# Lecture 1: Microscopic Theory of Radiation

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2 2 Section 2 Now let s take the continuum limit, L. Then total energy up to frequency ω in the blackbody is 2π ω ω n e ωnβ = nk cosθ dφ d nk 2 ω n 0 e ωnβ (5) =4π L3 ω 8π 0 3 dω ω 3 e ω β (6) nk ω E(ω) = d 3, the intensity of light as a function of frequency is I(ω) = V de(ω) dω = π 2 ω 3 e ωβ Which is Plank s showed in 900 correctly matches experiment. (NB: I threw in a factor of 2 for the 2 polarizations of light, which you shouldn t worry about yet.). What does this have to do with Quantum Field Theory? Well, in order for this derivation, which used equilibrium statistical mechanics, to make sense, light has to be able to equilibrate. That also implies that if we shine monochromatic light into the box, eventually it must reach all frequencies. But if different frequencies are different particles, equilibration must involve one kind of particle turning into another kind of particle. So particles must get created and destroyed. Quantum field theory tells us how that happens. (7) 2 Einstein Coefficients The easiest handle on the creation of light comes from the coefficient of spontaneous emission. This is the rate for an excited atom to emit light. Even by 900, this phenomenon had been observed in chemical reactions, and as a form of radioactivity. But at that time, it was only understood statistically. In 906, Einstein developed a beautiful proof of the relation between emission and absorption based on the existence of thermal equilibrium. In addition to being relevant to chemical phenomenology, his relation made explicit why a first principles quantum theory of fields was needed. Einstein said: Suppose we have a cavity full of atoms with energy levels E and E 2. Say there are n of the E atoms and n 2 of the E 2 atoms. Let ω = E 2 E. The probability for an E 2 atom to emit a photon of frequency ω and transition to state E is called the coefficient for spontaneous emission. The probability for a photon of frequency ω to induce a transition from 2 to is proportional to the coefficient of stimulated emission B and to the number of photons of frequency ω in the cavity, that is, the intensity I(ω). Then dn 2 = [+BI(ω)] n 2 (8) The probability for a photon to induce a transition from to 2 is called the coefficient of stimulated absorption B, which in general may be different from B. So, dn = B I(ω)n (9) Note, we computed I(ω) above for the equilibrium blackbody situation, but these equations hold for any I(ω), for example, if we shine a laser beam at our atoms in the lab. t this point, Einstein assumes the gas is in equilibrium. Then the rate going up must be the same as the rate going down, so dn 2 =dn. So, [ +BI(ω)] n 2 =B I(ω)n (0) In equilibrium the number densities are determined by Boltzmann distributions. n = Ne βe n 2 = Ne βe2 () where N is some normalization factor. Then [ B e βe Be βe2 ] I(ω)=e βe 2 (2)

3 Quantum Field Theory 3 and so, I(ω)= B e βω B (3) But we already know that in equilibrium I(ω)= π 2 ω 3 e βω (4) from Equation (7). Since equilibrium must be satisfied at any temperature, i.e. for any β, we must have and B = B (5) B = π 2ω3 (6) These are beautiful results. The first, B = B, says that the coefficient of stimulated absorption must be the same as the coefficient for stimulated emission. ctually, it is a straightforward exercise to compute B in quantum mechanics (not QFT!) using time-dependent perturbation theory, and to show that B = B that way. Then the second equation determines the coefficient for spontaneous emission from the result of this calculation. You might have noticed something odd in this derivation. Einstein needed to use an equilibrium result about the black-body spectrum to derive the /B relation. Does spontaneous emission from an atom have anything to do with equilibrium of a gas? It sure doesn t seem that way, since an atom radiates at the same rate no matter what s around it. s beautiful as Einstein s derivation was, it took another 20 years for someone to calculate /B from first principles. It took the invention of Quantum Field Theory. 3 Quantum Field Theory The basic idea behind the calculation of the spontaneous emission coefficient is to think of the photons of each energy as separate particles, and then to study the system with multiparticle quantum mechanics. Let s start by looking at just a single frequency (energy) mode of a photon, say of energy. This mode can be excited n times. Each excitation adds energy to the system. So the energy eigenstates have energies, 2, 3,. There is a quantum mechanical system with this property one you may remember from your quantum mechanics course the simple harmonic oscillator. We will review it for homework. For now, let me just remind you of how it works. The easiest way to study the system is with creation and annihilation operators. These satisfy: There is also the number operator which counts modes [a, a ] = (7) Nˆ =a a (8) Nˆ n =n n (9) Then, and so Nˆa n = a aa n =a n + a a a n = (n + a)a n (20) a n = n + n + (2) a n = n n (22) Where the normalization is set so that n n = (23)

4 4 Section 3 Let s look how a photon interacts with an atom. Remember Fermi s golden rule? It s the formula for the transition rate between states. It says that the transition rate between two states is proportional to the matrix element squared Γ M 2 δ(e f E i ) (24) where the δ-function serves to enforce energy conservations. In perturbation theory, this matrix element M is the projection of the initial and final states on the interaction Hamiltonian M = f H int + i (25) We don t know exactly what the interaction Hamiltonian H int is, but it s got to have something to connect the initial and final atomic states and some creation operator or annihilation operator to create the photon. So it must look like H int = H 0 a +H 0 a (26) For the 2 transition, the initial state is an excited atom with n photons of energy : i = atom ; n (27) The final state is a lower energy atom with n + photons of energy So, where f = atom; n + (28) M 2 = atom; n+ H 0 a + H 0 a atom ; n (29) = atom H 0 atom n + a n (30) = M 0 n + n + n + (3) = M 0 n + (32) M 0 = atom H 0 atom (33) M 2 2 = M 0 2 (n + ) (34) But if we are exciting an atom, then the initial state is an unexcited atom and n photons: and the final state we have lost a photon and have an excited atom Then i = atom; n (35) f = atom ; n (36) M 2 = atom ; n H 0 a + H 0 a atom; n (37) = atom H 0 atom n a n (38) = M 0 n (39) dn 2 = M 2 2 n 2 = M 0 2 (n ω + )n 2 (40) where we have replaced n by n ω for clarity with = ω. These are pretty close to Einstein s equations: dn = M 2 2 n = M 0 2 (n ω )n (4) dn 2 =(+BI(ω))n 2 (42) dn =B I(ω)n (43)

5 Quantum Field Theory 5 To get them to match exactly, we just need to relate the number of photon modes of frequency ω to the intensity I(ω). Since the energies are quantized by = ω = 2π nk, total energy is L ω E(ω)= d 3 nk( ω)n ω =(4π) L 3 ω dω (2π) 3ω3 n ω (44) nd then the intensity is 0 I(ω)= L 3 de dω = ω3 2π 2 n ω (45) This is just some standard statistical mechanical relation, independent of what n ω actually is. There is no mention of temperature or of equilibrium, just a phase space integral. [ ] π dn 2 = M ω 3I(ω)+ n 2 (46) [ ] π dn = M ω 3I(ω) n (47) and we can read off that B = B (48) B = 2π2 k 3 (49) Which are Einstein s relations. Beautiful! This derivation is due to a paper of Dirac s from 927. Note that we never needed to talk about temperature.

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