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Contents 1 Introduction : Phase transitions in 2D electron systems 2 An RSRG approach to lattice conductivity 10 A Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10 B The 1D lattice . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17 1 The conductor super-conductor lattice . . . . . . . . . . . . . . . . . . 17 2 The conductor insulator lattice . . . . . . . . . . . . . . . . . . . . . . 19 The Cayley tree . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22 1 The conductor super-conductor tree . . . . . . . . . . . . . . . . . . . 22 2 the conductor insulator tree . . . . . . . . . . . . . . . . . . . . . . . . 27 3 the mixed lattice . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33 The 2D square and triangular lattice . . . . . . . . . . . . . . . . . . . . . 37 1 The square lattice . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37 2 The triangular lattice . . . . . . . . . . . . . . . . . . . . . . . . . . . 41 C D 3 2 A Percolation description of the IQHE 43 A The model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43 B The calculation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46 1 1. INTRODUCTION : PHASE TRANSITIONS IN 2D ELECTRON SYSTEMS Two-dimensional (2D) electronic systems have been in the center of attention for the last couple of decades, both experimentally and theoretically. The main reason for that is the rich physics and welth of phenomenae which are unique for 2D systems. Such phenomenae are phase-transitions between different conduction states of the systems (i.e. insulating, metallic, super-conducting), triggered by externally controled parameters such as magnetic field, carrier density e.t.c. Since the transitions are not induced by temperature and are believed to hold even at zero temperature, it is the quantum mechanical nature of the system that governes its properties and that is the reason that such transitions are reffered to as quantum phase transitions. Although a Large number of experimental (and theoretical) results concerning these transitions is at hand, still many of the features are still not well understood, and the mechanisms that drive these transitions in many cases are still in debate. We begin by reviewing the main experimental results, along with some previous theoretical results. 1. The integer quantum Hall transition First reported in 1980 by Von-Klizting et al. [1], the Hall conductivity σ xy of a 2D electron gas in the presence of a strong perpendicular magnetic field and lattice impurities is quantized in units of e2/h , together with vanishing of the longtitudinal conductivity σ xx vanishes. The relevant parameter is the filling factor, defined by ν= N nhc 2 = 2πnlH = Φ/Φ0 eH (1.1) where n is the carrier (electrons) concetration, N the number of carriers, Φ the magnetic flux and Φ0 = hc e the flux quanta. In the experiment the filling factor is changed continously (by a change in the magnetic field) and the Hall conductivity remaines constant (with a value of an integer multiple of e2 /h) and changes (by the above quanta) only as the filling factor crosses half integer values, where at these values a finite longtitudinal conductivity is observed (Fig. 1). This is known as the Integer Quantum Hall Effect (IQHE) [2] [3]. 2 Although no analytic solutions for the IQHE are at hand, the phenomena is nevertheless believed to be well understood in terms of Anderson localization theory. In this theory the electron states are localized at (energy) regions far from the center of the Landau Levels (LL) and are only extended at the vicinity of the center of LLs. Thus, when the fermi energy lies far from the center of the LL, there will be no contribution to σ xx and σxy will remain constant. As the Fermi energy crosses over to the region of extended states there will be a contribution to both σxx , and σxy and as it crosses over again to another region of localized states, σxx will again vanish and σxy will stabilize on a new value. The measured quantity is usualy the resistance, which is connected to the conductance by tensor inversion (i.e. ρµν = (σ −1 )µν ) which yields for the longtitudal and tranverse resistivities, respectively, ρxx = σxx , 2 + σxy 2 σxx ρxy = σxy 2 + σxy 2 σxx (1.2) which are shown in Fig. 1. Fig. 1 : The Integer Quantum Hall Effect A simple description for the way this mechanism works is available in the framework of the semi-classical theory . The starting point is considering non-interacting electrons in a strong magnetic field and a smooth random potential V (r) which changes on a length scale λ which is much larger then the magnetic length lH . In this limit the motion of the electrons can be seperated into a rapid cyclotron orbiting , with frequency ω c around guiding centers, and a slow drift of the guiding center along equipotentials of V (r). The local energy of an electron will be (in this limit) E = En + V (r) , where En = h̄ωc (n + 12 ) is the LL energy. For each LL there will be an extended state at an energy Ec = En + Vc if the equipotential line at Vc percolates through the sample, while at energies where the equipotentials define closed 3 orbits the states will be localized. As the Fermi energy crosses from one region of localized states (states orbiting puddles of the potential) to another (states orbiting peaks at the potential) it crosses Ec and at that energy, where the potential contour percolates, the states are extended. Thus the electrons at the Fermi energy undergo a localization-delocalization (LD) transition, and as they are the ones contributing to the longtitudinal conductivity, the entire system goes through a phase transition between an Anderson insulator and a metal. This is the Quantum Hall phase transition ∗ . Many attempts have been made to understand the properties of this transition. We are mainly interested in the critical exponent of the localization length, defined by ξ ∝ |E − Ec |−ν (1.3) near Ec , where Ec is the critical threshold. This Exponent can be measured experimentally ( [4], [5]) and its value is known to be ν ≈ 2.3 ± 0.1 . A huristic derivation of this exponent is present in [6] [7] [8] and can be put in naive argument as follows. The basic assumption is the scaling assumption, L σ ∝ F( ) ξ (1.4) where σ is the conductance of the system, F is some unknown function and L the length of the sample. This assumption already containes the fact that the conductivity at the critical point (where ξ diverges) is independent of length of the sample. Next, the quantum effects can be regarded by an expression of the form [9] Ti ∝ exp(−|Ei − Ec |) ∗ There (1.5) is recent experimental evidence pointing that this picture is wrong and that the electron- electron interaction plays an important role in the localization of electronic states ( [10] and appendix 1). Still the nature of the transition into extended states remains a percolation transition and this is what dominates the behaviour at the critical point. 4 where Ti is the transition probability between two orbits seperated by a single saddle point with energy Ei . Since the system is quantum mechanical the conductivity is given by the Landauer formula and is proportional to the Transition probability for the entire sample, which is multiplicative (i.e. Ttotal ∼ ΠTi). Thus the conduction of the entire sample becomes σ ∼ ΠTi ∼ exp(−N (|Ei − Ec |) , (1.6) where N is the number of localized orbits, given by the classical percolation result N = L , ξ ξ ∝| Ei − Ec |−νclassic (1.7) νclassic being the classical percolation localization exponent. Combining (1.6) and (1.7) results in an effective exponent, νquantum = νclassical + 1 (1.8) which for the known result of νclassical ≈ 1.33 yields the above experimental result. A more quantitative approach was introduced by Chalker & Coddington [11] who introduced the chiral network model. In this model the lattice is mapped into a square lattice, each node representing a saddle-point between two neighbouring orbits, which are represented by a closed square of four links. The wave functions are supposed to travers along the orbits and scatter at the saddle points of the potential, represented by the nodes of the lattice. The disorder is modeled by random phases acuired by the wave function as they traverse a link and the Magnetic field causes chirality, which ensures that on each link the wave functions are traversing in only one way. Thus the transfer-matrix of each node is a 2 × 2 matrix. The transfer-matrix for the entire sample can then be calculated and the conductivity obtained using the Landauer formula. The network model then became a standard tool for calculating the properties of the IQH phase transition ( [12] [13]). 