Understanding a boundary condition on the density of probability

1. Jul 31, 2013

fluidistic

The book states that $P(x|y,t)$ represents the probability density that the potential has a value x at time t, knowing that it had the value y at t=0.
I understand this, the words are very clear. However I'd find much more intuitive the notation $P(x,t|y,0)$, but I guess that's just me?

Now comes the problem:
Then the book states that for a particular system where B is an absorbing potential and -infinity is a reflective potential, it states that the boundary conditions are:
(1)$P(x|y,0)=\delta (y-x)$
(2)$P(B|y,t)=0$
(3)$P(-\infty |y,t)=0$
Here is my understanding of (2): Using my own notation I rewrite it as $P(B,t|y,0)=0$ which means that the density of probability that the potential has a value of B at time t, knowing that it had a value of y at time t=0 is 0.
However the book states that (2) says: The second condition reflects the existence of a threshold potential B such that when x(t) takes the value B, the process is absorbed.
For the (1), my understanding is : The probability density that the potential has a value of x at time t=0 knowing that it had the value y at time t=0 is worth $\delta (y-x)$. Which makes perfect sense to me if this is the Dirac delta. However the book states that (1) : no change can occur in a zero time interval. A weird statement to me, but after all it's true.
So I understand this part and also the 3rd condition.
So I'm stuck at understanding the second condition.
Does this mean that the probability of finding the potential to be very close to B is approximately null, because it is "an absorbing potential"?

2. Aug 1, 2013

HallsofIvy

Staff Emeritus
This is exactly backwards. Saying that B is "absorbing" means the probability the potential has value y at time t, knowing it had value B at time t= 0, is 0. The "object" has been "absorbed" and can never have any other value in the future.

3. Aug 1, 2013

fluidistic

So if I understand you well, mathematically this would translate as $P(y,t|B,0)=0$ using my own notation or $P(y|B,t)=0$ using the book notation.
So in both cases the book is wrong by writting $P(B|y,t)=0$?