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Let $U(x)$ be the static potential and $\gamma$ the drag coefficient, so terminal Genetic drift under force $-U^{\prime }$ is $v=-U^{\prime }/\gamma$. The probability current is $J=-D{U}^{\prime }p/(k_{B}T)$ only after imposing the desired relation; instead start independently from $J=-D{p}^{\prime }-U^{\prime }p/\gamma$. At thermal equilibrium $p\propto e^{-U/(k_{B}T)}$ and detailed balance requires $J=0$. Substitution yieldsAt 298 K with $a=0.5\mu$m, this gives approximately $4.37\times 10^{-13}$ m${}^2$ s${}^{-1}$. The argument assumes an overdamped particle in an equilibrium thermal bath, rather than arbitrary driven diffusion.

The stationary occupation ratio is $p_1/p_2=2$. Balance of the two probability currents gives $p_1{k}_{12}=p_2{k}_{21}$, and Boltzmann occupation of the trapped macrostates gives $p_2/p_1=e^{-(G_2-G_1)/(k_{B}T)}$. ThereforeThe sketch has two wells separated by 1 micrometre, with the second free-energy level higher by $k_{B}T\log 2$ and an intervening escape barrier. Free energy includes the volume of each trapped basin; equal minimum potential heights alone would not ensure equal occupations if basin shapes differed.