Heisenberg Chain

The spin-$1/2$ Heisenberg chain is a one-dimensional array of two-level quantum spins coupled by rotationally invariant nearest-neighbor exchange, with an exact ferromagnetic ground multiplet for $J<0$ and a Bethe-ansatz-integrable, gapless spinon liquid for $J>0$.

The reference chain has uniform exchange, no applied field, and dimensionless spins. Small examples make the energy normalization and spin multiplets explicit before the thermodynamic antiferromagnet is discussed.

Helpful background. Spin-1/2 Chain introduces local operators and bond conventions; Heisenberg Model explains exchange on general graphs. Exact Solutions Preview introduces Bethe rapidities.

Isotropic Exchange in One Dimension

The model has the following physical choices.

Field	Baseline choice
sites	$L$ spin-$1/2$ degrees of freedom
local space	$\mathcal H_j\simeq\mathbb C^2$
spin convention	dimensionless $\mathbf s_j=\boldsymbol\sigma_j/2$
geometry	a uniform one-dimensional nearest-neighbor chain
exchange	isotropic $x$, $y$, and $z$ coupling
coupling	real $J$ with units of energy
sign	$J>0$ antiferromagnetic; $J<0$ ferromagnetic
boundaries	open or periodic, stated explicitly
field	absent
energy offset	none
anisotropy, disorder, and longer range	absent
thermodynamic antiferromagnetic reference	even periodic rings followed by $L\to\infty$

This baseline is narrower than “the Heisenberg model” as a family. Changing the onsite spin, graph, dimension, bond signs, unit cell, anisotropy, or field can change the low-energy phase and exact solution status. State these changes explicitly when comparing spectra.

Degrees of Freedom and Basis

The many-body Hilbert space is

$$\mathcal H_L = \bigotimes_{j=1}^{L} \mathbb C^2, \qquad \dim\mathcal H_L = 2^L.$$

Dimensionless local spins obey

$$\begin{aligned} [s_i^\alpha,s_j^\beta] &= i\delta_{ij} \epsilon_{\alpha\beta\gamma} s_i^\gamma, \\ \mathbf s_i^2 &= \frac34, \end{aligned}$$

with

$$\mathbf s_i = \frac12 \boldsymbol\sigma_i.$$

The computational basis diagonalizes every $s_j^z$:

$$s^z\lvert\uparrow\rangle = \frac12\lvert\uparrow\rangle, \qquad s^z\lvert\downarrow\rangle = -\frac12\lvert\downarrow\rangle.$$

A basis state can be encoded by a bit string, but the Hamiltonian is not diagonal in that basis. On one bond,

$$\mathbf s_i\cdot\mathbf s_j = s_i^zs_j^z + \frac12 \left( s_i^+s_j^- + s_i^-s_j^+ \right).$$

The first term measures parallel or antiparallel $z$ projections. The remaining terms exchange an up-down pair and are responsible for quantum fluctuations in the antiferromagnet.

Hamiltonian and Boundary Convention

For an open chain,

$$H_{\mathrm O} = J \sum_{j=1}^{L-1} \mathbf s_j\cdot\mathbf s_{j+1}.$$

For a periodic ring,

$$H_{\mathrm P} = J \sum_{j=1}^{L} \mathbf s_j\cdot\mathbf s_{j+1}, \qquad \mathbf s_{L+1} = \mathbf s_1.$$

The open chain has $L-1$ undirected bonds; an ordinary ring has $L$. Each physical bond appears once. The $L=2$ periodic shorthand is exceptional because the terms labeled $j=1$ and $j=2$ refer to the same undirected pair. A two-site ring must specify whether it contains one bond or a deliberately doubled multigraph edge.

Pauli-matrix form

Because $\mathbf s_j=\boldsymbol\sigma_j/2$,

$$H = \frac J4 \sum_{\langle j,k\rangle} \boldsymbol\sigma_j \mathbin{\cdot} \boldsymbol\sigma_k.$$

A spectrum four times too large usually comes from inserting Pauli matrices directly into the spin-operator Hamiltonian without the factor $1/4$.

