Chapter 2 Ising Model for Ferromagnetism

Transcription

1 Capter Ising Model for Ferromagnetism Abstract Tis capter presents te Ising model for ferromagnetism, wic is a standard simple model of a pase transition. Using te approximation of mean-field teory, te free energy is minimized, and ence te magnetization is calculated, as a function of temperature and applied field. Tis calculation demonstrates some fundamental concepts in statistical mecanics, including spontaneous symmetry breaking, an order parameter wit magnitude and direction, first- and second-order pase transitions, and a critical point caracterized by critical exponents. In te rest of te book, tese concepts will be applied to te teory of soft matter. In tis capter, I will introduce te Ising model for ferromagnetism. Tis is probably te single most commonly studied model in statistical mecanics one migt say tat te Ising model is to statistical mecanics as te fruit fly is to genetics. As you will see, te Ising model sows te essential concept of ow te balance between energy and entropy leads to a pase transition. Historical note: Tis model of ferromagnetism was developed in 194 by Professor Wilelm Lenz and is graduate student Ernst Ising. As far as I know, it is te only case in te istory of science were te work was named after te student, not after te professor. I promise you tat will NEVER appen again!.1 Model In te Ising model, we consider a lattice of magnetic moments, as sown in Fig..1. (In tis figure, I ave drawn a two-dimensional (D) square lattice, but in general, we could ave any lattice structure in any dimension.) On eac lattice site, te local magnetic moment is represented by a spin, drawn as an arrow in te figure. We assume tat te spin as just two possible states, eiter pointing up or pointing down. Matematically, we represent te spin at site i by te variable σ i =±1. In tis notation, +1 means tat te spin is pointing up, and 1 means tat it is pointing down. Te energy for te Ising model includes two contributions: te interaction between neigboring spins and te effect of an applied magnetic field on eac individual spin. Te interaction between neigboring spins tends to induce parallel alignment of te Springer International Publising Switzerland 016 J.V. Selinger, Introduction to te Teory of Soft Matter, Soft and Biological Matter, DOI / _ 7

2 8 Ising Model for Ferromagnetism Fig..1 Example of te Ising model on a D square lattice. Eac arrow is a spin, wic represents a magnetic moment tat can point eiter up or down neigbors, so it sould be favorable (negative) wen te neigbors are bot +1 or bot 1, and unfavorable (positive) wen te neigbors are +1 nextto 1. Hence, for eac pair of neigbors i and j, te interaction energy can be written as Jσ i σ j, were J is a positive coefficient giving te interaction strengt. 1 If te magnetic field is pointing up, it favors eac spin pointing up; if te field is pointing down, it favors eac spin pointing down. Hence, for eac site i, te field energy can be written as σ i. Putting tese pieces togeter, te total energy for te system becomes E = J σ i σ j i, j i σ i. (.1) Note te indices on tese two sums. In te first sum, te angle brackets i, j represent nearest-neigbor pairs (for example, nort-sout or east-west on te D square lattice); te sum is taken over all nearest-neigbor pairs. By comparison, te second sum is taken over all individual sites i, wic are eac affected by te magnetic field. Do not worry about te edges of te system; we will neglect tem in our discussion. Tat is a reasonable approximation if te system is very large, so tat only a tiny fraction of te sites are on te surface. Our goal is now to calculate ow muc magnetic order is in te system. Suppose tere are N spins in te lattice, wit N spins pointing up and N spins pointing down, 1 Te parameter J is sometimes called te excange constant, for reasons based on te quantum mecanics of magnetism. To be precise, is proportional to te magnetic field H, scaled by te magnetic moment μ per spin; people often disregard te factor of μ and refer to as a field.

