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

1. May 20, 2010

### yarospo

Hello,
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
$$V=V_0(r)+\frac{V_1}{r^3}L\cdot S$$
and V0 > 0.
We define the eigenvalue of the operator $$L\cdot S$$ in the basis |J,L,S> as aJ,L,S while J=L+S.
What is the necessary condition on V1a 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 V0 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 V0>0 the energy threshold for bound states depends solely on it. But if I know nothing about it, then for example, taking $$V_0(r)=r^2$$ there is absolutely no restriction on the factor of $$\frac{1}{r_3}$$ - 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:
$$E_{J,L,S}=<J,L,S|H|J,L,S> = \left.<J,L,S|V_0(r)\right|J,L,S>+a_{J,L,S}V_1<J,L,S\left|\frac{1}{r^3}\right|J,L,S>$$
$$=\int _0^{\infty }dr r^2V_0(r)\left[R_{\text{nl}}(r)\right]{}^2+\frac{a_{J,L,S}V_1}{a_0^3n^3l\left(l+\frac{1}{2}\right)(l+1)} < ??$$

Last edited: May 21, 2010