformaton Up to ths pont, A and Q are undetermned. Let s choose a such that:

16 4-16 () #$ A + % µ = t Lorentz condton then from (17): () A t # A + $ µ = µ J The RHS can be set to zero for no currents. From (19), we have: (4) # $ µ % & = # t $ Eqns. () and (4) are wave equatons for A and. Wthn the Lorentz gauge, we can stll arbtrarly add another (t must only satsfy ). If we substtute () and (1) nto (4), we see: (5) µ t # $% = So we can make further choces/constrants on A and as long as t obeys (5). For a feld far from charges and currents, J = and =. (6) (7) µ A t # A +$ = µ t #$ +% = We now choose = (Coulomb gauge), and from () we see: (8) A = So, the wave equaton for our vector potental s: (9) µ A t # A +$ =

17 4-17 The solutons to ths equaton are plane waves. () A A sn ( t k r ) = # $ + : phase (1) = A cos ( t $ k % r + # ) k s the wave vector whch ponts along the drecton of propagaton and has a magntude: () k = µ = / c Snce (8) A = ( ) # k $ A cos t # k $ r + = () k A = k A A s the drecton of the potental polarzaton. From (15) and (16), we see that for = : # A E = $ = $ A cos t $ k % r + # t ( ) cos( ) B = &' A = $ k ' A t $ k % r + ( k ) E ) B

18 Rate of Absorpton and Stmulated Emsson The rate of absorpton nduced by the feld s w E k ( ) = ( ) $% ˆ µ # ( &) k k (4.1) The rate s clearly dependent on the strength of the feld. The varable that you can most easly measure s the ntensty I (energy flux through a unt area), whch s the tme-averaged value of the Poyntng vector, S c S = E B 4 c c I = S = E = E 4 8 (4.) (4.) Another representaton of the ampltude of the feld s the energy densty I 1 U = = E c 8 (for a monochromatc feld) (4.4) Usng ths we can wrte 4 w k = U k #$µ % k & ˆ (4.5) or for an sotropc feld where 1 E xˆ = E yˆ = E zˆ = E w 4 = U µ # $ k k k (4.6) or more commonly w k = Bk U k (4.7) B 4 = µ k k Ensten B coeffcent (4.8)

19 4-19 (ths s sometmes wrtten as Bk ( ) k = µ when the energy densty s n ν). U can also be wrtten n a quantum form, by wrtng t n terms of the number of photons N E N = U = N (4.9) 8 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 A = total ncdent ntensty energy / unt tme / area # w I k = = U( k ) ( ) # B k c U k (4.1) A = c B k More generally you may have a frequency dependent absorpton coeffcent A ()# Bk () = Bk g( ) where g(ω) s a lneshape functon. 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 ( nm ) = B nm U ( nm ) snce U ( nm ) = U ( mn ) (4.11) 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, A nm SE mn

20 4- 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 I = N dx + N dx (4.1) n A m SE di Nn Nm a dx I = (4.1) N n, N m These are populaton of the upper and lower states, but expressed as a populaton denstes. If N s the molecule densty, N n En e # $ = N% & ' Z ( (4.14) N=N N s the thermal populaton dfference between states. n m Integratng over a pathlength L: I I N al e # = for hgh freq. N N (4.15) # N al e N : cm : cm L :cm n or wrtten as Beer s Law: I A = log = C L C : mol / lter :lter / mol cm (4.16) I = N A

21 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 obects. 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 mn / kt (4.17) For the system to be at equlbrum, the tme-averaged transtons up must equal those down. In the presence of a feld, we would want to wrte for an ensemble W nm ( ) ( ) m nm mn n mn mn Wmn? N B U = N B U (4.18) 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 ( + ( )) = ( ) m nm nm mn n mn mn mn N A B U N B U (4.19)

