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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 $0<r<a$ 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.

$r<b$: 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 $0<E<V_0$ for a bound state.

$b<r<a$: 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.

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