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Ising model at criticality

{$$ H = -\sum_{\langle i j \rangle} \sigma_i^z \sigma_{i+1}^z + \sum_i \sigma_i^x $$}

using the spinchain or spinchain-z2 models, the Hamiltonian is -4*H_zz + 2*H_x (or use the shortcut H_ITF)

Note that the sign of the H_zz term is important for translation symmetry. With positive sign, the ordered phase is antiferromagnetic, so only 2-site translationally invariant. WIth the minus sign, the ordered phase is ferromagnetic and the groundstate is everywhere 1-site translationally invariant. The groundstate is conformally invariant with central charge {$c=1/2$}.

  • Exact energy per site in the thermodynamic limit = {$-4 / \pi = -1.2732395447351626861510701\ldots$}
  • Energy for finite-size open boundary conditions = {$1 - 1 / \sin(\frac{\pi}{2(2L+1)})$}

{$S=1/2$} chain

The {$SU(2)$}-symmetric form of the Hamiltonian is {$$H = \sum_{\langle i j \rangle} s_i^z s_{i+1}^z + \frac{1}{2}\left( s_i^+ s_{i+1}^- + s_i^- s_{i+1}^+ \right) $$}

The groundstate has quasi-long-range antiferromagnetic correlations, and is 2-site translationally invariant, with central charge {$c=1$}. A {$\pi$} spin rotation on every other site produces a 1-site translationally invariant state at the cost of breaking SU(2) symmetry (and injectivity).

Hamiltonian H_J1

  • Exact energy per site in the thermodynamic limit = {$1/4 - \ln 2 = -0.4431471805599453094172\ldots$}
  • Exact energy for finite-size periodic boundary conditions is available in misc/heisenberg-energy.cpp
  • Spinon dispersion relation: {$\epsilon(k) = \frac{\pi}{2} \cos k$}, {$-\frac{\pi}{2} \leq k \leq \frac{\pi}{2}$}

Haldane--Shastry model, {$$H = \frac{\pi^2}{N^2} \sum_{i<j}^N \frac{\vec{S}_i \cdot \vec{S}_j}{\sin^2(\pi(j-i)/N)} .$$} In the thermodynamic limit {$N \rightarrow \infty$} this becomes {$$H = \sum_{i<j} \frac{1}{(j-i)^2} \vec{S}_i \cdot \vec{S}_j.$$}

{$S=1$} chain

  • AKLT model
    {$$ H = \sum_{i} \vec{S}_i \cdot \vec{S}_{i+1} + \frac{1}{3} \left( \vec{S}_i \cdot \vec{S}_{i+1} \right)^2 $$}
    Use the Hamiltonian H_J1 + (1/3) * H_B1, or the shortcut H_AKLT
    energy per site = {$-2/3$}
  • Heisenberg model energy per site (numerical) -1.401 484 038 971(4) from
    It is hard to do much better than this with double-precision floating point. The next couple of digits are probably -1.401 484 038 971 14 (?)
  • Haldane gap {$\Delta$} = 0.410 479 248 463(6), from
  • BT-point [Babujain, and Takhtajan
    analagous to the spin-1/2 chain, and Bethe-Ansatz solvable with the same approach), Hamiltonian is
    {$$ H = \sum_{i} \vec{S}_i \cdot \vec{S}_{i+1} - \left( \vec{S}_i \cdot \vec{S}_{i+1} \right)^2$$}
    Use Hamiltonian H_J1 - H_B1
    Energy per site is -4
    Central charge c=3/2
  • ULS-point: Lai-Sutherland model, exactly solvable point with {$SU(3)$} symmetry. See and
    {$$H = \sum_{i} \vec{S}_i \cdot \vec{S}_{i+1} + \left( \vec{S}_i \cdot \vec{S}_{i+1} \right)^2$$}
    Use Hamiltonian H_J1 + H_B1
    energy per site is {$-\ln 3 - \frac{\pi}{3\sqrt{3}} + 2 \simeq 0.296787923253818$}
    Central charge c=2
  • {$SU(3)$} chain: This is a transformation of the Lai-Sutherland model. In terms of the 8 generators of {$SU(3)$}, eg the Gell-Mann {$\lambda$} matrices, we write
    {$$H = \frac{1}{4} \sum_{i} \sum_{a=1}^8 \lambda^a_i \lambda^a_{i+1}$$}
    Relation to the bilinear-biquadratic model is {$H_{SU(3)} = \frac{1}{2} H_{BQ} - \sum_i \frac{2}{3}$}. The {$2/3$} is an energy shift per site. Energy per site -0.518272705039758
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Page last modified on February 20, 2024, at 11:58 AM