Differential Poisson's ratio of a crystalline two-dimensional membrane
I.S. Burmistrov, V. Yu. Kachorovskii, I.V. Gornyi, A.D. Mirlin
DDi ff erential Poisson’s ratio of a crystalline two-dimensionalmembrane I.S. Burmistrov a,b,c,d, ∗ , V. Yu. Kachorovskii e,c,a , I.V. Gornyi c,d,e,a , A.D. Mirlin c,d,f,a a L.D. Landau Institute for Theoretical Physics, Kosygina street 2, 119334 Moscow, Russia b Laboratory for Condensed Matter Physics, National Research University Higher School of Economics, 101000 Moscow, Russia c Institut f¨ur Nanotechnologie, Karlsruhe Institute of Technology, 76021 Karlsruhe, Germany d Institut f¨ur Theorie der kondensierten Materie, Karlsruhe Institute of Technology, 76128 Karlsruhe, Germany e A. F. Io ff e Physico-Technical Institute, 194021 St. Petersburg, Russia f Petersburg Nuclear Physics Institute, 188300, St.Petersburg, Russia
Abstract
We compute the di ff erential Poisson’s ratio of a suspended two-dimensional crystalline mem-brane embedded into a space of large dimensionality d (cid:29)
1. We demonstrate that, inthe regime of anomalous Hooke’s law, the di ff erential Poisson’s ratio approaches a univer-sal value determined solely by the spatial dimensionality d c , with a power-law expansion ν = − / + . / d c + O (1 / d c ), where d c = d −
2. Thus, the value − / d c → ∞ . Keywords: crystalline membrane, Poisson’s ratio
1. Introduction
Poisson’s ratio is defined as the ratio of a transverse compression to a longitudinal stretching.In the classical theory of elasticity, the Poisson’s ratio is given by ν cl = λ µ + ( D − λ , where µ and λ are the Lam´e coe ffi cients and D is the dimensionality of the elastic body [1].General conditions of thermodynamic stability restrict the Poisson’s ratio to the range between − / ( D − ∗ Corresponding author. Fax: + Email address: [email protected] (I.S. Burmistrov)
Preprint submitted to Annals of Physics January 17, 2018 a r X i v : . [ c ond - m a t . m e s - h a ll ] J a n n interesting example of auxetic material is a crystalline membrane of dimension D em-bedded into the space of dimension d > D . The self-consistent theory of such crystalline mem-branes [6] predicts the negative Poisson’s ratio in the thermodynamic limit. This limit is achievedin large membranes, when the membrane size L exceeds the Ginzburg length L ∗ ∼ κ / √ T µ ,where κ is the bending rigidity and T stands for the temperature. A crystalline membrane hosts d c = d − D soft out-of-plane modes, the so-called flexural phonons, which are characterizedby strong anharmonicity mediated by the coupling to conventional in-plane phonons [7]. As aconsequence of such anharmonicity, the elastic moduli show a nontrivial power-law scaling withthe system size, temperature, and tension. The scaling of all elastic moduli, λ, µ, κ is controlledby the universal exponent η which depends only on d c . The critical exponent η was determinedwithin several approximate analytical schemes[6, 8–11], none of which being controllable in thephysical case D = d =
3. Numerical simulations for the latter case yielded η = . ± . η = . ± .
04 [13], and η = .
85 [14]. It is because of the nontrivial scaling of the elasticmoduli that the linear Hooke’s law fails in the regime of small tension [9, 15–20].Le Doussal and Radzihovsky [6] found a negative Poisson’s ratio of a two-dimensional crys-talline membrane within the self-consistent screening approximation. More specifically, theyobtained an entirely universal value ν = − / d c .In Ref. [18], this result of the self-consistent membrane theory was reproduced by Kosmrljand Nelson by means of a renormalization-group analysis for a relatively large membrane size L (cid:29) L ∗ and not too strong tension, σ (cid:28) σ ∗ = κ L − ∗ . On the other hand, as shown by thepresent authors together with Katsnelson and Los in a parallel paper [22], the Poisson’s ratiois strongly dependent of boundary conditions in the range of lowest tensions (linear-responseregime), σ (cid:38) σ L = κ L η − L − η ∗ . An independence on boundary conditions is reached only atstronger tensions, σ (cid:29) σ L . However, also in this regime, one should exert a care when defin-ing the Poisson’s ratio. Specifically, emergence of the anomalous, non-linear Hooke’s law re-sults in an essential di ff erence between the absolute and di ff erential Poisson’s ratio, as shown inRef. [22].In this paper, we consider the non-linear regime σ L (cid:28) σ (cid:28) σ ∗ and focus on the di ff erentialPoisson’s ratio. In order to define the di ff erential Poisson’s ratio ν , one needs to consider theresponse to an infinitesimally small anisotropic tension: σ (cid:107) = σ + δσ and σ ⊥ = σ . Then,the ratio of the infinitesimally small change in transverse, δε ⊥ , and longitudinal, δε (cid:107) , stretchingdetermines the di ff erential Poisson’s ratio ν = − δε ⊥ δε (cid:107) . (1)We demonstrate that in the regime σ L (cid:28) σ (cid:28) σ ∗ the di ff erential Poisson’s ratio indeed acquiresa universal value. However, contrary to the result of the self-consistent membrane theory, thisuniversal value depends on the dimensionality d c of embedded space. We perform calculationswhich are controlled by the small parameter 1 / d c (cid:28) ff erential Poisson’sratio of the two-dimensional crystalline membrane is given by ν = − + . d c + O (cid:0) d − c (cid:1) , σ L (cid:28) σ (cid:28) σ ∗ . (2)Thus, the di ff erential Poisson’s ratio at σ L (cid:28) σ (cid:28) σ ∗ is universal (in the sense of independenceon material parameters) but represents a nontrivial function of d c .The paper is organized as follows. In Sec. 2 we present the general formalism for the compu-tation of the di ff erential Poisson’s ratio of a two-dimensional crystalline membrane. The details2f evaluation of the di ff erential Poisson’s ratio to the first order in 1 / d c are presented in Sec. 3.We end the paper with a summary of results, Sec. 4. Technical details are given in Appendices.
