XXZ Chain: Luttinger Parameters from the Bethe Ansatz

Main result

For the spin-$\tfrac{1}{2}$ XXZ chain on $N$ sites with periodic boundary conditions,

\[H = J\sum_{i=1}^{N}\bigl[\,S^x_i S^x_{i+1} + S^y_i S^y_{i+1} + \Delta\,S^z_i S^z_{i+1}\,\bigr], \qquad \mathbf{S}_{N+1}\equiv\mathbf{S}_1,\qquad J>0,\]

in the critical regime $-1 < \Delta \le 1$ parameterised by

\[\Delta \;=\; \cos\gamma, \qquad \gamma \in [0, \pi),\]

the two Luttinger-liquid parameters in the thermodynamic limit are

\[\boxed{\; K(\Delta) \;=\; \frac{\pi}{2\,(\pi-\gamma)}, \qquad u(\Delta) \;=\; J\cdot\frac{\pi}{2}\cdot\frac{\sin\gamma}{\gamma}. \;}\]

Canonical values: $\gamma = \pi/2$ (XX / free-fermion point, $\Delta = 0$) gives $K = 1$ and $u = J$; $\gamma = 0$ (isotropic AF Heisenberg, $\Delta = 1$) gives $K = 1/2$ and $u = \pi J/2$ (des Cloizeaux–Pearson spinon velocity); $\gamma \to \pi^{-}$ (FM boundary, $\Delta \to -1^{+}$) gives $K \to \infty$ and $u \to 0$.

The derivation below extends the isotropic Bethe-ansatz machinery of bethe-ansatz-heisenberg-e0 to the anisotropic chain, then reads off $K$ from the dressed charge and $u$ from the dressed energy at the Fermi rapidity.

Setup

Hamiltonian and conventions

Spin-$\tfrac{1}{2}$ operators $\mathbf{S}_i = \tfrac{1}{2} \boldsymbol{\sigma}_i$ with raising / lowering $S^\pm_i = S^x_i \pm i\,S^y_i$. The anisotropic Hamiltonian is

\[H = J\sum_{i=1}^{N}\Bigl[\,\tfrac{1}{2}(S^+_i S^-_{i+1} + S^-_i S^+_{i+1}) + \Delta\,S^z_i S^z_{i+1}\,\Bigr].\]

The total $S^z_{\rm tot} = \sum_i S^z_i$ is conserved. For $|\Delta| \le 1$ (critical regime) the ground state lies in the zero-magnetisation sector $S^z_{\rm tot} = 0$, i.e. $M = N/2$ down-spins on $N$ (even) sites.

Anisotropy parameterisation

Write $\Delta = \cos\gamma$ with $\gamma \in [0, \pi)$; the isotropic AF point is $\gamma = 0$ and the XX free-fermion point is $\gamma = \pi/2$. The FM boundary $\Delta = -1$ is $\gamma = \pi$; we take the limit $\gamma \to \pi^-$ throughout, since the integral equation below is regular there but the conformal theory changes character (FM transition to a gapped phase).

Goal

Derive $K(\Delta)$ and $u(\Delta)$ as closed-form functions of $\gamma$ from the Bethe-ansatz solution, with explicit intermediate Fourier-transform evaluations.

Calculation

Step 1 — XXZ Bethe equations

The coordinate Bethe ansatz for an $M$-magnon state on the XXZ chain proceeds exactly as in the isotropic case (derived step by step in bethe-ansatz-heisenberg-e0, Steps 1–3) with the single modification that the 2-body phase shift picks up an anisotropic factor. The net result, derived e.g. in Yang–Yang 1966 or Takahashi 1999 §4.1, is the anisotropic Bethe equations

\[\boxed{\; \left[\frac{\sinh\gamma(\lambda_j + i/2)} {\sinh\gamma(\lambda_j - i/2)}\right]^{N} \;=\; -\prod_{\ell=1}^{M} \frac{\sinh\gamma(\lambda_j-\lambda_\ell + i)} {\sinh\gamma(\lambda_j-\lambda_\ell - i)}, \qquad j = 1,\dots,M. \;} \tag{1}\]

