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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?

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    $\begingroup$ Which lecture notes? $\endgroup$ – Qmechanic Jan 4 '18 at 12:17
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    $\begingroup$ It would be helpful if you could either give a link to these ominous lecture notes or if you specified the way how the conclusion "For $E < V$ we have $|c_2|^2 + |d_1|^2 < 1$" is reached. You should also specify that $c_1$ is chosen to be $1$. Otherwise your probability statements would not make sense. $\endgroup$ – freecharly Jan 4 '18 at 17:16
  • $\begingroup$ Thanks for your comments! I have added that I assume $c_1 = 1, d_2 = 0$. Unfortunately the notes are in German and not publicly available, so I do not want to post them here. Furthermore no additional explanation is given. The only thing written there is: "For $E < V$ we get $|c_2|^2+|d_1|^2 < 1$." I know that these is not particularly helpful but hearing that the probabilities have to some up to $1$ is already very helpful for me. $\endgroup$ – blablablup Jan 4 '18 at 19:17
  • $\begingroup$ It is unclear what OP is asking (v3). Unitarity dictates that $R+T=|c_2|^2 + |d_1|^2 = 1$, cf. e.g. my Phys.SE answer here. $\endgroup$ – Qmechanic Jan 6 '18 at 14:22
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It is not clear why you come to the conclusion that for $E < V$ you have $|c_2|^2 + |d_1|^2 < 1$. From the conservation of probability current the reflection probability should always sum up with the transmission probability to one. Even for the purely damped electron waves in the potential barrier there is no loss of electrons there. Thus no electrons can "get stuck" in the barrier.

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