22 4- 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 ( mn ) e /kt mn ( 1) (4.) For the energy densty we wll use Planck s blackbody radaton dstrbuton: 1 U( ) = mn / kt c e # 1 #$%$& U N (4.1) U s the energy densty per photon of frequency ω. N s the mean number of photons at a frequency ω. Anm = # c B nm Ensten A coeffcent (4.) The total rate of emsson from the excted state s = ( ) + w B U A nm nm nm nm = + c B N 1 nm usng U = N C nm (4.) (4.4) Notce, even when the feld vanshes ( N ), we stll have emsson. Remember, for the semclasscal treatment, the total rate of stmulated emsson was w nm = B nm N c (4.5) If we use the statstcal analyss to calculate rates of absorpton we have w mn = B mnn c (4.6) The A coeffcent gves the rate of emsson n the absence of a feld, and thus s the nverse of the radatve lfetme: 1 rad = (4.7) A

23 Andre Tokmakoff, MIT Department of Chemstry, 5/19/ 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, and the photon of ths frequency s assocated wth a cavty mode wth wavevector k = / 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 f () = / kt 1 (4.1) e 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( ) = (4.) c Usng the energy of a photon, the energy densty per mode s g( ) = (4.) c and so the probablty dstrbuton that descrbes the quantum frequency dependent energy densty s 1 u( ) = g( ) f () = (4.4) / kt c e # 1

24 The Quantum Vector Potental So, for a quantzed feld, the feld wll be descrbed by a photon number N k, whch represents the number of photons n a partcular mode (k, ) wth frequency = ck n a cavty of volume v. For lght of a partcular frequency, the energy of the lght wll be electromagnetc feld can be wrtten: N. So, the state of the k = N, N, N, (4.5) EM k 1,1 k, k, If my matter absorbs a photon from mode k, then the state of my system would be = N, N # 1, N, (4.6) EM k 1, k, k, 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 = HEM + HM (4.7) If the matter s descrbed by M, then the total state of the E.M. feld and matter can be expressed as product states: = (4.8) EM M And we have egenenerges E = E + E (4.9) EM M

25 4-5 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: I = ; N 1, N, N, N, matter feld = k; N, N, N,, N ± 1, F 1 + : emsson # $ & % : absorpton ' (4.1), (4.11) Where I have abbrevated N N k,, the energes of these two states are: E I = E + N = ck (4.1) E F = E k + ± N So lookng at absorpton # states as: k $ &, we can wrte the Golden Rule Rate for transtons between % w E E V t k = k # # $ % F % I (4.1) Now, let s compare ths to the absorpton rate n terms of the classcal vector potental: q w A k p ' ˆ k = # k $ # % k, & v k m (4.14) must have part that looks lke ( ˆ ) that acts on the If these are to be the same, then clearly V t p 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:

26 4-6 q 1 * = ( k# ˆ ˆ + ˆ ˆ k, k# k, ) V t $ p A p A (4.15) m v k, where Aˆ and A ˆ are lowerng/rasng operators for photons n mode k. These are operators n k, k, 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 F V( t) ˆ ˆ I = k p k#$, N 1, A, N, m v 1 = % ˆ k k $#µ v A () (4.16) Comparng wth our Golden Rule expresson for absorpton, # k w k = # k $ # # E µ k (4.17) We see that the matrx element () E E A = but = N 4v 8# # = N v (4.18) So we can wrte   = a v k, k, = a v k, k, (4.19) where a,a are lowerng, rasng operators. So

27 ( k r t) ( k r t ˆ $ # # ) A = & % ˆ ( a e + a e k k ) v # k, 4-7 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 EM M 1 ( a a ) EM k k k k, k, k H = H + H + V t = H + V t p HM =, + V r, t m V t, = # + q = A p m q $ ( k r t ) ( k r t) = ( ˆ p) % a e # a e # &, ' + k, k, m v # (* ) + () ( + ) = V + V Let s look at the matrx elements for absorpton ( k > ) () q k, N 1 V, N = k, N 1 (#$ ˆ p) a, N m v % q = N ˆ k #$ p m v % % v = N #$µ k ˆ

