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Dirac Delta Potential

  1. Oct 22, 2006 #1
    Question:
    Consider the motion of a particle of mass m in a 1D potential [itex] V(x) = \lambda \delta (x)[/itex]. For [itex]\lambda > 0[/itex] (repulsive potential), obtain the reflection R and transmission T coefficients.

    [Hint] Integrate the Schordinger equation from [itex]-\eta[/itex] to [itex]\eta[/itex] i.e.
    [tex]\Psi^{'}(x=\epsilon )-\Psi^{'}(x=-\epsilon )=\frac{2m}{\hbar^{2}}\lambda\int^{\epsilon}_{-\epsilon}\delta (x)\Psi (x)dx = \frac{2m}{\hbar^{2}}\lambda\Psi (x > 0)[/tex]

    What I have so far:
    Inside the barrier, the wave function is:

    [tex]\psi (x)= Ae^{\kappa x}+Be^{-\kappa x}[/tex]

    where:

    [tex]\kappa = \sqrt{\frac{2m}{\hbar^{2}}\left(V-E\right)} [/tex]

    Outside we have wave function in the form of:

    [tex]\psi (x) = Ce^{ikx}+De^{-ikx} x < 0[/tex]
    [tex]\psi (x) = Ee^{ikx} x > a[/tex]

    and [itex]R = \frac{|D|^2}{|C|^2} [/itex] and [itex]T = \frac{|E|^2}{|C|^2}[/itex].

    I have in my notes how to get the ratio [itex]\frac{D}{C}[/itex] and [itex]\frac{E}{C}[/itex], but how does the hint that was given to me used for? where does the delta function come in play?

    I don't really get the hint itself either. How does integrating Schrodinger Equation give me that relation in the hint? I am very lost...:confused:
     
  2. jcsd
  3. Oct 23, 2006 #2

    George Jones

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    Staff Emeritus
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    Gold Member

    There is no "inside the barrier," since a delta function is a width-zero barrier.

    1) Write down Schrodinger's equation.

    2) Integrate it term-by-term over the interval [itex](-\epsilon, \epsilon)[/itex].

    3) Take the limit as [itex]\epsilon \rightarrow 0[/itex].

    Assume [itex]\psi[/itex] is continuous at [itex]x = 0[/itex]. This gives you a relationship between the three coefficients.
     
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