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[Basic QM] Potential step & particle flow, my answer is inconsistent ?

  1. Jan 6, 2009 #1
    1. The problem statement, all variables and given/known data
    A flow of particles collides with a potential step. The potential is zero for x < 0 and a certain value [tex]V_0[/tex] for x > 0.
    The flow of particles comes in from the left (x<0 area).

    One of the questions is to give the general form of the wavefunction in the area x > 0 and to find an expression for the wavenumber k.

    2. Relevant equations
    [tex]-\frac{\hbar^2}{2m} \frac{d^2 \psi}{dx^2} + V_0 \psi = E \psi[/tex]
    (For x > 0)

    3. The attempt at a solution

    I first write the schrodinger eq. into a more general form:
    [tex]-\frac{\hbar^2}{2m} \frac{d^2 \psi}{dx^2} + (V_0 - E) \psi = 0[/tex]
    [tex]- \frac{d^2 \psi}{dx^2} + \frac{2m (V_0-E)}{\hbar^2} \psi = 0[/tex]
    [tex]\frac{d^2 \psi}{dx^2} - \frac{2m (V_0-E)}{\hbar^2} \psi = 0[/tex]
    [tex]\frac{d^2 \psi}{dx^2} - k^2 \psi = 0[/tex]

    So
    [tex]k^2 = \frac{2m (V_0-E)}{\hbar^2}[/tex]

    And the solution are complex exponents due to the minus-sign (they would be real exponents if it was a plus sign):
    [tex]\psi (x) = A e^{ikx} + B e^{-ikx}[/tex]


    Now, when I look at the answer, I noticed that they also use complex exponents, but their k was different:
    [tex]k^2 = \frac{2m (E-V_0)}{\hbar^2}[/tex]
    (E-V instead of V-E)

    This is a pretty important error because:

    In my case: If E > V, the particle is 'above the potential step' and thus it should be a complex exp. (sin/cos). But, if E > V then k becomes complex, and the exponents become real!
    If E < V, the particle is 'inside the potential step' and thus it should be a real exponent (as usual with penetration in classically forbidden area). But, if E < V then k is real and the exponents stay complex!...

    Obviously I made an error with a minus sign somewhere, but I honestly can't find it..!?


    I also tried switching the minus sign for a plus sign (and swapping the E and V again obviously) to get:
    [tex]\frac{d^2 \psi}{dx^2} + \frac{2m (E-V_0)}{\hbar^2} \psi = 0[/tex]
    So now, my k is in accordance with the answer (E-V), but because of the + I get REAL exponents, and if E > V, k is real and the exponents stay real (wrong), and if E < V, k is complex and the exponents become complex (wrong)...???

    I can't figure it out...
     
  2. jcsd
  3. Jan 6, 2009 #2


    This is where things went backwards. The solution to
    [tex]\frac{d^2 \psi}{dx^2} = k^2 \psi[/tex]
    is
    [tex]\psi (x) = A e^{kx} + B e^{-kx}[/tex]
    with no [tex]i[/tex] in the exponents. In order to get the other form we should define
    [tex]k^2 = \frac{2m (E-V_0)}{\hbar^2}[/tex]
    so that
    [tex]\frac{d^2 \psi}{dx^2} = -k^2 \psi.[/tex]

    This is the answer "they" get (and has the right behavior for [tex]E > V_0[/tex], [tex]E < V_0[/tex]), but you can easily check the signs just by substituting [tex]\psi[/tex] back into the differential equation.
     
  4. Jan 6, 2009 #3
    Omg... You're completely right of course. I checked and re-checked that, and I was actually under the impression that the solution to
    [tex]y'' - k^2y = 0[/tex]
    had complex exponents and the solution to
    [tex]y'' + k^2y = 0[/tex]
    had real exponents... It's the other way around of course. Can't believe I mixed them up, must have been the holidays hehe...
     
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