Physical angular momenta are

$$\mathbf S_j = \hbar\mathbf s_j.$$

In that convention the same exchange is

$$J\mathbf s_i\cdot\mathbf s_j = \frac{J}{\hbar^2} \mathbf S_i\cdot\mathbf S_j.$$

Writing $J\mathbf S_i\cdot\mathbf S_j$ instead assigns different units to the symbol $J$.

Bond spin and permutation

For two spins,

$$\mathbf s_{ij} = \mathbf s_i+\mathbf s_j,$$

and

$$\mathbf s_i\cdot\mathbf s_j = \frac12 \left( \mathbf s_{ij}^2 - \mathbf s_i^2 - \mathbf s_j^2 \right).$$

Two spin-$1/2$ spaces decompose into a singlet and a triplet:

$$\frac12\otimes\frac12 = 0\oplus1.$$

The exchange eigenvalues are

$$\begin{array}{c|c|c} \text{bond spin} & \mathbf s_i\cdot\mathbf s_j & E/J \\ \hline 0 & -3/4 & -3/4 \\ 1 & +1/4 & +1/4 \end{array}$$

so $J>0$ favors the singlet and $J<0$ favors the triplet. Equivalently, the swap operator on two spin-$1/2$ sites satisfies

$$\mathsf P_{ij} = 2\mathbf s_i\cdot\mathbf s_j + \frac12\mathbb I.$$

The bond projectors are

$$\begin{aligned} \Pi_{S} &= \frac14 - \mathbf s_i\cdot\mathbf s_j, \\ \Pi_{T} &= \frac34 + \mathbf s_i\cdot\mathbf s_j. \end{aligned}$$

These formulas are exact local checks and do not imply that every bond in an extended antiferromagnet can simultaneously occupy a singlet.

Symmetries

Global spin rotations

Define total spin

$$\mathbf S_{\mathrm{tot}} = \sum_{j=1}^{L} \mathbf s_j.$$

Every exchange bond is a scalar under a common spin rotation, hence

$$\begin{aligned} [H,S_{\mathrm{tot}}^\alpha] &= 0, \\ [H,\mathbf S_{\mathrm{tot}}^2] &= 0. \end{aligned}$$

The internal continuous symmetry is $SU(2)$ at the Hilbert-space level, with physical rotations represented projectively through the corresponding $SO(3)$ action. Energy eigenstates can be organized into total-spin multiplets labeled by

$$S_{\mathrm{tot}}= \begin{cases} 0,1,\ldots,L/2,&L\text{ even},\\ 1/2,3/2,\ldots,L/2,&L\text{ odd}. \end{cases}$$

A spin-$S$ multiplet contains $2S+1$ states with

$$m = -S,-S+1,\ldots,S.$$

Numerically equal energies across all these $m$ values are a stringent symmetry check.

Fixed-magnetization sectors

The conserved component

$$S_{\mathrm{tot}}^z = \sum_j s_j^z$$

allows a product-basis block decomposition. If $M$ sites are down,

$$S_{\mathrm{tot}}^z = \frac L2-M,$$

and the block dimension is

$$\dim\mathcal H_M = \binom LM.$$

An $S_{\mathrm{tot}}^z$ block does not isolate one total-spin irreducible representation; it generally contains one state from every compatible multiplet. Recovering the complete spectrum requires all magnetization blocks or an explicit multiplet reconstruction.

Lattice and antiunitary symmetries

The uniform periodic chain has translation and reflection symmetry. The open chain retains reflection about its midpoint but not one-site translation symmetry.

With no magnetic field, physical time reversal sends

$$\mathbf s_j \mapsto -\mathbf s_j$$

and leaves every exchange bond invariant. For $L$ spin-$1/2$ sites, the many-body time-reversal operator obeys

$$\Theta^2 = (-1)^L.$$

Odd chains therefore carry Kramers degeneracy in the absence of time-reversal-breaking terms. This antiunitary statement is distinct from the unitary $SU(2)$ multiplet degeneracy, although both may act on the same levels.

