量子力学 Quantum mechanics. School of Physics and Information Technology

量子力学 Quantum mechanics School of Physics and Information Technology Shaanxi Normal University

Chapter 9 Time-dependent perturation theory

Chapter 9 Time-dependent perturation theory 9.1 Two-level systems 9.2 Emission and asorption of radiation 9.3 Spontaneous emission

9.1 Two-level systems 9.2 Emission and asorption of radiation 9.3 Spontaneous emission

9.1 Two-level systems Suppose that there are just two states of the system. They are eigenstates of the unpertured Hamiltonian,and they are eigenstates orthonormal: H ψ = Eψ and H ψ = Eψ a a a ψ ψ = δ a a Any state can e expressed as a linear comination of them () = ca a + c ψ ψ ψ In the asence of any perturation,each component evolves with its characteristic exponential factor: ψ ( t) = c ψ e + c ψ e ieat / h ieat / h a a 2 2 c + c = 1 a

9.1.1 The pertured system Now suppose we turn on a time-dependent perturation, then the wave function can expressed as follows: ψ ( t) = c ( t) ψ e + c ( t) ψ e ieat / h ieat / h a a If the particle started out in the state ψ a We solve for c ( t) and c ( t) y demanding that ψ (t) satisfy the tim-dependent a SchrÖdinger equation Hψ = So, we find that ih ψ, t where H = H + H ( t)

c [ H ψ ] e + c [ H ψ ] e + c [ H ψ ] e ieat / h iet / h ieat / h a a a a + c [ H ψ ] e = ih[ c& ψ e + c& ψ e iet / h ieat / h iet / h a a ie ie + c ψ e + c ψ e h h a ieat / h iet / h a a The first two terms on the left cancel the last two terms on the right, and hence [ ψ ] ie a t / h + [ ψ ] ie t / h a a c H e c H e ψ ie a t / h & ψ ie t / h a a = ih[ c& e + c e ] Take the inner product with ψ, and exploit the orthogonality of ψ and ψ a a ]

For short, we define H ψ H ψ ij i j * Note that the Hermiticity of H entails H ji = ( H ji ). M ultiplying - i h e ieat / h through y ( / ), we conclude that i i( E Ea ) t / h c & a = [ cahaa + chae ] h Similarly, the inner product with ψ picks out c& : i( E Ea ) t / c i [ ch cahae h & = + ] h Typically, the diagonal matrix elements of H vanish: H aa = H =

In that case the equations simplify where & i iω t a a c = H e c & h i iω t a c = H e c ω h E E h a (1) Well assume that E E, so ω a

9.1.2 Time-Dependent Perturation Theory So far, everything is exact: we have made no assumption aout the size of the perturation. But if H is "small", we can solve Equation (1) y a process of successive approximations, as follows. Suppose the particle starts out in the lower state: c a Zeroth Order: () = 1, c () =. If there were no perturation at all, they would stay this way foreever c ( t) = 1, c =. ( ) ( ) a To calculate the first-order approximation, we insert these values on the right side of Equation (1)

First Order: dca (1) = ca = 1; dt dc t i iωt (1) i iωt = Hae c = H ( ) a t e dt dt h h Now we insert these expressions on the right to otain the second-order approximation: Second-Order: dc t a i H e iωt H i t ( t ) e ω = dt ( i ) a h a dt h 1 t c t H t e ω H t e dt dt h t (2) iω t i t a ( ) = 1 ( ) ( ) 2 a a

Notice that in my notation Notes: c (2) a ( t) includes the zeroth order term; the second-order correction would e the integral term alone. In principle, we could continue this ritual indefinitely, always inserting the n th order approximation into the right side of Equation (1) and solving for the ( n+ 1) order. Notice that c is modified in every even order, and c in every odd order. a th

9.1.3 Sinusoidal Perturation Suppose the perturation has sinusoidal time dependence: v v H ( r, t) = V ( r ) cos( ωt) so that where a To first order we have H = V cos( ωt), V = ψ a V ψ a a i t iωt dt c ( t) V a cos( ωt ) e h iv t a = - e + e dt 2h i( ω + ω ) t i( ω ω ) t i( ω + ω ) t i( ω ω ) t V a e 1 e 1 = - + 2 h ω + ω ω ω This is the answer, ut it is a little cumersome to work with.

