Semiconductor of spinons: from Ising band insulator to orthogonal band insulator

Electron conduction in periodic structures can cease for two reasons. The simplest is to couple single-particle states across a reduced Brillouin zone by an off-diagonal matrix elements due to a reduction in periodicity. However, the second and more exciting way is to introduce strong electron correlations where, due to Coulomb interactions (as suggested by NF Mott), electron conduction in an otherwise conducting state is interrupted [1]. This may seem to suggest that strong correlation has its most dramatic effect on metals by transforming them into many-body Mott insulators. The canonical model within which the metal-to-insulator transition (MIT) problem is investigated is the Hubbard model [2]. Efforts to understand the nature of the MIT has lead to many technical [3–16] and conceptual [17–21] developments providing clues into the possible mechanisms of non-Fermi liquid formation.

But an even more challenging question is what happens when both mechanisms of gap formation are simultaneously present, i.e. what are the properties of strongly correlated band insulators or semiconductors? Let us formalize the problem as follows: imagine a staggered potential of strength Δ (the ionic potential) that can gap out the parent metallic state and sets the scale of the single-particle gap. When the Hubbard interaction U is turned on in such an already gapped state (band insulator), an interesting competition between the Hubbard U and the staggered potential Δ sets in. This is the simplest model addressing the competition between a 'many-body' gap parameter U and a 'single-particle' gap parameter Δ which is called the ionic Hubbard model [22]. The band insulating state at $U=0$ is adiabatically connected to the insulating state at non-zero but small values of $U\ll \Delta$ as the effect of weak Hubbard U is to renormalize the parent metallic state on top of which the ionic potential creates a band insulator (BI) state. In the opposite limit of strong correlations $U\gg \Delta$ one gets a Mott insulating (MI) state. Although these two extreme limits both represent insulating states, the origin of the gap in the former case is a simple one-particle scattering, while in the latter case the gap has a many-body character arising from projecting out doubly occupied configurations.

The nature of possible state(s) between the above two extreme insulating states has been the subject of debates in the past decade. In one dimension Fabrizio and coworkers [23] find an ordered state. In two dimensions Hafez-Torbati and coworkers find an orientational and bond ordering phase in between the Mott and band insulators [24]. Kancharla and coworkers find an insulating phase with possible bond order in two dimensions [25]. The topologically non-trivial variant of the model was considered by Prychynenko and coworkers [26] who find that, for the topologically trivial situation, two spin density wave states are sandwiched between the band and Mott insulating states. In the limit of infinite dimensions, however, Garg and coworkers using the dynamical mean field theory (DMFT) found that the competition between the tendency of the ionic potential Δ and the Hubbard term U gives rise to a metallic state [27]. Within a perturbative continuous unitary transformation, one finds a metallic state when the Hubbard U and the ionic potential Δ are comparable [28]. Similar results were obtained in two dimensions [29, 30]. The method of DMFT was also applied to study the quantum phase transitions of the ionic Hubbard model on the honeycomb lattice. Starting from massive Dirac fermions on the honeycomb lattice the competition between U and the single-particle gap parameter Δ (known as the mass term when it comes to Dirac fermions) gives rise to massless Dirac fermions [31]. This result is smoothly connected to the earlier DMFT study of the massless Dirac electrons [32]. A recent strong coupling expansion gives a quantum critical semi-metallic state [7]. The DMFT study of a somehow related model of a band insulator, i.e. the bilayer Hubbard model, predicts a direct and discontinuous transition between a correlated band insulator and the Mott state [33].

When electron correlations in a conductor are not strong enough to transform the metallic state to a Mott insulator, they give rise to possible non-Fermi liquid states. From this point of view one may now turn on the ionic potential Δ and ask the following question: how does this staggered potential interfere with possible non-Fermi liquid state of the parent metal? Can the ionic potential gap out a non-Fermi liquid state? With this motivation, let us summarize one of the simplest mechanisms of creating a non-Fermi liquid state, and then add the ionic potential Δ to it in a self-consistent way. This is the question on which we will be focused in this paper. We will use the language of slave rotor (section 2.1) and slave spins (section 3) to illustrate the main findings of this paper.

Recently, Nandkishore and coworkers have argued that starting from a metallic state, as one increases the Hubbard U beyond $U_\perp$ the Fermi liquid (FL) undergoes a phase transition to an exotic non-FL state termed orthogonal metal (OM). Upon a further increase in the interaction strength beyond U_Mott, the system is expected to become a Mott insulator [34, 35]. OM is an interesting—and perhaps the simplest—non-FL state for $U_\perp<U<U_{\rm Mott}$ that separates an FL from a Mott insulator. In the FL phase the rotor variable is ordered, i.e. the phase variable has small fluctuations meaning that the electric charge has large fluctuations. When the rotor variable disorders, an earlier interpretation would assume that wild fluctuations of the rotor field that is responsible for vanishing of the rotor order-parameter corresponds to freezing of charge fluctuations and hence makes the system a Mott insulator [18]. Likewise the disordering phase transition in Z₂ slave-spin theories at $U_\perp$ was also interpreted earlier as the Mott transition. However, in [35] it was argued that the Mott transition comes much later. For $U_\perp<U<U_{\rm Mott}$, one is in the OM phase. We find that for $U_b<U<U_\perp$ the rotor variable is disordered, i.e. $\newcommand{\la}{\langle} \newcommand{\ra}{\rangle} \la {\rm e}^{{\rm i}\theta}\ra=0$, meaning that the single boson $b\sim {\rm e}^{{\rm i}\theta}$ is not condensed, but a Z₂ slave variable can still have less severe sign (Ising type) fluctuations and get condensed, $\newcommand{\la}{\langle} \newcommand{\ra}{\rangle} \la \tau^x\ra\ne 0$. This presents a new chance to charge fluctuations to survive in the form controlled by an Ising variable [19]. This phase can be dubbed the Ising metal. The disordered phase of such an Ising variable for $U_\perp<U<U_{\rm Mott}$ will correspond to an OM state where, although quasiparticle weight corresponding to physical electrons is zero, its transport behavior is metallic. To see this, the following simple and powerful argument is due to Nandkishore and coworkers [35]: within the slave spin representation, the physical electron is represented as $c^\dagger_{\sigma}=f^\dagger_{\sigma} \tau^x$. The $U(1)$ transformation representing the conservation of the electric (Noether) charge can only be incorporated into the spinon $f^\dagger_\sigma$, simply because the Pauli matrix $\tau^x$ is purely real. Therefore the f spinon inherits the whole charge from the parent electron and after the Ising disordering transition at $U_\perp$ the spinons will continue to display metallic transport properties despite the fact that the quasiparticle weight of the physical electrons has already been lost at $U_\perp$ as it is proportional to Ising magnetization [35]. If, instead of a metallic state at $U=0$ one starts with a semi-metallic state (such as graphene), the state appearing after the disordering transition of Ising pseudo-spins will be an orthogonal semi-metal [36]. For our purpose in the present paper, we would like to show that, both the orthogonal metal phase for $U_\perp<U<U_{\rm Mott}$ and the Ising metal (IM) for $U_b<U<U_\perp$ are equally interesting when one gaps out the underlying metallic state.

Now let us turn on the ionic potential Δ that locally couples to the electric charge as $(-1){\hspace{0pt}}^i \Delta c^\dagger_{i\sigma} c_{i\sigma}$ at every site i of the lattice. The Ising term $\tau^x$ is 'eaten up' by the $U(1)$ invariance of this term and the spinon density directly couple to the ionic potential as $(-1){\hspace{0pt}}^i\Delta f^\dagger_{i\sigma}f_{i\sigma}$. This is because spinons carry the whole charge and hence their density couples to external electrostatic potential (including even random potentials). The above term clearly gaps out both the IM and the OM conducting state of spinons and creates a band insulator of Ising metal and spinon metal. These phases can be called the Ising band insulator (IBI) and orthogonal band insulator (OBI), respectively, as they are born out of an underlying IM and OM states. When the (thermal) gap of the OBI is small enough to comply with semiconducting gaps, we will have an orthogonal semiconductor, i.e. a semiconductor of spinons that is separated from an Ising semiconductor by an Ising disordering transition.

Our discussion is schematically summarized in figure 1 where the notion of Ising and orthogonal state before a Mott state has been illustrated for three weak coupling states: metal, semimetal, and band insulator. The weak coupling states at the bottom row (blue) turn into their Ising counterpart (green) by increasing U beyond U_b at which the single boson condensate (rotor order parameter) vanishes. In the Ising (green) phase, still the physical electrons are governing the transport properties, but the charge condensate is survived as an Ising order which eventually vanishes at $U_\perp$. From this point the orthogonal (yellow) phase starts. By further increasing the Hubbard U beyond U_Mott even the residual interaction between the spinons of the orthogonal phase become so strong that renders the system Mott insulating.

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Therefore the intuitive picture that emerges for the ionic Hubbard model (right column) is as follows: at small values of Hubbard U (blue region) the gapped state is adiabatically connected to a band insulator. When U crosses U_b, the single bosons 'quantum evaporate' from the condensate and we will have IBI where the charge is controlled by an Ising variable; its characteristic Ising-like properties are expected to give rise to unusual semiconducting properties (green region). By further increasing U up to $U_\perp$ where the Ising order parameter of IBI vanishes, the OBI phase (yellow region) starts, which is eventually gapped at U_Mott.

In the OBI phase for $U_{\perp}<U<U_{\rm Mott}$ (yellow region) the disordering of the underlying slaved Ising variables leaves the states at the bottom of the conduction and top of the valence bands inaccessible to ARPES, while accessible to any probe coupling to the electric current (charge), such as optical conductivity, cyclotron resonance and thermal probes. This plays a significant role in the experimental discrimination of the orthogonal insulators (semiconductors) from their Ising (green) or normal (blue) relatives. The bandwidth and hence the effective mass of the correlated semiconducting phase IBI is controlled by an Ising order parameter which will then possess Ising-like temperature dependencies and hence e.g. cyclotron frequency will acquire Ising-like temperature dependence, etc; while the OBI phase is characterized by a wider ARPES gap compared to the thermal gap.

The results of this paper apply to quite general correlated insulators on any two-dimensional lattice. However, in this paper we focus on a honeycomb system where the parent band insulator in the continuum limit is described by massive Dirac electrons [37]. At the end we critically contrast the above possibilities with our previous DMFT of the ionic Hubbard model on honeycomb lattice [31] to discuss the plethora of Ising and orthogonal phases that may be conducting or insulating. This will shed a new light: to get a chance to realize IBI, the ionic potential Δ must be large enough. To realize OBI, the ionic potential must be even larger than what is required to realize IBI. For very small values of Δ, only the Ising and orthogonal phase of massless Dirac fermions can be realized, which corresponds to an orthogonal semi-metal.

We are interested in the phase diagram of the Hubbard model augmented by a staggered ionic potential of strength Δ as follows:

$$\begin{align} \newcommand{\e}{{\rm e}} \newcommand{\la}{\langle} \displaystyle H=H_0 + \frac{U}{2}\sum_i (n_{i\uparrow}+ n_{i\downarrow}-1)^2 \label{hubbard.eqn} \nonumber \end{align} \tag{ 2.1 }$$

$$\begin{align} \newcommand{\e}{{\rm e}} \newcommand{\ra}{\rangle} \newcommand{\la}{\langle} \displaystyle H_0= -t\sum_{\langle i,j\rangle \sigma} c^\dagger_{i\sigma} c_{j\sigma} +\Delta\sum_{i\sigma} (-1)^i n_{i\sigma}-\mu\sum_{i\sigma} n_{i\sigma} \nonumber \end{align} \tag{ 2.2 }$$

where $c^\dagger_{i\sigma}$ creates an electron at a localized orbital in site i with spin σ, $n_{i\sigma}=c^\dagger_{i\sigma}c_{i\sigma}$ is the occupation number, U is the on-site Hubbard repulsion, t is the hopping amplitude which will be set as the unit of energy throughout the paper, Δ is the ionic potential, and μ is the chemical potential that at half-filling in the present representation turns out to be $\mu=0$.

2.1. Slave rotor method

The slave rotor formulation is one of the slave particle family methods employed in studying the Hubbard model that provides a very economical representation of the charge state of an orbital in terms of a rotor variable conjugate to an angular momentum operator locked to the charge [18, 38–42]. This method can also be applied to the study of the Anderson impurity problem [43]. In this approach the local Hilbert space is represented by a direct product of the Hilbert space of a fermion carrying the spin index (the so called spinon) and a rotor that controls the charge state of the system as $\newcommand{\ra}{\rangle} \left\vert \psi \right\rangle =\left\vert {{\psi }_{f}} \right\rangle \left\vert {{\psi }_{\theta }} \right\rangle$. In terms of the operator creating a particle from its vacuum the above equation can be represented as

$$\begin{align} \newcommand{\e}{{\rm e}} \newcommand{\la}{\langle} \displaystyle \hat{c}_{i\sigma }^{\dagger }=f_{i\sigma }^{\dagger }{{\rm e}^{-{\rm i}{{\theta }_{i}}}}, \label{slaverotor.eqn} \nonumber \end{align} \tag{ 2.3 }$$

where $f^\dagger_{i\sigma}$ is the spinon creation operator and $\theta_i$ is the rotor variable at site i. The physical Hilbert space of the electron in terms of the spinons and rotors is constructed from the following states:

$$\begin{align} \newcommand{\e}{{\rm e}} \newcommand{\ra}{\rangle} \newcommand{\la}{\langle} \displaystyle \label{states.eqn} & \left\vert 0 \right\rangle \equiv \left\vert 0 \right\rangle_f \left\vert -1 \right\rangle_\theta \nonumber \\ & \left\vert \uparrow \right\rangle \equiv \left\vert \uparrow \right\rangle_f \left\vert 0 \right\rangle_\theta \nonumber \\ & \left\vert \downarrow \right\rangle \equiv \left\vert \downarrow \right\rangle_f \left\vert 0 \right\rangle_\theta \nonumber \\ & \left\vert \uparrow \downarrow \right\rangle \equiv \left\vert \uparrow \downarrow \right\rangle_f \left\vert 1 \right\rangle_\theta \nonumber \end{align} \tag{ 2.4 }$$

where $\newcommand{\ra}{\rangle} \vert \ra_\theta$ represents a state in the rotor space, and $\newcommand{\ra}{\rangle} \vert \ra_f$ represents states in the spinon Hilbert space. As can be seen in the above representation, a state such as $\newcommand{\ra}{\rangle} \vert \uparrow\ra_f \vert -1\ra_\theta$ containing one $\uparrow$-spin spinon and corresponding to angular momentum eigen state of $-1$ does not correspond to any physical state. Such redundancy is a characteristic of auxiliary particle methods where the physical Hilbert space is enormously enlarged. What we gain in the enlarged Hilbert space is the gauge freedom. But the physically sensible states are obtained from those in the enlarged space by projecting them on to the physical space. In the present case such a projection amounts to the constraint,

$$\begin{align} \newcommand{\e}{{\rm e}} \newcommand{\la}{\langle} \displaystyle \sum\limits_{\sigma }{f_{i\sigma }^{\dagger } {{f}_{i\sigma }}}=L_i+1 \label{gheid.eqn} \nonumber \end{align} \tag{ 2.5 }$$

that locks the particle number to angular momentum. It can be seen that all of the states in equation (2.4) satisfy this constraint. If the above constraint can be implemented exactly, then either of the representations in terms of original electrons or in terms of spinons and rotors will describe the same physics. But the full-fledged implementation of the above projection requires taking complete care of the fluctuations of the internal gauge fields that glue spinons and rotor fields. In the present work we treat the constraint in the mean field via a space- and time-independent Lagrange multiplier. This is known as the slave rotor mean field approximation [18, 38].

Let us proceed with representing the electron operators in terms of spinons and rotors, equation (2.3),

$$\begin{align} \newcommand{\e}{{\rm e}} \newcommand{\ra}{\rangle} \newcommand{\la}{\langle} \displaystyle H=&\,-\sum\limits_{\left\langle ij \right\rangle,\sigma }{f_{i\sigma }^{\dagger } {{f}_{j\sigma }}{{\rm e}^{-{\rm i}{{\theta }_{ij}}}} +{\rm h.c.}}+\Delta \sum\limits_{i}{f_{i}^{\dagger } {{f}_{i}}}-\Delta \sum\limits_{j}{f_{j}^{\dagger }{{f}_{j}}}\nonumber \\ &+\frac{U}{2}\sum\limits_{i}{L_{i}^{2}-\mu \sum\limits_{i\sigma }{f_{i\sigma } ^{\dagger }{{f}_{i\sigma }} }} + \lambda\sum_i \left(\,f^\dagger_{i\sigma} f_{i\sigma}-L-1 \right), \nonumber \end{align} \tag{ 2.6 }$$

where $i, j$ are nearest neighbors in two different sub-lattices and the hopping amplitude t of original electrons is set as the unit of energy, $t=1$, $\theta_{ij}=\theta_i-\theta_j$, the Hubbard term of equation (2.1) has been transformed with the aid of the constraint (2.5), μ is the chemical potential that becomes zero at half-filling which is our focus in this paper, and λ is a position-independent Lagrange multiplier that implements the constraint (2.5) on average. The merit of representation in terms of auxiliary rotor variables is that the quartic interaction between the fermions $U/2{{\sum\nolimits_{i\sigma }{\left(n_{i\sigma }-1 \right)}}^{2}}$ is replaced by a simple rotor kinetic energy $U {{L}^{2}}/2$ at every site, where the angular momentum $L= -{\rm i}{{\partial }_{\theta }}$ is a variable that is associated with a $O(2)$ quantum rotor θ.

