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Delta Potential - Bound and Continuum States

  1. Apr 8, 2014 #1
    1. The problem statement, all variables and given/known data


    I am studying my lecturer's notes and in this part he uses a delta potential to illustrate a simple example of Fermi's golden rule, that the rate of excitation is ##\propto t##.


    2s8jkn5.png

    2. Relevant equations



    3. The attempt at a solution

    I've managed to get the bound states, by solving boundary conditions. They turn out to be ##\frac{\sqrt{Wm}}{\hbar}exp(-\frac {mW|x|}{\hbar^2})##.

    How are the continuum states derived? And why do only odd parity states contribute, when there is a perturbation ##-Fxe^{-i\omega t}##?
     
  2. jcsd
  3. Apr 8, 2014 #2

    TSny

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    The continuum states can be derived is a similar manner to the bound state. You can assume the solutions are either even or odd functions. Start by thinking about the general form of the solutions for E>0 for the regions x>0 and x<0.

    Fermi's golden rule contains a matrix element of the perturbation function between the initial and final states. Consider the overall parity of the integrand that expresses the matrix element.
     
  4. Apr 8, 2014 #3
    The matrix element in the integral is: ##|<p|V|i>|^2 = \left(\int \psi_p^* V \psi_i\right)^2##

    If ##\psi_p## is odd, multiplied by ##V## which is odd, together they form an even function so the integral ##\int_{-\infty}^{\infty} = 2\int_0^{\infty}## which gives a non-zero value.
     
  5. Apr 8, 2014 #4

    TSny

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    That's right. So the odd ##\psi_p## will generally have a nonzero matrix element. Similarly, you can show that the even ##\psi_p## must yield zero for the matrix element.
     
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