Total reflection but still a non-zero probability? Consider the standard problem of the step potential in Quantum Mechanics.
One writes the Ansatz and using the boundary conditins finds the coefficient amplitides.
If $A$ is the coefficient for the incoming wave and $B$ for the one reflected from the potential step. It is trivial to show that the co-efficient for reflection $R=\frac{|B|^2}{|A|^2}$ comes out to be unity.
Physically this means that the wave is totally reflected but we know that some part of the wave sneaks beyond the step decaying exponentially.
This decay being exponential never quite makes the escaped wave zero though negligible quite soon.
How is some part of the wave sneaking beyond when the wave is getting reflected totally?
I've been told by some friends that this is nothing new and in fact happens even in a non-quantum setting to any wave and it is natural that wavefunctions being 'waves' behave this way.
My issue though is not why quantum mechanically such a thing happens. This is coming out from Schrodinger's equation and the fact that there are waves.
My question is, in general, when we see this phenomenon with waves what are we exactly calculating physically with $R$ ?
Does $R$ coming out 1 not mean that no part of the wave can be present beyond the step? If not what does it tell us?
In other words, what do we mean when say that the whole wave is getting reflected? Certainly in the step potential case there is decaying wave function.
 A: The way I understand the question is the following: suppose we have a potential step, finite in height, but extending to infinity, e.g.,
$$
V(x)=\begin{cases}0, x<0,\\V_0>0, x\geq 0.\end{cases}$$
Then, an electron incident from the left with energy lower than the potential step
$$E=\frac{\hbar^2k^2}{2m}<V_0$$
will be fully reflected, i.e., the flux flowing from left to right is the same as the flux flowing from right to left. What the OP finds surprising is that we still have an exponentially decaying solution in the step itself, as
$$
\psi(x)\propto e^{-q x},\;q=\frac{1}{\hbar}\sqrt{2m(V_0-E)}.
$$
This decaying solution simply characterizes the manner of scattering - it will have different form for different shapes of the step potential and determine the phase of the reflected wave, but it does not change the fact that the wave is fully reflected.
This is true not only for waves (as the OP notes), but for any scattering event. If we think, e.g., about a tennis ball being scattered from a wall, we have no doubt that the wall is deformed by the ball, and this deformation may extend far to the wall (may be to infinity - depending on how one models the problem). Moreover, the ball penetrates to some depth into the wall before reversing the direction of its motion.
What is really different about wave-like phenomena, is that scattering emerges as a static solution of wave equations, extended in space and without variation in time. Although one can get closer to the mechanical analogy by considering scattering of wave packets.
Updates
Flux vs. probability density
The answer above was written assuming that the math is well understood, as implied by the OP. Still, for completeness it is necessary to point out that we distinguish here two things:

*

*non-zero probability of finding the particle in some region of space,
$$|\psi(x)|^2dx,$$

*non-zero flux
$$j(x)=\frac{-i\hbar}{2m}\left[\psi^*(x)\frac{d\psi(x)}{dx}-\frac{d\psi^*(x)}{dx}\psi(x)\right]$$
It can be easily shown that the eigenfunctions in this problem are real (up to a constant complex phase), which means zero flux, i.e.,

*

*non-propagation for $x>0$

*full reflection for $x<0$
Electromagnetic analogy
A close equivalent of this problem is reflection of an electromagnetic wave from a metal. If the metal is ideal (i.e., all the electric field inside is immediately screened), then there is no field inside the metal, and the wave is fully reflected. If the metal ahs finite conductivity, then there is an exponentially decaying field within the metal (skin effect), but the reflection is still complete. The equivalent of flux in this problem is the Poynting vector.
