Spin interaction term for a two-particle system of fermions with a potential that contains spin-spin interaction

Question

Consider a two-particle system consisting of two identical fermions in a potential $$V(\vec{r})\vec{\sigma_{1}}\cdot\vec{\sigma_{2}}$$ where $V(\vec{r})$ is the spatial part of the potential and the indices $1$ and $2$ correspond to each particle respectively and $\vec{\sigma}=\sigma_1i+\sigma_2j+\sigma_3k$ with $\sigma_{i}$ the $i$-th Pauli matrix.

I want to find the contribution of the spin interaction term to the energy of the system. Now, working with singlet and triplet states, I know that the energy will be analogous to the eigenvalue of each state respectively.
Working things out, I have found that the eigenvalue of the singlet state for $\vec{\sigma_{1}}\cdot\vec{\sigma_{2}}$ is $-3$ times the eigenvalue of the triplet state.
By also trying out as $V(\vec{r})$ a potential used in physics(Yukawa potential), I also found that the triplet state has the same energy as the case in which we have two bosons in that potential.

So, denoting the ground state energy of that potential for two identical bosons as $E$, the triplet state gives $E_{triplet}=E$ and the singlet state gives $E_{singlet}=-3E$.

So, the question is:
What is the physical explanation of this?.
And, why is the spin-spin interaction more "intense" for the singlet state than for the triplet state(by taking the absolute value of each state's eigenvalue)?

EDIT: Since I don't consider spin-orbit interaction, I think that the answer might have to do with spin-spin coupling because of the dependence on $\vec{\sigma_{1}}\cdot\vec{\sigma_{2}}$.

Answer 1

The answer to ' why is the spin-spin interaction more intense for a singlet than triplet' is due to the fact that any internal interaction cannot change the overall energy of the system. Put informally its a sort of 'centre of gravity' thing and applies in all cases. The total energy change is $-E_S + gE_T$, where the $g=3$ is the multiplicity of the triplet. The energy change is zero as $E_S = 3E_T$.

The spin-spin interaction has a standard calculation for example, for the hyperfine structure splitting of hydrogen atom ground $1$s $^2\mathrm S_{1/2}$ ground state, which produces the 21 cm line of radio astronomy and the hydrogen maser. The potential in which the spins exist (a proton and electron in this case) only comes into the calculation in determining the overall magnitude of the splitting and then as $|\psi(0)|^2$ sometimes called the Fermi contact interaction.

(By standard methods I mean the interaction is of the form $H=A\vec I\cdot \vec J$ where the magnetic moment is generated by nuclear spin angular momentum $\vec I$ and $\vec J$ that due to electron spin. The total angular momentum of the atom is $\vec F= \vec I + \vec J$ is constant and the expectation value is the energy $E=A<\vec I\cdot \vec J>= \frac{A}{2}(F(F+1)-J(J+1)-I(I+1))$. In the case of two spin $1/2$ particles $J=1/2$ and $I=1/2$ and the energies are $A/4$ for $F=1$ and $-3A/4$ for $F=0$. The scaling term A contains constants and $|\psi(0)|^2$ )