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QM - one-dimensional barrier - a simple step

  1. Apr 24, 2008 #1
    analyzing the simple-step scattering problem for E<V, we find that the solution to the schroedinger equation is:

    PHI(left) = Aexp(ikx)+Bexp(-ikx)
    PHI(right) = Cexp(-qx)

    Continuity of the function and it's derivative at x=0 gives the relations between the parameters A, B and C.

    Solving the appropriate equations we obtain

    R = |B\A|^2 = 1
    meaning that there is total reflection; hence the transmission must be zero.

    But if we calculate the probability of finding a particle, say in the interval a<x<2a, we get a non-zero probability (because PHI(right) does not venish).
    How is it possible, if every particle must be reflected?
     
  2. jcsd
  3. Apr 24, 2008 #2

    Hootenanny

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    If you obtain a reflection value of one then you have incorrectly determined the coefficients and/or the R value.
     
  4. Apr 24, 2008 #3
    The calculation of R is taken from a book, there shouldn't be a mistake there.

    And this is also a quotation from the book:
    "In classical physics region 2 is a "forbidden" domain. In quantum mechanics, however, it is possible for particles to penetrate the barrier."

    Region 2 is the region of the step (where V=V0).
     
  5. Apr 24, 2008 #4

    Hootenanny

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    Then you must have incorrectly determined the relationships between the coefficients and/or wave numbers. What is the formula you have for the R value?
    That is correct...
     
  6. Apr 24, 2008 #5
    R = |B\A|^2 = |(1-iq/k)/(1+iq/k)| = 1

    Anyway, if the particles can penetrate the barrier, it's logical that the probability to find a particle in a certain interval on the "other side" is non-zero;
    But still R=1. Does it mean that the particles are actually reflected from the inside of the barrier?
     
  7. Apr 24, 2008 #6

    Hootenanny

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    Sorry my bad, I thought you were talking about the case where E>V. In the case where E<V, then yes, the R value is unity.
    So yes, although the wavefunction is 'totally reflected' at the boundary, if we calculate the probability density inside the potential step, we find it to be non-zero. Therefore, even though the entire wavefunction is reflected, there is still some non-zero probability to find the particles inside the potential. This is what is referred to as tunnelling.
     
  8. Apr 24, 2008 #7
  9. Apr 24, 2008 #8

    George Jones

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    What Hootenanny said in different words:

    Very roughly, unlike classical reflection, the position at which quantum reflection occurs is uncertain; the particle can sometimes penetrate into the forbidden region before being reflected.
     
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