# Spin Polarization Curve of Fractional Quantum Hall States with Filling Factor Smaller than 2

Authors: Shosuke Sasaki
Journal: ISRN Condensed Matter Physics
 (2013)

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## Abstract

Kukushkin et al. have measured the electron spin polarization versus magnetic field in the fractional quantum Hall states. The polarization curves show wide plateaus and small shoulders. The 2D electron system is described by the total Hamiltonian (HD+HI). Therein, HD is the sum of the Landau energies and classical Coulomb energies. HI is the residual interaction yielding Coulomb transitions. It is proven for any filling factor that the most uniform electron configuration in the Landau states is only one. The configuration has the minimum energy of HD. When the magnetic field is weak, some electrons have up-spins and the others down-spins. Then, there are many spin arrangements. These spin arrangements give the degenerate ground states of HD. We consider the partial Hamiltonian only between the ground states. The partial Hamiltonian yields the Peierls instability and is diagonalized exactly. The sum of the classical Coulomb and spin exchange energies has minimum for an interval modulation between Landau orbitals. Using the solution with the minimum energy, the spin polarization is calculated which reproduces the wide plateaus and small shoulders. The theoretical result is in good agreement with the experimental data.


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## Body

## 1. Introduction

In this paper, we examine the electron spin polarization in the FQH states with the filling factorν<2. Before the examination, we see here the investigations on the FQHE briefly.The fractional quantum Hall effect was discovered by Pan et al. [1, 2]. The quasi particle with a fractional charge and its wave function were introduced by Laughlin using the variational method [3, 4]. Many physicists developed it [5–7]. Jain proposed the composite fermion theory [8, 9]. Thereafter, the FQH states with the nonstandard filling factors have been investigated by employing various methods as in the references [10–14]. These theories assume the various types of the quasi particles and their mixing. On the other hand, Tao and Thouless [15, 16] examined the case that the lowest Landau levels are partially filled with electrons. Their method is very important to investigate the FQH states. We have developed the Tao-Thouless theory and have found the most uniform configuration of electrons. It has been proven that the configuration is unique for any filling factor [17]. The configuration minimizes the expectation value of the total Hamiltonian.The Coulomb transitions conserve thex component of the total momentum where the x-direction indicates the current direction. The conservation law produces energy gaps for the specific filling factors. For the other filling factors, we have found the gapless structure and peak structure [17–23]. The theory can well explain the behaviors of the FQHE at ν<2 without any quasi particles.On the other hand, the electrons in FQH states withν>2 occupy the higher Landau levels. Therein, many experimental and theoretical investigations [24–43] are carried out and various interesting phenomena are discovered. Also, the electron spin polarization is investigated for the FQH states with ν>2 by the papers [44–49].Although the function form of the spin-polarization versus magnetic field is very important for investigating the FQHE, there is no theoretical calculation of the function shape quantitatively. In this paper, we calculate the spin-polarization versus magnetic field atν<2 by developing the previous method in the references [17–23].Kukushkin et al. have measured the electron spin-polarization versus magnetic field [50]. They clarified the function forms at twelve filling factors in ν<2. Although their experiments are rather old data, the obtained function forms give us the important knowledge.Their results are shown in Figure1. Hereafter, we describe the electron spin polarization by the symbol γe. Then, γe=1 means a fully polarized state. The experimental polarization curves have the following properties.(1)
The wide plateau appears atγe=0 for ν=2/3 and 2/5.The wide plateau appears atγe=1/3 for ν=3/5 and 3/7.The two wide plateaus appear atγe=0 and 1/2 for ν=4/7 and 4/9.(2)
The small shoulder appears atγe=1/2 for ν=2/3 and 2/5.The small shoulder appears atγe=2/3 for ν=3/5 and 3/7.The two small shoulder appear atγe=1/4 and 3/4 for ν=4/7 and 4/9.Figure 1
Experimental data of the polarization for the FQH states [50].As shown in Figure1, the ν=2/3 polarization curve resembles the ν=2/5 curve. The ν=3/5 polarization curve resembles the ν=3/7 curve. Also, the ν=4/7 curve resembles the ν=4/9 curve. The characters mentioned above indicate that the shape of the spin polarization curves depends mainly upon the numerator of the filling factor. The numerator means the electron number per unit configuration (see [17]). Therefore the polarization belongs to electrons (not holes). In this paper we clarify the origin of the polarization curve.We shortly describe the fundamental properties of the quasi-two-dimensional electron system below. We illustrate a quantum Hall device where the directions of the axesx, y, and z are taken, as in Figure 2. Then, the vector potential, A, has the components:
(1)A=(-yB,0,0),
where B is the strength of the magnetic field.Figure 2
Quantum Hall device.The Hamiltonian,H0, of a single electron in the absence of the Coulomb interaction between electrons is given by
(2)H0=(p+eA)22m*+U(y)+W(z),
where U(y) and W(z) indicate the potentials confining electrons to an ultrathin conducting layer in Figure 2. Therein m* is an effective mass of electron, and p=(px,py,pz) is the electron momentum. The effective mass m* differs from material to material and the value in GaAs is about 0.067 times the free electron mass. The eigenvalue problem of this Hamiltonian is solved and the single electron wave function ψL,J is expressed as follows:
(3)ψL,J(x,y,z)=1ℓexp(ipxℏ)uLHL(m*ωℏ(y-αJ))×exp(-m*ω2ℏ(y-αJ)2)ϕ(z),
where αJ is given by
(4)αJ=p(eB)=[2πℏ(ℓeB)]J,ℓ:lengthof2Delectronsystem.Therein,ϕ(z) is the wave function of the ground state along the z-direction, HL is the Hermite polynomial of Lth degree and uL is the normalization constant. We call L the Landau level number. Also, the eigenenergy is given by
(5)EL,J=λ+U(αJ)+ℏω(L+12),(L=0,1,2,3,…),
where λ is the ground state energy along the z-direction U(αJ) is the potential energy in the y-direction.When there are many electrons, the total Hamiltonian is given by(6a)HT=∑i=1NH0,i+∑i=1N-1∑j>iNe24πε(xi-xj)2+(yi-yj)2+(zi-zj)2,
where N is the total number of electrons and H0,i is the single particle Hamiltonian of the ith electron without the Coulomb interaction as
(6b)H0,i=(pi+eA)22m*+U(yi)+W(zi).The many-electron state is characterized by a set of Landau level numbersL1,L2,…,LN and a set of momenta p1,p2,…,pN. The complete set is composed of the Slater determinant as
(7)Ψ(L1,…,LN;p1,…,pN)=1N!|ψL1,p1(x1,y1,z1)⋯ψL1,p1(xN,yN,zN)⋮⋮ψLN,pN(x1,y1,z1)⋯ψLN,pN(xN,yN,zN)|.This state is the eigenstate of∑i=1NH0,i. The expectation value of the total Hamiltonian is denoted as W(L1,…,LN;p1,…,pN) which is given by
(8)W(L1,…,LN;p1,…,pN)=∑i=1NELi(pi)+C(L1,…,LN;p1,…,pN),
where C is the expectation value of the Coulomb interaction defined by
(9)C(L1,…,LN;p1,…,pN)=∫⋯∫Ψ(L1,…,LN;p1,…,pN)*×∑i=1N-1∑j>iNe24πε(xi-xj)2+(yi-yj)2+(zi-zj)2×Ψ(L1,…,LN;p1,…,pN)dx1dy1dz1,…,dxNdyNdzN.Hereafter, we callC(L1,…,LN;p1,…,pN) “classical Coulomb energy.” We divide the total Hamiltonian HT into two parts HD, and HI, as follows:
(10)HD=∑L1,…,LN∑p1,…,pN|Ψ(L1,…,LN;p1,…,pN)〉×W(L1,…,LN;p1,…,pN)×〈Ψ(L1,…,LN;p1,…,pN)|(11)HI=HT-HD,
where HI is constructed only by the off-diagonal elements and depends upon only the relative coordinate. Therefore, the total momentum of the x-direction conserves in this system. That is to say, the sum of the initial momenta pi and pj is equal to the sum of the final momenta pi′ and pj′ via Coulomb transition as follows:
(12)pi′+pj′=pi+pj.At a filling factor smaller than 1, the ground state ofHD satisfies the following properties.(1)
N electrons exist in the lowest Landau levels with L1=L2=⋯=LN=0.(2)
The electrons most uniformly occupy the lowest Landau levels. Then, the classical Coulomb energy takes the lowest one. The electron momentap1,p2,…,pN are related to each centre position as in (4).For any filling factor, we can find only one electron configuration in Landau states which is the most uniform and has the minimum energy ofHD. The proof is done in [17]. When the magnetic field is weak, there are many spin arrangements for a given configuration. These spin arrangements construct the degenerate ground states of HD. The interaction Hamiltonian HI yields the quantum transitions among the ground states. We examine the interaction in the next section.


