Boundary conditions for a step potential

In summary, the conversation discusses a particle with mass m and spin 1/2 in a spherical potential step with height V_0. The boundary conditions for the eigenfunctions are that the wavefunction and its radial derivative must be continuous at the potential step, and for bound states they must approach zero as r goes to infinity. The degeneracy level for the energy is found to be 2n^2 when E<V_0, with n being the principal quantum number. This is similar to the degeneracy of hydrogen atom energy eigenstates.
  • #1
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.
 
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  • #2
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}##
 

1. What are boundary conditions for a step potential?

Boundary conditions for a step potential refer to the conditions at the edges of the potential, where there is a sudden change in potential energy. These conditions determine the behavior of the wave function and the probability of finding a particle in a particular region.

2. What are the two types of boundary conditions for a step potential?

The two types of boundary conditions for a step potential are the continuity condition and the smoothness condition. The continuity condition requires that the wave function and its derivative are continuous at the boundary, while the smoothness condition requires that the wave function and its derivative are smooth (have no sharp changes) at the boundary.

3. How do boundary conditions affect the behavior of a particle in a step potential?

Boundary conditions play a crucial role in determining the behavior of a particle in a step potential. They determine the allowed energy levels and the probabilities of finding the particle in different regions. They also affect the shape and behavior of the wave function, which is a representation of the particle's probability amplitude.

4. Can boundary conditions be violated in a step potential?

No, boundary conditions cannot be violated in a step potential. If the conditions are not satisfied, the wave function will not be a valid solution to the Schrödinger equation and will not accurately describe the behavior of the particle.

5. How do boundary conditions for a step potential differ from those of a finite potential well?

The main difference between boundary conditions for a step potential and a finite potential well is that the step potential has a discontinuity in potential energy at the boundary, while the finite potential well has a finite potential energy in both regions. This results in different allowed energy levels and different behaviors of the particle in each potential.

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