28 and for stmulated emsson ( k < ) 4-8 ( + ) q k, N + 1 V, N = k, N + 1 (#$ ˆ p) a, N m v % q = N ˆ ˆ + 1 k #$ p m v % % v = N + 1 #$µ k ˆ We have spontaneous emsson Even f there are no photons n the mode ( N k = ), 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. ( + ) w = d # $ d% k, N + 1 V, N ' ( ) k k # 8 = ( #)( N + 1) µ k # ( c) $%&%' $%&%' number denstyper mode 4 N + 1 = µ k # c = B N + 1 # c k energydensty per mode ' & averageover polarzaton So we have the result we deduced before.

29 4-9 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 w k nduced by an sotropc, broadband lght source wk = # wk E d where, for a monochromatc lght source w E k ( ) = ( ) $% ˆ µ # ( &) k k For a broadband sotropc lght source ( )d represents a number densty of electromagnetc modes n a frequency range d ths s the number of standng electromagnetc waves n a unt volume. For one frequency we wrote: A = A ˆ ( k r #$t) e + c.c. but more generally: ( k r t) A = A $ ˆ + c.c. % k, k e # where the sum s over the k modes and s the polarzaton component. By summng over wave vectors for a box of fxed volume, the number densty of modes n a frequency range d radated nto a sold angle d s and we get ρ E by ntegratng over all 1 dn = d d c ( # ) 1 # E () d = d d d c % $ = c ( ) 4 number densty at ω

30 4- We can now wrte the total transton rate between two dscrete levels summed over all frequences, drecton, polarzatons 1 w k ( k ) ˆ = ) d E # $ d k ( % & 'µ c ) #$$$%$$$& E k = µ 6 c k 8 µ k We can wrte an energy densty whch s the number densty n a range d # of polarzaton components energy densty per mode. E U( k ) = # c 8# w = B U k k k rate of energy flow/c B 4# = µ k k 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 E N = U = N 8 c ( k ) 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 ( nm ) = B nm U ( nm ) snce U ( nm ) = U ( mn ) B nm = B mn The absorpton probablty per unt tme equals the stmulated emsson probablty per unt tme.

31 4-1 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. c c A ˆ E S = E B = k = 4# 8# # and the energy densty for ths sngle mode wave s the tme average of S/ c. The vector potental for a sngle mode s ( k r # t ˆ ) A = A $ + c.c. e wth = ck. More generally any wave can be expressed as a sum over Fourer components of the wave vector: ( k r # t) e A = % A $ ˆ k + c.c. V k, The factor of V normalzes for the energy densty of the wave whch depends on k. The nteracton Hamltonan for a sngle partcle s: V( t) = q m A # or for a collecton of partcles V( t) = # q m A p 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 r p r p r = p r r So f we assume that all partcles have the same mass and charge say electrons: V( t) = q # m d r A r,t p ( r)

32 4- The rate of transtons nduced by a sngle mode s: q V m k k, k k, ( w ) = (# $ #) A k % ˆ & p( r) And the total transton rate for an sotropc broadband source s: w = ( w ) k k k, k, We can replace the sum over modes for a fxed volume wth an ntegral over k : 1 V k ' d k ( # ) $ ' dk k d% ( # ) $ ' d& & d% # C So for the rate we have: d =sn d dø q w d d k p r A ˆ k = ) # k $ % ( &' k, m ) ( c) can be wrtten as k The matrx element can be evaluated n a manner smlar to before: q m q k ˆ # p r = k # p $ r # r & ˆ m = q& q # ˆ k r 1, H $ r r [ ] & = % q # ˆ k r k 1 & = % k #µ ˆ where µ = q r k For the feld k A k = k E k = E 4