2. The MIT A metal-to-insulator transition (MIT) occures in 2D electron systems of high quality (i.e. 5 clean samples with little disorder) , high mobility and low carrier density [14]. In these samples there exists a critical density nc such that for densities lower then nc the system is insulating (i.e. the resistance diverges with decreasing temperature) and for densities higher then nc the system exhibits metallic behavior and the resistance seems to saturate with decreasing temperature . Fig. 2 : The MIT (taken from Kravchenko et al. [15]) Some features of the MIT in presence of a weak magnetic field (both tilted and perpendicular) have been intensively investigated. For parallel fields, it has been shown that the metallic behaviour is destroyed and a large magneto-resistance is developed. The behavior of the magneto-resistance is the same wethear the density is higher then n c or lower then nc (i.e. on the metallic side or the insulating side) (see references within [14]). For perpendicular fields a shift in the critical density has been observed, initially decreasing with magnetic field up to a certain magnetic field, and then start to grow again until the insulating phase disappears and the QH state appears [16]. At these low densities, the electron-electron interaction is comparable (if not larger) then the Fermi energy. Thus, the central belief is that the e-e interaction causes a delocalization of the otherwise localized states of the Anderson insulator. This have been the basis for most of the theoretical models concerning 6 the MIT, but none of them fully encorporates all of the features mentioned above. Recently other mechanisems have been suggested, such as charging and discharging of traps on the 2D layer surface or scattering of electrons from the surface ions. A non-interacting electrons percolation model was suggested by Y.Meir [17], in which the states at the Fermi energy are localized within puddles (due to the disorder potential) and are conected via quantum point contacts (QPC’s). In this model each of the QPCs is characterized by its critical energy, such that the transmission through it is given by T (ε) = Θ(ε−ε c). The conductance through each QPC , calculated using the Landauer formula , is G(µ, T ) = 1 2e2 h 1 + exp((εc − µ)/kT ) , (1.9) where µ is the chemical potential and T the temperature. At low tempratures G(µ, T ) approximatelly assumes only the values 1 (if εc < µ ) or 0 (if εc > µ). Assuming that a dephasing process takes plase within the puddle (i.e. the dephasing time is shorter then the time spent by each electron in a puddle before tunneling on) and taking the distribution of conductances to be the sum of two δ-functions this leads to a percolation model for the transport through the entire sample. As this model has a direct relation to this work, it will be further discussed in chap.VI. 3. The insulator superconductor transition The investigation of superconducting materials in the 2D-limit, achieved by fabricating ultra-thin films of superconducting material on a normal substrate, has shown that disorder introduced into the system destroys superconductivity. When disorder is introduced gradually into the system then a superconductor-to-insulator transition (SIT) takes place . This transition can be viewed by measuring the temperature-dependence of the resistance. In the superconducting regime the resistance vanishes as T → 0 and in the insulating regime it divrges [18]. Fig. 3 shows the resistance as a function of temperature for different sample thickness. For thin samples , the disorder dominates the electornic behavior and the sample is insulating, while for thicker samples superconductivity evolves. 7 ************************* PICTURE OF SIT ************************* A large number of experimental results and characteristic features of the SIT are at hand, but unfortunatly they cannot be summerised within the scope of this introduction, maily because even in features that are supposed to be universal such as the critical exponents, there is still much disagreement. More generally, it is still not clear which of the characteristics of the SIT are intrinsic and which are sample- or measurement-dependent. Still , there are some relevant experimental results. One of these is the critical-thickness dependence on magnetic field. As the magnetic field is increased, the critical thickness also increases, behaving like a power-law with the magnetic field [19]. A more detailed discussion on this and other features of the SIT will be given in chapter V. The main theory that describes the SIT is the so-called ”dirty boson” model, in which the electronic Cooper-pairs are described by pointlike Bose particles in the presence of a random potential and long range Coloumb interaction. At sufficient disorder the phase of the superconducting order parameter begins to fluctuate strongly, destroying the creation of cooper pairs and superconductivity vanishes. A magnetic field introduces vortices. The vortices act as quantum Bose particles which can be Bose-Einstein-Condensated into a coherent state, which is insulating. The vortices are dual particles to the boson paires, and as the coherent state of the latter induce super-current, the coherent state of the vortices induces voltage and an insulating state is reached for some critical disorder, now with different value then in the zero-field case. Although there are many succeses to the dirty-boson model, still it is lacking in explaining some features of the experiment, specially reagrding the density of states near the transition [20]. In this work we shall try to answer some of the main questions arising from the experiments on 2D phase transitions : 1. What is thye origin of the QH universality class and can the critical exponent be calculated from a simple non-interacting electrons model ? 2. Can this model explain some features of the MIT, specifically the behavior in magnetic 8 field [16] and the aapparent symmetry between the MIT and the QH transition [21] ? 3. can a qualitative connection be made between the QHT & the MIT and between the SIT ? 9 2. AN RSRG APPROACH TO LATTICE CONDUCTIVITY Since the main tool in which we are going to examine the 2D phase transition is the real space renormalization group within the (quantum) percolation formalism, it is important to check whether this tool can be applied to the classical problem, where other methods of solution are available and a comparison with our reults can be made. This chapter is organized as follows. We begin with an introduction to the classical problem and the relation between lattice conductivity and percolation theory and an introduction of the Renormalization Group approach via finite-size scaling. the next sections are devoted to the 1D lattice and the Cayley tree, where analytic solutions are found, and in the last section a numerical solution for the 2D lattice is presented. A. Introduction The general problem of lattice conductivity in the context of percolation has a very simple representation.We consider a D-dimensional lattice of some shape, composed of bonds (links) which are connected by nodes (lattice sites). Each bond can be an insulator with conductance σ = 0 , a (classical) resistor with conductance σ = σ 0 (usually taken to be σ0 = 1) or a super-conductor ,with conductance σ = ∞, each with a different probability. The distribution function for the conductivity can thus be written as f(r) = pins δ(σ) + pres δ(σ − σ0) + psc lim δ(σ − S) S→∞ . (2.1) The goal of the calculation is to find the nature of the conductance of the entire sample (in the thermodynamic limit, i.e. as its size becomes infinite) as a function of the probabilities pins , pres and psc . The connection to the classical percolation problem can be demonstrated by looking at the simple case of pres = 0. We then have a lattice composed of bonds which are either supr-conducting or insulating ( which can be regarded just as absent bonds). Thus the entire lattice can either be super-conducting if there is a percolating cluster , else it is an insulator. The problem of lattice conductivity is now narrowed to whether there is a 10 percolating cluster or not, which is exactly the classical percolation problem. As this simple case is expanded to the more general problem of eq.(2.1) we can still expect that knowledge of percolation theory be applicable, which, as will be shown, is indeed the case. A brief description of the percolation problem is required [22]. We consider the simplest case of a D-dimensional lattice of some shape, where each bond has some probability p to exist and probability 1 − p for being absent ( p is just the fraction of existing sites of all availabe sites in each realization of the lattice). This problem is usualy refered to as ”dilute bond percolation”. When p is small (∼ 0) the lattice consists of scattered bonds. As p is increased, connected clusters of bonds are created. These clusters get larger and larger as p is further increased, until at some value pc the largest cluster percolates throughout the lattice and the lattice is said to be connected (fig.2). There is thus a distinction between two phases of the system - the disconnected phase and the connected phase, and a 2nd order phase transition between them occures at pc . p=0.2 p=0.5 p=0.8 Fig. 4 : Bond Percolation for different bond concentrations in a square lattice This transition is characterized by a diverging length ξ, defined by r g(r) ∼ exp(− ) ξ (2.2) where g(r) is the probability for two sites a distance r apart to belong to the same cluster. It can be shown that near the phase transition ξ is typically the avrage size of the largest cluster, and thus diverges at the phase transition, i.e. as p → p c . Its divergence can be characterized by the critical exponent ν, defind by ξ ∝ |pc − p|−ν , 11 < pc . p∼ (2.3) We can now ask the following questions: what are the values of p c and ν ? These, among other things, will be characters of the specific problem, i.e. dimension, lattice shape etc. We now turn to see the effect of the lattice size L. For this purpose we introduce P (p) to be the ratio between the size of the largest cluster and the lattice size ( P (p) is just the fraction of lattice bonds belonging to the largest cluster, and is sometimes reffered to as ‘cluster strength‘). For values p ≥ pc the largest cluster is infinite, thus as L → ∞ P can still be finite, while for values p < pc P vanishes as L → ∞ . It can be shown that for an infinite lattice just above pc , P has the form P ∝ |p − pc |β . (2.4) The form of P (p) for a finite lattice can be postulated , in the so-called finite-size scaling assumption, to be of the form L P = L−x F ( ) = L−x F (L(pc − p)ν ) . ξ (2.5) (ξ enters since it is the charecteristic length scale of the problem). If we are to recover eq.