# Quantum transmission resonances

1. Nov 10, 2006

### b2386

Hi all,

Here is my question:

In the $$E>U_0$$ potential barrier, there should be no reflected wave when the incident wave is at one of the transmisson resonances. Assuming that a beam of particles is incident at the first transmission resonance, $$E=U_0+(\frac{\pi^2 h^2}{2mL^2})$$, combine the continuity conditions to show that B=0. Here are the continuity conditions:

1st $$A+B=C+D$$

2nd $$k(A- B)=k^{'}(C-D)$$

3rd $$Ce^{ik^{'}L}+De^{-ik^{'}L}=Fe^{ikL}$$

4th $$k^{'}(Ce^{ik^{'}L}-De^{-ik^{'}L})=kFe^{ikL}$$

A couple more equations that we already know are $$k=\sqrt{\frac{2mE}{h^2}}$$ and $$k^'=\sqrt{\frac{2m(E-U_0)}{h^2}$$

Here is my attempted solution:

I divided the 4th equation by K and then set equation 3 and 4 equal to each other. I then used the new equation to solve for C in terms of D giving me

$$C=De^{-2i \pi}\frac{k^{'}+k}{k^{'}-K}$$ where I substituted $$\frac{\pi}{L}$$ in for k'.

I substituted this result into the first equation to now give me

$$A+B={De^{-2i \pi}\frac{k^{'}+k}{k^{'}-K} + D$$.

Solving for D gives me

$$\frac{A+B}{e^{-2i \pi}\frac{k^{'}+k}{k^{'}-K} + 1} = D$$

Now, plugging in our solutions for D and C into the 2nd equation

$$\frac{(A+B)e^{-2i \pi}\frac{k^{'}+k}{k^{'}-K}}{e^{-2i \pi}\frac{k^{'}+k}{k^{'}-K}+1}-\frac{A+B}{e^{-2i \pi}\frac{k^{'}+k}{k^{'}-K}+1}=\frac{k}{k^{'}}(A-B})$$

At this point, it seems impossible to simplify the equation to a point where it is obvious that B = 0. Am I on the right track or is there an easier way?

2. Nov 10, 2006

### TMFKAN64

There is an easier way.

I see that you've already evaluated k' at the transmission resonance. Plug this value into your 3rd and 4th equations and get rid of the exponentials on the left side. You should see that the left side of these two now look similar to the right side of your first two equations, and a little algebra should give you B=0.