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Up until now I was certain that a bound state is a state with energy below the minimum of the potential at infinity. However, in this question I don't know at all how to proceed.

1. The problem statement, all variables and given/known data

A spin 3/2 particle moves in a potential

[tex]V=V_0(r)+\frac{V_1}{r^3}L\cdot S[/tex]

and V_{0}> 0.

We define the eigenvalue of the operator [tex]L\cdot S[/tex] in the basis |J,L,S> as a_{J,L,S}while J=L+S.

What is the necessary condition on V_{1}a to get a bound state. Should it

be positive or negative?

2. Relevant equations

3. The attempt at a solution

Right from the start what puzzles me is that I have no idea how V_{0}looks aside from the fact that it's positive. I also note that the potential seems to depend on angular momentum and spin, but their effects cancel out when r goes to infinity it goes to 0, and since V_{0}>0 the energy threshold for bound states depends solely on it. But if I know nothing about it, then for example, taking [tex]V_0(r)=r^2[/tex] there is absolutely no restriction on the factor of [tex]\frac{1}{r_3}[/tex] - all the states will be bound.

Well, here's how I intended to continue, once I knew what I'm actually looking for:

Applying the Hamiltonian on the basis vectors (Using central potential formulas), I get:

[tex]E_{J,L,S}=<J,L,S|H|J,L,S> = \left.<J,L,S|V_0(r)\right|J,L,S>+a_{J,L,S}V_1<J,L,S\left|\frac{1}{r^3}\right|J,L,S>[/tex]

[tex]=\int _0^{\infty }dr r^2V_0(r)\left[R_{\text{nl}}(r)\right]{}^2+\frac{a_{J,L,S}V_1}{a_0^3n^3l\left(l+\frac{1}{2}\right)(l+1)} < ??[/tex]

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# Homework Help: Bound state of a spin 3/2 particle in a potential

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