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Quantum mechanics scattering problem

  1. Feb 27, 2009 #1
    Quantum mechanics scattering problem, Please help!

    1. The problem statement, all variables and given/known data

    Calculate the reflection and transmission probabilities for right-incident scattering from the potential
    V(x) = V0 for x<0
    V(x)= 0 for x>0.
    at an energy E <V0. Find the probability density and the probability current density in the region x<0.
    What can you say about where the reflection is taking place?


    2. Relevant equations

    Stationary Scrodinger Equation,

    [tex]E\psi(x)=-\frac{h^{2}}{8\pi^{2}m}*\frac{\partial^{2}\psi}{\partial\psi^{2}}[/tex]

    Probability density,

    [tex]\rho(x)=\bar{\psi}\psi[/tex]

    Probability current density,

    [tex]J_{x}=\frac{ih}{2\pi\m}(\psi\frac{\partial\bar{\psi}}{\partial x}-\bar{\psi}\frac{\partial\psi}{\partial x})[/tex]

    Transmission probability, [tex]T=\frac{J_{trans}}{J_{incident}}[/tex]

    Reflection probabilty, [tex]R=\frac{J_{reflected}}{J_{incident}}[/tex]

    Also T+R=1

    3. The attempt at a solution

    Hi, I have gone some way with this problem and I won't inculde all my calculations (although if you want to see them please just ask and I will post them). Lets start with my solutions to the schrodinger equation in both regions,

    [tex]\phi(x)=Be^{-K_{1}X}[/tex] for x<0.

    [tex]\phi(x)=Ce^{iK_{2}X}+De^{-iK{_2}X}[/tex] for x>0

    where,
    [tex]k_{1}=\frac{\sqrt{-2m(E-V_{0})}}{\frac{h}{2\pi}}[/tex]

    and,
    [tex]k_{2}=\frac{\sqrt{2mE}}{\frac{h}{2\pi}}[/tex]

    Now I have that the wave with the B coefecient is the transmitted wave moving to the left and the wave with the C coefficient is the reflected wave moving to the right. The wave with the D coefficient is the right incident wave and is moving to the left.

    Now if I calculate the probability current density for the transmitted wave I get zero, since the wave function is real. This means that the refelction probability must be 1. However this implies that no quantume tunneling can occur, which shouldn't be the case. Also it seems way too easy.

    Please could someone check through my working and see where I have gone wrong? I have been pondering for ages but can't see the gap in my logic.
     
    Last edited: Feb 27, 2009
  2. jcsd
  3. Mar 1, 2009 #2
    Please could someone help me? I am really stuck on this one and cannot see where I have gone wrong.

    Thanks
     
  4. Mar 1, 2009 #3

    TFM

    User Avatar

    Hi,

    Firstly, your phi(x) for x<0 should have a i on the exponential.


    This should change your values for the probability and probability density
     
  5. Mar 1, 2009 #4
    Hi TFM,

    Thankyou for replying I really appreciate it. Please could you explain why there should be an i on the exponential? Because if there is on that one surely there should be on the other term which we have disregarded, and if the other term has an i it's second differential will be negative, when it should be positive?

    Thanks again!
     
  6. Mar 1, 2009 #5

    TFM

    User Avatar

    Well I have:

    [tex] k\phi(x) = \frac{\partial^2 \phi}{\partial x^2} [/tex]

    [tex] \phi (x)= Ae^{ikx} + Be^{-ikx} [/tex]
     
  7. Mar 1, 2009 #6
    If you differentiate the above twice you get -k^2 *A or B*phi(x). Thats why I'm confused.
     
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