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:

Lecture 8:

The Ising modelA magnet at Cern.Credit:Cern 2D Ising model.Credit:

Sascha Wald

Ising model:Intro duction

I Build-on previous lectures (random walks, cluster growth, percolation) by adding interactions b etweenpa rticles i.e.between occupied sites of the regular grid discussed in Lecture 7 I

In addition, add a

random comp onent to mimic the eects of temp erature on t hesystem not surprisingly, this results in a system with more complex behaviour simulation mimics thermodynamic behaviour of a `real' system I

We will use this

Isi ngmo del

as a mo delfo r ferro-magnetismand study the phase transition associated with ferro magnetism

Ising model:Intro duction

Illustrating the relation between an electron's spin and its magnetic moment.

Credit: Princeton Universit y

The physics!Electrons have a quantum mechanical property calledspin. When spin is measured alonganyaxis, its

value is either~=2 or~=2, where 2~is Planck's constant. An electron's spin is closely related to its magnetic

moment - it is as if an electron is a tiny bar magnet with a North (N) and South (S) pole, with the N pole either

pointing up or down. Just as is the case for bar magnets, the magnetic moments of two electrons close together

create a force between them, such that they will preferentially line up anti-parallel. So we would expect the spins of

two electrons close together to each other to be preferentially anti-aligned. However, there is more to it than that,

because (i) electrons repel each other electro statically since they have the same charge, and (ii) the Pauli exclusion

principle, which states that no two electron can be in the same quantum mechanical state. So consider electrons

arranged on a regular grid, and focus on a nearest-neighbour pair. When anti-aligned, they can be close together

since they are in dierent quantum mechanical states, and hence they will repel each-other electro statically. In

contrast when aligned, they can never get close to each other because that would violate Pauli's exclusion principle,

therefore the electrostatic repulsion between them is not very strong. The upshot of this is, that it is energetically

favourable to be in the parallel spin state. The dierence in energy (between parallel and anti-parallel) is mostly

electrostatic in origin, and can be quite large (of ordereV). This is much larger than the energy associated with

the magnetic interaction. As a consequence, electron spins on a regular lattice will tend to be aligned in the same

direction, with the combined magnetic moment of each electron adding to a large net magnetic moment - this is

Ising's model for ferromagnetism, click

here if y ouw antto kno wmo re.I nsingle atoms, the same phenomenon gives rise to

Hund' srules

fo ro rderingo rbitalsin energ y.

Ising model:Intro duction

I Ising model a superb toy model to understand the physics of ferro-magnetism I

Subject of E. Ising's PhD thesis (1920's)

Ising model:Mathematical mo del

we will restrict ourselves to a two dimensional (2D) Ising model I Consider a 2D square lattice with spins at each lattice site I

Spins can have two values:si=1

our convention here - see below for correct units I Take into account only nearest neighbour interactions left-right, up-down. Nearest-neighbour interaction is good approximation because Pauli's exclusion principle only relevant if spins are close I Write total energy due to electron interactions as E=NX i=1E i;Ei=J2 X j=i1s isj; Sumiruns over allNlattice sites on the square lattice, sumjruns overneighboursofi; factor 1/2 to avoid double counting pairs. Unfo rtunately,the course b ookmisses this facto rof 2. I

Jis theexchange constant, J>0 for ferromagnets

Ising model:Mathematical mo del

I Look at units:~J(~s)2has dimension of energy, where ~s is physical spin with units~, and~Jis the exchange constant I

In our notation, ~s=~2

s, so ~s=~2 impliess=1 I

Therefore~J=~J(~2

)2s2J s2andJhas the dimension of energy I Energy of lattice depends on whether spins are mostly aligned, or mostly random I If all spins are aligned,E=2JN- lowest energy state

IIf spins are random,E0

Ising model:Mathematical mo del

Consider a 2D lattice of spins, at a given

temp erature,T. Temperature means electrons can jiggle about: ifTis suciently high, spins can ip randomly I

Probability

of spin ip from state 1 !state 2 e.gfrom up to down, or vice versa is theBoltzmann factor P

12/exp

E12k BT E

12E2E1, the dierence between the energy in the nal state (i.e.state 2) and initial state (i.e.

state 1);kBis Boltzmann's constant I ifE2P21more likely to ip to lower energy state I ifjE12j kBT,P12 P21at highT, ips in either direction equally likely

Ising model:Mathematical mo del

Suppose we have a spin lattice at a given value ofT. Spin may or may not ip. Whichmacroscopic quantitiescan we compute, and how are they related to the individual spin states? I

For agivenspin conguration, called `micro states'

I

Total energy:E=J2

P N i=1siP j=i1sj I

Magnetisation:M=PN

isi Mis dimensionless, get physical magnetization by multiplying with electron's magnetic moment I A given value ofTcan correspond to many micro state.Themacroscopic state 's properties are I E=P EP I M=P MP

Weigh each micro state by its probability,P.

