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What is the transmission amplitude of a wavefunction $\phi(x)=e^{ikx}(\tanh x -ik)$? I would have thought that it is $\tanh x -ik$ since this is the factor associated with the forward travelling $e^{ikx}$ but then since the reflection coefficient is $0$, we have that the reflection probability is $0$, but $|\tanh x-ik|^2$ is dependent on $x$ so not identically $=1$? Where have I gone wrong?

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up vote 2 down vote accepted

The clue here is that the coefficients of $\mathrm e^{\mathrm ikx}$ and $\mathrm e^{-\mathrm ikx}$ are to be evaluated at infinity. The terms and concepts you're using apply to a situation where we have an incoming wave proportional to $\mathrm e^{\mathrm ikx}$ for $x\to-\infty$. It interacts with a system around the origin and is partially reflected, transmitted and/or absorbed. The reflected component is an outgoing wave proportional to $\mathrm e^{-\mathrm ikx}$ for $x\to-\infty$, and the transmitted component is an outgoing wave proportional to $\mathrm e^{\mathrm ikx}$ for $x\to\infty$.

In your case, $\tanh x\to\pm1$ for $x\to\pm\infty$. Thus the incoming wave has amplitude $-1-\mathrm ik$, and the transmitted wave has amplitude $+1-\mathrm ik$, and the transmission coefficient is

$$\frac{+1-\mathrm ik}{-1-\mathrm ik}=\frac{\left(1-\mathrm ik\right)^2}{-1-k^2}=-\left(\frac{1-\mathrm ik}{|1-\mathrm ik|}\right)^2=\exp\left(\mathrm i\left(\pi-2\arctan k\right)\right)\;.$$

This has magnitude $1$, so probability is conserved, as there are no reflected or absorbed components. The interaction merely shifts the phase of the wave by a $k$-dependent angle $\pi-2\arctan k$.

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Thanks, joriki! Very good explanation. – danz Oct 25 '11 at 15:48

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