Two step (spherical) square well potential

Your name]In summary, the problem involves finding the energies for the ground state and the first excited state of a spherically symmetric potential. The approach is to use spherical Bessel and Neumann functions, and to solve for the energy eigenvalues by eliminating the three constants using three boundary conditions. To take derivatives of the functions, one can use the recursion formula or known identities. For large values of V_0, the approximation that the spherical Bessel function is equal to the spherical Neumann function can simplify the transcendental equation for the energy splitting.
  • #1
IHateMayonnaise
94
0

Homework Statement


Just need some rough guidance on this one, nothing specific is really needed. The problem:

Given a spherically symmetric potential (V(r))
[tex]
V\left( r \right) = \left\{\begin{gathered}
V_0 \hfill \hspace{2}r<a \\
0 \qquad a<r<b \\
\infty \hfill r>b \\
\end{gathered} \right
[/tex]

Find the energies for the ground state and the first excited state. Also find an (approximate) expression for the energy splitting of the levels if [itex]V_0[/itex] is very large compared to these energy levels.


Homework Equations



Must use spherical Bessel and Neumann functions.

The Attempt at a Solution



The wavefunctions:

[tex]\mathcal{U}_I(r)=A_rJ_{\ell}(k_1r)[/tex]
[tex]\mathcal{U}_{II}(r)=C_rJ_{\ell}(k_2r)+D_r n_{\ell}(k_2r)[/tex]
[tex]\mathcal{U}_{III}(r)=0[/tex]

[tex]k_I=\frac{\sqrt{2m(V_0-E)}}{\hbar}[/tex]

[tex]k_{II}=\frac{\sqrt{2mE}}{\hbar}[/tex]

where the spherical Neumann function ([tex]n_{\ell}[/tex]) goes away in the first function since it is singular at the origin.

At this point work with the following boundary conditions:

[tex]\mathcal{U}_I(a)=\mathcal{U}_{II}(a)[/tex]
[tex]\mathcal{U}_I^{'}(a)=\mathcal{U}_{II}^{'}(a)[/tex]
[tex]\mathcal{U}_II(b)=0[/tex]

From here you get a three equations, where the derivatives are taken with respect to K1 and K2 (respectfully). First question: How do I take a derivative of the spherical Bessel and Neumann functions? Should I use the recursion formula and solve for the derivative?

Actually I am just not quite sure where to go from here in terms of finding the ground state energy. If this were a regular potential (two step) barrier, all I would do is eliminate the three constants using the three equations, arriving at a transcendental equation which I would then graph graph each side of the equation separately and find the intersections which represent the energy eigenvalues. Same thing here? Any thoughts?

Thanks yall

IHateMayonnaise
 
Physics news on Phys.org
  • #2


Dear IHateMayonnaise,

Thank you for your post. It seems like you have made a good start on solving this problem. To take derivatives of the spherical Bessel and Neumann functions, you can use the recursion formula as you suggested. Another option would be to use the identities for the derivatives of these functions, which can be found in most mathematical physics textbooks or online resources.

As for finding the ground state energy, you are correct in saying that you can eliminate the three constants using the three boundary conditions and arrive at a transcendental equation. This is the standard approach for solving problems with a spherically symmetric potential. You can then graph both sides of the equation and find the intersections to determine the energy eigenvalues.

In terms of the energy splitting for large V_0, you can use the approximation that the spherical Bessel function is approximately equal to the spherical Neumann function when the argument is large. This can help simplify the transcendental equation and make it easier to solve for the energy splitting.

I hope this helps guide you in the right direction. If you have any further questions or need clarification, please don't hesitate to ask.

Best of luck with your calculations!
 

1. What is a two step (spherical) square well potential?

A two step (spherical) square well potential is a type of potential energy function used in physics and chemistry to model the interaction between particles. It consists of two distinct regions, with one region having a higher potential energy and the other having a lower potential energy. The potential energy is represented as a square well shape, hence the name.

2. How is a two step (spherical) square well potential calculated?

To calculate a two step (spherical) square well potential, the potential energy function is first defined as a function of the distance between two particles. This function is then integrated over all possible distances to obtain the total potential energy of the system. The potential energy function typically includes terms for the attractive and repulsive forces between particles.

3. What is the significance of a two step (spherical) square well potential?

A two step (spherical) square well potential is often used to model the behavior of particles in a system, such as atoms or molecules. It can provide insights into the stability and structure of a system, as well as the behavior of particles within the system. It is also commonly used in simulations and calculations to study the properties of materials.

4. How does the depth of the potential well affect particle behavior in a two step (spherical) square well potential?

The depth of the potential well in a two step (spherical) square well potential is directly related to the strength of the attractive and repulsive forces between particles. A deeper well indicates stronger interactions between particles, which can result in a more stable system. Conversely, a shallower well may lead to more movement and less stability within the system.

5. What are some real-world applications of a two step (spherical) square well potential?

A two step (spherical) square well potential has many applications in physics and chemistry. It is commonly used to model the interaction between atoms in a solid, the behavior of particles in a gas, and the stability of molecules in a chemical reaction. It is also used in simulations and calculations to study the properties of materials, such as the melting point and phase transitions.

Similar threads

  • Advanced Physics Homework Help
Replies
29
Views
156
Replies
7
Views
2K
  • Advanced Physics Homework Help
Replies
10
Views
2K
  • Advanced Physics Homework Help
Replies
4
Views
3K
  • Advanced Physics Homework Help
Replies
16
Views
2K
  • Advanced Physics Homework Help
Replies
2
Views
1K
Replies
1
Views
4K
  • Advanced Physics Homework Help
Replies
4
Views
1K
  • Advanced Physics Homework Help
Replies
1
Views
2K
  • Advanced Physics Homework Help
Replies
1
Views
1K
Back
Top