## 2. Coulomb Interaction between Up- and Down-Spin States (Equivalency between Coulomb Transition and Spin Exchange Interaction)

The degenerate ground states have the same momentum set corresponding to the most uniform electron configuration. The HamiltonianHI given by (11) acts between two electrons. We indicate the spin states by ↑ and ↓ for up- and down-spins, respectively. Then, all the initial spin states with the momentum pair p1, p2 are described as
(13)|p1↑,p2↑〉,|p1↑,p2↓〉,|p1↓,p2↑〉,|p1↓,p2↓〉.When these states transfer viaHI, their final states are described as follows:
(14)|p1′↑,p2′↑〉,|p1′↑,p2′↓〉,|p1′↓,p2′↑〉,|p1′↓,p2′↓〉,
where p1′ and p2′ indicate the final momenta via the Coulomb interaction. We consider the transitions only between the degenerate ground states. Therefore, the final momentum set should have the minimum energy of HD. Accordingly, the final momentum set is equivalent to the initial momentum set because of the uniqueness of the electron configuration for the ground state of HD:
(15)p1′=p2,p2′=p1.The case of(p1′=p1,p2′=p2) is removed because the diagonal matrix elements of HI are zero. Applying (15), the final state |p1′↑,p2′↑〉 becomes |p2↑,p1↑〉 which is the same as its initial state. Also |p1′↓,p2′↓〉 becomes |p2↓,p1↓〉. In the two cases, the final state is identical to the initial state, and therefore the matrix elements of HI are zero. Accordingly, nonzero matrix elements are (16a)〈p2↑,p1↓|HI|p1↑,p2↓〉,(16b)〈p2↓,p1↑|HI|p1↓,p2↑〉,where(17)〈p2↓,p1↑|HI|p1↓,p2↑〉=〈p2↑,p1↓|HI|p1↑,p2↓〉.We examine the following three cases.Case 1.
Considerp1-p2=±2πℏ/ℓ(18a)ξ=〈p2↑,p1↓|HI|p1↑,p2↓〉.Case 2.
Considerp3-p4=±4πℏ/ℓ(18b)η=〈p4↑,p3↓|HI|p3↑,p4↓〉.Case 3.
Considerp5-p6=±6πℏ/ℓ(18c)ς=〈p6↑,p5↓|HI|p5↑,p6↓〉.The Coulomb transition in Case1 is shown in Figure 3. The open circle indicates the up-spin state and the filled one the down-spin state. We describe the momenta after the transition by the symbols p1′ and p2′ which are given by
(19)p1′=p1+2πℏℓ=p2,p2′=p2-2πℏℓ=p1.Figure 3
Equivalence of a specific Coulomb transition and spin exchange interaction for Case1.This Coulomb transition is equivalent to the following process: The spin at site 1 flips from up to down and the spin at site 2 flips from down to up simultaneously without changing the momenta. Thus, the Coulomb transition of Case1 is equivalent to a spin exchange process which is described by the interaction ξσ1-σ2+. Therein, σ+ is the spin transformation operator from down- to up-spin state and σ- is the adjoint operator of σ+. There is another Coulomb transition given by (16b) which is equivalent to ξσ1+σ2-. Accordingly, the Coulomb transition between the two electrons at sites 1 and 2 is expressed as
(20)ξ(σ1-σ2++σ1+σ2-),
where ξ was already defined by (18a). In this Coulomb transition, the classical Coulomb energy in the initial state is exactly equal to the one in the final state.The Coulomb transition of Case2 is illustrated in Figure 4.Figure 4
Equivalence of a specific Coulomb transition and spin exchange interaction for Case2.The momenta after the transition are described by the symbolsp3′, p4′, the values of which are given by
(21)p3′=p3+2×2πℏℓ=p4,p4′=p4-2×2πℏℓ=p3.The Coulomb interaction of Case2 is equivalent to the following spin exchange interaction between an electron pair placed in second nearest-neighboring orbital pair:
(22)η(σ3-σ4++σ3+σ4-),
where η is the coupling constant defined by (18b).The Coulomb transition of Case3 is illustrated in Figure 5. The momenta after the transition are described by the symbols p5′, p6′:
(23)p5′=p5+3×2πℏℓ=p6,p6′=p6-3×2πℏℓ=p5.Figure 5
Equivalence of a specific Coulomb transition and spin exchange interaction for Case3.Figure5 shows that the Coulomb interaction of Case 3 is equivalent to the following spin exchange interaction between an electron pair placed in third nearest-neighboring orbital pair:
(24)ς(σ5-σ6++σ5+σ6-).We show the most uniform configuration of electrons for the two cases ofν=2/3 and ν=2/5 in Figures 6(a) and 6(b), respectively, where the spin-states are numbered sequentially from left to right, as indicated by the green color.(a) Coulomb transitions for first and second nearest pairs atν=2/3. (b) Coulomb transitions for the case of ν=2/5.
(a)
