9 Perturbation Theory II: Time Dependent Case

Transcription

1 9 Perturbation Theory II: Time Dependent Case In this chapter we ll study perturbation theory in the case that the perturbation varies in time. Unlike the time independent case, where we mostly wanted to know the bound state energy levels of the system including the perturbation, in the time dependent case our interest is usually in the rate at which a quantum system changes its state in response to the presence of the perturbation. 9.1 The Interaction Picture Consider the Hamiltonian H(t) =H + (t) (9.1) where H is a time independent Hamiltonian whose spectrum and eigenstates we understand, and (t) is a perturbation that we now allow to depend on time. The idea is that H models a system in quiescent state, and we want to understand how eigenstates of H respond as they start to notice the perturbation. In the absence of the time dependent perturbation, the time evolution operator would be U (t) =e iht/~ as appropriate for the model Hamiltonian H.Wedefinetheinteraction picture state I (t)i as I (t)i = U 1 (t) S(t)i, (9.2) where S (t)i is the Schrödinger picture state evolving in accordance with the TDSE for the full Hamiltonian (9.1). Thus, in the absence of (t), we would simply have I (t)i = U 1 (t) U (t) ()i = ()i so the state I ()i would in fact be independent of time. The presence of the perturbation means that our states do not evolve purely according to U (t), so I (t)i has non trivial time dependence. To calculate the evolution of I (t)i, we di I(t)i = H e ih t/~ S (t)i +e ih S(t)i = H e ih t/~ S (t)i +e ih t/~ (H + (t)) S (t)i = U 1 (t) (t) U (t) I (t)i, (9.3) where we have used the full TDSE to S i/@t, and in the last line we ve written our state back in terms of the interaction picture. Similarly, the expectation value of an operator A S in the Schrödinger picture becomes h S (t) A S S (t)i = h I (t) U 1 (t)a SU (t) I (t)i (9.4) in terms of the interaction picture states, so we define the interaction picture operator A I (t) by A I (t) =U 1 (t)a SU (t) (9.5) again using just the evolution operators of the unperturbed, rather than full, Hamiltonian. Infinitesimally, A I(t) = i ~ [H,A I (t)]. (9.6) 111

2 In this sense, the interaction picture is halfway between the Schrödinger picture in which states evolve and operators are time independent, and the Heisenberg picture in which states are always taken at their initial value ()i, but operators evolve according to the full Hamiltonian. The point of the interaction picture that we ve assumed we understand the eigenstates & eigenvalues of H, so at least in principle we can calculate how states evolve according to U (t), whereas the true evolution accounting for the perturbation is as yet unknown to us. The time dependence (9.3) of the interaction picture state can be written I(t)i = I (t) I (t)i (9.7) where I(t) =U 1 (t) (t) U (t) is the perturbation operator in the interaction picture. Defining the time evolution operator in the interaction picture U I (t) by I (t)i = U I (t) I ()i for any initial state I ()i 2H, (9.8) from (9.7) we find that U I (t) satisfies the di erential U I(t) = I (t) U I (t). (9.9) For many purposes, it is more convenient to integrate this equation over time, writing it as the integral equation i t U I (t) =1 I(t ) U I (t ) dt (9.1) ~ in which the operator U I (t) we wish to understand also appears on the rhs inside the integral. Note that we have used the initial condition U I () = 1. Up to now our considerations have been exact, but to make progress we must now approximate. The integral form (9.1) is well adapted to a perturbative treatment. Recall that the only reason I (t)i has any time dependence at all, and hence the only reason U I (t) is non trivial, is the presence of (t). Thus we can replace the U I (t ) in the integral (9.1) by its value according to its own equation, obtaining U I (t) =1 i t ~ I(t 1 ) dt 1 + i 2 t ~ apple t1 I(t 1 ) I(t 2 ) U I (t 2 ) dt 2 dt 1. (9.11) The term containing U I (t) on the rhs now has two explicit powers of the perturbation so may be hoped to be of lower order. Iterating this procedure gives U I (t) = 1X n= (t), i n t tn 1 dt 1 dt n I (t 1 ) I(t 2 ) I(t n ) (9.12) ~ in general. Note that the operators (t i ) at di erent times do not necessarily commute they appear in this expression in a time ordered sequence, with apple t n apple t n 1 apple applet 1 apple t. 112