Problematic because computational expensive: there areverymany possible micro states (in fact, 2N) I Need good way of calculating these macroscopic values - we discuss two of them

Ising model:Mean Field App roximation

MFA - fo r M ean F ield A pproximation I Elegant method - but its predictions are not very accurate is only an approximation I MFA : Replace individual spins with average spin, s i=1! hsi M=X is i!M=X ihsi=Nhsi Nhsii I Works well for innitely large system where all spins are equivalent I

How can we compute this in practise?

Ising model:Mean Field App roximation

I Add an external magnetic eldappears to be a detour, but wait & see! E=J2 P N i=1 P j=i1s isj! HP is i (External magnetic eldHinteracts with spins through their magnetic moment,.) I

Apply this to a system with just one spin:

E =H notice how ! : spin aligned withHhas less energy than anti-aligned IThis has two micro states,with probabilitiesP=Cexph Hk BTi I

Determine normalisationCby requiringP++P= 1

=)C=1exp Hk BT +exp Hk

BT=12cosh

Hk BT I

Therefore thermal average of the single spin:

hsii=P+ P= tanhHk BT

Ising model:Mean Field App roximation

I Having the solution for a single spin in a background eld, we replace the background eld with the average spins E=P i J2 P j=i1s j+H! s i HeP is i I

The eective magnetic eld is therefore

H e=J2P j=i1s j+H I

Mean eld approximation

: setsj! hsiandH!0: H

MFA=nJ2hsi

Here,nis the number of nearest neighbours,n= 4 in our 2D case I

Combining this withhsi= tanhHMFAk

BTyields a non-linear equation forhsi

hsi= tanhTcT hsi ;TcnJ2kB: T cis called thec riticaltemp erature

Ising model:Mean Field App roximation

numerical example, forT=Tc2 (`low'T, left panel) andT=Tc0:8(`high'T, right panel) I

Notice the two dierent regimes:

3 solutionsTTc, rightas expected: left panel: lowT, magnetization, can be up,hsi= 1, or down,hsi=1, or no magnetization

right panel: highT, no net magnetization,hsi= 0

Ising model:Mean Field App roximation

Magnetization as a function of temperature

I

Solve numericallyf(hsi) =hsi tanhTchsiT

= 00.000.250.500.751.001.251.501.752.00

T/Tc0.00.20.40.60.81.0

M/Mmax

Magnetization from mean field approximation

Ising model:Mean Field App roximation

I When making plots:plotMin units ofMmax=N, and setJ=kB for simplicity. I Plots illustrates a phase transition atT=Tc(Tc= 2J=kB= 2) ofsecond order meaning 1st derivative of o rderpa rameter , in this case magnetisation , is discontinuous at transition I

AroundTc:dMdT! 1.

I Exact form of singularity from Taylor expansion of tanh: tanhx=xx33 +O(x4) I

Therefore, aroundT=Tc:

hsi=TcT hsi 13 TcT 3hsi3

Ising model:Mean Field App roximation

I

Examine behaviour aroundT=Tc: deneTcT

1, with 0< 1 hsi= (3)1=2/(TcT)1=2/(TcT) I

Critical temperature

and critical exp onent T c=nJ2kB;=12 I

Exact analytical (non MFA) result is

T c=2:27Jk B;=18 for a square lattice withn= 4 I

Will now turn numerical/simulation treatment

Ising model:Numerical treatment

I

Strategy very similar to what's been done before:

Use a random number generator to dec idewhether to ip a spin spin ip probability is Boltzmann factor I Algorithm: loop over spins one at a time, decide whether it ips (compareP ipwith number from RNG), repeat untilMequilibrates I

To calculateP

ip: Use energy of the two micro-states (before and after ip) and Boltzmann factors. I While running, evaluate observables directly and take thermal average (average over many steps).

