Download 1 Introduction : Phase transitions in 2D electron systems 2

Contents
1
Introduction : Phase transitions in 2D electron systems
2
An RSRG approach to lattice conductivity
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A
Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
10
B
The 1D lattice . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
17
1 The conductor super-conductor lattice . . . . . . . . . . . . . . . . . .
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2 The conductor insulator lattice . . . . . . . . . . . . . . . . . . . . . .
19
The Cayley tree . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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1 The conductor super-conductor tree . . . . . . . . . . . . . . . . . . .
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2 the conductor insulator tree . . . . . . . . . . . . . . . . . . . . . . . .
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3 the mixed lattice . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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The 2D square and triangular lattice . . . . . . . . . . . . . . . . . . . . .
37
1 The square lattice . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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2 The triangular lattice . . . . . . . . . . . . . . . . . . . . . . . . . . .
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C
D
3
2
A Percolation description of the IQHE
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A
The model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
43
B
The calculation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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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].
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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
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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.
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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.
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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
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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.
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************************* 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
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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 ?
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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
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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 ,
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< 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
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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.
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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. The resulting exponent was ν = 2.66 ± 0.1,
somewhat different from that of the uncorrolated phases calculation , but still a mild correction in comparison to the difference from the classical exponent, a result which supports
the above conjecture.
3. the 1D case
In the case of a 1D linear chain, the critical probability for percolation is p = 1, and the
fixed distribution in that case is Gf ixed (T ) = δ(T − 1) . If we try to repeat the RG calculation for the 1D case we are then complied to choose a distribution for the saddle point
energies ( while for the 2D case such a distribution was obtained from the fixed distribution
of conductances ). It turns out that the critical exponent strongly depends on the detailes
of the chosen distribution, and thus for the 1D case this calculation is meaningless, and a
different scheme must be found .
51
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53