XXZ Chain

The spin-$1/2$ XXZ chain is an axially anisotropic nearest-neighbor Heisenberg chain whose single parameter $\Delta$ connects a polarized ferromagnet, a gapless $c=1$ Luttinger liquid, the isotropic antiferromagnet, and a gapped Néel phase while retaining Bethe-ansatz integrability.

The phase diagram below is at zero temperature and zero longitudinal field, with positive transverse exchange. The same parameter changes the interaction between Jordan–Wigner fermions, so only one point is a free-fermion model.

Helpful background. Heisenberg Chain explains the isotropic spin normalization and multiplets; XXZ Spin Chain develops the Bethe and Luttinger descriptions.

Axial Anisotropy and the XXZ Hamiltonian

The reference chain has these physical choices.

Field	Baseline choice
sites	$L$ spin-$1/2$ degrees of freedom
local space	$\mathcal H_j\simeq\mathbb C^2$
operators	dimensionless $\mathbf s_j=\boldsymbol\sigma_j/2$
geometry	uniform one-dimensional nearest-neighbor chain
transverse exchange	$J>0$
longitudinal exchange	$J\Delta$ with real dimensionless $\Delta$
field	$h=0$
boundaries	open or periodic, stated explicitly
energy offset	none
disorder, dimerization, and longer range	absent
thermodynamic reference	even periodic rings followed by $L\to\infty$

The zero-field phase labels below assume $J>0$. A longitudinal field is a useful integrable deformation, but its value must be stated because it changes magnetization, filling, and parts of the phase diagram. Negative $J$, transverse fields, bond alternation, and frustrated exchange are not silently absorbed into the baseline.

Degrees of Freedom and Basis

The Hilbert space is

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

Dimensionless spin operators satisfy

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

and

$$s_j^\alpha = \frac12 \sigma_j^\alpha.$$

The product basis uses

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

Writing

$$s_j^\pm = s_j^x \pm i s_j^y$$

gives the bond identity

$$s_j^xs_{j+1}^x + s_j^ys_{j+1}^y = \frac12 \left( s_j^+s_{j+1}^- + s_j^-s_{j+1}^+ \right).$$

The longitudinal term is diagonal in the product basis. The transverse term exchanges an adjacent up-down pair and conserves the total number of down spins.

Hamiltonian and Conventions

For an open chain,

$$H_{\mathrm O} = J \sum_{j=1}^{L-1} \big( s_j^xs_{j+1}^x + s_j^ys_{j+1}^y + \Delta s_j^zs_{j+1}^z \big).$$

For a periodic chain,

$$H_{\mathrm P} = J \sum_{j=1}^{L} \big( s_j^xs_{j+1}^x + s_j^ys_{j+1}^y + \Delta s_j^zs_{j+1}^z \big),$$

with

$$s_{L+1}^{\alpha} = s_1^\alpha.$$

The open chain has $L-1$ undirected bonds; an ordinary ring has $L$. As in other nearest-neighbor models, the $L=2$ periodic shorthand can count the same undirected pair twice. A two-site check therefore uses one explicitly open bond.

A longitudinal field measured in energy units adds

$$H_h = -h \sum_{j=1}^{L} s_j^z.$$

Pauli-matrix form

In terms of Pauli matrices,

$$H = \frac J4 \sum_{\langle j,k\rangle} \big( \sigma_j^x\sigma_k^x + \sigma_j^y\sigma_k^y + \Delta \sigma_j^z\sigma_k^z \big) - \frac h2 \sum_j\sigma_j^z.$$

A result four times too large in the exchange spectrum or twice too large in the field response indicates that Pauli and spin conventions were mixed.

Anisotropy tensor

The exchange constants are

$$J_x = J_y = J, \qquad J_z = J\Delta.$$

$\Delta$ is dimensionless. The labels easy plane and easy axis refer to this tensor, not to a classical orientation already chosen by a finite quantum ground state.

Symmetries

Axial spin rotations

For every $\Delta$ and longitudinal field $h$,

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

is conserved:

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

The internal continuous symmetry is generically $U(1)$. If $M$ sites are down relative to the all-up state,

$$S_{\mathrm{tot}}^z = \frac L2-M, \qquad \dim\mathcal H_M = \binom LM.$$

This is both a physical charge decomposition and the natural exact-diagonalization basis.

