MIT Department of Chemistry 5.74, Spring 2004: Introductory Quantum Mechanics II p. 28 Instructor: Prof. Andrei Tokmakoff THE RELATIONSHIP BETWEEN U(t,t0) AND cn(t) For a time-dependent Hamiltonian, we can often partition H = H0 + Vt ( ) H0 : time-independent; Vt(): time-dependent potential. We know the eigenkets and eigenvalues of H0: n n We describe the initial state of the system (t = t0 ) as a superposition of these eigenstates: ψ(t0 ) =EnH0 =∑c nn n t ψFor longer times , we would like to describe the evolution of in terms of an expansion in these kets: ψ(t) =∑cn (tn )n The expansion coefficients ck ()are given bytck ()=tkψt( ) = kU t,t0( )ψt0( ) Alternatively we can express the expansion coefficients in terms of the interaction picture wavefunctions ( )= kψ( ) bk tIt(This notation follows Cohen-Tannoudji.) Notice ct k ψ(t )= k U UI ψ(t0 )k ()= 0 it=e−ωk kUI ψ()t0 it=e−ωkbtk () bk ()2 =ttso that ck ()2. Also, bk 0 0()= ck ( ). It is easy to calculate bk (t) and then add in the extra oscillatory term at the end.p. 29 Now, starting with ∂ψIi = VI ψI∂t we can derive an equation of motion for bk ∂bk ψI () ψI ()=∑bn ni = k V UII t0 t0∂t n =1 =∑kVI n ninserting ∑nn UI ψI (t0 ) nn kVI n b (t )=∑ n n it −ωnkt bn ()i ∂bk =∑Vkn ()e t ∂tn This equation is an exact solution. It is a set of coupled differential equations that describe how probability amplitude moves through eigenstates due to a time-dependent potential. Except in simple cases, these equations can’t be solved analytically, but it’s often straightforward to integrate numerically. Exact Solution: Resonant Driving of Two-level System Let’s describe what happens when you drive a two-level system with an oscillating potential. (tVt )= V cos ω = Vf (t ) This is what you expect for an electromagnetic field interacting with charged particles: dipole transitions. The electric field is Et )=Ecos ωt(0 For a particle with charge q in a field E , the force on the particle is Fq E = which is the gradient of the potentialp. 30 ∂VF =− =qE ⇒ V =−qE x x x x∂x qx is just the x component of the dipole moment µ. So matrix elements in V look like: k| V(t) | =−qE k |x| cos ωtx More generally, V =−E ⋅µ. So, Vt t()=V cos ω=− E0 ⋅µcos ωt . ()=V cos ω=− E0 ⋅µ cos ωtVt k tk k We will now couple our two states k + with the oscillating field. Let’s ask if the system starts in what is the probability of finding it in k at time t ? The system of differential equations that describe this situation are: i ∂bk ()=∑bn () t −ωnktt t Vkn ()e ∂tn −ωnkt 1 −ωi it it=∑bt Vkne ×2 (e +e+ω)n () n kib =1 k2 b V ( )ki te ω−ω ( )ki te ω+ω + + 1 k2 b kk V ite ω ite−ω + (1) and (2) = ib b V 1 2 = it it 1 k k2 b Ve eω −ω + + ( )ki te ω−ω ( )ki te ω +ω + (3) and (4) = or ( )ki te−ω +ω ( )ki te−ω −ω + We can drop (2) and (3). For our case, Vii=0 . We also make the secular approximation (rotating wave approximation) in which the k nonresonant terms are dropped. When ωk ≈ω, terms like e±iωt or ei(ω+ω)t oscillate very rapidly and so don’t contribute much to change of cn .p. 31 So we have: −ikbk = 2 b Vk ei(ω−ω)t (1) −ii(kb = 2 bk Vke−ω −ω)t (2) Note that the coefficients are oscillating out of phase with one another. Now if we differentiate (1): −i i ( i(ω−ω)tkbk = 2 b Vk ei(ω−ω)t + ω −ω)b Ve k (3) k k Rewrite (1): b = 2i bk e−ω −ω)t i(k (4)Vk and substitute (4) and (2) into (3), we get linear second order equation for bk . 2Vk i ( bk −ω −ω)b + 2 bk = 0k k 4 This is just the second order differential equation for a damped harmonic oscillator: ax + bx + cx = 0 x =e−(b2a t )(A cos µ+ B sin µt )µ =1 4ac −b t 2a 2 12 With a little more work, we find (remember bk(0)=0 and b (0)=1) 2 2 Vk= t = sin2 Ω tbk ()Pk 2 2r + 2 (ω −ω)Vk k 2 + 2 (ω −ω)2 12 R Vk kΩ=21 2P1 =−bkp. 32 Amplitude oscillates back and forth between the two states at a frequency dictated by the coupling. Resonance: To get transfer of probability amplitude you need the driving field to be at the same frequency as the energy splitting. Note a result we will return to later: Electric fields couple states, creating coherences! On resonance, you always drive probability amplitude entirely from one state to another. 1 Vkl 0.5Pkl 0 large detuning 0 tkl/Vπ klω−ω= on resonance Efficiency of driving between and k states drops off with detuning. Pmax kl2V /
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