(2.4) from eq.(2.5), the function F (z) must obey lim F (z) ∼ z x z→∞ (2.6) so that P is independent of length as L → ∞. This then gives P ∝ |pc − p|νx (2.7) and in order to have eq.(2.4) we must have x = βν . thus the form of P (p) in the finite size scaling is given by P (p) = L−β/ν F (L(pc − p)ν ). At p = pc , L ξ (2.8) = 0 and F (0) = const., and we see that P (pc ) vanishes according to the power-law P ∝ L−β/ν 12 (2.9) which gives an accurate way of determining critical exponents. Similarly we can apply eq.(2.8) for any quantity varying as |p − pc |x , with β replaced by x and with a different shape of the scaling function F (z). We can now turn to the conductivity of the lattice, and distinguish between two cases. The first case is of a lattice composed of insulators and resistors. Here the conductivity σ → 0 as p → pc , and is assumed to have the form σ ∼ |pc − p|t (2.10) which leads to the form for the lattice conductivity at finite length σ ∼ L−t/ν F (L/ξ). (2.11) The second case is that of a lattice composed of resistors and super-conductors. In this case the conductivity diverges as p → pc and has the form σ ∼ |pc − p|−s (2.12) which gives for the lattice conductivity σ ∼ Ls/ν f(L/ξ). (2.13) there is no known connection between the exponents ν, s and t except for a few special cases (some will be mentioned later on), but it is conjectured that they do not depent on the lattice shape but only on dimensionality (the universality conjecture). We now introduce the renormalization group technique [23]. At the transition (critical region) the characteristic scale of the system, i.e. the corrolation length, diverges, as the system undergoes fluctuations at all length scales. In the absence of characteristic scale the system is said to be ”scale-invariant”, in the sense that the statistical characteristics of the system are independent of the scale by which the system is observed and the inner degrees of freedom (at a small scale) are unimportant. We can use this featur in the following way. First we divide the system into small pieces, each with some inner degrees of freedom and some 13 statistical characteristics. We then combine some of these pieces into a new piece, enlarged by a factor b (this is called ”coarse graining”), and rescale the length scale by the same factor b (this is called ”decimation”). We now have a new cell which has a smaller number of degrees of freedom then previouse to the procedure, but is assumed to have the same statistical nature. The transformation between the small cells and the coarse-grained cells is called the renormalization group (RG) transformation, and by fomulating it mathematically and demanding scale-invariance, one can learn about the characteristics of the system. As an example we look at the 2D square lattice in the dilute bonds problem, and are interested in finding the values of pc and ν. We start by dividing the lattice into cells of the form shown on the l.h.s. of fig. 3. We then decimate each such hyper-cell into a new, smaller cel , shown on the r.h.s. of fig. 3. The change of scale is by a factor b = 2. Fig. 5 : RG transformation for the dilute bond percolation on a 2D square lattice If we look at the connectivity of the lattice from one side to the other, then the 4-fold symmetry of the lattice allows us to consider only one direction, say the horizontal, as shown in fig.4. Fig. 6 : horizontal RG transformation for the 2D square lattice The probability of each bond to be occupied (i.e. not to be absent) is p. We say that a hyper-cell is occupied if the two sides are connected by bonds. Thus, the probability for a hyper cell to be occupied is given by 14 p0 (p) = p5 + 5p4 (1 − p) + 8p3 (1 − p)2 + 2p2 (1 − p)3 . (2.14) eq.(2.14) is the RG equation for the probability. At the critical point the system is scaleinvariant, and the probability for occupation of a hyper-cell and of a single bond (which is the hyper-cell coarse-grained) should be the same. This means that p c is a fixed point of eq.(2.14) , i.e. p0 (p∗ ) = p∗ . (2.15) Solving for pc = p∗ we find three solutions: p∗ = 0, 1 and p∗ = 0.5. The first two solutions are stable solutions, meaning that the soluitions of applying the RG equation iteratively for probability values close to them will flow toward these values. This means that they represent stable phases of the system (i.e. the fully occupied and fully absent phases) and are thus not the desired solutions. The third value, on the other hand, is an unstable solution and applying the RG equation for values close to it will result in a flow away from it tword one of the stable solutions. This means that this solution represents an unstable phase of the system and is the solution we are seeking. Thus, for this lattice p c = 0.5. Although this is the exact result obtained for this problem in other ways ( [22], [23]), this procedure is nevertheless still an approximation and does not yield the exact result (if it exists) for other cases. To see this more clearly we turn to calculate the critical exponent ν. We start from eq.(2.3), and now look at the correlation length of the coarse-grained lattice, which is just given by ξ0 = ξ b (2.16) where b is the change in scale. We can immediatly see that scale invariance at the phase transition, which means that ξ 0 = ξ , implies the divergence of the correlation length. In accordance the eq.(2.3) the coarse-grained corrolation length satisfies ξ 0 = |pc − p0 (p)|ν We now expand p0 (p) near pc , 15 (2.17) p0 (p) = pc + λ(pc − p) + O(|pc − p| ), λ= dp0 (p) | dp pc (2.18) substituting into eq.(2.17) gives ξ 0 ≈ λν |pc − p|ν = λν ξ (2.19) which gives the exponent in terms of b and λ, by ν= lnb lnλ . (2.20) Solving for ν we find that λ = 1.65, and ν= ln2 = 1.428 ln1.65 while the (believed to be exact) result is ν = 4/3. The origin of the error is the fact that our coarse-graining scheme allows situations where clusters which are disconnected at one scale will be connected at another scale, as in fig.5. Fig. 7 : Although these two clusters are disconnectes the coarse-grained bonds are connected Although the error can be reduced by introducing next-nearest neighbours connectivity [24] , we restrict our selves only to nearest-neighbours in order to keep the calculation of lattice conductivity (which will be described shortly) as simple as possible. We can now use the RSRG procedure to determine the conductivity of the lattice in the following way: We start with an initial distribution of the conductance for a single link. We then use Kirchoff‘s laws to express the distribution of conductance of a renormalized cell. Using self-similarity gives us an integral RG equation for the conductivity which can then be solved, either analitically or numerically. This is done in the following sections. 16 B. The 1D lattice The 1D lattice is a wire consistent of N links connected in series. The resistance is measured by putting the ends of the wire in a unit potential difference and measuring the current in the first link, which gives the conductance of the wire. We are mainly interested in the two cases of a conductor super-conductor lattice and of a conductor insulator lattice. 1. The conductor super-conductor lattice In this case the resistance distribution for a single link is given by g(r) = αδ(r) + (1 − α)δ(r − 1) (2.21) in units of link resistance r0 , meaning that each link can be a super-conductor or a regular conductor with probabilities α and 1 − α respectively. The two phases of the wire are the conducting phase and the super-conducting phase. Since the resistances are added linearly in such a lattice we expect that the average total resistance of the wire will be of the form hri = N (1 − α) = N , (2.22) which implies both a critical exponent s = 1 and a critical probability α = 1. These results will be obtained using RSRG. Our hyper-link is just two links connected in series, giving a scale factor b = 2. Fig. 8 : RG transformation in 1D The resistance of the n + 1 generation hyper-link is r = r 1 + r2 17 (2.23) where r1 and r2 are the resistances of the two nth generation hyper links. We now assume that the resistance of the nth generation lattice is given by the same distribution function P (n) (r). Using eq.(2.23) and the definition of the distribution of a function of random variables we obtain an integral equation for the moments of the distributions Z m r P (n) (r)dr = Z Z P (n−1) (r1 )P (n−1) (r2 )(r1 + r2 )m dr1 dr2 Now multiplying eq.(2.24) by a factor Z e −itr P (n) (r)dr = Z Z (−it)m m! . (2.24) and summing over m we obtain an equation e−it(r1 +r2 ) P (n−1) (r1 )P (n−1) (r2 )dr1 dr2 , (2.25) which, by using the definition of the Fourier transform, can be written as F (n) (t) = [F (n−1)(t)]2 (2.26) where F (n)(t) is the Fourier transform of P (n) (r). The initial function is given by eq.(2.21) F (0) = α + (1 − α)e−it . (2.27) The solution of the recursion equation (2.26) is n n F (n)(t) = [F (0)(t)]2 = [α + (1 − α)e−it ]2 , (2.28) and definition of the Fourier transform gives the relation hri = −i ∂F (t = 0) ∂t (2.29) which is easily obtained from eq.