# Central Potential + Coupled Angular Momenta

I am considering a two-state system with a Hamiltonian of the form

$$H = \frac{p_1^2}{2m_1} + \frac{p_2^2}{2m_2} + V_a(|r|) + \bigg(\frac{1}{4}-\frac{S_1\cdot S_2}{\hbar^2}\bigg)V_b(|r|),$$ where

$$V_a(|r|) = \begin{cases} 0 & r < a\\ V_0 & r > a \end{cases}, \qquad V_b(|r|) = \begin{cases} 0 & r < b\\ V_0 & r >b \end{cases},$$ and $$b < a$$. Essentially, what this looks like is a step potential from 0 to $$V_0/4$$ for $$0 and then a second step to $$5V_0/4$$ for all $$r>a$$. Additionally, the spin coupling activates when $$r>b$$. I am trying to analyze the eigenfunctions and energies for this problem.

Since the potential is central, we can rewrite this Hamiltonian as $$H = \frac{P^2}{2M} + \frac{p^2}{2\mu} + V_a(r) + \bigg(\frac{1}{4}-\frac{S_1\cdot S_2}{\hbar^2}\bigg)V_b(r),$$ where $$P$$ is the usual momentum of the COM and $$p$$ is the relative momentum. The total momentum eigenstates are just $$|P\rangle$$ with energies $$P^2/2M$$ so we can discard it. Now we have to break up the problem into three separate regions. We will always assume $$l = 0$$, so that we are solving for the ground state.

$$r: The Hamiltonian becomes $$H = \frac{p^2}{2\mu}$$, which is the same as a free particle. The radial equation solution for $$R_1(r) \equiv u_1(r)/r$$ is simply

$$u_1(r) = A\sin kr,$$ where $$k^2 \equiv 2\mu E/\hbar^2$$. Let us assume that $$V_0$$ is very large and $$0 for a bound state.

$$b: Now the Hamiltonian is $$H = \frac{p^2}{2\mu} + V_0(1/4-S_1\cdot S_2/\hbar^2) = H_{\text{spin}} + H_{\text{space}}$$, where

$$H_{\text{spin}} = -\frac{V_0}{4}S_1\cdot S_2, \qquad H_{\text{space}} = \frac{p^2}{2\mu} + \frac{V_0}{4} \approx \frac{p^2}{2\mu} + V_0,$$ since $$V_0$$ is very large, we can assume $$V_0/4$$ is also very large. The radial equation solution is then

$$u_2(r) = Be^{\alpha r} + Ce^{-\alpha r},$$ where $$\alpha^2 \equiv 2\mu E/\hbar^2(V_0-E)$$. We should also match the boundary conditions at $$r = b$$, i.e. $$u_1(b) = u_2(b)$$ and $$u_1'(b) = u_2'(b)$$.

For a Hamiltonian that involves the coupling of two spins, the eigenstates are

\begin{align} &|1,1\rangle = |\frac{1}{2},\frac{1}{2}\rangle\\ &|1,0\rangle = \frac{1}{\sqrt{2}}|\frac{1}{2},\frac{-1}{2}\rangle + \frac{1}{\sqrt{2}}|\frac{-1}{2},\frac{1}{2}\rangle\\ &|1,-1\rangle = |\frac{-1}{2},\frac{-1}{2}\rangle\\ &|0,0\rangle = \frac{1}{\sqrt{2}}|\frac{1}{2},\frac{-1}{2}\rangle - \frac{1}{\sqrt{2}}|\frac{-1}{2},\frac{1}{2}\rangle \end{align}

The first three states have an energy $$H^- = -V_0/4$$ and the last state has energy $$H^+ = 3V_0/4$$.

$$r>a$$: The Hamiltonian is now $$H = \frac{p^2}{2\mu} + V_0(5/4 - S_1\cdot S_2/\hbar^2)$$. Again, we can split up the Hamiltonian into spin and spacial terms, then make the assumption that $$5V_0/4 \approx V_0$$. The radial eigenfunctions are then

$$u_3(r) = De^{-\alpha r},$$ and the appropriate boundary conditions at $$r = a$$ are applied, i.e. $$u_2(a) = u_3(a)$$ and $$u_2'(a) = u_3'(a)$$. The spin states are the same with the same energies.

Applying the four boundary conditions gives $$B = 0$$, $$C = D$$, and

$$\alpha = -b\cot kb,$$ which is an equation that can be solved for the energy $$E$$.

This is all good, but what I am trying to understand are the energies of the system. This whole analysis is for the case of no orbital angular momentum, i.e. $$l = 0$$, which corresponds to the ground state wavefunction. The boundary conditions led to a constraint on the energy, but how do we know if there are multiply energy levels? If you expand out that expression, you will get a transcendental equation for the energies, but how do we know how many solutions there will be?

Secondly, the addition of the spin angular momentas gives a perturbation to the energy levels, either $$\Delta = - V_0/4$$ or $$\Delta = + 3V_0/4$$. Would it be correct to say that the ground state energy is therefore $$E_{\text{gs}} = E^0_{\text{space}} - V_0/4$$? Also, since $$V_b = 0$$ for $$r < b$$, there is no restriction on the spin states in that region because the coupling term does not appear in the Hamiltonian. Can we then have any spin state for $$r < b$$? Ultimately, what I am confused on is how the energy can seemingly jump depending on whether $$r < b$$ or $$r > b$$. Thanks.