Two-Spin Systems

Two spin-½ particles together live in a four-dimensional Hilbert space, the tensor product of two qubit spaces. The product basis is

But the more physical basis groups these states by their total spin $\mathbf{S} = \mathbf{S}_1 + \mathbf{S}_2$.

Singlet and triplet

Adding two spin-½'s ($s_1 = s_2 = \tfrac{1}{2}$) yields total spin $S = 1$ or $S = 0$:

The $S = 1$ sector is the triplet (three states, symmetric under particle exchange):

The $S = 0$ sector is the singlet (one state, antisymmetric):

Counting checks out: $3 + 1 = 4$ states, matching the dimension of the combined space.

The singlet is maximally entangled

The singlet cannot be written as a product $|\chi_1\rangle\otimes|\chi_2\rangle$ of single-spin states — it is entangled. Remarkably, it has the same form along every axis: measuring either spin gives a perfectly random result, but the two outcomes are always anti-correlated, whatever direction you choose. This rotational invariance is the signature of total spin zero, and it is exactly the state at the center of Bell-inequality experiments.

Total-spin operators

The squared total spin is

Acting on the singlet, $S^2|0,0\rangle = 0$, while on every triplet state $S^2 = 1(1+1)\hbar^2 = 2\hbar^2$. The dot product $\mathbf{S}_1\cdot\mathbf{S}_2$ therefore takes the value $-\tfrac{3}{4}\hbar^2$ in the singlet and $+\tfrac{1}{4}\hbar^2$ in the triplet — the energy difference that underlies the exchange interaction in magnetism.

Try it

Build the singlet state $(|01\rangle - |10\rangle)/\sqrt{2}$ on two qubits. The state vector panel should show equal magnitudes on $|01\rangle$ and $|10\rangle$ with opposite signs, and zero on $|00\rangle$ and $|11\rangle$.

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