(2.28) to be hri = 2n (1 − α) . (2.30) This is exactly the result obtained from intuitive considerations. In order to show that the critical probability is indeed αc = 1 we use the fact that at the critical probability the resistance distribution of the nth hyper-link is similar to that of the previous generation. Setting F (n)(t) = F (n−1) (t) will give a single solution for α, αc = 1. 18 2. The conductor insulator lattice Since an insulator is described by the limit r → ∞ it is convenient to adress to the conductance of single link, given by σ = r −1 . The conductance distribution for a single link is given by g(σ) = αδ(σ) + (1 − α)δ(σ − 1) . (2.31) Our RG transformation will be the same as in the previous subsection with the same scale factor b = 2 and the nth generation hyper-link will have in it 2 n single links. The conductance of the hyper-link is given by xy x+y σ= , (2.32) where x and y are the conductances of the two n − 1 generation hyper-links which constitute the nth generation hyper-link. Taking the mth power of eq.(2.32) we obtain an equation for the moments of the conductance probability of the nth generation hyper-link Z m σ P (n) (σ)dσ = Z Z P (n−1 We now multiply eq.(2.33) by a factor )(x)P (−it)m m! (n−1 xy )(y) x+y !m dxdy . (2.33) and sum over m. Using the definition of the Fourier transform This gives us a recursive integral equation for the Fourier transform of P (n) (σ) F (n) (t) = Z K(t, u, v)F (n−1)(u)F (n−1)(v)dudv , (2.34) where ! 1 Z xy K(t, u, v) = exp −it + iux + ivy dxdy (2π)2 x+y , the zeroth generation solution being F (0)(t) = α + (1 − α)e−it . This implies that a general form for the solution of eq.(2.34) can be chosen to be 19 (2.35) F (n) (t) = X j (n) (n) fj (α) exp −iγj (t) . (2.36) Using this ansazt and F (0)(t) one obtaines t F (1)(t) = α(α + 2(1 − α)) + (1 − α)2 e−i 2 which implies (since this is valid for all t) that only two elements from the sum of eq.(2.36) are non-zero, and one can write F (n)(t) = g (n) (α) + h(n) (α)e−itγ (n) , (2.37) where g (n) (α) and h(n) (α) are some polynomials of α. Inserting this into eq.(2.34) yields the following recursion equations g (n) (α) = g (n−1) (α)(g (n−1) (α) − 2h(n−1) (α)) h(n) (α) = [h(n−1) (α)]2 γ (n) = γ (n−1) 2 (2.38) which, by using F (0) and the normalization condition of F (n)(t) (which is just F (n)(0) = 1), can be solved exactly to give n g (n) (α) = (1 − (1 − α)2 ) n h(n) (α) = (1 − α)2 γ (n) = 1 2n (2.39) and thus n t n F (n)(t) = (1 − (1 − α)2 ) + (1 − α)2 e−i 2n . (2.40) We can now easily calculate the average conductance by means of n hσi = −i ∂F (n) (1 − α)2 (t = 0) = ∂t 2n 20 (2.41) This result demonstrates very well the physics of such a lattice, since such a lattice will n conduct only if all the links are conductors (with probability (1 − α) 2 ) and will then have a conductance σ = 1 . 2n This result shows that in this case there is no universal critical exponent. This can be explained by the fact that the two phases of the sample in this problem are not essentially different, since at the limit of an infinite sample there is only one phase, regardless of the value of the probability. 21 C. The Cayley tree In this section we consider a resistance network in the form of a Cayley tree with coordination number f = 3. This is a tree such that every branch is a two prong fork (fig.7) . For the Cayley tree the percolation problem can be solved exactly , and the critical probability for percolation is 1 f −1 αc = (2.42) ··· Fig. 9 : The f = 3 Cayley tree 1. The conductor super-conductor tree Let the resistance distribution function for a single link be g(r) = αδ(r) + (1 − α)δ(r − 1). (2.43) If we denote by r the resistance of the trunk and by R 1(R2 ) the resistance of the upper (lower) branch, then the resistance of the entire tree can be written as R=r+ R1 R2 R1 + R2 . (2.44) This gives us a recursion equation for the resistance distribution of the nth generation tree in terms of the n − 1 generation tree resistance distribution, P (n) (R) = = Z Z Z Z Z P g(r)P (n−1) (R1 )P (n−1) (R2 )δ R − r + (n−1) (R1 )P (n−1) R1 R2 R1 + R2 drdR1 dR2 = R1 R2 R 1 R2 (R2 ) αδ R − + (1 − α)δ R − 1 − R1 + R2 R1 + R2 22 (2.45) dR1 dR2 We now make the ansatz P (n) (R) = β (n) δ(R) + (1 − β (n))Λ(n) , (R) (2.46) where Λ(n) (R) satisfies the normalization condition Z Λ(n) (R)dR = 1 . This is reasonable since the conditions R = 0 and R 6= 0 represent the two different phases of the network. Placing this ansatz in eq.(2.45) and performing the trivial integrations on the δ-functions and then equating coefficients we get our RG equations β (n) = αβ (n−1) (2 − β (n−1)) Z Z (2.47) R1 R2 + R1 + R2 R1 R2 (n−1) 2 + (1 − α)(1 − β ) δ R−1− ]Λ(n−1) (R1 )Λ(n−1) (R2 )dR1 dR2 + R1 + R2 (1 − β (n))Λ(n) (R) = [α(1 − β (n−1))2 δ R − + (1 − α)β (n−1)(β (n−1) + 2(1 − β (n−1)))δ(R − 1) (2.48) We now use the renormalization assumption that at αc , β (n) = β (n−1) and Λ(n) = Λ(n−1) . This yields a value for β β =2− 1 αc (2.49) Although this does not define a specific α , the critical probability can be found by setting β = 1 (i.e. the demand that at the critical probability the tree is super conducting) which gives αc = 1. Notice that this value is different from the critical probability for percolation, given by eq.(2.42), which can be obtained by setting b = 0. We now substitute eq.(2.49) into eq.(2.48) to obtain Λ(R) = Z Z R1 R2 [(1 − α)δ R − + R1 + R2 (1 − α)2 R 1 R2 + δ R−1− ]Λ(R1 )Λ(R2)dR1 dR2 + α R1 + R2 2α − 1 + δ(R − 1) α 23 (2.50) We now substitute Λ(R) with δ(R−f(α)), where f(α) = R RΛ(R)dR. This approximation is valid since we are only interested in the value of the first moment (i.e. the average resistance) and not in the resistance distribution itself. Inserting this into eq.(2.50) gives δ(R − f(α)) = (1 − α)δ(R − (1 − α)2 f(α) (2α − 1) f(α) )+ δ(R − 1 − )+ δ(R − 1) (2.51) 2 α 2 α We now integrate over R and find an equation for f(α), f(α) = (1 − α)2 (1 − α)2 2α − 1 (1 − α) f(α) + f(α) + + 2 2α α α , (2.52) and solving for f(α) we find f(α) = 2α2 3α − 1 (2.53) This can be expanded around (1 − α) = (which is the critical probability) , which gives f(α) = 1+O(). We now use this result to integrate eq.(2.46) over R, for finding the average resistance hRi = Z RP (R)dR = (1 − α)f(α) ∼ (1 − α) , (2.54) which means that the critical exponent is s = 1. This result has an intuitive justification. If we calculate exactly (using Kirchoff’s laws) the average resistance of an n-generation tree we would find something which always has the form hRi = (1 − α) + Q n (1 − α) , where Qn (1−α) is some polynomial with order larger than 1. The first term always comes from the trunk. This is justified intuitively since in order to have a non-zero resistance at the right side of each node, all the links attached to it (from the right) must be ordinary conductors. The probability for this to happen on nodes which are further into the tree is the square of the probability that this will happen at the first node, since it is connected to only one link instead of two. Thus, as α → 1 the conductance mainly comes form the trunk, which is a single link. A different way of obtaining the critical exponent is by looking at the moments of the resistance distribution, by taking the mth power of both sides of eq.(2.44) and integrating 24 over the single link attached in series (for which the integration is trivial). Noting that the two terms of the resistance are uncorrolated, we obtain the equation Z m y P (y)dy = Z [(1 − α)(1 + yy 0 m yy 0 m ) + α( ) ]P (y)P (y 0)dydy 0 0 0 y+y y+y m = 0, 1, 2, ... (2.55) where it is assumed that near αc the distribution is unchanged by the RG transformation. To test the validity of eq.(2.55) we check it at the two extreme points where the resistance is trivially known, namely at α = 0 (identical resistors R i = r) and at α = 1 (all Ri = 0). It is easy to see that for α = 0 the solution of eq.(2.55) is P (y) = δ(y − 2) namely R = 2r. That is evident directly from eq.(2.44) which, due to the self similarity, now reads ρ ρ=r+ , 2 (2.56) with the solution ρ = 2r. For α = 1 the solution of equation (2.55) is P (y) = δ(y), that is hRi = 0 as asserted. Naturally, the averaged resistance hR(α)i is a monotonic decreasing function of α and the critical point is at α = 1. Unfortunately, perturbation expansion at small = 1 − α near α = 1 is not promising since at this point the dependence on α is non-analytic. Still it is of some value to expand P (x) around α = 1 in order to demonstrate the linear dependence of hRi in (1−α) as α → 1. We use the fact that at α = 1 , P (y) = δ(y) and use the Taylor expansion around α = 1. Taking the derivatives of equation (2.55) we get Z Z yy 0 m yy 0 m y Ṗ (y)dy = [[(1 + ) −( ) ]P (y)P (y 0) + 0 0 y+y y+y yy 0 m yy 0 m + (1 − α)(1 + ) + α( ) ]P (y)Ṗ (y 0)]dydy 0 0 0 y+y y+y m . (2.57) Setting α = 1 we get Z y m Ṗ (y)dy = 1 m = 0, 1, 2, ... (2.58) which has a single solution for all m, Ṗ (y) = δ(x − 1). In the same manner the results for further derivatives are obtained : 25 P (2)(y) = 2δ(y − 1) + 2δ(y − 12 ) P (3)(y) = 24δ(y − 1) + 12δ(y − 13 ) + 6δ(y − 32 ) + 6δ(y − 12 ) P (4)(y) = 8δ(y) + 168δ(y − 12 ) + 48 δ(y − 13 ) + δ(y − 35 ) + δ(y − 32 ) + δ(y − 43 ) (2.59) calculating hyi near α = 0 we get 20 341 3 hyi = (1 − α) + (1 − α)2 + α3 + (1 − α)4 + O((1 − α)5 ) 2 3 30 (2.60) which shows that hyi ∝ (1 − α) at α → 1. In attempt to find an analytic solution for P (y) it is useful to create a generating function for the moments. This is obtained by multiplication of eq.