Singlets, Triplets, and Two-Spin Entanglement

Singlet and triplet states are the standard coupled basis for two spin-$1/2$ systems. They sit at a useful crossroads: angular momentum addition explains why the basis exists, entanglement theory explains which states are product or nonproduct, and exchange symmetry explains why the same basis matters for identical spin-$1/2$ particles.

The full angular-momentum derivation begins in Two Spin-1/2 Particles, and the rotational scalar/vector interpretation is explained in Symmetry Singlet and Triplet States. This page uses the result as a composite-systems example: how the same four-dimensional Hilbert space can be read as a product basis, a total-spin basis, a Bell-state basis, and an exchange-symmetry decomposition.

Helpful background. Two spin-1/2 particles derives angular-momentum addition; Bell states explains reduced-state purity and two-qubit correlations.

Product spin basis

For two distinguishable spin-$1/2$ systems, the spin Hilbert space is

$$\mathcal H = \mathbb C^2\otimes\mathbb C^2.$$

Using the $z$-axis spin basis for each factor, a product basis is

$$\lvert\uparrow\uparrow\rangle,\qquad \lvert\uparrow\downarrow\rangle,\qquad \lvert\downarrow\uparrow\rangle,\qquad \lvert\downarrow\downarrow\rangle.$$

Here the first arrow refers to the first tensor factor and the second arrow to the second tensor factor. For distinguishable spins, those labels can represent two atoms, two quantum registers, two spatially separated particles, or two independently addressable spin degrees of freedom.

Here $S_{1\alpha}=s_\alpha\otimes I$ and $S_{2\alpha}=I\otimes s_\alpha$, with $s_\alpha=\hbar\sigma_\alpha/2$. The product basis diagonalizes $S_{1z}$ and $S_{2z}$. It answers the question: what is each spin’s $z$ component?

Coupled Singlet-Triplet Basis

The total spin is

$$\mathbf S = \mathbf S_1+\mathbf S_2.$$

The coupled basis diagonalizes $S^2$ and $S_z$:

$$S^2\lvert s,m\rangle = \hbar^2s(s+1)\lvert s,m\rangle, \qquad S_z\lvert s,m\rangle = \hbar m\lvert s,m\rangle.$$

For two spin-$1/2$ systems, the four-dimensional space decomposes as

$$\frac12\otimes\frac12 = 1\oplus0.$$

The spin-$1$ sector is the triplet:

$$\begin{aligned} \lvert1,1\rangle &= \lvert\uparrow\uparrow\rangle,\\ \lvert1,0\rangle &= \frac{1}{\sqrt2} \bigl( \lvert\uparrow\downarrow\rangle + \lvert\downarrow\uparrow\rangle \bigr),\\ \lvert1,-1\rangle &= \lvert\downarrow\downarrow\rangle. \end{aligned}$$

The spin-$0$ sector is the singlet:

$$\lvert0,0\rangle = \frac{1}{\sqrt2} \bigl( \lvert\uparrow\downarrow\rangle - \lvert\downarrow\uparrow\rangle \bigr).$$

The words “triplet” and “singlet” count the number of magnetic sublevels in the total-spin multiplet. A spin-$1$ multiplet has $m=1,0,-1$, while a spin-$0$ multiplet has only $m=0$.

Exchange Symmetry

Let $P_{12}$ exchange the two spin slots:

$$P_{12} \lvert\alpha\rangle_1\lvert\beta\rangle_2 = \lvert\beta\rangle_1\lvert\alpha\rangle_2.$$

The triplet states are symmetric under this exchange:

$$P_{12}\lvert1,m\rangle = \lvert1,m\rangle, \qquad m=1,0,-1.$$

The singlet is antisymmetric:

$$P_{12}\lvert0,0\rangle = -\lvert0,0\rangle.$$

This is exchange symmetry of the spin factor alone. For identical particles, the symmetrization postulate applies to the full state, including spatial and internal degrees of freedom. The spin symmetry tells only how the spin part must be paired with the rest of the wavefunction.