2. Formalism
We start with the partition function of a two-dimensional crystalline membrane written interms of the functional integral over in-plane, u = { u x , u y } , and out-of-plane, h = { h , . . . , h d c } phonons (see Refs. [20, 21, 23]): Z = (cid:90) D [ u , h ] exp( − S ) . (3)Here the action in the imaginary time is given by ( β = / T ) S = β (cid:90) d τ (cid:90) d x (cid:40)(cid:104) µ δ αβ + λ (cid:105)(cid:104)(cid:16) ξ α − + K α (cid:17) (cid:16) ξ β − + K β (cid:17) − K α K β (cid:105) + ρ (cid:104) ( ∂ τ u ) + ( ∂ τ h ) (cid:105) + κ (cid:104) ( ∆ h ) + ( ∆ u ) (cid:105) + µ u αβ u βα + λ u αα u ββ (cid:41) , (4)where u αβ = (cid:16) ξ β ∂ α u β + ξ α ∂ β u α + ∂ α u ∂ β u + ∂ α h ∂ β h (cid:17) , (5)and K α = β L β (cid:90) d τ (cid:90) d x K α , K α = ∂ α u ∂ α u + ∂ α h ∂ α h . (6)The free energy per unit area, f = − T L − ln Z , is a function of the stretching factors ξ x and ξ y , i.e. f ≡ f ( ξ x , ξ y ). With the function f ( ξ x , ξ y ), the diagonal components of the tension tensorcan be found as σ x = ξ x ∂ f ∂ξ x , σ y = ξ y ∂ f ∂ξ y . (7)We emphasize that Eq. (7) determines the dependence of the tension tensor { σ x , σ y } on thestretching tensor { ξ x , ξ y } , i.e., Eq. (7) is the equation of state.In order to find the di ff erential Poisson’s ratio ν , we consider the case of slightly anisotropicstretching factors, ξ α = ξ + δε α , and adjust the ratio ν = − δε y /δε x in such a way that thecomponents of the tension tensor, σ x = σ + δσ and σ y = σ , di ff er only by an infinitesimaladdition δσ in σ x . Then, we find ν = (cid:32) ∂σ y ∂ξ x (cid:33) ξ y (cid:32) ∂σ y ∂ξ y (cid:33) ξ x = ∂ f ∂ξ y ∂ξ x ∂ f ∂ξ y − σ . (8)Here the derivatives are taken at ξ x = ξ y = ξ . 3e note that instead of independent variables ξ x and ξ y , one can choose as independentvariables the components of the tension tensor, σ x and σ y . Equation (8) can be then rewritten inan alternative form: ν = − (cid:32) ∂ξ y ∂σ x (cid:33) σ y (cid:32) ∂ξ x ∂σ x (cid:33) σ y = − ∂ g ∂σ x ∂σ y ∂ g ∂σ x , (9)where the derivatives are assumed to be calculated for σ x = σ y = σ . As usual, the free energy g ( σ x , σ y ) is related to the free energy f ( ξ x , ξ y ) via the Legendre transform: g ( σ x , σ y ) = f ( ξ x , ξ y ) − σ x ( ξ x − / − σ y ( ξ y − / , (10)where ξ α is expressed in terms of σ α with the help of the equation of state (7). We note thatthe expression (9) has been used for the numerical evaluation of the Poisson’s ratio in Ref. [24](though with the di ff erent form of the free energy). Although, both formulations (8) and (9)are completely equivalent, in what follows we will use the formulation in which the stretchingfactors ξ α are the independent variables.Using the exact form (4) of the action, one finds the following expressions for the secondderivatives of the partition function f : ∂ f ∂ξ y ∂ξ x (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ξ x = ξ y = ξ = ξ λ − ξ (cid:90) d τ (cid:48) d x (cid:48) (cid:104)(cid:104) L y ( x , τ ) · L x ( x (cid:48) , τ (cid:48) ) (cid:105)(cid:105) , (11) ∂ f ∂ξ y (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ξ x = ξ y = ξ = σ + ξ (2 µ + λ ) + (2 µ + λ ) (cid:10) ( ∂ y u y ) (cid:11) + µ (cid:10) ( ∂ x u y ) (cid:11) − µ + λξ (cid:104) u yy ∂ y u y (cid:105)− µξ (cid:104) u xy ∂ x u y (cid:105) − ξ (cid:90) d τ (cid:48) d x (cid:48) (cid:104)(cid:104) L y ( x , τ ) · L y ( x (cid:48) , τ (cid:48) ) (cid:105)(cid:105) . (12)Here the average (cid:104) . . . (cid:105) is defined with respect to the action (4), (cid:104)(cid:104) A · B (cid:105)(cid:105) = (cid:104) AB (cid:105) − (cid:104) A (cid:105)(cid:104) B (cid:105) and L x = µ + λ K x + λ K y + µ + λ ξ u xx ∂ x u x + µξ u yx ∂ y u x + λξ u yy ∂ x u x , (13) L y = µ + λ K y + λ K x + µ + λ ξ u yy ∂ y u y + µξ u xy ∂ x u y + λξ u xx ∂ y u y . (14)Equations (8), (11), and (12) express the Poisson’s ratio in terms of correlation functions ofelastic deformations. The actual computation of these correlation functions of the in-plane andflexural phonons is complicated due to interaction between these phonon modes.Below we limit the analysis to the case of high temperature, T (cid:29) κ / ( µ L ) in which onecan consider the phonons to be quasistatic. In this regime, one can also neglect the term ∂ α u ∂ β u in comparison with ∂ α h ∂ β h in the expressions for ˜ u αβ and K α . Then we can simplify Eqs. (11)and (12). Indeed, by making the following change of variables: u α → ξ α u α , we can recast thepartition function (3) as: Z = (cid:90) D [ u , h ] exp( − ˜ E / T ) , (15)4here ˜ E = (cid:90) d x (cid:40)(cid:34) µ δ αβ + λ (cid:35)(cid:34)(cid:18) ξ α − + ˜ K α (cid:19) (cid:18) ξ β − + ˜ K β (cid:19) − ˜ K α ˜ K β (cid:35) + κ ∆ h ) + µ ˜ u αβ ˜ u βα + λ u αα ˜ u ββ (cid:41) . (16)Here we have introduced the following notations:˜ u αβ = (cid:16) ∂ α u β + ∂ β u α + ∂ α h ∂ β h (cid:17) , ˜ K α = L (cid:90) d x ˜ K α , ˜ K α = ∂ α h ∂ α h . (17)Since the action (16) becomes quadratic in the in-plane phonons, we can integrate them outand express the partition function as an integral over static flexural phonons, Z = (cid:90) D [ h ] exp( − E / T ) , (18)where the energy E for a given configuration of the flexural phonon field h ( x ) is given by [20] E = (cid:90) d x (cid:34) µ δ αβ + λ (cid:35)(cid:34)(cid:18) ξ α − + ˜ K α (cid:19) (cid:18) ξ β − + ˜ K β (cid:19)(cid:35) + µ L (cid:32)(cid:90) d x ∂ x h ∂ y h (cid:33) + κ (cid:90) d x ( ∆ h ) + µ ( µ + λ )4(2 µ + λ ) (cid:90) (cid:48) d k d k (cid:48) d q (2 π ) [ k × q ] q [ k (cid:48) × q ] q (cid:0) h k + q h − k (cid:1)(cid:0) h − k (cid:48) − q h k (cid:48) (cid:1) . (19)The ‘prime’ sign in the last integral means that the interaction with q = q = u αβ ˜ u βα and ˜ u αα ˜ u ββ to the energy ˜ E in Eq. (16) hasbeen combined with the term ˜ K α ˜ K β , yielding exactly the term with ∂ x h ∂ y h in Eq. (19).Since now ξ α does not enter the interaction part of the free energy which depends on u , weobtain a much simpler equation of state: (cid:32) σ x σ y (cid:33) = M (cid:32) ξ x − + (cid:104) ˜ K x (cid:105) ξ y − + (cid:104) ˜ K y (cid:105) (cid:33) , M = (cid:32) µ + λ λλ µ + λ (cid:33) . (20)Here the average (cid:104) . . . (cid:105) is with respect to the energy (19). The second derivatives of the freeenergy with respect to the stretching factors become ∂ f ∂ξ y ∂ξ x (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ξ x = ξ y = ξ = ξ λ − ξ β (cid:90) d x (cid:48) (cid:104)(cid:104) ˜ L y ( x ) · ˜ L x ( x (cid:48) ) (cid:105)(cid:105) , (21) ∂ f ∂ξ y (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ξ x = ξ y = ξ = σ + ξ (2 µ + λ ) − ξ β (cid:90) d x (cid:48) (cid:104)(cid:104) ˜ L y ( x ) · ˜ L y ( x (cid:48) ) (cid:105)(cid:105) , (22)where ˜ L α = M αβ ˜ K β / E involves two types of interaction of flexural phonons. The termsin the first line of Eq. (19) correspond to the interaction with zero momentum transfer (’zero5ode’). In the case of large membrane size, σ (cid:29) σ L , this interaction can be treated in therandom phase approximation. Then, we find β (cid:90) d x (cid:48) (cid:104)(cid:104) ˜ K α ( x ) · ˜ K β ( x (cid:48) ) (cid:105)(cid:105) = Π (cid:32) + M Π (cid:33) − αβ , (23)where Π denotes the polarization operator (at zero momentum) irreducible with respect to theinteraction with the zero-momentum transfer: Π αβ = β (cid:90) d x (cid:48) (cid:104)(cid:104) ˜ K α ( x ) · ˜ K β ( x (cid:48) ) (cid:105)(cid:105) irr . (24)We note that Π αβ has two independent components: Π xx = Π yy and Π xy = Π yx . Using Eqs. (21)and (22), we express the di ff erential Poisson’s ratio in terms of the components of Π : ν = ν − Y Π xy / + Y Π xx / . (25)Here ν = λ µ + λ , Y = µ ( µ + λ )2 µ + λ (26)denote the bare values of the Poisson’s ratio and Young modulus for the two-dimensional crys-talline membrane, respectively.In order to clarify the meaning of Π xx and Π xy , it is useful to consider a general form of thepolarization operator at finite momentum q :ˆ Π αβ,γδ ( q ) = (cid:90) d x (cid:48) e − i q ( x − x (cid:48) ) (cid:104)(cid:104) (cid:0) ∇ α h ( x ) ∇ β h ( x ) (cid:1) · (cid:0) ∇ γ h ( x (cid:48) ) ∇ δ h ( x (cid:48) ) (cid:1) (cid:105)(cid:105) . (27)Due to the rotation symmetry and the symmetry under permutation of the indices α and β (aswell as γ and δ ), the polarization operator at zero momentum is expressed as follows [16]ˆ Π αβ,γδ (0) = Π xy δ αβ δ γδ + (cid:0) Π xx − Π xy (cid:1)(cid:0) δ αγ δ βδ + δ αδ δ βγ (cid:1) . (28)We emphasize that, in general, there are no reasons for ˆ Π αβ,γδ to be fully symmetric with respectto permutations of all its indices as it is assumed in the self-consistent screening approximation[6, 25]. Therefore, Eq. (25) yields the most general expression for the di ff erential Poisson’s ratio.We also note that Eq. (25) can be written as (see Appendix A) ν = λ (cid:48) µ (cid:48) + λ (cid:48) , (29)where λ (cid:48) and µ (cid:48) are the screened Lam´e coe ffi cients: µ (cid:48) = µ + ( Π xx − Π xy ) µ , B (cid:48) = B + ( Π xx + Π xy ) B . (30)Here we have introduced bare and screened bulk moduli: B = µ + λ and B (cid:48) = µ (cid:48) + λ (cid:48) , respectively.In order to find how ν depends on parameters of the problem, e.g., on the number of flexuralphonon modes d c , one needs to compute Π xx and Π xy . In the next section we remind the reader6n the results of the self-consistent screening approximation and then compute corrections in1 / d c .Using the equation of state (20), we can express the stretching factors ξ α via tensions σ α .Then, with the help of Eq. (9), we find the following representation for the di ff erential Poisson’sratio ν = ν + Y (cid:32) ∂ (cid:104) ˜ K y (cid:105) ∂σ x (cid:33) σ y − Y (cid:32) ∂ (cid:104) ˜ K x (cid:105) ∂σ x (cid:33) σ y . (31)Here (cid:104) ˜ K α (cid:105) is expressed in terms of σ x and σ y . After taking derivatives in Eq. (31), one sets σ x = σ y = σ . Below we demonstrate how the two representations of the di ff erential Possionratio, (25) and (31), are related.The irreducible polarization operator at the zero momentum can be exactly expressed via thefull triangular vertex Γ β ( k , k ): Π αβ = (cid:90) d k (2 π ) k α G k Γ β ( k , k ) . (32)Here G k = T κ k + ( ξ α − M αβ k β / − Σ k denotes the propagator for the flexural phonons, with Σ k being the exact self-energy. The barevalue of the triangular vertex Γ β ( k , k ) is equal to k β / T . The full triangle vertex satisfies thefollowing identity: Γ β ( k , k ) = ∂ G − k ∂σ β . (33)As a consequence of this identity, we obtain Π αβ = − ∂∂σ β (cid:90) d k (2 π ) k α G k = − ∂ (cid:104) ˜ K α (cid:105) ∂σ β . (34)Therefore, expressions (25) and (31) are identical.
