Notes for Lectures on Coherent Excitation I. RATE EQUATIONS: 2-LEVEL SYSTEM Consider the 2-level system: A A 21 FIG. 1: Schematic of two-level system, showing excitation (left) and decay(right). If we denote the relative populations of levels 1 and 2 by n1 and n2, respectively, then the equations for the rates of change of these populations are given by: n_ 1 = n2A + n2BI ¡ n1BI (1a) n_ 2 = ¡n2A ¡ n2BI + n1BI (1b) n1 + n2 = 1: (1c) The constants A and B are the familiar Einstein A and B coe±cients for spontaneous and ¼2c2 stimulated emission/absorption. Here, Bab = ~!3¢º Aab, (for g1 = g2 = 1), and I is the intensity of light (in, say, mW/cm2). In order to allow direct use of the more common laboratory de¯nition of I, the linewidth of the light, ¢º, has been included in the de¯nition of B. The ¯rst two equations are clearly redundant. Combining the second two equations we get n_ 2 = ¡n2(A + 2BI) + BI: (2) BI First, consider the steady-state solution of (2). That is, we setn _ 2 = 0. Then, n2 = A+2BI . Note that if BI À A, then the steady-state excited population ! 1=2. Now, back to equation (2): Try a solution like n2(t) = f(t) exp[¡(A + 2BI)t]. Then, h i _ ¡(A+2BI)t n_ 2 = f ¡ f(A + 2BI) e : (3) Setting the right-hand side (RHS) of this expression equal to the RHS of equation (2), and applying the boundary condition n2(t = 0) = 0, leads to BI ¡ ¢ n (t) = 1 ¡ e¡(A+2BI)t : (4) 2 A + 2BI 2 BI Note that as t ! 1, n2 ! A+2BI , as in the steady-state case. HW1a: Set up the rate equations for the 3-level \ladder" system shown below. Here, A31 = 0, and for all levels, Bab = Bba. That is, g1 = g2 = g3 = 1. A 32 2 1 A 21 HW1b: Solve the equations analytically for steady-state conditions. What is the greatest fractional population that can be put into j3 >? Here, A31 = 0, and for all levels, Bab = Bba. That is, g1 = g2 = g3 = 1. HW1c: Numerically solve the coupled di®erential equations. Assume various values for 2 6 I1 and I2 (of order of a few mW/cm ). For the purposes of this problem, set A21 = 37:6£10 ¡1 6 ¡1 sec , and A32 = 12:5 £ 10 sec . Also assume ¸1 = 780 nm, and ¸2 = 1:5¹m. (These are parameters consistent with the 5s, 5p, and 4d levels of Rb.) II. COHERENCE: 2-LEVEL SYSTEM: THE RABI PROBLEM It is natural to ask \can we ever do better than putting only 1=2 of the population into the excited state of a 2-level system? If we had a 3-level system, could we ever put more than 1=3 of the population into the highest level?" Clearly, from our study of these systems based on rate equations, the answer is \no!" However, as we shall soon see, in these studies we left out a very important resource available to us: the possibility of making use of the coherent property of light. From here on, we consider the use of coherent light in exciting atoms or molecules. Consider again the 2-level system above but now, for the moment, set A = 0. The most appropriate vehicle for studying excitation of this system by coherent light is the time- @ª 0 0 ^ 0 dependent SchrÄodingerequation. That is, i~ @t = H ª, H = H0 + V (t), where H , H0, and ^ V are Hermitian. H0 is the Hamiltonian of the atomic (or molecular) system in the absence of external ¯elds, and V^ represents the time-dependent external radiation ¯eld, treated here as a classical electromagnetic ¯eld. Note that, in general, near resonant, narrow linewidth 3 radiation can not be treated as a perturbation. P ¡i»n(t) 0 Set ª(t) = n cn(t)Ãne , where Ãn satisfy the time-independent SE: H0Ãn = EnÃn. The »n(t) are time-dependent phases. Their values are arbitrary since, being just phases, they have no e®ect on any observable. Then, @ª X h i = à c_ ¡ i»_ c e¡i»n ; @t n n n n n and X ³ ´ 0 ^ ¡i»n H ª = H0 + V cnÃne n Now, V^ acting on à re-distributes the probability. That is, X ^ VÃn = Ã1V1n + Ã2V2n + ¢ ¢ ¢ = ÃmVmn: m ¤ Or, upon multiplying from the left by Ãq and integrating over space, Vqn = hÃqjV jÃni ´ hqjV jni , where we have made use of the orthonormality of Ã. Then, H0ª = P P 0 ¡i»n n [cnEnÃn + cn m ÃmVmn] e . Putting everything together gives X h i X X X _ ¡i»n 0 ¡i»n ¡i»n i~ Ãn c_n ¡ i»ncn e = cnEnÃne + cnÃmVmne : n n n m ¤ Multiplying through by Ãl , and integrating over space gives: Z X h i Z X Z X X ¤ _ ¡i»n 0 ¤ ¡i»n ¤ ¡i»n i~ Ãl