Important Limits and Boundary Effects

Regime	Exact or controlled picture	Main caution
$J=0$	all $2^L$ states are degenerate	no exchange scale remains
$L=2$ with one bond	exact singlet–triplet dimer	a periodic shorthand may double the bond
$J<0$	maximal-spin ground multiplet	low-energy magnons are quadratic
$J>0$, even ring	singlet ground state and gapless spinon continuum as $L\to\infty$	finite-size gaps vanish with logarithmic corrections
$J>0$, odd ring	unavoidable antiferromagnetic frustration	ground state cannot be a singlet for odd $L$
open chain	$L-1$ bonds and boundary corrections	momentum is not a good quantum number
strong uniform field	polarized state above the saturation field	field reduces $SU(2)$ to $U(1)$
higher onsite spin	integer and half-integer chains can differ qualitatively	not the spin-$1/2$ baseline

Ferromagnetic branch

For $J<0$ on a connected chain, every bond can attain its triplet value simultaneously. The ground states form the maximal-spin multiplet

$$S_{\mathrm{tot}} = \frac L2.$$

With $N_b$ bonds, its energy is

$$E_F = \frac{J N_b}{4}.$$

The fully polarized product state is one member. Applying $(S_{\mathrm{tot}}^-)^r$, for $r=0,\ldots,L$, and normalizing gives a basis for the $(L+1)$-dimensional multiplet. Rotations of the fully polarized state produce spin-coherent states; their linear span is the multiplet, but a single rotation cannot produce every pure state in it.

Antiferromagnetic branch

For $J>0$, a singlet minimizes one isolated bond, but adjacent bonds share sites. An extended chain cannot be a product of singlets on every nearest-neighbor bond while preserving one-site translation symmetry. The classical Néel product state also fails to be an eigenstate because the transverse exchange terms move spin deviations.

For the connected, even, uniform antiferromagnetic chain with open boundaries or an even periodic ring, the ground state is a total-spin singlet. It has strong staggered correlations but no nonzero staggered magnetization in the translation-invariant one-dimensional ground state. The thermodynamic chain is critical rather than Néel ordered.

Odd periodic rings

An odd antiferromagnetic ring cannot alternate up and down consistently around the boundary. Its graph is nonbipartite, its minimum total spin is half-integer, and its low-energy quantum numbers differ from an even ring. Odd and even sequences should not be mixed casually in finite-size extrapolations.

Bethe Equations and Their Scope

The chain contains several exact structures, but they have different scopes.

Quantity or regime	Status
two-site spectrum	exact by angular-momentum addition
ferromagnetic ground manifold	exact for uniform $J<0$
ferromagnetic one-magnon sector	exact plane-wave diagonalization
uniform spin-$1/2$ periodic chain	Bethe-ansatz integrable
thermodynamic antiferromagnetic energy density	exact
antiferromagnetic low-energy velocity and continuum edges	exact
finite-temperature bulk thermodynamics	exact integral-equation framework
generic spin correlations	exact representations exist, but evaluation and asymptotics are nontrivial
arbitrary higher-spin chain	not automatically solved by the spin-$1/2$ Bethe equations
higher dimensions or frustrated graphs	not generically Bethe-ansatz solvable
generic anisotropy, disorder, or longer-range exchange	integrability must be reassessed

One rapidity convention for the periodic spin-$1/2$ chain uses roots $\lambda_j$ satisfying

$$\left( \frac{ \lambda_j+i/2 }{ \lambda_j-i/2 } \right)^L = \prod_{\substack{k=1\\k\ne j}}^{M} \frac{ \lambda_j-\lambda_k+i }{ \lambda_j-\lambda_k-i }.$$

For the unshifted Hamiltonian defined above, the energy of a regular Bethe state is

$$E = \frac{JL}{4} - \frac J2 \sum_{j=1}^{M} \frac{1}{ \lambda_j^2+1/4 }.$$

Regular finite-root solutions describe highest-weight states with $M\le L/2$; their spin descendants can be generated by total-spin lowering, formally adding roots at infinity. Solutions containing $\lambda=\pm i/2$ require regularization and a physical-state condition: substituting singular roots directly into the energy expression is invalid. The XXZ Spin Chain and Exact Solutions Preview explain how roots represent physical states.

