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Finite Square Barrier

  1. Feb 19, 2010 #1
    Hello to all,

    I am looking at a scattering example in my book. A particle is incident from the left with energy E > Vo. The barrier is of width L, and located between x = 0 and x = L.

    The solutions to the time-independent Schrodinger equation in eacch of the regions comprising the left and right of the barrier and inside the barrier are:

    [tex]\psi(x) = G e^{ik_{1} x} + H e^{-ik_{1} x }.......... x < 0[/tex]

    [tex]\psi(x) = I e^{ik_{2} x} + J e^{-ik_{2} x }............0 \leq x \leq L[/tex]

    [tex]\psi(x) = K e^{ik_{1} x} + L e^{-ik_{1} x } ...........x > L [/tex]

    The example in the book gives the transmission coefficient [tex] T = \frac{|K|^{2}}{|G|^{2}} [/tex] but my question is what ratio gives the reflection coefficient?

    I thought [tex] R = \frac{|H|^{2}}{|G|^{2}} [/tex] but does the constant [tex]J[/tex] play any part in determining the reflection coefficient?

  2. jcsd
  3. Feb 19, 2010 #2
    I think L = 0
    G + H = I so H = I - G
    I + J = K
    so J = K - I

    I just used the continuity of the wave function.

    So reflection coefficient L = 0, H = I - G, J = K - I

    right? I think its simple algebra when dealing with a single mode incoming wave.
  4. Feb 19, 2010 #3

    I follow your algebra, but not sure I am understanding the whole picture. There is an example in my book with a finite square step and it has the reflection coefficient as

    [tex] R = \frac{|H|^{2}}{|G|^{2}}[/tex]

    Do I need to add the contributions from H and J in the square barrier example? I think I dont get the full picture of what you are saying. Can anyone shed any light on this. Thank you again.
  5. Feb 19, 2010 #4
    The reflection coefficient is just H. Can you scan the text book page?
  6. Feb 20, 2010 #5
    I'm afraid I dont have a scanner.
  7. Feb 21, 2010 #6
    Terminology problems abound. Using your original notation:

    1) the amplitude of the reflected wave is H
    2) the ratio of the amplitude of the reflected wave to the incoming wave is r=H/G (often called the reflection coefficient)
    3) the reflection probability is the ratio of the flux in the reflected wave to the flux in the incoming wave. (R = |r|^2 k_1/k_1) (also sometimes called the reflection coefficient)
    4) the amplitude of the transmitted wave is K
    5) the ratio of the amplitude of the transmitted wave the incoming wave is t=K/G (often called the transmission coefficient)
    6) the transmission probability is the ratio of the flux in the transmitted wave to the flux in the incoming wave. (T = |t|^2 k_3/k_1 where k_3/k_1 is the ratio of wavevectors in the different regions) (also sometimes called the transmission coefficient)

    For it to make sense, you don't want the L wave coming into the barrier from x>L. Moreover, it's crazy to use L twice in the same problem.

    H,I,J,K must be determined by boundary condition matching. Once you have done that, the reflection probability is |H|^2/|G|^2 (without any reference to J).

    The point is that something complicated goes on in the middle, but you don't need to look at what is going on there to determine the probability of reflection, you just count how many particles you get back, divided by how many you sent in.
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