33 4- k W ˆ k = ) d # k $ E d% ( k & ' µ 4 ( c) ) #$$$%$$$& 8 / µ k for sotropc = 6 c E µ k For a broadband source, the energy densty of the lght I E U = = c 8 c 4 W = B U ( ) B = µ k k k k k We can also wrte the ncdent energy densty n terms of the quantum energy per photon. For N photons n a sngle mode: N = B k N 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, A P total energy absorbed/unt tme A = = I total ntensty (energy/unt tme/area) P = W = B U I = cu # a = B c k k k k k or more generally, when you have a frequency-dependent absorpton coeffcent descrbed by a lneshape functon g a = k c B g unts of cm

34 Andre Tokmakoff, MIT Department of Chemstry, /8/ Appendx: Revew of Free Electromagnetc Feld Maxwell s Equatons (SI): (1) B = () # E = / $ () B # E = $ t µ µ E t (4) # B = J +$ E : electrc feld; B : magnetc feld; J : current densty; : charge densty; : electrcal permttvty; µ : magnetc permttvty We are nterested n descrbng E and B n terms of a scalar and vector potental. Ths s requred for our nteracton Hamltonan. Generally: A vector feld F assgns a vector to each pont n space. The dvergence of the feld (5) F F Fz x y z x y # F = + + s a scalar. For a scalar feld, the gradent (6) # = xˆ + yˆ + zˆ x y z s a vector for the rate of change at on pont n space. Here xˆ + yˆ + zˆ = rˆ are unt vectors. Also, the curl

35 4-5 (7) # F = xˆ yˆ zˆ x y z F F F x y z s a vector whose x, y, and z components are the crculaton of the feld about that component. Some useful denttes from vector calculus are: (8) (# F) = (9) #( ) = (1) ( F) = (# F) $ F We now ntroduce a vector potental A( r, t ) and a scalar potental ( r, t) relate to E and B, whch we wll = and ( A) Snce B (11) B = A Usng (), we have: A # E = $ # t or # A (1) $% & E + = # t ' = : From (9), we see that a scalar product exsts wth: A t (1) E + = # $ ( r, t) or conventon

36 4-6 (14) A E = # $ t So we see that the potentals A and determne the felds B and E : (15) B( r, t) = A( r, t) = # $ # t (16) E ( r, t) ( r, t) A( r, t) We are nterested n determnng the wave equaton for A and. Usng (15) and dfferentatng (16) and substtutng nto (4): (17) ( A) $ A $ # µ ( µ ) %& %& + ' +% = * $ t $ t + Usng (1): J (18) $ A # % $ & '( + ) µ + ( (* + ) µ = µ - A + A, J t. 1 $ / $ t From (14), we have: # $ A #$ E = % % # t and usng (): (19) #$ V % A $ t # & = / ' Notce from (15) and (16) that we only need to specfy four feld components ( Ax, Ay, A z, ) to determne all sx E and B components. But E and B do not unquely determne A and. So, we can construct A and n any number of ways wthout changng E and B. Notce that f we change A by addng where s any functon of r and t, ths won t change B ( ( B) ) can change to # $, but we # =. It wll change E by ( t ) $ = % #. Then E and B wll both be unchanged. Ths property of # t

37 4-7 changng representaton (gauge) wthout changng E and B s gauge nvarance. We can transform between gauges wth: () A ( r, t) = A( r, t) +#$ ( r, t) # $ = % # t (1) ( r, t) ( r, t) ( r, t) gauge transformaton Up to ths pont, A and Q are undetermned. Let s choose a such that: () #$ A + % µ = t Lorentz condton then from (17): () A t # A + $ µ = µ J The RHS can be set to zero for no currents. From (19), we have: (4) # $ µ % & = # t $ Eqns. () and (4) are wave equatons for A and. Wthn the Lorentz gauge, we can stll arbtrarly add another (t must only satsfy ). If we substtute () and (1) nto (4), we see: (5) µ t # $% = So we can make further choces/constrants on A and as long as t obeys (5). For a feld far from charges and currents, J = and =. (6) µ A t # A +$ =

