MIT OpenCourseWarehttp://ocw.mit.edu 5.74 Introductory Quantum Mechanics II Spring 2009 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms.3-1 Andrei Tokmakoff, MIT Department of Chemistry, 3/2/2007 3. IRREVERSIBLE RELAXATION1 It may not seem clear how irreversible behavior arises from the deterministic TDSE, although this is a hallmark of all chemical systems. To show how this comes about, we will describe the relaxation of an initially prepared state as a result of coupling to a continuum. We will show that first-order perturbation theory for transfer to a continuum leads to irreversible transfer—an exponential decay—when you include the depletion of the initial state. The Golden Rule gives the probability of transfer to a continuum as: 2 ρ(E E k = A )wkA =∂PkA = 2πVkA∂t =PkA = w tt A (− ) (3.1)k 0 PAA =−1 PkA The probability of being observed in k varies linearly in time. This will clearly only work for short times, which is no surprise since we said for first-order P.T. ()≈k0k ().bt bWhat long-time behavior do we expect? A time-independent rate is also expected for exponential relaxation. In fact, for exponential relaxation out of a state A , the short time behavior looks just like the first order result: P t P − 1 wt"AA ()= AA (0exp )(wkAt) (3.2)=− kA + So we might believe that wkA represents the tangent to the relaxation behavior att = 0. ∂PkAwkA= (3.3)∂tt0 The problem we had previously was we don’t account for depletion of initial state. From an exact solution to the two-level problem, we saw that probability oscillates sinusoidally between the two states with a frequency given by the coupling:3-2 2 2Δ+VΩ= kA R = But we don’t have a two-state system. Rather, we are relaxing to a continuum. We might imagine that coupling to a continuous distribution of states may in fact lead to exponential relaxation, if destructive interferences exist between oscillations at many frequencies representing exchange of amplitude between the intital state and continuum states. COUPLING TO CONTINUUM When we look at the long-time probability amplitude of the initial state (including depletion and feedback), we will find that we get exponential decay. The decay of the initial state is irreversible because there is feedback with a distribution of destructively interfering phases. Let’s look at transitions to a continuum of states {k} from an initial state A under constant perturbation. These form a complete set; so for ()= 0+ with Hn EHt H Vt() 0 = n .n1 =∑nn= AA+∑kk (3.4) n k initial continuum As we go on, you will see that we can identify A with the “system” and {k}with the “bath” when we partition HH = + H . We want a more accurate description of the occupation of the 0 S B initial and continuum states, for which we will use the interaction picture expansion coefficients bt()= kUt(,t0 )A (3.5)k I The exact solution to UIwas: Ut,t01 τ I τUτ,t0 (3.6)I( ) =−i∫tdV () ( ) I =t03-3 For first-order perturbation theory, we set the final term in this equation UI (τ,t0 )→1. Here we keep it as is. tbt kA − = i ∫dτ kV U τ,0 A (3.7)k ()= I () ( τ It )t0 nn =1, and recognizing k ≠ l,Inserting the projection operator ∑ n bt()=−i∑∫t τ i kn V b kn () (3.8)deωτ τk n =t0n Note, here Vkn is not a function of time. Equation (3.8) expresses the occupation of state k in terms of the full history of the system from t0 → t with amplitude flowing back and forth between the states n. Equation (3.8) is just the integral form of the coupled differential equations, that we used before: i=∂∂ btk =∑ n eiωkntVbt kn n () (3.9) These exact forms allow for feedback between all the states, in which the amplitudes bk depend on all other states. Now let’s make some simplifying assumptions. For transitions into the continuum, let’s assume that transitions in the continuum only occur from the initial state. That is, there are no interactions between the states of the continuum: k′kV =0 . This can be rationalized by thinking of this problem as a discrete set of states interacting with a continuum of normal modes. Moreover we will assume that the coupling of the initial to continuum states is a constant for all states k: A Vk = A Vk′ = constant . So since you only feed from A into k , we can remove the summation in (3.8) and express the complex amplitude of a state within the continuum as it τ ωτ bbk =−=VkA ∫t0 dei kAA ()τ (3.10)3-4 We want to calculate the rate of leaving A , including feeding from continuum back into initial state. From eq. (3.9) we can separate terms involving the continuum and the initial state: i=∂ bA =∑eiωAkt VAk bk + VAA bA (3.11)∂tk ≠A Now substituting (3.10) into (3.11), and setting t0 =0: ∂b 1 t kAA =− 2 ∑VkA 2 ∫bA ()τ eiωτ(−t) dτ− =iVAA bA ()t (3.12)0∂t = k ≠A This is an integro-differential equation that describes how the time-development of bA depends on entire history of the system. Note we have two time variables for the two propagation routes: τ: A → k (3.13)t : k → AThe next assumption is that bA varies slowly relative to ωkA , so we can remove it from integral. This is effectively a weak coupling statement: =ωkA >> VkA . b is a function of time, but since it is in the interaction picture it evolves slowly compared to the ωkA oscillations in the integral. ∂b ⎡ 1 kAA =bA ⎢− 2 ∑VkA 2 ∫teiωτ(−t) dτ− =iVAA ⎤⎥ (3.14)∂t ⎣ = k ≠A 0 ⎦ 1Now, we want the long time evolution of b, for times t >> , we will investigate the integration ωkA limit t →∞. Complex integration of (3.14): Defining t′=τ−t dt ′= dτ kA kA∫teiωτ(−t) dτ =∫teiω t′ dt ′ (3.15)0 0 The integral lim Te ω′dt′ is purely oscillatory and not well behaved. TheT →∞∫0 +it strategy to solve this is to integrate:3-5 lim ∞ (+)t′ dt′=lim 1 ε→0+∫0 eiωε ε→0+ iωε+ =lim ⎛ε +i ω⎞ (3.16)+⎜⎝ω ε2 +2 ωε 2 +2 ⎟⎠ε→0 () 1⇒+πδ ω −iPω In the final term we have used the Cauchy Principle Part: ⎛⎞ ⎧P 1 = 1 xx ≠ 0 (3.17)⎜⎟ ⎨ x 0 x =0⎝⎠ ⎩ This leads to ⎡ ⎤ ⎢ 2 ⎞⎥∂b πA =bA ⎢−2 ∑V δω( ) −⎜ V +P2 kA i ⎛⎜AA ∑VkA ⎟∂t ⎢= k ≠A kA= k ≠A Ek −EA ⎟⎥⎥ (3.18) ⎢ ⎝⎠⎥ ⎣ term 1 term 2 ⎦ Term 1 is just