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Consider a potential $V(x)$ which is zero when $x<0$ and $V_0>0$ when $x>0$. Suppose there is an incident particle with momentum $p=\hbar k$ and energy $E = \hbar^2 k^2 / 2m < V_0$ coming from $x= - \infty$. Now then after scattering ($t \to \infty$) the wavefunction should be $$\Psi = \psi_{reflected} + \psi_{tunneling}$$ where $$\psi_{reflected} = A e^{-ikx} \;\; ,x<0$$ $$\psi_{tunneling} = B e^{- \kappa x}\;\; ,x>0$$ Because the wavefunction at $x>0$ is real, there is no probability flow there. So, using the definition of reflection coefficient where it is the ratio of the reflected to incident probability current, the reflection coefficient comes out to be 1. But the reflection coefficient is also defined as $$R = \int |\psi_{reflected}|^2 dx \;\;\;, t \to \infty$$ which doesn't make sense because there should be some non-zero probability that the particle is in the region where $x$ is positive.

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There are two ways of thinking about it, one is strictly mathematical (shut up and calculate, in a way), and another is maybe not very precise but rather intuitive. The first one you did correctly, $R=1$. In a potential $V=V_0$ for all $x>0$ (continuing infinitely "long" to the right) there is no way for the particle with $E<V$ be "transmitted", so reflectivity is indeed 1. This means that all particles will be reflected sooner or later (simply because they cannot be transmitted). It does not mean that you cannot register a particle at some point under the potential; even it gets there, it will still be reflected. The apparent contradiction originates in that we tend to think that the act of reflection happens at the point $x=0$ only; this is not true.

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  • $\begingroup$ What do you mean here by transmitted? I thought something being transmitted would mean that there is some probability that I could detect that something ,in this case , in the region where $x>0$. $\endgroup$ Commented May 29, 2021 at 9:56
  • $\begingroup$ Consider a potential barrier $V=V_0$ at $0<x<a$, and $V=0$ everywhere else. Then, a particle with energy $E<V_0$ can tunnel under this barrier and be found at $x>a$ with a certain (non-zero) probability. Of course, after that its wavefunction will be restored to $e^{ikx}$, and it will be transmitted through the barrier. In your example, $a\rightarrow\infty$, and transmittance $T\rightarrow 0$. $\endgroup$
    – sleepy
    Commented May 29, 2021 at 15:40

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