ABA TLG

Some basic knowledge of ABA tri-layer graphene.

Introduction

Lattice Structure


The lattice structure of ABA tri-layer graphene is shown in the left figure.

It contains six sublattices in its unit cell,

A1,B1,A2,B2,A3,B3A_1, B_1, A_2, B_2, A_3, B_3

where A and B refer to the sublattice index while the Arabic number 1,2,3 represent the layer index.

In order to give a precise description of its band structure, the interlayer hopping and intralayer hopping between different sublattice sites need to be considered. Fortunately, it’s enough for most cases to take into account only several of them, which are listed in the table.

hopping direction AnBnA_n \leftrightarrows B_n B1A2B3B_1 \leftrightarrows A_2 \leftrightarrows B_3 A1A2A3A_1 \leftrightarrows A_2 \leftrightarrows A_3 A1B2A3A_1 \leftrightarrows B_2 \leftrightarrows A_3 A1A3A_1 \leftrightarrows A_3 B1B3B_1 \leftrightarrows B_3
nominal symbol γ0\gamma_0 γ1\gamma_1 γ4\gamma_4 γ3\gamma_3 γ22\frac{\gamma_2}{2} γ52\frac{\gamma_5}{2}

Besides these hopping parameters, the onsite potential also plays an important role in its electronic structure. The onsite potentials on A1,B1,A2,B2,A3,B3A_1, B_1, A_2, B_2, A_3, B_3 are

(Δ1+Δ2,Δ1+Δ2+δ,2Δ2+δ,2Δ2,Δ1+Δ2,Δ1+Δ2+δ)(\Delta_1 + \Delta_2,\quad \Delta_1 + \Delta_2 + \delta,\quad -2\Delta_2 + \delta,\quad-2\Delta_2,\quad-\Delta_1 + \Delta_2,\quad -\Delta_1 + \Delta_2 + \delta )

where Δ1\Delta_1 is the potential difference between the nearest layers which can be controlled by a perpendicular displacement field D\mathbf{D} through Δ1=αdD\Delta_1 = \alpha d\mathbf{D} , where α\alpha accounts for the imperfect screening.

Δ2\Delta_2 is the difference between the potential of the central layer and the averaged potential of outer layers, implying the charge density imblance between the inner and outer layers. It can be expressed as Δ2=4πe2d6εrn2\Delta_2 = - \frac{4\pi e^2 d}{6\varepsilon_r}n_2.

and δ\delta is the dimer site potential.

Therefore, the tight-binding Hamiltonian of ABA TLG in the sublattice basis can be written as

H(k)=αβα><αH(k)β><β=[A1>B1>A2>B2>A3>B3>](Δ1+Δ2γ0f(k)γ4f(k)γ3f(k)γ2/20γ0f(k)Δ1+Δ2+δ    γ1γ4f(k)   0γ5/2     γ4f(k)γ1 2Δ2+δγ0f(k) γ4f(k)γ1  γ3f(k)γ4f(k) γ0f(k)2Δ2 γ3f(k)γ4f(k)  γ2/20 γ4f(k)γ3f(k) Δ1+Δ2γ0f(k) 0γ5/2γ1γ4f(k)γ0f(k)Δ1+Δ2+δ) [<A1<B1<A2<B2<A3<B3]\small H(\mathbf{k})=\sum_{\alpha\beta}|\alpha><\alpha|H(\mathbf{k})|\beta><\beta| \\ \tiny =\begin{bmatrix}|A_1> & |B_1> & |A_2> & |B_2> & |A_3> & |B_3>\end{bmatrix} \begin{pmatrix} \Delta_1 + \Delta_2 & \gamma_0 f^{*}(\mathbf{k}) & -\gamma_4 f^{*}(\mathbf{k}) & \gamma_3 f(\mathbf{k}) & \gamma_2 / 2 & 0 \\[6pt] \gamma_0 f(\mathbf{k}) & \Delta_1 + \Delta_2 + \delta &         \gamma_1 & - \gamma_4 f^{*}(\mathbf{k}) &       0 & \gamma_5 / 2\\[6pt]           -\gamma_4 f(\mathbf{k}) & \gamma_1 &   -2\Delta_2 + \delta & \gamma_0 f^{*}(\mathbf{k}) &   -\gamma_4 f(\mathbf{k}) & \gamma_1\\[6pt]     \gamma_3 f^{*}(\mathbf{k}) & -\gamma_4 f(\mathbf{k}) &   \gamma_0 f(\mathbf{k}) & -2\Delta_2 &   \gamma_3 f^{*}(\mathbf{k}) & -\gamma_4 f(\mathbf{k})\\[6pt]     \gamma_2 / 2 & 0 &   -\gamma_4 f^{*}(\mathbf{k}) & \gamma_3 f(\mathbf{k}) &   -\Delta_1 + \Delta_2 & \gamma_0 f^{*}(\mathbf{k})\\[6pt]   0 & \gamma_5 / 2 & \gamma_1 & -\gamma_4 f^{*}(\mathbf{k}) & \gamma_0 f(\mathbf{k}) & -\Delta_1 + \Delta_2 + \delta \end{pmatrix}   \begin{bmatrix}<A_1| \\ <B_1| \\ <A_2| \\ <B_2| \\ <A_3| \\ <B_3|\end{bmatrix}

