Inverse problem for the Landau - Zener effect

– We consider the inverse Landau-Zener problem which consists in finding the energy-sweep functions W (t) ≡ ε1(t) − ε2(t) resulting in the required time dependences of the level populations for a two-level system crossing the resonance one or more times during the sweep. We find sweep functions of particular forms that let manipulate the system in a required way, including complete switching from the state 1 to the state 2 and preparing the system at the exact ground and excited states at resonance. The Landau-Zener (LZ) effect [1, 2, 3], which consists in the quantum-mechanical transition between the two time-dependent levels of a two-level system (states 1 and 2) caused by crossing the resonance, is a well known phenomenon in many areas of physics. Well known applications of the LZ effect are those to molecular collision and dissociation[4, 5]. Recently, LZ effect has been used to measure the tunnel splitting in molecular magnets whereby detecting topological interference effects [6] which were predicted theoretically [7, 8, 9]. If the system was in state 1 before crossing the resonance, then, for a constant sweep rate v, the probability to stay in this state after crossing the resonance is given by P = exp − π∆ 2 2¯ hv , (1) where v ≡ | ˙ W (t)|, W (t) ≡ ε 1 (t)−ε 2 (t), ε 1,2 (t) = ψ 1,2 | ˆ H|ψ 1,2 , ∆ ≡ 2ψ 1 | ˆ H|ψ 2 , W (t) satisfies W (±∞) = ±∞ and ∆ is the tunnel level splitting. For fast energy sweep rates v the system spends too little time in the vicinity of the resonance so that the tunneling matrix element ψ 1 | ˆ H|ψ 2 cannot bring the system into the state 2 (say, onto the other side of a potential barrier), and the system remains in the initial state 1. For slow sweep rates the probability to remain in the initial state becomes exponentially small, and the system travels into the state 2 remaining on the lower exact energy term ε − (t) of ε ± (t) = 1 2 ε 2 (t) + ε 1 (t) ± W 2 (t) + ∆ 2. (2) The corresponding adiabatic time dependence of the probability to be in state 1 is given by |c 1 (t)| 2 = |ψ 1 |ψ − (t)| 2 = 1 2 1 …