3. Evaluation of the di ff erential Poisson’s ratio The interaction between flexural phonons with finite momentum transfer results in renormal-ization of the bending rigidity at k (cid:28) q ∗ ≡ L − ∗ [6, 10, 16], κ → κ ( k ) = κ ( q ∗ / k ) η f ( k / q σ ) . (35)Here, q σ = q ∗ ( σ/σ ∗ ) / (2 − η ) and the function f ( x ) has the following asymptotic behavior: f ( x ) = , x (cid:29) , x − η , x (cid:28) . (36)7he simplest approach for computing the irreducible polarization operator is to neglect the vertexcorrections. As we shall see below, this can be justified for d c (cid:29)
1. Then, we find Π (0) αβ = d c (cid:90) d k (2 π ) T k α k β G k . (37)Independently of the form of the exact propagator G k , we find the irreducible polarization oper-ator as Π (0) αβ = d c γ (cid:104) n α n β (cid:105) n , γ = (cid:90) d k (2 π ) T k G k . (38)Here n stands for the two-dimensional unit vector and (cid:104) . . . (cid:105) n denotes the averaging over direc-tions of n . Thus, neglecting the vertex corrections yields the following relation: Π (0) xx = Π (0) xy . (39)Relation (39) implies that ˆ Π αβ,γδ is fully symmetric with respect to permutation of indices. Thisassumption is used in the self-consistent screening approximation.Motivated by the renormalization of bending rigidity (35) and the Ward identity (see Ap-pendix B), we use the following ansatz for the exact propagator: G k = T κ ( k ) k + σ k . (40)The integral over k in Eq. (37) is then dominated by k ∼ q σ and we obtain γ ∼ T κ σ (cid:18) σ ∗ σ (cid:19) η/ (2 − η ) . (41)Then, from Eq. (25) we find at σ (cid:28) σ ∗ that the di ff erential Poisson’s ratio becomes ν ≈ − Π (0) xy Π (0) xx = − . (42)It is exactly the result that was obtained within the self-consistent screening approximation [6]. Corrections to the result (42) stem from the violation of the relation (39). In order to refine thedi ff erential Poisson’s ratio, we expand the right-hand side of Eq. (25) in the di ff erence 3 Π xy − Π xx : ν ≈ − + Π xy − Π xx Π (0) xy (43)As we shall see below, the correction to the value − / / d c .There are three diagrams with non-trivial vertex corrections (see Fig. 1) that contribute to Π αβ at order d c . They yield the following corrections: Π (a) αβ = − d c (cid:90) d k d q (2 π ) T G k G k − q [ k × q ] q N (cid:48) q k α ( k β − q β ) , (44)8nd Π (b + c) αβ = d c (cid:90) d k d k (cid:48) d q (2 π ) T G k G k − q G k (cid:48) G k (cid:48) − q [ k × q ] q [ k (cid:48) × q ] q N (cid:48) q k α k (cid:48) β . (45)Here N (cid:48) q denotes the screened interaction between flexural phonons (see Appendix A), N (cid:48) q = Y / + Y Π (0) q / , (46)where Π (0) q stands for the polarization operator at finite momentum calculated without vertexcorrection. We note that Π (0) xx = Π (0) xy = Π (0) q = . The polarization operator Π (0) q is given byfollowing expression: Π (0) q = d c (cid:90) d k (2 π ) T [ k × q ] q G k G k − q . (47)Since we are interested in the regime q (cid:28) q ∗ , we can approximate N (cid:48) q by 1 / [3 Π (0) q ]. Then,combining both contributions together, we find3 Π xy − Π xx Π (0) xy = − d c Π (0) xy (cid:90) d k d k (cid:48) (2 π ) T G k G k (cid:48) (cid:40) [ k × k (cid:48) ] | k − k (cid:48) | Π (0) k − k (cid:48) − d c (cid:90) d q (2 π ) T G k − q G k (cid:48) − q × [ k × q ] q [ k (cid:48) × q ] q Π (0) q (cid:41)(cid:104) k x k (cid:48) y − k x k (cid:48) x (cid:105) . (48)We note that this expression can be written in a rotationaly invariant way. Indeed, in the first term,the expression under the integral sign depends on the angle θ between k and k (cid:48) only. Averagingover directions of k , we find (cid:90) π d φ π cos φ (cid:104) ( φ + θ ) − sin ( φ + θ ) (cid:105) = sin θ. (49)In the second term, the expression under the integral sign depends on the angles θ and θ (cid:48) between k and q , and between k (cid:48) and q , respectively. Averaging over directions of q , we find (cid:90) π d φ π cos ( φ + θ ) (cid:104) ( φ + θ (cid:48) ) − sin ( φ + θ (cid:48) ) (cid:105) = sin ( θ − θ (cid:48) ) . (50)Therefore, we obtain 3 Π xy − Π xx Π (0) xy = I ( a ) + I ( b + c ) , (51)where I ( a ) = − d c Π (0) xy (cid:90) d k d k (cid:48) (2 π ) T [ k × k (cid:48) ] | k − k (cid:48) | G k G k (cid:48) Π (0) k − k (cid:48) , I ( b + c ) = d c Π (0) xy (cid:90) d k d k (cid:48) d q (2 π ) T [ k × k (cid:48) ] [ k × q ] q Π (0) q [ k (cid:48) × q ] q Π (0) q G k − q G k (cid:48) − q G k G k (cid:48) . (52)9a) (b) (c) Figure 1: The diagrams of the first order in 1 / d c for the polarization operators Π xx and Π xy at zero momentum transfer.The solid line denotes the propagator G k . The wavy line depicts the screened interaction between flexural phonons,which is equal to 1 / [3 Π (0) q ] in the universal regime, q < q ∗ . The results (51) and (52) can be derived in a di ff erent way using the relation (33) betweenthe triangular vertex at zero momentum and the inverse Green’s function. In view of Eq. (34),in order to find the di ff erential Poisson’s ratio one needs to compute the change of the Green’sfunction upon applying an infinitesimally small tension δσ along the x direction.In the presence of δσ , the Green’s function can be written in terms of the self-energy Σ k : G k = T κ k + σ k − Σ k . (53)We mention that the ansatz (40) used above for δσ = Σ k = [ κ − κ ( k )] k . Wealso note that the trivial term δσ k x is included into Σ k for the sake of convenience.In order to find the change of Σ k induced by the infinitesimally small tension δσ , we use thelowest-order diagram for the self-energy (see Fig. 2): Σ k = (cid:90) d q (2 π ) [ k × q ] q G k − q Π (0) q , (54)We note that, as above, the dominant contribution comes from momenta q (cid:28) q ∗ such that theinteraction line is determined by the inverse polarization operator.As one can see from the diagram in Fig. 2, the variation of the self-energy in the presence of δσ arises from the variation of the Green’s function: δ G k − q = G k − q δ Σ k − q , (55)as well as from the the change of the polarization operator (see Eq. (47)) δ Π (0) q = d c (cid:90) d k (2 π ) [ k × q ] q δ G k G k − q = d c (cid:90) d k (2 π ) [ k × q ] q G k G k − q δ Σ k . (56)Now the correction δ Σ k can be found from the variation of Eq. (54): δ Σ k = − δσ k x + (cid:90) d q (2 π ) [ k × q ] q δ G k − q Π (0) q − G k − q δ Π (0) q (cid:2) Π (0) q (cid:3) . (57)Since the right-hand side of this equation is linear in δ Σ k , it can be rewritten as(1 + ˆ α ) δ Σ = − δσ k x , (58)10 igure 2: The diagram for the self energy (see text). where we formally introduce the linear integral operator ˆ α as:ˆ α δ Σ k = − (cid:90) d k (cid:48) (2 π ) G k (cid:48) ( k × k (cid:48) ) | k − k (cid:48) | Π (0) k − k (cid:48) − d c (cid:90) d q (2 π ) [ k × q ] q Π (0) q [ k (cid:48) × q ] q Π (0) q G k (cid:48) − q G k − q δ Σ k (cid:48) . (59)It is worthwhile to mention that the linear operator ˆ α conserves the angular momentum, as fol-lows from the rotational invariance of Eq. (59). Therefore, it is convenient to split ˆ α into thezeroth and second harmonics: ˆ α k x =
12 ˆ α + k + k x − k y k ˆ α − k . (60)The formal solution of Eq. (58) can be then written as δ Σ k = − δσ + ˆ α + + k x − k y k + ˆ α − k . (61)Although Eq. (61) yields a formal solution for δ Σ k , it is not justified to keep ˆ α ± beyond thelowest order: not all the terms of the order 1 / d c can be generated from the diagram in Fig. 2.After a straightforward calculation, we obtain ν ≈ − + (cid:10) ˆ α + − ˆ α − (cid:11) k , (62)where (cid:104) ˆ α ± (cid:105) k = (cid:82) d k k G k ˆ α ± k (cid:82) d k k G k . (63)Expressing the di ff erence (cid:104) ˆ α + − ˆ α − (cid:105) k in the rotationally invariant way, we obtain from Eq. (62)exactly the same expression as in Eqs. (51) and (52). As we shall see below, all the integrals determining the 1 / d c correction to the di ff erentialPoisson’s ratio are dominated by the momenta of the order of q σ . Since the dependence of thebending rigidity on q is controlled by η (cid:39) / d c , we can neglect this dependence in the calculationof the correction (51). Therefore, in what follows, we approximate the propagator of the flexuralphonons by Eq. (40) with the bare bending rigidity. Then, we find Π (0) q = d c T π κ q P (cid:32) q √ κ √ σ (cid:33) , (64)where the dimensionless function P ( Q ) is given as P ( Q ) = Q (cid:90) d K (2 π ) [ K × Q ] Q K ( K +
1) 1 | Q − K | ( | Q − K | + . (65)11he function P ( Q ) can be evaluated exactly with the help of the following set of transformations: P ( Q ) = Q ∞ (cid:90) dt dt (cid:104) − e − t (cid:105) (cid:104) − e − t (cid:105) (cid:90) d K (2 π ) [ K × Q ] Q e − t K − t | K − Q | = Q ∞ (cid:90) dt dt ( t + t ) × (cid:104) − e − t (cid:105) (cid:104) − e − t (cid:105) e − Q t t t + t = Q ∞ (cid:90) −∞ dz cosh z (cid:90) ∞ d ττ e − Q τ/ (cid:89) σ = ± (cid:104) − e − τ e σ z cosh z (cid:105) = Q ∞ (cid:90) −∞ dz cosh z (cid:40)(cid:32) + zQ (cid:33) ln (cid:32) + zQ (cid:33) − (cid:32) + e z cosh zQ (cid:33) ln (cid:32) + e z cosh zQ (cid:33)(cid:41) = (cid:40) + Q ln Q − (1 + Q ) ln(1 + Q ) Q + Q (4 + Q ) / ln (cid:112) + Q + Q (cid:41) . (66)Here we used the parameterization t , = τ e ± z cosh z . We note that the function P ( Q ) has thefollowing asymptotic behavior: P ( Q ) = Q − Q − Q ) , Q (cid:28) − Q (1 + Q ) , Q (cid:29) . (67)In particular, we find that Π (0) xy = d c T / (32 π κ σ ).Now we compute the contribution I ( a ) in Eq. (51) from the diagram in Fig. 1a. This contri-bution can be written as I ( a ) = − d c Π (0) xy (cid:90) d k d q (2 π ) T G k G k − q [ k × q ] q Π (0) q = − (32 π ) d c (cid:90) d Q (2 π ) Q P ( Q ) Y ( Q ) , (68)where the function Y ( Q ) is given by Y ( Q ) = ∞ (cid:90) dt dt (cid:89) j = , (cid:104) t j − + (2 + t j ) e − t j (cid:105) (cid:90) d K (2 π ) [ K × Q ] Q e − t K − t | K − Q | = π ∞ (cid:90) dt dt ( t + t ) (cid:89) j = , (cid:104) t j − + (2 + t j ) e − t j (cid:105) e − Q t t t + t . (69)Performing the transformation t , = τ e ± z cosh z and integrating over τ , we find Y ( Q ) = Q π ∞ (cid:90) −∞ dz cosh z (cid:40)(cid:16) (1 + Q ) cosh z + Q (cid:17) ln Q + + Q ) cosh zQ + Q cosh z − z (cid:41) = − π Q (cid:40) Q (5 + Q + Q ) ln Q + (1 + Q ) (cid:104) + (cid:0) − Q + Q + Q (cid:1) ln(1 + Q ) Q (cid:105) + Q (cid:112) + Q ( − − Q + Q + Q ) ln (cid:112) + Q + Q (cid:41) . (70)12he contribution I ( b + c ) in Eq. (51) from the diagrams in Fig. 1b and Fig. 1c can be computedin a similar way. We rewrite I ( b + c ) as follows: I ( b + c ) = (32 π ) d c ∞ (cid:90) dt dt (cid:48) dt dt (cid:48) (cid:89) j = , (cid:104) t j − + (2 + t j ) e − t j (cid:105) [1 − e − t (cid:48) j ] e − Q tjt (cid:48) jtj + t (cid:48) j (cid:90) d Q (2 π ) Q P ( Q ) × (cid:90) d K d K (2 π ) [ K × K ] (cid:89) j = , [ K j × Q ] Q e − ( t j + t (cid:48) j )( K j − Q t (cid:48) jtj + t (cid:48) j ) . (71)Then, integrating over K and K , we get I ( b + c ) = π ) d c (cid:90) d Q (2 π ) Q P ( Q ) Y ( Q ) (cid:104) Y ( Q ) + Y ( Q ) (cid:105) , (72)where Y ( Q ) = π ∞ (cid:90) dt dt (cid:48) ( t + t (cid:48) ) (cid:104) t − + (2 + t ) e − t (cid:105)(cid:104) − e − t (cid:48) (cid:105) e − Q t t (cid:48) t + t (cid:48) , (73)and Y ( Q ) = π ∞ (cid:90) dt dt (cid:48) ( t + t (cid:48) ) (cid:104) t − + (2 + t ) e − t (cid:105)(cid:104) − e − t (cid:48) (cid:105)(cid:104) − + Q t (cid:48) t + t (cid:48) (cid:105) e − Q t t (cid:48) t + t (cid:48) . (74)Using the parametrization t = τ e z cosh z and t (cid:48) = τ e − z cosh z , and integrating over τ , we obtain Y ( Q ) = π (cid:90) ∞−∞ dz cosh z (cid:40) − cosh z − Q Q + + e − z ) ln (cid:32) + + e − z Q (cid:33) + ( Q + + e − z ) (cid:16) Q + cosh z (cid:17)(cid:104) ln( Q + z ) − ln( Q + + e − z ) (cid:105)(cid:41) . (75)Integrating over z , we arrive at Y ( Q ) = π (cid:40) − Q (cid:0) + Q + Q (cid:1) − Q (cid:0) + Q (cid:1) (cid:0) Q + Q − (cid:1) ln (cid:0) + Q (cid:1) + Q (cid:0) Q + (cid:1) ln Q + Q (cid:0) Q + (cid:1)(cid:0) + Q (cid:1) / ln (cid:112) + Q + Q (cid:41) . (76)The function Y ( Q ) can be conveniently expressed as Y ( Q ) = − Y ( Q ) + ˜ Y ( Q ) , (77)where the function ˜ Y ( Q ) after the integration over τ acquires the following form:˜ Y ( Q ) = Q π ∞ (cid:90) −∞ dz cosh z e − z (cid:40) (2 Q + + e z ) ln (cid:32) + + e z Q (cid:33) − (2 Q + + e − z + z ) × (cid:104) ln( Q + z ) − ln( Q + + e − z ) (cid:105)(cid:41) . (78)13ntegration over z yields˜ Y ( Q ) = π (cid:40) − Q (5 + Q ) ln Q + (1 + Q ) (6 Q + Q + Q − Q ln(1 + Q ) + + Q + Q Q − (cid:112) + Q Q (26 + Q + Q ) ln (cid:112) + Q + Q (cid:41) . (79)Then, we obtain the following expression Y ( Q ) = − Q − π (cid:40) − + Q + Q ) + Q (1 + Q ) ln Q − − + Q + Q + Q ) × (1 + Q ) Q ln(1 + Q ) + Q (8 + Q + Q ) (cid:112) + Q ln (cid:112) + Q + Q (cid:41) . (80) ff erential Poisson’s ratio Combining together the results for the contributions I ( a ) and I ( b + c ) , we express the di ff erenceof the polarization operators responsible to the 1 / d c correction to ν through a single integral:3 Π xy − Π xx Π (0) xy = d c ∞ (cid:90) dQ H ( Q ) P ( Q ) , (81)where H ( Q ) = π Q (cid:110) −Y ( Q ) P ( Q ) + π Q Y ( Q )[ Y ( Q ) + Y ( Q )] (cid:111) . (82)Using Eqs. (66), (70), (76), and (80), we obtain the following lengthy explicit expression for H ( Q ): H ( Q ) = − Q (cid:40) − Q (5 + Q ) ln Q + + Q ) (9 − Q + Q ) ln (1 + Q ) Q − + Q ) ln(1 + Q ) Q (cid:104) + Q − Q − Q − Q − Q + Q (cid:112) + Q (cid:0) − − Q + Q + Q + Q (cid:1) ln (cid:112) + Q + Q (cid:105) + (cid:16) + Q + Q + Q + Q − Q (cid:112) + Q (cid:0) + Q + Q + Q + Q (cid:1) ln (cid:112) + Q + Q + Q (4 + Q ) (11 + Q ) ln (cid:112) + Q + Q (cid:17) + Q (cid:16) + Q ) (9 + Q − Q + Q ) ln(1 + Q ) − Q (cid:0) + Q + Q + Q − Q (cid:112) + Q (16 + Q ) ln (cid:112) + Q + Q (cid:1)(cid:17) ln Q (cid:41) . (83)The function H ( Q ) has the following asymptotic behavior: H ( Q ) = Q / , Q (cid:28) , (cid:32) −
653 ln Q +
10 ln Q (cid:33) / Q , Q (cid:29) . (84)14 (cid:45) (cid:45) (cid:45) (cid:72) (cid:72) Q (cid:76) (cid:144) (cid:80) (cid:72) Q (cid:76) Figure 3: The plot of the function H ( Q ) / P ( Q ) (see text). The function H ( Q ) / P ( Q ) is shown in Fig. 3. As one can see, it changes sign twice which leadsto a partial compensation of the corrections from diagrams on Fig. 1a-c. Numerically evaluatingthe integral in Eq. (81) and substituting it into Eq. (43), we find the result (2).