where f(k)f(\mathbf{k}) is the structure factor associated with the lattice geometry as :

f(k)=eikδ1+eikδ2+eikδ3{32a(kxiky)near K+ valley32a(kx+iky)near K valleyf(\mathbf{k}) = e^{-i\mathbf{k\cdot\delta_1}} + e^{-i\mathbf{k\cdot\delta_2}} + e^{-i\mathbf{k\cdot\delta_3}} \approx \left\{ \begin{aligned} -\frac{\sqrt{3}}{2}a(k_x - i k_y) & \quad \text{near } K_{+} \text{ valley}\\ \frac{\sqrt{3}}{2}a(k_x + i k_y) & \quad \text{near } K_{-} \text{ valley} \end{aligned} \right.

Mirror Symmetry

Because of its mirror symmetry(A1>A3>,B1>B3>|A_1>\longleftrightarrow|A_3>, |B_1>\longleftrightarrow|B_3>) in the absence of D\mathbf{D}, the Hamiltonian can be reduced into two blocks, each of which holds a irreducible representation of mirror symmetry, although itself reducible.

First step : basis transformation

[ϕ1>ϕ2>ϕ3>ϕ4>ϕ5>ϕ6>]=[A1>B1>A2>B2>A3>B3>](1/201/200001/201/2000000100000011/201/200001/201/200)\tiny \begin{bmatrix} |\phi_1> & |\phi_2> & |\phi_3> & |\phi_4> & |\phi_5> & |\phi_6> \end{bmatrix} = \begin{bmatrix}|A_1> & |B_1> & |A_2> & |B_2> & |A_3> & |B_3>\end{bmatrix} \begin{pmatrix} 1 / \sqrt{2} & 0 & 1 / \sqrt{2} & 0 & 0 & 0 \\[4pt] 0 & 1 / \sqrt{2} & 0 & 1 / \sqrt{2} & 0 & 0\\[4pt] 0 & 0 & 0 & 0 & 1 & 0 \\[4pt] 0 & 0 & 0 & 0 & 0 & 1 \\[4pt] - 1 / \sqrt{2} & 0 & 1 / \sqrt{2} & 0 & 0 & 0 \\[4pt] 0 & - 1 / \sqrt{2} & 0 & 1 / \sqrt{2} & 0 & 0\\ \end{pmatrix}

