Finding R and T for double delta function potential

  • Thread starter Thread starter jhosamelly
  • Start date Start date
  • Tags Tags
    Delta Potential
Join the discussion
Registration is free. Ask a follow-up in this thread, or start your own.
5 replies · 2K views
jhosamelly
Messages
125
Reaction score
0

Homework Statement



THIS IS THE QUESTION

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?



2. The attempt at a solution


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.
 
Physics news on Phys.org
The continuity condition A+B = T isn't correct. You need to set x to the location of the second delta function.
 
vela said:
The continuity condition A+B = T isn't correct. You need to set x to the location of the second delta function.

Hmmm.. ok. Thanks. What about the discontinuity equation? are they correct?
 
I didn't notice the other delta function is at x=-a. All four of your equations, as written, have errors in them.
 
vela said:
I didn't notice the other delta function is at x=-a. All four of your equations, as written, have errors in them.

really? how should the equation be then? Please help me. Only the signs are wrong or the whole equation?
 
The boundary between UI and UII is at x=-a, not x=0, so the boundary conditions are
\begin{align*}
U_I(-a) &= U_{II}(-a) \\
U'_{II}(-a) - U'_I(-a) &= \lim_{\varepsilon \to 0^+} \frac{2m}{\hbar^2}\int_{-a-\varepsilon}^{-a+\varepsilon} V(x)\psi(x)\,dx
\end{align*}Similarly, the other boundary conditions occur at x=+a.