3 .1 Model 9 so tat N = N + N. Te total magnetic moment of te system is μ(n N ), were μ is te magnetic moment of eac spin, and te largest possible magnetic moment is μn. Hence, it is natural to define te magnetic order parameter or magnetization M as te expectation value of te magnetic moment relative to te largest possible magnetic moment, N N M =. (.) N Hence, we want to calculate M as a function of te interaction strengt J, te magnetic field, and te temperature T. Note tat M can assume values from 1 to 1. Te absolute value of M indicates te magnitude of magnetic order. If M is close to 0, ten te system is igly disordered, wit approximately alf of te spins pointing up and alf pointing down. By comparison, if M is close to 1, te system is igly ordered, wit almost all of te spins pointing in te same direction. Te positive or negative signs of M indicate te direction of magnetic order if it is positive, ten te net order is pointing up; if it is negative, ten te net order is pointing down. In furter capters, we will see tat tese are very general features of order parameters; tey always sow te magnitude and direction of order.. Non-interacting Spins As a first step, just for practice, let us do te calculation for non-interacting spins wit J = 0. In tis case, we can solve for M exactly, and it will elp us get ready for te muc arder problem of interacting spins wit J > 0. In tis section, I will present two ways to solve te problem of non-interacting spins. Te first approac is a standard solution, wic you ave probably seen in courses on termal pysics or statistical termodynamics. Te second approac is a more interesting solution in terms of energy and entropy...1 Standard Solution For te standard solution, we begin wit te partition function Z = states e E state/k B T. (.3) Here, a state refers to te full list of te values of te spins σ 1, σ, etc., and te energy of a state is E = i σ i in te non-interacting model. Hence, te partition function becomes

4 10 Ising Model for Ferromagnetism Z = σ 1 =±1 σ =±1 σ N =±1 e (/k B T ) i σ i. (.4) Because te energy consists of separate terms for eac spin, wit no interactions, te sum factorizes into Z = e (/k B T )σ 1 e (/k B T )σ... e (/k B T )σ N. (.5) σ 1 =±1 σ =±1 σ N =±1 Because eac of tose factors is identical, te partition function for N spins factorizes into te product of single-spin partition functions, Z = Z N 1, (.6) were Z 1 = σ 1 =±1 e (/k B T )σ 1 = e +/k B T + e /k B T. (.7) For eac site, te probabilities of pointing up or down are e +/k B T Hence, te expectation value of any single spin is p = e +/k B T + e /k B T, (.8) e /k B T p = e +/k B T + e /k B T. (.9) σ i =(+1)p + ( 1)p = e+/k B T e /k B T ( ) e +/k B T + e /k B T = tan. (.10) k B T Likewise, because all te spins are identical, te magnetic order parameter is ( ) M = tan. (.11) k B T Figure. sows a plot of tis result for M as a function of /k B T. Note tat M is zero at = 0, i.e., tis non-interacting model as no magnetic order witout a field. Moreover, M saturates at its maximum value of +1 wen +, and at 1 wen. We migt ask: How large of a magnetic field is required to induce an order parameter of, say, 75 % of its maximum value? From Eq. (.11), we see tat it occurs at /k B T = tan 1 (0.75) 1, or in oter words, at k B T. Hence, te temperature determines ow sarply M saturates as a function of.only a small field is required at low temperature, but a muc larger field is required at ig temperature. A related question is: How strongly does M respond to a small applied

5 . Non-interacting Spins 11 (a) M 1.0 (b) M Fig.. Magnetic order parameter of te non-interacting Ising model, as a function of. a For k B T = 0.5. b For k B T = (Interactive version at ttp:// document/cda_downloaddocument/selinger+interactive+figures.zip?sgwid= p ) field, i.e., wat is te derivative of M wit respect to at = 0? Tis derivative is called te susceptibility χ, and it can be calculated as χ M = 1 =0 k B T. (.1) Hence, M responds extremely sensitively to small in te limit of low temperature... Solution in Terms of Energy and Entropy I would now like to present a different solution to te same problem of non-interacting spins in terms of energy and entropy. It will, of course, give te same answer! As we recall from te toy model of dust on a table, a key concept in statistical mecanics is classifying microstates into macrostates. In te Ising model, eac microstate refers to a particular configuration of up and down spins. Let us now classify tese microstates according to te value of M. For example, if we ave a system of 100 spins, we can ave one macrostate wit M = 0 (50 up, 50 down), anoter macrostate wit M = 0.0 (51 up, 49 down), etc. In general, if we ave a system of N spins, wit N up and N down, ten N + N = N, (.13) N N = NM. Tese equations imply ( ) 1 + M N = Np = N, (.14) ( ) 1 M N = Np = N.

