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Suppose a potential barrier of the form $$ V(x) = \begin{cases} V_0 & x>0 \\ 0 & x<0 \end{cases} $$ Then, for energy $E$ such that $E < V_0$, we have that the transmission and reflection coefficients for the probabilities are $R = 1, T = 0$. In case where $V_0$ is not enormously large with respect to $E$, wave function will decay in $x>0$, but in a somewhat not rapid fashion. This means there is some reasonable not negligible probability for the particle to tunnel through. How does this agree with the fact that $R=1$, which means all of the probability has been reflected?

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  • $\begingroup$ Apologies if I am wrong, but is what you described above an infinitely wide barrier? Which would explain the R equals 1. $\endgroup$ – user108787 Jul 9 '16 at 16:10
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    $\begingroup$ Yes, I understand mathematically why $R=1$, I am trying to understand the physical meaning of this, given that there is some probability for the particle to tunnel (even though $R = 1$). $\endgroup$ – JonTrav1 Jul 9 '16 at 16:11
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$T$ and $R$ are transmission and reflection coefficients for waves. They refer to the probability that an incident wave will penetrate the barrier and continue propagating infinitely far. Physically, you should think of it as sending a constant sine wave in from the far left and looking to see what amplitude of constant sine wave you get at the far right.

In this case, since $E < V_0$ no matter how far to the right you go, the amplitude keeps decaying exponentially for all positive $x$. It never recovers its sinusoidal form, as it would with a rectangular barrier of finite extent, so there is no transmission.

Now, if you think of this waves as probability waves, then yes, it seems like there is a chance that the particle will materialize inside the barrier. And there is. That doesn't count as transmission, though. The process of a wave "turning into a particle" (i.e. wavefunction collapse) is not taken into account by the calculation of the transmission and reflection coefficients.

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