Dropping the first term: We assume ω + ω>> ω ω. so i( ω ω) t/2 Va e c ( t) e e 2h ω ω V = i h a sin[( ω ω) t/2] e ω ω i( ω ω) t/2 i( ω ω) t/2 i( ω ω) t/2 The transition proa ility ------the proa ility that a particle which started out in the state ψ will e found, at time, in the state ψ : V sin [( ω ω) t/2] P ( t) = c ( t) (2) a a 2 2 2 a 2 2 h ( ω ω)

As a function of time, the transition proaility oscillates sinusoidally. P(t) V a h( ω ω ) 2 2π ω ω 4π ω ω ω 6π ω t Fig. 1 Transition proaility as a function of time, for a sinusoidal perturation

The proaility of a transition is greatest when the driving frequency is close to the natural frequency P(t) ( ω 2 πt) ω ( ω + 2 πt) ω Fig. 2 Transition proaility as a function of driving frequency

9.2 Emission and asorption of radiation 9.2.1 Electromagnetic Waves An atom, in the presence of a passing light wave, responds primarily ot the electic component. If the wavelenght is long, we can ignore the spatial variation in the field; the atom is exposed to a sinusoidally oscillating electric field v E = E cos( ωt) kˆ The perturing Hamiltonian is H = qe z cos( ωt) where q is the charge of the electron. Evidently H = E cos( ωt), where = q ψ z ψ a a

Chapter 9 TIME-DEPENDENT PERTURBATION THEORY 9.1 Two-level systems 9.2 Emission and asorption of radiation 9.3 Spontaneous emission

An electromagnetic wave consists of transverse oscillating electric and magnetic fields. Electric field H v E v o An electromagnetic wave u v x Magnetic field

Typically, ψ is an even or odd function of z; in 2 either case z is odd, and integrates to zero. This licenses our usual assumption that the diagonal ψ matrix elements of H vanish. Thus the interaction of light with matter is governed y precisely the kind of oscillatory perturation with Va = E

9.2.2 Asorption, Stimulated Emission, and Spontaneous Emission If an atom starts out in the "lower" state ψ a, and you shine a polarized monochromatic eam of light on it, the proaility of a transition to the "upper" state is given y Equation (2), which takes the form P a ( t) 2 2 E ω ω 2 2 ( ω ω ) sin [( ) t / 2] = h 2 2 E sin [( ω ω) t / 2] = h ( ω ω) ψ Of course, the proaility of a transition down to to the lower level: P a ( t)

Three ways in which light interacts with atoms: (a) Asorption In this process, the atom asors energy E - E a = hω from the electromagnetic field. we say that it has "asored a photon" (a) Asorption

() Stimulated emission If the particle is in the upper state, and you shine light on it, it can make a transition ot the lower state, and in fact the proaility of such a transition is exactly the same as for a transition upward from the lower state. which process, which was first discovered y Einstein, is called stimulated emission. () Stimulated emission

(c) Spontaneous emission If an atom in the excited state makes a transition downward, with the release of a photon ut without any applied electromagnetic field to initiate the process. Thi s process is called spontane ous emission. (c) Spontaneous emission

9.2.3 Incoherent Perturation The energy density in an electromagnetic wave is ε 2 u = E 2 where E is the amplitude of the electric field. So the transition proaility is proportional portional to the energy density of the fields: P a 2u sin [( ω ω) t / 2] ( t) = ε ω ω 2 2 2 2 h ( ) This is for monochromatic perturation, consisting of a single frequency ω.

In many applications the system is exposed to electromagnetic waves at a whole range of frequencies; in that case u ρ( ω) dω, and the net transition proality takes the form of an integral: P ( ) a t = 2u ε 2 2 ρ( ω) 2 2 h ( ω ω) sin [( ω ω) t / 2] dω replace ρ ( ω ) y ρ ( ω ) and take it outside the integral, we get 2 2 sin [( ω ω) t / 2] ρ ω 2 2 ( ω ω) 2 Pa ( t) ( ) dω ε h Changing variales to x ( ω ω) / 2, extending the limits of integration to x - 2 2 = ±, and looking up the define integral sin x x dx = π

we find 2 P t t 2 ( ) ( ) a ρ ω 2 εh This time the transition proaility is proportional ot t. When we hit the system with an incoherent spread of frequencies. The transition rate is Note : now 2 R = π a ρ( ω 2 ) εh the perturing wave is coming in along the x-direction and polarized in the z-directi on. a constant :

But we shall e interested in the case of an atom athed in radiation coming from all direction, and with all possile polarizations. What we 2 need, in place of,is th e average of where q ψ v r ψ and the average is over oth polarizations nˆ ( nˆ ) 2, and over all incident direction. This averaging can e carried out as follows: a

Polarization: For propagation in the z-direction, the two possile polarizations are iˆ and ˆj, so the polarization average is 2 1 2 2 (ˆ n ) p = [( i ˆ ) + ( ˆ j ) ] 2 1 2 1 = ( x + 2 2 2 y) = sin θ 2 2 where θ is the angle etween and the direction of propagation.

