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Home , Einstein coefficients, Stimulated emission

2012 Matthew Schwartz I-1: Microscopic Theory of

1 Blackbody Radiation Quantum mechanics began on October 19, 1900 with Max Planck’s explanation of the black- body radiation spectrum. Actually, although Planck’s result helped lead to the development of quantum mechanics, that is, the quantum mechanics of , his original observation was about the quantum nature of , which is a topic for quantum field theory. So, radiation is a great way to motivate the need for a quantum theory of fields. This introductory topic involves a little history, a little statistical mechanics, a little quantum mechanics, and a little quantum field theory. Since this material is mostly motivational, it is not critical to understanding quantum field theory, and can be skipped. In 1900 people were very confused about how to explain the spectrum of radiation from hot objects. The equipartition theorem says that a body in thermal equilibrium should have equally distributed among all possible modes. For a hot gas, this gives the Maxwell-Boltzmann distribution of thermal velocities, which is beautifully confirmed by data. But when you try to do the same thing for light, you get a very strange result. A blackbody can be approximated as a Jeans cube, which is a hot box of light in thermal equilibrium. Classically, a box of size L supports standing electromagnetic waves with angular frequencies 2π

ωn = nQ c (1) L | |

for integer 3-vectors nQ . Here c is the . Before 1900 people thought you could have as much or as little energy in each mode as you want. By the classical equipartition theorem, blackbodies should emit light equally in all modes with the intensity growing as the differential volume of phase space

1 d − 3 2 I( ω) E( ω) = const c ω kBT ( classical) (2) ≡ V dω × More simply, this classical result follows from dimensional analysis: it’s the only quantity with − 3 units of energy time distance you can construct out of ω, kBT and c. We will set c = 1 from now on, since× it can× be restored by dimensional analysis (see Appendix A). The classical spectrum implies there should be a lot of high frequency light – the ultravi- olet catastrophe. Experimentally, the distribution looks more like a Maxwell-Boltzmann dis- tribution, peaked at some finite ω, as shown in Figure 1. Clearly something is wrong, but why does the equipartition theorem not seem to work?

Classical

Observed

Figure 1. The ultraviolet catastrophe. The classical prediction for the intensity of radiation coming out of a blackbody disagrees with experimental observation at large frequencies.

This blackbody paradox led Planck to postulate that the energy of each electromagnetic mode in the cavity is quantized in units of frequency1

2π Q En = ~ωn = ~ nQ = p (3) L | | | n |

1 2 Section 2

h ~ h where is Planck’s constant and = 2 π . Einstein later interpreted this as saying light is made up of particles (later called , by the chemist Gilbert Lewis). Note that if the excitations are particles, then they are massless 2 2 2

m = E Qp = 0 (4) n n − | n | So if Planck and Einstein are right, then light is really a collection of massless photons. As we will see, there are a number of simple and direct experimental consequences of this hypothesis: quantizing light resolves the blackbody paradox; light having energy leads to the photoelectric effect; and light having momentum leads to Compton scattering. Most importantly for us, this was the key insight that led to the development of quantum field theory. With Planck’s energy hypothesis, the thermal distribution is easy to compute. Each mode of frequency ωn can be excited an integer number j times, giving energy jEn = j( ~ ωn) in that mode. The probability of finding that much energy in the mode is the same as the probability of finding energy in anything, proportional to the Boltzmann weight exp ( energy/kBT) . Thus the expectation value of the the energy in each mode is − d 1 ∞ − jEn β j=0 ( jEn) e dβ 1 − e − ~ ωn β ~ωn En = = − = (5) ∞ − jEn β 1 ~ωn β h i P =0 e e 1 j 1 − e − ~ ω n β − P where β = 1/kBT. (This simple derivation is due to Debye. The more modern one, using ensem- bles and statistical mechanics, was first given by Bose in 1924.) Now let’s take the continuum limit, L . Then the the sums turn into the integrals and the average total energy up to frequency ω→in ∞ the blackbody is ω 1 2 π ω 2

3 ~ωn nQ ~ωn Q E( ω) = d nQ = d cosθ dφ d n (6) ~ωn β | ~ω| n β Z e 1 Z− 1 Z0 Z0 | | e 1 − − 3 ω ′3 ~ L ′ ω =4π 3 dω ~ ′ (7) 8π Z0 e ω β 1 − Thus, the intensity of light as a function of frequency is (adding a factor of 2 for the two polar- izations of light) 1 dE( ω) ~ ω3 I( ω) = = (8) V dω π2 e ~ωβ 1 − Which is what Planck’s showed in 1900 correctly matches experiment. 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. For example, if we heat up a box with monochromatic light, eventually all frequencies must be excited. 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.

