Heisenberg Dimer: Singlet–Triplet Splitting

Main result

For two spin-$\tfrac{1}{2}$ particles coupled by the isotropic Heisenberg exchange

\[H \;=\; J\,\mathbf{S}_{1}\cdot\mathbf{S}_{2},\]

the Hilbert space $\mathbb{C}^{2}\otimes\mathbb{C}^{2}$ (dimension $4$) decomposes into a singlet ($S_{\rm tot} = 0$) and a triplet ($S_{\rm tot} = 1$), with spectrum and gap

\[\boxed{\; \mathrm{Spectrum}(H) \;=\; \Bigl\{\,-\tfrac{3 J}{4},\;\tfrac{J}{4},\; \tfrac{J}{4},\;\tfrac{J}{4}\,\Bigr\}, \qquad \Delta \;=\; E_{t} - E_{s} \;=\; J. \;}\]

For $J > 0$ (antiferromagnetic) the singlet is the ground state; for $J < 0$ (ferromagnetic) the three degenerate triplet states are the ground manifold.

The singlet and triplet eigenvectors are

\[\boxed{\; |s\rangle = \tfrac{1}{\sqrt{2}}\bigl(|{\uparrow\downarrow}\rangle - |{\downarrow\uparrow}\rangle\bigr), \qquad \{|t_{+1}\rangle,\,|t_{0}\rangle,\,|t_{-1}\rangle\} = \Bigl\{|{\uparrow\uparrow}\rangle,\;\tfrac{1}{\sqrt{2}}\bigl(|{\uparrow\downarrow}\rangle + |{\downarrow\uparrow}\rangle\bigr),\;|{\downarrow\downarrow}\rangle\Bigr\}. \;}\]

This is the $N = 2$ building block of every result for spin-1/2 Heisenberg chains and ladders; its exact spectrum is the reference point against which every lattice-ED and QAtlas cross-check is validated at small system sizes.

Setup

Operators and conventions

Spin-1/2 operators $\mathbf{S}_{i} = \tfrac{1}{2} \boldsymbol{\sigma}_{i}$ act on the $i$-th factor of $\mathcal{H} = \mathbb{C}^{2}\otimes\mathbb{C}^{2}$. The computational basis is

\[\bigl\{\,|{\uparrow\uparrow}\rangle,\, |{\uparrow\downarrow}\rangle,\, |{\downarrow\uparrow}\rangle,\, |{\downarrow\downarrow}\rangle\,\bigr\}.\]

Raising / lowering operators $S^{\pm}_{i} = S^{x}_{i} \pm i S^{y}_{i}$ satisfy $S^{+}_{i}|{\downarrow}\rangle_{i} = |{\uparrow}\rangle_{i}$, $S^{-}_{i}|{\uparrow}\rangle_{i} = |{\downarrow}\rangle_{i}$, all other actions zero. The Casimir

\[\mathbf{S}_{i}^{2} \;=\; \tfrac{3}{4}\,\mathbb{I}_{i}, \qquad s = \tfrac{1}{2},\ \ s(s + 1) = \tfrac{3}{4}.\]

Total-spin decomposition

Angular-momentum addition $1/2 \otimes 1/2 = 0 \oplus 1$ splits the four basis states into a singlet and a triplet. We will label eigenstates by the two commuting good quantum numbers $(S_{\rm tot}, S^{z}_{\rm tot})$.

Goal

Derive the four eigenvalues of $H$ from the $\mathbf{S}_{\rm tot}^{2}$ Casimir, identify the corresponding eigenvectors explicitly, and verify them by direct action of $H$ and $\mathbf{S}^{2}_{\rm tot}$ on the basis.

Calculation

Step 1 — Rewrite $H$ using $\mathbf{S}_{\rm tot}^{2}$

Define $\mathbf{S}_{\rm tot} = \mathbf{S}_{1} + \mathbf{S}_{2}$ and square:

\[\mathbf{S}_{\rm tot}^{2} \;=\; (\mathbf{S}_{1} + \mathbf{S}_{2})^{2} \;=\; \mathbf{S}_{1}^{2} + \mathbf{S}_{2}^{2} + 2\,\mathbf{S}_{1}\cdot\mathbf{S}_{2}.\]