Symmetry enhancement

At

$$\Delta=1, \qquad h=0,$$

the Hamiltonian becomes isotropic:

$$H = J \sum_{\langle j,k\rangle} \mathbf s_j\cdot\mathbf s_k.$$

All three components of total spin commute with $H$, enhancing $U(1)$ to $SU(2)$. Heisenberg Chain describes the resulting multiplets and exact finite spectra.

Spin reversal and time reversal

At $h=0$, a global $\pi$ rotation about $x$ maps

$$s_j^z \mapsto -s_j^z$$

and exchanges magnetization sectors $M\leftrightarrow L-M$. It leaves the XXZ exchange invariant.

Physical time reversal sends every spin component to its negative and also preserves the zero-field Hamiltonian. A nonzero longitudinal field breaks both magnetization reversal and time reversal, while retaining the axial $U(1)$ symmetry.

Lattice symmetries

The uniform periodic chain has one-site translation and reflection symmetry. The open chain retains reflection about its midpoint but not translation. Bond alternation, disorder, twists, and boundary fields can change the spatial symmetry without necessarily breaking magnetization conservation.

Zero-Field Phase Map

For $J>0$, $h=0$, and the thermodynamic ground state:

Anisotropy	Ground-state regime	Bulk gap	Long-distance character
$\Delta<-1$	$z$-polarized ferromagnet	nonzero away from $-1$	saturated magnetization
$\Delta=-1$	first-order endpoint	quadratic soft mode	enlarged ground space on an even bipartite chain
$-1<\Delta<1$	easy-plane Luttinger liquid	zero	algebraic correlations and $c=1$
$\Delta=1$	isotropic antiferromagnet	zero	$c=1$ with logarithmic corrections
$\Delta>1$	easy-axis antiferromagnet	nonzero	thermodynamic Néel order

The critical interval is an entire line of fixed points, not one isolated critical point. Its correlation exponents vary continuously with $\Delta$. The endpoint at $\Delta=-1$ is first order and has a vanishing linear-mode velocity; the transition at $\Delta=1$ is Berezinskii–Kosterlitz–Thouless and opens a gap through an essential singularity.

A finite chain does not literally display all thermodynamic order parameters. Phase assignments require size sequences, long-distance correlations, sector information, or a symmetry-breaking source taken to zero after the thermodynamic limit.

Important Limits

XX point

At

$$\Delta=0,$$

Jordan–Wigner maps the chain to free spinless fermions. At zero field and half filling,

$$\frac{e_0}{J} = -\frac1\pi, \qquad v = \frac{Ja}{\hbar}.$$

The fermions are free, but transverse spin operators contain nonlocal strings. Their correlations are not reduced to a one-body spin observable.

Isotropic point

At

$$\Delta=1,$$

the model is the antiferromagnetic spin-$1/2$ Heisenberg chain. Exact bulk fingerprints include

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

and

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

The point remains gapless, but marginally irrelevant interactions produce slow logarithmic corrections.

Ferromagnetic endpoint

On an even bipartite chain, a staggered $\pi$ rotation about $z$ maps the $\Delta=-1$ Hamiltonian to an isotropic ferromagnetic Heisenberg chain. Its ground-space dimension is therefore

$$L+1$$

for spin $1/2$. The low-energy dispersion is quadratic, so the endpoint is not an ordinary finite-velocity conformal point.

Easy-axis limits

For $\Delta\gg1$, the longitudinal exchange dominates and the two alternating $z$ configurations organize the low-energy antiferromagnetic sector. Transverse exchange dresses them with quantum fluctuations. For $\Delta\ll-1$, the two fully polarized $z$ product states are exact zero-field ground states.

Integrability and Its Scope

The uniform nearest-neighbor spin-$1/2$ chain is Bethe-ansatz integrable for the standard anisotropies and a longitudinal field. Exactness remains quantity and boundary dependent.

Quantity or regime	Status
finite-volume spectrum	encoded by Bethe roots after sector and boundary conventions are fixed
$\Delta=0$	exactly quadratic after Jordan–Wigner
one-magnon sector	exact plane-wave result
zero-field phase boundaries	exact
bulk energy and dressed thermodynamics	exact integral-equation framework
zero-field $K$ and $v$ in the critical interval	exact
finite-size conformal spectrum	universal leading form with model-specific corrections
correlation exponents in the critical interval	exact once the bosonization convention is fixed
correlation amplitudes and dynamical functions	substantially harder; exact representations need further evaluation
generic next-neighbor, disordered, or transversely driven chain	not generically integrable

Bethe solvability does not make the model free. For $\Delta\ne0$, quasiparticle momenta or rapidities are coupled by nontrivial scattering phases, and spin correlations require matrix elements as well as energies.