Second step : convert the Hamiltonian into this basis

where

Hm(k)=(γ2/2+Δ2γ0f(k)γ0f(k)δγ5/2+Δ2)Hb(k)=(γ2/2+Δ2γ0f(k)2γ4f(k)2γ3f(k)γ0f(k)δ+γ5/2+Δ22γ12γ4f(k)2γ4f(k)2γ1δ2Δ2γ0f(k)2γ3fk)2γ4f(k)γ0f(k)2Δ2)D=(Δ10000Δ100)H_m(\mathbf{k}) = \begin{pmatrix} -\gamma_2/2 + \Delta_2 & \gamma_0 f^{*}(\mathbf{k}) \\ \gamma_0 f(\mathbf{k}) & \delta - \gamma_5 / 2 + \Delta_2 \\ \end{pmatrix}\\[20pt] H_b(\mathbf{k}) = \begin{pmatrix} \gamma_2/2 + \Delta_2 & \gamma_0 f^{*}(\mathbf{k}) & -\sqrt{2}\gamma_4f^{*}(\mathbf{k}) & \sqrt{2}\gamma_3f(\mathbf{k}) \\ \gamma_0 f(\mathbf{k}) & \delta + \gamma_5 / 2 + \Delta_2 & \sqrt{2}\gamma_1 & -\sqrt{2}\gamma_4f^{*}(\mathbf{k}) \\ -\sqrt{2}\gamma_4f(\mathbf{k}) & \sqrt{2}\gamma_1 & \delta - 2 \Delta_2 & \gamma_0 f^{*}(\mathbf{k}) \\ \sqrt{2}\gamma_3f^{*}\mathbf{k}) & -\sqrt{2}\gamma_4f(\mathbf{k}) & \gamma_0 f(\mathbf{k}) & -2\Delta_2 \end{pmatrix} \\[20pt] D = \begin{pmatrix} \Delta_1 & 0 & 0 & 0 \\ 0 & \Delta_1 & 0 & 0 \end{pmatrix}

Continuum Limit

In the continuum limit, the Hamiltonian can be expanded around the two Dirac points K+K_{+} and KK_{-}. It makes sense to treat K±K_{\pm} as a discrete degree of freedom with binary values, called “valley”, which is similar to the spin.

Introduce two operators : πpx+ipy=(kx+iky),πpxipy\pi \equiv p_x + i p_y = \hbar (k_x + i k_y), \pi^{\dagger} \equiv p_x - i p_y

Under a perpendicular magnetic field, px,pyp_x , p_y should be replaced by px+eAx,py+eAyp_x + eA_x , p_y + eA_y (e < 0), then π/π\pi^{\dagger}/ \pi act like creation / annihilation operator, which can be useful in the calculation of Landau level spectra :

πψn=ilB2(n+1)ψn+1πψn=ilB2nψn1\pi^{\dagger} \psi_n = i\frac{\hbar}{l_B}\sqrt{2(n+1)}\psi_{n+1} \\ \pi \psi_n = - i\frac{\hbar}{l_B}\sqrt{2n}\psi_{n-1}

Introduce other symbols : vi=32aγi for i=0,3,4v_i = \frac{\sqrt{3}}{2}\frac{a\gamma_i}{\hbar} \text{ for } i = 0, 3, 4

near K+K_{+} :

Hm+=(γ2/2+ Δ2v0πv0πδγ5/2+Δ2)(δγ5/2+Δ2v0πv0πγ2/2+ Δ2)H_{m+}= \begin{pmatrix} -\gamma_2 / 2 + \Delta_2 & -v_0 \pi \\ -v_0 \pi^{\dagger} & \delta - \gamma_5 / 2 + \Delta_2 \end{pmatrix} \longrightarrow \begin{pmatrix} \delta - \gamma_5 / 2 + \Delta_2 & -v_0 \pi^{\dagger} \\ -v_0 \pi & -\gamma_2 / 2 + \Delta_2 \end{pmatrix} \\

exchange the order of basis !!

