Probability of finding a particle in a region even if the transmission into that region is zero We've a potential given as:
$V(x)=\left\{\begin{array}{ll}0, & x<0 \\ V_{0}, & x \geq 0\end{array}\right.$.
We've got particles coming in from the left towards the step and getting reflected from it. Further, if we assume the energy of the particles coming towards the barrier have lesser energy than the step i.e $E<V$, then it's said that

$T = 0$,
The transmission coefficient at a potential step with $E < V$ is zero.

Further it's said that,

The wavefunction in the second region, where $V(x)=V_0$ is $ψ_2(x) = 
e^{αx}$ where $\alpha=2m(V−E)$.

So we have a probability of finding a particle in the second region even though the transmission at the step is zero!? But this is contradictory. Can anyone please help.
 A: This has to do with the definition of the transmission coefficient $T$ in quantum scattering experiment. What we want to know is how a particle will behave when it's "shot" at the barrier. In a general situation, part of the wave function probability will reflect off the barrier, and part will be transmitted. However, for your potential, as $t\xrightarrow{}\infty$, all of the probability will be reflected. Note that we are NOT talking about a static situation. The wave function changes with time, so there will be some points in time when there is a non-zero probability for the particle to be found in the region where $V(x)=V_0$. However, after a long time has past, all of the probability will be reflected.
If the barrier were not infinitely thick, there would be some probability of transmission. For example, a particle certainly could tunnel through the potential
$$V=\begin{cases}
0 & x<0 \\
V_0 & 0\leq x < a \\
0 & a\leq x.
\end{cases}$$
Therefore, part of the probability could escape toward large $x$.
Edit:
For those who want more information, I'll do an explicit calculation. $R$ and $T$ are defined in the following way. We take a particle, localized on one side of the barrier and sharply peaked in momentum space, and allow it to travel toward the barrier. We then define $T$ as the probability of finding the particle on the opposite side of the barrier after waiting an infinitely long time. $R$ is defined as the probability that the particle will be found on the original side.
A standard way to approach the problem of calculating $R$ and $T$ (which is covered in most introductory textbooks) is to consider a slightly different situation. We look at the stationary state solutions for this potential (the energy eigenstates), and use $R=\frac{j_R}{j_I}$ and $T=\frac{j_T}{j_I}$, where $j_I$ is the probability current of the part of the wave function that is incident on the barrier, $j_R$ is the current of the reflected wave function, and $j_T$ is the current of the transmitted wave function.
The solution in the region where $V=0$ contains both $e^{ikx}$ and $e^{-ikx}$ terms, which give probability currents moving both to the left and to the right. The solution in the region where $V=V_0$ is a falling exponential $\psi_T(x)=e^{-\alpha x}$. The associated probability current is proportional to
$$j_T\propto \psi^*\frac{\partial\psi}{\partial x}-\psi\frac{\partial\psi^*}{\partial x}=\psi\frac{\partial\psi}{\partial x}-\psi\frac{\partial\psi}{\partial x}=0$$
because the wave function is purely real ($\psi_T=\psi^*_T$). If you add time dependence, $\psi_T$ will rotate in phase space and acquire an imaginary part, but this imaginary part has no dependence on $x$, so it passes through the derivatives and does not affect the identity above. Therefore, $T=0$.
A: Filling in the $V$, as given in the problem (the "one step up $V$"), in the Schrödinger equation, you'll get the wavefunction $ψ_2(x)=e^{αx}$ for the positive values of $x$. This means there is a quantum mechanical probability to find a particle with less kinetic energy than the height of the potential ($E\lt{V}$). The classical transmission coefficient will be zero obviously.
And also, $T$ will be zero in the quantum mechanical case. There is no transmission, as the particle doesn't travel up to infinity with a constant velocity. It will reside near the boundary between $V=0$ and $V=V_0$ and probably bounce back. Probably, which QM is all about.
I don't know what more I should write, so...I hope it helped you!