Apart from the kinetic term that involves spinons and rotors on neighbouring sites, the above Hamiltonian is decoupled into rotor (θ-only) and spinon (f-only) terms. As for the first term we introduce mean field variables,

$$\begin{align} \newcommand{\e}{{\rm e}} \newcommand{\ra}{\rangle} \newcommand{\la}{\langle} \displaystyle \chi _\theta =\la {\rm e}^{{\rm i}\theta_i} {\rm e}^{-{\rm i}\theta_j} \ra_\theta, \label{chitheta.eqn} \nonumber \end{align} \tag{ 2.7 }$$

$$\begin{align} \newcommand{\e}{{\rm e}} \newcommand{\ra}{\rangle} \newcommand{\la}{\langle} \displaystyle \label{chif.eqn} \chi_f= \la \sum\nolimits_\sigma f_{\sigma i}^\dagger{\,f}_{\sigma j} \ra_f, \nonumber \end{align} \tag{ 2.8 }$$

to decouple the kinetic term, which eventually gives decoupled rotor, $H_\theta$ and spinon H_f Hamiltonians,

$$\begin{align} \newcommand{\e}{{\rm e}} \newcommand{\ra}{\rangle} \newcommand{\la}{\langle} \displaystyle H_f=&\,-\sum\limits_{\la ij \ra,\sigma}f_{i\sigma }^\dagger f_{j\sigma }\chi_\theta + {\rm h.c} +(\lambda-\mu)\sum_{j\sigma} f^\dagger_{j\sigma} f_{j\sigma} \nonumber \\ &+\Delta \sum_{i\in A} f_{i\sigma}^\dagger f_{i\sigma}-\Delta\sum_{j\in B} f_{j\sigma}^\dagger f_{j\sigma}, \label{spinon.eqn} \nonumber \end{align} \tag{ 2.9 }$$

and

$$\begin{align} \newcommand{\e}{{\rm e}} \newcommand{\ra}{\rangle} \newcommand{\la}{\langle} \displaystyle H_\theta=\sum_{\la i,j\ra} \chi_f {\rm e}^{{\rm i}\theta_i-{\rm i}\theta_j}+{\rm h.c.} +\sum_j\left(\frac{U}{2} L_j^2 - \lambda L_j\right), \label{rotor.eqn} \nonumber \end{align} \tag{ 2.10 }$$

where A and B are the two sub-lattices on the honeycomb lattice. In the above equation, solution of the H_f requires a knowledge of $\chi_\theta$ which according to equation (2.7) can only be calculated after having diagonalized the rotor sector $H_\theta$. The latter itself depends on unknown quantity $\chi_f$ that according to equation (2.8) can be obtained from the H_f Hamiltonian. This provides a self-consistency loop, i.e. the rotor and the spinon sectors talk to each other via the mean-field self-consistency equations (2.7) and (2.8).

It is interesting to note that in equation (2.9) the ionic potential Δ, being a local potential couples only to the spinon density. This considerably simplifies the analysis of the problem. Indeed in the absence of ionic potential Δ, the spinon sector would have been described by a spinon hopping Hamiltonian with renormalized hopping amplitudes whose reduced kinetic energy is encoded in $\chi_\theta$ which is self-consistently determined by the rotor sector. When the ionic potential is turned on, the spinon sector describes spinons hopping with renormalized hopping parameters $\chi_\theta$, plus an additional Bragg reflection due to doubling of the unit cell that always gaps out the spinon sector and the spectrum in the spinon-sector becomes

$$\begin{align} \newcommand{\e}{{\rm e}} \newcommand{\vk}{\vec{k}} \newcommand{\eps}{\varepsilon} \displaystyle \eps_f(\vk)=\pm \sqrt{\chi_\theta^2\vert \phi(\vk)\vert ^2+\Delta^2} \nonumber \end{align} \tag{ 2.11 }$$

where $\newcommand{\cd}{c^\dagger} \newcommand{\vk}{\vec{k}} \phi(\vk)=1+{\rm e}^{{\rm i}\vk\cdot\vec a_1}+{\rm e}^{{\rm i}\vk \cdot \vec a_2}$ with $\vec a_1$ and $\vec a_2$ unit being translation vectors of the honeycomb lattice.

Now let us turn to determination of chemical potential μ and Lagrange multiplier field λ. Since we are interested in the competition between U and Δ, we stay at half-filling where, even in the $U=0$ limit, the system is described by a simple band insulator. The particle-hole symmetry of the original Hamiltonian in terms of physical electrons simply implies that at half-filling $\mu=0$. Since we want to fix the average occupation at $n=1$, the constraint (2.5) implies that on average one should have $\newcommand{\la}{\langle} \newcommand{\ra}{\rangle} \la L_j \ra=0$ for every lattice site j. Now imagine that in a rotor Hilbert space corresponding to a given total angular momentum $\newcommand{\e}{{\rm e}} \ell$, we construct the local angular momentum operator $\newcommand{\la}{\langle} UL_j^2/2-\lambda L_j$ that would take $\newcommand{\e}{{\rm e}} 2\ell+1$ diagonal values $\newcommand{\la}{\langle} U(n^\theta_j ){\hspace{0pt}}^2-\lambda n^\theta_j$, where $\newcommand{\e}{{\rm e}} n^\theta_j=-\ell \ldots \ell$ represents all possible values of the magnetic quantum number. Obviously any non-zero value of λ breaks the symmetry between the states corresponding to positive and negative values of $n^\theta_j$ and therefore makes the expectation value $\newcommand{\la}{\langle} \newcommand{\ra}{\rangle} \la L_j\ra$ non-zero that places the system away from half-filling. Therefore half-filling corresponds to the time-reversal symmetry for the rotor dynamics which pins down the Lagrange multiplier λ to zero.

Now the remaining challenge is to solve the rotor problem posed by the Hamiltonian (2.10). One simple way to think of the rotor Hamiltonian is to fix a value for $\newcommand{\e}{{\rm e}} \ell$ that gives a local Hilbert space dimension of $\newcommand{\e}{{\rm e}} 2\ell+1$ for each rotor. This Hilbert space grows as $\newcommand{\e}{{\rm e}} N^{2\ell+1}$ where N is the number of lattice sites. However, if we further decouple the nearest neighbour terms as

$$\begin{align} \newcommand{\e}{{\rm e}} \newcommand{\ra}{\rangle} \newcommand{\la}{\langle} \displaystyle {\rm e}^{-{\rm i}\theta_i+i\theta_j}\approx {\rm e}^{-{\rm i}\theta_i}\la {\rm e}^{{\rm i}\theta_j}\ra +{\rm h.c.} ={\rm e}^{-{\rm i}\theta_i} \Phi_j +{\rm h.c.}, \nonumber \end{align} \tag{ 2.12 }$$

where a mean field rotor variable $\newcommand{\la}{\langle} \newcommand{\ra}{\rangle} \Phi_j=\la {\rm e}^{{\rm i}\theta_j}\ra$ is introduced, the rotor problem is considerably simplified. This mean field decomposition can be implemented on various clusters. We consider two types of clusters with a finite number of sites denoted in figure 2. For the single-site cluster the mean field variables connect a given site to its neighbors on the lattice, and the single-site Hamiltonian has no structure. In this case the mean field rotor Hamiltonian for every site is given by

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle H_\theta^{\rm 1\mbox{-}site~MF}=-3\chi_f\Phi \left({\rm e}^{-{\rm i}\theta}+{\rm e}^{{\rm i}\theta} \right) +\frac{U}{2}(n^\theta)^2 \nonumber \end{align} \tag{ 2.13 }$$

where the coefficient of 3 is due to three neighbours of every single-site (figure 2, left side). The explicit matrix representation of the above single-site mean field Hamiltonian is

$$\begin{align} \newcommand{\e}{{\rm e}} \newcommand{\vd}{v^\dagger} \newcommand{\la}{\langle} \displaystyle \left[ \begin{array}{@{}ccccc@{}} u\ell^2& -3\chi_f\Phi & 0 &\ldots & 0 \nonumber \\ -3\chi_f\Phi^* &u(\ell-1)^2 & -3\chi_f\Phi &\ldots & 0 \nonumber \\ 0 & -3\chi_f\Phi^* &\ddots &\ldots &\vdots \nonumber \\ \vdots & \vdots & -3\chi_f\Phi^* &u(-\ell+1)^2 & -3\chi_f\Phi \nonumber \\ 0 & 0 &\ldots &-3\chi_f\Phi^* & u(-\ell)^2 \end{array} \right] \label{matrix.eqn} \nonumber \end{align} \tag{ 2.14 }$$

where for notational brevity we have introduced $u=U/2$. Within the mean field a two-site cluster can also be adopted (see figure 2) for which the Hamiltonian becomes

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle {H}_\theta^{\rm 2\mbox{-}site~MF}=&\,-\chi_f\left({{\rm e}^{-{\rm i}{{\theta }_{1}}}} {{\rm e}^{{\rm i}{{\theta }_{2}}}}+{{\rm e}^{-{\rm i}{{\theta }_{2}}}}{{\rm e}^{{\rm i}{{\theta }_{1}}}} \right)\nonumber \\& -4{{\chi }_{f}}\Phi \left(\cos {{\theta }_{1}}+\cos {{\theta }_{2}} \right)+u{{\left(n_{1}^{\theta } \right)}^{2}}+u{{\left(n_{2}^{\theta } \right)}^{2}}. \nonumber \end{align} \tag{ 2.15 }$$

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The above Hamiltonian operates in the two-particle space represented by $\newcommand{\ra}{\rangle} \newcommand{\nt}{n^\theta} \vert \nt_1, \nt_2\ra$ where $\newcommand{\nt}{n^\theta} \newcommand{\e}{{\rm e}} \nt_i=-\ell_i\ldots\ell_i$ for $i=1, 2$. The effect of the two-site MF Hamiltonian on every such state is given by

$$\begin{align} \newcommand{\e}{{\rm e}} \newcommand{\nt}{n^\theta} \newcommand{\ra}{\rangle} \displaystyle &-\chi_f(\vert \nt_1-1,\nt_2+1\ra+ \vert \nt_1-1,\nt_2+1\ra)\nonumber \\ &-2\chi_f\Phi \sum_{a=\pm1}(\vert \nt_1+a,\nt_2\ra+\vert \nt_1,\nt_2+a\ra )\nonumber \\ &+u\left({\nt_1}^2+{\nt_2}^2\right)\vert \nt_1,\nt_2\ra. \nonumber \end{align} \tag{ 2.16 }$$

The following algorithm self-consistently determines all the mean field parameters: for a given set of external parameters such as U: (I) Start with an initial guess for $\chi_f$. (II-a) For the above $\chi_f$, guess a Φ. (II-b) Diagonalize the matrix 2.14 and obtain its ground state. (II-c) In the obtained ground state update the $\newcommand{\la}{\langle} \newcommand{\ra}{\rangle} \Phi=\la {\rm e}^{{\rm i}\theta}\ra$ and keep repeating until Φ is self-consistently determined. (III) For the present value of $\chi_f$ and Φ, use equation (2.7) to obtain $\chi_\theta$. (IV) Plug in the $\chi_\theta$ into the spinon Hamiltonian (2.9) and diagonalize it. (V) For the ground state of the above spinon Hamiltonian use equation (2.8) to update the initial guess $\chi_f$. This procedure is repeated until mean field parameters $\chi_f, \chi_\theta, \Phi$ are self-consistently determined. It must be noted that the above procedure is done for a fixed value of $\newcommand{\e}{{\rm e}} \ell$. One has to repeat the procedure for larger values of $\newcommand{\e}{{\rm e}} \ell$ to ensure that the final converged results do not change much upon a further increase in the dimension of the rotor space. We confirm that, as previously noted [38], the choice $\newcommand{\e}{{\rm e}} \ell=2$ is accurate enough. Let us emphasize that $\newcommand{\e}{{\rm e}} \ell$ here is not a physical angular momentum, but rather a fictitious angular momentum corresponding to a quantum rotor that controls the charge configuration.

Upon increasing the Hubbard interaction U beyond a critical point U_b the mean field parameter Φ vanishes that corresponds to strong fluctuations in the phase variable, and hence frozen fluctuations of the corresponding number operators, i.e. putative Mott state. However, as will be discussed in the next section the phase fluctuations still have the chance to survive in the form of sign fluctuations that can be captured by enslaving an Ising variable.

The charge degree of freedom at every site can be described by variety of methods. A description in terms of a rotor variable whose conjugate variable controls the charge state is one possibility. Another appealing possibility is to use an Ising variable to denote the charge state. Using an Ising variable to specify the electric charge can be implemented in various ways [19, 20, 44, 45]. In this section we briefly review the presentation of [19] and adopt it in our investigation of the ionic Hubbard model on the honeycomb lattice.

In this representation, an Ising variable τ is introduced to take care of the charge configuration of every site. Empty and doubly occupied configuration are represented by $\newcommand{\ra}{\rangle} \vert +\ra$, while singly occupied configurations possessing local moment both are represented by $\newcommand{\ra}{\rangle} \vert -\ra$ where,

$$\begin{align} \newcommand{\e}{{\rm e}} \newcommand{\ra}{\rangle} \displaystyle \tau^z \vert \pm\ra = (\pm 1) \vert \pm\ra. \nonumber \end{align} \tag{ 3.1 }$$

The correspondence between physical states and those in the Hilbert space extended by introduction of Ising variables is:

$$\begin{align} \newcommand{\e}{{\rm e}} \newcommand{\ra}{\rangle} \newcommand{\la}{\langle} \displaystyle \left\vert {\rm empty} \right\rangle =\left\vert + \right\rangle \left\vert 0 \right\rangle \nonumber \\ \left\vert {\rm singly~occupied},\uparrow \right\rangle =\left\vert - \right\rangle \left\vert \uparrow \right\rangle \nonumber \\ \left\vert {\rm singly~occupied},\downarrow \right\rangle =\left\vert - \right\rangle \left\vert \downarrow \right\rangle \nonumber \\ \left\vert {\rm doubly~occupied} \right\rangle =\left\vert + \right\rangle \left\vert \uparrow\downarrow \right\rangle \label{IsingStates.eqn} \nonumber \end{align} \tag{ 3.2 }$$

where states on the left hand correspond to physical electrons, and those in the right hand are product of states corresponding to Ising variables, $\newcommand{\ra}{\rangle} \vert \pm\ra$, and those corresponding to spinon. The creation of each electron has almost a parallel on the right side, implying that $c^\dagger\sim f^\dagger$. However, each time an electron is created, the charge state flips between the one with a local moment and the one with no local moment. This corresponds to a flip in the Ising variable that can be achieved with the action of Pauli matrix $\tau^x$. Therefore the physical electron at every site j can be represented as,

$$\begin{align} \newcommand{\e}{{\rm e}} \newcommand{\la}{\langle} \displaystyle c_{j\sigma }^{\dagger }\equiv \tau^x_j f_{j\sigma }^{\dagger }, \label{slaveIsing.eqn} \nonumber \end{align} \tag{ 3.3 }$$

where $f^\dagger_{j\sigma}$ creates a spinon of spin σ at site j. The fact that the creation of each physical charge is synonymous to the creation of a spinon implies that the physical charge is basically carried by spinons. This was a key observation made by Nandkishore and coworkers [35] that is formally reflected in equation (3.3) as the fact that the Pauli matrix $\tau^x$, being a real matrix, cannot absorb the $U(1)$ phase transformation that generates the conservation of charge, and hence all the electron charge is carried by the spinon.

The Hilbert space represented by the product of Ising pseudo-spin and spinon spaces is larger than the physical space and includes states such as $\newcommand{\ra}{\rangle} \vert +\ra\vert \uparrow\ra$, etc that do not correspond to any physical state. Inspection shows that those states can be eliminated by the following constraint:

$$\begin{align} \newcommand{\e}{{\rm e}} \newcommand{\la}{\langle} \displaystyle \tau^z_j+1-2(n_j-1)^2=0. \label{Isinggheid.eqn} \nonumber \end{align} \tag{ 3.4 }$$

In this new representation, the ionic Hubbard Hamiltonian becomes

$$\begin{align} \newcommand{\e}{{\rm e}} \newcommand{\ra}{\rangle} \newcommand{\la}{\langle} \displaystyle H=-\sum\limits_{\la ij \ra}\tau_i^x \tau_j^x f_{i\sigma}^{\dagger} f_{j\sigma} +\frac{U}{4}\sum_j{\left(\tau _{j}^{z}+1 \right)}+\Delta \sum_j f^\dagger_{j\sigma} f_{j\sigma} (-1)^{\,j} \nonumber \end{align} \tag{ 3.5 }$$

where as before the hopping t of the physical electrons is set as the unit of energy and the constraint equation (3.4) has enabled us to cast the Hubbard U into a form involving only Ising variable $\tau^z$. Again, in the ionic potential term, the Ising pseudo-spins being squared to unit matrix cancel each other and hence the staggered ionic potential is only coupled to the spinons. Mean-field decoupling of the Ising and spinon variables leads to two separate Hamiltonians governing the dynamics of spinons f and Ising variables τ as follows:

$$\begin{align} \newcommand{\e}{{\rm e}} \newcommand{\la}{\langle} \displaystyle H_f=&-\sum\limits_{ij}{{{\chi }_{I}}f_{i\sigma }^{\dagger }{{f}_{j\sigma }}} +\Delta \sum_j f^\dagger_{j\sigma} f_{j\sigma} (-1)^{\,j} \nonumber \\ &-2\lambda^{\prime}\sum_j\left(\,f^\dagger_{j\uparrow}f_{j\uparrow}+f^\dagger_{j\downarrow}f_{j\downarrow}-1\right)^2, \label{HfIsing.eqn} \nonumber \end{align} \tag{ 3.6 }$$

and

$$\begin{align} \newcommand{\e}{{\rm e}} \newcommand{\la}{\langle} \displaystyle H_{\rm ITF}=-\chi^{\prime}_f\sum\limits_{ij}\tau _{i}^{x}\tau _{j}^{x} +\left(\frac{U}{4}+\lambda^{\prime}\right)\sum\limits_{i}{\tau _{i}^{z}} \label{Htau.eqn}, \nonumber \end{align} \tag{ 3.7 }$$

where the two Hamiltonians are coupled to each other through the following self-consistency equations:

$$\begin{align} \newcommand{\e}{{\rm e}} \newcommand{\ra}{\rangle} \newcommand{\la}{\langle} \displaystyle \chi_I= \la \tau^x_i \tau^x_j\ra_I,\label{chiI.eqn} \nonumber \end{align} \tag{ 3.8 }$$

$$\begin{align} \newcommand{\e}{{\rm e}} \newcommand{\ra}{\rangle} \newcommand{\la}{\langle} \displaystyle \chi^{\prime}_f=\la\sum\nolimits_{\sigma} f^\dagger_{i\sigma} f_{j\sigma}\ra_f, \nonumber \end{align} \tag{ 3.9 }$$

and the Lagrange multiplier $\newcommand{\la}{\langle} \lambda^{\prime}$ is introduced to implement the constraint (3.4) on average. As can be seen, within the present representation, the fermion part, (3.6), still remains an interacting problem of the Hubbard type, where the scale of on-site interaction among the spinons is set by the Lagrange multiplier $\newcommand{\la}{\langle} \lambda^{\prime}$. The mean-field approximation lends itself to the assumption that the system in the enlarged Hilbert space is a product state composed of a spinon part and an Ising part. The spinon part of the wave function is expected to have a form close to a Slater determinant and hence the parameter $\newcommand{\la}{\langle} \lambda^{\prime}$ is expected to represent small residual interactions between the spinons. Approximate strategies to handle the interacting spinons have been suggested and discussed in [19]. With this argument we think that it is reasonable to assume $\newcommand{\la}{\langle} \lambda^{\prime}=0$ as an approximate strategy to obtain the simplest possible solution [45].