3 We often write the series (9.12) as U I (t) =Texp i t ~ (t ) dt (9.13) for short, where the time ordered exponential 64 Texp( )isdefinedbytheseries(9.12). It s a good exercise to check that this time ordered exponential is equal to the usual exponential in the case that [ (t 1 ), (t 2 )] = for all t 1 and t 2. The interaction picture perturbation is I(t) = U 1 (t) U (t), so we can also write (9.12) as U I (t) = 1X n= i n t ~ dt 1 t n 1 dt n U 1 (t 1) (t 1 ) U (t 1 t 2 ) (t 2 ) U (t n 1 t n ) (t n ) U (t n ) (9.14) using the fact that U (t) provides a homomorphism from the Abelian group of time translations. In this form we see that the perturbative expansion of U I (t) treats the e ect of the perturbation as a sequence of impulses: we evolve according to H for some time t n then feel the e ect (t n ) of the perturbation just at this time, before evolving according to H for a further time (t n 1 t n ) and feeling the next impulse from the perturbation, and so on. Finally, we integrate over all the possible intermediate times at which the e ects of the perturbation were felt, and sum over how many times it acted. This method, closely related to Green s functions in pdes, is also the basis of Feynman diagrams in QFT, where H is usually taken to be a free Hamiltonian. Now let s put these ideas to work. Suppose that at t = we re in the state ()i = P n a n() ni, described in terms of some superposition of the basis 65 { ni} of eigenstates of the unperturbed Hamiltonian. Then a later time t the interaction picture state will be I (t)i = U I (t) ()i and can again be described by a superposition P n a n(t) ni where the amplitudes a n (t) 6= e ient/~ a n () in general. We have that a k (t) =hk U I (t) I ()i (9.15) in terms of the interaction picture time evolution operator, so using (9.12) to first non trivial order we have i X t a k (t) a k () a n () hk I(t ) ni dt ~ = a k () = a k () i ~ i ~ n X a n () n X a n () n t t hk U 1 (t ) (t ) U (t ) ni dt e i(e k E n)t /~ hk (t ) ni dt, (9.16) 64 Time ordered exponentials were introduced by Freeman Dyson and play a prominent role in QFT. They re also familiar in di erential geometry in the context of computing the holonomy of a connection around a closed path. (This is not a coincidence, but it would take us too far afield to explain here.) 65 In this chapter, we shall label eigenstates & eigenvalues of the unperturbed Hamiltonian simply as ni and E n, dropping the superscripts in n () i and E n (). We hope this won t cause confusion, because we re not going to be so interested in the new energy levels, but rather in the time dependence of the states. 113

