# Why are two-electron systems usually described in singlet-triplet basis?

Why are the two-electron system usually described in singlet-triplet basis, but not computational basis $$\uparrow\uparrow$$,$$\uparrow\downarrow$$,$$\downarrow\uparrow$$,$$\downarrow\downarrow$$? What is the advantage of that?

Typical spin-spin coupling has form: $$H_J=J\mathbf{S}_1\cdot \mathbf{S}_2.$$ Thus, if we take two-spin Hamiltonian $$H=\Delta_1 S_1^z+\Delta_2 S_2^z +J\mathbf{S}_1\cdot \mathbf{S}_2,$$ it will be diagonal in the singlet-triplet basis, but not in the computational basis. And working in the diagonal basis is nearly always an advantage.
This is also true for more general coupling of angular momenta, such as, e.g., the spin-orbit coupling $$H_{LS}\propto \mathbf{L}\cdot\mathbf{S},$$ or even for system containing many particles with various spin and orbital momenta. The reason for this that the Hamiltonian of the system as a whole often possesses symmetry in respect to rotations in 3D (or at least around the quantization axis), so the total angular momentum is a good quantum number, regardless of the interactions within the system.
• Did you mean $\mathbf{S}_1^2$ and $\mathbf{S}_2^2$ in your two-spin Hamiltonian? I don't think that $\mathbf{S}_1$ is well-defined as an operator. Nov 17, 2021 at 20:02
Electrons are quanta of the same fermionic field, so they are indistinguishable. A two-electron state like $$|\uparrow\downarrow\;\rangle$$ is unphysical, since in this state the two electrons are distinguishable: an exchange of electrons will yield a state differing from the original by more than just a phase factor. A proper symmetrization (for triplet) or antisymmetrization (for singlet) will yield physically admissible states (to be augmented by orbital part, to get antisymmetric total state).
The number state representation can be as good as the singlet-triplet representation in some situations (or even better). As always, it depends on what your specific problem is. However, the singlet-triplet representation is important because they are states with different total-spin angular momentum: the total spin quantum number of the system is $$S=0$$ in the singlet state, and $$S=1$$ in the triplet state. This means that if the system is described by an Hamiltonian $$\hat{H}$$ which commutes with the total spin operator $$\hat{S}$$, the eigenstates of the system are either on a singlet state or on a triplet state.