A particle with spin 1/2 in a potential well

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The discussion focuses on finding the energy levels of a spin 1/2 particle in a potential well defined by V(r) and W(r) with an ls interaction. Participants emphasize the need to solve the radial Schrödinger equation while addressing the complexities introduced by the ls interaction at the delta layer. A suggested approach involves transforming the stitching condition to account for the interaction. Additionally, there is a reminder for users to post equations in text format instead of images for clarity. The conversation highlights the importance of understanding both the potential and the interaction terms in solving the problem effectively.
kisdrA
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Hello everyone. Help me solve the problem. I don't understand how to handle this type of task.

Find the energy levels of a spin 1/2 particle in a potential well: V(r)+W(r)*(l,s), where V(r<a)=-U, V(r>a)=0, W(r) = q*δ(r-a)
 
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@kisdrA you will need to post any relevant equations and show your attempt at a solution.
 
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It is clear that the problem in the potential ##V(r)+W(r)*(l,s)##, but without the (l,s) term: ##V(r)+W(r)##, simply reduces to solving the radial Schrodinger equation. And I understand how to find energy levels in such a task. But what to do when ls interaction also appears in the delta layer?
 
A vector product can be written like this.
##(\overrightarrow{l}, \overrightarrow{s}) = \frac{1}{2} (\overrightarrow{j}^2 - \overrightarrow{l}^2 - \overrightarrow{s}^2)##
Then maybe just transform the stitching condition?
##\Psi^{'}_{II}(a+0) -\Psi^{'}_{I}(a-0) = \frac{2m}{\hbar^{2}}*q*\frac{1}{2}(j(j+1)-l(l+1)-s(s+1)) \Psi_{I,II}(a\pm 0)##
 
kisdrA said:
@kisdrA posting text and equations in images is not allowed here. Please post text and equations directly; use the PF LaTeX feature for equations. There is a LaTeX Guide link at the bottom left of each post window.
 
At first, I derived that: $$\nabla \frac 1{\mu}=-\frac 1{{\mu}^3}\left((1-\beta^2)+\frac{\dot{\vec\beta}\cdot\vec R}c\right)\vec R$$ (dot means differentiation with respect to ##t'##). I assume this result is true because it gives valid result for magnetic field. To find electric field one should also derive partial derivative of ##\vec A## with respect to ##t##. I've used chain rule, substituted ##\vec A## and used derivative of product formula. $$\frac {\partial \vec A}{\partial t}=\frac...