4 using the fact that H ni = E n ni. In particular, if at t = we begin in an eigenstate mi of H, so that ( 1 if n = m a n () =, (9.17) else then the amplitude for our perturbation to caused the system to make a transition into a di erent H eigenstate ki 6= mi is a k (t) = i ~ to first order in the perturbation. t e i(e k E m)t /~ hk (t ) mi dt (9.18) To go further, we must specify how the perturbation (t) actually depends on time. There are two main cases of interest: monochromatic perturbations where (t) oscillates with a single, fixed frequency, and perturbations where (t) fluctuates randomly, but with statistical properties that do not change in time. Let s consider each of these cases in turn. 9.2 Fermi s Golden Rule We ll first consider small perturbations of the form (t) = e i!t + e +i!t (9.19) for some time independent operator, so that (t) oscillates with fixed frequency!. Note that (t), being part of the Hamiltonian, should be Hermitian so we have included both an e i!t and e +i!t term, and wlog we take!>. We ll see later that these two terms are responsible for di erent physics. We will call such perturbations monochromatic in anticipation of the case that they describe an applied electromagnetic field, corresponding to light of frequency!. Performing the time integral in (9.18) in this monochromatic case is trivial and yields hk mi h i a k (t) = e i(e k E m ~!)t/~ 1 + hk mi h i e i(e k E m+~!)t/~ 1. E k E m ~! E k E m + ~! (9.2) Each of these terms vanishes at t = and then grow linearly in t for times small compared to either E k E m ~! 1 or E k E m + ~! 1, respectively, where the oscillatory nature of the exponentials reveals itself. Now suppose the final state ki into which we are making a transition has energy very close to E m +~!, so that in particular E k >E m corresponding to the atom absorbing energy from the perturbation. In this case, the timescale E k E m ~! 1 is very long and the first term on the rhs of (9.2) grows for a long time. On the other hand, if the final state has E k very close to E m ~!, corresponding to the atom losing energy to the perturbation, then it is the second term which grows for a long time. In the case of absorption, the small denominator of the first term means that this term will dominate at late times. Thus hk mi a k (t) E k E m ~! h e i(e k E m ~!)t/~ i 1 for large t, (9.21) 114

5 so that the probability that the system has made a transition to the k th eigenstate of H after a large time t is P k (t) = a k (t) 2 hk mi 2 apple (Ek E m ~!)t (E k E m ~!) cos ~ (9.22) 4 hk mi 2 (Ek E m ~!)t = (E k E m ~!) 2 sin2. 2~ correct to order 2. Now consider the function sin 2 (xt)/x 2 t. For x 6= wehave whilst sin 2 (xt) 1 lim t!1 x 2 apple lim t t!1 x 2 =, (9.23) t lim x! sin2 (xt)/x 2 t = t and R sin 2 (xt) x 2 t dx =. (9.24) Thus, for large t we have sin 2 (xt) x 2! (x) t (9.25) (see figure??). Using this in (9.22) gives t a k (t) 2 =4 hk mi 2 2~ (E k E m ~!) (9.26) and the corresponding rate of transition into the k th state is a k (t) 2 (m! k) = lim t!1 t = 2 ~ hk mi 2 (E k E m ~!). (9.27a) For the case of emission, with E k <E m, we d instead have found (m! k) = 2 ~ hm ki 2 (E k E m + ~!). (9.27b) with an opposite sign in the function. These results were essentially obtained by Dirac and proved to be so useful in describing atomic transitions that Fermi dubbed them golden rules. The name stuck and now (9.27a) &(9.27b) are known as Fermi s golden rule. The function in (9.27a) of course says that the perturbation must supply the energy di erence E k E m between the two energy levels of our atomic transition. This function has the e ect that, in practice, a truly monochromatic perturbation cannot cause transitions between two discrete energy levels of a bound system such as an atom, because the frequency of the perturbation would need to be perfectly tuned to the energy di erence 66. For this reason, we re often more interested in the probability that the system has transitioned not to some specific state ki, but rather to any one of a large number of states of H. 66 At times that are of order E k E m ~! 1, before our replacement of sin 2 (xt)/x 2 t by a -function was valid, no tuning was needed. However, in this case the probability P k (t) does not continue to increase with time once the perturbation is ended, and there is no overall transition rate. 115