(2.55) by a factor of (−it)m m! and summing over m. The sums converge into an exponent and we get Z e−ity P (y)dy = Z Z [(1 − α)e 0 yy −it(1+ y+y 0) + αe 0 yy −it y+y 0 ]P (y)P (y 0)dydy 0 , (2.61) the integration here is from −∞ to ∞ and the distributions are expanded on the negative side using the Heaviside step-function, i.e. P (x) → H(x)P (x). the l.h.s of eq.(2.61) is just the Fourier transform of P (y). Using the inverse Fourier transform formula P (y) = 1 2π R eity F (t)dt for P (y) and P (y 0) on the l.h.s of eq.(2.61) we have a nonlinear integral equation for the Fourier transform of P (y) F (t) = ((1 − α)e−it + α) Z Z dudvK(t, u, v)F (u)F (v) , (2.62) where K(t, u, v) = yy 0 1 Z ∞Z ∞ 0 −i(t y+y 0 −uy−vy 0 ) dydy e (2π)2 0 0 (2.63) is the Kernel of the equation. Although the integration in eq.(2.63) can be performed (using a transformation to polar coordinates) the result is a non-continuous singular function, which creates great difficulties in solving eq.(2.62) analytically. Still, eq.(2.62) can supply us with the proper results for the edge probabilities α = 1 and α = 0. For the first case P (x) = δ(x) and F (t) = 1. inserting this in eq.(2.62) we get 26 1= Z dudvK(t, u, v) (2.64) which is obviously true for the kernel of eq.(2.63). For the latter case we will use the ansatz F (t) = e−itγ which will yield e −itγ =e = = −it e−it Z2πZ Z Z dudve−iuγ−ivγ K(t, u, v) = Z Z Z Z dudvdydy 0e dydy 0δ(y 0 − γ)δ(y − γ)e γ = e−it(1+ 2 ) 0 yy 0 −it y+y 0 +iu(y−γ)iv(y −γ) 0 yy −it y+y 0 = = . (2.65) This gives γ = 1 + γ2 or γ = 2 , which gives f(x) = δ(x − 2) and for the resistance hRi = 2r as expected from self-similarity considerations. 2. the conductor insulator tree We now consider a distribution for the conductance of a single link g(σ) = αδ(σ) + (1 − α)δ(σ − 1) . (2.66) If we denote by σ1 the conductance of the trunk and by x(y) the conductance of the upper(lower) branch, then the conductance of the entire tree is σ= 1 1 + σ1 x + y !−1 = σ1(x + y) σ+x+y , (2.67) which yields a recursion equation for the conductance distribution of the nth generation tree ! Z Z Z σ1 (x + y) P (n) (σ) = g(σ1)P (n−1) (x)P (n−1) (y)δ σ − dσ1 dxdy = σ1 + x + y ! Z Z x+y (n−1) (n−1) = αδ(σ) + (1 − α) P (x)P (y)δ σ − dxdy 1+x+y (2.68) after the integration over σ1 was performed. We again make an ansatz P (n) (σ) = β (n)δ(σ) + (1 − β (n))Λ(n) (σ) . 27 (2.69) Inserting this to eq.(2.68) and comparing the coefficients of the δ-function we get our RG equations β (n) = α + (1 − α)[β (n−1)]2 (1 − β (n) (n) )Λ Z (2.70) x (σ) = 2(1 − α)β (1 − β ) Λ (x)δ σ − dx + 1+x ! Z Z x+y (n−1) 2 (n−1) (n−1) + (1 − α)(1 − β ) Λ (x)Λ (y)δ σ − dxdy (2.71) 1+x+y (n) (n) (n−1) We now use the RG assumption that at α → αc , β (n) = β (n−1) and Λ(n) = Λ(n−1) . This gives for β β = α + (1 − α)β 2 . (2.72) This equation has two solutions, β = 1 (which is the trivial solution) and β = two solutions merge only at the critical probability, which is thus given by α . 1−α αc 1−αc These = 1 or αc = 12 . We can now insert eq.(2.72) into eq.(2.71) and obtain an integral equation Λ(σ) = 2α Z Z Z x x+y Λ(x)δ(σ − )dx + (1 − 2α) Λ(x)Λ(y)δ(σ − )dxdy 1+x 1+x+y (2.73) Again, since we are only interested with the first moment of the distribution we replace Λ(x) by δ(x − f(α)). Inserting this to eq.(2.73) and integrating over σ gives f(α) = 2α f(α) 2f(α) + (1 − 2α) f(α) + 1 2f(α) + 1 or, after a little algebra, f(α)2 + f(α) 1 + (α − ) = 0 2 2 with the solutions f = At α = 1 2 1 1 (−1 ± (1 − 8(α − )))1 . 4 2 the second solution gives a negative result , which means that the only physical solution is 28 1 1 f = (−1 − (1 − 8(α − ))) = (2α − 1) 4 2 . (2.74) α 1 − 2α δ(σ) + δ(σ − (2α − 1)) 1−α 1−α (2.75) We now insert this result into the ansatz and get P (σ) = and calculating the average conductance we get hσi = and expanding around α = 1 2 1 − 2α (2α − 1) , 1−α (2.76) we get 1 hσi ∼ (α − )2 2 (2.77) which means that the critical exponent is t = 2. As in the previouse section we introduce an equation for the moments of the distribution for the conductance by raising eq.(2.67) to the mth power, Z y mP (y)dy = (1 − α) Z Z x+y 1+x+y !m P (x)P (y)dxdy m = 1, 2, ... notice that eq.(2.78) is not valid for m = 0. By Multiplying eq.(2.78) by a factor of (2.78) (−it)m m! and adding and subtracting 1 (corresponding to the m=0 case), and summing over m we get Z e−ity f(y)dy = α + (1 − α) Z Z x+y e−it x+y+1 P (x)P (y)dxdy (2.79) The l.h.s of eq.(2.79) is the Fourier transform of P (y). Using the inversion formula for the Fourier transform we now get a nonlinear integral equation F (t) = α + (1 − α) Z Z K(t, u, v)F (u)F (v)dudv , (2.80) with the kernel 1 K(t, u, v) = (2π)2 Z Z x+y e−it x+y+1 +iux+ivy dxdy setting new variables z = x + y and w = x − y we get 29 (2.81) Z Z z 1 −it z+1 +iz u+v +iw u−v 2 2 dzdw = e (2π)2 Z z u+v 1 = e−it z+1 +iz 2 δ(u − v)dz . 2π K(t, u, v) = (2.82) Inserting this kernel into eq.(2.80) and performing the integration with respect to v yields the equation F (t) = α + (1 − α) Z K(t, u)F 2(u)du (2.83) where z 1 Z −it z+1 +izu e dz 2π K(t, u) = , (2.84) is a normalized kernel, i.e. Z K(t, u)du = 1 (2.85) We can now check the validity of eq.(2.83) for the edge probabilities. for the case α = 1 eq.(2.83) becomes F (t) = 1 that gives P (x) = δ(x), as expected for a lattice which has only insulating links. For the case α = 0, eq.(2.83) is F (t) = Z K(t, u)F 2(u)du . (2.86) Using the ansatz F (t) = e−itγ we obtain z 1 Z Z −it z+1 +iuz−2iuγ e dudz = 2π Z Z z 1 = e−it z+1 +iu(z−2γ) dudz = Z2π e−itγ = = z e−it z+1 δ(z − 2γ)dz = 2γ = e−it 2γ+1 which is an equation for γ, γ = 2γ 1+2γ (2.87) with the solutions γ = 1 2 and γ = 0. The first solution corresponds to the case of all links present and gives P (x) = δ(x − 12 ) which yields the conductance hσi = 1 2 (in units of the single link contuctance), as expected from considerations of self-similarity. The second solution is γ = 0. It is easily seen that such 30 a solution is valid for all values of α because setting F (t) = C in eq.(2.83) gives C = α + (1 − α)C 2 which has only one constant solution C = 1. We now use the fact that the critical probability for the Cayley tree is p c = 1 f −1 where f is the coordination number, which for our case gives α c = 12 . This means that for α ≥ F (t) = 1 (and P (x) = δ(x)) is the only solution, while for α < solution. Since the two solutions must be equal at α = 1 2 1 2 1 2 there must be another an expansion around small = α− 12 is possible. Starting with F 1 (t) = 1 and using the normalization of the kernel we get, by 2 taking the derivative of eq.(2.83) Ḟ (t) = Z K(t, u)Ḟ (t) , (2.88) with the solution Ḟ (t) = 1. This means that for first order in α − 12 = ,P (x) = δ(x)(1 + ) so hσi = o(2 ) at least. Now taking the second derivative of eq.(2.83) and substituting for F (t) and Ḟ (t) we get F̈ (t) = 5 + Z K(t, u)F̈ (t)du , (2.89) which has no constant solution. This assures us that there must be a non-trivial dependence of hσi on 2 and that the critical exponent in this case is indeed t = 2. Using this procedure a generalization to coordination number f > 3 is easily obtained. Consider a conductor-insulator Cayley tree with coordination number f, then eq.(2.67)takes the following form σ(Σ1, Σ2 , . . .) = Σ1 (σ1 + σ2 + . . . + σf −1) Σ1 + σ1 + σ2 + . . . + σf −1 (2.90) where σi represents the conductance of the ith branch. Following the same procedure as before we obtain an equation similar to eq.(2.80) F (t) = α + (1 − α) Z ··· Z K(t, u1, . . . , uf −1 )F (u1) · · · F (uf −1)du1 · · · duf −1 , (2.91) with the kernel Pf −1 Z Z fX −1 1 i=1 xi K(t, u1, . . . , uf −1) = · · · exp −it Pf −1 + ui xi dx1 · · · dxf −1 (2.92) (2π)f −1 x + 1 i=1 i=1 i 31 in order to simplify this kernel we introduce the change of variables z1 . .. zf −1 = 1 1 ··· ··· 1 x1 1 −1 1 · · · 1 . .. .. .. . . x f −1 1 1 · · · −1 1 (2.93) with the inverse transformation given by x1 . .. xf −1 1 = 2 1 1 .. . −1 0 · · · 0 −1 · · · 0 z1 . ... .. · · · −1 0 zf −1 0 1 0 −(f − 4) 1 · · · 1 1 . (2.94) Changing the variables in eq.(2.92) gives K(t, u1, . . . , uf −1) = Z Z z1 z1 − z2 + iu1( ) + ...+ z1 + 1 2 z1 − zf −1 −(f − 4)z1 + z2 + . . . + zf −1 ) + iuf −1( ) = +iuf −2( 2 2 Z Z z1 (u1 + . . . + uf −2 − (f − 4)uf −1 ) = · · · dz1 . . . dzf −1 exp −it + iz1 + ...+ z1 + 1 2 (uf −1 − u1) (uf −1 − uf −2) +iz2 + . . . + izf −1 = 2 2 Z Z z1 = · · · dz1 . . . dzf −1 exp −it + z1 + 1 u1 + . . . + uf −2 − (f − 4)uf −1 +iz1( ) δ(u1 − uf −1 ) · · · δ(uf −2 − uf −1 ) (2.95) 2 = ··· dz1 . . . dzf −1 exp −it and inserting this kernel into eq.(2.91) and integrating over u 1 , . . . , uf −2 gives the equation F (t) = α + (1 − α) Z K(t, u)F f −1(u)du , 32 (2.96) with K(t, u) the same as in eq.(2.84). Eq.(2.96) has a solution F (t) = 1 for all α. Again, another solution must exist for α ≤ αc = 1 f −1 and can be found by expanding F (t) around αc . Differentiating eq.(2.96) with respect to (1 − α) and substituting (1 − α) = 1 f −1 and F (t) = 1 gives exactly eq.(2.88) with its solution Ḟ (t) = 1. Taking the second derivative of eq.(2.96) will again give an in homogeneous equation, thus assuring that the conductivity hσi depends on (α − αc )2 as α → αc , i.e. t=2. 3. the mixed lattice In this section we consider the case in which a link can either an insulator, a conductor or a super-conductor. This is represented by the conductance distribution P (σ) = αδ(σ − σ0 ) + βδ(σ) + γδ(σ − M) ,α + β + γ = 1 (2.97) where M is a large number, eventually taken to infinity. Using eq.