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I General Solution of Dirac Delta Potential Well

  1. Mar 10, 2016 #1
    We know that the solutions of time-independent Dirac delta potential well contain bound and scattering states:
    $$\psi_b(x)=\frac{\sqrt{mu}}{\hbar}e^{-\frac{mu|x|}{\hbar^2}}\text{ with energy }E_b=-\frac{mu^2}{2\hbar^2}$$
    A(e^{ikx}+\frac{i\beta}{1-i\beta}e^{-ikx}) \quad &x \leq0\\
    \frac{A}{1-i\beta}\quad &x\geq0
    \text{ with energy }E_k=\frac{\hbar^2k^2}{2m}.
    My question is how to construct a normalizable time-dependent general solution like in free particle case
    \Psi(x,t)=\frac{1}{\sqrt{2\pi}}\int\limits_{-\infty}^\infty\phi(k) e^{ikx-\frac{\hbar k^2t}{mt}}\mathrm d k,
    $$\phi(k)=\frac{1}{\sqrt{2\pi}}\int\limits_{-\infty}^\infty\Psi(x,0)e^{-ikx}\mathrm dx?$$
  2. jcsd
  3. Mar 10, 2016 #2


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    By doing the integral ?
  4. Mar 10, 2016 #3
    I suppose the result is
    \int\limits_{-\infty}^0\phi(k)(e^{ikx}+\frac{i\beta}{1-i\beta}e^{-ikx})e^{-i\frac{\hbar k^2t}{2m}} \mathrm dk+\int\limits_{0}^\infty\phi(k)\frac{1}{1-i\beta}e^{ikx}e^{-i\frac{\hbar k^2t}{2m}} \mathrm d k,$$
    where $$\beta=\frac{mu}{\hbar^2 k}.$$
    The question becomes how to find $$c_b$$ and $$\phi(k).$$
    I don't know if my general solution is correct of not.
    If yes, how to determine the coefficients?
  5. Mar 10, 2016 #4


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  6. Mar 10, 2016 #5


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    Looks good. You only have to normalize the scattering states to a ##\delta## distribution,
    $$\int_{\mathbb{R}} \mathrm{d} x \psi_k^*(x) \psi_{k'}(x)= \delta(k-k'),$$
    and the scattering state should have a factor ##\exp(\mathrm{i} k x)## for ##x>0##.

    Then the coefficients are given by
    $$c_b=\int_{\mathbb{R}} \mathrm{d} x \psi_b^*(x) \psi(x,0), \quad \phi(k) = \int_{\mathbb{R}} \mathrm{d} x \psi_{k}^*(x) \psi(x,0).$$
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