To understand the nature of the approximation $\newcommand{\la}{\langle} \lambda^{\prime}\approx 0$, let us focus on some special lucky situations where it can be proven that $\newcommand{\la}{\langle} \lambda^{\prime}=0$ is exact. When the ionic term is absent, i.e. for the pure Hubbard model, one may have situations where the partition function Z happens to be an even function of U, such as the particle-hole symmetric case considered here [46, 47]. In this case, first of all the Hubbard interaction at half-filling can be written as

$$\begin{align} \newcommand{\e}{{\rm e}} \newcommand{\la}{\langle} \displaystyle Un_{j\uparrow} n_{j\downarrow} -\frac{U}{2}\left(n_{j\uparrow}+n_{j\downarrow} \right), \label{phsh.eqn} \nonumber \end{align} \tag{ 3.10 }$$

where due to half-filling conditions the chemical potential $\mu=U/2$ is used. Under a particle-hole transformation in one spin sector only (let us call it PHσ transformation in this paper), namely,

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle c^\dagger_{j\uparrow}=\tilde c^\dagger_{j\uparrow},~~c^\dagger_{j\downarrow}=(-1)^{\,j} \tilde c_{j\downarrow}, \nonumber \end{align} \tag{ 3.11 }$$

where $(-1){\hspace{0pt}}^{\,j}=\pm 1$ depends on whether it is on sublattice A, or B. Under PHσ the role of charge and spin density are exchanged, namely $n_\uparrow+n_\downarrow \to \tilde n_\uparrow-\tilde n_\downarrow$. This transformation maps the particle-hole symmetric Hubbard interaction of equation (3.10) to,

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \tilde U\tilde n_{j\uparrow} \tilde n_{j\downarrow} - \frac{\tilde U}{2}\left(\tilde n_{j\uparrow}+\tilde n_{j\downarrow} \right) \nonumber \end{align} \tag{ 3.12 }$$

which is nothing but the original Hubbard interaction at half-filling with the only difference that $\tilde U=-U$. At half-filling under the PHσ transformation is a symmetry of the Hubbard model, which implies that the properties of system are even with respect to Hubbard U.

Let us now examine how the constraint (3.4) behaves under the PHσ transformation. As indicated in the figure 3 the role of PHσ transformation in terms of Ising pseudo-spins is to exchange the role of $\newcommand{\ra}{\rangle} \vert +\ra$ and $\newcommand{\ra}{\rangle} \vert -\ra$ states of Ising variables. Therefore the effect of PHσ transformation on any operator ${{\mathcal O}}$ that contains Ising variables is to change ${{\mathcal O}}\to \tau^x {{\mathcal O}}\tau^x$. This transformation leaves the first term in the ITF Hamiltonian (3.7) intact, but it changes the second term at a given site as follows:

$$\begin{align} \newcommand{\e}{{\rm e}} \newcommand{\la}{\langle} \displaystyle & \left(\frac{U}{4}+\lambda^{\prime}\right) \tau^z \to \left(\frac{U}{4}+\lambda^{\prime}\right) \tau^x\tau^z\tau^x\nonumber \\ &= -\left(\frac{U}{4}+\lambda^{\prime}\right)\tau^z =-\left(-\frac{U}{4}+\lambda^{\prime}\right)\tau^z, \nonumber \end{align} \tag{ 3.13 }$$

where in the last equality we have used the fact that the Hubbard model at half-filling is even with respect to Hubbard U. Therefore we have proven that at half-filling for the Hubbard model, the Lagrange multiplier $\newcommand{\la}{\langle} \lambda^{\prime}$ is exactly zero.

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Turning on the ionic potential $(-){\hspace{0pt}}^{\,j} \Delta (n_\uparrow+n_\downarrow)$, the PHσ transformation maps to $(-){\hspace{0pt}}^{\,j}\Delta (\tilde n_{j\uparrow}-\tilde n_{j\downarrow})$, i.e. the charge density is mapped to spin-density. Therefore the PHσ is not a symmetry of ionic Hubbard model at half-filling. Hence the partition function has both even and odd parts as a function of U. This prevents the Lagrange multiplier $\newcommand{\la}{\langle} \lambda^{\prime}$ from becoming zero. However, the above symmetry consideration suggests that the physics of ionic (and repulsive) Hubbard model maps onto the physics of attractive Hubbard model in the presence of a staggered magnetization field (since it is coupled to spin density in a staggered way). Another merit of the above symmetry discussion is that based on the following argument in [46], we can infer what is precisely missed by the approximation $\newcommand{\la}{\langle} \lambda^{\prime}=0$: Let us rewrite the constraint (3.4) as

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle {{\mathcal P}}_+=1+\tau^z_j\Omega_j=0,~~\Omega_i=1-2(n_j-1)^2, \nonumber \end{align} \tag{ 3.14 }$$

which identifies the operator $\Omega_j$ as the fluctuations of the charge away from half-filling. It was shown in [46] that to all orders in perturbation theory, the term 1 in ${{\mathcal P}}_+$ contributes only in even powers of U while the second term contributes only in odd powers of U. Therefore for a half-filled situation of the pure Hubbard model, the effect of a second term is nullified, and basically one need not worry about projection. This is another way of saying that the Lagrange multiplier $\newcommand{\la}{\langle} \lambda^{\prime}$ at half-filling becomes zero. By adding the ionic term to the half-filled Hubbard model, or placing the Hubbard model itself away from the half-filling, $\newcommand{\la}{\langle} \lambda^{\prime}$ is expected to be small based on our earlier argument on small residual interactions. In this case using the approximation $\newcommand{\la}{\langle} \lambda^{\prime}\approx 0$ amounts to missing the effects that are odd functions of the Hubbard U.

Within the approximation of $\newcommand{\la}{\langle} \lambda^{\prime}=0$, the spinon part describes a non-interacting band of spinons whose bandwidth is renormalized by the $\chi_I$ parameter obtained from the Ising part. The physics of transition to orthogonal state is then captured by the disordering transition of the Ising sector (3.7) where vanishing quasi-particle weight of the physical electrons are characterized by $\newcommand{\la}{\langle} \newcommand{\ra}{\rangle} \la \tau^x\ra =0$ [35, 47].

In the absence of the ionic term, i.e. when $\Delta=0$, both the ordered and disordered side of the ITF Hamiltonian (3.7) are conducting state: (i) if the underlying rotors are ordered, namely $\newcommand{\la}{\langle} \newcommand{\ra}{\rangle} \la {\rm e}^{{\rm i}\theta}\ra \ne 0$ the conducting state within the Hubbard model is a Fermi liquid. This holds for $U<U_b$. (ii) When the rotor order vanishes beyond U_b, i.e. $\newcommand{\la}{\langle} \newcommand{\ra}{\rangle} \la {\rm e}^{{\rm i}\theta}\ra =0$ still we may have $\newcommand{\la}{\langle} \newcommand{\ra}{\rangle} \la \tau^x\ra\ne 0$ which again is a conducting state (Ising metal) for the pure Hubbard model. This state persists until $U_\perp$ where the Ising order vanishes and the pure Hubbard model describes the orthogonal metallic state. By adding the ionic term Δ, it is crucial to note that on-site ionic potential does not couple to either slave Ising variables or to the salve rotor variables. However, the effect of the ionic potential Δ is to modify the order parameters through mean-field self-consistency equations, but their order-disorder physics remains the same as the Hubbard model as the ionic term does not explicitly appear in the Ising or rotor sectors. The essential role of the ionic term is to create Bragg reflections in the spinon Hamiltonian and to gap them out, which corresponds to rendering Fermi liquid, Ising metal and orthogonal metal phase of the pure Hubbard model to band insulator, Ising band insulator, and orthogonal insulator, respectively.

The disordering phase transition of the ITF Hamiltonian (3.7) can be captured within a simple cluster mean field approximation. Let us decompose the lattice to clusters Γ labeled by integer I whose internal sites are labeled by integers $a, b$ etc. Then the Ising variables are denoted by $\vec{\tau}_{\Gamma, a}$. With this re-arrangement, and after mean field decoupling of the cluster with its surrounding sites via a mean field order parameter $\newcommand{\la}{\langle} \newcommand{\ra}{\rangle} m=\la \tau^x\ra$ (see figure 4), the Ising part becomes

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle H=\sum_\Gamma H_\Gamma, \nonumber \end{align} \tag{ 3.15 }$$

where the Hamiltonian for cluster Γ is,

$$\begin{align} \newcommand{\e}{{\rm e}} \newcommand{\la}{\langle} \displaystyle H_\Gamma= -\chi^{\prime}_f\sum_{a,b\in \Gamma} \tau_{\Gamma,a}^x\tau_{\Gamma,b}^x -\frac{mz}{2}\sum_{a\in\Gamma}\tau_{\Gamma,a}^x+\frac{U}{4}\sum_a{\tau _{\Gamma,a}^{z}}. \label{Hcluster.eqn} \nonumber \end{align} \tag{ 3.16 }$$

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In this cluster Hamiltonian, the z denotes the number of bonds crossing the boundary of the cluster. The factor $1/2$ avoids double counting and $\newcommand{\la}{\langle} \newcommand{\ra}{\rangle} m=\la\tau^x\ra$ is the Ising order parameter that at the mean field level decouples the cluster Γ from its surroundings, but the interactions within the cluster Γ are treated with exact diagonalization. In figure 4 we have depicted clusters used in the present work. For more details on the construction of the Hilbert space and diagonalization of the Hamiltonian, please refer to the appendix of [48]. The Ising disordering considered here for four-site and six-site clusters show no appreciable difference. In this work we report the critical values U_c of the Ising disordering transition that is obtained from 6-site clusters.

In figure 5 we have plotted the results of a slave-rotor mean field for a two-site cluster. The rotor Hilbert space in this plot has been constructed for the angular momentum $\newcommand{\e}{{\rm e}} \ell=2$. We have checked that the results are not sensitive to increase in the size of the rotor Hilbert space beyond $\newcommand{\e}{{\rm e}} \ell=2$. The left panel shows the evolution of order parameter Φ of rotors as a function of Hubbard U for a selected set of ionic potentials indicated in the legend. As can be seen for $\Delta=0$ the critical value U_b starts around $3.5$ and decreases by increasing Δ. At $\Delta=1$ the critical value for the disordering of single-boson is around $2.8$. This means that in the presence of a staggered potential it becomes easier to lose the single-boson condensate whereby the Ising phase (paired boson superfluid) starts. Right panel in the figure provides an intensity map of the rotor order parameter in the $(U, \Delta)$ plane. The blue region corresponds to zero single-boson condensation amplitude, and the red corresponds to maximal (i.e. 1) condensation amplitude for the single-bosons $\newcommand{\la}{\langle} \newcommand{\ra}{\rangle} \la {\rm e}^{{\rm i}\theta}\ra$.

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In figure 6 we present the cluster mean field results for the Ising order parameter. The left panel shows the Ising magnetization $\newcommand{\la}{\langle} \newcommand{\ra}{\rangle} m=\la \tau^x\ra$ as a function of U for selected values of the staggered potential Δ indicated in the figure. The right panel provides an intensity map of the Ising order parameter in the $(U, \Delta)$ plane. The color code is the same as in figure 5. Once the Ising order goes away, we are in the orthogonal phase. By increasing the staggered ionic potential from $\Delta=0$ to $\Delta=1$, the critical value U_c decreases from $\sim 6.9$ to $5.7$. This trend is similar to the behavior of the rotor order parameter, i.e. the effort of U to destroy the Ising order parameter $\newcommand{\la}{\langle} \newcommand{\ra}{\rangle} m=\la \tau^x\ra$ is assisted by the ionic potential Δ. This points to the fact that obtaining both IBI and OBI is facilitated by ionic potential Δ. The larger the ionic potential, the easier to 'quantum melt' the 1- and 2- boson condensates that correspond to entering Ising and orthogonal phases.

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Figure 7 combines figures 5 and 6 and shows that there is a clear region $U_b \leqslant U < U_\perp$ where the rotor order parameter is zero, i.e. the single boson condensate has vanished, while the the Ising order parameter is non-zero. In the region $U<U_b$ the underlying metallic state is a normal Fermi liquid which is gapped out by directly coupling to the staggered potential and therefore the underlying Fermi liquid state becomes a normal band insulator. For $U_b<U<U_\perp$ the the rotor variable undergoes wild fluctuations and vanishes; however, the Z₂ slave variable remains ordered. It should be noted that the $U(1)$ slave rotor can be considered as the $n\to\infty$ limit of a Z_n slave-particle theory. Therefore using any Z_n theory with finite n is expected to reduce the $U_\perp$ from its Z₂ value, but still keeps room for the 'Ising-like' phase. The Ising order parameter vanishes at $U_\perp$ beyond which the semiconducting transport will be entirely done by spinons. For largest values of $U>U_{\rm Mott}$ the system eventually becomes Mott insulating [31].

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The existence of a region where the Ising variable is ordered but the rotor variable is disordered endows the non-Mott phase of the ionic Hubbard model with a condensate of Z₂ charge bosons where charge fluctuations survive in the form controlled by Ising variables. Due to Ising form of charge fluctuations, this is still a Fermi liquid proceeding to the OM phase whose quasiparticle weight is determined by the Ising order parameter. In the ionic Hubbard model this corresponds to an Ising band insulating phase where the kinetic energy of spinon Hamiltonian is controlled by an Ising order parameter, and hence the band properties of such a semiconducting phase inherits characteristic temperature and field dependence from the underlying Ising model. Across the Ising transition, the quasi-particle weight of the physical electron is lost, and the electric charge is carried by spinons which corresponds to losing the Ising condensate. This is the orthogonal phase which in the ionic Hubbard model corresponds to OBI. The quasiparticle weight of the physical electrons in this phase vanishes as it was being controlled by the now-vanished Ising order parameter [35], and hence in the OBI phase the states at the bottom of the conduction and top of the valence band are not visible by ARPES. However, since the current operators are solely constructed by the spinons, the optical conductivity measurement (i.e. the current-current correlation function) does couple to the states near the bottom of conduction band and those near the top of the valence band of the resulting OBI. Therefore an important characteristic property of OBI is that the optical conductivity gap is expected to be less than the ARPES gap. For the two-dimensional semiconductors or insulators the ability to tune the chemical potential into the conduction band provides a chance to examine such an 'ARPES-dark state' by quantum oscillations experiments. Once the chemical potential is tuned to the conduction band, the thermally excited carriers into the ARPES-invisible states at the bottom of the conduction band would display quantum oscillations. The 'ARPES-dark' states of the OBI phase would couple to thermal probes as well, which means that the gap extracted from thermal measurements will be less than the ARPES gap.

