Total Spin of Two Qubits

The smallest addition problem

The simplest non-trivial case of adding angular momenta is two spin-$\tfrac12$ particles, with $j_1 = j_2 = \tfrac12$. The combined space has dimension $(2\cdot\tfrac12+1)^2 = 4$, spanned by the four uncoupled states

where the first arrow is particle 1 and the second is particle 2. In qubit language these are the computational basis states $|00\rangle, |01\rangle, |10\rangle, |11\rangle$ with $|\!\uparrow\rangle = |0\rangle$ and $|\!\downarrow\rangle = |1\rangle$.

The total-spin-squared operator

The total spin is $\mathbf{S} = \mathbf{S}_1 + \mathbf{S}_2$. Squaring,

The dot product is most usefully written using the $z$-component and the ladder operators $S_\pm = S_x \pm i S_y$:

The ladder form is the practical engine for the whole subject. Recall the action on a single spin-$\tfrac12$:

with $S_+|\!\uparrow\rangle = 0$ and $S_-|\!\downarrow\rangle = 0$ (you cannot raise the top or lower the bottom).

Total spin of the aligned state

Apply $\mathbf{S}^2$ to $|\!\uparrow\uparrow\rangle$. Each single spin gives $\mathbf{S}_i^2 = \tfrac34\hbar^2$. The $S_{1z}S_{2z}$ piece gives $(\tfrac{\hbar}{2})(\tfrac{\hbar}{2}) = \tfrac14\hbar^2$. Both ladder cross terms vanish, since $S_{2-}$ followed by $S_{2+}$ would need to raise particle 2 above the top, and symmetrically for particle 1. Therefore

So $|\!\uparrow\uparrow\rangle$ is an eigenstate of $\mathbf{S}^2$ with eigenvalue $2\hbar^2$. Matching to $\mathbf{S}^2 = \hbar^2 s(s+1)$ gives $s(s+1) = 2$, hence $s = 1$. The fully aligned two-spin state therefore has total spin $s = 1$.

Try it

This is a numerical exercise — your code should return a number. Compute $\langle\!\uparrow\uparrow| \mathbf{S}^2 |\!\uparrow\uparrow\rangle / \hbar^2$ by summing the three contributions worked out above.

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