6 Suppose the number of states with energy in the interval (E k,e k + E k ) is denoted by (E k ) E k. The function (E k ) which needs to be calculated in each case is known as the density of states. Note that (E k ) must have dimensions of 1/(energy) in order for (E k ) E k to be a number of states. The total rate we seek is then (m! k) (E k ) de k = 2 hk mi 2 (E k E m ~!) (E k ) de k ~ = 2 (9.28) ~ he out mi 2 (E out ) where E out = E m + ~!. This is the form of Fermi s golden rule that is most often used. It s valid at times late enough that our replacement sin 2 (xt)/x 2 t! (x) above is valid, provided the e ect of the perturbation (i.e. the total decay probability) remains su ciently small that our use of first order perturbation theory in (9.18) isjustified Ionization by Monochromatic Light A common example of the use of Fermi s golden rule is to calculate the rate at which an applied monochromatic electromagnetic field can ionize an atom, say hydrogen, stripping o the electron from a bound state into the continuum of ionized states. We ll assume that the electromagnetic field itself may be treated classically this assumption is valid e.g. in a laser, or at the focus of the antenna of a radio telescope where the field is large. In the vacuum, the electromagnetic field is divergence free and is entirely generated by Faraday s Law r E The whole electromagnetic field can be described by a vector potential A via B = r A and E = 1 c In the monochromatic case, Faraday s equation is (9.29) A(x,t)= cos(k x!t), (9.3) where is a constant vector describing the polarization of the wave and k is the wavevector. From the vacuum Maxwell equation r E =, we learn that k =, (9.31) saying that the wave is transverse. The coupling of an elementary particle of charge q to such a vector potential is given by the minimal coupling rule P! P qa(x,t) (9.32) in the Hamiltonian 67,so H = (P qa)2 2µ + V (X) =H q 2µ (P A + A P)+q2 A 2 2µ. (9.33) 67 This is called minimal coupling because more complicated modifications are possible if the charged particle is not elementary, or in other more complicated circumstances. You ll learn more about this if you take the Part II course on AQM next term. 116

7 In our case of atomic transitions, an electron has charge q = e so to first order in e the perturbing Hamiltonian is (t) = e (P cos(k X 2µ!t) + cos(k X!t) P). (9.34) The component P of the momentum operator in the direction of the polarization of the electromagnetic wave commutes with cos(k X!t) by the transversality condition (9.31) (recall that itself is a constant vector). Thus (t) = e µ P cos(k X!t) = e 2µ P e i(k X!t) +e i(k X!t), (9.35) simplifying the perturbation. Suppose that electromagnetic wavelength is much larger than the Bohr radius of the atom. This is a good approximation provided the atomic transition occurs between states that are separated in energy by much less that µc 2, as will be the case for waves with frequencies that are less than those of soft X-rays. In this approximation, k X 1 for all locations in the atom or molecule at which their is significant probability of finding the electron. To lowest order, we can thus approximate (t) e 2µ P e i!t +e +i!t. (9.36) We ve retained the full time dependence, because large values of t cannot be ruled out in the way we excluded large values of the electron s position. Finally, since the momentum operator occurs in H only in the kinetic term, we can write this perturbation as (t) = ie 2~ [H, X] e i!t +e +i!t (9.37) The purpose of rewriting the perturbation in this way is that we know how the unperturbed Hamiltonian H acts on the original eigenstates. Note that the perturbation is proportional to the dipole operator D = e X whose matrix elements between the initial and final states we will need to take. This result, known as the dipole approximation, arose from our approximation of the vector potential as constant over the scale of the atom. Since the electric field E /! we can interpret the transition as occuring due to the interaction between the applied electric field and this dipole. We must now calculate the matrix element hf ii = ie 2~ (E f E i ) hf X ii. Suppose that the hydrogen atom is initially in the ground state so that ii = 1i, with electronic wavefunction hx 1i = p 1 e r/a. (9.38) a 3 If the energy of the ejected electron is su ciently great, we can ignore its Coulomb attraction back to the ion and so treat the final electron as a free particle of momentum p, so fi = pi with energy E = p 2 /2µ and wavefunction 1 i hx pi = exp (2 ~) 3/2 ~ p x (9.39) 117