(b)In theν=2/3 state, the nearest electron pairs have the coupling constant ξ and the second nearest electron pairs have the coupling constant η. At ν=2/5, the nearest electron pair is placed in the second neighboring orbital pair, and so the coupling constant is η. The second nearest electron pair is placed in the third neighboring orbital pair and so the coupling constant is ζ as shown in Figure 6(b). Thus, it is noteworthy that the site number (namely, electron number) is different from the orbital number.In the third or further nearest electron pair, another electron is inserted as in Figures6(a) and 6(b). Therefore the interaction between the third nearest electrons is quite weak due to the screening effect of the interposing electron, and so may be neglected.Atν=2/3 the most effective interaction is obtained as follows:
(25)Heffective=∑j=1,2,3,…[ξ(σ2j-1+σ2j-+σ2j-1-σ2j+)+η(σ2j+σ2j+1-+σ2j-σ2j+1+)],
where the operator σ2j-1+ indicates the transformation from a down- to up-spin state of the electron at the (2j-1)th site. This Hamiltonian, (25), yields the quantum transition between the degenerate ground states. When the external magnetic field is applied in the z-direction, the Hamiltonian becomes
(26)H=∑j=1,2,3,…[ξ(σ2j-1+σ2j-+σ2j-1-σ2j+)+η(σ2j+σ2j+1-+σ2j-σ2j+1+)]+∑i=1,2,3,…μBg*B(12)σiz,
where g* is the effective g-factor, B is the magnetic field, (1/2)σz is the electron spin operator in the z-direction, and μB is the Bohr magneton. The matrices (the Pauli spin matrices) are explicitly indicated below:
(27)σ+=(0100),σ-=(0010),σz=(100-1).We can obtain the HamiltonianH for other filling factors. The Hamiltonian for ν=3/5 and ν=4/7 is given, respectively by;
(28)H=∑j=1,2,3,…[ξ(σ3j-2+σ3j-1-+σ3j-2-σ3j-1+)+η(σ3j-1+σ3j-+σ3j-1-σ3j+)+η(σ3j+σ3j+1-+σ3j-σ3j+1+)]+∑i=1,2,3,…μBg*B(12)σiz,forν=35,H=∑j=1,2,3,…[ξ(σ4j-3+σ4j-2-+σ4j-3-σ4j-2+)+η(σ4j-2+σ4j-1-+σ4j-2-σ4j-1+)+η(σ4j-1+σ4j-+σ4j-1-σ4j+)+η(σ4j+σ4j+1-+σ4j-σ4j+1+)]+∑i=1,2,3,…μBg*B(12)σiz,forν=47.The three Hamiltonians, (26) and (28), can be exactly diagonalized by using the method of [51].


## 3. Isomorphic Mapping from the FQH State to One-Dimensional Fermion State

We examine the following mapping from a single spin state to a fermion state. The down-spin state|↓〉 is mapped to the vacuum state |0〉, and the up-spin state |↑〉 is mapped to the one fermion state c*|0〉, where c* is the creation operator
(29)|↑〉⟶c*|0〉,|↓〉⟶|0〉.Then the spin operatorsσ+, σ-, and σz are mapped to the operators of the fermion system as
(30)σ+⟶c*,σ-⟶c,σz⟶(2c*c-1).Next, we find the isomorphic mapping from many-spin states to many-fermion states. Two examples of the mapping are written as follows:(31)|↑1,↑2,↓3,↑4,↓5,↓6,↑7,↑8,↓9,↑10〉⟷c1*c2*c4*c7*c8*c10*|0〉,|↓1,↑2,↑3,↓4,↓5,↑6,↓7,↓8,↓9,↑10〉⟷c2*c3*c6*c10*|0〉.It is noteworthy that the multiplying order of creation operators is the same as the order of the up-spins. The operatorsci and ci* satisfy the anticommutation relations as follows: (32a){ci,cj*}=ci×cj*+cj*×ci=δi,j,(32b){ci,cj}=ci×cj+cj×ci=0,{ci*,cj*}=ci*×cj*+cj*×ci*=0.The products of the spin operatorsσ2j-1+σ2j-,  σ2j-1-σ2j+,  σ2j+σ2j+1- and σ2j-σ2j+1+ are mapped to the products of fermion operators as follows: (33a)σ2j-1+σ2j-⟷c2j-1*c2j,σ2j-1-σ2j+⟷-c2j-1c2j*,(33b)σ2j+σ2j+1-⟷c2j*c2j+1,σ2j-σ2j+1+⟷-c2jc2j+1*,(33c)σiz⟷2ci*ci-1.It has been verified that the mapping ((33a), (33b), (33c)) is isomorphic (see [51]). Accordingly Hamiltonian (26) is equivalent to the following form:
(34)H=∑j=1,2,3,…[ξ(c2j-1*c2j-c2j-1c2j*)+η(c2j*c2j+1-c2jc2j+1*)]+∑i=1,2,3,…μBg*B(12)(2ci*ci-1).We exactly solve the eigenvalue problem of this Hamiltonian.New operatorsaj and bj are introduced as follows:
(35)aj=c2j-1,bj=c2j,aj*=c2j-1*,bj*=c2j*,
where j is the cell number. Then the Hamiltonian (34) becomes
(36)H=∑j=1J[ξ(aj*bj-ajbj*)+η(bj*aj+1-bjaj+1*)]+∑j=1JμBg*B(12)(2aj*aj+2bj*bj-2),
where J is the total number of cells given by J=N/2 for the total number of electrons N. We apply a Fourier transformation given by (38) to the operators aj, aj*, bj, and bj* and obtain
(37)H=∑p[(eipb*(p)a(p)+e-ipa*(p)b(p))ξ(a*(p)b(p)+b*(p)a(p))+η(eipb*(p)a(p)+e-ipa*(p)b(p))]+∑pμBg*B(12)(2a*(p)a(p)+2b*(p)b(p)-2),(38)wherean=1J∑peipna(p),bn=1J∑peipnb(p),
and p=(2π/J)×integer, -π<p≤π.In the summation in (37), the single term with a value of p is expressed by the following matrix:
(39)(μBg*Bξ+ηe-ipξ+ηeipμBg*B).This matrix has two eigenvaluesλ1(p) and λ2(p) which are given by (40a)λ1(p)=μBg*B-ξ2+η2+2ξηcosp,(40b)λ2(p)=μBg*B+ξ2+η2+2ξηcosp.Using new annihilation operatorsA1(p) and A2(p)(41a)A1(p)=12a(p)-ξ+ηe-ip2(ξ2+η2+2ξηcosp)b(p),(41b)A2(p)=12a(p)+ξ+ηe-ip2(ξ2+η2+2ξηcosp)b(p),the Hamiltonian (37) is expressed as follows:
(42)H=∑p(λ1(p)A1*(p)A1(p)+λ2(p)A2*(p)A2(p)-μBg*B).Thus, we have succeeded in diagonalizing the Hamiltonian (26).


