Double Delta Fuinction Potential - Tell me if Im correct please

  • Thread starter jhosamelly
  • Start date
  • #1
128
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.
 

Answers and Replies

  • #2
vela
Staff Emeritus
Science Advisor
Homework Helper
Education Advisor
14,874
1,449
The continuity condition A+B = T isn't correct. You need to set x to the location of the second delta function.
 
  • #3
128
0
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?
 
  • #4
vela
Staff Emeritus
Science Advisor
Homework Helper
Education Advisor
14,874
1,449
I didn't notice the other delta function is at x=-a. All four of your equations, as written, have errors in them.
 
  • #5
128
0
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?
 
  • #6
vela
Staff Emeritus
Science Advisor
Homework Helper
Education Advisor
14,874
1,449
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.
 

Related Threads on Double Delta Fuinction Potential - Tell me if Im correct please

  • Last Post
Replies
6
Views
6K
  • Last Post
Replies
7
Views
25K
  • Last Post
Replies
2
Views
3K
  • Last Post
Replies
1
Views
3K
Replies
2
Views
1K
  • Last Post
Replies
16
Views
9K
  • Last Post
Replies
0
Views
2K
Top