1 Time-Dependent Two-State Systems: Rabi Oscillations

Transcription

1 Advanced kinetics Solution 7 April, 16 1 Time-Dependent Two-State Systems: Rabi Oscillations a In order to show how Ĥintt affects a bound state system in first-order time-dependent perturbation theory we have to write down the expression for Ψt in terms of the approximated time-evolution operator: Ψt = Ût,t Ψt Û1 t,t Ψt = Ût, t Ψt ī h t t Û t,t Ĥintt Ût, t Ψt dt. 1 We can now insert the prototypical bound-state wavefunction Ψ bound = c n φ n where the φ n are the solutions to the time-independent problem. By projecting on the individual φ n we can obtain expressions for time-dependent coefficients c n t = φ n Ψt for compactness we switch to Dirac notation: φ m Ψt n c n t φ m Û1 t,t φ n [ = c n t φ m Ût, t φ n n ] i t φ m Ût,t Ĥintt Ût, t φ n dt t = [ c n t exp i E n t t nm ī t exp i E m h n t t t ] exp i E n t t φ m Ĥintt φ n dt = c m t exp i E m t t exp i E t n t exp t n [c n t īh exp i E m t i E n E m t φ m Ĥintt φ n dt ]. Choosing t =, substituting ω nm = En Em finds and inserting the explicit form for Ĥint one c m t = exp i E [ m t c m t n ic n t V nm t exp iω nm t cosω t dt ], 3 where V mn = ee φ m ˆx φ n. Substituting cosω t = 1 expiω t +exp iω t into the integral in equation 3 becomes 1

2 Advanced kinetics Solution 7 April, 16 t t exp iω nm t cosω t dt = 1 exp iω nm ω t dt + 1 t exp iω nm +ω t dt = 1 i [exp iω nm ω t 1]+ ω nm ω 1 i [exp iω nm +ω t 1]. 4 ω nm +ω We have therefore learned two things. First, we can see directly from equation 3 that the time evolution of c m t is the same as for the field-free evolution with an additional factor depending on the integrals from equation 4. Second, the integrals from equation 4 are strongly peaking for ω nm = ω and ω nm = ω however they are not diverging. We are therefore expecting a strong effect of Ĥ int on all those states for which there is a second state such that ω = ω nm = En Em. It is implicit in this argument that V mn has to be nonzero as well. b We start by writing down the TDSE: i Ψt = ĤtΨt. 5 t With the suggested ansatz Ψt = c 1 tφ 1 + c tφ, equation 5 takes the following form: i t c 1t φ 1 +c t φ = [Ĥ +Ĥintt] c 1 t φ 1 +c t φ. 6 We can now obtain the desired equations by projection on the two time-independent eigenstates. For example i φ 1 ] t c 1t φ 1 +c t φ = φ 1 [Ĥ +Ĥintt c 1 t φ 1 +c t φ. 7 This expression is equivalent to i t c 1t = E 1 c 1 t+c 1 t φ 1 Ĥintt φ 1 +c t φ 1 Ĥintt φ. 8 From symmetry considerations it can be shown that for the specific case of the dipole operator or any other operator that does not contain the totally symmetric element the matrix element φ n ˆx φ n =. This allows us to write i t c 1t = E 1 c 1 t+c tee φ 1 ˆx φ cosω t = E 1 c 1 t+c tv 1 cosω t. 9

3 Advanced kinetics Solution 7 April, 16 c The starting point is to substitute b 1 = c 1 and b = c expiω t into the differential equations from the previous task: i t b 1t = E 1 b 1 t+v 1 cosω b texp iω t i t b texp iω t = E b texp iω t+v 1 cosω b 1 t. 1 Wecannowapplythetimederivativeontheleft-handsideofequation1andsubstitute cosω t = 1 expiω t+exp iω t: i t b 1t = E 1 b 1 t+v exp iω tb t i t b t = E ω b t+v 1 1 expiω t+1b 1 t. 11 By dropping the exponential ω terms, choosing E 1 =, identifying ω = E and defining = V 1 the desired expressions are obtained: i t b 1t = b t 1 i t b t = ω ω b t+ b 1t. 13 d Equations 1 and 13 can be solved by differentiating equation 13 in time and inserting equation 1 in the resulting expression. We obtain = t b t+i t b t+ b t, 14 where we have substituted = ω ω. This type of homogeneous second-order partial differential equations are solved by a linear combination of particular solutions of the form yt = Aexpαt. The coefficient α can be determined by substituting this particular solution into equation 14. We find = α Aexpαt+iαAexpαt+ Aexpαt = α +iα+, 15 with the two solutions of this quadratic equation being α 1/ = i ±

