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Homework Help: Double Potential Well

  1. Apr 24, 2010 #1
    1. The problem statement, all variables and given/known data

    V(x) = [tex]\left\{\infty \textrm{ for } x<0[/tex]
    [tex]\left\{0 \textrm{ for } 0<x<a[/tex]
    [tex]\left\{V_0 \textrm{ for } a<x<a+2b[/tex]
    [tex]\left\{0 \textrm{ for } a+2bx<2a+2b[/tex]
    [tex]\left\{\infty \textrm{ for } 2a+2b<x[/tex]

    Set up the relevant equations in each region, write down the appropriate solution and then show that the wavenumber of the wave functions inside and outside the barrier satisfy a transcendental equation.

    2. Relevant equations

    3. The attempt at a solution

    I have basically used Schrödinger equations for the energy for regions 1 2 and 3. but this is where i get stuck. I have to show the energies that work (hope that makes sense). I'm not suppose to solve for [tex]E>V_0, E=V_0 or E<V_0[/tex] but find the energies that work. I am a bit confused about the boundary conditions also. I set them up the same as a finite potential barrier, but do i need boundary conditions for x<0 and x>2a+2b? as these sections are infinite, we should just get 100% reflection?
    I am also confused about the barrier in the middle. When solving for [tex]E>V_0, E=V_0 or E<V_0[/tex] we have different Schrödinger equations and different boundary conditions. How do i do it for any value of E?

    I guess you all know by now i am pretty stuck lol.
    Hope someone can help =)
    and hope my LaTeX and writing is easy to understand =P
  2. jcsd
  3. Apr 25, 2010 #2


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    I'm not exactly sure what that means, but I'd try considering each case separately even if you're not going to have them in your final solution, just so you can develop an intuition for the problem and its solutions.
    Yes, you'll get 100% reflection and no penetration into those regions. You know that the wavefunction vanishes when x<0 or x>2a+2b because the potential is infinite in those regions. You still need continuity of the wavefunction at the boundaries, but because you're dealing with an infinite potential, the derivative doesn't need to be continuous.
  4. Apr 25, 2010 #3
    V(x)= \left\{ \begin{array}{ccccc}
    \infty & \textrm{ for } x<0 \\
    0 & \textrm{ for } 0 < x < a \\
    Vo & \textrm{ for } a < x < a+2b \\
    0 & \textrm{ for } a+2bx<2a+2b \\
    \infty & \textrm{ for } 2a+2b

    \end{array} \right

    Last edited: Apr 25, 2010
  5. Apr 25, 2010 #4
    you can not have the wavefunction and the potential infinity for the same region! .. please try to solve the question step by step..
  6. Apr 25, 2010 #5
    whoops I just copyed some latex, forgot to replace Psi with v(x)
    my bad
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