Singlet and Triplet States

Splitting four states into 3 + 1

Two spin- $\tfrac12$ particles live in a four-dimensional space. The previous lesson showed that the aligned state $|\!\uparrow\uparrow\rangle$ has total spin $s = 1$ . The full space organizes into a triplet ( $s=1$ , three states) and a singlet ( $s=0$ , one state). The names count the $2s+1$ values of $m$ in each family.

The triplet ( $s = 1$ )

The three triplet states are symmetric under exchange of the two particles:

\begin{aligned} |1, +1\rangle &= |\!\uparrow\uparrow\rangle, \\ |1, 0\rangle &= \tfrac{1}{\sqrt{2}}\big(|\!\uparrow\downarrow\rangle + |\!\downarrow\uparrow\rangle\big), \\ |1, -1\rangle &= |\!\downarrow\downarrow\rangle . \end{aligned}

We can see how the middle state arises by applying the total lowering operator $S_- = S_{1-} + S_{2-}$ to the top state $|1,+1\rangle = |\!\uparrow\uparrow\rangle$ :

S_-\,|\!\uparrow\uparrow\rangle = (S_{1-}|\!\uparrow\rangle)|\!\uparrow\rangle + |\!\uparrow\rangle(S_{2-}|\!\uparrow\rangle) = \hbar\big(|\!\downarrow\uparrow\rangle + |\!\uparrow\downarrow\rangle\big).

Normalizing gives exactly $|1,0\rangle$ above. Applying $S_-$ once more lands on $|\!\downarrow\downarrow\rangle = |1,-1\rangle$ . All three share $\mathbf{S}^2 = 2\hbar^2$ .

The singlet ( $s = 0$ )

The remaining, antisymmetric combination is orthogonal to the triplet's $m=0$ member:

|0, 0\rangle = \tfrac{1}{\sqrt{2}}\big(|\!\uparrow\downarrow\rangle - |\!\downarrow\uparrow\rangle\big).

Acting with $\mathbf{S}^2$ gives zero: it has total spin $s = 0$ , the unique state with no total angular momentum at all. Swapping the two particles multiplies the singlet by $-1$ , in contrast to the symmetric triplet.

Why the singlet is special

Because $|0,0\rangle$ carries zero total angular momentum and is rotationally invariant, it looks the same along every axis: measuring the two spins along any common direction always yields opposite results. This perfect anticorrelation is the heart of the Einstein–Podolsky–Rosen setup and of Bell-inequality tests.

Try it

Build the singlet $|0,0\rangle = \tfrac{1}{\sqrt2}(|01\rangle - |10\rangle)$ as a two-qubit statevector. Remember qubit 0 is the left bit, so the amplitude is $+\tfrac{1}{\sqrt2}$ on $|01\rangle$ and $-\tfrac{1}{\sqrt2}$ on $|10\rangle$ .

Run your code to see the quantum state.