Fermions, Magnons, and Luttinger Parameters

Jordan–Wigner form

With

$$s_j^z = n_j-\frac12,$$

the open chain maps to

$$H_{\mathrm O} ={} \frac J2 \sum_{j=1}^{L-1} \left( c_j^\dagger c_{j+1} + c_{j+1}^\dagger c_j \right) + J\Delta \sum_{j=1}^{L-1} \left( n_j-\frac12 \right) \left( n_{j+1}-\frac12 \right) - h \sum_{j=1}^{L} \left( n_j-\frac12 \right).$$

Thus the spinless-fermion hopping and nearest-neighbor interaction are

$$t = \frac J2, \qquad V = J\Delta,$$

up to a removable sign of $t$ from a staggered phase convention on this open chain. On a periodic odd ring, the same transformation changes the boundary twist and cannot silently remove the sign. $\Delta=0$ is free; $\Delta\ne0$ retains a density interaction.

One-magnon and saturation check

For a periodic chain in a longitudinal field, the all-up energy is

$$E_F = \frac{J\Delta L}{4} - \frac{hL}{2}.$$

One down spin with physical wave number $q$ has excitation energy

$$\varepsilon(q) = h - J\Delta + J\cos(qa).$$

For an even periodic ring, or in the thermodynamic limit, the minimum lies at $q=\pi/a$. For $\Delta\ge-1$ the saturation field is

$$h \ge h_{\mathrm{sat}}, \qquad h_{\mathrm{sat}} = J(1+\Delta).$$

This is both a necessary one-magnon condition and a sufficient ground-state condition in that regime. To see sufficiency, write each bond with half of the field on each end. Its four energies are

$$J\Delta/4-h/2,\quad J\Delta/4+h/2,\quad -J\Delta/4\pm J/2.$$

For $h\ge J(1+\Delta)\ge0$, the all-up state minimizes every bond, hence their sum. At equality other states can become degenerate. A finite odd ring need not contain $q=\pi/a$ and has a different finite-size threshold.

For $\Delta<-1$, the thermodynamic equilibrium positive saturation threshold is instead $h=0$: any $h>0$ selects all-up over all-down. The negative value $J(1+\Delta)$ is merely a one-magnon stability bound for the all-up branch, not its global ground-state boundary. At zero field in that regime, the magnon gap is

$$\Delta_{\mathrm{mag}} = J(-\Delta-1).$$

It closes at $\Delta=-1$.

Luttinger parameters

In the critical interval, write

$$\Delta = \cos\gamma, \qquad 0<\gamma<\pi.$$

At zero magnetization, one common bosonization convention gives

$$K = \frac{\pi}{ 2(\pi-\gamma) }$$

and

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

The limiting checks are

$$\begin{array}{c|c|c} \Delta & K & v \\ \hline 0 & 1 & Ja/\hbar \\ 1^- & 1/2 & \pi Ja/(2\hbar) \\ -1^+ & \infty & 0 \end{array}$$

and

$$c = 1$$

throughout $-1<\Delta\le1$, with endpoint and logarithmic qualifications as stated above.

Correlation exponents

With the same $K$ convention, leading zero-magnetization correlations behave schematically as

$$\begin{aligned} \langle s_0^zs_r^z\rangle \sim {}& -\frac{K}{2\pi^2r^2} + A_z \frac{(-1)^r}{r^{2K}}, \\ \langle s_0^+s_r^-\rangle \sim {}& A_\perp \frac{(-1)^r}{r^{1/(2K)}}. \end{aligned}$$

Here $r$ is an integer site separation. The amplitudes are nonuniversal. At $\Delta=1$, multiplicative logarithms modify the pure powers. At finite magnetization, both $K$ and the oscillation wavevectors change.

Boundary Conditions and Finite-Size Effects

Three boundary statements must be kept distinct:

1. open and periodic spin chains have different bond counts;
2. periodic Jordan–Wigner fermions acquire a boundary sign fixed by total fermion parity;
3. Bethe equations depend on periodicity, twists, magnetization sector, and rapidity convention.