5.1. Four types of insulating states

5.2. Experimental signatures of Ising and orthogonal insulating states

5.3. Foes of the Ising and orthogonal phases

Let us briefly discuss the connection of the present work to other works on the ionic Hubbard model and speculate under what conditions the present picture of IBI and OBI are challenged. Investigations of the nature of intermediate phase in the ionic Hubbard model fall into two major groups: the first group suggests that the intermediate phase is gapped [25, 51], while the second group suggests gapless intermediate state [27, 29–31]. The present work finds a gapped intermediate phase. However, the gap in the present case is due to symmetry breaking in a fractional degree of freedom. This order does not correspond to any ordered spin or charge density profiles as no form of density operator depends on the Ising pseudo-spin simply because $\tau^x$ is Hermitian and squares to unit matrix. The gap in the Ising and orthogonal semiconductor is caused by the Bragg reflection of spinons, and as such there are no low-energy Goldstone modes associated with our present proposal. The second group of investigations suggest a gapless state for a region $\Delta\sim U$. Within the present mean field approach, we get three gapped phases depicted in figure 7: BI (blue), IBI (green) and OBI (yellow). The present approach does not capture the Mott phase as in the mean field and within the half-filled Hubbard model we do not take interactions among spinons of the IBI phase into account, i.e. in equation (3.6), we set the Lagrange multiplier $\newcommand{\la}{\langle} \lambda^{\prime}=0$. For the pure Hubbard model with $\Delta=0$ this was proven exactly to be the case. However for the ionic Hubbard model, $\newcommand{\la}{\langle} \lambda^{\prime}$ in general is not zero, and may eventually give rise to a Mott transition of spinons where the orthogonal phase ends. This poses the question of how realistic the Ising and orthogonal phases are. Along the lines of the [35] very straightforward generalization of a model of fermions coupled to a Z₂ gauge field can have the orthogonal phase as its exact ground states. Consider a honeycomb lattice whose neighboring sites are labeled by $i, j$. The model basically consists of matter, H_mat and field $H_{Z_2}$ parts. The matter part is

$$\begin{align*} \newcommand{\e}{{\rm e}} \newcommand{\ra}{\rangle} \newcommand{\la}{\langle} \displaystyle H_{\rm mat} = -t \sum_{\langle ij\rangle} \left(c^{\dagger}_i \sigma^{z}_{ij}c_{j} + {\rm h.c.} \right) \nonumber \end{align*}$$

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle + \Delta \sum_{j} (-1)^{\,j} c_j^\dagger c_j - \mu \sum_{j}c^\dagger_j c_j \nonumber \end{align} \tag{ 5.1 }$$

and the dynamic of the Z₂ variables is given by [35]

$$\begin{align} \newcommand{\e}{{\rm e}} \newcommand{\ra}{\rangle} \newcommand{\la}{\langle} \displaystyle H_{Z_2}= -J \sum_{\langle i,j \rangle} \sigma^z_{ij} -h \sum_i (-1)^{c^\dagger_i c_i} \prod_{j \in nn(i)} \sigma^x_{ij} \nonumber \end{align} \tag{ 5.2 }$$

where $j\in nn(i)$ emphasizes that j is nearest neighbor of i. The Ising and orthogonal insulating phases are exact ground states of the above Hamiltonian. But for the ionic model with Hubbard type of interactions they remain approximate ground states. A proper handling of the fluctuations of the gauge fields is needed to assess the quality of such an approximation.

One may argue that critical values of the Mott transition in sophisticated treatments of the Mott transition [10] give a critical value for Mott transition which can shadow the whole orthogonal phase. However, the present results being a spatial mean field can be compared to a dynamic form of mean field [31]. The inclusion of spatial fluctuations beyond the mean field is expected to reduce all critical values, and hence the present ordering of the phases is expected to be qualitatively robust. Within the DMFT approach, the battle between U and Δ to close the gap takes place. The dashed lines in figure 8 represent the phase boundaries from DFMT. The left branch of the dashed line separates the band insulator from the semi metal (SM) while the right branch of the dashed phase boundary separates the SM from the Mott insulator (MI). The intermediate phase is a massless Dirac phase within the DMFT approach. When we superimpose the DMFT phase diagram [31] with that of figure 7 we find that the phase boundaries obtained from present study (bold lines) partition the BI and SM phase of the DMFT phase diagram into three phases corresponding to normal, Ising and orthogonal variants. Although these are two different mean field methods, and critical values obtained from DMFT and present study may correspond to different mechanisms, that does not concern us here. Improvements in both mean field approximations may push the bold lines slightly away, but does not change the fact that the bold phase boundaries cross the left branch of the DMFT (dashed boundary). Moreover, the inclusion of spatial fluctuations is expected to push the solid boundaries to the left as well. This comparison with dynamical mean field theory sheds a new light: Realization of IBI and OBI phases requires large enough ionic potential. For very small ionic potentials, the DMFT based scenario of the battle between U and Δ can possibly kill the insulating phase and give a massless SM. Then an increase in Hubbard U will give rise to an Ising semi metal (ISM) or orthogonal semi metal (OSM) [36]. If the ionic potential grows further, the green phase in the band insulating side also gets a chance and therefore IBI could be realized if Δ is larger than about  ∼$0.1$ (in units of hopping t of course). If we keep increasing Δ beyond  ∼$0.38$ the OBI phase also gets a chance. However the DMFT scenario of the battle between U and Δ suggests that the OBI phase does not directly transform into Mott phase, but instead goes through an orthogonal semi metal which is appealing: The battle will continue in the fractionalized OBI phase of spinons and can presumably close the spinon gap in OBI to render it OSM before getting into the Mott phase [31].

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5.4. Outlook

We thank A G Moghaddam for useful comments and R Ghadimi for very helpful discussions. TF appreciates the Ministry of Science, Research and Technology (MSRT) of Iran for financial support during a visit to Sharif University of Technology. SAJ was supported by the Alexander von Humboldt fellowship for experienced researchers.

In this appendix we present details of the exact diagonalization for the ITF Hamiltonian on a finite cluster for 4 and 6 site clusters. We employ group theory methods to reduce the dimension of ensuing matrices.

A.1. Y-shaped four-site cluster

To solve the equation (3.16) first we choose a Y-shaped four-site cluster Γ shown in figure A1. The spin variables at every site have two possible states giving a total of $2^4-16$ possible states for the cluster Γ. Each state of this cluster is of the form $\newcommand{\ra}{\rangle} \vert \sigma_3, \sigma_2, \sigma_1, \sigma_0\ra$ where $\sigma_a$ can take two possible values $\uparrow, \downarrow$ and the site indices $a=0, 1, 2, 3$ are indicated in figure A1. The basis in this 16-dimensional Hilbert space are as follows (for brevity we have dropped $\newcommand{\ra}{\rangle} \vert \ra$ from the representation of basis states):

$$\begin{align} \newcommand{\e}{{\rm e}} \newcommand{\ra}{\rangle} \displaystyle &\left\vert 1 \right\rangle =\uparrow \uparrow \uparrow \uparrow~ \left\vert 2 \right\rangle =\uparrow \uparrow \uparrow \downarrow~ \left\vert 3 \right\rangle =\uparrow \uparrow \downarrow \uparrow~ \left\vert 4 \right\rangle =\uparrow \downarrow \uparrow \uparrow \nonumber \\ &\left\vert 5 \right\rangle =\downarrow \uparrow \uparrow \uparrow~\left\vert 6 \right\rangle =\uparrow \uparrow \downarrow \downarrow ~\left\vert 7 \right\rangle =\uparrow \downarrow \downarrow \uparrow~\left\vert 8 \right\rangle =\downarrow \downarrow \uparrow \uparrow \nonumber \\ &\left\vert 9 \right\rangle =\downarrow \uparrow \uparrow \downarrow~\left\vert 10 \right\rangle =\uparrow \downarrow \uparrow \downarrow~\left\vert 11 \right\rangle =\downarrow \uparrow \downarrow \uparrow~\left\vert 12 \right\rangle =\uparrow \downarrow \downarrow \downarrow~\nonumber \\ &\left\vert 13 \right\rangle =\downarrow \uparrow \downarrow \downarrow~ \left\vert 14 \right\rangle =\downarrow \downarrow \uparrow \downarrow~ \left\vert 15 \right\rangle =\downarrow \downarrow \downarrow \uparrow~ \left\vert 16 \right\rangle =\downarrow \downarrow \downarrow \downarrow\nonumber . \nonumber \end{align} \tag{ A.1 }$$

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In the four-site cluster of figure A1 the positions $1, 2, 3$ are not nearest neighbors of each other, while they are all neighbors of the site 0. So the exchange interaction in the cluster takes place only between the site 0 and the above three sites. Hence the first term of equation (3.16) for the four-site cluster is

$$\begin{align} \newcommand{\e}{{\rm e}} \newcommand{\ra}{\rangle} \newcommand{\la}{\langle} \displaystyle {{H}_{1}}=-\sum\limits_{\la a,b\ra\in \Gamma}{\tau _{a}^{x}\tau _{b}^{x}=-\left\{\tau _{0}^{x}\tau _{1}^{x}+\tau _{0}^{x}\tau _{2}^{x}+\tau _{0}^{x}\tau _{3}^{x} \right\}}. \nonumber \end{align} \tag{ A.2 }$$

The effect of the above term on the bases is:

$$\begin{align} \newcommand{\e}{{\rm e}} \newcommand{\ra}{\rangle} \displaystyle &{{H}_{1}}\left\vert 1 \right\rangle =-\left\{\left\vert 6 \right\rangle +\left\vert 9 \right\rangle +\left\vert 10 \right\rangle \right\}\nonumber \\ &{{H}_{1}}\left\vert 2 \right\rangle =-\left\{\left\vert 3 \right\rangle +\left\vert 4 \right\rangle +\left\vert 5 \right\rangle \right\}\nonumber \\ &{{H}_{1}}\left\vert 3 \right\rangle =-\left\{\left\vert 2 \right\rangle +\left\vert 12 \right\rangle +\left\vert 13 \right\rangle \right\}\nonumber \\ &{{H}_{1}}\left\vert 4 \right\rangle =-\left\{\left\vert 2 \right\rangle +\left\vert 12 \right\rangle +\left\vert 14 \right\rangle \right\}\nonumber \\ &{{H}_{1}}\left\vert 5 \right\rangle =-\left\{\left\vert 2 \right\rangle +\left\vert 13 \right\rangle +\left\vert 14 \right\rangle \right\}\nonumber \\ &{{H}_{1}}\left\vert 6 \right\rangle =-\left\{\left\vert 1 \right\rangle +\left\vert 7 \right\rangle +\left\vert 11 \right\rangle \right\}\nonumber \\ &{{H}_{1}}\left\vert 7 \right\rangle =-\left\{\left\vert 6 \right\rangle +\left\vert 10 \right\rangle +\left\vert 16 \right\rangle \right\}\nonumber \\ &{{H}_{1}}\left\vert 8 \right\rangle =-\left\{\left\vert 9 \right\rangle +\left\vert 10 \right\rangle +\left\vert 16 \right\rangle \right\}\nonumber \\ &{{H}_{1}}\left\vert 9 \right\rangle =-\left\{\left\vert 1 \right\rangle +\left\vert 8 \right\rangle +\left\vert 11 \right\rangle \right\}\nonumber \\ &{{H}_{1}}\left\vert 10 \right\rangle =-\left\{\left\vert 1 \right\rangle +\left\vert 7 \right\rangle +\left\vert 8 \right\rangle \right\}\nonumber \\ &{{H}_{1}}\left\vert 11 \right\rangle =-\left\{\left\vert 6 \right\rangle +\left\vert 9 \right\rangle +\left\vert 16 \right\rangle \right\}\nonumber \\ &{{H}_{1}}\left\vert 12 \right\rangle =-\left\{\left\vert 3 \right\rangle +\left\vert 4 \right\rangle +\left\vert 15 \right\rangle \right\}\nonumber \\ &{{H}_{1}}\left\vert 13 \right\rangle =-\left\{\left\vert 3 \right\rangle +\left\vert 5 \right\rangle +\left\vert 15 \right\rangle \right\}\nonumber \\ &{{H}_{1}}\left\vert 14 \right\rangle =-\left\{\left\vert 4 \right\rangle +\left\vert 5 \right\rangle +\left\vert 15 \right\rangle \right\}\nonumber \\ &{{H}_{1}}\left\vert 15 \right\rangle =-\left\{\left\vert 12 \right\rangle +\left\vert 13 \right\rangle +\left\vert 14 \right\rangle \right\}\nonumber \\ &{{H}_{1}}\left\vert 16 \right\rangle =-\left\{\left\vert 8 \right\rangle +\left\vert 11 \right\rangle +\left\vert 12 \right\rangle \right\}. \nonumber \end{align} \tag{ A.3 }$$

The second term of the cluster Hamiltonian (3.16) is

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle {{H}_{2}}=-\frac{mz}{2}\sum_{a\in\Gamma}{\tau _{a}^{x}}=-\frac{mz}{2}\left\{\tau _{0}^{x}+\tau _{1}^{x}+\tau _{2}^{x}+\tau _{3}^{x} \right\} \nonumber \end{align} \tag{ A.4 }$$

where m is the Ising magnetization coupling the boundary sites of the cluster to boundary sites of neighboring clusters and z is the number of bonds connecting boundary sites $1, 2, 3$ to other clusters which for this cluster is $z=2$. The effect of H₂ on the bases is given by

$$\begin{align} \newcommand{\e}{{\rm e}} \newcommand{\ra}{\rangle} \displaystyle &{{H}_{2}}\left\vert 1 \right\rangle =-m\left\{\left\vert 2 \right\rangle +\left\vert 3 \right\rangle +\left\vert 4 \right\rangle +\left\vert 5 \right\rangle \right\}\nonumber \\ &{{H}_{2}}\left\vert 2 \right\rangle =-m\left\{\left\vert 1 \right\rangle +\left\vert 6 \right\rangle +\left\vert 9 \right\rangle +\left\vert 10 \right\rangle \right\}\nonumber \\ &{{H}_{2}}\left\vert 3 \right\rangle =-m\left\{\left\vert 1 \right\rangle +\left\vert 6 \right\rangle +\left\vert 7 \right\rangle +\left\vert 11 \right\rangle \right\}\nonumber \\ &{{H}_{2}}\left\vert 4 \right\rangle =-m\left\{\left\vert 1 \right\rangle +\left\vert 7 \right\rangle +\left\vert 8 \right\rangle +\left\vert 10 \right\rangle \right\}\nonumber \\ &{{H}_{2}}\left\vert 5 \right\rangle =-m\left\{\left\vert 1 \right\rangle +\left\vert 8 \right\rangle +\left\vert 9 \right\rangle +\left\vert 11 \right\rangle \right\}\nonumber \\ &{{H}_{2}}\left\vert 6 \right\rangle =-m\left\{\left\vert 2 \right\rangle +\left\vert 3 \right\rangle +\left\vert 12 \right\rangle +\left\vert 13 \right\rangle \right\}\nonumber \\ &{{H}_{2}}\left\vert 7 \right\rangle =-m\left\{\left\vert 3 \right\rangle +\left\vert 4 \right\rangle +\left\vert 12 \right\rangle +\left\vert 15 \right\rangle \right\}\nonumber \\ &{{H}_{2}}\left\vert 8 \right\rangle =-m\left\{\left\vert 4 \right\rangle +\left\vert 5 \right\rangle +\left\vert 14 \right\rangle +\left\vert 15 \right\rangle \right\}\nonumber \\ &{{H}_{2}}\left\vert 9 \right\rangle =-m\left\{\left\vert 2 \right\rangle +\left\vert 5 \right\rangle +\left\vert 13 \right\rangle +\left\vert 14 \right\rangle \right\}\nonumber \\ &{{H}_{2}}\left\vert 10 \right\rangle =-m\left\{\left\vert 2 \right\rangle +\left\vert 4 \right\rangle +\left\vert 12 \right\rangle +\left\vert 14 \right\rangle \right\}\nonumber \\ &{{H}_{2}}\left\vert 11 \right\rangle =-m\left\{\left\vert 3 \right\rangle +\left\vert 5 \right\rangle +\left\vert 13 \right\rangle +\left\vert 15 \right\rangle \right\}\nonumber \\ &{{H}_{2}}\left\vert 12 \right\rangle =-m\left\{\left\vert 6 \right\rangle +\left\vert 7 \right\rangle +\left\vert 10 \right\rangle +\left\vert 16 \right\rangle \right\}\nonumber \\ &{{H}_{2}}\left\vert 13 \right\rangle =-m\left\{\left\vert 6 \right\rangle +\left\vert 9 \right\rangle +\left\vert 11 \right\rangle +\left\vert 16 \right\rangle \right\}\nonumber \\ &{{H}_{2}}\left\vert 14 \right\rangle =-m\left\{\left\vert 8 \right\rangle +\left\vert 9 \right\rangle +\left\vert 10 \right\rangle +\left\vert 16 \right\rangle \right\}\nonumber \\ &{{H}_{2}}\left\vert 15 \right\rangle =-m\left\{\left\vert 7 \right\rangle +\left\vert 8 \right\rangle +\left\vert 11 \right\rangle +\left\vert 16 \right\rangle \right\}\nonumber \\ &{{H}_{2}}\left\vert 16 \right\rangle =-m\left\{\left\vert 12 \right\rangle +\left\vert 13 \right\rangle +\left\vert 14 \right\rangle +\left\vert 15 \right\rangle \right\}. \nonumber \end{align} \tag{ A.5 }$$

The last term of the cluster Hamiltonian (3.16) is the transverse field term,

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle {{H}_{3}}=h\sum\limits_{a\in\Gamma}{\tau _{a}^{z}}=h\left\{\tau_0^z+\tau_1^z+\tau_2^z+\tau_3^z\right\}. \nonumber \end{align} \tag{ A.6 }$$

The above term acts on the 16 bases as follows:

$$\begin{align} \newcommand{\e}{{\rm e}} \newcommand{\ra}{\rangle} \displaystyle &{{H}_{3}}\left\vert 1 \right\rangle =4h\left\vert 1 \right\rangle, {{H}_{3}}\left\vert 2 \right\rangle =2h\left\vert 2 \right\rangle \nonumber \\ &{{H}_{3}}\left\vert 3 \right\rangle =2h\left\vert 3 \right\rangle, {{H}_{3}}\left\vert 4 \right\rangle =2h\left\vert 4 \right\rangle \nonumber \\ &{{H}_{3}}\left\vert 5 \right\rangle =2h\left\vert 5 \right\rangle\nonumber \\ &{{H}_{3}}\left\vert 6 \right\rangle =0, {{H}_{3}}\left\vert 7 \right\rangle =0, {{H}_{3}}\left\vert 8 \right\rangle =0 \nonumber \\ &{{H}_{3}}\left\vert 9 \right\rangle =0, {{H}_{3}}\left\vert 10 \right\rangle =0, {{H}_{3}}\left\vert 11 \right\rangle =0 \nonumber \\ &{{H}_{3}}\left\vert 12 \right\rangle =-2h\left\vert 12 \right\rangle, {{H}_{3}}\left\vert 13 \right\rangle =-2h\left\vert 13 \right\rangle \nonumber \\ &{{H}_{3}}\left\vert 14 \right\rangle =-2h\left\vert 14 \right\rangle, {{H}_{3}}\left\vert 15 \right\rangle =-2h\left\vert 15 \right\rangle \nonumber \\ &{{H}_{3}}\left\vert 16 \right\rangle =-4h\left\vert 16 \right\rangle. \nonumber \end{align} \tag{ A.7 }$$