(2.67) and the same procedure as that of the last section we now have an equation for the moments Z y m f(y)dy = Z Z x+y m α( ) +γ 1+x+y M(x + y) M +x+y !m ! f(x)f(y)dxdy (2.98) and using the same method as in the previous section we get the integral equation F (t) = β + Z (αK1 (t, u) + γK2 (t, u)) F 2 (u)du , (2.99) where z 1 Z −it z+1 +izu K1 (t, u) = e dz 2π (2.100) Mz 1 Z −it z+M +izu K2 (t, u) = e dz 2π (2.101) and which, by taking the limit M → ∞ , is just K2 (t, u) = δ(t − u) 33 (2.102) and thus eq.(2.99) has the form 2 F (t) = β + γF (t) + α Z K1 (t, u)F 2(u)du . (2.103) The cases of the previous sections are readily obtained by setting γ = 0 and β = 0 respectively. Notice that F (t) = 1 is a solution of eq.(2.103) for all probabilities, although it is clearly not always a physical solution, as will be discussed later on. We first examine the case in which α = 0, describing the case of a network of insulators and super-conductors. A real lattice of this kind will have only two conductance values possible: if there is a connected cluster of super conductors then it will be a super conductor, else it will be an insulator (with the critical probability γ c = 1 ). Thus, one would expect a step-function-like behavior of the conductance as a function of probabilities and not a critical exponent behavior, and this is indeed the case at hand. for values of β 6= 0, i.e. when the system is still insulating, the only solution of eq.(2.103) is F (t) ≡ 1 which means that the conductance distribution is P (σ) = δ(σ) . At β = 0 we can search for another solution of the form F (t) = exp(iγt). This solution exists only if γ → ∞, i.e. the system is super-conducting. † Next we consider the case in which α 6= 0. I this case there are three kinds of links in the lattice. Clearly at β > 1 2 the lattice will be an insulator (again, since the value of the conductanse does not influence the critical behaviour) the critical probability for an insulator conductor transition is still βc = 12 . In analogy with the previous sections, in order to expand F (t) around this point we now have to differentiate with respect to two probabilities since there are two independent probabilities. These will be chosen for convenience to be β and γ. Differentiating eq.(2.99) (with α = 1 − β − γ) with respect to β and γ gives † This solution demonstrates the difficulty in concidering both super-conducting and insulating links on the same lattice. This difficulty is overcome in the next chapter by introducing quantum conductance, which is confined between 0 and 1, and so this difficulty does not arrise. 34 ∂β F (t) = 1 − + ∂γ F (t) = Z Z Z 1 2 [2(1 − β − γ)K1 (t, u) + 2γK2 (t, u)]F (u)∂β F (u)du (2.104) [ −K1 (t, u) + K2 (t, u)]F 2(u)du + +2 we now set β = K1 (t, u)F 2(u)du + Z [(1 − β − γ)K1 (t, u) + γK2 (t, u)]F (u)∂γ F (u)du (2.105) , γ = 0 and F (t) = 1, thus expanding F (t) around the point ( 12 , 0) in (β, γ) phase space. This gives the equations ∂β F (t) = ∂γ F (t) = Z Z with the trivial solutions ∂β F (t) = 1, K1 (t, u)∂β F (u)du (2.106) K2 (t, u)∂γ F (u)du (2.107) ∂γ F (t) = 1. Since taking the second derivatives of eq.(2.99) yield a non-homogeneous equation, this result again indicates a critical exponent t = 2. Notice that this is valid for every approach to the point ( 12 , 0), since the route will not affect eq.(2.104) and (2.105) and will only appear in the expansion factor , which will be changed according to the route, but this will not affect the resulting conductance hσi. In order to check whether this exponent is valid along the line β = 1 2 we simply set β = 1 2 in eq.(2.104) and (2.105), but allow γ to vary between 0 and 12 . The resulting equations are ∂β F (t) = (1 − 2γ) Z ∂γ F (t) = (1 − 2γ) K1 (t, u)∂β F (u)du + 2γ Z Z K1 (t, u)∂γ F (u)du + 2γ K2 (t, u)∂β F (u)du (2.108) Z (2.109) K2 (t, u)∂γ F (u) and using the normalization of K1 (t, u) and K2 (t, u) we again have the solutions ∂β F (t) = 1 and ∂γ F (t) = 1, which shows that the critical behavior along the β = 1 2 line is independent of the position on that line. We can now present our results in the form of a phase diagram in (γ, β) phase space. 35 γ SC C I β 1 2 Fig. 10 : The 3 species Cayley tree phase diagram As shown, the Cayley tree is only super-conducting if all the links are super-conducting (γ = 1) and the conductivity diverges with a critical exponent s = 1, and it is insulating if most of the links are insulators (α ≥ 0.5), the conductivity now vanishes with an exponent t = 2. 36 D. The 2D square and triangular lattice 1. The square lattice In this section we follow the work of Bernasconi [26] and Stinchcombe et.al. [25]. Since they have already obtained the result of this section we describe the calculation briefly. We treat the case of the 2D square lattice. It is known (from symmetry considerations, and as shown numerically [26]) that in this lattice the critical exponents for the conductor super-conductor case and for the conductor insulator case are the same, i.e. t = s , Thus we only have to deal with one of the cases. our transformation is the link-RG transformation shown in fig.3 which gives rise to a scale factor b = 2. The conductivity of the hyper-cell ,calculated using Kirchoff’s equations, is given by σeq = σ4(σ3 σ5 + σ2 (σ3 + σ5 )) + σ1((σ3 + σ4)σ5 + σ2(σ3 + σ4 + σ5)) σ3 (σ4 + σ5) + σ1(σ3 + σ4 + σ5) + σ2 (σ3 + σ4 + σ5) (2.110) and the conductance distribution of the nth iteration is given by G (n) (σ) = Z ··· Z dσ1 · · · dσ5G(n−1) (σ1) · · · G(n−1) (σ5 )δ(σ − σeq ) . (2.111) We now make the ansatz that G(n) (σ) = β (n) δ(σ) + (1 − β (n))P (n) (σ) , with the normalization condition R (2.112) P (n) (σ)dσ = 1. Placing this ansatz in eq.(2.111) and equating the coefficients of the δ-function gives the RG equations 5 4 β (n) = β (n−1) + 5β (n−1) (1 − β (n−1)) + 3 2 + 8β (n−1) (1 − β (n−1))2 + 2β (n−1) (1 − β (n−1))3 (2.113) P (n) (σ) = W {P (n−1) (σ)} where W is a 5-dimensional integral operator. 37 (2.114) Using the RG assumption that the critical probability is the unstable fixed-point solution of eq.(2.113) we find that βc = β ∗ = 1 2 (where β ∗ is the unstable fixed point) , which is indeed the known exact result for the critical probability of the 2D square lattice. The eigenvalue of the linearized RG equation, λ1 = dβ (n) | ∗ dβ (n−1) β has the value λ1 = 1.625 . In order to find the critical exponent we use the result [25] that at βc a fixed-point solution of eq.(2.114) exists , which satisfies λt P (λt σ) = W {P (σ)} (2.115) i.e. the RG transformation does not change the shape of the conductance distribution but only the scale change. This is justified by the assumption that very close to the critical probability the conductance behaves as < σ >= σ 0 (p − pc )t and near the critical point σ0 is unchanged by a change of scale, so after the RG trasformation the new conductance is < σ >0 = σ0(p0 − pc )t = σ0 λt1 (p − pc )t , and since a fixed solution in the form of eq.(2.115) will satisfy < σ >0= λt < σ > the critical exponent is given by λt = λ1 t (2.116) (notice that this method was not applicable for the Cayley tree since in the Cayley tree there is no typical scale factor for the RG transformation ). Thus, in order to find the critical exponent it is enough to find λt . We use the following scheme. Starting with the distribution P (0)(σ) = δ(σ − 1) (corresponding to the binary distribution G (0) (σ), in units of the singlelink conductance) we obtain a new distribution P (1)(σ) = W {P (0)(σ)}. We then find a value λ(0) such that λ(0) P (0)(λ(0) σ) gives the same average conductance as P (1)(σ). This can then be done iteratively, starting with the distributions P (1)(σ), . . . , P (n) (σ) , The set {λ(n) } converging to λt (by this procedure we actually calculate the distribution of conductance rather then approximate it, and the approximation is done by equating the exponent we get by only a few iterations to the real exponent). Calculating directly we obtain 38 1 1 1 1 1 1 3 P (1) (σ) = δ(σ − ) + δ(σ − ) + δ(σ − 1) + δ(σ − ) 2 2 8 3 8 4 5 (2.117) which gives hσi(1) = 0.566, and a second iteration gives hσi(2) = 0.3078. We can now calculate the exponents directly , using Z σλδ(λσ − σ0) = σ0 λ and obtain 1 = 1.764 0.566 0.566 = = 1.841 0.3078 λ(0) = λ(1) (2.118) and from this we calculate the critical exponent t≈ ln1.841 lnλ(2) = 1.257 . = lnλ1 ln1.625 In order to improve the approximation we wish to solve eq.(2.115) numerically. This is done by evaluating the integrals of eq.(2.115) using a Monte-Carlo technique. The integral operator has the form W {P (σ)} = X i pi × Z ··· Z dσ1 · · · dσni P (σ1) · · · P (σni )δ(σ − σ̃i (σ1, . . . , σni )) (2.119) where the sum is over all configurations (5 of them) , n i the number of links creating the ith configuration and σ̃i the Kirchopf conductivity of the ith configuration. By integrating over σni explicitly each integral can be made to be of the form Z ··· Z dσ1 · · · dσni −1 P (σ1 ) · · · P (σni −1 )F (σ1, . . . , σn1 −1 , σ) (2.120) where −1 ∂σ̃ i F (σ1, . . . , σn1 −1 , σ) = (σ̂ni ) P (σ̂ni ) ∂σn i where σ̂ni is the solution of σ − σ̃i(σ1 , . . . , σni ) = 0 . 