Solve for $\mathbf{S}_{1}\cdot\mathbf{S}_{2}$:

\[\mathbf{S}_{1}\cdot\mathbf{S}_{2} \;=\; \tfrac{1}{2}\bigl[\mathbf{S}_{\rm tot}^{2} - \mathbf{S}_{1}^{2} - \mathbf{S}_{2}^{2}\bigr] \;=\; \tfrac{1}{2}\bigl[\,S_{\rm tot}(S_{\rm tot} + 1) - \tfrac{3}{2}\,\bigr]. \tag{1}\]

Multiplying by $J$, we read off the two eigenvalues of $H$:

\[E(S_{\rm tot} = 0) = J\cdot\tfrac{1}{2}\bigl[\,0 - \tfrac{3}{2}\,\bigr] = -\tfrac{3 J}{4},\]

\[E(S_{\rm tot} = 1) = J\cdot\tfrac{1}{2}\bigl[\,2 - \tfrac{3}{2}\,\bigr] = \tfrac{J}{4}.\]

The singlet is one-dimensional ($2S_{\rm tot} + 1 = 1$ state) and the triplet is three-dimensional ($2S_{\rm tot} + 1 = 3$ states), matching the Hilbert-space dimension $1 + 3 = 4$. The gap is

\[\Delta \;=\; E(1) - E(0) \;=\; \tfrac{J}{4} - \bigl(-\tfrac{3 J}{4}\bigr) \;=\; J.\]

Step 2 — Singlet eigenvector by explicit $\mathbf{S}^{2}_{\rm tot}$ verification

Claim: |s\rangle = \tfrac{1}{\sqrt{2}}(|{\uparrow\downarrow}\rangle

• |{\downarrow\uparrow}\rangle)$ has $S_{\rm tot} = 0$ and

\[S^{z}_{\rm tot} = 0\]

.

Verify $S^{z}_{\rm tot}|s\rangle = 0$. Acting on each term:

\[S^{z}_{\rm tot}|{\uparrow\downarrow}\rangle = (\tfrac{1}{2} + (-\tfrac{1}{2}))\,|{\uparrow\downarrow}\rangle = 0,\]

\[S^{z}_{\rm tot}|{\downarrow\uparrow}\rangle = (-\tfrac{1}{2} + \tfrac{1}{2})\,|{\downarrow\uparrow}\rangle = 0.\]

So $S^{z}_{\rm tot}|s\rangle = 0$. ✓

Verify $\mathbf{S}^{2}_{\rm tot}|s\rangle = 0$. Use $\mathbf{S}^{2}_{\rm tot} = (S^{z}_{\rm tot})^{2} + \tfrac{1}{2} (S^{+}_{\rm tot}S^{-}_{\rm tot} + S^{-}_{\rm tot}S^{+}_{\rm tot})$ with $S^{\pm}_{\rm tot} = S^{\pm}_{1} + S^{\pm}_{2}$:

\[S^{+}_{\rm tot}|{\uparrow\downarrow}\rangle = S^{+}_{1}|{\uparrow\downarrow}\rangle + S^{+}_{2}|{\uparrow\downarrow}\rangle = 0 + |{\uparrow\uparrow}\rangle = |{\uparrow\uparrow}\rangle,\]

\[S^{+}_{\rm tot}|{\downarrow\uparrow}\rangle = |{\uparrow\uparrow}\rangle + 0 = |{\uparrow\uparrow}\rangle,\]

so $S^{+}_{\rm tot}|s\rangle = \tfrac{1}{\sqrt{2}} (|{\uparrow\uparrow}\rangle - |{\uparrow\uparrow}\rangle) = 0$. Similarly $S^{-}_{\rm tot}|s\rangle = 0$. Combined with $(S^{z}_{\rm tot})^{2}|s\rangle = 0$, this gives $\mathbf{S}^{2}_{\rm tot}|s\rangle = 0$, confirming $S_{\rm tot} (S_{\rm tot} + 1) = 0 \Rightarrow S_{\rm tot} = 0$. ✓

Step 3 — Triplet eigenvectors by $S^{\pm}_{\rm tot}$ ladder

The $|t_{+1}\rangle = |{\uparrow\uparrow}\rangle$ state has $S^{z}_{\rm tot}|{\uparrow\uparrow}\rangle = +1 \cdot |{\uparrow\uparrow}\rangle$ (maximal $S^z$), so it is the "highest-weight" vector of the triplet. Apply $S^{-}_{\rm tot}$:

\[S^{-}_{\rm tot}|{\uparrow\uparrow}\rangle = S^{-}_{1}|{\uparrow\uparrow}\rangle + S^{-}_{2}|{\uparrow\uparrow}\rangle = |{\downarrow\uparrow}\rangle + |{\uparrow\downarrow}\rangle = \sqrt{2}\cdot\tfrac{1}{\sqrt{2}}\bigl(|{\uparrow\downarrow}\rangle + |{\downarrow\uparrow}\rangle\bigr),\]

