Double Delta Function Potential

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jhosamelly
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I have

V (x) = [itex]\sqrt{((h-bar ^{2})V_{0})/2m}[/itex] [[itex]\delta(x-a)[/itex]+ [itex]\delta(x+a)[/itex]]

How do I find R and T?

Under what condition is there resonant transmission?
 
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Matterwave said:
How did you solve the 1 delta function potential? Do something similar here.

well, Yes I have the solution for that but I don't understand it that much. Please help. Thanks
 
So, for the scattering states, you can assume plane waves. You can, for example, have an incident plane wave from the left, and then you can have waves going in both directions in the middle and on the right. And then you would need to match boundary conditions.
 
ok. I got these answers. Are these correct? Someone please tell me.

General Equations

[itex]U_{I}[/itex] = [itex]e^{ikx}[/itex] + R [itex]e^{-ikx}[/itex]


[itex]U_{II}[/itex] = A [itex]e^{ikx}[/itex] + B [itex]e^{-ikx}[/itex]


[itex]U_{III}[/itex] =T [itex]e^{-ikx}[/itex]


Boundary Conditions

if a = 0

[itex]U_{I}[/itex] = [itex]U_{II}[/itex]

1 + R = A + B

[itex]U_{II}[/itex] = [itex]U_{III}[/itex]

A + B = T



discontinuity equation

[itex]U'_{I}[/itex] - [itex]U'_{II}[/itex] = - [itex]\sqrt{\frac{2m V_{o}}{h-bar^{2}}}[/itex] [itex]U_{a}[/itex]

ik (1 - R) - ik (A - B) = - [itex]\sqrt{\frac{2m V_{o}}{h-bar^{2}}}[/itex] R


[itex]U'_{II}[/itex] - [itex]U'_{III}[/itex] = - [itex]\sqrt{\frac{2m V_{o}}{h-bar^{2}}}[/itex][itex]U_{a}[/itex]


ik (A-B) - ikT = - [itex]\sqrt{\frac{2m V_{o}}{h-bar^{2}}}[/itex] T


/// i hope someone can tell me if these are correct so I can continue my calculations. Thanks.