Let us proceed by employing symmetry considerations to reduce the above 16-dimensional Hamiltonian to smaller blocks. The Y-shaped cluster in figure A1 is invariant under rotations by $2\pi/3$ which is denoted by C and the group of rotation is formed by $\{C^0, C^1, C^2\}$. The effect of this operation on the site labels is

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle C=\left\{\begin{array}{@{}l@{}} 1\to 2 \nonumber \\ 2\to 3 \nonumber \\ 3\to 1 \nonumber \\ \end{array} \right.. \nonumber \end{align} \tag{ A.8 }$$

Successive operations of C on a prototypical state, e.g. $\newcommand{\ra}{\rangle} \vert 12\ra$ gives the following pattern:

$$\begin{align} \newcommand{\e}{{\rm e}} \newcommand{\ra}{\rangle} \displaystyle \vert 12\rangle\mathop{\longrightarrow}\limits^{C}\vert 14\rangle\mathop{\longrightarrow}\limits^{C}\vert 13\rangle\mathop{\longrightarrow}\limits^{C}\vert 12\rangle \nonumber \end{align} \tag{ A.9 }$$

which is a concise representation of

$$\begin{align} \newcommand{\e}{{\rm e}} \newcommand{\ra}{\rangle} \newcommand{\la}{\langle} \displaystyle C^0\vert 12\ra=\vert 12\ra,~~C\vert 12\ra = \vert 14\ra,~~C^2\vert 12\ra=\vert 13\ra. \label{state12.eqn} \nonumber \end{align} \tag{ A.10 }$$

According to projection theorem of group theory a symmetry adopted state in representation labeled by n can be constructed from an arbitrary state $\newcommand{\ra}{\rangle} \vert \phi\ra$ as

$$\begin{align} \newcommand{\e}{{\rm e}} \newcommand{\ra}{\rangle} \displaystyle \vert \psi^{(n)}\ra\sim \left(\sum_{g} g \Gamma_n[g]\right)\vert \phi\ra \nonumber \end{align} \tag{ A.11 }$$

where g denotes a member of the group, and $\Gamma_n(g)$ is the n'th irreducible representation of element g of the group. In the case of rotation group the irreducible representations of the cyclic group are labeled by three integer (angular momenta) $n=0, \pm 1$ and are represented by $\{\omega^0, \omega^n, \omega^{2n}\}$ where $\newcommand{\e}{{\rm e}} \omega=\exp(2\pi i/3)$. A compact way of expressing the above representations for the cyclic group is $\Gamma_n(C^{\,p})=\omega^{\,pn}$. This gives a symmetry adopted state build from e.g. basis state $\newcommand{\ra}{\rangle} \vert 12\ra$ as

$$\begin{align} \newcommand{\e}{{\rm e}} \newcommand{\ra}{\rangle} \displaystyle \left(C^0\omega^0+C^1\omega^n+C^2\omega^{2n}\right)\vert 12\ra, \nonumber \end{align} \tag{ A.12 }$$

which after using (A.10) gives the following state

$$\begin{align} \newcommand{\e}{{\rm e}} \newcommand{\ra}{\rangle} \displaystyle \vert 12\ra+\omega^n\vert 14\ra+\omega^{2n}\vert 13\ra, \nonumber \end{align} \tag{ A.13 }$$

with definite discrete 'angular momentum' n.

The same symmetry consideration could be applied to every other state which is summarized as

$$\begin{align} \newcommand{\e}{{\rm e}} \newcommand{\ra}{\rangle} \displaystyle &\vert 6\rangle\mathop{\longrightarrow}\limits^{C}\vert 10\rangle\mathop{\longrightarrow}\limits^{C}\vert 9\rangle\mathop{\longrightarrow}\limits^{C}\vert 6\rangle\nonumber \\ &\vert 3\rangle\mathop{\longrightarrow}\limits^{C}\vert 4\rangle\mathop{\longrightarrow}\limits^{C}\vert 5\rangle\mathop{\longrightarrow}\limits^{C}\vert 3\rangle\nonumber \\ &\vert 7\rangle\mathop{\longrightarrow}\limits^{C}\vert 8\rangle\mathop{\longrightarrow}\limits^{C}\vert 11\rangle\mathop{\longrightarrow}\limits^{C}\vert 7\rangle\nonumber \\ &\vert 1\rangle\mathop{\longrightarrow}\limits^{C}\vert 1\rangle,~~~\vert 2\rangle\mathop{\longrightarrow}\limits^{C}\vert 2\rangle\nonumber \\ &\vert 15\rangle\mathop{\longrightarrow}\limits^{C}\vert 15\rangle,~~~\vert 16\rangle\mathop{\longrightarrow}\limits^{C}\vert 16\rangle. \nonumber \end{align} \tag{ A.14 }$$

Let us now focus on the $n=+1$ sector. The $n=-1$ sector has identical spectrum by time-reversal symmetry. The $n=+1$ sector is spanned by the following normalized states,

$$\begin{align} \newcommand{\e}{{\rm e}} \newcommand{\ra}{\rangle} \displaystyle &\left\vert {{\alpha }_{1}} \right\rangle =\frac{1}{\sqrt{3}}\left(\left\vert 12 \right\rangle +\omega \left\vert 14 \right\rangle +{{\omega }^{2}}\left\vert 13 \right\rangle \right) \nonumber \\ &\left\vert {{\alpha }_{2}} \right\rangle =\frac{1}{\sqrt{3}}\left(\left\vert 3 \right\rangle +\omega \left\vert 4 \right\rangle +{{\omega }^{2}}\left\vert 5 \right\rangle \right) \nonumber \\ &\left\vert {{\alpha }_{3}} \right\rangle =\frac{1}{\sqrt{3}}\left(\left\vert 6 \right\rangle +\omega \left\vert 10 \right\rangle +{{\omega }^{2}}\left\vert 9 \right\rangle \right) \nonumber \\ &\left\vert {{\alpha }_{4}} \right\rangle =\frac{1}{\sqrt{3}}\left(\left\vert 7 \right\rangle +\omega \left\vert 8 \right\rangle +{{\omega }^{2}}\left\vert 11 \right\rangle \right) \nonumber \end{align} \tag{ A.15 }$$

where $\omega={\rm e}^{2{\rm i}\pi/3}$. The first term of the cluster Hamiltonian on the above states has the following effect:

$$\begin{align} \newcommand{\e}{{\rm e}} \newcommand{\ra}{\rangle} \displaystyle {{H}_{1}}\left\vert {{\alpha }_{1}} \right\rangle =-\Omega_-\left\vert {{\alpha }_{2}} \right\rangle\nonumber \\ {{H}_{1}}\left\vert {{\alpha }_{2}} \right\rangle =-\Omega_+\left\vert {{\alpha }_{1}} \right\rangle \nonumber \\ {{H}_{1}}\left\vert {{\alpha }_{3}} \right\rangle =-\Omega_+\left\vert {{\alpha }_{4}} \right\rangle \nonumber \\ {{H}_{1}}\left\vert {{\alpha }_{4}} \right\rangle =-\Omega_-\left\vert {{\alpha }_{3}} \right\rangle \nonumber \end{align} \tag{ A.16 }$$

where $\newcommand{\e}{{\rm e}} \Omega_\pm=1+\omega^{\pm 1}=\exp(\pm i\pi/3)$. The matrix elements $\newcommand{\la}{\langle} \newcommand{\ra}{\rangle} \la \alpha_i\vert H_1\alpha_j\ra$ are represented in matrix form as,

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle {{\tilde{H}}_{1}}=-\left(\begin{array}{@{}cccc@{}} 0 & \Omega_- & 0 & 0 \nonumber \\ \Omega_+ & 0 & 0 & 0 \nonumber \\ 0 & 0 & 0 & \Omega_+ \nonumber \\ 0 & 0 & \Omega_- & 0 \end{array} \right). \nonumber \end{align} \tag{ A.17 }$$

Similarly the H₂ term in this basis becomes

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \tilde{H}_2=-m\left(\begin{array}{@{}cccc@{}} 0 & 0 & \Omega_- & 1 \nonumber \\ 0 & 0 & 1 & \Omega_+ \nonumber \\ \Omega_+ & 1 & 0 & 0 \nonumber \\ 1 & \Omega_- & 0 & 0 \end{array} \right). \nonumber \end{align} \tag{ A.18 }$$

The transverse field term was already diagonal in the original basis, and remains so in the symmetry adopted basis for the $n=+1$ sector,

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle {{\tilde{H}}_{3}}=-\left(\begin{array}{@{}cccc@{}} 2h & {} & {} & {} \nonumber \\ {} & -2h & {} & {} \nonumber \\ {} & {} & 0 & {} \nonumber \\ {} & {} & {} & 0 \end{array} \right). \nonumber \end{align} \tag{ A.19 }$$

Adding up the above matrices we obtain the matrix representation of the ITF Hamiltonian for Y-shaped cluster in the $n=+1$ sector as,

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle &{{\tilde{H}}}={{\tilde{H}}_{1}}+{{\tilde{H}}_{2}}+ {{\tilde{H}}_{3}}\nonumber \\ &=-\left(\begin{array}{@{}llll@{}} 2h & {{\rm e}^{{{\rm i}\pi }/{3}}} & m{{\rm e}^{{{\rm i}\pi }/{3}}} & m \nonumber \\ {{\rm e}^{{-{\rm i}\pi }/{3}}} & -2h & m & m{{\rm e}^{{-{\rm i}\pi }/{3}}} \nonumber \\ m{{\rm e}^{{-{\rm i}\pi }/{3}}} & m & 0 & {{\rm e}^{{-{\rm i}\pi }/{3}}} \nonumber \\ m & m{{\rm e}^{{{\rm i}\pi }/{3}}} & {{\rm e}^{{{\rm i}\pi }/{3}}} & 0 \nonumber \\ \end{array} \right). \nonumber \end{align} \tag{ A.20 }$$

The matrix representation for the $n=-1$ sector is simply obtained from the above equation by complex conjugation $i\to -i$ corresponding to time reversal operation. The matrix representation in the $n=0$ sector can be constructed in similar way. The ground state is the least eigen-value among all sectors with various n values. Here it turns out that the ground state belongs to $n=0$ sector.

A.2. Hexagonal cluster

The details of the group theory consideration for larger clusters is similar to the Y-shaped cluster. In this section for reference we only provide explicit representation of all $2^6=64$ basis states and the effects of cluster ITF Hamiltonian on it. As can be seen in figure 2 each hexagonal cluster Γ is connected to the rest of the lattice by $z=1$ neighbor. The basis is labeled as before by $\newcommand{\ra}{\rangle} \vert \sigma_5, \sigma_4, \sigma_3, \sigma_2, \sigma_1, \sigma_0\ra$ where the site index a in $\sigma_a$ varies from 0 to 5 as depicted in right panel of figure 2.

$$\begin{align} ~~\vert 1 \rangle =\,\,\downarrow \downarrow \downarrow \downarrow \downarrow \downarrow~~\vert 2 \rangle =\,\downarrow \downarrow \downarrow \downarrow \downarrow \uparrow~~\vert 3 \rangle =\,\downarrow \downarrow \downarrow \downarrow \uparrow \downarrow~~\vert 4 \rangle =\,\downarrow \downarrow \downarrow \downarrow \uparrow \uparrow \nonumber \\ ~~\vert 5 \rangle =\,\downarrow \downarrow \downarrow \uparrow \downarrow \downarrow~~\vert 6 \rangle =\,\downarrow \downarrow \downarrow \uparrow \downarrow \uparrow~~\vert 7 \rangle =\,\downarrow \downarrow \downarrow \uparrow \uparrow \downarrow~~\vert 8 \rangle =\,\downarrow \downarrow \downarrow \uparrow \uparrow \uparrow \nonumber \\ ~\vert 9 \rangle =\,\downarrow \downarrow \uparrow \downarrow \downarrow \downarrow~~\vert 10 \rangle =\,\downarrow \downarrow \uparrow \downarrow \downarrow \uparrow~\vert 11 \rangle =\,\downarrow \downarrow \uparrow \downarrow \uparrow \downarrow~\vert 12 \rangle =\,\downarrow \downarrow \uparrow \downarrow \uparrow \uparrow \nonumber \\ \vert 13 \rangle =\,\downarrow \downarrow \uparrow \uparrow \downarrow \downarrow~\vert 14 \rangle =\,\downarrow \downarrow \uparrow \uparrow \downarrow \uparrow ~\vert 15 \rangle =\,\downarrow \downarrow \uparrow \uparrow \uparrow \downarrow ~\vert 16 \rangle =\,\downarrow \downarrow \uparrow \uparrow \uparrow \uparrow \nonumber \\ \vert 17 \rangle =\,\downarrow \uparrow \downarrow \downarrow \downarrow \downarrow ~\vert 18 \rangle =\,\downarrow \uparrow \downarrow \downarrow \downarrow \uparrow ~\vert 19 \rangle =\,\downarrow \uparrow \downarrow \downarrow \uparrow \downarrow ~\vert 20 \rangle =\,\downarrow \uparrow \downarrow \downarrow \uparrow \uparrow \nonumber \\ \vert 21 \rangle =\,\downarrow \uparrow \downarrow \uparrow \downarrow \downarrow ~\vert 22 \rangle =\,\downarrow \uparrow \downarrow \uparrow \downarrow \uparrow ~\vert 23 \rangle =\,\downarrow \uparrow \downarrow \uparrow \uparrow \downarrow ~\vert 24 \rangle =\,\downarrow \uparrow \downarrow \uparrow \uparrow \uparrow \nonumber \\ \vert 25 \rangle =\,\downarrow \uparrow \uparrow \downarrow \downarrow \downarrow ~\vert 26 \rangle =\,\downarrow \uparrow \uparrow \downarrow \downarrow \uparrow ~\vert 27 \rangle =\,\downarrow \uparrow \uparrow \downarrow \uparrow \downarrow ~\vert 28 \rangle =\,\downarrow \uparrow \uparrow \downarrow \uparrow \uparrow \nonumber \\ \vert 29 \rangle =\,\downarrow \uparrow \uparrow \uparrow \downarrow \downarrow ~\vert 30 \rangle =\,\downarrow \uparrow \uparrow \uparrow \downarrow \uparrow ~\vert 31 \rangle =\,\downarrow \uparrow \uparrow \uparrow \uparrow \downarrow ~\vert 32 \rangle =\,\downarrow \uparrow \uparrow \uparrow \uparrow \uparrow \nonumber \\ \vert 33 \rangle =\,\uparrow \downarrow \downarrow \downarrow \downarrow \downarrow ~\vert 34 \rangle =\,\uparrow \downarrow \downarrow \downarrow \downarrow \uparrow ~\vert 35 \rangle =\,\uparrow \downarrow \downarrow \downarrow \uparrow \downarrow ~\vert 36 \rangle =\,\uparrow \downarrow \downarrow \downarrow \uparrow \uparrow \nonumber \\ \vert 37 \rangle =\,\uparrow \downarrow \downarrow \uparrow \downarrow \downarrow ~\vert 38 \rangle =\,\uparrow \downarrow \downarrow \uparrow \downarrow \uparrow~\vert 39 \rangle =\,\uparrow \downarrow \downarrow \uparrow \uparrow \downarrow~\vert 40 \rangle =\,\uparrow \downarrow \downarrow \uparrow \uparrow \uparrow \nonumber \\ \vert 41 \rangle =\,\uparrow \downarrow \uparrow \downarrow \downarrow \downarrow~\vert 42 \rangle =\,\uparrow \downarrow \uparrow \downarrow \downarrow \uparrow~\vert 43 \rangle =\,\uparrow \downarrow \uparrow \downarrow \uparrow \downarrow~\vert 44 \rangle =\,\uparrow \downarrow \uparrow \downarrow \uparrow \uparrow \nonumber \\ \vert 45 \rangle =\,\uparrow \downarrow \uparrow \uparrow \downarrow \downarrow~\vert 46 \rangle =\,\uparrow \downarrow \uparrow \uparrow \downarrow \uparrow~\vert 47 \rangle =\,\uparrow \downarrow \uparrow \uparrow \uparrow \downarrow~\vert 48 \rangle =\,\uparrow \downarrow \uparrow \uparrow \uparrow \uparrow \nonumber \\ \vert 49 \rangle =\,\uparrow \uparrow \downarrow \downarrow \downarrow \downarrow~\vert 50 \rangle =\,\uparrow \uparrow \downarrow \downarrow \downarrow \uparrow~\vert 51 \rangle =\,\uparrow \uparrow \downarrow \downarrow \uparrow \downarrow~\vert 52 \rangle =\,\uparrow \uparrow \downarrow \downarrow \uparrow \uparrow \nonumber \\ \vert 53 \rangle =\,\uparrow \uparrow \downarrow \uparrow \downarrow \downarrow~\vert 54 \rangle =\,\uparrow \uparrow \downarrow \uparrow \downarrow \uparrow~\vert 55 \rangle =\,\uparrow \uparrow \downarrow \uparrow \uparrow \downarrow~\vert 56 \rangle =\,\uparrow \uparrow \downarrow \uparrow \uparrow \uparrow \nonumber \\ \vert 57 \rangle =\,\uparrow \uparrow \uparrow \downarrow \downarrow \downarrow~\vert 58 \rangle =\,\uparrow \uparrow \uparrow \downarrow \downarrow \uparrow~\vert 59 \rangle =\,\uparrow \uparrow \uparrow \downarrow \uparrow \downarrow~\vert 60 \rangle =\,\uparrow \uparrow \uparrow \downarrow \uparrow \uparrow \nonumber \\ \vert 61 \rangle =\,\uparrow \uparrow \uparrow \uparrow \downarrow \downarrow~\vert 62 \rangle =\,\uparrow \uparrow \uparrow \uparrow \downarrow \uparrow~\vert 63 \rangle =\,\uparrow \uparrow \uparrow \uparrow \uparrow \downarrow~\vert 64 \rangle =\,\,\uparrow \uparrow \uparrow \uparrow \uparrow \uparrow \end{align} \tag{ A.21 }$$

the first term of equation (3.16) for 6-site cluster is

$$\begin{align} \newcommand{\e}{{\rm e}} \newcommand{\ra}{\rangle} \newcommand{\la}{\langle} \displaystyle {{H}_{1}}&=-\sum_{\la a,b\ra\in \Gamma} \tau^x_a \tau^x_b \nonumber \\ &\quad-\left(\tau _{0}^{x}\tau _{1}^{x}+ \tau _{1}^{x}\tau _{2}^{x}+\tau _{2}^{x}\tau _{3}^{x}+\tau _{3}^{x}\tau _{4}^{x}+\tau _{4}^{x}\tau _{5}^{x}+\tau _{5}^{x}\tau _{0}^{x}\right).\nonumber \end{align} \tag{ A.22 }$$

