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Full reflection of wave function for E = V0

  1. Oct 6, 2013 #1
    I have seen discussions which suggests that there is no solution for the interval after the step in a step potential where E = V0.

    The set up is a potential step where E = V0, with an interval 1 defined as x < 0 before the step and an interval 2 as x > 0 after the step.

    Is the following correct?

    Wave equation for interval 1: Ψι = A1eikx + B1e-ikx

    Wave equation for interval 2:

    2/2m d2/dx2Ψ2(x) + v(x)Ψ(x)2 = EΨ(x)

    v(x)Ψ(x)2 = EΨ(x)

    So, d2/dx2Ψ(x)2 = 0

    In order for d2/dx2Ψ(x)2 to equal 0

    d/dx Ψ(x)2 = A2 and Ψ(x)2 = A2x + C


    Ψι = A1eikx + B1e-ikx

    Ψ2 = A2x

    Continuity conditions:

    Ψι = Ψ2

    d/dx Ψ1 = d/dx Ψ2


    A1eikx + B1e-ikx = A2x

    at interval boundary x = 0


    A1 + B1 = 0

    A1 = B1

    i.e. full reflection of the incident wave function.

    d/dx [A1eikx + B1e-ikx] = d/dx [A2x]

    ikA1eikx + ikB1e-ikx = A2

    at interval boundary x = 0

    ikA1 - ikB1 = A2

    A1 = B1


    ik[A1 - A1] = A2

    0 = A2

    i.e. no transmission of the incident wave function.
  2. jcsd
  3. Oct 7, 2013 #2

    Simon Bridge

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    Homework Helper

    ##\psi_I=Ae^{ik_1x}+Be^{-ik_1x}\\ \psi_{II}=Ce^{ik_2x}+De^{-ik_2x}##

    The wave-numbers are related to the energy of the wave:
    So for a wave of energy E incoming:

    So, for ##E=E_0##, ##k_2=0## and ##\psi_{II}## becomes a constant.
    The next step is to apply the boundary conditions and check the normalization.
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