## 4. Magnetic Field Dependence of the Spin Polarization

The electron spin polarizationγe depends upon the temperature. We calculate its thermodynamic mean value as follows:
(43)γe=1N〈-∑i=1Nσiz〉=-1N〈∑i=1N(2ci*ci-1)〉=-1N〈∑j=1J(2aj*aj+2bj*bj-2)〉=-12J〈∑p(2a*(p)a(p)+2b*(p)b(p)-2)〉=-12J〈∑p(∑s=12(2As*(p)As(p)-1))〉,
where 〈·〉 is the thermal average and the minus sign comes from the negative charge of an electron. In deriving (43), we have utilized (33c), (35), (38), (41a), and (41b). The diagonal form, (42), indicates that all the eigenenergy states are expressed by the direct product of the creation operators As*(p). Therefore, the eigenstate is identified by the eigenvalue of the fermion number operator ns(p)=As*(p)As(p). Equation (42) shows that the eigenenergy for ns(p)=1 is λs(p) and the eigenenergy for ns(p)=0 is zero. Then, the Boltzmann factor is exp(-λs(p)/kBT) for ns(p)=1 and exp(-0/kBT) for ns(p)=0, where kB is the Boltzmann constant and T is the temperature. Accordingly, the probability for ns(p)=1 is given by exp(-λs(p)/kBT)/(1+exp(-λs(p)/kBT)) and the probability for ns(p)=0 is given by 1/(1+exp(-λs(p)/kBT)). These probabilities yield the thermal average of As*(p)As(p) as
(44)〈As*(p)As(p)〉=exp(-λs(p)/kBT)1+exp(-λs(p)/kBT),
which gives
(45)〈2As*(p)As(p)-1〉=exp(-λs(p)/kBT)-11+exp(-λs(p)/kBT)=-tanh(λs(p)(2kBT)).Substitution of (45) into (43) derives
(46)γe=12J∑p(∑s=12tanh(λs(p)2kBT)).Since the total number of electrons is a macroscopic value, we can replace the summation by integration as(47)γe=14π∫-ππdp(∑s=12tanh(λs(p)2kBT)).Thus, the electron-spin polarization at the filling factor of 2/3 is expressed by (47).First, we study the low field behavior of the spin polarization.Equation (40a) indicates that λ1(p) is restricted to the following region:(48a)μBg*B-|ξ+η|≤λ1(p)≤μBg*B-|ξ-η|.Also,λ2(p) is in the region:
(48b)μBg*B+|ξ-η|≤λ2(p)≤μBg*B+|ξ+η|.When the magnetic field takes a value between 0 and|ξ-η|/(μBg*),  λ1(p) is negative and λ2(p) is positive for any value of p:
(49)λ1(p)<0,λ2(p)>0for0<B<|ξ-η|(μBg*).Therefore,tanh(λ1(p)/2kBT) is nearly equal to -1 and tanh(λ2(p)/2kBT) is nearly equal to 1 at very low temperatures (T≈0). Then, the spin polarization is almost zero because the summation in the right hand side of (47) is nearly equal to zero. When the magnetic field increases beyond the value |ξ-η|/(μBg*), the spin polarization increases continuously until it reaches the maximum value of 1. This behavior is in agreement with the experimental data in Figure 1.When the quality of a quantum Hall device is bad, many random potentials are produced by the impurities and lattice defects. Then, the plateau in the Hall resistance curve is rounded at both ends. Also, the wide plateau in the polarization curve is rounded by the random potentials. The effect resembles that of the thermal vibration. Therefore, we include the random potential effect into the value ofT.The spin polarization is evaluated by integrating the right hand side of (47). The integration has been done by using a computer program. The result is shown in Figure 7 for the parameter η/ξ=0.2 and kBT/ξ=0.1. Experimental data [50] are plotted by red dots in the figure. We find that the theoretical result reproduces the experimental data without the small shoulder.Figure 7
Calculated spin polarization curve forν=2/3  η/ξ=0.2  kBT/ξ=0.1. Red dots are experimental data [50].