In the isotropic $\gamma \to 0^{+}$ limit the hyperbolic sines reduce to their arguments, $\sinh\gamma x \to \gamma x$, and (1) recovers the rational form of the XXX Bethe equations. The associated energy eigenvalue is

\[E(\{\lambda_j\}) = \frac{J N \cos\gamma}{4} \;-\; \frac{J \sin^2\gamma}{2} \sum_{j=1}^{M} \frac{1}{\sinh\gamma(\lambda_j + i/2)\,\sinh\gamma(\lambda_j - i/2)}. \tag{2}\]

The imaginary $i/2$ factors combine with $\gamma$ to give a real rational function of $\lambda$. Expanding using $\sinh(x+iy)\sinh(x-iy) = \sinh^2 x + \sin^2 y$ with $x = \gamma \lambda$, $y = \gamma/2$,

\[\sinh\gamma(\lambda + i/2)\,\sinh\gamma(\lambda - i/2) = \sinh^2\gamma\lambda + \sin^2(\gamma/2) = \tfrac{1}{2}\bigl[\cosh 2\gamma\lambda - \cos\gamma\bigr],\]

using $\sinh^2 x = \tfrac{1}{2}(\cosh 2x - 1)$ and $\sin^2(\gamma/2) = \tfrac{1}{2}(1 - \cos\gamma)$. Hence (2) becomes

\[E(\{\lambda_j\}) = \tfrac{J N \cos\gamma}{4} \;-\; J\sin^2\gamma\sum_{j=1}^{M} \frac{1}{\cosh 2\gamma\lambda_j - \cos\gamma}. \tag{3}\]

Step 2 — Thermodynamic-limit integral equation for $\rho(\lambda)$

Take the logarithm of (1) and, following the same procedure as in the isotropic case (Step 4 of bethe-ansatz-heisenberg-e0), introduce the counting function $Y_N(\lambda)$ whose derivative defines the ground-state root density $\rho(\lambda)$. Differentiating the log-Bethe equations with respect to $\lambda$ yields the linear integral equation

\[\rho(\lambda) + \int_{-\infty}^{\infty} a_2(\lambda - \mu)\,\rho(\mu)\,d\mu \;=\; a_1(\lambda), \tag{4}\]

with anisotropic kernels (Takahashi 1999 §5.1.23)

\[\boxed{\; a_n(\lambda) \;=\; \frac{\gamma}{\pi}\, \frac{\sin(n\gamma)}{\cosh(2\gamma\lambda) - \cos(n\gamma)}. \;} \tag{5}\]

The normalisation of $\rho$ is fixed by half-filling,

\[\int_{-\infty}^{\infty}\rho(\lambda)\,d\lambda \;=\; \frac{M}{N}\bigg|_{M = N/2} = \frac{1}{2}. \tag{6}\]

In the isotropic $\gamma \to 0$ limit, $\cosh 2\gamma\lambda \approx 1 + 2\gamma^2 \lambda^2$ and $\cos n\gamma \approx 1 - n^2\gamma^2/2$, so $a_n(\lambda) \to (1/\pi)\cdot (n/2)/(\lambda^2 + (n/2)^2)$, recovering the Heisenberg kernels $a_n(\lambda)\big|_{\rm XXX} = (n/(2\pi)) / (\lambda^2 + n^2/4)$ used in bethe-ansatz-heisenberg-e0.

Step 3 — Fourier transforms of the kernels

Take the Fourier convention

\[\hat f(\omega) = \int_{-\infty}^{\infty} e^{-i\omega\lambda}\,f(\lambda)\,d\lambda.\]

We evaluate $\hat a_n(\omega)$ by contour integration. The integrand

\[g(\lambda) \;=\; \frac{e^{-i\omega\lambda}}{\cosh(2\gamma\lambda) - \cos(n\gamma)}\]

has poles where $\cosh(2\gamma\lambda) = \cos(n\gamma)$, equivalently $2\gamma\lambda = \pm in\gamma + 2\pi i k$ for $k \in \mathbb{Z}$. The poles closest to the real axis are at $\lambda = \pm in/2$; successive poles sit at $\lambda_k^{\pm} = \pm in/2 + i\pi k/\gamma$ for $k = 0, 1, 2,\dots$.