38 4-8 (7) µ t #$ +% = We now choose = (Coulomb gauge), and from () we see: (8) A = So, the wave equaton for our vector potental s: (9) µ A t # A +$ = The solutons to ths equaton are plane waves. () A A sn ( t k r ) = # $ + : phase (1) = A cos ( t $ k % r + # ) k s the wave vector whch ponts along the drecton of propagaton and has a magntude: () k = µ = / c Snce (8) A = ( ) # k $ A cos t # k $ r + = () k A = k A A s the drecton of the potental polarzaton. From (15) and (16), we see that for = : # A E = $ = $ A cos t $ k % r + # t ( ) cos( ) B = &' A = $ k ' A t $ k % r + ( k ) E ) B

39 4-9 Rate of Absorpton and Stmulated Emsson The rate of absorpton nduced by the feld s w E k ( ) = ( ) $% ˆ µ # ( &) k k (4.1) The rate s clearly dependent on the strength of the feld. The varable that you can most easly measure s the ntensty I (energy flux through a unt area), whch s the tme-averaged value of the Poyntng vector, S c S = E B 4 c c I = S = E = E 4 8 (4.) (4.) Another representaton of the ampltude of the feld s the energy densty I 1 U = = E c 8 (for a monochromatc feld) (4.4) Usng ths we can wrte 4 w k = U k #$µ % k & ˆ (4.5) or for an sotropc feld where 1 E xˆ = E yˆ = E zˆ = E w 4 = U µ # $ k k k (4.6) or more commonly w k = Bk U k (4.7) B 4 = µ k k Ensten B coeffcent (4.8) (ths s sometmes wrtten as Bk ( ) k = µ when the energy densty s n ν).

40 4-4 U can also be wrtten n a quantum form, by wrtng t n terms of the number of photons N E N = U = N (4.9) 8 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 A = total ncdent ntensty energy / unt tme / area # w I k = = U( k ) ( ) # B k c U k (4.1) A = c B k More generally you may have a frequency dependent absorpton coeffcent A ()# Bk () = Bk g( ) where g(ω) s a lneshape functon. 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 ( nm ) = B nm U ( nm ) snce U ( nm ) = U ( mn ) (4.11) 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, A nm SE mn

41 4-41 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 I = N dx + N dx (4.1) n A m SE di Nn Nm a dx I = (4.1) N n, N m These are populaton of the upper and lower states, but expressed as a populaton denstes. If N s the molecule densty, N n En e # $ = N% & ' Z ( (4.14) N=N N s the thermal populaton dfference between states. n m Integratng over a pathlength L: I I N al e # = for hgh freq. N N (4.15) # N al e N : cm : cm L :cm n or wrtten as Beer s Law: I A = log = C L C : mol / lter :lter / mol cm (4.16) I = N A

42 4-4 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 obects. 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 mn / kt (4.17) For the system to be at equlbrum, the tme-averaged transtons up must equal those down. In the presence of a feld, we would want to wrte for an ensemble W nm ( ) ( ) m nm mn n mn mn Wmn? N B U = N B U (4.18) 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 ( + ( )) = ( ) m nm nm mn n mn mn mn N A B U N B U (4.19)

43 4-4 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 ( mn ) e /kt mn ( 1) (4.) For the energy densty we wll use Planck s blackbody radaton dstrbuton: 1 U( ) = mn / kt c e # 1 #$%$& U N (4.1) U s the energy densty per photon of frequency ω. N s the mean number of photons at a frequency ω. Anm = # c B nm Ensten A coeffcent (4.) The total rate of emsson from the excted state s = ( ) + w B U A nm nm nm nm = + c B N 1 nm usng U = N C nm (4.) (4.4) Notce, even when the feld vanshes ( N ), we stll have emsson. Remember, for the semclasscal treatment, the total rate of stmulated emsson was w nm = B nm N c (4.5) If we use the statstcal analyss to calculate rates of absorpton we have w mn = B mnn c (4.6) The A coeffcent gves the rate of emsson n the absence of a feld, and thus s the nverse of the radatve lfetme: 1 rad = (4.7) A

