Transverse field Ising model

Mon, 14 Aug 2023 00:00:00 +0000

To cite this page

@misc{swpkim2026transverse,
   author={P. Kim, Sun Woo},
   title={Transverse field Ising model},
   year={2023},
   howpublished={\url{https://sunwoo-kim.github.io/ko/posts/tfi-model/}},
   note={Accessed: 2026-06-13}
}

In condensed matter physics, there are myriads of phenomenological models out there, with information about them scattered throughout different papers and resources. In the field of machine learning, there are efforts such as the Model Zoo which tries to organise these models in one place.

In this new series, I wanted to write down an overview of some of the models I looked at in my studies, with, where possible, the precise definition of what physicists mean when they say things like ’this model has a $U(1)$ symmetry’, or ’the FM case is dual to the AFM case’, etc., which were confusing to me when I was first reading about them. I will also try to support the discussion with some simulations from exact diagonalisation (ED) or Monté Carlo results. Since websites are interactive, I’ll try to nest the finer details in collapsibles. Lastly, this page may be updated periodically with new information.

차례

Introduction

The transverse-field Ising (TFI) model is a prototypical model of a quantum magnet, where quantum fluctuations are used to generate paramagnetism instead of temperature. Another common name is ‘quantum Ising model’, or ’transverse Ising model’ (TIM). There is already a pretty good Wikipedia article about this model, but I wanted to just write it down in a more technical way, and fill in some missing gaps.

Hamiltonian

Broadly, I will use the following conventions for the TFI model,

\begin{equation} \hat{H} = J\sum_{\langle i,j \rangle} \hat{\sigma}^z_i \hat{\sigma}^z_j - g \sum_i \hat{\sigma}^x_i - b \sum_i \hat{\sigma}^z_i, \end{equation} where $\langle i, j\rangle$ denotes nearest neighbours, and $\sigma^\alpha_i$ are Pauli matrices. We consider periodic boundary conditions (PBC). Without a loss of generalisation, we can set $J=\pm 1$. Then $J=-1$ corresponds to ferrmomagnetic (FM) couplings, $J=1$ to antiferromagnetic couplings (AFM), respectively. $g$ is the transverse field strength, and $b$ the longitudinal field strength, which we will take as $b=0$ unless stated otherwise.

Symmetries

For $b=0$, the Hamiltonian is $\mathbb{Z}_2$ symmetric, in the sense that we can globally send $\hat{\sigma}^z_i \rightarrow - \hat{\sigma}^z_i$ and keep the Hamiltonian invariant, or concretely, $[\hat{H}, \hat{U}] = 0$ for $\hat{U} = \vec{\prod_i} \hat{\sigma}^x_i$.

Phases of the ground state

The TFI model admits 3 phases: the ordered phase $g<g_c$, the critical/gapless phase $g=g_c$, and the disordered phase $g>g_c$.

Quantum to classical mapping

The TFI model in spatial dimensions $d$ can be explicitly mapped to a $d+1$ classical Ising model, in the sense that the zero-temperature (ground-state) density matrix can be explicitly mapped to the partition function of the classical Ising model. (click to open)

It is well known that $d$-dimensional quantum systems map to $(d+1)$-dimensional classical systems. Here, I’ll show an explicit mapping between $d$-dimensional quantum Ising model of length $L_Q$ with a $(d+1)$-dimensional Ising Hamiltonian, which is very similar to¹ but is for general $d$ and with longitudinal field $b$.

Let $\hat{H}_0$ be the part of $\hat{H}$ that only contain $\hat{\sigma}^z$’s, and $\hat{H}_1$ be the rest of the Hamiltonian. Then the quantum partition function is given by

\begin{align} Z_Q = \mathrm{tr} [e^{-\beta_Q \hat{H}_0 - \beta_Q \hat{H}_1}]. \end{align}

From Trotter’s theorem, for any two Hermitian operators bounded from below, $\hat{A}$, $\hat{B}$, we have $e^{\hat{A} + \hat{B}}= \lim_{L \rightarrow \infty} \left(e^{-\hat{A}/L} e^{-\hat{B}/L}\right)^L$, therefore. defining $\tau = \beta_Q / L$,

\begin{align} Z_Q = \sum_\sigma \langle \sigma \rvert \lim_{L \rightarrow \infty} \left(e^{-\tau\hat{H}_0} e^{-\tau \hat{H}_1}\right)^L \lvert \sigma \rangle. \end{align}

