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No solution with E=v0 for step potential

  1. Apr 27, 2014 #1
    Is it the case that there is no general solution for the step potential with E = V0?

    This seems as though it must be the case as there are only certain energy eigenvalues which would be allowed and, therefore, we cannot for all possible values for the potential barrier find allowed energy eigenvalues.

    This would be appear to be reflected in the inconsistent solutions for the wave function in interval 2 (right of the step potential x > 0).

    Wave equation for interval 2: Ψ2 = A2eik2x + B2e-ik2x 0 < x < ∞

    Wave number for interval 2

    k2 = [2m/ħ2 (E - V2)]1/2 Interval 2 ; x > 0 V2 = E

    k2 = [2m/ħ2 (0)]1/2 = 0

    The angular frequency in this eigenfunction corresponds to the wave number k2 and is:

    ω2 = 0 = 0

    Thus, the terms in the eigenfunction correspond to the following:

    A2 – incident wave (momentum 0)
    B2 – reflected wave (momentum 0)

    Producing the following wave equation for interval 2: Ψ2 = A2 + B2

    0 < x < ∞

    This result is inconsistent with the solution to the TISE for interval 2

    (-ħ2/2m)d2/dx2Ψ2(x) + v(x)Ψ(x)2 = EΨ(x)

    where v(x)Ψ(x)2 = EΨ(x)

    So, d2/dx2Ψ2(x) = 0

    In order for d2/dx2Ψ2(x) to equal 0

    d/dx Ψ(x)2 = A2 and Ψ(x)2 = A2x + B2
  2. jcsd
  3. Apr 28, 2014 #2
    What exactly are you asking? You give the general solution to the Schrodinger equation for this situation:

    As you found, the general solution to this differential equation is *not* given by a linear combination of exponentials.

    As another example, you could try solving

    [tex]\frac{d^2 y}{dx^2} + 2 \frac{dy}{dx} + y = 0[/tex]

    Well, for any E you can find a pair of linearly independent solutions to the Schrodinger equation, as you did above. Whether you want to call that E "allowed" then depends on what boundary conditions you want to impose on the wave function. We pretty much always require that the wave function be bounded as ##x \to \pm \infty##. This leads to quantization of energy levels--rules out most E's--if E is less than the potential at infinity. But all E's greater than or equal to the potential at infinity are allowed; these are the "scattering states." For example in your case for ##E=V_0## there is the constant solution, which does not blow up at infinity.

    Sometimes we require the solutions to be square-normalizable as part of an especially rigorous approach. This effectively requires a rapid falloff of the wave function as ##x \to \pm \infty##. Then the only true energy eigenstates are bound states.
    Last edited: Apr 28, 2014
  4. Apr 29, 2014 #3
    Thank you for the response.

    I am still confused as to the wave function

    Ψ2 = A2 + B2

    is this also a general solution to the Schrodinger equation for this situation?

    Is this equivalent to Ψ2 = A2x + B2

    with appropriate boundary conditions?
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