8 as in (4.34). Thus the matrix element we seek is determined by the integral 1 hp X 1i = (2 ~) 3/2 ( a 3 e ip x/~ x e r/a d 3 x. (9.4) )1/2 R 3 Performing this integral is rather messy 68 ; when the dust settles one finds hp X 1i = 8p 2i a 7/2 ~ 5/2 (1 + p 2 a 2 /~2 ) 3 p. (9.41) The fact that this result had to be proportional to p is clear: the only direction in which the expectation value hp X 1i can point is that of p. We define the di erential rate d 1i! pi by d 1i! pi = 2 ~ hp 1i 2 (E p E 1 ~!) p 2 dp d (9.42) where d =sin d d. For a free particle of mass µ we have p 2 dp = p 2 (de p /dp) 1 de p = µp de p,so d 1i! pi d = 2 ~ hp 1i 2 (E p E 1 ~!) µp de p = 26 µp 3 e 2 2 ~ 3 a 7 (E p E 1 ) 2 (1 + p 2 a 2 /~2 ) 6 cos2 (E p E 1 ~!) de p, (9.43) where =. On the support of the -function we can replace E p E 1 by ~!, and the remaining e ect of the function is to soak up the de p integral. Since we re assuming the energy of the final state electron is so great that we can neglect it s attraction back to the proton, we must have p 2 ~ 2 /a 2. Thus we simplify our di erential rate (9.43) to d 1i! pi d 28 µ~ 2 (p/~) 9 a 5 e 2 E 2 cos 2. (9.44) where E =!/2 is the magnitude of the electric field and p 2 =2µE p is frozen to 2µ(~! + E 1 ) 2µ~!. 9.3 Randomly Fluctuating Perturbations In our derivation of Fermi s golden rule (9.28), we took the frequency! of the perturbation to be fixed and assumed a continuum of final states. Now let s consider the opposite case of transitions into a discrete set of final states, such as the di erent bound state energy 68 If you insist: First note that for any function f(r) of the radius, e ik x f(r) d 3 x = 4 k 1 rf(r) sinkr dr with r = x and k = k. Di erentiating this result gives i e ik x x e r/a d 3 x = 4 1 k k r e r/a 32 a 5 ( sin kr + kr cos kr) dr = 3 (1 + k 2 a k, 2 )3 where the final radial integral is done by parts. 118

9 levels of an atom, but where we have a continuous range of perturbing frequencies. This case is relevant e.g. when our atom is immersed not in a monochromatic source such as a laser, but in a thermal bath of radiation. We ll assume that, while the actual value of the perturbation experienced by our system may be subject to rapid fluctuations, the statistical properties of these fluctuations remain constant in time so that hk (t 1 ) mihm (t 2 ) ki = f km (t 1 t 2 ) (9.45) where the bar indicates the average over fluctuations. We can write the function f km (t 1 t 2 ) as a Fourier transform f km (t 1 t 2 )= F km (!)e i!(t 1 t 2 ) d! (9.46) where F km (!) represents the average strength of the frequency-! contribution of our perturbation. From equation (9.18) for the case that a n () = nm (so that the system is initially in state mi), we have to first order a k (t) 2 = 1 ~ 2 t t hk (t 1 ) mihm (t 2 ) ki e i(e k E m)(t 1 t 2 )/~ dt 1 dt 2. (9.47) Averaging over the fluctuations, equation (9.45) gives a k (t) 2 = 1 ~ 2 = 1 ~ 2 =4 apple t F km (!) F km (!) t t e i(e k E m ~!)(t 1 t 2 )/~ dt 1 dt 2 d! 2 e i(e k E m ~!)t 1 /~ dt 1 d! F km (!) sin2 ((E k E m ~!)t/2~) (E k E m ~!) 2 d!. (9.48) Just as before, for large times we can replace sin 2 (xt)/x 2! t (x) in this integral. Thus we find the rate of transitions ( mi! ki) = a k(t) 2 t = 2 ~ 2 F km ((E k E m )/~) (9.49) in the case that the states mi and ki are discrete but we have a continuous range of perturbing frequencies Emission and Absorption of Radiation The classic application of the above results is to an atom in a bath of radiation. As before, we ll assume the radiation has wavelengths than the typical atomic size, so that in the dipole approximation the Hamiltonian may be taken to be H = H + e X r E(t) X r, (9.5) 119

