A box containing a particle is divided into right and left compartments by a partition. If the particle is known to be on the left side with certainty, we call the state |L >; if on the right, we call the state |R >. Of all possible |L > and |R > states, we consider only the ones with the lowest energy, thus we have a two-state problem.
Assume that the box is symmetric and shift the zero of energy so that < L|H|L >=< R|H|R >= 0. The Hamiltonian H does not vanish, however, because the particle can tunnel through the partition, as allowed by the coupling
H = iη(|L >< R| − |R >< L|) (1)
where η is a real number with dimensions of energy. Furthermore: < L|R >= 0.
Suppose at time t=0 the state vector is given by,
|Ψ(0) >= a|L > +b|R > (2)
a) Find |Ψ(t) > by applying the time evolution operator and express your answer in the form
|Ψ(t) >= fL(t)|L > +fR(t)|R > (3)
Determine the functions fL(t) and f fR(t).
b) Suppose that at time t=0 the particle is on the right side with certainty. What is the probability for observing the particle on the left side as a function of time?
c) What is wrong with this Hamiltonian:
H’ = iη(|L >< R| + |R >< L|)
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