Inserting identities, we have

\begin{align} Z_Q & = \lim_{L \rightarrow \infty} \sum_{\sigma^1, \dots, \sigma^L} \prod_{l=1}^{L} \langle \sigma^{l+1} \rvert e^{-\tau \hat{H}_1} e^{-\tau \hat{H}_0} \lvert \sigma^l \rangle \end{align}

\begin{align} Z_Q & = \lim_{L \rightarrow \infty} \sum_{\sigma^1, \dots, \sigma^L} \prod_{l=1}^{L} e^{-\tau H_0(\sigma^l)} \langle \sigma^{l+1} \rvert e^{\tau g \sum_i \hat{\sigma}^x_i } \lvert \sigma^l \rangle, \end{align}

where $H_0(\sigma^l) = - J \sum_{\langle i, j\rangle} \sigma^l_i \sigma^l_j - b \sum_i \sigma_i^l$. Now we can use the identity that $\langle \sigma^{l+1}_i \rvert e^{\tau g \hat{\sigma}^x_i} \lvert \sigma^l_i \rangle = \Lambda e^{\gamma \sigma^{l+1}_i \sigma^{l}_i}$, where $\Lambda = \sqrt{\sinh(\tau g)\cosh(\tau g)}$ and $\gamma = -\frac{1}{2} \ln \tanh(\tau g)$. Therefore we have

\begin{align} Z_Q = \lim_{L \rightarrow \infty} \Lambda^{L_Q^d L} \sum_{\sigma^1, \dots, \sigma^L} e^{\tau J \sum_{\langle i, j \rangle, l} \sigma^l_i \sigma^l_j + \tau b \sum_{i, l} \sigma^l_i + \gamma \sum_{i, l} \sigma^l_i \sigma^{l+1}_i}, \end{align}

so we can interpret index $l$ as an additional dimension, and write $Z_Q = \lim_{L \rightarrow \infty} \Lambda^{L^d_Q L} Z_C$,

where the classical partition function is

\begin{align} Z_C = \lim_{L_{d+1} \rightarrow \infty} \sum_{\sigma} e^{-\beta_C(-\sum_i \sum_{a=1}^{d+1} J_a \sigma_i \sigma_{i+e_a} -h \sum_i \sigma_i)}, \end{align}

where $e_a$ is the $a$^th basis vector. Therefore we have the identification $L_Q = L_a$ for $a = 1, \dots d$, $\beta_C J_a = J \tau$ for $L=L_{d+1}$, $\tau b = \beta_C h$ and $\beta_C J_{d+1} = \gamma$. We are free to set ratio $\tau=\beta_Q/L$ to a value of our choosing, and send $\beta_Q \rightarrow \infty$. To map to an isotropic $(d+1)$-dim Ising model, we can set $\tau = \beta_C$ and assert $J_a = J \forall a$, and have the mapping

\begin{align} \boxed{\lim_{\beta_Q \rightarrow \infty} Z_Q[\beta_Q, J, g, b] = \lim_{L_{d+1} \rightarrow \infty} \Lambda^{L_Q^d L_{d+1}} Z_C[\beta_C, J, b]}, \end{align}

where $g = \frac{1}{\beta_C} \mathrm{arctanh}(e^{-2 \beta_C J})$ and $\Lambda = \frac{1}{2} \mathrm{csch}(2 \beta_C J)$.

Since $\beta_C \rightarrow \infty \leftrightarrow g \rightarrow 0$, $\beta_C \rightarrow 0 \leftrightarrow g \rightarrow \infty$, we can see that $g$ plays the role of temperature.

Lechner, D. (2014). Quantum Phase Transitions-The Quantum-Classical Mapping. ↩︎

$d=1$ case

Mapping to free fermions

Pfeuty¹ showed that the TFI model can be mapped to free fermions, using the Jordan-Wigner transformation.

Ignoring the boundary terms, one can show that the ground state energy is equal to

From Pfeuty, ignoring the terms at $k=0$,

$$e_0 \approx -\frac{g}{L}\sum_{m=-L/2}^{L/2} \sqrt{1+\lambda^2+2\lambda\cos(2\pi m/L)},$$ where $e_0=E_0/L$ is the ground state energy density, and $\lambda = J/g$. As $L \rightarrow \infty$ this tends to

$$e_0 = -\frac{1}{2\pi} \int_{-\pi}^{\pi} dk \sqrt{1+2g\cos(k)+g^2}$$

Upon integration, this is equal to

$$e_0 = -\frac{2}{\pi} \lvert g-1\rvert E\left(\frac{-4g}{(g-1)^2}\right)$$

$$E_0 = -\frac{2L}{\pi} \lvert g-1\rvert E\left(\frac{-4g}{(g-1)^2}\right),$$ and from this or from the quantum-classical mapping, we can show that the second derivative of the ground state energy density diverges as $\sim \ln L$. We also know that $g_c=1$.