Ising model:Numerical treatment

this is called theMetropolisalgorithmLayout of programme: 1.

Initialise

the lattice, i.e.choosesifor each spin (either at random, orsi= 18i, or similar) 2.

S weep

over all spins

At each step, decide whether or not to

ip spin: I

Calculate the system's energyE=J=2Psisj

Ifor current spin state, energyE1

Iif spin were

ipped, energyE2

ICalculate E=E2E1

IE<0 :

ip spin IE0: ip spin if exp Ek BT >R whereRis a random number2[0;1] 3.

Re peat

step 2 until magnetization in equilib riumatTc, never in equilibrium

Ising model:Numerical treatment

why does Metropolis algorithm work: Detailed balance I

Consider spin

ips between states 1 and 2energyE1, andE2>E1 I

Metropolis algorithm:

I

Probability spin

ip 1!2 isP1!2= 1 I

Probability spin

ip 2!1 isP2!1= exp E1E2k BT 1

Does this give the right answer?

I

Analysis:LetW1be the fraction of spins in state 1&W2for state 2The rate of transitions from 1!2 and vice versa is

R

1!2=W1P1!2=W1;R2!1=W2P2!1=W2exp

E1E2k BT! the product of the fraction of spins in a given state times the probability that a spin ips

In thermal equilibrium,R1!2=R2!1, in which case

W

1=W2= exp((E1E2)=kBT)that is, states are occupied according to the

Boltzmann distribution. Applying the Metropolis algorithm therefore drives systems to thermal equilibrium

Ising model:Numerical treatment

I

Metropolis algorithm drives system to

thermal equilib rium W 1W

2= exp((E1E2)=kBT)

W

1andW2fraction of spin in states 1, and 2, with energiesE1andE2

I In principle, all systems in thermal equilibrium can be studied with Metropolis - just need to write transition probabilities in accordance with detailed balance, as above. I Metropolis algorithm simulates thecanonical ensembleby summing over micro-states with a Monte Carlo method.

Ising model:Numerical treatment

Sketch of Metropolis codeInitialiseanLLlattice with spinssi. Set allispins constant,si= 1, or at random,si=1Sweepover al lspins Sweep (meaning go) systematically through the lattice, line by line, column by column, and decide for each spin in turn whether it ips or not. Note that, to compute Efor a given ip, youdo not need to sum over all spinsImpose periodic boundary conditions spin at (0;j) has neighbours at (1;j)and(L1;j), in addition to (0;j+ 1) and (0;j1). Such a treatment reduces nite-size eects, but one should keep in mind that correlations with a length larger than p2Lcannot be simulatedComputeMso that you can plotMversus number of sweeps

Ising model:Numerical treatment

I In workshop:red-black sweepssubtlety in sweeping over spin I

Consider a chess-board: has red and black squares

Iwhen sweeping over spins:

I sweep over red spins rstrst horizontally, then vertically Ithen sweep over all black spinrst horizontally, then vertically I this improves the rate at which the system thermalises

Ising model:Numerical treatment

Example ofMas a function of sweep number

I

At choosenT, sweeps on an 1010 lattice

Ising model:Numerical treatment

Analysis of result

I

At low temperature (T= 1 orT= 1:5Tc2):

system quite stable, with small uctuations around

M=Mmax

I

At high temperature (T= 4Tc2): system has

M0, with relatively large

uctuations aroundM= 0 I At intermediate temperatures (T= 2) we see very large uctuations I Close to the critical value (T= 2:25Tc) see even large uctuations, withM1 for a large number of sweeps, followed by a jump toM=1 for a large number of sweeps

Ising model:Numerical treatment

Phase transition - the MC look at things

I

Analyse 1010 lattice as function of temperatureAs expected from MFA: whenTTc: spins are aligned,MMmax

WhenTTcspins are not aligned,M0. Second-order phase-transition aroundT=Tc2J=kB

Ising model:Numerical treatment

Discussion

I Results above plotted when system is in equilibrium I critical slowdown a roundcritical p oint: The system's time to equilibrate diverges (never in equilibrium) I Independent of this: Monte Carlo results in agreement with exact calculationand MFA calculation not very accurate but does describe generic behaviour correctlyquotesdbs_dbs19.pdfusesText_25
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