4. Conclusions
To summarize, we have computed the di ff erential Poisson’s ratio of a suspended two-dimensional crystalline membrane embedded into a space of large dimensionality d (cid:29)
1. Ourresult (2) demonstrates that, for σ L (cid:28) σ (cid:28) σ ∗ , the di ff erential Poisson’s ratio of a crystallinemembrane is a universal but non-trivial function of d c . This results invalidates a common belief(based on results of the self-consistent screening approximation) that the Poisson’s ratio is equalto − / d c .In the physical case of a two-dimensional membrane ( d c = ff erential Poisson’s ratio is not too far from the value − / / d c in Eq. (2) is numerically small. Clearly, a comparison with computational results wouldbe of great interest. Unfortunately, the existing numerical results the Poisson’s ratio of two-dimensional membranes (including graphene) are, however, quite controversial. This may bepartly related with a very delicate character of the problem, see a detailed analysis in Ref. [22].As has been mentioned in Sec. 1, the Poisson ratio in the linear-response regime σ (cid:28) σ L de-pends on boundary conditions. In order to get rid of such finite-size e ff ects but still to be in theregime of universal elasticity, the stress should be in the intermediate range σ L (cid:28) σ (cid:28) σ ∗ . Toresolve well this regime in numerical simulations, su ffi ciently large systems should be consid-ered. Furthermore, in this regime, a care should be exerted in order to distinguish between thedi ff erential and the absolute Poisson ratio [22].Finally, we mention that it would be interesting to extend our analytical result for the 1 / d c -expansion of the di ff erential Poisson’s ratio of a two-dimensional membrane in two directions.First, one can address in a similar way the absolute Poisson ratio. (In this case, the zeroth-orderterm corresponding to the limit d c = ∞ is equal to −
1, see Ref. [22].) Second, the case of adisordered membrane [21] is of interest. 15 . Acknowledgements
We are grateful to E. Kats, M. Katsnelson, I. Kolokolov, V. Lebedev, and J. Los for use-ful discussions. The work was funded in part by Deutsche Forschungsgemeinschaft, by theAlexander von Humboldt Foundation, and by Russian Science Foundation under the grant No.14-42-00044.
Appendix A. Screening of the elastic modulus µ and λ In this Appendix, we present technical details of the calculation of screening of elastic modu-lus. We start from rewriting the term in Eq. (19) which describes the interaction between flexuralphonons in a symmetric form [6]:14 (cid:90) (cid:89) j = d D k j (2 π ) δ (cid:88) j = k j R αβ,γδ ( k + k ) (cid:0) h k h k (cid:1)(cid:0) h k h k (cid:1) . (A.1)Here we consider a membrane of dimensionality D . The interaction kernel reads R αβ,γδ ( q ) = ND − P αβ P γδ + µ (cid:32) P αγ P βδ + P αδ P βγ − P αβ P γδ D − (cid:33) , (A.2)where N = µ (2 µ + D λ ) / (2 µ + λ ). The projection operator is given as P αβ = δ αβ − q α q β q . (A.3)The screened interaction kernel obeys [6]:˜ R αβ,γδ ( q ) = R αβ,γδ ( q ) − R αβ,γ (cid:48) δ (cid:48) ( q ) ˆ Π γ (cid:48) δ (cid:48) ,α (cid:48) β (cid:48) ( q ) ˜ R α (cid:48) β (cid:48) ,γδ ( q ) . (A.4)The polarization operator at finite momenta can be written as [16]ˆ Π γ (cid:48) δ (cid:48) ,α (cid:48) β (cid:48) ( q ) = Π xy ( q ) δ γ (cid:48) δ (cid:48) δ α (cid:48) β (cid:48) + D (cid:0) Π xx ( q ) − Π xy ( q ) (cid:1)(cid:16) δ γ (cid:48) α (cid:48) δ δ (cid:48) β (cid:48) + δ γ (cid:48) β (cid:48) δ δ (cid:48) α (cid:48) (cid:17) + Π ( q ) (cid:16) δ γ (cid:48) δ (cid:48) q α (cid:48) q β (cid:48) + δ α (cid:48) β (cid:48) q γ (cid:48) q δ (cid:48) (cid:17) + Π ( q ) (cid:16) δ γ (cid:48) β (cid:48) q δ (cid:48) q α (cid:48) + δ γ (cid:48) α (cid:48) q δ (cid:48) q β (cid:48) + δ δ (cid:48) α (cid:48) q γ (cid:48) q β (cid:48) + δ δ (cid:48) β (cid:48) q γ (cid:48) q α (cid:48) (cid:17) + Π ( q ) q α (cid:48) q β (cid:48) q γ (cid:48) q δ (cid:48) . (A.5)Because of the projection operators entering R αβ,γ (cid:48) δ (cid:48) , the components Π ( q ), Π ( q ), and Π ( q )of the polarization operator drop from Eq. (A.4). This equation can be solved by ˜ R αβ,γδ whichhas exactly the same structure as R αβ,γδ , Eq. (A.2), but with the screened coe ffi cients N (cid:48) and µ (cid:48) instead of N and µ , respectively: µ (cid:48) ( q ) = µ + (cid:16) Π xx ( q ) − Π xy ( q ) (cid:17) µ , N (cid:48) ( q ) = N + (cid:16) Π xx ( q ) + ( D − D + Π xy ( q ) (cid:17) N / D . (A.6)Within the self-consistent screening approximation the following relation holds: Π xx ( q ) = ( D + Π xy ( q ) ≡ ( D + Π (0) q , and we reproduce the results of Ref. [6].For D =
