Singlet and Triplet States

The singlet and triplet states are the two irreducible rotational sectors of a pair of spin-$1/2$ systems. They are the concrete content of

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

This page owns the symmetry meaning of the decomposition: the singlet is a scalar under joint rotations, while the triplet is a spin-one vector multiplet. Entanglement properties, Bell-state comparisons, and identical-particle applications are developed in the composite-systems page on Singlet and Triplet States.

Spin Product Space

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

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

Choose the $z$-axis basis for each spin. The uncoupled product basis is

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

It diagonalizes $S_{1z}$ and $S_{2z}$. The total spin operators are

$$\mathbf S = \mathbf S_1+\mathbf S_2, \qquad S_z = S_{1z}+S_{2z}.$$

The coupled basis instead diagonalizes $S^2$ and $S_z$. This is the natural basis for rotationally invariant two-spin interactions and for questions about total angular momentum.

Triplet Multiplet

The spin-$1$ sector has three magnetic sublevels. In the standard Condon-Shortley phase convention,

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

These states satisfy

$$S^2\lvert1,m\rangle = 2\hbar^2\lvert1,m\rangle, \qquad S_z\lvert1,m\rangle = \hbar m\lvert1,m\rangle.$$

The word “triplet” refers to the three $m$ values. It does not mean that every state in the triplet subspace is entangled, nor that triplet states are three-particle states.

Under joint rotations, the triplet subspace maps into itself. It carries the same irreducible spin-$1$ representation that appears for ordinary vector angular momentum.

Singlet State

The remaining state is the spin-$0$ singlet:

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

It obeys

$$S^2\lvert0,0\rangle=0, \qquad S_z\lvert0,0\rangle=0.$$

The stronger statement is that every component of the total spin annihilates the singlet:

$$S_x\lvert0,0\rangle = S_y\lvert0,0\rangle = S_z\lvert0,0\rangle = 0.$$

Therefore a joint spin rotation

$$U(R) = \exp\!\left( -\frac{i}{\hbar}\theta\,\hat{\mathbf n}\cdot\mathbf S \right)$$

leaves the singlet invariant:

$$U(R)\lvert0,0\rangle = \lvert0,0\rangle.$$

This rotational invariance is the symmetry reason the singlet has isotropic spin correlations. It is not just a special property of the $z$-axis basis.

How the Decomposition Is Found

The highest $m$ state in the product basis is $\lvert\uparrow\uparrow\rangle$. Since it has $m=1$, it must be the highest-weight state of a spin-$1$ multiplet:

$$\lvert1,1\rangle = \lvert\uparrow\uparrow\rangle.$$

Apply the total lowering operator

$$S_- = S_{1-}+S_{2-}.$$

Because

$$S_-\lvert1,1\rangle = \hbar\sqrt2\,\lvert1,0\rangle,$$

and

$$S_-\lvert\uparrow\uparrow\rangle = \hbar \left( \lvert\downarrow\uparrow\rangle + \lvert\uparrow\downarrow\rangle \right),$$

normalization gives the $m=0$ triplet state. The other independent $m=0$ product-space vector must be orthogonal to it, which fixes the singlet up to an overall phase.

This construction is the first nontrivial example of Clebsch–Gordan coefficients.

Projectors Onto the Sectors

The scalar product of the two spins can be written in terms of the total spin:

$$\mathbf S_1\cdot\mathbf S_2 = \frac12 \left( S^2-S_1^2-S_2^2 \right).$$

For two spin-$1/2$ systems,

$$S_1^2=S_2^2=\frac34\hbar^2.$$

Therefore

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

and

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

The corresponding projectors are

$$P_{\mathrm{singlet}} = \frac14 I - \frac{1}{\hbar^2}\mathbf S_1\cdot\mathbf S_2,$$

and

$$P_{\mathrm{triplet}} = \frac34 I + \frac{1}{\hbar^2}\mathbf S_1\cdot\mathbf S_2.$$

Equivalently, using $\mathbf S_i=\hbar\boldsymbol\sigma_i/2$,

$$P_{\mathrm{singlet}} = \frac14 \left( I-\boldsymbol\sigma_1\cdot\boldsymbol\sigma_2 \right), \qquad P_{\mathrm{triplet}} = \frac14 \left( 3I+\boldsymbol\sigma_1\cdot\boldsymbol\sigma_2 \right).$$

These projectors are often the cleanest way to identify singlet and triplet content without choosing a particular spin quantization axis.

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 subspace is symmetric:

$$P_{12}\lvert1,m\rangle = \lvert1,m\rangle.$$

The singlet is antisymmetric:

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

This is exchange symmetry of the spin factor. For identical particles, the symmetrization postulate applies to the full state, including spatial degrees of freedom. The spin-space pairing rules are developed in Spin and Spatial Wavefunctions.

Spin Correlations

The singlet has no preferred axis. In Pauli-matrix notation,

$$\langle\sigma_i\otimes\sigma_j\rangle_{\mathrm{singlet}} = -\delta_{ij}.$$

For any two unit vectors $\mathbf a$ and $\mathbf b$,

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

This formula is the rotationally invariant version of perfect anticorrelation along the same axis. It is central in spin-correlation experiments and in Bell-inequality discussions, but Bell’s theorem itself requires a separate assumptions-and-inequalities analysis.

Triplet correlations are not isotropic in the same way for a fixed $m$ state. For example, $\lvert1,1\rangle=\lvert\uparrow\uparrow\rangle$ gives perfect $z$-axis correlation, while $\lvert1,0\rangle$ gives perfect $z$-axis anticorrelation. The triplet is a multiplet, not a single rotationally invariant state.