## 5. Modulation of the Intervals between Landau Orbitals

If we observe carefully the experimental spin polarization curves, namely, Figure1, we find small shoulders in it. The structure has not been considered in the previous sections. We examine the origin of the small shoulders in this section.Peierls studied an electron system in a one-dimensional crystal and considered the lattice distortion with the period doubling the unit cell. This lattice distortion produces new band gaps and the energy becomes lower than that without the distortion. This effect is called spin Peierls effect [52].In the present theory, the spin polarization of FQH states is derived from the Hamiltonians (26), (28), and so on. If we consider a new modulation of the intervals between the nearest Landau orbitals with the period doubling the unit configuration, the spin chain Hamiltonian of FQH system resembles the one with the spin Peierls effect.As an example, forν=2/3 we change the distance between nearest orbitals in the first unit-configuration longer, the one in the second unit-configuration shorter, and so on. Then, we have the four kinds of the coupling constants ξ, ξ′, η, and η′ as shown in Figure 8.Figure 8
Coupling constants of interactions caused by distortion with double period.The value ofξ′ is larger than that of ξ because the distance for the ξ′ interaction path is shorter than that for ξ. Also, η′>η holds. This distortion with the double period of the unit-configuration produces additional energies via the classical Coulomb and spin exchange interactions. We call it “interval modulation.” We calculate the total energy of this system.We express the distance between the nearest orbitals by the symbolr0 for nondistortion case. We consider the distortion that one of the distances between the nearest Landau orbitals becomes r0+d for an odd cell number and the other one becomes r0-d for an even cell number. Then, the classical Coulomb energy W increases. The increasing value per electron is proportional to d2 as
(50)ΔWN=fC(dr0)2,
where fC is the constant parameter.Next, we examined-dependence of the coupling constants ξ and ξ′. When d>0, the coupling constant ξ is weaker than ξ′ because the ξ interaction path is longer than that of ξ′. When the other case d<0, ξ is stronger than ξ′ because the ξ interaction path becomes shorter than that of ξ′. Therefore, there is a linear term of d in ξ and ξ′ as follows:
(51)ξ=ξ0-fξ(dr0),ξ′=ξ0+fξ(dr0),
where ξ0 is the coupling constant in the non-distortion case and fξ is the proportionality constant. In order to simplify (50) and (51), we define a new dimensionless quantity t as
(52)t=(fξξ0)(dr0).Then the coupling constantsξ and ξ′ are expressed as
(53)ξ=ξ0(1-t),ξ′=ξ0(1+t).The increasing value of the classical Coulomb energyΔW is also expressed by this dimensionless parameter t as follows:(54a)ΔWN=ξ0Ct2,
where C is the dimensionless coefficient as
(54b)C=ξ0fCfξ2.