using the standard ladder normalisation $S^{-}|S, m\rangle = \sqrt{S(S + 1) - m(m - 1)}\,|S, m - 1\rangle$ which for $S = 1, m = 1$ gives the factor $\sqrt{2}$. Hence

\[|t_{0}\rangle \;=\; \tfrac{1}{\sqrt{2}}\bigl(|{\uparrow\downarrow}\rangle + |{\downarrow\uparrow}\rangle\bigr).\]

Applying $S^{-}_{\rm tot}$ once more:

\[S^{-}_{\rm tot}|t_{0}\rangle = \tfrac{1}{\sqrt{2}}\bigl(S^{-}_{\rm tot}|{\uparrow\downarrow}\rangle + S^{-}_{\rm tot}|{\downarrow\uparrow}\rangle\bigr) = \tfrac{1}{\sqrt{2}}(|{\downarrow\downarrow}\rangle + 0 + 0 + |{\downarrow\downarrow}\rangle) = \sqrt{2}\,|{\downarrow\downarrow}\rangle,\]

confirming $|t_{-1}\rangle = |{\downarrow\downarrow}\rangle$. The three triplet states form a Clebsch–Gordan-standard spin-1 multiplet.

Verify triplet is eigenspace with $\mathbf{S}^{2}_{\rm tot} = 2$. For $|t_{+1}\rangle$, $S^{z}|t_{+1}\rangle = +1 |t_{+1}\rangle$ and $S^{+}|t_{+1}\rangle = 0$ (highest weight). Using $\mathbf{S}^{2} = (S^{z})^{2} + S^{z} + S^{-}S^{+}$:

\[\mathbf{S}^{2}|t_{+1}\rangle = (1 + 1 + 0)|t_{+1}\rangle = 2\,|t_{+1}\rangle,\]

i.e. $S_{\rm tot}(S_{\rm tot} + 1) = 2 \Rightarrow S_{\rm tot} = 1$. The other two triplet states follow by ladder-preservation of the $\mathbf{S}^{2}$ eigenvalue.

Step 4 — Direct $4\times 4$ diagonalisation cross-check

Write $H = J\,\mathbf{S}_{1}\cdot\mathbf{S}_{2}$ as an explicit $4 \times 4$ matrix in the computational basis $\{|{\uparrow\uparrow} \rangle, |{\uparrow\downarrow}\rangle, |{\downarrow\uparrow} \rangle, |{\downarrow\downarrow}\rangle\}$. Expand

\[\mathbf{S}_{1}\cdot\mathbf{S}_{2} = S^{z}_{1}S^{z}_{2} + \tfrac{1}{2}\bigl(S^{+}_{1}S^{-}_{2} + S^{-}_{1}S^{+}_{2}\bigr).\]

Compute matrix elements:

\[\begin{aligned} \langle{\uparrow\uparrow}|\mathbf{S}_{1}\cdot\mathbf{S}_{2}|{\uparrow\uparrow}\rangle &= \tfrac{1}{2}\cdot\tfrac{1}{2} = \tfrac{1}{4},\\ \langle{\uparrow\downarrow}|\mathbf{S}_{1}\cdot\mathbf{S}_{2}|{\uparrow\downarrow}\rangle &= \tfrac{1}{2}\cdot(-\tfrac{1}{2}) = -\tfrac{1}{4},\\ \langle{\downarrow\uparrow}|\mathbf{S}_{1}\cdot\mathbf{S}_{2}|{\downarrow\uparrow}\rangle &= -\tfrac{1}{4},\\ \langle{\downarrow\downarrow}|\mathbf{S}_{1}\cdot\mathbf{S}_{2}|{\downarrow\downarrow}\rangle &= \tfrac{1}{4},\\ \langle{\uparrow\downarrow}|\mathbf{S}_{1}\cdot\mathbf{S}_{2}|{\downarrow\uparrow}\rangle &= \tfrac{1}{2},\\ \langle{\downarrow\uparrow}|\mathbf{S}_{1}\cdot\mathbf{S}_{2}|{\uparrow\downarrow}\rangle &= \tfrac{1}{2}. \end{aligned}\]