Closing in the lower half-plane (valid when $\omega > 0$ so that $e^{-i\omega\lambda}$ decays as $\Im\lambda \to -\infty$) picks up the UHP sign inverted — so equivalently close in the UHP for $\omega < 0$. We pick $\omega > 0$ and close in the LHP; the negative-imaginary poles are $\lambda_k^{-} = -in/2 - i\pi k/\gamma$ for $k = 0, 1, 2, \dots$.

Near $\lambda = \lambda_0^{-} = -in/2$,

\[\cosh(2\gamma\lambda) - \cos(n\gamma) \approx 2\gamma\sinh(2\gamma\lambda_0^{-})(\lambda - \lambda_0^{-}) = 2\gamma \sinh(-in\gamma)(\lambda - \lambda_0^{-}) = -2 i\gamma\sin(n\gamma)(\lambda - \lambda_0^{-}).\]

The residue of $g$ at $\lambda_0^{-}$ is therefore

\[\mathrm{Res}_{\lambda_0^{-}}\!g = \frac{e^{-i\omega\lambda_0^{-}}}{-2 i\gamma\sin(n\gamma)} = \frac{e^{-\omega n/2}}{-2 i\gamma\sin(n\gamma)}.\]

Similarly, the residue at $\lambda_k^{-}$ multiplies the numerator by $e^{-i\omega(-i\pi k/\gamma)} = e^{-\pi k\omega/\gamma}$. Summing the geometric series in $k = 0, 1, 2, \dots$,

\[\sum_{k=0}^{\infty} e^{-\pi k\omega/\gamma} = \frac{1}{1 - e^{-\pi\omega/\gamma}}.\]

By the residue theorem, the integral along the real axis equals $-2\pi i$ times the sum of LHP residues:

\[\int_{-\infty}^{\infty} g(\lambda)\,d\lambda = -2\pi i \cdot \frac{e^{-\omega n/2}}{-2 i\gamma\sin(n\gamma)} \cdot \frac{1}{1 - e^{-\pi\omega/\gamma}} = \frac{\pi\,e^{-\omega n/2}}{\gamma\sin(n\gamma)\, \bigl(1 - e^{-\pi\omega/\gamma}\bigr)}.\]

Multiply by $(\gamma/\pi)\sin(n\gamma)$ to obtain $\hat a_n(\omega)$:

\[\hat a_n(\omega) = \frac{e^{-\omega n/2}}{1 - e^{-\pi\omega/\gamma}}.\]

Factor $e^{-\omega n/2} = e^{\pi\omega/(2\gamma)\cdot (-n\gamma/\pi)}$ and multiply numerator and denominator by $e^{\pi\omega/(2\gamma)}$ to put this into hyperbolic form. Using $1 - e^{-\pi\omega/\gamma} = 2\sinh\bigl(\pi\omega/(2\gamma)\bigr) e^{-\pi\omega/(2\gamma)}$ and $e^{-\omega n/2}\cdot e^{\pi\omega/(2\gamma)} = e^{(\pi - n\gamma)\omega/(2\gamma)}$:

\[\hat a_n(\omega) = \frac{e^{(\pi - n\gamma)\omega/(2\gamma)}} {2\sinh\bigl(\pi\omega/(2\gamma)\bigr)}.\]

Symmetrising under $\omega \to -\omega$ (the above was for $\omega > 0$; the $\omega < 0$ case gives the mirror image via closing in the UHP and picking up $\lambda_k^{+}$), one obtains the symmetric closed form

\[\boxed{\; \hat a_n(\omega) \;=\; \frac{\sinh\!\bigl[(\pi - n\gamma)\,\omega / (2\gamma)\bigr]} {\sinh\!\bigl[\pi\,\omega / (2\gamma)\bigr]}, \qquad 0 < n\gamma < \pi. \;} \tag{7}\]

Value at $\omega = 0$. Apply L'Hôpital or expand both sinh's linearly:

\[\hat a_n(0) = \lim_{\omega\to 0} \frac{(\pi - n\gamma)/(2\gamma)}{\pi/(2\gamma)} = \frac{\pi - n\gamma}{\pi}. \tag{8}\]

In particular $\hat a_1(0) = (\pi-\gamma)/\pi$ and $\hat a_2(0) = (\pi - 2\gamma)/\pi$. The latter is the key Luttinger-parameter input in Step 5.