## 6. Total Energy due to the Interval Modulation

Now we calculate the spin exchange energy. Using the coupling constants in Figure8, the spin exchange Hamiltonian for ν=2/3 is given by
(55)H=∑j=1,2,3,…[ξ(σ4j-3+σ4j-2-+σ4j-3-σ4j-2+)+η(σ4j-2+σ4j-1-+σ4j-2-σ4j-1+)]+∑j=1,2,3,…[ξ′(σ4j-1+σ4j-+σ4j-1-σ4j+)+η′(σ4j+σ4j+1-+σ4j-σ4j+1+)]+∑j=1,2,3,…μBg*B(12)(σ4j-3z+σ4j-2z+σ4j-1z+σ4jz).This Hamiltonian equation (55) is rewritten from (33a), (33b), and (33c) as
(56)H=∑j=1,2,3,…[ξ(c4j-3*c4j-2-c4j-3c4j-2*)+η(c4j-2*c4j-1-c4j-2c4j-1*)]+∑j=1,2,3,…[ξ′(c4j-1*c4j-c4j-1c4j*)+η′(c4j*c4j+1-c4jc4j+1*)]+∑i=1,2,3,…μBg*B(12)(2ci*ci-1).Using the cell numberj, we introduce new operators a1,j, a2,j, a3,j, and a4,j as follows:
(57)a1,j=c4j-3,a2,j=c4j-2,a3,j=c4j-1,a4,j=c4j.Fourier transformation yields(58)a1,j=1J∑peipja1(p),a2,j=1J∑peipja2(p),a3,j=1J∑peipja3(p),a4,j=1J∑peipja4(p),
where J is the total number of unit cells (unit configurations), namely, J=N/4, and p=(2π/J)×integer(-π<p≤π). Substitution of (57) and (58) into (56) gives
(59)H=∑p[{(eipa4*(p)a1(p)+e-ipa1*(p)a4(p))}+η′(eipa4*(p)a1(p)+e-ipa1*(p)a4(p))ξ(a1*(p)a2(p)+a2*(p)a1(p))+η(a2*(p)a3(p)+a3*(p)a2(p))+ξ′(a3*(p)a4(p)+a4*(p)a3(p))+η′(eipa4*(p)a1(p)+e-ipa1*(p)a4(p))]+∑pμBg*B(12)(2(a1*(p)a1(p)+a2*(p)a2(p)+a3*(p)a3(p)+a4*(p)a4(p))-4).For one value ofp, (59) is expressed by the following matrix M:
(60)M=(μBg*Bξ0η′e-ipξμBg*Bη00ημBg*Bξ′η′eip0ξ′μBg*B).The four eigenvalues ofM are denoted by the symbols λ1(p), λ2(p), λ3(p), and λ4(p)(λ1≤λ2≤λ3≤λ4) which are given by(61)λ1(p)=μBg*B-12((1+α2)(ξ2+ξ′2)+(1+α2)2(ξ2+ξ′2)2-4((1+α4)ξ2ξ′2-2α2ξ2ξ′2cosp)),λ2(p)=μBg*B-12((1+α2)(ξ2+ξ′2)-(1+α2)2(ξ2+ξ′2)2-4((1+α4)ξ2ξ′2-2α2ξ2ξ′2cosp)),λ3(p)=μBg*B+12((1+α2)(ξ2+ξ′2)-(1+α2)2(ξ2+ξ′2)2-4((1+α4)ξ2ξ′2-2α2ξ2ξ′2cosp)),λ4(p)=μBg*B+12((1+α2)(ξ2+ξ′2)+(1+α2)2(ξ2+ξ′2)2-4((1+α4)ξ2ξ′2-2α2ξ2ξ′2cosp)).Therein, we have assumedξ′/ξ=η′/η because the interval modulation is expected to give almost the same effect to the coupling constants ξ and η. Then, ratios between the coupling constants are expressed as follows:(62a)α=ηξ=η′ξ′=η0ξ0,(62b)β=ξ′ξ=η′η.Equation (53) derives the following relation:
(62c)β=ξ′ξ=η′η=(1+t)(1-t).We show the two eigenvaluesλ1(p) and λ2(p) by red and black curves for β=1, blue curves for β=1.2 and green curves for β=1.4 in Figure 9.Figure 9
Eigenvalues of spin interaction via interval modulation. The red and black curves indicate the eigenvaluesλ1/(μBg*) and λ2/(μBg*) at β=1.0. The blue curves indicate the eigenvalues λ1/(μBg*) and λ2/(μBg*) at β=1.2. The green curves indicate the eigenvalues λ1/(μBg*) and λ2/(μBg*) at β=1.4.The differenceλ2(p)-λ1(p) is minimal at p=π as seen in Figure 9. When β is equal to 1, namely, nondistorted case, the energy gap λ2(π)-λ1(π) disappears. Equation (61) gives the difference between λ2(p) and λ1(p) at p=π as
(63)λ2(π)-λ1(π)=1+α2|ξ′-ξ|=1+α2ξ02t,
where we have used (53). Thus, the difference is proportional to t.The eigenenergies (61) give the diagonal form of the Hamiltonian as follows:
(64)H=∑p(λ1(p)A1*(p)A1(p)+λ2(p)A2*(p)A2(p)+λ3(p)A3*(p)A3(p)+λ4(p)A4*(p)A4(p)-2μBg*B).We calculate the thermal average of this Hamiltonian. The thermal average has been already examined in (44) which gives
(65)〈As*(p)As(p)〉=exp(-λs(p)/kBT)1+exp(-λs(p)/kBT)=12(1-tanh(λs(p)(2kBT))).Then the thermal average of the spin exchange energy is(66)〈H〉=∑p[{∑s=14(λs(p)12(1-tanh(λs(p)(2kBT))))}-2μBg*B{{∑s=14(λs(p)12(1-tanh(λs(p)(2kBT))))}}].Since the total number of electrons is a macroscopic value, we can replace the summation by integration as(67a)〈H〉N=14×12π∫p=02π[{∑s=14(λs(p)12(1-tanh(λs(p)2kBT)))-2μBg*B{(λs(p)12(1-tanh(λs(p)2kBT)))}}]dp,(67b)ΔETotalN=(ΔWN)+Δ(〈H〉N).We numerically calculate the integration in (67a). The dashed red curve in Figure 10 shows the calculated result of the spin exchange energy for B=2.2[T]. The dashed black curve shows the classical Coulomb energy for the parameter C=0.5 (see (54a), (54b)). Their sum ΔETotal/N is the total energy which is expressed by the blue curve. Then, the total energy has a minimum at a nonzero t as shown in Figure 10. Consequently, the interval modulation actually occurs.Figure 10
Dependence of total energy upont.Further calculations are carried out for various values of the magnetic field in the case ofC=0.5. The results are shown in Figure 11. The interval modulation occurs in the region of 1.86[T]<B<2.51[T].Figure 11
Dependence of total energy upont.We calculate the total energy for another caseν=3/5. The coupling constants are illustrated in Figure 12.Figure 12
Coupling constants forν=3/5.These coupling constants in Figure12 produce the ν=3/5 Hamiltonian which is described by the following matrix:
(68)(μBg*Bξ000η′e-ipξμBg*Bη0000ημBg*Bη0000ημBg*Bξ′0000ξ′μBg*Bη′η′eip000η′μBg*B)(forν=35).This matrix has the six eigenvaluesλ1(p), λ2(p), λ3(p), λ4(p), λ5(p), and λ6(p), the p-dependences of which are shown in Figure 13.Figure 13
p-Dependence of the six eigen values for ν=3/5.Figure13 indicates the energy gap between λ1(p) and λ2(p) at p=π. Then the interval modulation with the double period of the unit-configuration produces additional energies for the classical-Coulomb and spin-exchange interactions, respectively, as follows: (69a)ΔWN=ξ0Ct2,(69b)〈H〉N=16×12π∫p=02π[{∑s=16(λs(p)12(1-tanh(-λs(p)2kBT)))-3μBg*B{(1-tanh(-λs(p)2kBT))}}]dp,(69c)ΔETotalN=(ΔWN)+Δ(〈H〉N).The sum ofΔW/N and Δ(〈H〉/N) is numerically calculated and the result is shown in Figure 14.Figure 14
Dependence of the total energy upont at ν=3/5.Therein, we have used the parameterC=0.5 which is the same as in the case of ν=2/3. This value of C affects the shape of the polarization curve near the small shoulder as will be discussed in the next section. We have shown the dependence of the total energy upon t for two examples. Therein, the total energy has a minimum at a nonzero t in some region, of the magnetic field. The nonzero t yields the stable state with the distortion (interval modulation). This mathematical mechanism is the one resembling the spin Peierls effect.