These effects are subextensive in a bulk energy density but decisive for exact finite spectra, momenta, and gaps.

An even ring is the clean zero-field thermodynamic sequence. In the easy-axis antiferromagnet, an odd periodic ring frustrates the alternating order. Near the BKT point, very large correlation lengths and logarithmic corrections make modest system sizes especially deceptive.

A twist can be imposed through

$$s_{L+1}^{\pm} = e^{\pm i\Phi} s_1^\pm.$$

The curvature of the ground energy with $\Phi$ probes spin stiffness, but conventions differ by factors of $L$, $\pi$, and $\hbar$.

Typical Observables

Magnetization and order

The uniform magnetization per site is

$$m_z = \frac1L \sum_j \langle s_j^z\rangle.$$

For an even chain, a staggered operator is

$$M_{\mathrm{st}}^z = \sum_j (-1)^j s_j^z.$$

In a finite spin-reversal eigenstate, $\langle M_{\mathrm{st}}^z\rangle$ can vanish even in the easy-axis regime. Useful diagnostics include $\langle(M_{\mathrm{st}}^z)^2\rangle$, the staggered structure factor, long-distance $zz$ correlations, and a controlled symmetry-breaking source.

Correlations and structure factors

The anisotropy makes longitudinal and transverse correlators inequivalent:

$$C_{zz}(r) = \langle s_j^z s_{j+r}^z\rangle,$$

and

$$C_{+-}(r) = \langle s_j^+s_{j+r}^-\rangle.$$

Their Fourier transforms distinguish $z$-ferromagnetic, critical easy-plane, and Néel regimes. Structure Factors explains normalization and scattering conventions.

Energy gaps and entanglement

The finite gap

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

must identify magnetization, momentum, parity, and boundary sectors. In the critical phase, low gaps scale as $1/L$ with corrections. In either gapped phase they approach nonzero bulk scales, aside from symmetry-partner or boundary splittings.

At criticality, interval entanglement has the leading $c=1$ logarithm. In the gapped phases it saturates beyond the correlation length, with possible finite-cat or boundary contributions.

Spin current

The continuity equation for $s_j^z$ defines one current convention:

$$\mathcal J_j^z = \frac J\hbar \left( s_j^x s_{j+1}^y - s_j^y s_{j+1}^x \right).$$

Transport claims require temperature, magnetization, frequency, system-size, and order-of-limits conventions. Integrability can protect ballistic contributions in selected regimes but does not make every transport coefficient trivial.

Physical Phenomena

Interacting criticality

For $-1<\Delta<1$ away from the free point $\Delta=0$, the chain is both interacting and exactly integrable. Its low-energy physics is collective: the Luttinger parameter continuously changes correlation exponents even though the central charge remains $1$.

Fractionalization and variable descriptions

The same microscopic system can be described as local spins, interacting spinless fermions, Bethe quasiparticles, or a compact boson at low energy. These descriptions are complementary rather than interchangeable. A local spin operator may be nonlocal in fermions, and the continuum theory omits lattice-scale observables.

Distinct endpoints

At $\Delta=-1$, a quadratic soft mode and enlarged ground space accompany a first-order boundary. At $\Delta=1$, a marginal interaction drives a BKT transition whose gap opens essentially rather than through a simple power. Treating both as ordinary critical endpoints loses the central physics.

Finite versus thermodynamic order

The easy-axis phase has two symmetry-related Néel states in the thermodynamic description, while a finite ring can retain translation or spin-reversal symmetry. Conversely, large finite staggered correlations inside the critical phase do not imply a nonzero order parameter.

No nonzero-temperature order

A short-range one-dimensional XXZ chain has no nonzero-temperature transition into true ferromagnetic or Néel long-range order. Low-temperature quantum-critical windows and long correlation lengths are crossovers, not additional equilibrium phase transitions.