44 4-44 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, and the photon of ths frequency s assocated wth a cavty mode wth wavevector k = / 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 f () = / kt 1 (4.1) e 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( ) = (4.) c Usng the energy of a photon, the energy densty per mode s g( ) = (4.) c and so the probablty dstrbuton that descrbes the quantum frequency dependent energy densty s 1 u( ) = g( ) f () = (4.4) / kt c e # 1

45 4-45 The Quantum Vector Potental So, for a quantzed feld, the feld wll be descrbed by a photon number N k, whch represents the number of photons n a partcular mode (k, ) wth frequency = ck n a cavty of volume v. For lght of a partcular frequency, the energy of the lght wll be electromagnetc feld can be wrtten: N. So, the state of the k = N, N, N, (4.5) EM k 1,1 k, k, If my matter absorbs a photon from mode k, then the state of my system would be = N, N # 1, N, (4.6) EM k 1, k, k, 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 = HEM + HM (4.7) If the matter s descrbed by M, then the total state of the E.M. feld and matter can be expressed as product states: = (4.8) EM M And we have egenenerges E = E + E (4.9) EM M

46 4-46 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: I = ; N 1, N, N, N, matter feld = k; N, N, N,, N ± 1, F 1 + : emsson # $ & % : absorpton ' (4.1), (4.11) Where I have abbrevated N N k,, the energes of these two states are: E I = E + N = ck (4.1) E F = E k + ± N So lookng at absorpton # states as: k $ &, we can wrte the Golden Rule Rate for transtons between % w E E V t k = k # # $ % F % I (4.1) Now, let s compare ths to the absorpton rate n terms of the classcal vector potental: q w A k p ' ˆ k = # k $ # % k, & v k m (4.14) must have part that looks lke ( ˆ ) that acts on the If these are to be the same, then clearly V t p 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:

47 4-47 q 1 * = ( k# ˆ ˆ + ˆ ˆ k, k# k, ) V t $ p A p A (4.15) m v k, where Aˆ and A ˆ are lowerng/rasng operators for photons n mode k. These are operators n k, k, 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 F V( t) ˆ ˆ I = k p k#$, N 1, A, N, m v 1 = % ˆ k k $#µ v A () (4.16) Comparng wth our Golden Rule expresson for absorpton, # k w k = # k $ # # E µ k (4.17) We see that the matrx element () E E A = but = N 4v 8# # = N v (4.18) So we can wrte   = a v k, k, = a v k, k, (4.19) where a,a are lowerng, rasng operators. So

48 ( k r t) ( k r t ˆ $ # # ) A = & % ˆ ( a e + a e k k ) v # k, 4-48 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 EM M 1 ( a a ) EM k k k k, k, k H = H + H + V t = H + V t p HM =, + V r, t m V t, = # + q = A p m q $ ( k r t ) ( k r t) = ( ˆ p) % a e # a e # &, ' + k, k, m v # (* ) + () ( + ) = V + V Let s look at the matrx elements for absorpton ( k > ) () q k, N 1 V, N = k, N 1 (#$ ˆ p) a, N m v % q = N ˆ k #$ p m v % % v = N #$µ k ˆ

49 and for stmulated emsson ( k < ) 4-49 ( + ) q k, N + 1 V, N = k, N + 1 (#$ ˆ p) a, N m v % q = N ˆ ˆ + 1 k #$ p m v % % v = N + 1 #$µ k ˆ We have spontaneous emsson Even f there are no photons n the mode ( N k = ), 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. ( + ) w = d # $ d% k, N + 1 V, N ' ( ) k k # 8 = ( #)( N + 1) µ k # ( c) $%&%' $%&%' number denstyper mode 4 N + 1 = µ k # c = B N + 1 # c k energydensty per mode ' & averageover polarzaton So we have the result we deduced before.

