What Is the Transmission Probability Through a Potential Barrier?

[NavalChicken]

Homework Statement

A beam of particles, each of mass m and kinetic energy E, is incident on a potential barrier

V(x) = V_0 \; \; for \; \; 0 \leq x \leq a
\; \; \; \; \; \; \; \; \; = 0 \; \; for \; \; x < 0 \; \; and \; \; x > a
E = V_0 \; \mbox{is the special case}

The part of the problem I'm on is finding the transmission probability

The Attempt at a Solution

I've solved the Schrodinger Equation and equated the solutions at the two boundaries which gave me

C + D = B
ik(C - D) = A
Aa + B = Ge^{ika}
A = kiGe^{ika}

A, B, C, D, G \; \mbox{constants}
I feel like I am just going round in circles finding the transmission probability, in my notes I have transmission prob as (\frac{G}{A})^2. However, a hint at the bottom says once the 4 continuity equations have been found, eliminate A and B, which I've tried and doesn't seem to get me any where!

If anyone has some advice or could push me in the right direction that would be really appreciated. Thanks

 

[nickjer]

You definitely need to give more info that just that. We have no idea what the constants you posted are. Also, wikipedia can be your friend:

http://en.wikipedia.org/wiki/Rectangular_potential_barrier

Please read that site first to check your work.

 

[NavalChicken]

My constants correspond to those on the wikipedia article. So I have,
A = B_2a \\;<br /> B = B_1 \\;<br /> C = A_r\\;<br /> D = A_l\\;<br /> G = C_r\\;

I understant that the constant I would've had F disappears because there is no particle from the right, but I don't understand why D \; (or \; A_l) becomes 1. By the looks of it's something quite simple that I'm missing!

 

[nickjer]

They set it to 1 to normalize it. They assume the incident wave is 100%. So the transmission and reflection will be less than 100% and add up to it.

 

[NavalChicken]

I realized I mixed up A with C in my first post when I gave the transmission probability. Will it still work if I keep the transmission as a ratio with C as the denominator rather than normalizing it?

 

[nickjer]

You can keep it as ratios if you like. That works just as well.