The effect of the above term on the 64-basis states is

$$\begin{align*} \newcommand{\e}{{\rm e}} \newcommand{\ra}{\rangle} \displaystyle &{{H}_{1}}\left\vert 1 \right\rangle =-\left\{\left\vert 4 \right\rangle +\left\vert 7 \right\rangle +\left\vert 13 \right\rangle +\left\vert 25 \right\rangle +\left\vert 34 \right\rangle +\left\vert 49 \right\rangle \right\}\nonumber \\ &{{H}_{1}}\left\vert 2 \right\rangle =-\left\{\left\vert 3 \right\rangle +\left\vert 8 \right\rangle +\left\vert 14 \right\rangle +\left\vert 26 \right\rangle +\left\vert 33 \right\rangle +\left\vert 50 \right\rangle \right\}\nonumber \\ &{{H}_{1}}\left\vert 3 \right\rangle =-\left\{\left\vert 2 \right\rangle +\left\vert 5 \right\rangle +\left\vert 15 \right\rangle +\left\vert 27 \right\rangle +\left\vert 36 \right\rangle +\left\vert 51 \right\rangle \right\}\nonumber \\ &{{H}_{1}}\left\vert 4 \right\rangle =-\left\{\left\vert 1 \right\rangle +\left\vert 6 \right\rangle +\left\vert 16 \right\rangle +\left\vert 28 \right\rangle +\left\vert 35 \right\rangle +\left\vert 52 \right\rangle \right\}\nonumber \\ &{{H}_{1}}\left\vert 5 \right\rangle =-\left\{\left\vert 8 \right\rangle +\left\vert 3 \right\rangle +\left\vert 9 \right\rangle +\left\vert 29 \right\rangle +\left\vert 38 \right\rangle +\left\vert 53 \right\rangle \right\}\nonumber \\ &{{H}_{1}}\left\vert 6 \right\rangle =-\left\{\left\vert 7 \right\rangle +\left\vert 4 \right\rangle +\left\vert 10 \right\rangle +\left\vert 30 \right\rangle +\left\vert 37 \right\rangle +\left\vert 54 \right\rangle \right\}\nonumber \\ &{{H}_{1}}\left\vert 7 \right\rangle =-\left\{\left\vert 6 \right\rangle +\left\vert 1 \right\rangle +\left\vert 11 \right\rangle +\left\vert 31 \right\rangle +\left\vert 40 \right\rangle +\left\vert 55 \right\rangle \right\}\nonumber \\ &{{H}_{1}}\left\vert 8 \right\rangle =-\left\{\left\vert 5 \right\rangle +\left\vert 2 \right\rangle +\left\vert 12 \right\rangle +\left\vert 32 \right\rangle +\left\vert 39 \right\rangle +\left\vert 56 \right\rangle \right\}\nonumber \\ &{{H}_{1}}\left\vert 9 \right\rangle =-\left\{\left\vert 12 \right\rangle +\left\vert 15 \right\rangle +\left\vert 5 \right\rangle +\left\vert 17 \right\rangle +\left\vert 42 \right\rangle +\left\vert 57 \right\rangle \right\}\nonumber \\ &{{H}_{1}}\left\vert 10 \right\rangle =-\left\{\left\vert 11 \right\rangle +\left\vert 16 \right\rangle +\left\vert 6 \right\rangle +\left\vert 18 \right\rangle +\left\vert 41 \right\rangle +\left\vert 58 \right\rangle \right\}\nonumber \end{align*}$$

$$\begin{align*} \newcommand{\e}{{\rm e}} \newcommand{\ra}{\rangle} \displaystyle &{{H}_{1}}\left\vert 11 \right\rangle =-\left\{\left\vert 10 \right\rangle +\left\vert 13 \right\rangle +\left\vert 7 \right\rangle +\left\vert 19 \right\rangle +\left\vert 44 \right\rangle +\left\vert 59 \right\rangle \right\}\nonumber \\ &{{H}_{1}}\left\vert 12 \right\rangle =-\left\{\left\vert 9 \right\rangle +\left\vert 14 \right\rangle +\left\vert 8 \right\rangle +\left\vert 20 \right\rangle +\left\vert 43 \right\rangle +\left\vert 60 \right\rangle \right\}\nonumber \\ &{{H}_{1}}\left\vert 13 \right\rangle =-\left\{\left\vert 16 \right\rangle +\left\vert 11 \right\rangle +\left\vert 1 \right\rangle +\left\vert 21 \right\rangle +\left\vert 46 \right\rangle +\left\vert 61 \right\rangle \right\}\nonumber \\ &{{H}_{1}}\left\vert 14 \right\rangle =-\left\{\left\vert 15 \right\rangle +\left\vert 12 \right\rangle +\left\vert 2 \right\rangle +\left\vert 22 \right\rangle +\left\vert 45 \right\rangle +\left\vert 62 \right\rangle \right\}\nonumber \\ &{{H}_{1}}\left\vert 15 \right\rangle =-\left\{\left\vert 14 \right\rangle +\left\vert 9 \right\rangle +\left\vert 3 \right\rangle +\left\vert 23 \right\rangle +\left\vert 48 \right\rangle +\left\vert 63 \right\rangle \right\}\nonumber \\ &{{H}_{1}}\left\vert 16 \right\rangle =-\left\{\left\vert 13 \right\rangle +\left\vert 10 \right\rangle +\left\vert 4 \right\rangle +\left\vert 24 \right\rangle +\left\vert 47 \right\rangle +\left\vert 64 \right\rangle \right\}\nonumber \\ &{{H}_{1}}\left\vert 17 \right\rangle =-\left\{\left\vert 20 \right\rangle +\left\vert 23 \right\rangle +\left\vert 29 \right\rangle +\left\vert 9 \right\rangle +\left\vert 50 \right\rangle +\left\vert 33 \right\rangle \right\}\nonumber \\ &{{H}_{1}}\left\vert 18 \right\rangle =-\left\{\left\vert 19 \right\rangle +\left\vert 24 \right\rangle +\left\vert 30 \right\rangle +\left\vert 10 \right\rangle +\left\vert 49 \right\rangle +\left\vert 34 \right\rangle \right\}\nonumber \\ &{{H}_{1}}\left\vert 19 \right\rangle =-\left\{\left\vert 18 \right\rangle +\left\vert 21 \right\rangle +\left\vert 31 \right\rangle +\left\vert 11 \right\rangle +\left\vert 52 \right\rangle +\left\vert 35 \right\rangle \right\}\nonumber \\ &{{H}_{1}}\left\vert 20 \right\rangle =-\left\{\left\vert 17 \right\rangle +\left\vert 22 \right\rangle +\left\vert 32 \right\rangle +\left\vert 12 \right\rangle +\left\vert 51 \right\rangle +\left\vert 36 \right\rangle \right\}\nonumber \end{align*}$$

$$\begin{align*} \newcommand{\e}{{\rm e}} \newcommand{\ra}{\rangle} \displaystyle &{{H}_{1}}\left\vert 21 \right\rangle =-\left\{\left\vert 24 \right\rangle +\left\vert 19 \right\rangle +\left\vert 25 \right\rangle +\left\vert 13 \right\rangle +\left\vert 54 \right\rangle +\left\vert 37 \right\rangle \right\}\nonumber \\ &{{H}_{1}}\left\vert 22 \right\rangle =-\left\{\left\vert 23 \right\rangle +\left\vert 20 \right\rangle +\left\vert 26 \right\rangle +\left\vert 14 \right\rangle +\left\vert 53 \right\rangle +\left\vert 38 \right\rangle \right\}\nonumber \\ &{{H}_{1}}\left\vert 23 \right\rangle =-\left\{\left\vert 22 \right\rangle +\left\vert 17 \right\rangle +\left\vert 27 \right\rangle +\left\vert 15 \right\rangle +\left\vert 56 \right\rangle +\left\vert 39 \right\rangle \right\}\nonumber \\ &{{H}_{1}}\left\vert 24 \right\rangle =-\left\{\left\vert 21 \right\rangle +\left\vert 18 \right\rangle +\left\vert 28 \right\rangle +\left\vert 16 \right\rangle +\left\vert 55 \right\rangle +\left\vert 40 \right\rangle \right\}\nonumber \\ &{{H}_{1}}\left\vert 25 \right\rangle =-\left\{\left\vert 28 \right\rangle +\left\vert 15 \right\rangle +\left\vert 21 \right\rangle +\left\vert 1 \right\rangle +\left\vert 58 \right\rangle +\left\vert 41 \right\rangle \right\}\nonumber \\ &{{H}_{1}}\left\vert 26 \right\rangle =-\left\{\left\vert 27 \right\rangle +\left\vert 16 \right\rangle +\left\vert 22 \right\rangle +\left\vert 2 \right\rangle +\left\vert 57 \right\rangle +\left\vert 42 \right\rangle \right\}\nonumber \\ &{{H}_{1}}\left\vert 27 \right\rangle =-\left\{\left\vert 26 \right\rangle +\left\vert 29 \right\rangle +\left\vert 23 \right\rangle +\left\vert 3 \right\rangle +\left\vert 60 \right\rangle +\left\vert 43 \right\rangle \right\}\nonumber \\ &{{H}_{1}}\left\vert 28 \right\rangle =-\left\{\left\vert 25 \right\rangle +\left\vert 30 \right\rangle +\left\vert 24 \right\rangle +\left\vert 4 \right\rangle +\left\vert 59 \right\rangle +\left\vert 44 \right\rangle \right\}\nonumber \\ &{{H}_{1}}\left\vert 29 \right\rangle =-\left\{\left\vert 32 \right\rangle +\left\vert 27 \right\rangle +\left\vert 17 \right\rangle +\left\vert 5 \right\rangle +\left\vert 62 \right\rangle +\left\vert 45 \right\rangle \right\}\nonumber \\ &{{H}_{1}}\left\vert 30 \right\rangle =-\left\{\left\vert 31 \right\rangle +\left\vert 28 \right\rangle +\left\vert 18 \right\rangle +\left\vert 6 \right\rangle +\left\vert 61 \right\rangle +\left\vert 46 \right\rangle \right\}\nonumber \end{align*}$$

$$\begin{align*} \newcommand{\e}{{\rm e}} \newcommand{\ra}{\rangle} \displaystyle &{{H}_{1}}\left\vert 31 \right\rangle =-\left\{\left\vert 30 \right\rangle +\left\vert 25 \right\rangle +\left\vert 19 \right\rangle +\left\vert 7 \right\rangle +\left\vert 64 \right\rangle +\left\vert 47 \right\rangle \right\}\nonumber \\ &{{H}_{1}}\left\vert 32 \right\rangle =-\left\{\left\vert 29 \right\rangle +\left\vert 26 \right\rangle +\left\vert 20 \right\rangle +\left\vert 8 \right\rangle +\left\vert 63 \right\rangle +\left\vert 48 \right\rangle \right\}\nonumber \\ &{{H}_{1}}\left\vert 33 \right\rangle =-\left\{\left\vert 36 \right\rangle +\left\vert 39 \right\rangle +\left\vert 45 \right\rangle +\left\vert 57 \right\rangle +\left\vert 2 \right\rangle +\left\vert 17 \right\rangle \right\}\nonumber \\ &{{H}_{1}}\left\vert 34 \right\rangle =-\left\{\left\vert 35 \right\rangle +\left\vert 40 \right\rangle +\left\vert 46 \right\rangle +\left\vert 58 \right\rangle +\left\vert 1 \right\rangle +\left\vert 18 \right\rangle \right\}\nonumber \\ &{{H}_{1}}\left\vert 35 \right\rangle =-\left\{\left\vert 34 \right\rangle +\left\vert 37 \right\rangle +\left\vert 47 \right\rangle +\left\vert 59 \right\rangle +\left\vert 4 \right\rangle +\left\vert 19 \right\rangle \right\}\nonumber \\ &{{H}_{1}}\left\vert 36 \right\rangle =-\left\{\left\vert 33 \right\rangle +\left\vert 38 \right\rangle +\left\vert 48 \right\rangle +\left\vert 60 \right\rangle +\left\vert 3 \right\rangle +\left\vert 20 \right\rangle \right\}\nonumber \\ &{{H}_{1}}\left\vert 37 \right\rangle =-\left\{\left\vert 40 \right\rangle +\left\vert 35 \right\rangle +\left\vert 41 \right\rangle +\left\vert 61 \right\rangle +\left\vert 6 \right\rangle +\left\vert 21 \right\rangle \right\}\nonumber \\ &{{H}_{1}}\left\vert 38 \right\rangle =-\left\{\left\vert 39 \right\rangle +\left\vert 36 \right\rangle +\left\vert 42 \right\rangle +\left\vert 62 \right\rangle +\left\vert 5 \right\rangle +\left\vert 22 \right\rangle \right\}\nonumber \\ &{{H}_{1}}\left\vert 39 \right\rangle =-\left\{\left\vert 38 \right\rangle +\left\vert 33 \right\rangle +\left\vert 43 \right\rangle +\left\vert 63 \right\rangle +\left\vert 8 \right\rangle +\left\vert 23 \right\rangle \right\}\nonumber \\ &{{H}_{1}}\left\vert 40 \right\rangle =-\left\{\left\vert 37 \right\rangle +\left\vert 34 \right\rangle +\left\vert 44 \right\rangle +\left\vert 64 \right\rangle +\left\vert 7 \right\rangle +\left\vert 24 \right\rangle \right\}\nonumber \end{align*}$$

$$\begin{align*} \newcommand{\e}{{\rm e}} \newcommand{\ra}{\rangle} \displaystyle &{{H}_{1}}\left\vert 41 \right\rangle =-\left\{\left\vert 44 \right\rangle +\left\vert 47 \right\rangle +\left\vert 37 \right\rangle +\left\vert 49 \right\rangle +\left\vert 10 \right\rangle +\left\vert 25 \right\rangle \right\}\nonumber \\ &{{H}_{1}}\left\vert 42 \right\rangle =-\left\{\left\vert 43 \right\rangle +\left\vert 48 \right\rangle +\left\vert 38 \right\rangle +\left\vert 50 \right\rangle +\left\vert 9 \right\rangle +\left\vert 26 \right\rangle \right\}\nonumber \\ &{{H}_{1}}\left\vert 43 \right\rangle =-\left\{\left\vert 42 \right\rangle +\left\vert 45 \right\rangle +\left\vert 39 \right\rangle +\left\vert 51 \right\rangle +\left\vert 12 \right\rangle +\left\vert 27 \right\rangle \right\}\nonumber \\ &{{H}_{1}}\left\vert 44 \right\rangle =-\left\{\left\vert 41 \right\rangle +\left\vert 46 \right\rangle +\left\vert 40 \right\rangle +\left\vert 52 \right\rangle +\left\vert 11 \right\rangle +\left\vert 28 \right\rangle \right\}\nonumber \\ &{{H}_{1}}\left\vert 45 \right\rangle =-\left\{\left\vert 48 \right\rangle +\left\vert 43 \right\rangle +\left\vert 33 \right\rangle +\left\vert 53 \right\rangle +\left\vert 14 \right\rangle +\left\vert 29 \right\rangle \right\}\nonumber \\ &{{H}_{1}}\left\vert 46 \right\rangle =-\left\{\left\vert 47 \right\rangle +\left\vert 44 \right\rangle +\left\vert 34 \right\rangle +\left\vert 54 \right\rangle +\left\vert 13 \right\rangle +\left\vert 30 \right\rangle \right\}\nonumber \\ &{{H}_{1}}\left\vert 47 \right\rangle =-\left\{\left\vert 46 \right\rangle +\left\vert 41 \right\rangle +\left\vert 35 \right\rangle +\left\vert 55 \right\rangle +\left\vert 16 \right\rangle +\left\vert 31 \right\rangle \right\}\nonumber \\ &{{H}_{1}}\left\vert 48 \right\rangle =-\left\{\left\vert 45 \right\rangle +\left\vert 42 \right\rangle +\left\vert 36 \right\rangle +\left\vert 56 \right\rangle +\left\vert 15 \right\rangle +\left\vert 32 \right\rangle \right\}\nonumber \\ &{{H}_{1}}\left\vert 49 \right\rangle =-\left\{\left\vert 52 \right\rangle +\left\vert 55 \right\rangle +\left\vert 61 \right\rangle +\left\vert 41 \right\rangle +\left\vert 18 \right\rangle +\left\vert 1 \right\rangle \right\}\nonumber \\ &{{H}_{1}}\left\vert 50 \right\rangle =-\left\{\left\vert 51 \right\rangle +\left\vert 56 \right\rangle +\left\vert 62 \right\rangle +\left\vert 42 \right\rangle +\left\vert 17 \right\rangle +\left\vert 2 \right\rangle \right\}\nonumber \end{align*}$$

