- #1

b2386

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Here is my question:

In the [tex]E>U_0[/tex] 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, [tex]E=U_0+(\frac{\pi^2 h^2}{2mL^2})[/tex], combine the continuity conditions to show that B=0. Here are the continuity conditions:

1st [tex]A+B=C+D[/tex]

2nd [tex]k(A- B)=k^{'}(C-D)[/tex]

3rd [tex]Ce^{ik^{'}L}+De^{-ik^{'}L}=Fe^{ikL}[/tex]

4th [tex]k^{'}(Ce^{ik^{'}L}-De^{-ik^{'}L})=kFe^{ikL}[/tex]

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

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

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

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

[tex]A+B={De^{-2i \pi}\frac{k^{'}+k}{k^{'}-K} + D[/tex].

Solving for D gives me

[tex]\frac{A+B}{e^{-2i \pi}\frac{k^{'}+k}{k^{'}-K} + 1} = D[/tex]

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

[tex]\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})[/tex]

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?