## 7. Spin Polarization in the Case with the Interval Modulation

We calculate the spin-polarizationγe for the Hamiltonian (64) of ν=2/3. γe is obtained by the integration as
(70)γe=14×12π∫-ππdp(∑s=14tanh(λs(p)2kBT)),
where the four eigen-values λ1(p), λ2(p), λ3(p), and λ4(p) are given in (61). We numerically calculate the spin-polarization γe by the following two methods, namely, easy method A and precise method B. Method A is the rough calculation under the fixed value of the distortion parameter t. This method has been studied in the previous papers [53, 54] where the ratio ξ′/ξ=η′/η is treated to be a constant value as in (62c). Method B is newly performed in this paper.Method A. We use the fixed value ξ′/ξ=η′/η=1.4. The other parameters are adopted to be η/ξ=η′/ξ′=0.1 and (kBT/ξ0)=0.05. Then, a small energy gap appears between λ1 and λ2. We numerically calculate the integration in (70) by using the Mathematica program and draw the graph of spin-polarization versus magnetic field. A small shoulder appears in the theoretical curve of the electron spin-polarization as seen in Figure 15.Figure 15
Method A: theoretical curve of the spin polarization forν=2/3. Blue dots are experimental data [50].This curve is slightly different from the experimental data near the sharp cornersP and Q. So we choose the different value as (kBT/ξ0)=0.1 in order to make the curvature  small in the corners P and Q. Then the small shoulder disappears. In order to maintain the small shoulder, we take a larger value ξ′/ξ=η′/η=1.8. The calculated curve is shown in Figure 16. The result is also different from the experimental data. Accordingly, method A has some difficulty in explaining the experimental data. This inadequacy is improved by using the precise method B.Figure 16
Method A: calculated curve of the spin polarization forν=2/3 (the parameters are chosen as (kBT/ξ0)=0.1 and ξ′/ξ=η′/η=1.8).Method B. We carry out more precise method B where we calculate the t-dependence of the total energy per electron. We find the minimum point of the total energy for various values of the magnetic field. Some examples have been already shown in Figures 11 and 14. Therein, we obtain the t-value of the minimum point. The t-value gives ξ′/ξ=η′/η from (62c). We numerically calculate the spin-polarization by using the magnetic field dependence of the minimum point. The theoretical curve is shown in Figure 17.Figure 17
Method B: calculated curve of the spin-polarization forν=2/3.Therein, we have used the parameter valuesη/ξ=η′/ξ′=0.2, (kBT/ξ0)=0.1 and C=0.5. Thus, the calculated result is in good agreement with the experimental data. The reason is simply discussed below.The magnetic field is sufficiently strong near the cornerP in Figure 15. In this region, almost all the spins have a down direction. Then the number of up- and down-spin pairs decreases and the spin exchange energy becomes small. So the increment of the total energy (67b) is nearly equal to that of the classical Coulomb energy. Then, the energy minimum appears at t=0, namely, non-distortion (noninterval modulation). Thus, the distortion appears only near the small shoulder as in Figure 11. Due to this situation, the theoretical curve via method B is in good agreement with the experimental data.It is examined how the shape of the polarization curve depends on the parameterC. We calculate the polarization curve for the following two cases: C=0.4 and 0.65 in Figure 18. These calculations make it clear that the shape of the curve varies only in the neighborhood of the small shoulder when changing the parameter C.Figure 18
Spin-polarization via method B for two cases withC=0.4 and 0.65.We study the case ofν=3/5. The spin-polarization γe for ν=3/5 is given by
(71)γe=16×12π∫-ππdp(∑s=16tanh(λs(p)2kBT)),
where the eigen-energies λs for s=1,2,3,4,5,6 are numerically obtained from the matrix (68). The two calculated curves via methods A and B are shown in Figure 19. In the method B we have applied C=0.5, same as in ν=2/3. As seen in the right panel of Figure 19, the theoretical result via method B is in good agreement with the experimental data.Figure 19
Spin-polarization forν=3/5. Blue dots are experimental data [50].We next examine the case ofν=4/7. The most uniform electron configuration is illustrated in Figure 20.Figure 20
Coupling constants forν=4/7.The electron configuration yields theν=4/7 Hamiltonian described by the following matrix (72): (72)(μBg*Bξ00000η′e-ipξμBg*Bη000000ημBg*Bη000000ημBg*Bη000000ημBg*Bξ′000000ξ′μBg*Bη′000000η′μBg*Bη′η′eip00000η′μBg*B),(forν=47).The spin polarizationγe is given by
(73)γe=18×12π∫-ππdp(∑s=18tanh(λs(p)2kBT)).Therein, the eight valuesλs for s=1,2,3,4,5,6,7,8 indicate the eigenenergies of the matrix (72).The spin polarization can be evaluated from the eigen-energies. The results are shown in Figure21. Method B has used the same value 0.5 for the parameter C.Figure 21
Spin-polarization forν=4/7. Blue dots are experimental data [50].


