Boundary conditions for eigenfunctions in a potential step

[Dario SLC]

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

2. 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)$3. 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.

Thanks a lot!

 

[mark726]

Thank you for your post. I would like to clarify a few points and provide some guidance for your problem.

Firstly, the general form for the eigenfunctions in this case is indeed given by the equation you have provided, where the spin function is specified by the values of $m_s$. However, it should be noted that the value of $m_s$ is not necessarily restricted to only $+1/2$ and $-1/2$. It can take on any value between $-s$ and $+s$, as you have correctly stated in your attempt. This is important to keep in mind when considering the degeneracy of the energy levels.

As for the boundary conditions, you are on the right track. For the radial function, the boundary conditions will depend on the potential. In the case of a spherical potential step, the boundary conditions would indeed be as you have stated, with the additional condition that the radial function must be continuous at the boundary of the potential step. This means that the radial function must have the same value at the boundary from both sides.

For the angular part, the boundary conditions will depend on the values of $l$ and $m$. As you have correctly stated, $l$ can take on values from 0 to $n-1$, and $m$ can take on values from $-l$ to $+l$. Therefore, the boundary conditions for $Y(\theta,\varphi)$ will depend on the specific values of $l$ and $m$ for each energy level. This is because different values of $l$ and $m$ will result in different angular functions, and the boundary conditions will depend on the behavior of these functions.

In terms of the degeneracy of the energy levels, you have correctly stated that without spin, the degeneracy would be $n^2$ for a bound state with $l=0$. However, with spin, the degeneracy would be $2n^2$ for a bound state with $l=0$, as you have also stated. This is because each energy level can accommodate two different spin states, corresponding to the two possible values of $m_s$.

I hope this helps clarify some of your doubts and provides some guidance for your problem. Keep up the good work!