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 $\vec{\sigma_{1}}\cdot\vec{\sigma_{2}}$.