2, we can rewrite these equations in the following way: µ (cid:48) ( q ) = µ + (cid:16) Π xx ( q ) − Π xy ( q ) (cid:17) µ , B (cid:48) ( q ) = B + (cid:16) Π xx ( q ) + Π xy ( q ) (cid:17) B . (A.7)The result (A.7) generalizes Eq. (30) to a finite momentum transfer.16 ppendix B. Ward identity In this Appendix we discuss the Ward identity for the elastic action and its consequencesfor small-momentum behaviour of exact propagators of flexural phonons. While the main textfocuses on the high-temperature regime, here we discuss a more general case of arbitrary tem-peratures. For the sake of simplicity, we consider the case d = Appendix B.1. Basic equations
We start from the following imaginary-time Lagrangian written in terms of the 3-dimensionalvector r : L [ r ] = ρ ( ∂ τ r ) + κ (cid:52) r ) + µ (cid:16) ∂ α r ∂ β r − δ αβ (cid:17) + λ (cid:16) ∂ α r ∂ α r − (cid:17) . (B.1)Here Greek indices correspond to the 2D coordinates ( x , y ) ≡ x parameterizing the membrane.We note that substituting r = { ξ x x + u x , ξ y y + u y , h } into Eq. (B.1) yields the membrane action (4).The Lagrangian (B.1) is manifestly invariant under O (3) rotations of the vector r . Theserotations can be parameterized as r j → r j + ε a t ajk r k , (B.2)where ε a → t ajk = (cid:15) a jk are generators of O (3) group. In order to exploreimplications of this symmetry, we shall follow a standard approach [26, 27]. Let us consider thefunctional Φ [ ˆ Σ ] defined as followsexp (cid:16) − Φ [ ˆ Σ ] (cid:17) = (cid:90) D [ r ] exp (cid:40) − β (cid:90) d τ (cid:90) d x (cid:16) L [ r ] − Σ j α ∂ α r j (cid:17)(cid:41) . (B.3)At this stage, Σ j α are arbitrary functions of x and y ; as will become clear soon, they have ameaning of components of the stress tensor σ j α [9, 16]. The average deformation ∂ α R j = (cid:104) ∂ α r j (cid:105) (B.4)can be found as ∂ α R j = − δ Φ [ ˆ Σ ] δ Σ j α . (B.5)Evidently, R j transform according to Eq. (B.2) under rotation.Let us now consider the Legendre transform of Φ [ ˆ Σ ]: F [ R ] = Φ [ ˆ Σ ] + β (cid:90) d τ (cid:90) d x Σ j α ∂ α R j . (B.6)Here Σ j α should be found from the solution of Eq. (B.5). There is also the reciprocal relationbetween Σ α j and ∂ α R j : Σ j α = δ F [ R ] δ∂ α R j . (B.7)17e note that F [ R ] coincides with the free energy evaluated under the constraint (cid:104) ∂ α r j (cid:105) = ∂ α R j ,where R is a given function of x and y .Now let us introduce the two-point correlation function S αβ jk ( q , ω ) as the second variation ofthe functional F [ R ]: S αβ jk ( x τ, x (cid:48) τ (cid:48) ) = δ F [ R ] δ∂ α R j ( x τ ) δ∂ β R k ( x (cid:48) τ (cid:48) ) . (B.8)We note that the propagator of displacements, G jk ( x τ, x (cid:48) τ (cid:48) ) = − (cid:10) T τ r j ( x τ ) r k ( x (cid:48) τ (cid:48) ) (cid:11) Σ , (B.9)where (cid:104)· · · (cid:105) is defined with respect to the Lagrangian L [ r ] − Σ j α ∂ α r j and T τ denotes the orderingalong the imaginary time contour, is related with the two-point correlation function S αβ jk ( q , ω ) inthe following way: G − jk ( x τ, x (cid:48) τ (cid:48) ) = ∂ ∂ x α ∂ x (cid:48) β S αβ jk ( x τ, x (cid:48) τ (cid:48) ) . (B.10)The rotation symmetry (B.2) implies that Φ [ ˆ Σ ] = Φ [ ˆ Σ (cid:48) ] , (B.11)where Σ (cid:48) j α = Σ j α − ε a t ajk Σ k α . Expanding this equation to the lowest order in ε a , we find the Wardidentity: 0 = ε a t ajk β (cid:90) d τ (cid:90) d x Σ k α δ Φ [ ˆ Σ ] δ Σ j α = − ε a t ajk β (cid:90) d τ (cid:90) d x ∂ α R j δ F [ R ] δ∂ α R k . (B.12)In order to use the Ward identity for analysis of the two-point correlation function, it it convenientto perform a variation of the last part of Eq. (B.12) with respect to ∂ γ R l ( x (cid:48) , τ (cid:48) ). This yields ε a t alk Σ k γ ( x (cid:48) , τ (cid:48) ) + ε a t ajk β (cid:90) d τ (cid:90) d x ∂ α R j ( x , τ ) S αγ kl ( x τ, x (cid:48) τ (cid:48) ) = . (B.13) Appendix B.2. The propagator of flexural phonons
With the choice ε = { ε, , } , Eq. (B.13) reduces to t xzy Σ y γ ( x (cid:48) , τ (cid:48) ) + t xyz β (cid:90) d τ (cid:90) d x ∂ α R y ( x , τ ) S αγ zz ( x τ, x (cid:48) τ (cid:48) ) = . (B.14)Now we consider the function R ( x , τ ) which has the following form: R ( x , τ ) = R ( ξ ) = { ξ x x , ξ y y , } , (B.15)where ξ x and ξ y are arbitrary constants. The functional F [ R ( ξ ) ] corresponds to the free energyevaluated under the constraint (cid:104) ∂ α r j (cid:105) = ξ α δ α j , where the average is taken with respect to the18agrangian L [ r ]. This is exactly the action S (see Eq. (4)) discussed in the main text. Using Eq.(B.14), we find ξ y lim ω, q → S yyzz ( q , ω ) = Σ yy = ∂ f ∂ξ y , ξ y lim ω, q → S yxzz ( q , ω ) = ,ξ x lim ω, q → S xxzz ( q , ω ) = Σ xx = ∂ f ∂ξ x , ξ x lim ω, q → S xyzz ( q , ω ) = . (B.16)We recall that the physical stress is defined by Eq. (7). Therefore, we obtainlim ω, q → S αγ zz ( q , ω ) = σ α δ αγ . (B.17)By virtue of Eq. (B.10), this implies that the inverse propagator of the flexural phonons for q → ω → G − zz ( q , ω ) = σ x q x + σ y q y + . . . (B.18)We note that Eq. (B.18) extends the statement of Refs. [9, 16] to the case of σ x (cid:44) σ y . References [1] L. D. Landau and E. M. Lifshitz,
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