Exact Energies and Excitation Scales

Ferromagnetic one-magnon dispersion

Let $J<0$ and $\lvert F\rangle$ be fully polarized. A normalized localized spin deviation is

$$\lvert j\rangle = s_j^-\lvert F\rangle$$

for spin $1/2$. Within the one-deviation subspace,

$$\left( H-E_F \right) \lvert j\rangle = \frac J2 \left( \lvert j-1\rangle + \lvert j+1\rangle - 2\lvert j\rangle \right).$$

Momentum states have exact excitation energy

$$\varepsilon_F(k) = \lvert J\rvert \left[ 1-\cos(ka) \right].$$

Near $k=0$,

$$\varepsilon_F(k) \simeq \frac{\lvert J\rvert}{2} (ka)^2.$$

The $k=0$ state belongs to the degenerate ground multiplet. The first dispersing mode of a periodic ring has $k=2\pi/(La)$ and an energy of order $L^{-2}$.

Antiferromagnetic ground-state energy

For the even periodic spin-$1/2$ chain with $J>0$,

$$\lim_{L\to\infty} \frac{E_0(L)}{LJ} = \frac14-\ln2 \simeq -0.443147180559945.$$

The Néel product state has exchange expectation $-J/4$ per bond. The exact value is lower because transverse exchange produces correlated quantum fluctuations.

Spinons and continuum boundaries

At zero temperature and zero field, the thermodynamic antiferromagnetic chain is gapless. Its elementary fractional excitations are spinons with spin $1/2$. A local spin operator acting on the singlet ground state reaches total-spin-one states, including two-spinon and higher even-spinon continua.

For physical wave number $q$, the exact lower boundary of the two-spinon continuum is

$$\varepsilon_{\mathrm L}(q) = \frac{\pi J}{2} \left| \sin(qa) \right|.$$

The upper boundary of the two-spinon contribution is

$$\varepsilon_{\mathrm U}(q) = \pi J \left| \sin\frac{qa}{2} \right|.$$

The upper expression bounds only the two-spinon contribution. Four-spinon and higher contributions can carry intensity above it, so it is not an upper bound on the entire dynamic structure factor. The lower boundary is linear near a gapless wavevector, giving spin velocity

$$v = \frac{\pi Ja}{2\hbar}.$$

The low-energy continuum theory has central charge

$$c = 1.$$

Marginally irrelevant interactions generate logarithmic corrections to simple scaling forms. Finite-size gaps, correlations, and entanglement should therefore not be fit blindly to pure power laws over short size ranges.

Typical Observables

Total spin and magnetization

The most basic labels are

$$\mathbf S_{\mathrm{tot}}^2 \quad\text{and}\quad S_{\mathrm{tot}}^z.$$

A uniform magnetization per site is

$$\mathbf m = \frac1L \sum_j\mathbf s_j.$$

On a bipartite even chain, a staggered operator is

$$\mathbf m_{\mathrm{st}} = \frac1L \sum_j (-1)^j \mathbf s_j.$$

In an $SU(2)$-invariant finite-system state, $\langle\mathbf m_{\mathrm{st}}\rangle$ vanishes. Rotationally invariant diagnostics include $\langle\mathbf m_{\mathrm{st}}^2\rangle$ and long-distance spin correlations.

Equal-time correlations

Define

$$C^{\alpha\beta}(r) = \left\langle s_j^\alpha s_{j+r}^\beta \right\rangle.$$

In an $SU(2)$-invariant state,

$$\left\langle s_i^\alpha s_j^\beta \right\rangle = \frac{ \delta_{\alpha\beta} }{3} \left\langle \mathbf s_i\cdot\mathbf s_j \right\rangle.$$

This tensor relation is a useful numerical symmetry test. It need not hold after choosing a symmetry-broken state, applying a field, or introducing anisotropy.

Static and dynamic structure factors

For a translation-invariant ring, one common static convention is

$$S(q) = \frac1L \sum_{j,k} e^{-iq a(j-k)} \left\langle \mathbf s_j\cdot\mathbf s_k \right\rangle.$$

Ferromagnetic correlations accumulate near $q=0$; antiferromagnetic correlations accumulate near $q=\pi/a$. The peak must be scaled with $L$ before it is interpreted as long-range order.

A dynamic structure factor additionally resolves energy transfer:

$$S^{\alpha\beta}(q,\omega) ={} \frac{1}{2\pi L} \int_{-\infty}^{\infty} \mathrm dt\, e^{i\omega t} \times \sum_{j,k} e^{-iqa(j-k)} \left\langle s_j^\alpha(t) s_k^\beta(0) \right\rangle.$$

It distinguishes a sharp ferromagnetic magnon from the antiferromagnetic spinon continuum. Structure Factors explains Fourier normalization, detailed balance, and experimental conventions.

