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Dirac Delta potential

  1. Apr 22, 2015 #1

    dyn

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    When talking about the strength of a delta potential , the delta potential is multiplied by a parameter ie α but how does a delta potential have a strength ? It is zero everywhere and infinite at x = 0. The parameter makes no difference to zero or infinity.
     
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  3. Apr 22, 2015 #2

    stevendaryl

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    You can think of [itex]f(x) = - \alpha \delta(x)[/itex] as a limiting case of a step function:

    [itex]x < \frac{-\epsilon}{2} \longrightarrow f_\epsilon(x) = 0[/itex]
    [itex]\frac{-\epsilon}{2} < x < \frac{+\epsilon}{2} \longrightarrow f_\epsilon(x) = -\frac{\alpha}{\epsilon}[/itex]
    [itex]+\epsilon < x \longrightarrow f_\epsilon(x) = 0[/itex]

    Then for any other smooth function [itex]\psi(x)[/itex], you will have: (for any [itex]A, B[/itex] such that [itex]A < B[/itex])

    1. [itex]lim_{\epsilon \rightarrow 0} \int_{A}^{B} f_\epsilon(x) \psi(x) dx = -\alpha \psi(0)[/itex] (if [itex]A < 0 < B[/itex])
    2. [itex]lim_{\epsilon \rightarrow 0} \int_{A}^{B} f_\epsilon(x) \psi(x) dx = 0 [/itex](otherwise)
    So you can think of solving the wave equation for a delta-function potential as solving it for a square well, and then taking the limit as the width [itex]\epsilon[/itex] goes to zero.
     
  4. Apr 22, 2015 #3

    dyn

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    So does the potential have a value at x=0 ?
     
  5. Apr 22, 2015 #4

    stevendaryl

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    The delta function is undefined at [itex]x=0[/itex], except indirectly through the integral law:

    [itex]\int \delta(x) \psi(x) dx = \psi(0)[/itex]

    No real function has that property, but as I said, you can think of it as some kind of limiting case of a step function.
     
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