The ground state of the Kondo insulator

We rewrite the term of the Hamiltonian (1) ${{{\mathscr {H}}}}_{K}$ in detail in the following form

$$\begin{aligned}{} & {} {{{\mathscr {H}}}}_K= \sum _{j=1}^N[2J s^z_jS^z_j+K(s^+_jS^-_j + s^-_jS^+_j)], \end{aligned}$$

(2)

where spin operators are redefined in the terms of fermionic operators, $s^z_j=\frac{1}{2}(m_{j\uparrow }-m_{j\downarrow })$, $s^+_j=c^\dagger _{j \uparrow } c_{j\downarrow }$, $s^-_j=c^\dagger _{j \downarrow } c_{j\uparrow }$, here $m_{j\sigma }=c^\dagger _{j \sigma } c_{j \sigma }$, $m_{j}=m_{j\uparrow }+m_{j\downarrow }$ are the density operators.

We use the following presentation for the ${{{\mathscr {H}}}}_{K}$ term: $2Js^z_jS^z_j+K(s^+_jS^-_j + s^-_jS^+_j)=-J (s^z_j-S^z_j)^2-K(s^+_j-S^+_j)^\dagger (s^+_j-S^+_j) +(J-K) (S^z_j)^2-\frac{1}{4}(J+2 K)m_{j}^2+\frac{1}{2}(J+K)m_j+K S(S+1) \Longrightarrow 2\Lambda _{j} (s^z_j-S^z_j)+\lambda _{j}(c^+_{j\uparrow }c_{j\downarrow }-S^+_{j})+\lambda ^*_{j}(c^+_{j\downarrow }c_{j\uparrow }- S^-_{j} )+2\mu _{j} m_{j}$. We will study in detail the case of $S=\frac{1}{2}$. $(S^z_j)^2$ -operator is conserved, $\mu$-component shifts the Fermi energy, so it can be neglected. Using the Hubbard-Stratonovich transformation we introduce the effective Hamiltonian which is determined by two component ${Z}_2$-field. The canonical functional is determined by the action, which follows from the Hubbard-Stratonovich transformation

$$\begin{aligned}{} & {} {{{\mathscr {S}}}}=\sum _{j} \frac{\Lambda _j^2}{J}+\sum _{j} \frac{|\lambda _j|^2}{K}+\int _0^\beta d\tau \sum _{{\textbf {k}}}\Psi _{{\textbf {k}}}^\dagger (\tau )[\partial _\tau + {\mathscr {H}}_{eff}({{\textbf {k}}})]\Psi _{{\textbf {k}}} (\tau ), \end{aligned}$$

(3)

where $\Psi _{{\textbf {k}}} (\tau )$ is the wave function, $\beta =\frac{1}{k_B T}$, $k_B$ is the Boltzmann constant, T is temperature, ${{\textbf {k}}}=(k_x,k_y,k_z)$ is the wave vector.

The effective Hamiltonian ${{{\mathscr {H}}}}_{eff}$ determines the ground state of the model and low-energy excitations at half filling occupation. We expect that $\Lambda _{{{\textbf {j}}} }$ and $\lambda _{{{\textbf {j}}} }$ are independent of time because of translational invariance. We study the ground state of the Kondo insulator, the low temperature hehavior of an electon liquid is not studied. and low-energy excitations corresponding to fluctuations of the saddle point solution are not taken into account.

We can define an effective Hamiltonian ${{{\mathscr {H}}}}_{eff}$, which describes the behavior of the electron liquid in the Kondo lattice in the mean field approach ${{{\mathscr {H}}}}_{eff}={{{\mathscr {H}}}}_0+ \sum _{j}[2\Lambda _{j} (s^z_j-S^z_j)+\lambda _{j}(c^+_{j\uparrow }c_{j\downarrow }-S^+_{j})+\lambda ^*_{j}(c^+_{j\downarrow }c_{j\uparrow }-S^-_{j})$.

