- #1
b2386
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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?
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?