Entanglement Content

The singlet and the $m=0$ triplet are entangled across the first-spin versus second-spin split:

$$\lvert0,0\rangle = \frac{1}{\sqrt2} \bigl( \lvert\uparrow\downarrow\rangle - \lvert\downarrow\uparrow\rangle \bigr),$$

and

$$\lvert1,0\rangle = \frac{1}{\sqrt2} \bigl( \lvert\uparrow\downarrow\rangle + \lvert\downarrow\uparrow\rangle \bigr).$$

Both have Schmidt coefficients $1/\sqrt2$ and $1/\sqrt2$, so the reduced state of either spin is maximally mixed:

$$\rho_1 = \rho_2 = \frac12 I.$$

The $m=\pm1$ triplet states are product states:

$$\lvert1,1\rangle = \lvert\uparrow\rangle_1\lvert\uparrow\rangle_2, \qquad \lvert1,-1\rangle = \lvert\downarrow\rangle_1\lvert\downarrow\rangle_2.$$

Therefore “triplet” does not specify an entanglement class. A normalized triplet superposition

$$|t\rangle=a|1,1\rangle+b|1,0\rangle+c|1,-1\rangle$$

has coefficient matrix

$$C_t=\begin{pmatrix}a&b/\sqrt2\\b/\sqrt2&c\end{pmatrix}.$$

It is product exactly when $ac=b^2/2$. In particular, two spins pointing along the same axis form a symmetric product state, even when their expansion contains all three $z$-basis triplet vectors.

Relation to Bell States

If the computational qubit basis is identified with

$$\lvert0\rangle=\lvert\uparrow\rangle, \qquad \lvert1\rangle=\lvert\downarrow\rangle,$$

then the $m=0$ triplet and the singlet are two Bell states:

$$\lvert1,0\rangle = \lvert\Psi^+\rangle, \qquad \lvert0,0\rangle = \lvert\Psi^-\rangle.$$

The other two standard Bell states,

$$\lvert\Phi^\pm\rangle = \frac{1}{\sqrt2} \bigl( \lvert\uparrow\uparrow\rangle \pm \lvert\downarrow\downarrow\rangle \bigr),$$

are superpositions inside the triplet subspace. They are eigenstates of $S^2$ with $s=1$, but they are not eigenstates of $S_z$ because they superpose $m=1$ and $m=-1$.

Thus the Bell basis and the total-spin basis overlap but are not identical. The Bell basis is organized by two-qubit correlation and phase labels. The singlet-triplet basis is organized by total angular momentum and exchange symmetry.

Spin Correlations

The singlet is invariant under the same spin rotation on both factors, $U\otimes U$ with $U\in SU(2)$. Independent rotations $U_A\otimes U_B$ preserve its entanglement but generally change the vector and its correlation axes. Its Pauli correlations are

$$\langle\sigma_i\otimes\sigma_j\rangle_{\mathrm{singlet}} = -\delta_{ij}, \qquad i,j\in\{x,y,z\}.$$

Equivalently, for unit vectors $\mathbf a$ and $\mathbf b$,

$$\left\langle (\boldsymbol\sigma\cdot\mathbf a) \otimes (\boldsymbol\sigma\cdot\mathbf b) \right\rangle_{\mathrm{singlet}} = -\mathbf a\cdot\mathbf b.$$

This formula says that measurements along the same axis are perfectly anticorrelated, and the prediction is independent of which common axis is chosen.

Triplet correlations depend on the triplet state. For example, in $\lvert1,0\rangle=\lvert\Psi^+\rangle$,

$$\langle\sigma_x\otimes\sigma_x\rangle = +1, \qquad \langle\sigma_y\otimes\sigma_y\rangle = +1, \qquad \langle\sigma_z\otimes\sigma_z\rangle = -1.$$

This state has perfect anticorrelation for $z$-axis spin measurements but perfect correlation for $x$- and $y$-axis Pauli measurements. The singlet is special because the anticorrelation is isotropic.

Identical Spin-1/2 Particles

For identical spin-$1/2$ fermions, the total state must be antisymmetric under particle exchange. If the state factors into a spatial part and a spin part,

$$\Psi(q_1,q_2) = \psi(\mathbf x_1,\mathbf x_2)\, \chi(s_1,s_2),$$

then the symmetry of the spatial factor and spin factor must multiply to an antisymmetric total state.

Thus:

• a symmetric spatial wavefunction must be paired with the antisymmetric spin singlet;
• an antisymmetric spatial wavefunction must be paired with a symmetric spin triplet.

This symmetry constraint explains the two-electron ground state of helium and the singlet/triplet splitting in many two-electron problems. The detailed exchange pairing rules are developed in Spin and Spatial Wavefunctions. It is also the simplest way to see why the Pauli exclusion principle is a statement about the full one-particle state, not about spin alone.

When the fermions are identical, the formal particle slots are not two independently addressable subsystems. The singlet’s factorization across labeled spin slots is useful algebra, but an operational entanglement claim must identify accessible modes or separated regions. Antisymmetrization alone is not a resource of entanglement between two laboratories.