Gaps and entanglement

A finite-size gap must state the compared sectors:

$$\Delta_L = E_{\mathrm{target}}(L) - E_0(L).$$

The antiferromagnetic bulk gap vanishes, but finite rings have nonzero level spacings. The ferromagnetic ground multiplet is already degenerate, so the gap to the first state outside that multiplet is the meaningful dispersing scale.

For a contiguous interval, the ground-state entanglement entropy is a sensitive distinction between the gapped variants and the critical spin-$1/2$ antiferromagnet. At criticality its leading logarithm is governed by $c=1$, with boundary-dependent coefficients and subleading corrections.

Physical Phenomena

Exchange and multiplets

Rotational invariance organizes levels by total spin. The sign of $J$ selects whether a bond prefers low or high combined spin, but that local preference does not by itself establish a thermodynamic phase.

Exact ferromagnetic order

All ferromagnetic bonds can minimize their energy in the same maximal-spin state. The resulting product-state representative is exact, and the quadratic magnon follows from the broken nonrelativistic spin-rotation generators. At any nonzero temperature, however, a short-range one-dimensional isotropic chain has no true ferromagnetic long-range order.

Quantum antiferromagnetism

Antiferromagnetic bonds compete because each site is shared. The ground state lowers its energy through a coherent superposition of spin configurations rather than a classical alternating product. In one dimension the outcome is a critical singlet with algebraic correlations and fractional spinons, not a Néel-ordered state with a single sharp magnon branch.

Symmetry without integrability

$SU(2)$ symmetry and Bethe integrability are distinct structures. Many higher-dimensional or perturbed Heisenberg Hamiltonians remain rotationally invariant but lose the commuting transfer-matrix hierarchy that makes the uniform chain solvable.

Dimensional and spin dependence

The spin-$1/2$ antiferromagnetic chain is gapless, while the uniform spin-1 chain has a nonzero Haldane gap and boundary spin-$1/2$ modes under open boundaries. In a finite long spin-1 chain, the edge spins are coupled with an exponentially small splitting rather than being exactly independent. Higher-dimensional bipartite antiferromagnets can order at zero temperature. The spin and spatial dimension are therefore essential to interpreting the low-energy spectrum.

Minimal Worked Example: One Bond

For two spin-$1/2$ sites with one bond,

$$H_2 = J \mathbf s_1\cdot\mathbf s_2.$$

Let

$$\mathbf S = \mathbf s_1+\mathbf s_2.$$

Since

$$\mathbf s_1\cdot\mathbf s_2 = \frac12 \left( \mathbf S^2 - \frac32 \right),$$

the total-spin singlet has

$$E_S = -\frac{3J}{4},$$

and the triplet has

$$E_T = +\frac{J}{4}.$$

The singlet is

$$\lvert S\rangle = \frac{ \lvert\uparrow\downarrow\rangle - \lvert\downarrow\uparrow\rangle }{\sqrt2},$$

while the triplet is spanned by

$$\begin{aligned} \lvert T_+\rangle &= \lvert\uparrow\uparrow\rangle, \\ \lvert T_0\rangle &= \frac{ \lvert\uparrow\downarrow\rangle + \lvert\downarrow\uparrow\rangle }{\sqrt2}, \\ \lvert T_-\rangle &= \lvert\downarrow\downarrow\rangle. \end{aligned}$$

The singlet–triplet separation is

$$E_T-E_S = J.$$

Thus antiferromagnetic exchange selects an entangled singlet, while ferromagnetic exchange selects the triplet multiplet. The trace vanishes:

$$E_S+3E_T = 0,$$

as expected because every nonidentity Pauli string has zero trace.