Wavefunction

$g=0$

For $J=-1$, the ground state is degenerate; it can be any linear superposition $\lvert \psi_0 \rangle = \alpha \lvert \uparrow \rangle^{\otimes L} + \beta \lvert \downarrow \rangle^{\otimes L}$. This is ‘unphysical’ in that if an infinitesimal longitudinal field $b$ is added, the degeneracy is immediately broken.

For $J=1$ the ground state is any superposition of the Neél state, $\lvert \psi_0 \rangle = \alpha \lvert \uparrow \downarrow \rangle^{\otimes L/2} + \beta \lvert \downarrow \uparrow \rangle^{\otimes L/2}$.

$\lvert g\rvert=\infty$

Here we the spins would align in the $x$-direction, meaning that for $g>0$ we have $\lvert \psi_0 \rangle = \lvert + \rangle^{\otimes L}$ and for $g<0$ we have $\lvert \psi_0 \rangle = \lvert - \rangle^{\otimes L}$, where $\lvert \pm \rangle = \frac{1}{\sqrt{2}} \left(\lvert \uparrow \rangle \pm \rvert \downarrow \rangle\right)$.

Entanglement

The entanglement entropy for the $d=1$ TFI model is an area law away from the critical point $g=1$, meaning that it tends to a constant as $L \rightarrow \infty$. This can be confirmed for the limits $g=0$, $\lvert g \rvert = \infty$ above. For $g=0$, we have due to the macroscopic superposition, we have an entanglement of $S_l^{(1)} = \ln 2$. For $g=\infty$, we have a product state in the $\sigma^x$-basis, and therefore have an entanglement entropy of 0.

On the other hand, from results of conformal field theory (CFT) at the critical point², the bipartite entanglement entropy of scales as logarithm of system size with factor proportional to the central charge $c$,

\begin{equation} S_l^{(1)} = B + \frac{c}{3}\ln\left[\frac{L}{\pi} \sin \frac{\pi L_A}{L}\right], \end{equation} where $B$ is some constant.

We can check this using results from ED. We use netket as they have a nice Hamiltonian representation builder.

Code

import numpy as np
import netket as nk
from netket.operator.spin import sigmax,sigmaz, sigmay
from scipy.sparse.linalg import eigsh

def bipartite_ee(psi, alphas=[1]):
   L = int(np.log2(psi.shape[0]))
   A_dim =  2**(L//2)
   psi_block = np.reshape(psi, (A_dim, A_dim))
   s = np.linalg.svd(psi_block, compute_uv=0)
   s=s[s>10**(-20)]**2

   entropies = []
   for alpha in alphas:
      if alpha == 1:
         entropies.append(-np.inner(np.log(s),s))
      else:
         entropies.append(np.log(np.sum(s**alpha))/(1-alpha))

   return entropies

Ls = [4, 6, 8, 10, 12, 14]
J = -1
d = 1
gs = np.linspace(0, 2, 41)

ees = dict()

for L in Ls:
   es[L] = []
   ees[L] = []
   res[L] = []
   graph = nk.graph.Hypercube(length=L, n_dim=d, pbc=True)
   hi = nk.hilbert.Spin(s=1/2, N=L**d)
   
   for i, g in enumerate(gs):
      print(f"{L}, {delta:.2f}", end='\r')
      H = []
      for i in graph.nodes():
            H.append(-delta*sigmax(hi, i))
      for i, j in graph.edges():
            H.append(-sigmaz(hi, i)*sigmaz(hi, j))
      H = sum(H)

      w, v = eigsh(H.to_sparse(), k=1, which="SA", return_eigenvectors=True, tol=0)

      ee, re = f.bipartite_ee(v[:, 0], alphas=[1, 2])

      ees[L].append(ee)

   ees[L] = np.array(ees[L])

Entanglement entropy for the $d=1$ TFI model, calculated through exact diagonalisation.

$d=2$ case

In 2d there are multiple ways of building the lattice. The simplest is to have a square lattice. In that case, the critical transverse field is approximately $g_c \approx 3.045$³.

Pfeuty, P. (1970). The one-dimensional Ising model with a transverse field. ANNALS of Physics, 57(1), 79–90. ↩︎
Calabrese, P., & Cardy, J. (2009). Entanglement entropy and conformal field theory. Journal of physics a: mathematical and theoretical, 42(50), 504005. ↩︎
de Jongh, M. S. L., & van Leeuwen, J. M. J. (1997). The critical behaviour of the 2D Ising model in Transverse Field; a Density Matrix Renormalization calculation. arXiv preprint cond-mat/9709103. ↩︎

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