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I am trying to formulate boundary conditions and it occurred to me that I never had to implement a reflective boundary before.

The example is a one dimensional diffusion, where at $x=0$ the particles are reflected and at $x=L$ they instantly disappear in a sink.

My question now is:

Why can this be implemented by saying the flux must be zero at $x=0$ and with that due to Fick's first law the spatial derivative of the probability density? (this is what my research tells me so far)

In QM, we can confine a particle in a box by saying the wave function and with that the probability density are zero at the boundaries. Why does this not work here to create reflection?

And, secondly, I would like to get a better understanding on why the condition on the flux creates the conditions for reflection.

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Consider the general diffusion equation of the form, $$ \frac{\partial \psi}{\partial t}=D\frac{\partial\Phi}{\partial x}=D\frac{\partial^2\psi}{\partial x^2}\tag{1} $$ where $\Phi=\partial_x\psi$ is the probability current (flux).

In order to replicate a reflecting boundary, we establish an infinite potential at the boundaries, $V\left(x=0\right)=V\left(x=L\right)=\infty$. The particles are then unable to diffuse past $x=0$ and $x=L$ due to this boundary (containing all the particles within $0\leq x\leq L$), which must mean that the probability current, $\Phi$, must vanish at those points. The total population is then conserved in this case: $$ \int\psi(x,t)\,dx=1 $$ Assuming, of course, that the population is normalized initially.

Setting $\psi=0$ at $x=0,L$ means that the boundary absorbs the population $\psi$. Again assuming that $\int\psi(x,t=0)\,dx=1$, then for $t>0$, the total population is not conserved: $$ \int\psi(x,t)\,dx\neq1 $$ Hence, this cannot replicate a reflecting boundary. In fact, this integral returns the fraction of particles that have not been absorbed at time $t$ and is sometimes called the survival probability.

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  • $\begingroup$ thanks for the answer. This is so far what i thought, it does not explain though why in QM in case of the infinite potential well for example a boundary cnondition of $\Psi(L)=0$ does not work as a sink or are these just not comparable cases? If so why? $\endgroup$
    – pindakaas
    Apr 16 '15 at 7:32
  • $\begingroup$ Because $\psi$ in the QM case is not a probability density as $\psi$ is in my answer above. $\endgroup$
    – Kyle Kanos
    Apr 16 '15 at 12:49
  • $\begingroup$ Sorry for the delay. But if you set $\Psi=0$ in QM this implies $|\Psi|^2=0$ of course if $\Psi=0$ the probability steam density is zero then aswell. This is the same in the diffusion case. Setting the Probability density zero automatically sets the probability current to zero. So my question remains in a way why does this in one case mean reflection and in the other mean absorbtion? $\endgroup$
    – pindakaas
    Apr 17 '15 at 8:46
  • $\begingroup$ In QM, do we evolve $|\psi|^2$ or $\psi$ via the Schrodinger equation? $\endgroup$
    – Kyle Kanos
    Apr 17 '15 at 13:00
  • $\begingroup$ The answer is $\Psi§ of course. But to get back to something constuctive: I really don't understand what you are trying to tell me. Is the explanation possibly what i wrote in the comment before? "Setting the Probability density zero automatically sets the probability current to zero" so there is no need to treat these cases separately in QM? $\endgroup$
    – pindakaas
    Apr 17 '15 at 14:45

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