Minimal Worked Example: One Anisotropic Bond

For one open bond at zero field,

$$H_2 = J \left( s_1^xs_2^x + s_1^ys_2^y + \Delta s_1^zs_2^z \right).$$

The parallel states have

$$E_{\uparrow\uparrow} = E_{\downarrow\downarrow} = \frac{J\Delta}{4}.$$

In the zero-magnetization sector, define

$$\begin{aligned} \lvert t_0\rangle &= \frac{ \lvert\uparrow\downarrow\rangle + \lvert\downarrow\uparrow\rangle }{\sqrt2}, \\ \lvert s\rangle &= \frac{ \lvert\uparrow\downarrow\rangle - \lvert\downarrow\uparrow\rangle }{\sqrt2}. \end{aligned}$$

Their energies are

$$\begin{aligned} E_{t_0} &= \frac{J(2-\Delta)}{4}, \\ E_s &= -\frac{J(2+\Delta)}{4}. \end{aligned}$$

At $\Delta=1$, the three triplet states become degenerate at $J/4$ and the singlet lies at $-3J/4$. At $\Delta=-1$, the singlet crosses the two parallel states. The trace is zero for every $\Delta$:

$$2E_{\uparrow\uparrow} + E_{t_0} + E_s = 0.$$

The normalized second moment is

$$\frac14 \operatorname{Tr}H_2^2 = \frac{J^2}{16} \left( 2+\Delta^2 \right).$$

The dimer catches normalization and the $\Delta=-1$ crossing, but it does not reproduce the thermodynamic Luttinger phase or BKT transition.

Numerical Benchmark

Interacting four-site ring

Use $L=4$, periodic boundaries, four distinct undirected bonds,

$$J=1, \qquad \Delta=\frac12, \qquad h=0,$$

and no constant offset. The anisotropy is interacting after Jordan–Wigner and is not an $SU(2)$ point.

The complete spectrum is

$E/J$	degeneracy
$(-1-\sqrt{33})/4$	1
$-1$	2
$-1/2$	1
$0$	7
$+1/2$	2
$+1$	2
$(-1+\sqrt{33})/4$	1

The ground energy is

$$\frac{E_0}{J} = -1.686140661634507.$$

The degeneracies sum to $16$, and the fixed-$M$ block dimensions are

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

Useful independent comparisons are

$$\begin{aligned} \operatorname{Tr}H &= 0, \\ \frac1{16} \operatorname{Tr}H^2 &= \frac9{16}J^2, \\ [H,S_{\mathrm{tot}}^z] &= 0. \end{aligned}$$

At $h=0$, spin reversal also maps the $M$ block to the $L-M$ block with the same spectrum.

General trace-moment identity

For a simple graph with $N_b$ distinct XXZ bonds,

$$\frac1{2^L} \operatorname{Tr}H^2 = \frac{ N_bJ^2 }{16} \left( 2+\Delta^2 \right)$$

at zero field. Distinct Pauli strings are orthogonal under the full trace, so cross-bond terms vanish even for adjacent bonds. With $N_b=4$ and $\Delta=1/2$, this gives $9J^2/16$.

Bulk endpoint checks

After the finite matrix passes, separate thermodynamic calculations can test

$$\left. \frac{e_0}{J} \right|_{\Delta=0} = -\frac1\pi$$

and

$$\left. \frac{e_0}{J} \right|_{\Delta=1} = \frac14-\ln2.$$

These values use the unshifted spin-operator Hamiltonian. Adding $-J\Delta/4$ per bond changes energy densities but not eigenstates or gaps.

Other Anisotropic Chains

Variant	What changes	Further calculation
isotropic point	set $\Delta=1$ and recover $SU(2)$	Heisenberg Chain dossier
XX model	set $\Delta=0$ and obtain free fermions	Spinless Fermion Chains
longitudinal field	change magnetization and fermion filling	XXZ Spin Chain
twist	probe stiffness through boundary phase $\Phi$	Boundary Conditions on Lattices
dimerized chain	alternate bond strengths	translation and phase structure change
$J_1$–$J_2$ chain	add frustrating next-neighbor exchange	generic Bethe integrability is lost
random-field chain	break translation and study disorder dynamics	Many-Body Localization Preview
higher-spin XXZ chain	enlarge the onsite representation	phase and integrability claims must be re-established

For the isotropic limit, the Heisenberg Chain Hamiltonian card gives a compact operator formula.

Exercises

1. Verify magnetization conservation

Show directly that the exchange Hamiltonian commutes with $S_{\mathrm{tot}}^z$.