50 4-5 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 w k nduced by an sotropc, broadband lght source wk = # wk E d where, for a monochromatc lght source w E k ( ) = ( ) $% ˆ µ # ( &) k k For a broadband sotropc lght source ( )d represents a number densty of electromagnetc modes n a frequency range d ths s the number of standng electromagnetc waves n a unt volume. For one frequency we wrote: A = A ˆ ( k r #$t) e + c.c. but more generally: ( k r t) A = A $ ˆ + c.c. % k, k e # where the sum s over the k modes and s the polarzaton component. By summng over wave vectors for a box of fxed volume, the number densty of modes n a frequency range d radated nto a sold angle d s and we get ρ E by ntegratng over all 1 dn = d d c ( # ) 1 # E () d = d d d c % $ = c ( ) 4 number densty at ω

51 4-51 We can now wrte the total transton rate between two dscrete levels summed over all frequences, drecton, polarzatons 1 w k ( k ) ˆ = ) d E # $ d k ( % & 'µ c ) #$$$%$$$& E k = µ 6 c k 8 µ k We can wrte an energy densty whch s the number densty n a range d # of polarzaton components energy densty per mode. E U( k ) = # c 8# w = B U k k k rate of energy flow/c B 4# = µ k k 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 E N = U = N 8 c ( k ) 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 ( nm ) = B nm U ( nm ) snce U ( nm ) = U ( mn ) B nm = B mn The absorpton probablty per unt tme equals the stmulated emsson probablty per unt tme.

52 4-5 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. c c A ˆ E S = E B = k = 4# 8# # and the energy densty for ths sngle mode wave s the tme average of S/ c. The vector potental for a sngle mode s ( k r # t ˆ ) A = A $ + c.c. e wth = ck. More generally any wave can be expressed as a sum over Fourer components of the wave vector: ( k r # t) e A = % A $ ˆ k + c.c. V k, The factor of V normalzes for the energy densty of the wave whch depends on k. The nteracton Hamltonan for a sngle partcle s: V( t) = q m A # or for a collecton of partcles V( t) = # q m A p 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 r p r p r = p r r So f we assume that all partcles have the same mass and charge say electrons: V( t) = q # m d r A r,t p ( r)

53 4-5 The rate of transtons nduced by a sngle mode s: q V m k k, k k, ( w ) = (# $ #) A k % ˆ & p( r) And the total transton rate for an sotropc broadband source s: w = ( w ) k k k, k, We can replace the sum over modes for a fxed volume wth an ntegral over k : 1 V k ' d k ( # ) $ ' dk k d% ( # ) $ ' d& & d% # C So for the rate we have: d =sn d dø q w d d k p r A ˆ k = ) # k $ % ( &' k, m ) ( c) can be wrtten as k The matrx element can be evaluated n a manner smlar to before: q m q k ˆ # p r = k # p $ r # r & ˆ m = q& q # ˆ k r 1, H $ r r [ ] & = % q # ˆ k r k 1 & = % k #µ ˆ where µ = q r k For the feld k A k = k E k = E 4

54 4-54 k W ˆ k = ) d # k $ E d% ( k & ' µ 4 ( c) ) #$$$%$$$& 8 / µ k for sotropc = 6 c E µ k For a broadband source, the energy densty of the lght I E U = = c 8 c 4 W = B U ( ) B = µ k k k k k We can also wrte the ncdent energy densty n terms of the quantum energy per photon. For N photons n a sngle mode: N = B k N 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, A P total energy absorbed/unt tme A = = I total ntensty (energy/unt tme/area) P = W = B U I = cu # a = B c k k k k k or more generally, when you have a frequency-dependent absorpton coeffcent descrbed by a lneshape functon g a = k c B g unts of cm

55 4-55 Readngs 1 Jackson, J. D. Classcal Electrodynamcs (John Wley and Sons, New York, 1975). Cohen-Tannoud, C., Du, B. & Lalöe, F. Quantum Mechancs (Wley-Interscence, Pars, 1977), Appendx III. Cohen-Tannoud, et al. app. III, p. 149.