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A Step potential question

  1. Jun 4, 2004 #1
    Hi...I've come across this question on a past exam, and I can't seem to resolve it!

    A particle is incident on a step potential at x=0. The total energy of the particle E, is less then the height of the step, U. The particle has wavefunctions
    [tex]$\begin{array}{l}
    \psi (x) = \frac{1}{2}\{ (1 + i)e^{ikx} + (1 - i)e^{ - ikx} \} ,x \ge 0 \\
    \psi (x) = e^{ - kx} ,x < 0 \\
    \end{array}$
    [/tex]

    Note that k is the same on both sides of the step
    a) ….
    b.) How must k be related to E and U on both sides of the step and determine the ration E/U

    Well on the left, it is just a free wave-function where k=[tex]$k = \frac{{\sqrt {2mE} }}{\hbar }$
    [/tex]
    On the right k= [tex]$k = i\kappa = \frac{{i\sqrt {2m(U - E)} }}{\hbar }$
    [/tex]
    Since the k’s are equal this immediately implies that U=0, does it not? In which case the ratio E/U is infinite…
     
    Last edited: Jun 4, 2004
  2. jcsd
  3. Jun 4, 2004 #2

    turin

    User Avatar
    Homework Helper

    No, but you are thinking along the right lines. You have recognized that the requirement on k puts strong restriction on ... something. Attack the problem firstly as if k were different in the two regions, and then see what conditions have to be met so that the wavefunction can be put in the given form. (Biggest hint: Notice that there is an i in the exponent in one region and not in the other.)

    One thing I just noticed: it appears to me that the particle is incident from the right so you may have your suggestions backwards.
     
    Last edited: Jun 4, 2004
  4. Jun 5, 2004 #3
    Ok so if I call the region II solution k'
    [tex]$k' = \frac{{\sqrt {2m(E - U)} }}{\hbar }$[/tex]

    Since U > E this is complex.
    [tex]$k' = \frac{{i\sqrt {2m(U - E)} }}{\hbar } = i\kappa $
    [/tex]

    This leads to the exponential solution
    [tex]$e^{ - \kappa x} $
    [/tex]

    If I then impose
    [tex]$\kappa = k$ [/tex] i get
    [tex]$\begin{array}{l}
    \frac{{\sqrt {2mE} }}{\hbar } = \frac{{\sqrt {2m(U - E)} }}{\hbar } \\
    \sqrt E = \sqrt {U - E} \\
    E = U - E \\
    2E = U \\
    E = \frac{U}{2} \\
    \end{array}$
    [/tex]

    oh hey...i definatly did that yesterday...musta been dropping a negative somewhere!

    Ok, seems my latex sucks but how is E=U/2??

    Cheers!
     
  5. Jun 5, 2004 #4

    turin

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    Homework Helper

    That was my initial gut feeling about it. Are you asking how it is possible, or are you asking if I think this is correct? I think it is probably correct (without actually combing through your calculations).
     
  6. Jun 6, 2004 #5
    Turin - I don't think I would ever ask if something is possible in QM, per se, since things that seem impossible always seem to fall out of QM! So much more to learn aswell - makes me wonder what other things (classically impossible) will occur
     
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