Delta Potential - Bound and Continuum States

In summary, the conversation discusses the use of a delta potential to demonstrate Fermi's golden rule and the derivation of bound and continuum states. It is determined that only odd parity states contribute due to the perturbation function being odd and the overall parity of the matrix element. The matrix element is also shown to yield a non-zero value for odd parity states and zero value for even parity states.
  • #1
unscientific
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Homework Statement

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##.
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Homework Equations


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}##?
 
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  • #2
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
TSny said:
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.

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
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.
 
  • #5


The continuum states for a delta potential can be derived by considering the scattering solutions for the Schrödinger equation. These solutions will have a plane wave form ##e^{ikx}## with a scattering amplitude ##f(k)##, and the energy will be given by ##E = \frac{\hbar^2k^2}{2m}##. The scattering amplitude can then be found by solving the Schrödinger equation in the region where the potential is non-zero. This will result in a continuum of states with a continuous range of energies.

As for the odd parity states, this is due to the symmetry of the delta potential. Since the potential is symmetric about the origin, only states with odd parity will have non-zero overlap with the perturbation ##-Fxe^{-i\omega t}##. This means that only these states will contribute to the transition rate.
 

1. What is a delta potential?

A delta potential is a type of potential energy function commonly used in quantum mechanics to model the behavior of particles in a confined space. It is a localized potential that takes the form of a Dirac delta function, which is a mathematical function that is zero everywhere except at a single point. In this context, the delta potential represents a point-like impurity or defect in the system.

2. What are bound states in a delta potential?

Bound states in a delta potential refer to the energy states that are allowed for a particle confined within the potential. These states have discrete energy levels and are characterized by a wave function that is confined to a finite region around the delta function. Bound states are associated with particles that are trapped within the potential and cannot escape.

3. How are continuum states different from bound states in a delta potential?

Continuum states in a delta potential refer to energy states that are unbound and have a continuous energy spectrum. In contrast to bound states, the wave function for continuum states is not confined to a finite region and extends to infinity. These states are associated with particles that are not trapped by the potential and can escape to infinity.

4. How is the energy spectrum of a delta potential related to the strength of the potential?

The energy spectrum of a delta potential is directly related to the strength of the potential. As the strength of the potential increases, the energy levels for both bound and continuum states also increase. This means that particles in a stronger delta potential require more energy to overcome the potential barrier and escape, resulting in a higher energy spectrum.

5. What are some real-world applications of the delta potential?

The delta potential has many real-world applications, particularly in the fields of solid-state physics and quantum mechanics. It is commonly used to model the behavior of electrons in a crystal lattice, the motion of particles in a semiconductor, and the interaction between atoms and molecules. It is also used in the study of quantum tunneling, where particles can pass through a potential barrier even if they do not have enough energy to overcome it.

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