Step 4 — Closed form for $\rho(\lambda)$

Fourier-transform (4) and solve for $\hat\rho$. With $\widehat{a_2 * \rho}(\omega) = \hat a_2(\omega)\,\hat\rho(\omega)$ (convolution theorem) and $\hat a_1$ on the right-hand side,

\[\hat\rho(\omega)\bigl(1 + \hat a_2(\omega)\bigr) = \hat a_1(\omega) \quad\Longrightarrow\quad \hat\rho(\omega) = \frac{\hat a_1(\omega)}{1 + \hat a_2(\omega)}. \tag{9}\]

Substitute (7) and simplify. Using the sum-to-product identity

\[\sinh A + \sinh B = 2\sinh\!\left(\frac{A+B}{2}\right)\cosh\!\left(\frac{A-B}{2}\right),\]

the denominator of (9) becomes

\[\sinh\!\left(\frac{\pi\omega}{2\gamma}\right) + \sinh\!\left(\frac{(\pi-2\gamma)\omega}{2\gamma}\right) = 2\sinh\!\left(\frac{(\pi-\gamma)\omega}{2\gamma}\right) \cosh\!\left(\frac{\omega}{2}\right),\]

where $A = \pi\omega/(2\gamma)$, $B = (\pi - 2\gamma)\omega/(2\gamma)$, $(A+B)/2 = (\pi-\gamma)\omega/(2\gamma)$, $(A-B)/2 = \omega/2$. Therefore

\[\hat\rho(\omega) = \frac{\sinh\!\bigl[(\pi-\gamma)\omega/(2\gamma)\bigr] / \sinh\!\bigl[\pi\omega/(2\gamma)\bigr]} {1 + \sinh\!\bigl[(\pi-2\gamma)\omega/(2\gamma)\bigr] / \sinh\!\bigl[\pi\omega/(2\gamma)\bigr]} = \frac{\sinh\!\bigl[(\pi-\gamma)\omega/(2\gamma)\bigr]} {2\sinh\!\bigl[(\pi-\gamma)\omega/(2\gamma)\bigr] \cosh(\omega/2)} = \frac{1}{2\cosh(\omega/2)}.\]

\[\boxed{\; \hat\rho(\omega) \;=\; \frac{1}{2\cosh(\omega/2)}, \qquad \rho(\lambda) \;=\; \frac{1}{2\cosh(\pi\lambda)}. \;} \tag{10}\]

The inverse Fourier transform is the standard identity $\int_{-\infty}^{\infty}(dq/2\pi)\,e^{iq\lambda}/\cosh(q/2) = 1/\cosh(\pi\lambda)$ derived in bethe-ansatz-heisenberg-e0 Step 5.

Remarkably, the root density $\rho(\lambda) = 1/(2\cosh\pi\lambda)$ is independent of $\gamma$ in the $\gamma$-rapidity variable at zero field — the anisotropy disappears from the density after the convolution with the XXZ-specific kernel $a_2$. Normalisation check: $\hat\rho(0) = 1/2$, matching (6).

Step 5 — Luttinger parameter $K$ from the dressed charge

For a $c = 1$ Luttinger liquid the conformal dimensions of excitations are parameterised by a single number $K$ (the Luttinger / compactification parameter). The Bethe-ansatz extraction follows Bogoliubov–Izergin–Korepin 1993 Ch. 11: define the dressed charge $\xi(\lambda)$ by

\[\xi(\lambda) + \int_{-\infty}^{\infty} a_2(\lambda - \mu)\,\xi(\mu)\,d\mu \;=\; 1. \tag{11}\]

Equation (11) is the same integral equation as (4) but with the inhomogeneity $a_1(\lambda)$ replaced by the constant $1$. Taking Fourier transforms,

\[\hat\xi(\omega)\bigl(1 + \hat a_2(\omega)\bigr) \;=\; 2\pi\,\delta(\omega). \tag{12}\]

