# Delta Potential - Bound and Continuum States

1. Apr 8, 2014

### unscientific

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 $\propto t$.

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. Apr 8, 2014

### TSny

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.

3. Apr 8, 2014

### unscientific

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.

4. Apr 8, 2014

### TSny

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.