Question

a) Using the quantum theory of radiation we derived in class in which we gave that the vector potential could be written as an operator of the form ĀCE, 1) = V Patie [-istan + Ge-iz+iutat] where a represents an annhilation of a photon in a mode (k. A) and a represents a creation of a photon in a mode (k.). Present an argument that the vector potential operator has to appear at least twice in a matrix element calculation of elastic scattering of a photo off of electrons in an atom in order for there to be a non-zero element in perturbation theory. You can assume that the incoming photon energies are large compared to the atomic binding energy but that the wavelength of the photon remains larger than the atomic radius. b) Using the A A term in the interaction Hamiltonian for the elastic scattering of a photo off of electrons in an atom, calculate the differential cross-section in the dipole approximation (you can use first-order perturbation theory here) and show that (10)

Answer 1

When a photon hits a free electron, it cannot be absorbed and continue as a free electron. This follows from energy and momentum conservation. On the other hand, if the absorption is followed by re-emission of a new photon, we have an allowed process. Then we say that the photon has been scattered by the electron.This low-energy, elastic scattering of photons is called Thomson scattering.The field operator of photon and its relation is mentioned below

() The quantized electromagnete vector potential is . given by . à (2,4)= / i just téte. ..and c-kc. Eht LOE i-jut t atil Reluk, (v. É.nl According to elastic scattering of photon i r Era) te → [E',x)tē, .Where El=Rilek. The lattor implies that Since above equation implies that one. photon in the mode CF, is annihilated and one phaton in the mode CE, A) is created; the à field operator must appear atleast twice in the matric element which can be ... evaluated using fermi Golden rule. To see why. this is happening, consider the relevant matrin element, is of the form e <filex, Oral Histli , OR", A, 1 R,x), Eqh (1) .. where lis is the initial state of the electron and It is the final state of election. Here we have specified explicitly the pholon occupation numbers for the modes (R, A) and Rd1). All other photon lobich are suppressed in Egin U are absent, and Thus occupation numbers are Zero. their photon the only way for the matria. element exhibited. Hint contains the in Eq.n. 1 to be non- zero is if CS @eamed on this area, That is CamScanner'

< 15",2,07 x | ateix, àrial021,41,12,)=4 → Eqnlay (184, 's Ofis lag, ate y, OP 'x', 18.%78q0C6) This is according to the commutation relations for the creation and annihilation operato.as Tarja s atel, N = Spe; fra e Thus, for Ę + E!, it follows that L a Eid atalar = a Tarot, ai While all other matrix elements vanish it operators that appear consist of other combinat of a and at taken erthe singly or in pairs. we can conelude 'that Hist must contain A exactly twoice. (6) The interaction Hamiltonian is given by Hinta forj (7). R(2,1)+2 (exp(3)Ã22,), cobere p17) = 83(3-52), 347) = 2m (P8 ()+ 8367-)*7 ** To finst code ini perturbation theory, only ÅZ team of interaction Hamiltonian contributes a non zero result to the matrini element given by Eqr (1). Using Fermi's Golden rule, CS Sštinine follows that . CamScanner

Tipo 2kf this li* S(Ef - Ei),... where the superscript (1) inclicates that this is a first-order perturbation theory result. The initial and final energies are given by Ei= Eetto , Ef= Fethio Wher Ee is the election energy and to and to are the initial and final state photon energies. Thus, .. Tif" - 21622) Kép 12 82',02,0) 777,40 , Vix] . . x 8(50_50) [qul2 In part (a) only the ato , arix and arrat Cross terms arising from Ã. Â contribute to the matrix element above. By inserting the eppansion of ☆ (72,) yields only one non-zero leom <10'Oriola 2.0)|08|'>12.a) = 2.2 TtCZ ET £2, 51714Deial ate, maria loend) 18x) Wow & Ceil elle-E)?!(co-ost/ei). = 4 Tihc? Ene < effel(R-RD./e;>, Eqn (3) where we have used the fact that the energy consering delta function in Ean (2) sets w'aw. The overall factor So fcamnein werin (s) is included since the matrimo elements EX wou CamScanner

4 ic t cz Y Ja 11 52 of atap x and arxatir, yield the same result is R & R [as shown in Eqnia) and Eqr() 7. Therefore Tipo ) (1642) 11.2/kezdir-pboy ay x Siw-co), after noting that Stow* -tx) = +15(włos), Here we assume the photon wavelength is bath is substantial larger than a typical atomic radius. This means it can coork in the dipole approximation in which case · Ile-R').72 wiki Since liz and H) represent the same normalized atomic States Csince electron energy is unchanged in eleastre scattering), i't follows that atli)=1. Hence. TORT (092) * ( 4 KC3) 13 - E/? $(W-c) : ) Eqn (4) To compute the cross section, dif = { Tf - Eqr(s) . . tincidint flex Incident flux of photos having velocity c, then flex is given by clv. The sum over final states f is a sum over the outgoing photon momento. Tel. Taleing infinite volume limit, we take I kedide=v. I eo adole (Qc) 3 ). Scanned with CamScanner

2 > 2 - 2 where we have used w's k'c.. Ince in Equis) and taking infinite volume limon tanly) ongel ) ) | E-Eller fordelos de S(coles) The volume factors cancel out. The integral over Wo is trivial. fit enforces energy conservation), finally 067a! =(92) | Ex-E x1/2 The classical radius of the election is % = e/m (2), the differential cross section can be written as oor = 82 107.571/2 - - da Scanned with CamScanner