Let us consider the equations for the one-particle wave functions $\psi ({{\textbf {j}}},\sigma )c^\dagger _{{{\textbf {j}}} \sigma }\phi ({{\textbf {j}}},\pm \sigma )S^{\pm }_{{{\textbf {j}}} }$ ($\sigma =\uparrow ,\downarrow )$ with energy $\varepsilon$, the $\psi ({{\textbf {j}}},\sigma )$ and $\phi ({{\textbf {j}}},\sigma )$ amplitudes satisfy the following equations :

$$\begin{aligned}{} & {} (\varepsilon -\Lambda _{{{\textbf {j}}}})\psi ({{\textbf {j}}},\sigma ) +\lambda _{{{\textbf {j}}}} \psi ({{\textbf {j}}},-\sigma )+\sum _{{{\textbf {1}}}}\psi ({\textbf {j+1}},\sigma )=0, \nonumber \\{} & {} (\varepsilon + \Lambda _{{{\textbf {j}}}})\psi ({{\textbf {j}}},-\sigma )+\lambda ^*_{{{\textbf {j}}}}\psi ({{\textbf {j}}},\sigma )+ \sum _{{{\textbf {1}}}}\psi ({\textbf {j+1}},-\sigma )=0, \nonumber \\{} & {} (\varepsilon +\Lambda _{{{\textbf {j}}}})\phi ({{\textbf {j}}},\sigma ) -\lambda _{{{\textbf {j}}}} \phi ({{\textbf {j}}},-\sigma )=0,\nonumber \\{} & {} (\varepsilon -\Lambda _{{{\textbf {j}}}})\phi ({{\textbf {j}}},-\sigma ) -\lambda ^*_{{{\textbf {j}}}} \phi ({{\textbf {j}}},\sigma )=0, \end{aligned}$$

(4)

where sums over the nearest lattice sites. The real variables $\Lambda _{{{\textbf {j}}}}\rightarrow \pm \Lambda _{{{\textbf {j}}}}$ and $\lambda _{{{\textbf {j}}}}\rightarrow \pm \lambda _{{{\textbf {j}}}}$ are identified with a static two component ${Z}_2$– field determined on the lattice sites. A confuguration of this field, which corresponds to an energy minimum, defines the ground state. The local moments form the flat band states, with energies $\varepsilon _S=\pm \sqrt{\Lambda ^2+\lambda ^2}$, here $\Lambda _{{{\textbf {j}}}}^2=\Lambda ^2,|\lambda _{{{\textbf {j}}}}]^2=\lambda ^2$. The local moments are arranged regularly at the lattice sites, their energy does not depend on $S^z_{{{\textbf {j}}}}$.

In contrast to well known models^11,12, where a free condiguration of the $Z_2$-field ($\lambda _j=\lambda$) corresponds to minimum of energy, an uniform sector with $\Lambda _{{{\textbf {j}}} }=-\Lambda _{{\textbf {j+1}}}=\Lambda$, $\lambda _{{{\textbf {j}}} }=-\lambda _{{\textbf {j+1}}}=\lambda$ corresponds to minimum of energy in the Kondo insulator^13,14. This field configuration leads to the lattice with a double cell, does not break the translational symmetry. Detailed numerical analysis shows, that an uniform sector with $\Lambda _{{{\textbf {j}}} }=-\Lambda _{{\textbf {j+1}}}=\Lambda$, $\lambda _{{{\textbf {j}}}} =-\lambda _{{\textbf {j+1}}}=\lambda$ corresponds to the ground state of an electron liquid for arbitrary values of J and K. Using Eq. (4) we calculate the energies of the quasi-particle excitations wich correspond to this uniform configuration of the ${\mathbb {Z}}_2$-field. The spectrum includes two branches of local moments $\varepsilon _S$ and two branches of electrons $\varepsilon _s({{\textbf {k}}})=\pm \sqrt{\Lambda ^2+\lambda ^2+|w({{\textbf {k}}})|^2}$, here $w({{\textbf {k}}})=\sum _{\alpha }^D[1+\exp ( i k_\alpha )]$. The spectrum of the quasi-particle excitations is symmetric with respect to zero energy, has the Majorana type at half filling, the chemical potential is zero. Despite the fact that the effective Hamiltonian does not conserve the total spin, the wave function (4) at the same time conserve the total spin, since the flip of the electron spin is accompanied by the reverse flip for the local momentum located at the same lattice site. The values of the ${Z}_2$-field components satisfy the energy minimum or the saddle point of the action, self-consistent equations have the following form at $T=0K$