Hb+=(γ2/2+ Δ2v0π2v4π2v3πv0πδ+γ5/2+Δ22γ12v4π2v4π2γ1δ2Δ2v0π2v3π2v4πv0π2Δ2)(2Δ22v3π2v4πv0π2v3πγ2/2+ Δ2v0π2v4π2v4πv0πδ+γ5/2+Δ22γ1v0π2v4π2γ1δ2Δ2)H_{b+}= \begin{pmatrix} \gamma_2 / 2 + \Delta_2 & -v_0 \pi & \sqrt{2}v_4\pi & -\sqrt{2}v_3\pi^{\dagger}\\ -v_0\pi^{\dagger} & \delta+\gamma_5/2+\Delta_2 & \sqrt{2}\gamma_1 &\sqrt{2}v_4\pi\\ \sqrt{2}v_4\pi^{\dagger} & \sqrt{2}\gamma_1 & \delta - 2\Delta_2 & -v_0\pi \\ -\sqrt{2}v_3\pi & \sqrt{2}v_4\pi^{\dagger} & -v_0\pi^{\dagger} & -2\Delta_2 \end{pmatrix} \\ \longrightarrow \begin{pmatrix} -2\Delta_2 & -\sqrt{2}v_3\pi & \sqrt{2}v_4\pi^{\dagger} & -v_0\pi^{\dagger}\\ -\sqrt{2}v_3\pi^{\dagger} & \gamma_2 / 2 + \Delta_2 & -v_0 \pi & \sqrt{2}v_4\pi\\ \sqrt{2}v_4\pi&-v_0\pi^{\dagger} & \delta+\gamma_5/2+\Delta_2 & \sqrt{2}\gamma_1\\ -v_0\pi&\sqrt{2}v_4\pi^{\dagger} & \sqrt{2}\gamma_1 & \delta - 2\Delta_2 \end{pmatrix}

exchange the order of basis !!

near KK_{-} :

Hm=(γ2/2+Δ2v0πv0πδγ5/2+Δ2)H_{m-}= \begin{pmatrix} -\gamma_2 / 2 +\Delta_2 & v_0 \pi^{\dagger} \\ v_0 \pi & \delta - \gamma_5 / 2 + \Delta_2 \end{pmatrix}

don't need to exchange the order of basis !!

Hb=(γ2/2+Δ2v0π2v4π2v3πv0πδ+γ5/2+Δ22γ12v4π2v4π2γ1δ2Δ2v0π2v3π2v4πv0π2Δ2)(γ2/2+Δ22v3π2v4πv0π2v3π2Δ2v0π2v4π2v4πv0πδ2Δ22γ1v0π2v4π2γ1δ+γ5/2+Δ2)H_{b-}= \begin{pmatrix} \gamma_2/2+\Delta_2&v_0\pi^{\dagger}&-\sqrt{2}v_4\pi^{\dagger}&\sqrt{2}v_3\pi\\ v_0\pi & \delta+\gamma_5/2+\Delta_2 & \sqrt{2}\gamma_1 &-\sqrt{2}v_4\pi^{\dagger}\\ -\sqrt{2}v_4\pi & \sqrt{2}\gamma_1 & \delta - 2\Delta_2 & v_0\pi^{\dagger} \\ \sqrt{2}v_3\pi^{\dagger}&-\sqrt{2}v_4\pi&v_0\pi&-2\Delta_2 \end{pmatrix} \\ \longrightarrow \begin{pmatrix} \gamma_2/2+\Delta_2&\sqrt{2}v_3\pi&-\sqrt{2}v_4\pi^{\dagger}&v_0\pi^{\dagger}\\ \sqrt{2}v_3\pi^{\dagger}&-2\Delta_2&v_0\pi&-\sqrt{2}v_4\pi\\ -\sqrt{2}v_4\pi & v_0\pi^{\dagger} & \delta - 2\Delta_2 & \sqrt{2}\gamma_1 \\ v_0\pi &-\sqrt{2}v_4\pi^{\dagger}& \sqrt{2}\gamma_1 &\delta+\gamma_5/2+\Delta_2\\ \end{pmatrix}

exchange the order of basis !!

Schrieffer-Wolf Transformation

Although the continuum model describes the low energy regime around charge neutrality point very well, it is still a six-band model. In order to capture the essential of low-energy physics, the Hb±H_{b\pm} can be reduced to a 2×22\times 2 low-energy effective Hamiltonian :

