Boundary conditions for a step potential

Click For Summary
SUMMARY

The discussion focuses on the boundary conditions for eigenfunctions of a particle with mass m and spin 1/2 in a spherical potential step with height V0. The boundary conditions for the spherical harmonics Y(θ, φ) are defined by the quantum numbers l and m, where l = 0, 1, ..., n-1 and -l ≤ m ≤ +l. For the radial function u(r), it is established that u(0) approaches 0 and u approaches 0 as r approaches infinity, ensuring continuity of u and its derivative at the potential step. The degeneracy level for energy when E < 0 is determined to be 2n², accounting for spin.

PREREQUISITES
  • Understanding of quantum mechanics, specifically spherical potentials and eigenfunctions.
  • Familiarity with the radial Schrödinger equation and its boundary conditions.
  • Knowledge of angular momentum operators, including L² and Lz.
  • Concept of degeneracy in quantum systems, particularly for bound states.
NEXT STEPS
  • Study the radial Schrödinger equation in detail, focusing on boundary conditions for spherical potentials.
  • Explore the implications of angular momentum quantum numbers on eigenfunctions and degeneracy.
  • Research the properties of bound states in quantum mechanics, particularly in relation to potential wells.
  • Examine the degeneracy of energy eigenstates in the hydrogen atom and compare with other spherically symmetric potentials.
USEFUL FOR

This discussion is beneficial for physics students, quantum mechanics researchers, and educators seeking to deepen their understanding of boundary conditions in quantum systems and the implications for energy eigenstates and degeneracy.

Dario SLC

Homework Statement



A particle with mass m and spin 1/2, it is subject in a spherical potencial step with height ##V_0##.
What is the boundary conditions for this eigenfunctions?
Find the degeneracy level for the energy, when it is ##E<V_0##

Homework Equations


Radial equation
\begin{equation}
-\frac{\hbar^2}{2m}\left(\frac{\partial^2u}{\partial r^2}-\frac{l(l+1)}{r^2}u\right)+V(r)u(r)=Eu(r)
\end{equation}
with ##u(r)=rR(r)##

The Attempt at a Solution


Well, first the constant of motion it ##\hat{L_z}##, ##\hat{L^2}## and ##\hat{S_z}## because conmute with ##\hat{H}##, also ##\hat{H}##, and it former a complete set of observables that conmute, therefore the form of eigenfunctions are:
\begin{equation}
\psi(r,\theta,\varphi,\sigma)=R(r)Y(\theta,\varphi)\chi(\sigma)
\end{equation}
when ##\chi(\sigma)## is a spin function, and only take the values ##m_s={+1/2,-1/2}## because eigenvalues of operator ##\hat{S^2}## is ##3/4\hbar^2## with ##s=1/2##.

Here my doubt:
Boundary conditions to ##Y(\theta,\varphi)## this is referred to:
##l=0, 1, \dots, n-1## and ##-l\leq m\leq +l##?

For radial function, it must be:
because ##\hat{H}## is hermitian, then for ##r=0\Longrightarrow u(0)\longrightarrow0##
and for ##r\longrightarrow\infty \Longrightarrow u\longrightarrow0##

For ##\chi## the boundary conditions it is ##-s\leq m_s\leq+s##?

In the case for the degeneracy, when ##E<0## corresponds to ##l=0## because it is a bound state, then without spin the level of degeneracy is ##n^2## but with spin is ##2n^2##.

Up to here my poor attemp to solution, i am really puzzled for explain the boundary condition to Y and ##\chi##, and i am not sure that the
##r\longrightarrow\infty \Longrightarrow u\longrightarrow0## for this potential.
 
Physics news on Phys.org
The boundary conditions are that the wavefunction ##u(r)## and its radial derivative ##\frac{\partial u}{\partial r}## must be continuous at the potential step, and for bound states ##u## and ##\frac{\partial u}{\partial r}## must approach zero when going to ##r\rightarrow\infty## (for unbound states they must only remain finite). What do you know about the degeneracy of hydrogen atom energy eigenstates? The idea is the same for all spherically symmetric potentials, not only ##V\propto r^{-1}##
 

Similar threads

  • · Replies 12 ·
Replies
12
Views
2K
Replies
5
Views
2K
  • · Replies 13 ·
Replies
13
Views
2K
Replies
2
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 19 ·
Replies
19
Views
3K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 4 ·
Replies
4
Views
888