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.