39 (2.121) The evaluation was done for conductance instead of resistance for reasons of computational convenience. Since they are just one inverse of the other the critical exponent is the same (up to a sign). Starting from a test distribution, obtained by a fit to the exact conductance distribution (smoothed by coarse-graining) we sampled random numbers σ1, . . . , σni −1 from this distribution (by the Rejection Sampling Method [27]). We then evaluated F (σ1, . . . , σn1 −1 , σ) (for a specific point σ) at the sampled numbers. This procedure is repeated 3000 times, an arithematic mean is then taken to be the value of the integral for the specific σ. Summing for all configurations we then have the renormalized distribution function. The whole procedure is then repeated with the new distribution, at each iteration the avrage is calculated and the ratio between the avrages is used for evaluating the critical exponent as in eq.(2.116). 5 4 3 2 1 0.1 0.2 0.3 0.4 Fig. 11 : exact ant test distribution The exact distribution is ploted in dots, the fit in a line. Notice the drop from zero seen in the exact result, but not taken into account in the test distribution. This approximation is proved to be valid as the distribution remains in shape along the iterations. 2000.00308 <s>=0.00373 150 s7 =1.199 100 50 0 0.00308 a 0.00172 350 <s>=0.00208 300 250 s8 =1.201 200 150 100 50 0 0.00172 b 0.00096 600 500 400 300 200 100 0 <s>=0.00116 s9 =1.207 0.00096 c Fig. 12 : Iterations of the MC calculation of the conductance 40 Fig.10 shows the conductance distributions for the 7th, 8th and 9th iteration respectively. Clearly as the sample size goes to infinity the conductance distribution becomes a δ- function centerd at σ = 0 as expected at the critical point. The critical exponent converges neatly to t ≈ 1.2. This result is in fair agreement with other methods used, which give t ≈ 1.3 [22]. Here we note that the case of a conductor-super conductor lattice gave, within this model, exactly the same result, since the RG equations for the C-SC case are exactly the same as eq.(2.113) and eq.(2.114). 2. The triangular lattice We now apply the same method of the previous subsection to the triangular lattice. Our RG transformation is in the same spirit as that of the last subsection, maintaining the 6-fold symmetry of the triangular lattice Fig. 13 : the triangle lattice link-RG transformation with the scale factor b = 2 . Again, a direction for the conductance can be chosen, now between the 3 lattice direction vectors, and it is convenient to choose the horizontal direction, in which the transformation is as in fig.12 : Fig. 14 : the triangle lattice horizontal transformation We can now calculate the conductance of the renormalized cell using Kirchoff’s laws to obtain σeq = (σ3 + σ5 )(σ4σ6 + σ2(σ4 + σ6)) + σ1[(σ3 + σ4 + σ5)σ6 + σ2(σ3 + σ4 + σ5 + σ6)] (2.122) σ4 (σ3 + σ5 + σ6) + (σ1 + σ2)(σ3 + σ4 + σ5 + σ6) 41 an the conductance distribution is again given by a recursive integral relation similar to eq.(2.111) G(n) (σ) = Z ··· Z dσ1 · · · dσ6 G(n−1) (σ1) · · · G(n−1) (σ6)δ(σ − σeq ) (2.123) and again an ansatz is made, similar to eq.(2.112) which now gives the RG equations 6 5 β (n) = β (n−1) + 6β (n−1) (1 − β (n−1)) + 4 3 + 12β (n−1) (1 − β (n−1))2 + 6β (n−1) (1 − β (n−1))3 2 + β (n−1) (1 − β (n−1))4 P (n) (σ) = W {P (n−1) (σ)} (2.124) (2.125) where now W is a 6-dimensional integral operator. Using the RG assumption of scale invariance we get β c = β ∗ = 0.6372, which is the probability that the hyper-cell will be an insulator, thus 1 − β c = 0.3628 is the probability for it to be a conductor, which corresponds to the percolation critical probability since it requires a percolating cluster (notice that in the square lattice we need not separate the two cases because of the unique symmetry). This result deviates by ≈ 4.5% from the known exact result of pc = 0.3473 [22]. The eigenvalue of the linearized RG equation is now λ 1 = 1.6739. We now use the same method of the previous subsection to calculate the conductance distribution. Starting from P (0) = δ(σ − 1) we obtain 1 1 2 P (1)(σ) = 0.5183δ(σ − ) + 0.1017δ(σ − ) + 0.0725δ(σ − ) 2 3 3 3 2 1 + 0.135δ(σ − ) + 0.0385δ(σ − ) + 0.0192δ(σ − ) 5 5 4 5 7 + 0.0413δ(σ − 1) + 0.022δ(σ − ) + 0.0110δ(σ − ) 7 6 5 13 + 0.0110δ(σ − ) + 0.006δ(σ − ) 8 11 (2.126) which yields hσi(1) = 0.5267 . This, in turn, gives λ(1) t = 2.0272 and a critical exponent t ≈ 1.37414 , which deviates only by ≈ 5% from the t = 1.3 , and clearly supports dimensionalityonly dependence of the critical exponent. 42 3. A PERCOLATION DESCRIPTION OF THE IQHE In this chapter we describe in detail our model for the IQH phase transition which was briefly introduced in chap.I . The model is then solved numerically using the RSRG method presented in chap.II and the critical exponent for the transition is obtained. A. The model As mentioned in chap.I, the nature of the localized states at the insulating phase is still in debate. Nevertheless, the model prestsnted here can describe both approaches , so we present the model in terms of the non-interacting semi-classical picture (a discussion on the newly presented interacting-electrons alternative approach is given in appendix 1). In the semi-classical picture, the electrons (at the Fermi energy) are localized to within the magnetic length around equipotential lines defined by f = En + V (r) (3.1) where f is the Fermi energy and En = h̄ωc (n + 12 ) is the LL energy. Concider two neighbouring orbits, seperated by a potential barier of height . The tunneling amplitude between the two orbits (and hence the transmition between them) will in general depend at both the barrier height and the Fermi energy. As the Fermi energy is increased the orbits circumference becomes larger and the barier heigth between them decreases, resulting in an increase in the tunneling probability between them. As the Fermi energy crosses the barrier height, the two orbits are merged . Due to the magnetic field, the electron motion in both orbits is in the same direction and no back-scattering occurs, and so the two orbits can be concidered as one single orbital. Since we are interested in the conductance, we use the Landauer formula [28], which connects the conductance of a 1D system to the transmission probability through the system, G= ! e2 T πh̄ 43 , (3.2) where G is the conductance and T is the transmition amplitude (in the following we take the proportionality coefficient to be unity). To calculate the transmition of such a 1D system, which in our model will describe a single conducting link (bond), we use the Tight Binding hamiltonian H= X vij c†i cj + H.c. , (3.3) <i,j> where vij is the hopping matrix element (hme) between the ith site and the jth site, and the link is embedded in a 1D ideal linear chain which is connected to a reservoir from both sides, such that all hmes are equal to unity except for the one which characterises the saddle height. Since for Fermi energies larger then the barrier (saddle) height the transmition of the link is unity (since it in fact vanishes and leaves the linear chain unperturbed) a reasonable ansatz for the link hme is given by [9] v = exp(−( − f )) , ≥ f . (3.4) A calculation of the transmition (see appendix 2) gives T = sin2 k sin2 k+sinh(f −) , ≥ f 1 , ≤ f (3.5) where k is given by the dispersion E = f − 2 cos k. An analytic form of the conductance is given by means of the Heavyside step function, sin2 k T = Θ(f − ) + Θ( − f ) 2 sin k + sinh(f − ) . (3.6) For a sample with a smooth random potential, the saddle point heights are given by some distribution g() (we assume that the heights can vary between ±∞ and have zero mean, hi = 0). The conductance distribution can be calculated from g() end eq.(3.6) using the formula for variable change of a random variable, G̃(T ) = Z ∞ −∞ Z ∞ δ(T − T ())g()d = sin2 k = δ T − Θ(f − ) + Θ( − f ) 2 sin k + sinh(f − ) −∞ 44 !! g()d = = Z f −∞ g()dδ(T − 1) + Z ∞ f g()δ(T − = pδ(T − 1) + (1 − p)G(T ) , sin2 k )d = sin2 k + sinh(f − ) (3.7) where p= Z f −∞ g()d (3.8) and G(T ) = (1 − p)T 3 2 sin(k)2 q (1 − T ) sin(k) ×g(f − arcsinh( 2 q 2 r 1 + −1 + 1 T sin(k)2 × sin(k)2 − T sin(k)2 √ )) . T (3.9) Eq.(3.7) has a similar form to that of of eq.(2.21) with two differences. The first is that the super-conducting phase is described in terms of transmission, i.e. T = 1 whereas in the classical case the conductance diverges. The connection between the transmission and the classical conductance is just σ = T , 1−T and in the case of the QHE the transition between them is just a formal change of variables (this problem will arise again in the case where classical conductors are introduced to the system, in sec.4 of this chapter). The second difference is that in the conducting phase the conductance of each link is not unique (as in the classical case), but is given by some probability distribution. This distribution of transmission is dominated by two factors : the distribution of saddle point heights throughout the sample and the Fermi energy. The saddle point height distribution is a character of the disorder potential, and is taken to be the same for all disorder realizations, say a Gaussian. The Fermi energy, on the other hand, is an external parameter, since it is controlled by the filling factor (and thus for a sample with given electron density, by the external magnetic field). The Quantum Hall system is describrd (within the semi-classical description) by such orbits, each connected by a saddle point to its neighbours. If each of these pairs is described as above by two orbits connected by a link with a transmission distribution given by eq.