## 8. Effect Resembling Spin Peierls One in Various Filling Factors

We examine the cases ofν=2/5, ν=3/7, and ν=4/9 with the interval modulation, the most uniform electron configurations of which are illustrated in Figures 22(a), 22(b), and 22(c), respectively.(a) Coupling constants forν=2/5. (b) Coupling constants for ν=3/7. (c) Coupling constants for ν=4/9.
(a)
(b)
(c)These electron configurations and their coupling constants yield the spin-exchange Hamiltonians forν=2/5, ν=3/7, and ν=4/9, which are represented by the following matrices, respectively:(74a)
(
μ
B
g
*
B
η
0
ς
′
e
-
i
p
η
μ
B
g
*
B
ς
0
0
ς
μ
B
g
*
B
η
′
ς
′
e
i
p
0
η
′
μ
B
g
*
B
)
,
(
for
ν
=
2
5
)
,
(74b)
(
μ
B
g
*
B
η
0
0
0
ς
′
e
-
i
p
η
μ
B
g
*
B
η
0
0
0
0
η
μ
B
g
*
B
ς
0
0
0
0
ς
μ
B
g
*
B
η
′
0
0
0
0
η
′
μ
B
g
*
B
η
′
ς
′
e
i
p
0
0
0
η
′
μ
B
g
*
B
)
,
(
for
ν
=
3
7
)
,
(74c)
(
μ
B
g
*
B
η
0
0
0
0
0
ς
′
e
-
i
p
η
μ
B
g
*
B
η
0
0
0
0
0
0
η
μ
B
g
*
B
η
0
0
0
0
0
0
η
μ
B
g
*
B
ς
0
0
0
0
0
0
ς
μ
B
g
*
B
η
′
0
0
0
0
0
0
η
′
μ
B
g
*
B
η
′
0
0
0
0
0
0
η
′
μ
B
g
*
B
η
′
ς
′
e
i
p
0
0
0
0
0
η
′
μ
B
g
*
B
)
,
(
for
ν
=
4
9
)
.The average value ofξ and ξ′ is equal to ξ0. Also η0=(η+η′)/2. Accordingly, we obtain
(75)η=η0(1-t),η′=η0(1+t).The ratios between the coupling constants satisfy the following relations:(76a)ςη=ς′η′=ς0η0,(76b)η′η=ς′ς=(1+t)(1-t).In order to compare the spin exchange and classical Coulomb energies, we reexpress thet-dependence of the classical Coulomb energy by using the coupling constant η0,
(77)ΔWN=η0Dt2,
where D is a new coefficient. Using the eigenvalues of the matrices, the spin polarization is given by(78a)γe=14×12π∫-ππdp(∑s=14tanh(λs(p)2kBT)),(forν=25),(78b)γe=16×12π∫-ππdp(∑s=16tanh(λs(p)2kBT)),(forν=37),(78c)γe=18×12π∫-ππdp(∑s=18tanh(λs(p)2kBT)),(forν=49).We numerically calculate the spin-polarization curves via method B, the results of which are shown in Figure23.Figure 23
Spin-polarization forν=2/5, 3/7, 4/9. Blue dots are experimental data [50].It is found in these figures that the small shoulders originate from the interval modulation (distortion with double period). The theoretical curve via method B is in better agreement with the experimental data than the one via method A. Here we shortly discuss the parameter valuesC and D. The increasing value of the classical Coulomb energy is expressed in (69a) and (77) as ΔW/N=ξ0Ct2 and ΔW/N=η0Dt2, respectively. So the parameter D may be almost equal to(79a)D≈C×ξ0η0.The fitting values ofD are 0.65, 1.5, and 1.2 for ν=2/5, 3/7, and 4/9, respectively, as shown in Figure 23. We cannot understand why the parameter D is small at ν=2/5.We examine the remaining cases ofν=3/7 and 4/9. The ratio η/ξ=η′/ξ′=η0/ξ0 is 0.35 for ν=3/5 and 4/7. Substituting this value 0.35 for η0/ξ0 into (79a) we obtain the value D as
(79b)D≈0.50.35≈1.43which is the predicted value from the casesν=3/5 and 4/7. The fitting values of D are 1.5 and 1.2 for ν=3/7 and 4/9 as seen in Figure 23. These fitting values are consistent with 1.43 derived from the cases of ν=3/5 and 4/7.Next, we examine the spin polarization atν=4/3, 7/5, and 8/5. The most uniform electron-configuration and the coupling constants are shown in Figures 24(a), 24(b), and 24(c), respectively.(a) Coupling constants forν=4/3. Double line indicates a Landau orbital occupied by an electron pair with up- and down-spins. (b) Coupling constants for ν=7/5. Double line indicates a Landau orbital occupied by an electron pair with up- and down-spins. (c) Coupling constants for ν=8/5. Double line indicates a Landau orbital occupied by an electron pair with up- and down-spins.
(a)
(b)
(c)There are doubly occupied orbitals in these electron configurations of Figures24(a), 24(b), and 24(c). The spin exchange forces act between electrons in singly occupied orbitals. The electron pairs in doubly occupied orbitals have no polarization because of cancellation by up- and down-spin pairs. Therefore, the electron spin polarization is given by the following equations:
(80)γe=24×14×12π∫-ππdp(∑s=14tanh(λs(p)2kBT)),(forν=43),(81)γe=37×16×12π∫-ππdp(∑s=16tanh(λs(p)2kBT)),(forν=75),(82)γe=28×14×12π∫-ππdp(∑s=14tanh(λs(p)2kBT)),(forν=85),
where the coefficients 2/4, 3/7, and 2/8 in (80), (81) and (82) come from the fact that the up- and down-spin pairs cancel the polarization. That is to says, two electrons per four electrons at ν=4/3 have no polarization, four electrons per seven electrons at ν=7/5 have no polarization, and also six electrons per eight electrons at ν=8/5 have no polarization.We numerically calculate the spin-polarization curves via method B, the results of which are shown in Figure25.Figure 25
Spin-polarization forν=4/3, 7/5, and 8/5. Blue dots are experimental data [50].The polarization data atν=4/3 has a very sharp change of the curvature from B=8.5[T] to B=11.5[T]. Therefore, it is very difficult for the function to fit the experimental data. The theoretical result via method B is in good agreement with the experimental data at ν=4/3, as seen in the left panel of Figure 25.The coupling constants atν=8/5 are τ,κ,τ′,κ′ as in Figure 24(c). Accordingly, the coupling constants and the classical Coulomb energy are reexpressed by using τ0 as follows:(83a)τ=τ0(1-t),τ′=τ0(1+t),(83b)ΔWN=τ0Et2,whereE is a new coefficient. The fitting value is E=0.35 for ν=8/5. This value is different from C and D. Probably the reason is the shielding effect of the electron pair in doubly occupied Landau orbitals in Figure 24(c).In the case ofν=4/3 and ν=7/5, the fitting values of C are 0.7 and 0.5, respectively. These values are nearly equal to C=0.5 in ν=2/3, 3/5, and 4/7. The parameters η/ξ, (kBT/ξ0), C and others may be dependent upon the gate voltage, sample, temperature, and so on. We have used almost same value for C. If we use different values of C, we can find a better fitting to the experimental data than the present results. As seen in Figures 17, 19, 21, 23, and 25, the small shoulders are caused by the interval modulation which comes from the Peierls instability.


## 9. Conclusions

(1)
Theν=2/3 polarization curve resembles the ν=2/5 curve, the ν=3/5 one resembles the ν=3/7 one, and the ν=4/7 one resembles the ν=4/9 one. Thus, the shape of the spin polarization curves depends mainly upon the numerator of the filling factor. Therefore, the polarization belongs to electrons (not holes).
(2)
At low field, up- and down-spins coexist. Then, there are many degenerate ground states which are composed of different spin arrangements for a given electron-configuration in the Landau orbitals. These many electron states have the same eigenenergy ofHD. We have succeeded to diagonalize the partial Hamiltonian describing the Coulomb transitions among the degenerate ground states. The calculated results reproduce the wide plateaus in the spin polarization curves of the experimental data [50].
(3)
The experimental curve of the polarization versus magnetic field exhibits small shoulders. These small shoulders are derived from the following mechanism. We have exactly solved the partial Hamiltonian and also have minimized the total energy (sum of the spin exchange and classical Coulomb energies). The total energy decreases by modulating the intervals between Landau orbitals with the doubly period. Calculating thet-value with the minimum energy, we have found that the interval modulation actually occurs. Then, the theoretical polarization curve reproduces the small shoulder and the wide plateau. The results of the present theory are in good agreement with the experimental data.



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*Source: 489519-2013-10-07.xml*