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Homework Help: Double Delta Fuinction Potential - Tell me if Im correct please

  1. Oct 26, 2011 #1
    1. The problem statement, all variables and given/known data

    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.
     
  2. jcsd
  3. Oct 27, 2011 #2

    vela

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    The continuity condition A+B = T isn't correct. You need to set x to the location of the second delta function.
     
  4. Oct 27, 2011 #3
    Hmmm.. ok. Thanks. What about the discontinuity equation? are they correct?
     
  5. Oct 27, 2011 #4

    vela

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    I didn't notice the other delta function is at x=-a. All four of your equations, as written, have errors in them.
     
  6. Oct 27, 2011 #5
    really? how should the equation be then? Please help me. Only the signs are wrong or the whole equation?
     
  7. Oct 27, 2011 #6

    vela

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    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.
     
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