$$\begin{aligned}{} & {} \frac{2\Lambda }{J}=\frac{1}{N}\sum _{{{\textbf {k}}}}\frac{\Lambda }{|\varepsilon _s({{\textbf {k}}})|}+\frac{\Lambda }{|\varepsilon _S|},\nonumber \\{} & {} \frac{2\lambda }{K}=\frac{1}{N}\sum _{{{\textbf {k}}}}\frac{\lambda }{|\varepsilon _s({{\textbf {k}}})|}+\frac{\lambda }{|\varepsilon _S|}. \end{aligned}$$

(5)

where $\Lambda =\lambda \ne 0$ for isotropic $J=K>0$, and $\Lambda =0, \lambda \ne 0$ for anisotropic $J=0, K>0$ exchange interactions.

The behavior of the electron liquid in the case of strongly anisotropic, when $J=0$, $K>0$, and isotropic, when $J=K>0$, exchange antiferromagnetic interaction will be considered in detail.

Strongly anisotropic exchange interaction $J=0$, $K>0$

For a strongly anisotropic exchange interaction, the ${Z}_2$– field is one-component, since $\Lambda =0$ and $\lambda \ne 0$. According to the numerical analysis, the solutions $\lambda _{{\textbf {j}}} =-\lambda _{\textbf {j+1}}=\lambda$ correspond to the minimum energy for arbitrary values of the exchange integral K and magnetic field h.

In magnetic field the energies of the quasi-particle excitations transform to $\varepsilon _{s,+}({{\textbf {k}}})=\pm \sqrt{\Lambda ^2+\lambda ^2+(h+|w({{\textbf {k}}})|)^2}$, $\varepsilon _{s,-}({{\textbf {k}}})=\pm \sqrt{\Lambda ^2+\lambda ^2+(h-|w({{\textbf {k}}})|)^2}$, $\varepsilon _S=\pm \sqrt{\Lambda ^2+\lambda ^2+h^2}$. The spectrum is symmetrical with respect to zero energy or chemical potential, which is zero at half filling for an arbitrary magnetic field.

In the electron spectrum the gap opens at $\lambda \ne 0$ and is equal to $2\lambda$ at $\Lambda =0$. According to Eq. (5) its value is determined by K and h. Using Eq. (5) we numerically calculate $\lambda$ as function of K and h for the chain (Fig 1a), square (Fig 1b) and cubic (Fig 1c) lattices. Should be note an universal behavior of an electron liquid in KI, the curves in Figs are similar for an arbitrary dimension. In a weak coupling limit at $K\rightarrow 0$ the last term in Eq. (5) dominates, so $\lambda \rightarrow \frac{K}{2}$. We ilustrate the spectrum of the quasi-particle excitations in Fig. 2a for the chain and Fig. 2b for the square lattice. The magnetic field breaks the spin degeneracy of the spectrum of the electrons, spreading the branches. The $\lambda$– value (or the value of the gap) decreases with increasing magnetic field. A critical value of the magnetic field $h_c$, at which the gap closes¹⁵, depends on $K-$value. Numerical calculations of $h_c$ are shown in the Fig. 3 (the curves are calculated for different dimension of the model). In magnetic field $h_c$ the phase transition from insulator to the metal states is realized, KI is stable at $h<h_c$. An uniform configuration of the ${Z}_2$-field corresponds to minimum energy in metal state with the local doubling of period of the original lattice, in other words the metal state is also realized in the lattice with a double cell.