10 where H is the Hamiltonian of the unperturbed atom, E(t) is the fluctuating electric field due to the radiation and X r is the position operator for the r th electron in the atom. As in (9.45) take the fluctuating electric field to be correlated as E i (t 1 )E j (t 2 )= ij P(!)e i!(t 1 t 2 ) d!. (9.51) R The presence of ij on the rhs means that fluctuations in di erent directions are uncorrelated, whilst fluctuations that are aligned are correlated equally no matter their direction. This just says that there is no preferred direction for the electric field and holds e.g. for a thermal bath of radiation. We can understand the meaning of the function P(!) as follows. As you ll learn if you re taking the Part II Electrodynamics course, the energy density of an electromagnetic field is (x,t)= 1 8 E2 (x,t)+b 2 (x,t) (9.52) and in a purely radiative (source free) field E 2 = B 2 so = E 2 /4. In our case, equation (9.51) gives = 1 4 E2 (t) = 3 P(!) d! (9.53) 4 R so 3P(!)/2 represents the average energy density in the radiation (at any time) due to frequencies in the range (!,! + d! ). From (9.45) we have that perturbation itself is correlated as hk (t 1 ) mihm (t 2 ) ki = e 2 X r hk X r mi 2 P(!)e i!(t 1 t 2 ) d! (9.54) R where ni and mi are states of the unperturbed atom and the mod-square includes the Euclidean inner product (dot product) of the two hk X r mi matrix elements. Thus F km (!) =e 2 X r hk X r mi 2 P(!) (9.55) and the rate at which the radiation stimulates the atom to make a transition from state mi to a state ki 2 ( mi! ki) = 2 e2 X ~ 2 hk X r mi P(! km ) (9.56) where! km =(E k E m )/~. Now, if the fluctuating electric field is real, then the power function must obey P(!) = P(!) and P(!) =P (!). In particular, the rate ( mi! ki)in(9.56) is an even function of! km and consequently, for fixed E k E m, this rate is the same whether E k >E m or E k <E m. For E k >E m, the atom absorbs energy from the radiation, exciting to a higher energy level, whilst if E k <E m the atom emits energy back into the radiation as it decays to a lower energy state. The important result we ve obtained is that the rate for stimulated emission is the same as the rate of absorption. r 12

11 9.3.2 Einstein s Statistical Argument Our treatment above calculated the rate at which an atom would decay (or excite) due to being immersed in a bath of radiation. However, we currently cannot understand decays of an isolated atom: if the atom is prepared in any eigenstate of H, quantum mechanics says that in the absence of a perturbation it will just stay there. Remarkably, Einstein was able to relate the rate of spontaneous decay of an atom to the rate of stimulated decay that we calculated above. Einstein defined A m!k (!) as the rate at which an atom would spontaneously decay from a state of energy E m to a state of lower energy E k,where! =(E m E k )/~ is the energy of the emitted photon. If the atom is immersed in a bath of photons with (!) d! modes of frequency in the range (!,!+d!), we let (!) B k!m (!) denote the rate at which energy is absorbed from the radiation bath, exciting the atom from ki! mi. Similarly, let (!) B m!k denote the rate of decay of the excited atom due to stimulation by the presence of the radiation bath. We calculated values for B m!k and B k!m above, but let s pretend that (like Einstein) we do not know this yet. In thermodynamic equilibrium these rates must balance, so if there are n k atoms in state ki and n m in state mi we must have n m (A m!k (!)+ (!) B m!k (!)) = n k (!) B k!m (!). (9.57) In the Part II Statistical Mechanics course, you ll learn that in equilibrium at temperature T, the relative populations of atoms in states with energies E m and E k is given by the Boltzmann distribution n k n m = e Ek/kBT e Em/k BT =e~!/k BT, (9.58) where k B JK 1 is Boltzmann s constant. Furthermore, if the radiation is in equilibrium at the same temperature then it turns out that (!) = ~!3 2 c 3 1 e ~!/k BT 1 (9.59) as you ll again see next term. Using these in (9.57) gives A m!k (!) = ~!3 1 2 c 3 e ~!/k e ~!/kbt B k!m (!) B m!k (!). (9.6) BT 1 However, A m!k is supposed to be the rate of spontaneous emission of radiation from the atom, so it cannot depend on any properties of the radiation bath. In particular, it must be independent of the temperature T. This is possible i B k!m (!) =B m!k (!) (9.61) so that the rates of stimulated emission and absorption agree. In this case, we have A m!k (!) = ~!3 2 c 3 B m!k(!) (9.62) 121

