PHYS6562 F2 Daily Quaetion

Absorption boundary conditions

First observe that the probability density at the origin at any time is zero.

$\rho(0,t) = 0$

And the probability density on $$x=x'$$ at $$t=0$$ is

$\rho(x,0) = \delta(x-x')$

We can assume that there is a image probability density at $$t=0$$ on $$x=-x'$$. Therefore the total probability at $$t=0$$ should be

$\rho(x,0) = \delta(x-x')-\delta(x+x')$

The Green’s function for each delta function is a time-dependent Gaussian distribution,

$G(x,t) = \frac{1}{\sqrt{4\pi Dt}} e^{-x^2/4Dt}$

with the particle originally at the origin. For this case, both delta functions diffuse, yielding the same Gaussian distribution.

$G(x,t) = \frac{1}{\sqrt{4\pi Dt}} \left( e^{-(x-x')^2/4Dt} - e^{-(x+x')^2/4Dt}\right)$

The time evolution probability distribution is shown in fig.1.

Above: particle starts at $$x'=-4$$. Plotted at $$t=1$$, $$t=15$$, $$t=30$$. Below: particle starts at $$x'=2$$. Plotted at $$t=1$$, $$t=4$$, $$t=10$$.

For arbitrary probability distribution, we can transform the $$\rho(x,t_0)$$ into the combination of infinite delta functions and use the same method to have infinite Gassian distributions.