$$\begin{align*} \newcommand{\e}{{\rm e}} \newcommand{\ra}{\rangle} \displaystyle &{{H}_{1}}\left\vert 51 \right\rangle =-\left\{\left\vert 50 \right\rangle +\left\vert 53 \right\rangle +\left\vert 63 \right\rangle +\left\vert 43 \right\rangle +\left\vert 20 \right\rangle +\left\vert 3 \right\rangle \right\}\nonumber \\ &{{H}_{1}}\left\vert 52 \right\rangle =-\left\{\left\vert 49 \right\rangle +\left\vert 54 \right\rangle +\left\vert 64 \right\rangle +\left\vert 44 \right\rangle +\left\vert 19 \right\rangle +\left\vert 4 \right\rangle \right\}\nonumber \\ &{{H}_{1}}\left\vert 53 \right\rangle =-\left\{\left\vert 56 \right\rangle +\left\vert 51 \right\rangle +\left\vert 57 \right\rangle +\left\vert 45 \right\rangle +\left\vert 22 \right\rangle +\left\vert 5 \right\rangle \right\}\nonumber \\ &{{H}_{1}}\left\vert 54 \right\rangle =-\left\{\left\vert 55 \right\rangle +\left\vert 52 \right\rangle +\left\vert 58 \right\rangle +\left\vert 46 \right\rangle +\left\vert 21 \right\rangle +\left\vert 6 \right\rangle \right\}\nonumber \\ &{{H}_{1}}\left\vert 55 \right\rangle =-\left\{\left\vert 54 \right\rangle +\left\vert 49 \right\rangle +\left\vert 59 \right\rangle +\left\vert 47 \right\rangle +\left\vert 24 \right\rangle +\left\vert 7 \right\rangle \right\}\nonumber \\ &{{H}_{1}}\left\vert 56 \right\rangle =-\left\{\left\vert 53 \right\rangle +\left\vert 50 \right\rangle +\left\vert 60 \right\rangle +\left\vert 48 \right\rangle +\left\vert 23 \right\rangle +\left\vert 8 \right\rangle \right\}\nonumber \\ &{{H}_{1}}\left\vert 57 \right\rangle =-\left\{\left\vert 60 \right\rangle +\left\vert 63 \right\rangle +\left\vert 53 \right\rangle +\left\vert 33 \right\rangle +\left\vert 26 \right\rangle +\left\vert 9 \right\rangle \right\}\nonumber \\ &{{H}_{1}}\left\vert 58 \right\rangle =-\left\{\left\vert 59 \right\rangle +\left\vert 64 \right\rangle +\left\vert 54 \right\rangle +\left\vert 34 \right\rangle +\left\vert 25 \right\rangle +\left\vert 10 \right\rangle \right\}\nonumber \\ &{{H}_{1}}\left\vert 59 \right\rangle =-\left\{\left\vert 58 \right\rangle +\left\vert 61 \right\rangle +\left\vert 55 \right\rangle +\left\vert 35 \right\rangle +\left\vert 28 \right\rangle +\left\vert 11 \right\rangle \right\}\nonumber \\ &{{H}_{1}}\left\vert 60 \right\rangle =-\left\{\left\vert 57 \right\rangle +\left\vert 62 \right\rangle +\left\vert 56 \right\rangle +\left\vert 36 \right\rangle +\left\vert 27 \right\rangle +\left\vert 12 \right\rangle \right\}\nonumber \end{align*}$$

$$\begin{align} \newcommand{\e}{{\rm e}} \newcommand{\ra}{\rangle} \displaystyle &{{H}_{1}}\left\vert 61 \right\rangle =-\left\{\left\vert 64 \right\rangle +\left\vert 59 \right\rangle +\left\vert 49 \right\rangle +\left\vert 37 \right\rangle +\left\vert 30 \right\rangle +\left\vert 13 \right\rangle \right\}\nonumber \\ &{{H}_{1}}\left\vert 62 \right\rangle =-\left\{\left\vert 63 \right\rangle +\left\vert 60 \right\rangle +\left\vert 50 \right\rangle +\left\vert 38 \right\rangle +\left\vert 29 \right\rangle +\left\vert 14 \right\rangle \right\}\nonumber \\ &{{H}_{1}}\left\vert 63 \right\rangle =-\left\{\left\vert 62 \right\rangle +\left\vert 57 \right\rangle +\left\vert 51 \right\rangle +\left\vert 39 \right\rangle +\left\vert 32 \right\rangle +\left\vert 15 \right\rangle \right\}\nonumber \\ &{{H}_{1}}\left\vert 64 \right\rangle =-\left\{\left\vert 61 \right\rangle +\left\vert 58 \right\rangle +\left\vert 52 \right\rangle +\left\vert 40 \right\rangle +\left\vert 31 \right\rangle +\left\vert 16 \right\rangle \right\}. \nonumber \end{align} \tag{ A.23 }$$

The second term of the cluster Hamiltonian (3.16) is

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle {{H}_{2}}=-\frac{m}{2}\left\{\tau_0^x+\tau_1^x+\tau_2^x+\tau_3^x+\tau_4^x+\tau_5^x \right\} \nonumber \end{align} \tag{ A.24 }$$

for the above equation the number of bonds connecting to other clusters is $z=1$. The effect of H₂ on the basis is given by:

$$\begin{align*} \newcommand{\e}{{\rm e}} \newcommand{\ra}{\rangle} \displaystyle &{{H}_{2}}\left\vert 1 \right\rangle =\frac{-m}{2}\left\{\left\vert 2 \right\rangle +\left\vert 3 \right\rangle +\left\vert 5 \right\rangle +\left\vert 9 \right\rangle +\left\vert 17 \right\rangle +\left\vert 33 \right\rangle \right\}\nonumber \\ &{{H}_{2}}\left\vert 2 \right\rangle =\frac{-m}{2}\left\{\left\vert 1 \right\rangle +\left\vert 4 \right\rangle +\left\vert 6 \right\rangle +\left\vert 10 \right\rangle +\left\vert 18 \right\rangle +\left\vert 34 \right\rangle \right\}\nonumber \\ &{{H}_{2}}\left\vert 3 \right\rangle =\frac{-m}{2}\left\{\left\vert 4 \right\rangle +\left\vert 1 \right\rangle +\left\vert 7 \right\rangle +\left\vert 11 \right\rangle +\left\vert 19 \right\rangle +\left\vert 35 \right\rangle \right\}\nonumber \\ &{{H}_{2}}\left\vert 4 \right\rangle =\frac{-m}{2}\left\{\left\vert 3 \right\rangle +\left\vert 2 \right\rangle +\left\vert 8 \right\rangle +\left\vert 12 \right\rangle +\left\vert 20 \right\rangle +\left\vert 36 \right\rangle \right\}\nonumber \\ &{{H}_{2}}\left\vert 5 \right\rangle =\frac{-m}{2}\left\{\left\vert 6 \right\rangle +\left\vert 7 \right\rangle +\left\vert 1 \right\rangle +\left\vert 13 \right\rangle +\left\vert 21 \right\rangle +\left\vert 37 \right\rangle \right\}\nonumber \\ &{{H}_{2}}\left\vert 6 \right\rangle =\frac{-m}{2}\left\{\left\vert 5 \right\rangle +\left\vert 8 \right\rangle +\left\vert 2 \right\rangle +\left\vert 14 \right\rangle +\left\vert 22 \right\rangle +\left\vert 38 \right\rangle \right\}\nonumber \\ &{{H}_{2}}\left\vert 7 \right\rangle =\frac{-m}{2}\left\{\left\vert 8 \right\rangle +\left\vert 5 \right\rangle +\left\vert 3 \right\rangle +\left\vert 15 \right\rangle +\left\vert 23 \right\rangle +\left\vert 39 \right\rangle \right\}\nonumber \\ &{{H}_{2}}\left\vert 8 \right\rangle =\frac{-m}{2}\left\{\left\vert 7 \right\rangle +\left\vert 6 \right\rangle +\left\vert 4 \right\rangle +\left\vert 16 \right\rangle +\left\vert 24 \right\rangle +\left\vert 40 \right\rangle \right\}\nonumber \\ &{{H}_{2}}\left\vert 9 \right\rangle =\frac{-m}{2}\left\{\left\vert 10 \right\rangle +\left\vert 11 \right\rangle +\left\vert 13 \right\rangle +\left\vert 1 \right\rangle +\left\vert 25 \right\rangle +\left\vert 41 \right\rangle \right\}\nonumber \\ &{{H}_{2}}\left\vert 10 \right\rangle =\frac{-m}{2}\left\{\left\vert 9 \right\rangle +\left\vert 12 \right\rangle +\left\vert 14 \right\rangle +\left\vert 2 \right\rangle +\left\vert 26 \right\rangle +\left\vert 42 \right\rangle \right\} \nonumber \end{align*}$$

$$\begin{align*} \newcommand{\e}{{\rm e}} \newcommand{\ra}{\rangle} \displaystyle &{{H}_{2}}\left\vert 11 \right\rangle =\frac{-m}{2}\left\{\left\vert 12 \right\rangle +\left\vert 9 \right\rangle +\left\vert 15 \right\rangle +\left\vert 3 \right\rangle +\left\vert 27 \right\rangle +\left\vert 43 \right\rangle \right\}\nonumber \\ &{{H}_{2}}\left\vert 12 \right\rangle =\frac{-m}{2}\left\{\left\vert 11 \right\rangle +\left\vert 10 \right\rangle +\left\vert 16 \right\rangle +\left\vert 4 \right\rangle +\left\vert 28 \right\rangle +\left\vert 44 \right\rangle \right\}\nonumber \\ &{{H}_{2}}\left\vert 13 \right\rangle =\frac{-m}{2}\left\{\left\vert 14 \right\rangle +\left\vert 15 \right\rangle +\left\vert 9 \right\rangle +\left\vert 5 \right\rangle +\left\vert 29 \right\rangle +\left\vert 45 \right\rangle \right\}\nonumber \\ &{{H}_{2}}\left\vert 14 \right\rangle =\frac{-m}{2}\left\{\left\vert 13 \right\rangle +\left\vert 16 \right\rangle +\left\vert 10 \right\rangle +\left\vert 6 \right\rangle +\left\vert 30 \right\rangle +\left\vert 46 \right\rangle \right\}\nonumber \\ &{{H}_{2}}\left\vert 15 \right\rangle =\frac{-m}{2}\left\{\left\vert 16 \right\rangle +\left\vert 13 \right\rangle +\left\vert 11 \right\rangle +\left\vert 7 \right\rangle +\left\vert 31 \right\rangle +\left\vert 47 \right\rangle \right\}\nonumber \\ &{{H}_{2}}\left\vert 16 \right\rangle =\frac{-m}{2}\left\{\left\vert 15 \right\rangle +\left\vert 14 \right\rangle +\left\vert 12 \right\rangle +\left\vert 8 \right\rangle +\left\vert 32 \right\rangle +\left\vert 48 \right\rangle \right\}\nonumber \\ &{{H}_{2}}\left\vert 17 \right\rangle =\frac{-m}{2}\left\{\left\vert 18 \right\rangle +\left\vert 19 \right\rangle +\left\vert 21 \right\rangle +\left\vert 25 \right\rangle +\left\vert 1 \right\rangle +\left\vert 49 \right\rangle \right\}\nonumber \\ &{{H}_{2}}\left\vert 18 \right\rangle =\frac{-m}{2}\left\{\left\vert 17 \right\rangle +\left\vert 20 \right\rangle +\left\vert 22 \right\rangle +\left\vert 26 \right\rangle +\left\vert 2 \right\rangle +\left\vert 50 \right\rangle \right\}\nonumber \\ &{{H}_{2}}\left\vert 19 \right\rangle =\frac{-m}{2}\left\{\left\vert 20 \right\rangle +\left\vert 17 \right\rangle +\left\vert 23 \right\rangle +\left\vert 27 \right\rangle +\left\vert 3 \right\rangle +\left\vert 51 \right\rangle \right\}\nonumber \\ &{{H}_{2}}\left\vert 20 \right\rangle =\frac{-m}{2}\left\{\left\vert 19 \right\rangle +\left\vert 18 \right\rangle +\left\vert 24 \right\rangle +\left\vert 28 \right\rangle +\left\vert 4 \right\rangle +\left\vert 52 \right\rangle \right\} \nonumber \end{align*}$$

$$\begin{align*} \newcommand{\e}{{\rm e}} \newcommand{\ra}{\rangle} \displaystyle &{{H}_{2}}\left\vert 21 \right\rangle =\frac{-m}{2}\left\{\left\vert 22 \right\rangle +\left\vert 23 \right\rangle +\left\vert 17 \right\rangle +\left\vert 29 \right\rangle +\left\vert 5 \right\rangle +\left\vert 53 \right\rangle \right\}\nonumber \\ &{{H}_{2}}\left\vert 22 \right\rangle =\frac{-m}{2}\left\{\left\vert 21 \right\rangle +\left\vert 24 \right\rangle +\left\vert 18 \right\rangle +\left\vert 30 \right\rangle +\left\vert 6 \right\rangle +\left\vert 54 \right\rangle \right\}\nonumber \\ &{{H}_{2}}\left\vert 23 \right\rangle =\frac{-m}{2}\left\{\left\vert 24 \right\rangle +\left\vert 21 \right\rangle +\left\vert 19 \right\rangle +\left\vert 31 \right\rangle +\left\vert 7 \right\rangle +\left\vert 55 \right\rangle \right\}\nonumber \\ &{{H}_{2}}\left\vert 24 \right\rangle =\frac{-m}{2}\left\{\left\vert 23 \right\rangle +\left\vert 22 \right\rangle +\left\vert 20 \right\rangle +\left\vert 32 \right\rangle +\left\vert 8 \right\rangle +\left\vert 56 \right\rangle \right\}\nonumber \\ &{{H}_{2}}\left\vert 25 \right\rangle =\frac{-m}{2}\left\{\left\vert 26 \right\rangle +\left\vert 27 \right\rangle +\left\vert 29 \right\rangle +\left\vert 17 \right\rangle +\left\vert 9 \right\rangle +\left\vert 57 \right\rangle \right\}\nonumber \\ &{{H}_{2}}\left\vert 26 \right\rangle =\frac{-m}{2}\left\{\left\vert 25 \right\rangle +\left\vert 28 \right\rangle +\left\vert 30 \right\rangle +\left\vert 18 \right\rangle +\left\vert 10 \right\rangle +\left\vert 58 \right\rangle \right\}\nonumber \\ &{{H}_{2}}\left\vert 27 \right\rangle =\frac{-m}{2}\left\{\left\vert 28 \right\rangle +\left\vert 25 \right\rangle +\left\vert 31 \right\rangle +\left\vert 19 \right\rangle +\left\vert 11 \right\rangle +\left\vert 59 \right\rangle \right\}\nonumber \\ &{{H}_{2}}\left\vert 28 \right\rangle =\frac{-m}{2}\left\{\left\vert 27 \right\rangle +\left\vert 26 \right\rangle +\left\vert 32 \right\rangle +\left\vert 20 \right\rangle +\left\vert 12 \right\rangle +\left\vert 60 \right\rangle \right\}\nonumber \\ &{{H}_{2}}\left\vert 29 \right\rangle =\frac{-m}{2}\left\{\left\vert 30 \right\rangle +\left\vert 31 \right\rangle +\left\vert 25 \right\rangle +\left\vert 21 \right\rangle +\left\vert 13 \right\rangle +\left\vert 61 \right\rangle \right\}\nonumber \\ &{{H}_{2}}\left\vert 30 \right\rangle =\frac{-m}{2}\left\{\left\vert 29 \right\rangle +\left\vert 32 \right\rangle +\left\vert 26 \right\rangle +\left\vert 22 \right\rangle +\left\vert 14 \right\rangle +\left\vert 62 \right\rangle \right\} \nonumber \end{align*}$$

$$\begin{align*} \newcommand{\e}{{\rm e}} \newcommand{\ra}{\rangle} \displaystyle &{{H}_{2}}\left\vert 31 \right\rangle =\frac{-m}{2}\left\{\left\vert 32 \right\rangle +\left\vert 29 \right\rangle +\left\vert 27 \right\rangle +\left\vert 23 \right\rangle +\left\vert 15 \right\rangle +\left\vert 63 \right\rangle \right\}\nonumber \\ &{{H}_{2}}\left\vert 32 \right\rangle =\frac{-m}{2}\left\{\left\vert 31 \right\rangle +\left\vert 30 \right\rangle +\left\vert 28 \right\rangle +\left\vert 24 \right\rangle +\left\vert 16 \right\rangle +\left\vert 64 \right\rangle \right\}\nonumber \\ &{{H}_{2}}\left\vert 33 \right\rangle =\frac{-m}{2}\left\{\left\vert 34 \right\rangle +\left\vert 35 \right\rangle +\left\vert 37 \right\rangle +\left\vert 41 \right\rangle +\left\vert 49 \right\rangle +\left\vert 1 \right\rangle \right\}\nonumber \\ &{{H}_{2}}\left\vert 34 \right\rangle =\frac{-m}{2}\left\{\left\vert 33 \right\rangle +\left\vert 36 \right\rangle +\left\vert 38 \right\rangle +\left\vert 42 \right\rangle +\left\vert 50 \right\rangle +\left\vert 2 \right\rangle \right\}\nonumber \\ &{{H}_{2}}\left\vert 35 \right\rangle =\frac{-m}{2}\left\{\left\vert 36 \right\rangle +\left\vert 33 \right\rangle +\left\vert 39 \right\rangle +\left\vert 43 \right\rangle +\left\vert 51 \right\rangle +\left\vert 3 \right\rangle \right\}\nonumber \\ &{{H}_{2}}\left\vert 36 \right\rangle =\frac{-m}{2}\left\{\left\vert 35 \right\rangle +\left\vert 34 \right\rangle +\left\vert 40 \right\rangle +\left\vert 44 \right\rangle +\left\vert 52 \right\rangle +\left\vert 4 \right\rangle \right\}\nonumber \\ &{{H}_{2}}\left\vert 37 \right\rangle =\frac{-m}{2}\left\{\left\vert 38 \right\rangle +\left\vert 39 \right\rangle +\left\vert 33 \right\rangle +\left\vert 45 \right\rangle +\left\vert 53 \right\rangle +\left\vert 5 \right\rangle \right\}\nonumber \\ &{{H}_{2}}\left\vert 38 \right\rangle =\frac{-m}{2}\left\{\left\vert 37 \right\rangle +\left\vert 40 \right\rangle +\left\vert 34 \right\rangle +\left\vert 46 \right\rangle +\left\vert 54 \right\rangle +\left\vert 6 \right\rangle \right\}\nonumber \\ &{{H}_{2}}\left\vert 39 \right\rangle =\frac{-m}{2}\left\{\left\vert 40 \right\rangle +\left\vert 37 \right\rangle +\left\vert 35 \right\rangle +\left\vert 47 \right\rangle +\left\vert 55 \right\rangle +\left\vert 7 \right\rangle \right\}\nonumber \\ &{{H}_{2}}\left\vert 40 \right\rangle =\frac{-m}{2}\left\{\left\vert 39 \right\rangle +\left\vert 38 \right\rangle +\left\vert 36 \right\rangle +\left\vert 48 \right\rangle +\left\vert 56 \right\rangle +\left\vert 8 \right\rangle \right\} \nonumber \end{align*}$$