12 so that knowing the rate B k!m (!) at which a classical electromagnetic wave of frequency! =(E m E k )/~ is absorbed by an atom, we can also calculate the rate A m!k (!) at which the atom spontaneously decays, even in the absence of radiation. Now, in equation (9.56) we found from a first principles calculation that ( mi! ki) = 2 e2 X ~ 2 hk X r mi r = (2 e)2 X 3~ 2 hk X r mi r 2 P(! km ) 2 (! km ) (9.63) where in the second line we ve used the fact that P(!) =2 (!)/3 as obtained in (9.53). Consequently, for our atom Einstein s B coe cient is B k!m (!) =B m!k (!) = (2 e)2 X 3~ 2 hk X r mi r 2. (9.64) for stimulated emission or absorption. From Einstein s statistical argument we thus learn that A m!k (!) = 4e2! 3 2 X 3~c 3 hk X r mi (9.65) is the rate at which an excited atom must spontaneously decay, even in the absence of any applied radiation. Ingenious though it is, there s something puzzling about this calculation of the rate of spontaneous decay. Where does our quantum mechanical calculation stating that isolated excited atoms do not decay go wrong? The answer is that while we treated the energy levels of the atom fully quantum mechanically, the electromagnetic field itself was treated classically. Spontaneous emission is only possible once one also quantizes the electromagnetic field, since it is fluctuations in the zero point value of this field that allow the atom to emit a photon and decay; heuristically, the fluctuating value of the quantum electromagnetic field even in the vacuum tickles the excited atom prompting the decay. Treating the EM field classically is appropriate if the radiation comes from a high intensity laser, where the energy density of the field is so high that the terms B(!) (!) dominate in (9.57). By contrast, emission of light from the atoms in a humble candle is an inherently quantum phenomenon, occurring purely by spontaneous emission once the flame is lit candlelight shines even in ambient darkness. Einstein gave his statistical argument in 1916, when Bohr s model of the atom (the first thing you met in 1B QM) was still the deepest understanding physicists had of quantum theory. This was at the same time as he was revolutionising our understanding of gravity, space and time, which obviously wasn t enough to keep him fully occupied. The quantum treatment of stimulated emission given in the previous section is due to Dirac in 1926, the same year as Schrödinger first published his famous equation. A first principles calculation of the spontaneous emission rate A m!k, requiring the quantization of the electromagnetic r 122

