I am thinking about a problem in quantum mechanics. Given a rectangular potential barrier, say the potential $U(x) = V$ for $0 < x < L$ and $U(x) = 0$ otherwise, I found the solution for the Schroedinger equation $E \psi = -A \psi'' + U \psi$: $$ \psi(x) = \begin{cases} c_1 e^{i k_1 x} + c_2 e^{-i k_1 x} & x < 0 \\ b_1 e^{i k_2 x} + b_2 e^{-i k_2 x} & 0 < x < L \\ d_1 e^{i k_1 x} + d_2 e^{-i k_1 x} & x > L \end{cases} $$ with $k_1 = \sqrt{\frac{E}{A}}, k_2 = \sqrt{\frac{E-V}{A}}$. With $\psi$ being continuous it is possible to find the coefficients. Assuming $c_1 = 1$ and $d_2=0$ we get a unique solution.
Now to my question: I found in some lecture notes that for $E > V$ (the energy of the particle is greater than the energy of the barrier) we have $|c_2|^2 + |d_1|^2 = 1$ and interpret $|c_2|^2$ as the probability that the particle gets reflected and $|d_1|^2$ as the probability that the particle gets transmitted. So the particle is either reflected or transmitted.
For $E < V$ we have $|c_2|^2 + |d_1|^2 < 1$. If we now do the same interpretation we get that there has to be a third possibility for the particle. Can it also happen that the particle gets somehow stuck in the barrier, so the probabilities of reflection and transmission don't add up to $1$? How can we interpret this result physically? Why do we not have a similar behaviour as in the $E > V$ case?