Propagation direction: Now lets set polar axis along and integrate over all propagation directions to get the polarization-propagation average: 2 1 1 2 2 ( n ˆ ) = sin θ sinθdθdφ pp 4π 2 = sin 2 2 π 3 θdθ = 4 3 So the transition rate for stimulated emission from state to state a, under the influence of incoherent, unpolarized light incident from all directions, is

So the transition rate for stimulated emission from state to state a, under the influence of incoherent, unpolarized light incident from al l directions, is = π a 2 2 3 ρ( ω ) ε h R a Where is the matrix element of the electric dipole moment etween the two states and ρ( ω ) is the energy density in the fields, per unit frequency, evaluated at ω = ( E E )/ h. a

Chapter 9 TIME-DEPENDENT PERTURBATION THEORY 9.1 Two-level systems 9.2 Emission and asorption of radiation 9.3 Spontaneous emission

9.3 Spontaneous emission 9.3.1 Einsteins A and B coefficients Picture a container of atoms, N of them in the lower state ( ψ ), and N of them in the upper state ( ψ ). a Let A e the spontaneous emission rate, so that the numer of particle leaving the upper state y this process, per unit time, is N A. The transition rate for stimulated emission is proportional to the energy density of the electromagnetic field---call it B ρ ( ω ). The numer of particles leaving the upper state y this mechanism, per unit time, is N B ρ ( ω ). a a a

The asorption rate is likewise proportional to ρ( ω )---call it B ρ( ω ); a The numer of particles per unit joining the upper level is therefore N B ρ( ω a a ). All told, then, dn = N A NB ρ( ω a ) + NaB ρ( ω a ) dt Suppose that these atoms are in thermal equilirium with the amient field, so that the numer of particle in each level is constant.

In that case dn / dt =, ρ ( ω ) = and it follows that A ( N / N ) B B a a a The numer of particles with energy E, in thermal equilirium at temperature the Boltzmann factor, so N N and hence a ρ ( ω ) Ea / kbt e = = Ea / kbt e = hω / k T e B A B e a hω / k T T, is proportional to B B a

But Plancks lackody formula tells us the energy density of thermal radiation: ρ ( ω ) = h π 3 2 3 hω / kt c e B a Comparing the two expressions, we conclude that and ω 3 ω h Ba = Ba and A = 2 3 π c π 2 Ba = 2 3ε h and it follows that the spontaneous emission rateis A = 3 2 ω 3πε hc 3 B a

9.3.2 The Lifetime of an Excited State Suppose, now, that you have a ottle full of atoms, with N ( t), of them in the excited state. As a result of spontaneous emission, this numer w ill d ecrease as time goes on; specifically, in a time interval dt you will lose a fraction of them: Solving for N dn = AN dt ( t), we find N ( t) = N () e At Adt

evidently the numer remaining in the excited state decreases exponentially, with a time constant 1 τ = A We call this the lifetime of the state ---- technically, it i s the time it takes for 1/ e.368 initial value. a ( t) to reach of its This spontaneous emission formula gives the transition rate for other allowed states. ψ ψ N regardless of any

evidently the numer remaining in the excited state decreases exponentially, with a time constant 1 τ = A We call this the lifetime of the state ---- technically, it i s the time it takes for 1/ e.368 initial value. a ( t) to reach of its This spontaneous emission formula gives the transition rate for other allowed states. ψ ψ N regardless of any

Typically, an excited atom has many different decay modes ( that is, ψ can decay to a large numer of different lower-energy states, ψ, ψ, ψ L). In that a1 a2 a3 case the transition rates add, and the net lifetime is τ = 1 A + A + A + L 1 2 3 While h n x n = ( n δ + nδ ) n, n 1 n, n 1 2mω where ω is the natural frequency of the oscillator.