(3.7), 45 then the whole sample can be viewed as a 2D lattice, its vortices are the electron orbits and the links are the saddle points. We take the lattice to be a square lattice for convenience. The random potential is manifested in two ways. The saddle point heights enter the model via the hmes of each link, and the puddle length and shape enters through a random (Aharonov Bohm) phase aqquired by each orbital. This is the essence of our model. The nature of the QH phase transition which comes out of this model is that of a percolation transition. As the fermi energy increases and moves closer to the critical energy (which is the mean potential energy , taken to be that of the band center, i.e. hV i = 0) the concentration p of super-conducting links, given by eq.(3.8), increases. Since this is a square lattice, at p = 0.5 ( or f = 0) there is a percolationg cluster of super-conducting links and the sample is super-conducting. The physical meaning of this is just that at this energy there exists an orbit that spans the entire sample length. Since this transition is similar to that of the classical percolation transition one would expect that they belong to the same universality class, but both the experimental and theoretical evidence shows that this is not the case [4]. Due to the quantum nature of the links, the critical exponent is changed, and the goal of the calculation to follow is to determine, by using the quantum description of the links, the critical exponent. B. The calculation The calculation presented here is based on the same principal as that of the calculation of sec.4 in chap.II . We look at the conductance of the system at small scale (i.e. single links). Then we renormalize the length scale and construct cells from the lattice. Coarse-graining of the cells then takes place, assuming scale invariance near the phase transition. Each cell of the square lattice is composed of eight links as in Fig.3, and we look at the only one direction (say horizontal) such that the horizontal cell looks like that of Fig.4. The first step is to calculate the transmission of the hyper-cell in terms of the individual links. This is done by a tight-binding hamltonian with random hmes for each link , where the 46 hyper-cell is embeded in a 1D linear chain (a detailed calculation is described in appendix 2). In order to model the merging of two orbis into a single orbit (i.e. a super-conducting link) we say that if a link is super-conducting, its hme is given a ”magic value” that joins the two sites connected to the link into a single site. There is no numerical value that can be given to it that will have that effect. Even in the case of a single link with hme v = 1, although the transmission is unity, still the wave functions at the adjacent sites are not identical but have a different phase. In the case of a 2D cell with random phases attached to the hme, the transmition of the cell is different from unity even with all hmes equal to one (in absolute value), and the transition to super-conductivity must be treated in a somewhat artifficial way. Inserting this ”magic value” to the conductance of the hyper-cell, we are left with five possible configurations for the hyper-cell, depending on the number of super-conducting links in it. The configurations are shown in Fig.13 along with the probability for each cell configuration, where p is the probability for a link to be super-conducting, given by eq.(3.8). Fig. 15 : Five cell configurations Each of the configurations has a conductance T̂i, (i = 1, ..., 5) which depends on the hmes of its links. To write it in terms of the link conductances we use the connection of eq.(3.4) and inverse the upper side of eq.(3.5) and obtain √ 2 1−T = f − Arcsinh( √ ) T 47 (3.10) where we take the energies from the center of the tight-bindibg band, i.e. k = π . 2 This is taken throughout the calculation. The resulting integral equation for the transmission of a cell is G̃0 (T ) = p5 + 5p4 (1 − p) + 8p3 (1 − p)2 + 2p2 (1 − p)3 δ(T − 1) + + 5 X i=1 pi × Z ··· Z dT1 · · · dTni G(T1) · · · G(Tni )Λ(φ1) · · · Λ(φni ) × ×δ T − T̂i (T1, ..., Tni ) = = p0 δ(T − 1) + (1 − p0 )G0 (T ) , (3.11) where pi is the probability of each cell configuration , n i the number of links in the nth configuration and φi the random phase of the ith bond, taken from the distribution Λ (taken to be a uniform distribution). The RG equation for the probability is the same as that for the classical case (eq.(2.113)) and so the critical probability is indeed pc = 0.5. The RG equation for the transmission distribution is G0 (T ) = W{G(T )} (3.12) where W is a 5-dimensional integral operator. The numerical solution of eq.(3.12) is done using the following Monte-Carlo scheme. Starting with some initial distribution of transmission, we select (from that distribution) the transmissions of the five links constructing the cell and from a uniform distribution the five phases of the hme. The value of the resulting configuration transmission T̂i is then calculated. This is repeated a large number of times (usually 10000 times) and a histogram of the resulting transmission of each configuration is then ploted. the weighted sum of these functions is the next-generation transmission distribution. The procedure is then repeated with the new distribution function instead of the initial distribution. The first calculation is aimed at determining the length-dependence of the conductance at the critical point, i.e. the the value of the exponent s in eq.(2.13), and to see whether the assumption of eq.(1.4) is valid, which will mean that s = 0 and that at the critical point the 48 conductance in unchanged by change of scale. Starting from an initial uniform distribution the distribution indeed flow to a fixed distribution, given in Fig. 14. 2 10 8 6 4 2 1.5 1 0.5 0 0.2 0.4 0.6 0.8 Transmission 0 -4 1 -2 0 2 saddle height 4 Fig. 16 : The transmission fixed distribution (numeric) and resulting saddle height distribution The saddle heigth distribution was obtained by inversing eq.(3.7) on the > f = 0 side, and reflecting around = 0. This is because we have no information on the distribution for energies lower then the Fermi energy (where the transmission of a link is always unity ) but it is reasonable to assume that the distribution is symmetric around = 0. Notice that it vanishes at = 0, and is not peaked around zero. We belive this is just a nummerical artifact due to the fact that the numeric value of the transmission distribution at T = 1 does not diverge, although on would expect such divergence. This is supported by the fact that as the number of histories in each iteration of the numerical calculation is increased the transmission distribution value at T = 1 increases, suggesting a divergence in the exact distribution. If we fit the numerical distribution to a function which diverges at T = 1 (the best fit is G(T ) = 0.64(1 − T )−0.36 ) the resulting saddle height distribution indeed has a peak at = 0. Next we turn to determining the localization length exponent ν . We calculate the length dependence of the conductance near the critical point by applying the iterative RG procedure for some initial value of δ = (Ec −f ) = f , each iteration multiplies the length and renormalize p (and hence the fermi energy) according to eq.(3.11). The mean conductance is then evaluated for each iteration by the integral σ(L = 2n ) = Z T G(n) (T )dT 49 . (3.13) This is done for different values of δ, and a function σ(δ, L) is obtained. If undeed the scaling assumption is true, then L σ(δ, L) = f ν δ , (3.14) and rescaling of the length for each value of δ by L → Lδ ν would cause all the defferent plots of σ(δ, L) for different values of δ to collapse onto a single curve. Evaluating and minimizing the difference between the different curves will determine the best ν. This is shown in Fig.15, for the best value of ν given by ν = 2.84 ± 0.15 (the error taken as an over estimate, since many things , such as the number of Monte-Carlo histories, e.t.c. contribute to the error and it is hard to calculate the exact overall error) . The scaling function, defined by eq.(3.14) was found to be f(x) ' 0.845 exp(−4.05x0.7 ) . (3.15) 1 0.8 0.8 Conductance 0.6 0.4 0.6 0.2 100 0.4 200 300 400 0.2 0 0 0.02 0.04 0.06 0.08 scaled length 0.1 0.12 0.14 Fig. 17 : Data collapse for the conductance (inset : conductance Vs. unscaled sample length) Although our result for the critical exponent largly deviates from the experimental value of ν ≈ 2.3 (a deviation of ∼ 19%), still this exponent clearly shows that the quantum effects change the universality class from the classical percolation problem to the QHE. 50 We conclude this chapter with a few remarks concerning some aspects of the model. 1. dangling bonds the RG scheme presented here is similar to the classical scheme presented in Chap.II in the sence that the dangling bonds do not play a role in the calculation of the cell conductances. But in the quantum case they might cause additional interference within the cell. We checked the validity of ignoring dangling bonds by repeating the calculation, now taking them into account (at the stage of calculating cell conductances). Indeed we found no dramatic change (within numerical error) in the critical exponent nor in the conductance distribution. 2. phases The importance of the phases was examined in orer to test the hypothesis that tunneling is more importent that interference in determining the universality class of the transition [8]. This was done by repeating the calculation, now taking the phases from a δ-function distribution, i.e. taking constant identical phases. 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