The $4\times 4$ matrix of $H/J$ (in the order listed above) is

\[\frac{H}{J} \;=\; \begin{pmatrix} 1/4 & 0 & 0 & 0 \\ 0 & -1/4 & 1/2 & 0 \\ 0 & 1/2 & -1/4 & 0 \\ 0 & 0 & 0 & 1/4 \end{pmatrix}.\]

The $(|{\uparrow\uparrow}\rangle, |{\downarrow\downarrow}\rangle)$ states are already eigenvectors with eigenvalue $1/4$. The middle $2\times 2$ block

\[M \;=\; \begin{pmatrix}-1/4 & 1/2 \\ 1/2 & -1/4\end{pmatrix}\]

has eigenvalues $-1/4 \pm 1/2 = \{1/4, -3/4\}$ with eigenvectors $(1, 1)^{T}/\sqrt{2}$ and $(1, -1)^{T}/\sqrt{2}$. Multiplying by $J$ recovers

\[\mathrm{spec}(H) = \{1/4, 1/4, 1/4, -3/4\}\cdot J = \{J/4, J/4, J/4, -3J/4\},\]

exactly the Main-result spectrum. The $(+, +)$ eigenvector is $|t_{0}\rangle$ at eigenvalue $J/4$, and the $(+, -)$ eigenvector is $|s\rangle$ at eigenvalue $-3J/4$ — matching the total-spin identification.

Step 5 — Limiting-case checks

(i) AF ground state $J > 0$. The singlet at $E_{s} = -3J/4$ is the unique ground state. The first-excited triplet at $E_{t} = J/4$ is three-fold degenerate, with gap $\Delta = J$.

(ii) FM ground state $J < 0$. The triplet is now the three-fold-degenerate ground manifold at $E_{t} = J/4 < 0$, and the singlet at $E_{s} = -3J/4 > 0$ is an excited state. The sign convention is reversed but the spectrum is the same four eigenvalues up to reordering.

(iii) Trace. $\mathrm{Tr}\,H = J\,\mathrm{Tr}\,(\mathbf{S}_{1} \cdot\mathbf{S}_{2}) = 0$ identically (since $\mathbf{S}_{i}$ are traceless). Check: $-3 J/4 + 3 \cdot J/4 = 0$. ✓

(iv) Cross-reference with the spin-1/2 Heisenberg chain thermodynamic limit. The infinite AF Heisenberg chain has ground-state energy per bond $e_{0} = J(1/4 - \ln 2) \approx -0.443\,J$, derived in bethe-ansatz-heisenberg-e0. For $N = 2$ the per-bond energy is $E_{s}/1 = -3 J/4 = -0.75\,J$ (only one bond on a 2-site chain with open boundary condition; for PBC there are two "bonds" that double-count, giving $-3 J/2$ total and $-3 J/4$ per bond after division by 2). The finite-$N = 2$ value overshoots the thermodynamic limit because the dimer is maximally "entangled" locally without the spinon-fluctuation dilution that occurs at larger $N$.

Step 6 — QAtlas lattice verification

QAtlas confirms the dimer spectrum against independent lattice ED:

λ = QAtlas.fetch(Heisenberg1D(), ExactSpectrum(); N=2, J=1.0, bc=:OBC)
# → [-0.75, 0.25, 0.25, 0.25]

matching $\{-3J/4, J/4, J/4, J/4\}$ for $J = 1$. test/verification/test_heisenberg_dimer.jl cross-checks this against a Lattice2D-based spin-1/2 ED build — both constructions give identical eigenvalues to machine precision.

References

• A. Auerbach, Interacting Electrons and Quantum Magnetism (Springer, 1994), §2. Textbook derivation of (1) and the singlet– triplet decomposition.
• J. J. Sakurai, Modern Quantum Mechanics, 2nd ed. (Addison- Wesley, 2011), Ch. 3. Clebsch–Gordan addition for two spin-1/2 particles.
• K. Huang, Statistical Mechanics, 2nd ed. (Wiley, 1987), §13.2. Heisenberg dimer as the simplest interacting magnetic system.

Used by

• Heisenberg model page — $N = 2$ ExactSpectrum returns exactly the Main-result set.
• bethe-ansatz-heisenberg-e0 — the $N = 2$ dimer result is the smallest-$N$ limiting-case check for the thermodynamic-limit Bethe-ansatz energy $e_{0} = J(1/4 - \ln 2)$; the dimer overshoots $e_{0}$ from below at finite $N$ and converges from finite-size corrections.
• XXZ model page — dimer limit at the isotropic point $\Delta = 1$.