Hb+eff=12m(0(π)2π20)2v3(0ππ0)+(2Δ200γ2/2+Δ2)Hbeff=12m(0(π)2π20)+2v3(0ππ0)+(γ2/2+Δ2002Δ2)where1/2m=v02/(2γ1)[1+O(γ4/γ0)2]H^{eff}_{b+} = -\frac{1}{2m} \begin{pmatrix} 0 & (\pi^{\dagger})^2 \\ \pi^2 & 0 \end{pmatrix} - \sqrt{2}v_3 \begin{pmatrix} 0 & \pi \\ \pi^{\dagger} & 0 \end{pmatrix} + \begin{pmatrix} -2\Delta_2 & 0 \\ 0 & \gamma_2/2 + \Delta_2 \end{pmatrix} \\[10pt] H^{eff}_{b-} = -\frac{1}{2m} \begin{pmatrix} 0 & (\pi^{\dagger})^2 \\ \pi^2 & 0 \end{pmatrix} + \sqrt{2}v_3 \begin{pmatrix} 0 & \pi \\ \pi^{\dagger} & 0 \end{pmatrix} + \begin{pmatrix} \gamma_2/2 + \Delta_2 & 0 \\ 0 & -2\Delta_2 \end{pmatrix} \\[10pt] \text{where} \quad 1 / 2m = v_0^2 / (\sqrt{2}\gamma_1)[1 + O(\gamma_4 / \gamma_0)^2]

Benchmark

The typical values of parameters (in the unit of eV) :

γ0\gamma_0 γ1\gamma_1 γ2\gamma_2 γ3\gamma_3 γ4\gamma_4 γ5\gamma_5 δ\delta Δ2\Delta_2 α\alpha
3.16 0.39 -0.028 0.315 0.12 0.018 0.0355 0.0035 0.3

(α\alpha describes the imperfect screening effect)

Brief Review of ABA TLG

Here are some papers on ABA TLG

  • Band Structure

    1 : Landau level crossing points in ABA TLG at D=0V/nm\mathbf{D=0V/nm} can be reflected in the longitudinal magneto-resistance, and they can be used to determine the SWMC parameters.

    2 : Δ2\Delta_2 is an important parameter in the low-energy physics of ABA TLG

    9 : this theoretical paper tells me that Δ1=edD+4πe2d2εr(n1n3)\Delta_1 = ed\mathbf{D} + \frac{4\pi e^2 d}{2\varepsilon_r}(n_1 - n_3)(screened by outer layers) and Δ2=4πe2d6εrn2\Delta_2 = - \frac{4\pi e^2 d}{6\varepsilon_r}n_2(caused by nonzero charge density on the middle layer), and how to tell whether the disorder strengthen is strong or not (by scanning D\mathbf{D} at B=0T\mathbf{B} = 0 T to see how sharp the peak at D=0V/nm\mathbf{D} = 0 V/nm is).

    3 : hBN substrate can have influence on γ2,Δ2\gamma_2, \Delta_2, the picture is that with substrate, charges from TLG would move towards hBN, which makes the wave functions of the top and bottom layers of TLG shift away from each other and increases the effective distance between them. Therefore, γ2\gamma_2 is suppressed while Δ2\Delta_2 is enhanced.

    4 : ABA TLG at charge neutrality and D=0V/nm\mathbf{D=0V/nm} is a semimetal with zero bandgap, but under very high pressure (up to 60GPa), the bandgap can be several eV. To get the gap size, the author uses the following equation

    1R(T)=1R1(T)exp[E12kBT]+1R2(T)exp[E22kBT]+1R3(T)exp[(T3T)1/3]\frac{1}{R(T)} = \frac{1}{R_1(T)}exp[-\frac{E_1}{2k_BT}] + \frac{1}{R_2(T)}exp[-\frac{E_2}{2k_BT}] + \frac{1}{R_3(T)}exp[-(\frac{T_3}{T})^{1/3}]

    (the first two terms describe the Arrhenius-like thermally activated conduction process, and the third term describes a Mott variable-range hopping process) to fit the R-T curves.

    5 : This theoretical paper tells me that the screening coefficient α0.3\alpha \approx 0.3 at not very strong D\mathbf{D}, through first principle calculation.


    NOTE

    Q : Why Δ1=edD+4πe2d2εr(n1n3)\Delta_1 = ed\mathbf{D} + \frac{4\pi e^2 d}{2\varepsilon_r}(n_1 - n_3) and Δ2=4πe2d6εrn2\Delta_2 = - \frac{4\pi e^2 d}{6\varepsilon_r}n_2 ?

    The density on each layer is n1,n2,n3n_1, n_2, n_3. In the absence of D\mathbf{D}, the mirror symmetry requires n1n3n_1 \equiv n_3. As a result, the electric field from these two layers cancel each other, so only the electric field from middle layer should be considered. However, in the presence of non-zero D\mathbf{D}, n1n_1 doesn’t need to be equal with n3n_3, there is a charge imbalance which leads to another net internal electric field (n1n3)\propto (n_1 - n_3).