Ãn c_n ¡ i»ncn e d~r = cnEnÃl Ãne d~r+ cnÃl ÃmVmne d~r: n n n m ³ ´ _ ¡i»l 0 ¡i»l Making use of the orthonormality of à yields i~ c_l ¡ i»lcl e = clEl e + P P ¡i»n _ 0 ¡i(»n¡»l) n cnVlne , or, i~c_l = ¡~»lcl + El cl + n cnVlne . Finally, " # ³ ´ X 0 _ ¡i(»n¡»l) ~c_l = ¡i El ¡ ~»l cl + cnVlne : (5) n All of this is exact so far. Now, for the ¯rst approximation, we truncate the sum in Eq. (5) to just the number of levels in the system, here, 2. That is, we neglect the population of states far from resonance. Under this approximation we end up with: h³ ´ i 0 _ ¡i(»2¡»1) ~c_1 = ¡i E1 ¡ ~»1 + V11 c1 + V12c2e h³ ´ i 0 _ ¡i(»2¡»1) ~c_2 = ¡i E2 ¡ ~»2 + V22 c2 + V21c1e ; 4 or, more compactly, as: 0 1 0 _ ¡i(»2¡»1) E1 + V11 ¡ ~»1 V12e ~c_ = ¡i @ A c; (6) ¤ i(»2¡»1) 0 _ V12e E2 + V22 ¡ ~»2 with 0 1 c1 c ´ @ A : c2 ^ ~ ~ i!t ¡i!t Now, in the dipole approximation, V = ¡eE (~r; t) ¢ ~r, E ="E ^ 0 (e + e ) =2. ~ e i!t ¡i!t So, Vmn = ¡eE ¢ hÃmj~rjÃni = ¡ 2 E0 (e + e ) hmjrjni. Then, de¯ning ­ ´ ¡eE0 ~­ i!t ¡i!t ~ h1jrj2i, we get V12 = 2 (e + e ). Note that because E0 can, in general, have a phase term, ­ is, in general, a complex quantity. In many cases we can choose the phase of our electric ¯eld to be 0, in which case, ­ will be pure real. Equation (6) then becomes: 0 1 ~­ ¡ ¢ E ¡ ~»_ e¡i(»2¡»1¡!t) + e¡i(»2¡»1+!t) B 1 1 2 C ~c_ = ¡i @ A ; (7) ~­¤ ¡ ¢ ei(»2¡»1¡!t) + ei(»2¡»1+!t) E ¡ ~»_ 2 2 2 0 0 where E1 ´ E1 + V11 and E2 ´ E2 + V22. At this point, we could just solve the coupled di®erential equations represented by Eq. (7). First, however, we take advantage of the arbitrariness in the values of the time-dependent phases, »n; we assign values to them in such a way as to render Eq. (7) into as simple a form _ _ as possible. That is, we set »2 ¡ »1 = !t ) »2 ¡ »1 = !. We now make our third approximation. We say that if a term oscillates at a frequency signi¯cantly higher than the optical frequency, ! , we can ignore that term since it averages to 0. This is called the rotating wave approximation or RWA. Applying our de¯nitions of phase, and the RWA to Eq. (7) (whereupon we set terms exp(§2i!) = 0) we obtain: 0 1 ¡i 2~¢1 ~­ ~c_ = @ A c; (8) 2 ¤ ~­ 2~¢2 _ _ where we have de¯ned ~¢1 ´ E1 ¡ ~»1 and ~¢2 ´ E2 ¡ ~»2 = E2 ¡ E1 + ~¢1 ¡ ~!. We can then set the 0-point of our potential energy scale by setting ¢1 = 0. Putting all of this together with Eq. (8) we ¯nally obtain 0 1 ¡i 0 ­ c_ = @ A c; (9) 2 ¤ ­ 2¢2 5 with ~¢2 = E2 ¡ E1 ¡ ~!. (Throughout these notes I have chosen to use the unfortunate convention that red detuning is positive, and blue detuning is negative. The physical picture of this system is: { That is, ~¢2 is the energy by which the laser is detuned from resonance. Because ­ is related to the electric ¯eld amplitude it is, in that sense, related to the laser intensity. However, since it is also related to the dipole matrix element coupling two states, the value of ­ for two di®erent transitions are in general di®erent, even if the two laser intensities are identical. Without going through the (trivial) derivation, ­ is related to more easily obtained system parameters by: r q 3¸3Iγ ­ = , or ­[MHz] = 1:55 £ 10¡7 ¸3[nm]I[mW/cm2]γ[s¡1]; 2¼hc where γ is the spontaneous decay rate from the upper state. For example, in the case of the rubidium 5s-5p transition, ¸ = 780 nm, and 1/γ = 26:6 nsec. Then, for a laser intensity of 1.0 mW/cm2, the Rabi frequency is 21 MHz. Note that ­ is an angular frequency. Equation (9) is pretty simple, so let's try to solve it. Breaking up the matrix equation into its constituent parts, then taking the derivative of each of these we obtain: ¡i ¡i c_ = ­c ) cÄ = c_ (10a) 1 2 2 1 2 1 ¡i ¡i c_ = (­¤c + 2¢c ) ) cÄ = (­¤c_ + 2¢_c ) : (10b) 2 2 1 2 2 2 1 2 Combining these equations gives the single, second order di®erential equation: ­2 cÄ + i¢_c + c = 0: 2 2 4 2 With the boundary condition c2(0) = 0, we can readily show that this has the solution 0 p ­ ­ t ¡i ¢t 0 c = ¡i sin e 2 , ­ ´ ­2 + ¢2: (11) 2 ­0 2 Now, the probability of ¯nding the atom in state j2i is simply µ ¶ ­ 2 ­0t ­2 P (t) = c¤c = sin2 = (1 ¡ cos ­0t) : (12) 2 2 2 ­0 2 2 (­2 + ¢2) 6 = 0 1.0 = = 3 0.8 0.6 2 P 0.4 0.2 0.0 0 5 10 15 20 -1 t( ) Plots of P2(t) are shown for a few di®erent values of ¢.