Exchange Hamiltonians

The singlet-triplet basis diagonalizes any rotationally invariant two-spin Hamiltonian built from $\mathbf S_1\cdot\mathbf S_2$. Since

$$\mathbf S_1\cdot\mathbf S_2 = \frac12 \bigl( S^2-S_1^2-S_2^2 \bigr),$$

one finds

$$\mathbf S_1\cdot\mathbf S_2 = \frac14\hbar^2 \quad \text{on the triplet subspace},$$

and

$$\mathbf S_1\cdot\mathbf S_2 = -\frac34\hbar^2 \quad \text{on the singlet subspace}.$$

For a model Hamiltonian

$$H = \frac{J}{\hbar^2}\,\mathbf S_1\cdot\mathbf S_2,$$

the energies are $E_s=-3J/4$ and $E_t=J/4$, with $J$ in energy units. Positive $J$ favors the singlet and negative $J$ favors the triplet. Authors also use dimensionless spins or absorb a minus sign into $J$; compare the full Hamiltonian before comparing coupling constants.

Common pitfalls

Triplet symmetry does not imply entanglement. The triplet subspace contains both product and entangled vectors. Its total-spin label alone cannot distinguish them.

The singlet’s relative minus sign matters. Multiplying the entire vector by $-1$ changes no prediction, but changing the minus between its two terms into a plus makes the $m=0$ triplet.

Fermionic antisymmetry applies to the full state. Spin and spatial factors must have opposite exchange parity when such a factorization exists. Correlations among formal particle labels need a physical subsystem interpretation.

References

• Michael A. Nielsen and Isaac L. Chuang, Quantum Computation and Quantum Information, 10th anniversary ed., Cambridge University Press, 2010, section 2.5.
• J. J. Sakurai and Jim Napolitano, Modern Quantum Mechanics, 3rd ed., Cambridge University Press, 2020, chapter on angular momentum. Publisher.
• Ramamurti Shankar, Principles of Quantum Mechanics, 2nd ed., Springer, 1994, spin and identical particles.

Exercises

1. Verify the exchange symmetry of $\lvert1,0\rangle$ and $\lvert0,0\rangle$.

Solution

Exchange gives

$$P_{12}\lvert\uparrow\downarrow\rangle = \lvert\downarrow\uparrow\rangle, \qquad P_{12}\lvert\downarrow\uparrow\rangle = \lvert\uparrow\downarrow\rangle.$$

Therefore

$$P_{12}\lvert1,0\rangle = \frac{1}{\sqrt2} \bigl( \lvert\downarrow\uparrow\rangle + \lvert\uparrow\downarrow\rangle \bigr) = \lvert1,0\rangle,$$

while

$$P_{12}\lvert0,0\rangle = \frac{1}{\sqrt2} \bigl( \lvert\downarrow\uparrow\rangle - \lvert\uparrow\downarrow\rangle \bigr) = -\lvert0,0\rangle.$$

2. Compute the reduced density operator of either spin in the singlet.

Solution

For

$$\lvert0,0\rangle = \frac{1}{\sqrt2} \bigl( \lvert\uparrow\downarrow\rangle - \lvert\downarrow\uparrow\rangle \bigr),$$

the density operator is

$$\rho = \frac12 \Bigl( \lvert\uparrow\downarrow\rangle\langle\uparrow\downarrow\rvert - \lvert\uparrow\downarrow\rangle\langle\downarrow\uparrow\rvert - \lvert\downarrow\uparrow\rangle\langle\uparrow\downarrow\rvert + \lvert\downarrow\uparrow\rangle\langle\downarrow\uparrow\rvert \Bigr).$$

Tracing over the second spin removes the cross terms because $\langle\downarrow\vert\uparrow\rangle=0$, leaving

$$\rho_1 = \frac12 \bigl( \lvert\uparrow\rangle\langle\uparrow\rvert + \lvert\downarrow\rangle\langle\downarrow\rvert \bigr) = \frac12 I.$$

The same result holds for the second spin.

3. Expand $|+\rangle|+\rangle$ in the triplet basis and check the product condition $ac=b^2/2$.

Solution

$$|++\rangle=\frac12|1,1\rangle+\frac1{\sqrt2}|1,0\rangle+\frac12|1,-1\rangle.$$

Thus $ac=1/4=b^2/2$. The superposition is product despite its nonzero component along the entangled $m=0$ triplet. Entanglement is a property of the complete vector, not of individual terms in an expansion.

4. Two identical electrons occupy the same spatial orbital. Which spin state is allowed?

Solution

If the spatial orbital is the same for both electrons, the spatial factor is symmetric under exchange. Electrons are fermions, so the total state must be antisymmetric. Therefore the spin factor must be antisymmetric, which means the allowed spin state is the singlet:

$$\frac{1}{\sqrt2} \bigl( \lvert\uparrow\downarrow\rangle - \lvert\downarrow\uparrow\rangle \bigr).$$

Triplet spin states are symmetric and would require an antisymmetric spatial factor.