But were talking aout emission, so must e lower than n; for our purpose, then, nh = q ˆ l n, n 1 2m δ ω Evidently transitions occur only to states one step lower on the "ladder", and the frequency of the photon emitted is En E n n + hω n + ω = = h h = ( n n) ω = ω ( 1/ 2) ( 1/ 2) Not surprisingly, the system radiates at the classical oscillator frequency. n hω

But the transition rate is A = 2 2 nq ω 6πε mc th and the lifetime of the n stationary state is τ n 6πε mc = 2 2 nq ω Meanwhile, each radiate photon carries an energy so the power radiated is Ahω: P = 2 2 q ω 6πε mc 3 3 3 ( nhω) hω,

or, since the energy of an oscillator th in the n state is E = ( n + 1) hω, P 2 2 q ω 1 = ( E hω) πε 3 6 mc 2 This is the average power radiated y a quntum oscillator with (initial) energy E. For comparison, lets determine the average power radiated y a classical oscillator with the same energy According to Larmor formula: P = 2 2 q a 6πε c 3

For a harmonic oscillator with amplitude x, x( t) = x cos( ω t), and the acceleration is a = x 2 ω cos( ω t). Averaging over a full cycle, then P = q x 12 ω 2 4 3 πε c 2 2 But the energy of the oscillator is E = (1 / 2) mω x, so x = 2 E / mω,and herenc 2 2 P = q x 6πε ω 2 4 3 mc This is the average power radiated y a classical oscillator with (initial) energy E. E

9.3.3 Selective Rules The calculation of spontaneous emission rates has een reduced to a matter of evaluating matrix elements of the form v ψ r ψ a Suppose we are interested in systems like hydrogen, for which the Hamiltonian is spherically symmetrical. In that case we may specify the states with the usual quantum numers n,l, and m, and the matrix elements are nl m r v nlm

a) Selection rules involving m and m : Consider first the commutations of L with x, y, and z, which we worked out in Chapter 4: [ L,x] = ihy, [ L,y] = ihx, [ L,z] =. z z z From the third of these it follows that = nl m [ L z,z] nlm = nl m ( Lz z zlz ) nlm = h nl m [( m ) z z( m )) nlm h = ( m m) h nl m z nlm z

Conclusion : Either m = m, o r else nl m z nlm =. (1) So unless always zero. m = m, the matrix elements of z are Meanwhile, from the commutator of L with we get z x [ ] = ( z z z nl m L,x nlm nl m L x xl ) nlm = ( m m) h nl m x nlm = ih nl m y nlm

Conclusion : ( m m) n l m x = nlm i n l m y nlm Finally, the commutator of L with y yields n l m [ L, y] nlm = n l m ( L y yl ) nlm z z z = = ( m m) h n l m y nlm ih n l m x Conclusion : and ( m m) n l m y nlm z i n l m nlm x nlm 2 ( ) ( ) m m n l m x nlm = i m m n l m y nlm = = n l m x nlm

and hence either 2 ( m m) = 1, or else n l m x nlm n l m y nlm = = (2) From Equations (1) and (2), we otain the selection rule for m: No transitions occur unless m = ± 1 or This is an easy result ot understand if you rememer that the photon carries spin 1, and henc its value of m is 1,, or -1; conservation of angular momentum requires that the atom give up whatever the photon takes away.

) Selection rules involving l and l : n l m The commutation relation: v v v 2 2 2 2 2 [ L,[ L, r ]] = 2 h ( rl + L r ) As efore, we sandwich this commutation etween and nlm to derive the selection rule: 2 n l m [ L,[ L v z,r ]] nlm 2 v 2 2v = 2 h n l m ( rl L r ) nlm 4 v = 2 h [ l( l + 1) + l ( l + 1)] n l m r nlm 2 2 v 2 v 2 = n l m ( L [ L, r ] [ L, r ] L ) nlm 4 2 v = h [ l ( l + 1) + l( l + 1)] n l m r nlm

Conclusion Either 2[ l( l + 1) + l ( l + 1)] = [ l ( l + 1) l( l + 1)] v or else n l m r nlm = But and so : 2 [ l ( l + 1) l( l + 1)]=( l + l + 1)( l + 1) 2 2 2[ l( l + 1) + l ( l + 1)]=( l + l + 1) + ( l l) 1 2 2 ( l + l + 1) + ( l l) 1= The first factor cannot e zero, so the condition simplifies to l = l ± 1.

Thus we otain the selection rule for l: No transitions occur unless l = ± 1 Again, this result is easy to interpret: The photon carries spin 1, so the rules for addition of angular momentum woul d l = l 1. l = l + 1, l = l or

Allowed decays for the first four Bohr levels in hydrogen