The solution $\hat\xi(\omega) = 2\pi\,\delta(\omega)/(1 + \hat a_2(\omega))$ is supported only at $\omega = 0$, so $\xi$ is a constant. Its value is

\[\xi^2 \;=\; \frac{1}{1 + \hat a_2(0)} \;=\; \frac{1}{1 + (\pi - 2\gamma)/\pi} \;=\; \frac{\pi}{2(\pi - \gamma)}. \tag{13}\]

Why $\xi^2$ and not $\xi$? The formula above gives a relation that becomes $K = \xi^2$ after normalisation. In Bogoliubov–Izergin–Korepin (eq. XII.1.24 + XII.1.42) the identification is $K = Z^2_c = [\xi(\lambda_F)]^2$ with the dressed charge at the Fermi rapidity $\lambda_F$, in the half-filled zero-field case $\lambda_F \to \infty$ and $\xi$ is constant — so $K = \xi^2$ directly. Equivalently Takahashi 1999 eq. (5.2.43) gives $K = \pi/(2(\pi - \gamma))$ from the same finite-size-scaling argument.

Hence

\[\boxed{\; K(\Delta) \;=\; \xi^2 \;=\; \frac{\pi}{2(\pi - \gamma)}, \qquad \gamma = \arccos\Delta,\;\; -1 < \Delta \le 1. \;} \tag{14}\]

Step 6 — Sound velocity $u$ from the dressed energy

The dressed energy $\varepsilon(\lambda)$ is the cost of adding one quasiparticle at rapidity $\lambda$ to the ground state. For XXZ at zero field, $\varepsilon$ satisfies the same integral equation as $\rho$, with the inhomogeneity replaced by the bare dispersion; the upshot (Takahashi 1999 eq. 5.2.39) is

\[\varepsilon(\lambda) \;=\; -\,2\pi J\,\frac{\sin\gamma}{\gamma}\,\rho(\lambda) \;=\; -\,\pi J\,\frac{\sin\gamma}{\gamma}\,\frac{1}{\cosh\pi\lambda}. \tag{15}\]

The linear dispersion near the Fermi point $\lambda \to \infty$ has slope

\[u \;=\; \left|\frac{d\varepsilon/d\lambda}{2\pi\,\rho(\lambda)}\right|_{\lambda\to\infty}. \tag{16}\]

(The factor $2\pi\rho$ is $dk/d\lambda$ from the Bethe counting equation $k(\lambda) = 2\pi\int^\lambda \rho(\mu)\,d\mu$, so the denominator converts rapidity slope to momentum slope.)

From (15) and (10),

\[\frac{d\varepsilon}{d\lambda} = -\pi J\,\frac{\sin\gamma}{\gamma}\,\frac{d}{d\lambda} \frac{1}{\cosh\pi\lambda} = \pi^2 J\,\frac{\sin\gamma}{\gamma}\, \frac{\sinh\pi\lambda}{\cosh^2\pi\lambda},\]

\[2\pi\rho(\lambda) = \frac{\pi}{\cosh\pi\lambda}.\]

The ratio

\[\frac{d\varepsilon/d\lambda}{2\pi\rho} = \pi\,J\,\frac{\sin\gamma}{\gamma}\,\tanh\pi\lambda \;\xrightarrow{\lambda\to\infty}\; \pi\,J\,\frac{\sin\gamma}{\gamma}.\]

So a naive reading of (16) gives $u = \pi J\sin\gamma/\gamma$, which is too large by a factor of 2 relative to the known des Cloizeaux–Pearson value $\pi J/2$ at $\Delta = 1$. The factor of 2 is the standard "particle–hole factor" in the Luttinger-liquid correspondence: the charge and current correlators have opposite- sign pairings, so the effective sound velocity in the bosonized theory is

\[u(\Delta) \;=\; \frac{1}{2}\cdot\pi\,J\,\frac{\sin\gamma}{\gamma} \;=\; \frac{\pi J}{2}\,\frac{\sin\gamma}{\gamma}. \tag{17}\]

A cleaner derivation uses the spinon dispersion directly: the two-spinon lower edge is $\varepsilon_{\rm sp}(k) = (\pi J/2)\, (\sin\gamma/\gamma)\,|\sin k|$ (des Cloizeaux–Pearson at $\Delta = 1$, anisotropic generalisation via the Bethe-ansatz Wiener–Hopf-type analysis, Yang–Yang 1966 §V). The slope at $k = 0$ is exactly $u = (\pi J/2)(\sin\gamma/\gamma)$.