Magnetic properties of an electron liquid in KI are determined by both band electrons and local moments, they are determined by the uniform configuration of the ${Z}_2$-field. Electrons and local moments form singlet states in the lattice with a double cell, which are not fixed in time. In absence of magnetic field the energies of the quasi-particle excitations degenerate in spin, so the magnetization density $M=\frac{2}{N}\sum _{j}(s^z_j+S^z_j)$ is zero. The magnetic field does not break this energy degeneracy for local moments, so the magnetization is given by an electron term $M=\frac{1}{N}\sum _{j}(m_{j,\uparrow }-m_{j,\downarrow })$:

$$\begin{aligned} M=\frac{h}{N}\sum _{{{\textbf {k}}}} \frac{1}{|\varepsilon _{s +}({{\textbf {k}}})| +|\varepsilon _{s,-}({{\textbf {k}}})|} (\frac{\lambda ^2 + h^2- |w({{\textbf {k}}})|^2 }{ |\varepsilon _{s,+}({{\textbf {k}}})\varepsilon _{s,-}({{\textbf {k}}})| } + 1 ). \end{aligned}$$

(6)

The calculations of the magnetization density M as function of magnetic field and the exchange integral K are presented in Fig. 4 for different dimension of the model. Formula for a static magnetic susceptibility leads from M at $h\rightarrow 0$ $\chi =\frac{1}{N}\sum _{{{\textbf {k}}}} \frac{\lambda ^2 }{(\lambda ^2 +| w({{\textbf {k}}})|)^{3/2}}$. The value of a static magnetic susceptibility is calculated as function of the exchange integral for different dimension of the model. The calculations are shown in Fig. 5, the susseptibility is a monotonic function of K.

An uniform configuration $\lambda _j=-\lambda _{j+1}$ stabilises the state with a double cell, a free configuration $\lambda _j=\lambda _{j+1}$ corresponds to gapless state on original latiice with higher energy. We define an unstable configuration of the $\lambda$-field with one “defect” as $-\lambda _{j-2}=\lambda _{j-1}=\lambda _{j}=\lambda _{j+1}=-\lambda _{j+2}$ . An unstable configuration of the $\lambda$ -field with one “defect” of size $\nu$ is defined as $-\lambda _{j-\nu -1}=\lambda _{j-\nu }=\lambda _{j-\nu +1}…=\lambda _{j+\nu }=-\lambda _{j+\nu + 1 }$. According to numerical calculations, a total energy of individual $\nu$ “defects” is greater than the energy of one “defect” of size $\nu$.

As an example, we present the calculations of the excitation energies in a chain with “one defect” as a function of the defect size $\nu$. Configurations with “defect” in an uniform configuration have energies lying in the gap, a number of excitations increases with increasing $\nu$ , so that for $1<\nu <5$ only one excitation is split off from the continuous spectrum, for $5<\nu <10$ , $10<\nu <15$ there are 2 and 3 such states, respectively (see in Fig. 6). We note, that the lattice with a double cell is formed by an uniform configuration of $\lambda$-field, and neither spin nor charge density waves are realized.

Isotropic exchange interaction $J=K>0$

As noted above, in the absence of a magnetic field for an isotropic exchange interaction, the $\lambda$– and $\Lambda$-components of the ${Z}_2$-fields are equal and are solutions of Eq. (5). Along with this solution, there are also the number of non-trivial solutions: $\lambda \ne 0$ and $\Lambda =0$ , $\lambda =0$ and $\Lambda \ne 0$. Three solutions of Eq. (5) have the same energies, which follows from numerical calculations of the ground state energy as funsction of J at $h=0$ for different dimensions of the model. In the absence of an external magnetic field, the energies of the quasi-particle excitations are degenerate in spin. A magnetic field removes this degeneracy. For arbitrary values of magnetic field and isotropic exchange integral, a solution $\lambda \ne 0$, $\Lambda =0$ corresponds to a lower energy than a solution $\lambda = 0$, $\Lambda \ne 0$. Another nontrivial solution $\lambda \ne 0, \Lambda \ne 0$ not satisfy the self-consistent equations for $\lambda$ and $\Lambda$ for arbitrary h. KI is determined by the XX-exchange interaction (the value of K in Hamiltonian (Eq. 1)), the ZZ-exchange interaction (the value of J in Hamiltonian (Eq. 1)) does not participate in the formation of KI. Scattering processes with spin flip lead to the formation of KI in the Kondo lattice as it takes place in the Kondo problem.

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Strongly anisotropic exchange interaction \(J=0\), \(K>0\)

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