Phys402, II-Semester 2017/18, Assignment 4, solution Atom-light coupling with magnetic field: ^ Consider a hydrogenic atom in a homogeneous magnetic field B = B0k within the an- omalous Zeeman regime. We now use a mono-chromatic laser to couple the electronic ground state j g i = j 1s1=2[mj = −1=2] i to an excited state j e i = j 5p1=2[mj = 1=2] i. The ^ quantization axis is k. Let the laser frequency be !0 < !eg, where !eg = (Ee − Eg)=~, defined in the absence of magnetic field. Assume in total you have three control knobs, (i) adjusting the magnetic field strength B0, (ii) the laser frequency !0 and (iii) the laser output power P . ^ (a) [1 point] Write the effective Hamiltonian Heff for these two levels in the presence of magnetic field and laser light in the dipole and rotating wave approximation. Define and discuss all variables. [Hint: You can just add the two features together] In the end, use the ^ fact that we can arbitrarily define our zero of energy to enforce h g jHeffj g i = 0, this will make it easier in the later parts to use results from the lecture. Solution: The effective Hamiltonian after adjusting the zero of energy is Ω ^ 0 2 Heff = Ω ; (1) 2 −∆ + gsµBB0 this combines Eq. (3.50) from the lecture with Eq. (2.62) for the anomalous Zeeman shifts. Ω and ∆ are defined as in section 3.3. of the lecture. (b) [2 points] Verify the solution (3.51) of the Rabi problem (3.50) given in the lecture. (Thus here B0 = 0) [You may use mathematica. If you do please hand in a printout of your script with hand-written annotation as to what exactly each element of each command is doing. If you solve it manually, include all intermediate steps] Solution: See script \assignment4 rabi.nb\ p (c) [2 points] Let us now chose !eg − !0 = 3Ω, where Ω is the Rabi frequency of the coupling with B0 = 0. What is the maximum probability pe at any time for the atom to be in state j e i, if it is initially in j g i? At what time is this probability first reached? From whatever setting we now have, which adjustments (list three different options), can we make using our three control knobs to make the excited state probability reach pe = 1 or pe ≈ 1? Having made either of these adjustments, what can we do to reach pe ≈ 1 at an earlier time? 1 2 1 Solution: We have Ωeff = 2Ω, hence pb = 4 sin (Ωt). Thus the maximum pe = 4 , which is reached first at t = π=2=Ω. If we want to get pe = 1, we either increase the laser frequency such that ∆ = 0, or we adjust the magnetic field such that gsµBB0 = ∆. Then the effective detuning ∆eff = −∆ + gsµBB0 = 0. We could also massively increase the Rabi frequency Ω such that Ω ∆ [at this point we should however not fix ∆ = 2Ω], this yields pb ≈ 1 1 but not quite. For this we need to increase the laser power. From whatever setting we now have, to reach pe = 1 faster, we have to increase Ω through the light intensity and hence increase the laser power (further). (d) [3 points] Assume now a resonant laser !0 = !eg and the atom initially in j g i. Consider the two sequences of using the control knobs: (a) First 0 < t ≤ T3π=2: Ω = Ω0, B0 = 0, where T3π=2 = 3π=2=Ω0. Then T3π=2 < t ≤ T3π=2 + τ: Ω = 0, laser off for a duration τ, but magnetic field on with B0 = π~=[2gsµBτ]. Finally for T3π=2 + τ < t ≤ T3π=2 + τ + Tπ=2 with Tπ=2 = π=2=Ω0 again: Ω = Ω0, B0 = 0. For sequence (b) all is the same, except B0 = 0 always. For both sequences write the quantum state at the end of each given time interval. p Solution: Sequence (a): We write φini ! φfin for each segment. (i) j g i ! −(j g i+ij e i)= 2 by inserting t = T3π=2 into thep solution of the Rabi problem Eq. (3.51) [Note: Ωpeff = Ω0 since ∆ = 0]. (ii) −(j g i+ij e i)= 2 ! −(exp [−iEgτ=~]j g i+i exp [−iEeτ=~]j e i)= 2 from the usual time-dependent solution for a time-independent Hamiltonian, Eq. (1.47), since the Hamiltonian is diagonal in segment (ii). We can read off Eg = 0 and Ee = gsµBB0 from Eq. (1). Inserting also the value ofpB0 we obtain exp[−iEeτ=~] = exp[−iπ=2] = −ipand hence the final state: −(j g i+j e i)= 2. (iii) Now we realize that φini = −(j g i+j e i)= 2 is ^ an eigenstate of the Hamiltonian since in the last part Heff j φini i = Eφini, with E = Ω=2. The only timep evolution for an eigenstate is φfin =p exp [−iEt=~]φini, hence in this last part −(j g i + j e i)= 2 ! − exp [−iπ=4](j g i + j e i)= 2. Sequence (b): Since interval (ii) does not do anything, the sequence is equivalent to having Ω = Ω0, B0 = 0 for time T2π = 2π=Ω0. Inserting this into Eq. (3.51) gives −| g i as the final state. (All other states that have been asked for can be read-off from the final state of [a(i)]. (e) [2 points] Draw the states and the evolution into one Bloch sphere for (a) and a separate one for (b). Describe what your results could be useful for [8 lines]. Solution: The final atomic state depends on whether or not a magnetic field was present in the second step (even though first and last step are the same). Hence if we can measure the atomic state at the end of the sequence, we can learn about the presence (and value) of the magnetic field. 2.