Solution

The longitudinal term is built from $s_j^z$ operators and therefore commutes with $S_{\mathrm{tot}}^z$. For one transverse term,

$$\begin{aligned} \left[ s_j^z+s_{j+1}^z, s_j^+s_{j+1}^- \right] ={}& s_j^+s_{j+1}^- - s_j^+s_{j+1}^- \\ = {}& 0. \end{aligned}$$

The Hermitian-conjugate term also commutes, so

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

Each transverse exchange moves a down spin but does not create or destroy one.

2. Recover the dimer spectrum

Diagonalize the zero-magnetization block of the one-bond Hamiltonian in the basis $\{\lvert\uparrow\downarrow\rangle,\lvert\downarrow\uparrow\rangle\}$.

Solution

The block is

$$H_{M=1} = J \begin{pmatrix} -\Delta/4 & 1/2 \\ 1/2 & -\Delta/4 \end{pmatrix}.$$

Its symmetric and antisymmetric eigenvectors are $\lvert t_0\rangle$ and $\lvert s\rangle$, with eigenvalues

$$E_{t_0} = \frac{J(2-\Delta)}4$$

and

$$E_s = -\frac{J(2+\Delta)}4.$$

The parallel states are separate one-dimensional magnetization sectors with energy $J\Delta/4$.

3. Check the Pauli normalization

Write the one-bond Hamiltonian entirely in Pauli matrices. If a code omits the factor $1/4$, how do the dimer energies and the trace moment change?

Solution

The correct Pauli form is

$$H_2 = \frac J4 \left( \sigma_1^x\sigma_2^x + \sigma_1^y\sigma_2^y + \Delta\sigma_1^z\sigma_2^z \right).$$

Omitting $1/4$ multiplies the Hamiltonian and every energy by four. Since the second trace moment is quadratic in $H$, it becomes sixteen times too large:

$$\frac14\operatorname{Tr}\widetilde H_2^2 = 16 \left[ \frac{J^2}{16} \left( 2+\Delta^2 \right) \right].$$

4. Derive the saturation field

Use the one-magnon dispersion to derive the positive saturation field for $J>0$ and $\Delta>-1$.

Solution

The excitation energy above the all-up state is

$$\varepsilon(q) = h-J\Delta+J\cos(qa).$$

For $J>0$, its minimum occurs at $qa=\pi$, where $\cos(qa)=-1$. Stability requires

$$\varepsilon_{\min} = h-J(\Delta+1) \ge 0.$$

Therefore

$$h_{\mathrm{sat}} = J(1+\Delta).$$

For an even periodic chain, the bond-energy argument above also proves sufficiency. On a finite odd ring, minimize over the actually allowed wave numbers. The result depends on the spin-operator normalization and the field term being $-h\sum_js_j^z$.

5. Prove the trace-moment identity

Derive the zero-field normalized second moment for a simple graph of $N_b$ XXZ bonds.

Solution

Each bond contributes three Pauli strings with coefficients

$$\frac J4, \qquad \frac J4, \qquad \frac{J\Delta}{4}.$$

Every string is traceless, and distinct full-system Pauli strings are orthogonal. Therefore only identical-string squares survive:

$$\begin{aligned} \frac1{2^L} \operatorname{Tr}H^2 &= N_b \left[ \left( \frac J4 \right)^2 + \left( \frac J4 \right)^2 + \left( \frac{J\Delta}{4} \right)^2 \right] \\ &= \frac{N_bJ^2}{16} \left( 2+\Delta^2 \right). \end{aligned}$$

6. Check the Luttinger limits

Evaluate $K$ and $v$ at $\Delta=0$, and take their limits as $\Delta\to1^-$ and $\Delta\to-1^+$.

Solution

With $\Delta=\cos\gamma$,

$$K = \frac{\pi}{2(\pi-\gamma)}, \qquad v = \frac{Ja}{\hbar} \frac{\pi\sin\gamma}{2\gamma}.$$

At $\Delta=0$, $\gamma=\pi/2$, so

$$K=1, \qquad v=\frac{Ja}{\hbar}.$$

As $\Delta\to1^-$, $\gamma\to0$ and $\sin\gamma/\gamma\to1$, giving

$$K\to\frac12, \qquad v\to\frac{\pi Ja}{2\hbar}.$$

As $\Delta\to-1^+$, $\gamma\to\pi$, so

$$K\to\infty, \qquad v\to0.$$

The last limit signals the singular ferromagnetic endpoint rather than an ordinary finite-velocity conformal point.

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