$$\begin{align*} \newcommand{\e}{{\rm e}} \newcommand{\ra}{\rangle} \displaystyle &{{H}_{2}}\left\vert 41 \right\rangle =\frac{-m}{2}\left\{\left\vert 42 \right\rangle +\left\vert 43 \right\rangle +\left\vert 45 \right\rangle +\left\vert 49 \right\rangle +\left\vert 57 \right\rangle +\left\vert 9 \right\rangle \right\}\nonumber \\ &{{H}_{2}}\left\vert 42 \right\rangle =\frac{-m}{2}\left\{\left\vert 41 \right\rangle +\left\vert 44 \right\rangle +\left\vert 46 \right\rangle +\left\vert 34 \right\rangle +\left\vert 58 \right\rangle +\left\vert 10 \right\rangle \right\}\nonumber \\ &{{H}_{2}}\left\vert 43 \right\rangle =\frac{-m}{2}\left\{\left\vert 44 \right\rangle +\left\vert 41 \right\rangle +\left\vert 47 \right\rangle +\left\vert 35 \right\rangle +\left\vert 59 \right\rangle +\left\vert 11 \right\rangle \right\}\nonumber \\ &{{H}_{2}}\left\vert 44 \right\rangle =\frac{-m}{2}\left\{\left\vert 43 \right\rangle +\left\vert 42 \right\rangle +\left\vert 48 \right\rangle +\left\vert 36 \right\rangle +\left\vert 60 \right\rangle +\left\vert 12 \right\rangle \right\}\nonumber \\ &{{H}_{2}}\left\vert 45\right\rangle =\frac{-m}{2}\left\{\left\vert 46 \right\rangle +\left\vert 47 \right\rangle +\left\vert 41 \right\rangle +\left\vert 37 \right\rangle +\left\vert 61 \right\rangle +\left\vert 13 \right\rangle \right\}\nonumber \\ &{{H}_{2}}\left\vert 46 \right\rangle =\frac{-m}{2}\left\{\left\vert 45 \right\rangle +\left\vert 48 \right\rangle +\left\vert 42 \right\rangle +\left\vert 38 \right\rangle +\left\vert 62 \right\rangle +\left\vert 14 \right\rangle \right\}\nonumber \\ &{{H}_{2}}\left\vert 47 \right\rangle =\frac{-m}{2}\left\{\left\vert 48 \right\rangle +\left\vert 45 \right\rangle +\left\vert 43 \right\rangle +\left\vert 39 \right\rangle +\left\vert 63 \right\rangle +\left\vert 15 \right\rangle \right\}\nonumber \\ &{{H}_{2}}\left\vert 48 \right\rangle =\frac{-m}{2}\left\{\left\vert 47 \right\rangle +\left\vert 46 \right\rangle +\left\vert 44 \right\rangle +\left\vert 40 \right\rangle +\left\vert 64 \right\rangle +\left\vert 16 \right\rangle \right\}\nonumber \\ &{{H}_{2}}\left\vert 49 \right\rangle =\frac{-m}{2}\left\{\left\vert 50 \right\rangle +\left\vert 51 \right\rangle +\left\vert 53 \right\rangle +\left\vert 57 \right\rangle +\left\vert 33 \right\rangle +\left\vert 17 \right\rangle \right\}\nonumber \\ &{{H}_{2}}\left\vert 50 \right\rangle =\frac{-m}{2}\left\{\left\vert 49 \right\rangle +\left\vert 52 \right\rangle +\left\vert 54 \right\rangle +\left\vert 58 \right\rangle +\left\vert 34 \right\rangle +\left\vert 18 \right\rangle \right\} \nonumber \end{align*}$$

$$\begin{align*} \newcommand{\e}{{\rm e}} \newcommand{\ra}{\rangle} \displaystyle &{{H}_{2}}\left\vert 51 \right\rangle =\frac{-m}{2}\left\{\left\vert 52 \right\rangle +\left\vert 49 \right\rangle +\left\vert 55 \right\rangle +\left\vert 59 \right\rangle +\left\vert 35 \right\rangle +\left\vert 19 \right\rangle \right\}\nonumber \\ &{{H}_{2}}\left\vert 52 \right\rangle =\frac{-m}{2}\left\{\left\vert 51 \right\rangle +\left\vert 50 \right\rangle +\left\vert 56 \right\rangle +\left\vert 60 \right\rangle +\left\vert 36 \right\rangle +\left\vert 20 \right\rangle \right\}\nonumber \\ &{{H}_{2}}\left\vert 53 \right\rangle =\frac{-m}{2}\left\{\left\vert 54 \right\rangle +\left\vert 55 \right\rangle +\left\vert 49 \right\rangle +\left\vert 61 \right\rangle +\left\vert 37 \right\rangle +\left\vert 21 \right\rangle \right\}\nonumber \\ &{{H}_{2}}\left\vert 54 \right\rangle =\frac{-m}{2}\left\{\left\vert 53 \right\rangle +\left\vert 56 \right\rangle +\left\vert 50 \right\rangle +\left\vert 62 \right\rangle +\left\vert 38 \right\rangle +\left\vert 22 \right\rangle \right\}\nonumber \\ &{{H}_{2}}\left\vert 55 \right\rangle =\frac{-m}{2}\left\{\left\vert 56 \right\rangle +\left\vert 53 \right\rangle +\left\vert 51 \right\rangle +\left\vert 63 \right\rangle +\left\vert 39 \right\rangle +\left\vert 23 \right\rangle \right\}\nonumber \\ &{{H}_{2}}\left\vert 56 \right\rangle =\frac{-m}{2}\left\{\left\vert 55 \right\rangle +\left\vert 54 \right\rangle +\left\vert 52 \right\rangle +\left\vert 64 \right\rangle +\left\vert 40 \right\rangle +\left\vert 24 \right\rangle \right\}\nonumber \\ &{{H}_{2}}\left\vert 57 \right\rangle =\frac{-m}{2}\left\{\left\vert 58 \right\rangle +\left\vert 59 \right\rangle +\left\vert 61 \right\rangle +\left\vert 49 \right\rangle +\left\vert 41 \right\rangle +\left\vert 25 \right\rangle \right\}\nonumber \\ &{{H}_{2}}\left\vert 58 \right\rangle =\frac{-m}{2}\left\{\left\vert 57 \right\rangle +\left\vert 60 \right\rangle +\left\vert 62 \right\rangle +\left\vert 50 \right\rangle +\left\vert 42 \right\rangle +\left\vert 26 \right\rangle \right\}\nonumber \\ &{{H}_{2}}\left\vert 59 \right\rangle =\frac{-m}{2}\left\{\left\vert 60 \right\rangle +\left\vert 57 \right\rangle +\left\vert 63 \right\rangle +\left\vert 51 \right\rangle +\left\vert 43 \right\rangle +\left\vert 27 \right\rangle \right\}\nonumber \\ &{{H}_{2}}\left\vert 60 \right\rangle =\frac{-m}{2}\left\{\left\vert 59 \right\rangle +\left\vert 58 \right\rangle +\left\vert 64 \right\rangle +\left\vert 52 \right\rangle +\left\vert 44 \right\rangle +\left\vert 28 \right\rangle \right\}\nonumber \\ &{{H}_{2}}\left\vert 61 \right\rangle =\frac{-m}{2}\left\{\left\vert 62 \right\rangle +\left\vert 63 \right\rangle +\left\vert 57 \right\rangle +\left\vert 53 \right\rangle +\left\vert 45 \right\rangle +\left\vert 29 \right\rangle \right\}\nonumber \\ &{{H}_{2}}\left\vert 62 \right\rangle =\frac{-m}{2}\left\{\left\vert 61 \right\rangle +\left\vert 64 \right\rangle +\left\vert 58 \right\rangle +\left\vert 54 \right\rangle +\left\vert 46 \right\rangle +\left\vert 30 \right\rangle \right\}\nonumber \\ &{{H}_{2}}\left\vert 63 \right\rangle =\frac{-m}{2}\left\{\left\vert 64 \right\rangle +\left\vert 61 \right\rangle +\left\vert 59 \right\rangle +\left\vert 55 \right\rangle +\left\vert 47 \right\rangle +\left\vert 31 \right\rangle \right\}\nonumber \\ &{{H}_{2}}\left\vert 64 \right\rangle =\frac{-m}{2}\left\{\left\vert 63 \right\rangle +\left\vert 62 \right\rangle +\left\vert 60 \right\rangle +\left\vert 56 \right\rangle +\left\vert 48 \right\rangle +\left\vert 32 \right\rangle \right\}.\nonumber \\ \nonumber \end{align*}$$

Finally the last term of the cluster Hamiltonian (3.16) is the transverse field term

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle H_3=h\left\{\tau_0^z\tau_1^z+\tau_2^z+\tau_3^z+\tau_4^z+\tau_5^z\right\}. \nonumber \end{align} \tag{ A.25 }$$

The effect of above term on the 64 bases is:

$$\begin{align} \newcommand{\e}{{\rm e}} \newcommand{\ra}{\rangle} \displaystyle &{{H}_{3}}\left\vert 1 \right\rangle =-6h\left\vert 1 \right\rangle,~~~~~~~~~~{{H}_{3}}\left\vert 2 \right\rangle =-4h\left\vert 2 \right\rangle,\nonumber \\ &{{H}_{3}}\left\vert 3 \right\rangle =-4h\left\vert 3 \right\rangle,~~~~~~~~~~ {{H}_{3}}\left\vert 4 \right\rangle =-2h\left\vert 4 \right\rangle,\nonumber \\ &{{H}_{3}}\left\vert 5 \right\rangle =-4h\left\vert 5 \right\rangle,~~~~~~~~~~{{H}_{3}}\left\vert 6 \right\rangle =-2h\left\vert 6 \right\rangle\nonumber \\ &{{H}_{3}}\left\vert 7 \right\rangle =-4h\left\vert 7 \right\rangle,~~~~~~~~~~{{H}_{3}}\left\vert 8 \right\rangle =0,\nonumber \\ &{{H}_{3}}\left\vert 9 \right\rangle =-4h\left\vert 9 \right\rangle,~~~~~~~~~~{{H}_{3}}\left\vert 10 \right\rangle =-2h\left\vert 10 \right\rangle,\nonumber \\ &{{H}_{3}}\left\vert 11 \right\rangle =-2h\left\vert 11 \right\rangle,~~~~~~~~{{H}_{3}}\left\vert 12 \right\rangle =0\nonumber \\ &{{H}_{3}}\left\vert 13 \right\rangle =-2h\left\vert 13 \right\rangle,~~~~~~~~{{H}_{3}}\left\vert 14 \right\rangle =0,\nonumber \\ &{{H}_{3}}\left\vert 15 \right\rangle =0,~~~~~~~~~~~~~~~{{H}_{3}}\left\vert 16 \right\rangle =2h\left\vert 16 \right\rangle,\nonumber \\ &{{H}_{3}}\left\vert 17 \right\rangle =-4h\left\vert 17 \right\rangle,~~~~~~~~{{H}_{3}}\left\vert 18 \right\rangle =-2h\left\vert 18 \right\rangle,\nonumber \\ &{{H}_{3}}\left\vert 19 \right\rangle =-2h\left\vert 19 \right\rangle,~~~~~~~~{{H}_{3}}\left\vert 20 \right\rangle =0,\nonumber \\ &{{H}_{3}}\left\vert 21 \right\rangle =-2h\left\vert 21 \right\rangle,~~~~~~~~{{H}_{3}}\left\vert 22 \right\rangle =0,\nonumber \\ &{{H}_{3}}\left\vert 23 \right\rangle =0,~~~~~~~~~~~~~~~{{H}_{3}}\left\vert 24 \right\rangle =2h\left\vert 24 \right\rangle\nonumber \\ &{{H}_{3}}\left\vert 25 \right\rangle =-2h\left\vert 25 \right\rangle,~~~~~~~~{{H}_{3}}\left\vert 26 \right\rangle =0,\nonumber \\ &{{H}_{3}}\left\vert 27 \right\rangle =0,~~~~~~~~~~~~~~~{{H}_{3}}\left\vert 28 \right\rangle =2h\left\vert 28 \right\rangle,\nonumber \\ &{{H}_{3}}\left\vert 29 \right\rangle =0,~~~~~~~~~~~~~~~{{H}_{3}}\left\vert 30 \right\rangle =2h\left\vert 30 \right\rangle\nonumber \\ &{{H}_{3}}\left\vert 31 \right\rangle =2h\left\vert 31 \right\rangle,~~~~~~~~~~{{H}_{3}}\left\vert 32 \right\rangle =4h\left\vert 32\right\rangle,\nonumber \\ &{{H}_{3}}\left\vert 33 \right\rangle =-4h\left\vert 33 \right\rangle,~~~~~~~~{{H}_{3}}\left\vert 34 \right\rangle =-2h\left\vert 34 \right\rangle\nonumber \\ &{{H}_{3}}\left\vert 35 \right\rangle =-2h\left\vert 35 \right\rangle,~~~~~~~~{{H}_{3}}\left\vert 36 \right\rangle =0\nonumber \\ &{{H}_{3}}\left\vert 37 \right\rangle =-2h\left\vert 37 \right\rangle,~~~~~~~~{{H}_{3}}\left\vert 38 \right\rangle =0\nonumber \\ &{{H}_{3}}\left\vert 39 \right\rangle =0,~~~~~~~~~~~~~~~{{H}_{3}}\left\vert 40 \right\rangle =2h\left\vert 40 \right\rangle,\nonumber \\ &{{H}_{3}}\left\vert 41 \right\rangle =-2h\left\vert 41 \right\rangle,~~~~~~~~{{H}_{3}}\left\vert 42 \right\rangle =0\nonumber \\ &{{H}_{3}}\left\vert 43 \right\rangle =0,~~~~~~~~~~~~~~~{{H}_{3}}\left\vert 44 \right\rangle =2h\left\vert 44 \right\rangle,\nonumber \\ &{{H}_{3}}\left\vert 45 \right\rangle =0,~~~~~~~~~~~~~~~{{H}_{3}}\left\vert 46 \right\rangle =2h\left\vert 46 \right\rangle,\nonumber \\ &{{H}_{3}}\left\vert 47 \right\rangle =2h\left\vert 47 \right\rangle,~~~~~~~~~~{{H}_{3}}\left\vert 48 \right\rangle =4h\left\vert 48 \right\rangle\nonumber \\ &{{H}_{3}}\left\vert 49 \right\rangle =-2h\left\vert 49 \right\rangle,~~~~~~~~{{H}_{3}}\left\vert 50 \right\rangle =0,\nonumber \\ &{{H}_{3}}\left\vert 51 \right\rangle =0,~~~~~~~~~~~~~~~{{H}_{3}}\left\vert 52 \right\rangle =2h\left\vert 52 \right\rangle,\nonumber \\ &{{H}_{3}}\left\vert 53 \right\rangle =0,~~~~~~~~~~~~~~~{{H}_{3}}\left\vert 54 \right\rangle =2h\left\vert 54 \right\rangle,\nonumber \\ &{{H}_{3}}\left\vert 55 \right\rangle =2h\left\vert 55 \right\rangle,~~~~~~~~~{{H}_{3}}\left\vert 56 \right\rangle =4h\left\vert 56 \right\rangle,\nonumber \\ &{{H}_{3}}\left\vert 57 \right\rangle =0,~~~~~~~~~~~~~~~{{H}_{3}}\left\vert 58 \right\rangle =2h\left\vert 58 \right\rangle,\nonumber \\ &{{H}_{3}}\left\vert 59 \right\rangle =2h\left\vert 59 \right\rangle,~~~~~~~~~~{{H}_{3}}\left\vert 60 \right\rangle =4h\left\vert 60 \right\rangle,\nonumber \\ &{{H}_{3}}\left\vert 61 \right\rangle =2h\left\vert 61 \right\rangle,~~~~~~~~~~{{H}_{3}}\left\vert 62 \right\rangle =4h\left\vert 62 \right\rangle,\nonumber \\ &{{H}_{3}}\left\vert 63 \right\rangle =4h\left\vert 63 \right\rangle,~~~~~~~~~~{{H}_{3}}\left\vert 64 \right\rangle =6h\left\vert 64 \right\rangle. \nonumber \end{align} \tag{ A.26 }$$

The above $64\times 64$ Hamiltonian can be diagonalized on a computer even without resorting to group theory methods.