Transmission Coefficient for two step potential

[th5418]

Homework Statement

$$E>V_1 & V_2$$
So it's a step potential wave, setup. Let's set it up along the x axis. At x=0, there is the first step where it is $$V_1$$. At x=a, there is another step, where $$V_2>V_1$$. Show that the transmission coefficient is...
$$T=\frac{4k_1k_2^{2}k_3}{k_2^{2}(k_1+k_2)^{2}+(k_3^{2}-k_2^{2})(k_1^{2}-k_2^{2})sin^{2}k_2a}$$

Homework Equations

So the usual transmission is just $$T=(\frac{A}{F})^{2}$$ where A is the coefficient for incoming wave, and F is coefficient for leaving wave. However, one of my friend says that the transmission coefficient has extra terms in it because the velocity of the wave is different. Please help!

The Attempt at a Solution

I have the eigenfunctions..
$$\varphi_1 = Ae^{ik_1x} + Be^{-ik_1x}$$

$$k_1 = \sqrt{\frac{2mE}{h}}$$

$$\varphi_2 = Ce^{ik_2x} + De^{-ik_2x}$$

$$k_2 = \sqrt{\frac{2m(E-V_1)}{h}}$$

$$\varphi_3 = Fe^{ik_3x}$$

$$k_3 = \sqrt{\frac{2m(E-V_2)}{h}}$$

 

[turin]

The velocity of a complex plane wave has no effect on its (average) probability density.¹ However, I think that you have the expression upside down.



¹I say average to account for the small-scale fluctuations that can result from interference.

 

[sith]

The velocity of the wave affects the probability flux of the wave. The transmission probability is properly defined as the ratio of the probability flux of the transmitted and the incident wave. In the potential step case the transmission probability is then

$$T = \frac{|\boldsymbol{k}'|}{|\boldsymbol{k}|}\frac{|F|^2}{|A|^2}$$

if $$A$$ is the amplitude of the incident wave, $$\boldsymbol{k}$$ is its wave vector, $$F$$ the amplitude of the transmitted wave and $$\boldsymbol{k}'$$ its wave vector. It is described in more detail http://quantummechanics.ucsd.edu/ph130a/130_notes/node149.html" . (On the bottom of the page there is a link to the probability flux discussion).