More Electron Traps — Question 9

Question 9

Detuning two traps suppresses the maximum tunneling transfer probability

For two localized states use $H=(\Delta/2)(|L\rangle\langle L|-|R\rangle\langle R|)-J(|L\rangle\langle R|+|R\rangle\langle L|)$, with $J>0$. If the electron starts on the left, $P_R(t)=[J^2/\Omega^2]\sin^2(\Omega t/\hbar)$, where $\Omega=\sqrt{J^2+\Delta^2/4}$.

1. Find the two energies and the maximum transfer probability.

2. Find the smallest magnitude of detuning that makes the maximum probability no greater than .200.

Question 9 - Solution

Detuning two traps suppresses the maximum tunneling transfer probability

(a) Method and calculation

$$\det(H-EI)=E^2-J^2-\Delta^2/4=0\Rightarrow\boxed{E_\pm=\pm\Omega}.$$$$\boxed{P_{R,\max}=\frac{4J^2}{4J^2+\Delta^2}}.$$

(b) Interpretation and further calculation

$$\frac{4J^2}{4J^2+\Delta^2}\le.200\Rightarrow\Delta^2\ge16J^2.$$$$\boxed{|\Delta|_{\min}=4J}.$$

Physical check

Increasing detuning raises the oscillation frequency while reducing transfer contrast. Faster oscillation therefore does not imply more complete transport.