  • Symmetry Breaking

    6 : This theoretical paper discusses the strong-magnetic-field physics of duodectet subspace of N=0N = 0 Landau level in ABC TLG and ABA TLG, using the Hartree-Fock simulation. A Hund’s rules of LL filling order is predicted in ABA TLG.

    7 : At low magnetic field, single particle Quantum Hall picture works well while at higher magnetic field, complete filling of the degeneracy of Lowest Landau Levels (LLL) appears, indicating the physics of quantum hall ferromagnetism. The Dn\mathbf{D-n} measurement at fixed B\mathbf{B} also suggests the importance of Coulomb interaction at higher B\mathbf{B}.

    8 : Quantum Hall Ferromagnetism at finite B\mathbf{B}, and Fractional QH phases at D=0V/nm\mathbf{D} = 0V/nm which disappears at non-zero D\mathbf{D}.

    10 : Quantum Hall Ferromagnetism at finite B\mathbf{B} and hysteresis behavior in longitudinal resistance.

    11 : At low magnetic field B=1.5T\mathbf{B} = 1.5T, a symmetry-breaking state at ν=1\nu = 1 emerges which mainly concentrate in one gully, breaking the C3C_3 symmetry. (penetration field capacitance measurement Cp=c22c+nμC_p = \frac{c^2}{2c + \frac{\partial n}{\partial \mu}})

  • Edge State

    12 : At low magnetic field B0.5T\mathbf{B} \approx 0.5T, a helical edge state, named “quantum parity hall state” emerges at D=0V/nm\mathbf{D} = 0V/nm and ν=0\nu = 0. The phase diagram at ν=0\nu = 0 as a function of D,B,B\mathbf{D, B_{\perp}, B_{\parallel}} is also discussed.

Reference

[1] Quantum Hall effect and Landau-level crossing of Dirac fermions in trilayer graphene | Nature Physics

[2] Phys. Rev. Lett. 117, 066601 (2016) - Landau Level Splittings, Phase Transitions, and Nonuniform Charge Distribution in Trilayer Graphene (aps.org)

[3] Phys. Rev. Lett. 125, 246401 (2020) - Substrate-Dependent Band Structures in Trilayer \mathrm{Graphene}/h\text{\ensuremath{-}}\mathrm{BN} Heterostructures (aps.org)

[4] Large bandgap of pressurized trilayer graphene | PNAS

[5] Phys. Rev. B 92, 245416 (2015) - Density functional theory calculations of the electric-field-induced Dirac cones and quantum valley Hall state in ABA-stacked trilayer graphene (aps.org)

[6] Phys. Rev. B 85, 165139 (2012) - Hund’s rules for the N=0N=0 Landau levels of trilayer graphene (aps.org)

[7] Broken Symmetry Quantum Hall States in Dual-Gated ABA Trilayer Graphene | Nano Letters (acs.org)

[8] Phys. Rev. Lett. 117, 076807 (2016) - Tunable Symmetries of Integer and Fractional Quantum Hall Phases in Heterostructures with Multiple Dirac Bands (aps.org)

[9] Phys. Rev. B 79, 125443 (2009) - Gate-induced interlayer asymmetry in ABA-stacked trilayer graphene (aps.org)

[10] Strong electronic interaction and multiple quantum Hall ferromagnetic phases in trilayer graphene | Nature Communications

[11] Phys. Rev. Lett. 121, 167601 (2018) - Emergent Dirac Gullies and Gully-Symmetry-Breaking Quantum Hall States in ABAABA Trilayer Graphene (aps.org)

[12] Quantum parity Hall effect in Bernal-stacked trilayer graphene | PNAS

Simulation Codes

Band Structure

Landau Level

Hartree Fock Simulation

Above introduce the single-particle simulation of tri-layer ABA graphene, if you are interested in the Hartree Fock simulation of it, please click the link to the page.


ABA TLG
http://example.com/2023/04/26/ABA_TLG/
Author
Shijie Fang
Posted on
April 26, 2023
Licensed under