2 Einstein Coefficients A straightforward way to quantify the creation of light is through the coefficient of . This is the rate at which an excited emits light. Even by 1900, this phenomenon had been observed in chemical reactions, and as a form of radioactivity. But at that time it was only understood statistically. In 1916, Einstein developed a simple proof of the relation between emission and absorption based on the existence of thermal equilibrium. In addition to being rel- evant to chemical phenomenology, his relation made explicit why a first principles quantum theory of fields was needed.

1. Planck’s energy quantization actually did not come from trying to explain the UV catastrophe, which he was not particularly worried about since there was no fundamental reason for the equipartition theorem to hold universally. But he was trying to explain the observed spectrum. He first came up with a mathematical curve which fit data, generalizing previous work of Wien and Rayleigh, then wrote down a toy model which generated this curve. The interpretation of his model as referring to photons and the proper statistical mechanics derivation of the blackbody spectrum did not come until years later. 3

Einstein said: Suppose we have a cavity full of with energy levels E1 and E2 . Say there are n1 of the E1 atoms and n2 of the E2 atoms. Let ~ω = E2 E1 . The probability for an − E2 atom to emit a of frequency ω and transition to state E1 is called the coefficient for spontaneous emission A. The probability for a photon of frequency ω to induce a transition from 2 to 1 is proportional to the coefficient of B and to the number of photons of frequency ω in the cavity, that is, the intensity I( ω) . These contribute to a change in n2 of the form dn2 = [ A + BI( ω)] n2 (9) − The probability for a photon to induce a transition from 1 to 2 is called the coefficient of ′ absorption B , which in general may be different from B. This contributes to d n1 = ′ − B I( ω) n1 . Since the total number of atoms is conserved in this two-state system, dn1 + dn2 = 0 and therefore ′ dn2 = dn1 = [ A + BI( ω)] n2 + B I( ω) n1 (10) − − Even though we computed I( ω) above for the equilibrium blackbody situation, these equations should hold for any I( ω) , for example, if we shine a beam at some atoms in the lab. At this point, Einstein assumes the gas is in equilibrium. In equilibrium the number densi- ties are constant, dn1 = dn2 = 0 and determined by Boltzmann distributions.

− βE1 − βE2 n1 = Ne n2 = Ne (11) where N is some normalization factor. Then [ B ′e− βE1 Be− βE2 ] I( ω) = Ae − βE2 (12) − and so, A I( ω) = (13) B ′e~βω B − But we already know that in equilibrium ~ ω3 I( ω) = (14) π2 e~βω 1 − from Eq. (8). Since equilibrium must be satisfied at any , i.e. for any β, we must have B ′ = B (15) and A ~ = ω3 (16) B π2 These are beautiful results. The first, B = B ′, says that the coefficient of absorption must be the same as the coefficient for stimulated emission. Actually, it is an enjoyable exercise to compute B and B ′ in quantum mechanics (not quantum field theory!) using time-dependent perturbation theory, and show that B = B ′ that way. After that, Eq. (16) determines the remaining coeffi- cient, for spontaneous emission. Thus all the Einstein coefficients A, B and B ′ can be computed without using quantum field theory. You might have noticed something odd in the derivation of Eqs. (15) and (16). We, and Ein- stein, needed to use an equilibrium result about the blackbody spectrum to derive the A/B rela- tion. 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. As beautiful as Einstein’s derivation was, it took another 10 years for someone to calculate A/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. The following treatment comes from a paper of Paul Dirac from 1927, which introduced the idea of second quantization. This paper is often credited for initiating quantum field theory. 4 Section 3

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 ∆, 2∆, 3∆, . There is a quantum mechanical system with this prop- erty – one you may remember from your quantum mechanics course – the simple harmonic oscil- lator. Problems .. review it. For now, let me just remind you of how it works. If the following is not familiar to you, you are probably not ready for quantum field theory. The easiest way to study a quantum harmonic oscillator is with creation and annihilation operators, a † and a. These satisfy: [ a, a † ] = 1 (17) There is also the number operator Nˆ which counts modes Nˆ = a †a (18) Nˆ n = n n (19)

Then, Nˆ a † n = a †aa † n = a † n + a † a †a n = ( n + 1) a † n (20)