Hamiltonian Uses

Any Hamiltonian of the form

$$H = J\,\mathbf S_1\cdot\mathbf S_2$$

is diagonal in the singlet-triplet decomposition. Its energies are

$$E_{\mathrm{triplet}} = \frac14J\hbar^2, \qquad E_{\mathrm{singlet}} = -\frac34J\hbar^2.$$

With this convention, positive $J$ favors the singlet and negative $J$ favors the triplet. Other fields may define $J$ with the opposite sign, so the Hamiltonian convention should always be stated.

This same algebra appears in exchange models, two-electron spin splitting, hyperfine coupling, positronium, and effective spin Hamiltonians. The common structure is not the microscopic origin of $J$, but the fact that $\mathbf S_1\cdot\mathbf S_2$ is a rotational scalar whose eigenvalues are fixed by total spin.

What Is Canonical Here

This page is the canonical home for the symmetry structure of the singlet-triplet split. Use it for:

• the $0\oplus1$ rotational decomposition of two spin-$1/2$ systems;
• singlet rotational invariance;
• triplet transformation as a spin-one multiplet;
• projectors onto singlet and triplet sectors;
• scalar interactions proportional to $\mathbf S_1\cdot\mathbf S_2$.

For entanglement measures, Bell states, reduced density matrices, and identical-particle spin-spatial bookkeeping, use the linked composite-systems pages. The angular-momentum bridge to exchange symmetry is Identical Particles and Exchange Symmetry Preview.

Common Mistakes

• Treating $\frac12\otimes\frac12=1\oplus0$ as ordinary arithmetic rather than representation decomposition.
• Calling every $m=0$ state a singlet. The triplet also has an $m=0$ state.
• Forgetting that the singlet is invariant under joint rotations, not under rotating only one spin.
• Assuming that “triplet” means “entangled.” The states $\lvert1,1\rangle$ and $\lvert1,-1\rangle$ are product states in the usual two-spin split.
• Applying spin exchange symmetry to identical particles without including the spatial part of the state.
• Treating the sign of $J$ in $J\,\mathbf S_1\cdot\mathbf S_2$ as universal across communities.

Cross-Links

• Tensor Product Representations in Angular Momentum
• Total Angular Momentum
• Coupled and Uncoupled Bases
• Two Spin-1/2 Particles
• Identical Particles and Exchange Symmetry Preview
• Clebsch–Gordan Coefficients
• Spin-1/2 Hilbert Space
• Spin Rotations
• Pauli Matrices
• Tensor Products
• Composite-Systems Singlet and Triplet States
• Bell States
• Spin and Spatial Wavefunctions
• Symmetrization Postulate
• Bell Theorem
• CHSH Inequality

References

• J. J. Sakurai and J. Napolitano, Modern Quantum Mechanics, 3rd ed., Cambridge University Press, 2020.
• R. Shankar, Principles of Quantum Mechanics, 2nd ed., Springer, 1994.
• A. R. Edmonds, Angular Momentum in Quantum Mechanics, Princeton University Press, 1957.
• C. Cohen-Tannoudji, B. Diu, and F. Laloe, Quantum Mechanics, Wiley, 1977.
• D. A. Varshalovich, A. N. Moskalev, and V. K. Khersonskii, Quantum Theory of Angular Momentum, World Scientific, 1988.
• A. Peres, Quantum Theory: Concepts and Methods, Kluwer, 1995.

Exercises

1. Verify that $\lvert0,0\rangle$ is annihilated by $S_z$.

Solution

Use

$$S_z = S_{1z}+S_{2z}.$$

The two product states in the singlet each have total $m=0$:

$$S_z\lvert\uparrow\downarrow\rangle=0, \qquad S_z\lvert\downarrow\uparrow\rangle=0.$$

Therefore

$$S_z\lvert0,0\rangle = \frac{1}{\sqrt2} \left( 0-0 \right) =0.$$

2. Derive the eigenvalues of $\mathbf S_1\cdot\mathbf S_2$ on the triplet and singlet sectors.

Solution

Use

$$\mathbf S_1\cdot\mathbf S_2 = \frac12 \left( S^2-S_1^2-S_2^2 \right).$$

For two spin-$1/2$ systems,

$$S_1^2=S_2^2=\frac34\hbar^2.$$

On the triplet, $s=1$, so $S^2=2\hbar^2$ and

$$\mathbf S_1\cdot\mathbf S_2 = \frac12 \left( 2-\frac34-\frac34 \right)\hbar^2 = \frac14\hbar^2.$$

On the singlet, $s=0$, so

$$\mathbf S_1\cdot\mathbf S_2 = \frac12 \left( 0-\frac34-\frac34 \right)\hbar^2 = -\frac34\hbar^2.$$

3. Check that $P_{\mathrm{singlet}}$ is a projector by evaluating it on singlet and triplet eigenstates.

Solution

The operator is

$$P_{\mathrm{singlet}} = \frac14 I - \frac{1}{\hbar^2}\mathbf S_1\cdot\mathbf S_2.$$

On the singlet,

$$P_{\mathrm{singlet}} = \frac14 - \left( -\frac34 \right) = 1.$$

On the triplet,

$$P_{\mathrm{singlet}} = \frac14 - \frac14 = 0.$$

Thus it acts as the identity on the singlet subspace and zero on the triplet subspace, so $P_{\mathrm{singlet}}^2=P_{\mathrm{singlet}}$.

4. Explain why the singlet is invariant under joint rotations but not under rotating only one spin.

Solution

Joint rotations are generated by

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

Because the singlet has total spin $s=0$, every component of $\mathbf S$ annihilates it, so a joint rotation leaves it unchanged. A rotation of only the first spin is generated by $\mathbf S_1$, not by $\mathbf S_1+\mathbf S_2$. The singlet is not annihilated by $\mathbf S_1$ alone, so a one-spin rotation generally turns it into another two-spin state.