Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

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?

share|improve this question
add comment

1 Answer 1

up vote 1 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 $\pi-2\arctan k$.

share|improve this answer
    
Thanks, joriki! Very good explanation. –  danz Oct 25 '11 at 15:48
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.