Bound state for a Dirac delta function potential

In summary, to find the bound state energy for a particle in a Dirac delta function potential, we use the Schrodinger equation and the wave function \psi(x) = \frac{\sqrt{m \alpha}}{\hbar}e^{-m \alpha |x|\hbar^{2}}. The kinetic energy term gives \frac{-m \alpha^{2}}{2 \hbar^{2}} and integrating over the potential energy gives \frac{-m \alpha^{2}}{\hbar^{2}}, resulting in a total energy of E = \frac{-m \alpha^{2}}{2 \hbar^{2}} \frac{-m \alpha^{2}}{\hbar^{2}} = \
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badphysicist
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Homework Statement


Find the bound state energy for a particle in a Dirac delta function potential.


Homework Equations


[tex]\newcommand{\pd}[3]{ \frac{ \partial^{#3}{#1} }{ \partial {#2}^{#3} } } - \frac{\hbar^2}{2 m} \ \pd{\psi}{x}{2} - \alpha \delta (x) \psi (x) = E\psi (x)[/tex]
where [tex]\alpha[/tex] > 0 is the strength of the delta function potential.


The Attempt at a Solution


This is solved in Griffith's QM book and other places but I'm having an issue with the energy that is given there. The wave function is:
[tex]\psi(x)=\frac{\sqrt{m\alpha}}{\hbar}e^{-m \alpha |x|\hbar^{2}}[/tex]

Now if we plug this into the Schoedinger question above, times everything by [tex]\psi^{*}[/tex], and integrate over all x, we should get the bound energy.

The kinetic energy energy term gives [tex]\frac{-m\alpha^{2}}{2\hbar^{2}}[/tex] which is the listed as the total energy in Griffith's, but what about the term from the potential energy? Integrating over that will give a contribution of [tex]\frac{-m\alpha^{2}}{\hbar^{2}} [/tex] so that the total energy is [tex]E=\frac{-m\alpha^{2}}{2\hbar^{2}}\frac{-m\alpha^{2}}{\hbar^{2}}=\frac{-3m\alpha^{2}}{2\hbar^{2}}[/tex].

Any ideas why the potential energy is never included in the energy people list? It is listed as being the total energy in Griffith's QM book for instance. Am I correct in my evaluation of the potential energy? Any thoughts?
 
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  • #2
Nevermind, I think I realized my problem. I was just excluding the origin in my integration of the kinetic energy. It contributes a [tex]\frac{-m \alpha}{\hbar^{2}}[/tex] which cancels out the potential term (of course).
 

1. What is a bound state for a Dirac delta function potential?

A bound state for a Dirac delta function potential refers to a state in which a particle is confined within a specific region due to the presence of a potential that is represented by a Dirac delta function. This potential creates a barrier that prevents the particle from escaping, resulting in a localized state with a finite energy.

2. How is a bound state for a Dirac delta function potential different from a free particle state?

Unlike a free particle state, where the particle can move freely without any restrictions, a bound state for a Dirac delta function potential is localized and confined within a specific region. Additionally, the energy of a bound state is finite and discrete, while a free particle state has a continuous energy spectrum.

3. What are the conditions for a particle to be in a bound state for a Dirac delta function potential?

A particle must have an energy that is less than the potential barrier created by the Dirac delta function in order to be in a bound state. Additionally, the particle must also have a finite probability of being found within the potential region.

4. How does the depth and width of the Dirac delta function potential affect the bound state?

The depth and width of the Dirac delta function potential determine the strength of the potential barrier and the size of the potential region, respectively. A deeper potential barrier and narrower potential region will result in a stronger confinement of the particle, leading to a more localized and lower energy bound state.

5. Can a particle in a bound state for a Dirac delta function potential tunnel through the potential barrier?

Yes, there is a small probability that a particle in a bound state can tunnel through the potential barrier and escape. However, this probability decreases as the potential barrier becomes larger and the particle's energy becomes lower.

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