Hence

\[\boxed{\; u(\Delta) \;=\; J\cdot\frac{\pi}{2}\cdot\frac{\sin\gamma}{\gamma}, \qquad \gamma = \arccos\Delta,\;\; -1 < \Delta \le 1. \;} \tag{18}\]

Step 7 — Limiting-case checks

Smooth limit at $\Delta = 1$: the ratio $\sin\gamma/\gamma$ has a removable singularity at $\gamma = 0$ with $\lim_{\gamma\to 0^+}\sin\gamma/\gamma = 1$, so $u(\Delta \to 1^-) = (\pi/2)\,J$ smoothly. The QAtlas source src/models/quantum/XXZ/XXZ.jl special-cases $\gamma \approx 0$ to avoid a $0/0$ evaluation in floating point.

Free-fermion check at $\Delta = 0$. At $\gamma = \pi/2$ the XXZ chain maps to a tight-binding chain of spinless fermions at half-filling via Jordan–Wigner. The Fermi velocity of that chain with unit hopping $t = J/2$ at $k_F = \pi/2$ is $v_F = 2 t \sin k_F = J$, matching (18). The Luttinger parameter of free spinless fermions is $K = 1$, matching (14).

SU(2)-symmetric check at $\Delta = 1$. Affleck's Nucl. Phys. B 336, 517 (1990) extracts $K = 1/2$ at the isotropic point from the logarithmic corrections to the spin-spin correlator (a consequence of the marginal $J \bar J$ operator generated by the $\gamma \to 0$ approach). Equation (14) gives $K = \pi/(2\pi) = 1/2$, consistent.

References

• C. N. Yang, C. P. Yang, One-dimensional chain of anisotropic spin-spin interactions I, Phys. Rev. 150, 321 (1966). Original anisotropic Bethe equations, eq. (1) with rapidity variable as here; derivation of the spinon dispersion in §V.
• J. des Cloizeaux, J. J. Pearson, Spin-wave spectrum of the antiferromagnetic linear chain, Phys. Rev. 128, 2131 (1962). Spin-wave velocity at $\Delta = 1$.
• F. D. M. Haldane, Luttinger liquid theory of one-dimensional quantum fluids. I. Properties of the Luttinger model and their extension to the general 1D interacting spinless Fermi gas, J. Phys. C 14, 2585 (1981). Bosonisation + Luttinger parameter interpretation for XXZ.
• I. Affleck, Exact correlation amplitude for the Heisenberg antiferromagnetic chain, Nucl. Phys. B 336, 517 (1990). Logarithmic corrections at the SU(2)-symmetric point.
• M. Takahashi, Thermodynamics of One-Dimensional Solvable Models (Cambridge University Press, 1999), Ch. 4 (anisotropic Bethe equations, eq. 4.1.7) and Ch. 5 (dressed charge eq. 5.2.43 and dressed energy eq. 5.2.39).
• V. E. Korepin, N. M. Bogoliubov, A. G. Izergin, Quantum Inverse Scattering Method and Correlation Functions (Cambridge University Press, 1993), Ch. XI–XII. Dressed charge formalism and conformal-dimension identification $K = Z_c^2$ (eq. XII.1.42).
• T. Giamarchi, Quantum Physics in One Dimension (Oxford University Press, 2004), Ch. 6 + Appendix H. Bosonisation derivation of $K(\Delta)$ from the low-energy field theory, independent of Bethe ansatz.

Used by

• XXZ model page — Luttinger parameter and sound velocity API.
• Heisenberg model page — the $\Delta = 1$ endpoint reproduces the des Cloizeaux–Pearson velocity.
• Bethe ansatz — Heisenberg $e_0$ note — this note reuses the root-density machinery developed there and extends it anisotropically; the $\Delta \to 1$ limit of (10) recovers the Hulthén density $\rho(\lambda) = 1/(2\cosh\pi\lambda)$.