Question

4. If two potential wells are separated by a repulsive delta-function barrier, the two lowest energy states can be shown to be symmetric and antisymmetric superpositions of the ground states IL) and ㈣ of the left and right wells, IS)-흖 (IL) + IR)) and A Ξ Suppose that HIS) = EsIS) and HIA) EAIA) If at t 0 the particle is in state |L), solve for the state as a function of time. Calculate KLV(t))12. Comment on the meaning of this result.

Answer 1

Since dependent wave function Vertical-bar psi (t) > = 1s/sqw 2 e6 - i Ec/k^t + 1A > /squareroot 2 e^- t E_A/k^t But Vertical-bar L > = 15 > + 1A > /squareroot 2 Hence, < : Vertical-bar psi (t) > = e^- i E_s/k^t/(squareroot 20^2 + e6 - i E_A/k^t/(squareroot 20^2 Now, let E_s/k^t = implies_1 and E_A/t^t - emptyset_2 then Vertical-bar < L Vertical-bar psi (t) > Vertical-bar ^2 = Vertical-bar e^- i emptyset_1 + e^- i emptyset_2/2)^2 = 1/4 Vertical-bar e^- i(emptyset_1 + emptyset_2/2[e^i(emptyset_2 - emptyset_2)/2 + e6 - (emptyset_2 - emptyset_2)/2] Vertical-bar ^2 = 1/4 Vertical-bar 2 cos(emptyset_2 - emptyset_1)/2 Vertical-bar ^2 = cos^2 (emptyset_2 emptyset_1)/2 = cos^2(E_A - E_s)/h t Therefore the probability as particle is in one side of the well oscillates between 0 and 1 with period 2 pi/E_A = E_s/hwithstroke = 2 pi k/E_A - E_s