Numerical Benchmark

Four-site periodic ring

MB-B002 fixes

$$H = J \sum_{j=1}^{4} \mathbf s_j\cdot\mathbf s_{j+1}, \qquad \mathbf s_5 = \mathbf s_1,$$

with $J=1$, four undirected bonds, and no offset. The complete spectrum is

$E/J$	degeneracy	$SU(2)$ content
$-2$	1	one singlet
$-1$	3	one triplet
$0$	7	one singlet and two triplets
$+1$	5	one spin-two multiplet

The degeneracies sum to

$$1+3+7+5 = 16 = 2^4.$$

The fixed-$M$ product-basis block dimensions are

$$1,\ 4,\ 6,\ 4,\ 1.$$

Basis-independent checks are

$$\begin{aligned} \operatorname{Tr}H &= 0, \\ \frac1{16} \operatorname{Tr}H^2 &= \frac34J^2, \\ [H,S_{\mathrm{tot}}^z] &= 0, \\ [H,\mathbf S_{\mathrm{tot}}^2] &= 0. \end{aligned}$$

The fully polarized state has one-quarter unit of exchange on each bond, so

$$E_{\uparrow\uparrow\uparrow\uparrow} = J.$$

A result larger by a factor of four indicates Pauli matrices were used as the dimensionless spin operators. A wrong polarized energy with otherwise plausible low levels indicates a bond-list error. Actual energy splittings within an expected multiplet indicate a Hamiltonian, sector-comparison, or eigensolver problem. If the energy spectrum is correct but assigned spin labels are wrong, inspect the total-spin operator and tensor-factor ordering.

Trace-Moment Check

For a simple spin-$1/2$ graph with $N_b$ distinct bonds and uniform $J$, orthogonality of Pauli strings gives

$$\frac1{2^L} \operatorname{Tr}H^2 = \frac{3N_b}{16} J^2.$$

Cross terms between distinct bonds vanish under the full trace, including bonds that share one site. For the four-site ring $N_b=4$, reproducing $3J^2/4$ tests a sum over the whole matrix. This is stronger than one eigenvalue check but does not uniquely determine the Hamiltonian.

Thermodynamic Extrapolation

After the finite-ring test passes, an antiferromagnetic sequence can be compared with

$$\frac{e_0}{J} = \frac14-\ln2.$$

A small ring is not expected to equal this thermodynamic value. Even and odd rings carry different frustration and total-spin structure, so a controlled extrapolation should normally use one parity of $L$.

Other Exchange Chains

Variant	What changes	Further calculation
general spin-$s$ Heisenberg model	local dimension becomes $2s+1$	Heisenberg Model
XXZ chain	axial exchange anisotropy $\Delta$ reduces $SU(2)$ to $U(1)$	XXZ Chain dossier
XYZ chain	all three exchange couplings differ	Spin-$1/2$ Chain dossier
chain in a field	add $-hS_{\mathrm{tot}}^z$ and track magnetization sectors	Heisenberg Model
spin-1 chain	Haldane gap and open-chain edge degrees of freedom	Heisenberg Model
frustrated $J_1$–$J_2$ chain	competing next-neighbor exchange	generic Bethe integrability is lost
ladder or higher dimension	change graph and coordination	spin waves, tensor networks, Monte Carlo, or other methods may apply
Hubbard strong-coupling limit	exchange emerges as $J=4t^2/U$ under stated assumptions	Hubbard Chain

The Heisenberg Chain model card and Hamiltonian card provide compact formulas.

Exercises

1. Convert between Pauli and spin conventions

Suppose a code assembles

$$\widetilde H = \widetilde J \sum_{\langle j,k\rangle} \boldsymbol\sigma_j \mathbin{\cdot} \boldsymbol\sigma_k.$$

What value of $\widetilde J$ reproduces the baseline Hamiltonian with exchange $J$? What factor error appears if one sets $\widetilde J=J$?

Solution

Since

$$\mathbf s_j\cdot\mathbf s_k = \frac14 \boldsymbol\sigma_j \mathbin{\cdot} \boldsymbol\sigma_k,$$

the matching coefficient is

$$\widetilde J = \frac J4.$$

Setting $\widetilde J=J$ multiplies every energy and every gap by four. Degeneracies and dimensionless eigenvectors may still look correct, which is why the dimer energies and trace moment are essential normalization checks.

2. Derive the bond projectors

Use the two exchange eigenvalues to show that

$$\Pi_S = \frac14-\mathbf s_1\cdot\mathbf s_2, \qquad \Pi_T = \frac34+\mathbf s_1\cdot\mathbf s_2.$$

Verify projector completeness and orthogonality.