13 field, was also given by Dirac just one year later. Things move fast. Dirac s 1927 results agreed with those we ve found via Einstein s argument. The necessity also to quantize the electromagnetic field heralded the arrival of Quantum Field Theory. Let s use these results to calculate the typical lifetime of an excited state of an isolated atom. When the radiation density (!) is very small, the number of atoms n m in the excited state m = A m!k n m so the atom decays exponentially with a characteristic timescale A 1 m!k. For hydrogen or an alkali atom, typically the outermost electron changes its energy level in the transition, so we can drop the sum in (9.65) tofind A m!k (!) = 4e2! 3 3~c 3 hk X mi 2 (9.67) where X is the position of the outermost electron. Unless hk X mi vanishes for some reason (we ll consider possible reasons momentarily) we d expect this matrix element to be of the typical size a characteristic of the atom, so hk X mi 2 a 2. Hence the characteristic lifetime of the excited atomic state is roughly = A 1 m!k 3~c3 4e 2! 3 a 2 = 3c3 µ 4~! 3 a (9.68) where µ is the mass of the electron. Optical light has a frequency of order! 1 15 Hz, so the timescale is around 1 7 times longer than the timescale ~/E 2 associated to the energy E 2 of the first excited state of hydrogen. Thus around 1 7 oscillations of the hydrogen atom occur before it radiates a photon, decaying down into the ground state Selection Rules In the previous sections we ve found that, in the dipole approximation that the wavelength of any radiation is large compared to the scale of the atom, transition amplitudes in hydrogenic atoms are proportional to the matrix elements hn,`,m X n, `, mi. We now consider a set of conditions that are necessary if this matrix element is not to vanish. Firstly, in the dipole approximation the perturbing operator is independent of any details of the electron s spin, so the initial and final spin states must be the same. (We ve been tacitly assuming this by suppressing any mention of these spins thus far.) Next, the parity operator acts as 69 so since 1 = wehave n, `, mi =( 1)` n, `, mi and 1 X = X, (9.69) hn,`,m X n, `, mi = hn,`,m 1 X 1 n, `, mi =( 1)`+`+1 hn,`,m X n, `, mi. (9.7) 69 The electron has intrinsic parity e = +1, but our result (9.71) wouldholdevenifitdidn tasthe electron appears in both the initial and final states. 123

14 Consequently, parity imposes hn,`,m X n, `, mi = unless ` + ` 2 odd (9.71) as we saw in section 4.6. This is known as a selection rule on allowed transitions. We can obtain further selection rules by considering angular momentum. In the second problem set you showed that [L 2, [L 2, X]] = 2~ 2 (L 2 X + XL 2 ) (9.72) and hence that ( ) 2 2( + ) hn,`,m X n, `, mi = (9.73) where = `(` + 1) and similarly for. Thus the matrix element hn,`,m X n, `, mi vanishes unless ( ) 2 = 2( + ), which implies ` ` = 1 or ` = ` =. The parity selection rule already rules out this latter case, so we have hn,`,m X n, `, mi = unless ` ` =1, (9.74) strengthening our previous rule (9.71). A final selection rule comes from noticing that, since X is a vector operator independent of spin, [J i,x j ]=[L i,x j ]=i~ P k ijkx k. In particular, if we define X ± = X 1 ± ix 2 (9.75) then [J 3,X 3 ] = and [L 3,X ± ]=±~X ±. (9.76) We therefore find =hn, `,m [L 3,X 3 ] n, `, mi =(m m)~ hn,`,m X 3 n, `, mi (9.77a) and, since X ± n, `, mi is an eigenstate of L 3 with eigenvalue (m ± 1)~, =(m m ± 1)~ hn,`,m X ± n, `, mi. (9.77b) Combining these two we have the further selection rule hn,`,m X n, `, mi = unless m m apple1, (9.78) as you also derived in the second problem sheet. More specifically, for transitions with m = m only X 3 can contribute, so the matrix element hn,`,m X n, `, mi points in the ẑ direction. On the other hand, transitions with m m = 1 cannot receive any contribution from X 3, so the matrix element lies in the xy-plane and in fact hn B C,`,m+1 X n, `, mi /@ ia while hn B C,`,m+1 X n, `, mi /@ +ia. (9.79) 124

15 Finally, we remark that these selection rules were derived under the assumption that the transition is just due to the dipole approximation where the perturbation involves matrix elements of the dipole moment operator ex. The e ects of both higher orders in perturbation theory, and inclusion of interactions beyond those of an electric field with wavelength a mean that transitions that are forbidden according to the above selection rules may in fact occur. They will typically do so, however, at a much smaller rate. 125