Thus a † n = C n + 1 for some constant C which can be chosen real. We can determine C from the normalization| i | n ni = 1 : h | i C2 = n + 1 C2 n + 1 = n aa † n = n ( a † a + 1) n = n + 1 (21) h | | i h | | i h | | i so C = √n + 1 . Similarly, a n = C ′ n 1 and | i | − i 2 2 C ′ = n 1 C ′ n 1 = n a †a n = n (22) h − | | − i h | | i so C ′ = √n. The result is that a † n = √n + 1 n + 1 , a n = √n n 1 (23) | i | i | i | − i We will now see that although these normalization factors are simple to derive they have impor- tant implications. Consider how a photon interacts with an atom. Remember Fermi’s golden rule? It’s the for- mula for the transition rate between states. It says that the transition rate between two states is proportional to the matrix element squared 2 Γ δ( Ef Ei) (24) ∼ |M| − where the δ-function serves to enforce energy conservation. We’ll spend a lot of time with the relativistic version of formulas like this, which we’ll derive in quantum field theory. For now, all we need to recall about this formula is that the matrix element is the projection of the initial and final states on the interaction Hamiltonian M = f H i (25) M h | int| i In this case, we don’t know exactly what the interaction Hamiltonian Hint is, but it must have something to connect the initial and final atomic states and some creation operator or annihila- tion operator to create the photon. Since it is Hermitian, it must look like † † Hint = HI a + HIa (26)

Dirac derived HI from the canonical introduction of the vector potential into the Hamiltonian:

1 2 1 2 e

Q Q

Q Q H = Qp p +eA . This leads to H A p representing the photon interacting with 2 m → 2 m int ∼ m · the atom’s electric dipole moment. In our coarse approximation, the photon field AQ is repre-

sented by a and so HI must be related to the momentum operator Qp . Fortunately, all that is needed to derive the Einstein relations is that HI is something with non-zero matrix elements between different atomic states, thus we can be vague about its precise definition. (For more details consult Dirac’s paper or Sections 61-64 of his classic text The Principles of Quantum Mechanics, Oxford 1930.) For the 2 1 transition, the initial state is an excited atom (we call atom2 ) and n photons of energy ∆: → i = atom2 ; n (27) | i | i Quantum Field Theory 5

The final state is a lower energy atom, we call atom1 with n + 1 photons of energy ∆

f = atom1 ; n + 1 (28) h | h | So,

† † 2 → 1 = atom1 ; n + 1 HI a + HIa atom2 ; n M † | | † = atom1 HI atom2 n + 1 a n + atom 1 HI atom2 n + 1 a n † | | h | | i h | | ih | | i = 0 n + 1 n + 1 √ n + 1 + 0 M † h | i = 0 √n + 1 M † † where 0 = atom1 H atom2 . Thus M | I | 2 2 2 → 1 = 0 ( n + 1) (29) |M | |M |

If instead we are exciting an atom, then the initial state is an unexcited atom and n photons:

i = atom1 ; n (30) | i | i and in the final state we have lost a photon and have an excited atom

f = atom2 ; n 1 (31) h | h − | then

† † 1 → 2 = atom2 ; n 1 H a + HIa atom1 ; n M − | I | = atom2 HI atom1 n 1 a n h | | ih − | | i = 0√n M and therefore 2 2 2 2 dn2 = dn1 = 2 → 1 n2 + 1 → 2 n1 = 0 ( nω + 1) n2 + 0 ( nω) n1 (32) − −|M | |M | −|M | |M | where we have replaced n by nω for clarity with ∆ = ~ω. This is pretty close to Einstein’s equation, Eq. (10)

′ dn2 = dn1 = [ A + BI( ω)] n2 + B I( ω) n1 (33) − − To get them to match exactly, we just need to relate the number of photon modes of frequency 2 π

ω to the intensity I( ω) . Since the energies are quantized by ∆ = ~ω = ~ nQ , the total energy is L | | ω ω

3 3 dω 3 Q ~ ~ E( ω) = d n ( ω) nω = (4π) L 3 ω nω (34) Z Z0 (2π) We should multiply this by 2 for the two polarizations of light. Dirac actually missed this factor in his 1927 paper, since was not understood at the time. Including the factor of 2, the intensity is 1 dE ~ω3 I( ω) = = n (35) L3 dω π2 ω

This equation is just some standard statistical mechanical relation, independent of what nω actually is; its derivation required no mention of temperature or of equilibrium, just a phase space integral. So now we have 2 2 2 π 2 π dn2 = dn1 = 0 1 + I( ω) n2 + 0 I( ω) n1 (36) − −|M |  ~ω3  |M |  ~ω3  and can read off Einstein’s relations A ~ B ′ = B, = ω3 (37) B π2 without ever having to assume thermal equilibrium. This beautiful derivation was one of the first ever results in quantum field theory.