Solution

On the singlet, $\mathbf s_1\cdot\mathbf s_2=-3/4$, so $\Pi_S$ has eigenvalue $1$ and $\Pi_T$ has eigenvalue $0$. On the triplet, the exchange eigenvalue is $1/4$, so the roles reverse. The singlet and triplet span the full four-dimensional two-spin space, hence

$$\Pi_S+\Pi_T = \mathbb I.$$

Because both are spectral projectors of the Hermitian exchange operator, they obey

$$\Pi_S^2=\Pi_S, \qquad \Pi_T^2=\Pi_T, \qquad \Pi_S\Pi_T=0.$$

3. Explain the multiplet degeneracy

Let $\lvert E,S,m\rangle$ be an eigenstate of $H$, $\mathbf S_{\mathrm{tot}}^2$, and $S_{\mathrm{tot}}^z$. Show why all nonzero states obtained by applying $S_{\mathrm{tot}}^\pm$ have the same energy.

Solution

Rotational invariance gives

$$[H,S_{\mathrm{tot}}^\pm] = 0.$$

Therefore

$$\begin{aligned} H S_{\mathrm{tot}}^\pm \lvert E,S,m\rangle &= S_{\mathrm{tot}}^\pm H \lvert E,S,m\rangle \\ &= E S_{\mathrm{tot}}^\pm \lvert E,S,m\rangle. \end{aligned}$$

Whenever the ladder operation is nonzero, it produces another state at energy $E$ with $m$ shifted by one. Repeating the operation generates all $2S+1$ members of the multiplet. A numerical spectrum that splits these levels violates the baseline $SU(2)$ symmetry.

4. Derive the trace moment

For a simple graph with $N_b$ distinct spin-$1/2$ bonds, prove

$$\frac1{2^L} \operatorname{Tr}H^2 = \frac{3N_b}{16}J^2.$$

Solution

Write each bond as

$$h_{jk} = \frac J4 \sum_{\alpha=x,y,z} \sigma_j^\alpha\sigma_k^\alpha.$$

The $3N_b$ Pauli strings are distinct and nonidentity. Pauli-string orthogonality gives

$$\frac1{2^L} \operatorname{Tr} \left( Q_aQ_b \right) = \delta_{ab}.$$

Thus only the squares of identical strings survive in the full trace:

$$\frac1{2^L} \operatorname{Tr}H^2 = 3N_b \left( \frac J4 \right)^2 = \frac{3N_b}{16}J^2.$$

The argument also shows why cross terms from adjacent bonds vanish despite sharing one site.

5. Obtain the ferromagnetic finite-size scale

Use

$$\varepsilon_F(k) = \lvert J\rvert \left[ 1-\cos(ka) \right]$$

to find the first nonzero one-magnon energy on a periodic ring and its large-$L$ behavior.

Solution

Allowed momenta are

$$k_n = \frac{2\pi n}{La}.$$

The $n=0$ state is a member of the ground multiplet. The smallest nonzero magnitude is $2\pi/(La)$, so

$$\Delta_L^{\mathrm{mag}} = \lvert J\rvert \left[ 1-\cos \left( \frac{2\pi}{L} \right) \right].$$

Using $1-\cos x=x^2/2+O(x^4)$,

$$\Delta_L^{\mathrm{mag}} = \frac{2\pi^2\lvert J\rvert}{L^2} + O(L^{-4}).$$

The quadratic scaling is the finite-size signature of the ferromagnetic magnon near $k=0$.

6. Compare the Néel estimate with the exact energy

For the antiferromagnetic even chain, compute the exchange expectation per bond in the alternating product state and compare it with the exact thermodynamic energy density.

Solution

On an antiparallel product bond,

$$\left\langle s_i^zs_{i+1}^z \right\rangle = -\frac14,$$

while the transverse terms have zero expectation. Thus

$$\frac{ \langle H\rangle_{\mathrm{N\acute eel}} }{ LJ } = -\frac14$$

for a periodic even ring. The exact thermodynamic value is

$$\frac{e_0}{J} = \frac14-\ln2 \simeq -0.443147.$$

It is lower, as the variational principle requires. The difference quantifies energy gained from quantum correlations that the classical product state omits.

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