4 Advanced kinetics Solution 7 April, 16 The solution to equation 14 is therefore b t = A 1 expα 1 t+a expα t = exp i [ t A 1 exp i + t+a exp i ] + t 17 = exp i [ t B 1 cos 1 + t+b sin 1 ] + t. 18 In order to do the step from equation 17 to equation 18 we used the Euler identity expiωt = cosωt + isinωt. The new coefficients are then B 1 = A 1 + A and B = A 1 A. ThecoefficientsB 1 andb cannowbedeterminedbythegiveninitialconditions requiring that b 1 t = = 1 and b t = =. From the second condition follows that also b t = = and therefore that B 1 =. Inserting the resulting expression into equation 13, b 1 t is determined: b 1 t = [i ] t b t b t = [ B iκcos exp i t 1 ] κt sin1 κt, 19 where κ = +. With the initial condition b 1 t = = 1, equation 19 reduces to We have therefore proven that and therefore 1 = B + or B = ± +. b t = ± + exp i tsin1 + t 1 b t = [ + sin 1 + t]. e For the discussion of the results let s rewrite equation 4

5 Advanced kinetics Solution 7 April, 16 b t = 1 1 / +1 = 1 1 β +1 [ 1 cos [ 1 cos ] / +1t ] β +1t, 3 where β =. We have found that the two states of our two-state system show an oscillatory behavior in their respective populations with an oscillation period of T =. We have also found that the absolute value of the population transfer between π β +1 the two states depends on the ratio β and not on the absolute value of alone. This is surprising since one would expect intuitively that the population transfer should be dependent on the coupling strength V 1 which is proportional to. Figure 1 shows the behavior of b t for different values of β. 1.8 β = β =.5 β = 1 β = b t time [π/] Figure 1: Illustration of the time-dependent population b t for different values of β. The populations are calculated according to equation 3. 5

6 Advanced kinetics Solution 7 April, 16 Time-Dependent Two-State Systems: The A.C. Stark Effect a The matrix representation of equations 1 and 13 from this document multiplied by yields the desired form i t b = E1 V 1 V 1 E ω b. 4 b Using the definition of we can write E = E 1 +ω +. By inserting this into equation 4 one obtains i t b = E1 V 1 V 1 E 1 + We can then choose the zero energy such that E 1 = i t b = V 1 V 1 b b. 5 and therefore. 6 The eigenvalues of this Hamiltonian are obtained in the usual way of looking for the solutions to the eigenvalue equation detĥ λ1 =. Therefore It follows directly det λ V 1 V 1 λ = λ V 1 =. 7 λ 1/ = ± V1 + = ± V c The result of the previous task shows an interesting behavior. If the ratio γ = V 1 is small the eigenenergies of the time independent system are recovered. However if V 1 becomes comparable to the new eigenenergies are shifted relative to each other. See figure. d The time-independent unitary transformation matrix which diagonalizes 6 is nothing butthematrixrepresentationoftheprojectionoperator ˆP ontheneweigenbasis{φ 1,φ }. We can therefore write which is equivalent to i t i t ˆP a1 a b = ˆP V 1 = V1 + V 1 V1 + ˆP 1 ˆP, 9 b a1 a. 3 6

7 Advanced kinetics Solution 7 April, 16 positive detuning > negative detuning < E 1 +ω E E energy E 1 +ω energy E1 E1 1 V / for = const 1 1 V / for = const 1 Figure : Illustration of the energy shift due to the presence of the oscillating electric field. The time dependent solution given in the new basis is Ψt = a 1 exp = a 1 exp t φ 1 +a exp + i E 1 i t i E t φ φ 1 +a exp i + t φ. 31 e In order to solve this task we have to project Ψt = = b 1 φ 1 onto the new basis functions: a 1 = φ 1 Ψt = = cosθ φ 1 φ 1 +sinθ φ φ 1 a = φ Ψt = = sinθ φ 1 φ 1 +cosθ φ φ. 3 It follows that a 1 = cosθ and a = sinθ. The wavefunction is hence Ψt = cosθexp i + t φ 1 sinθexp i + t φ. 33 If we want to calculate b t we have to do the projection on φ, which leads to 7

8 Advanced kinetics Solution 7 April, 16 φ Ψt = cosθexp i + t φ φ 1 sinθ exp i + t φ φ = cosθ sinθ exp i + t sinθ cosθ exp i + t = isinθcosθsin + t. 34 It follows that b t = 4sinΘcosΘ sin + t = + sin + t, 35 which is what we set out to prove. What we have therefore shown is that the quantity b t calculated in task and b t calculated in task 1 are identical. If both formalisms are equivalent for the description of the quantum system the observables given by expectation values of a corresponding operator should remain unchanged. It is not per se clear how such an observable is related to the quantity b t. Note: Consider the following experiment. We study an atom or molecule with two states and two coinciding laser fields. The first laser field is a strong cw field with a small detuning such that the two bound states shift. The second laser field is very weak, pulsed and has a large detuning allowing no coupling with the unperturbed bound states. In the A.C. Stark picture it is clear that for a particular strength of the cw field the bound states shift into resonance with the pulsed field allowing for a population transfer. It is not at all straightforward to describe the same experiment in the Rabi picture. 8