6 1 Ising Model for Ferromagnetism We can now ask: How many microstates correspond to te macrostate wit a certain value of M? In oter words, in ow many ways can we divide N total spins into N spins pointing up and N spins pointing down, were N and N are given by Eq. (.14)? Tis is a standard combinatorial problem, wic you migt ave studied in a class on probability and statistics. Te answer is given by te binomial coefficient # of microstates = ( N N ) Hence, te entropy associated wit tat value of M is = N! N!N! (.15) S(M) = k B [log(n!) log(n!) log(n!)]. (.16) For large N, we can approximate N! by Stirling s formula log(n!) N log N N, (.17) and likewise for log(n!) and log(n!). Hence, te entropy simplifies to S(M) = k B [N log N N log N N log N ] (.18) = Nk B [p log p + p log p ] [( ) ( ) ( ) ( )] 1 + M 1 + M 1 M 1 M = Nk B log + log. Wat about te energy? For te non-interacting system, te energy is just te sum of te field energy terms for eac of te N spins: E(M) = (N )( ) + (N )(+) = NM. (.19) Hence, te free energy as a function of M is F(M) = E(M) TS(M) (.0) = NM [( ) ( ) ( ) ( )] 1 + M 1 + M 1 M 1 M +Nk B T log + log. Note tat te free energy is proportional to te number of spins N, as it sould be. Tis implies tat te free energy is extensive; if we double te number of spins, ten we double te free energy. Hence, we can factor out N to obtain te free energy per spin: F(M) N [( 1 + M = M + k B T ) log ( 1 + M ) + ( 1 M ) log ( 1 M )]. (.1)

7 . Non-interacting Spins 13 (a) F M (b) F M Fig..3 Free energy of te non-interacting Ising model, as a function of M. a For /k B T = 0. b For /k B T = 0.3 (Interactive version at ttp:// downloaddocument/selinger+interactive+figures.zip?sgwid= p ) Wat does tis free energy look like, as a function of M? Well, te function involves two parameters, and k B T. It would be easier to analyze te sape of te function if it ad only one parameter. For tat reason, we divide bot sides of te equation by k B T to obtain ( ) F(M) Nk B T = M + k B T ( 1 + M ) log ( 1 + M ) + ( ) ( ) 1 M 1 M log. (.) Now we can plot it for several values of te single parameter /k B T. Te results are sown in Fig..3. Note tat te minimum depends on te value of /k B T.If /k B T = 0, te minimum is exactly at M = 0. If /k B T > 0, te minimum sifts to positive values of M. As/k B T, te minimum sifts toward M = 1. Likewise, if /k B T < 0, te minimum is at negative values of M, and it approaces M = 1as/k B T. To find te minimum algebraically, we just calculate te derivative and set it equal to zero: M ( F(M) Nk B T ) = k B T + 1 ( ) 1 + M log = 0. (.3) 1 M Te solution is ( ) M = tan, (.4) k B T wic is exactly te same as Eq. (.11)! Tis solution is already plotted in Fig.., and we discussed it tere. Hence, tese solutions are consistent. Wat do we learn from tis second version of te solution? Well, we see tat order parameter M is controlled by a competition between energy and entropy. Te entropy favors te macrostate wit M = 0, because tis macrostate as te most microstates. By contrast, te energy favors te largest possible value of M aligned wit te field (positive M if > 0, negative M if < 0). By minimizing te free energy, we can find te equilibrium value of M. We will use tat concept trougout tis book.

8 14 Ising Model for Ferromagnetism.3 Interacting Spins Now let us consider te case were te spins are interacting. In oter words, we are back to te Ising energy of Eq. (.1) wit te interaction strengt J > 0. We would like to calculate te magnetic order parameter M as a function of J,, and T in tis case. Unfortunately, tis problem is muc arder tan te non-interacting spins. It is not just arder in te sense tat I need to look up a tricky integral, or tat I ave to get Matematica to calculate someting numerically. It is arder in te sense tat it consists of a uge number of variables tat are all coupled togeter. Instead of te non-interacting partition function of Eq. (.4), we now ave Z = e (J/k B T ) i, j σ i σ j +(/k B T ) i σ i. (.5) σ 1 =±1 σ =±1 σ N =±1 If we try to factorize tis partition function into separate terms for eac spin, as in Eq. (.5), it does not work! Te J term couples spin #1 to its neigbors, and tose spins to teir neigbors, and so fort, until all te spins are coupled togeter. For tat reason, we cannot solve N copies of te same single-spin problem; we must solve a single N-spin problem. For a macroscopic system, N is a very large number, peraps It is almost always impossible to solve a problem like tat exactly. Altoug we cannot solve te problem exactly, tere is a very useful approximation called mean-field teory, wic provides a lot of insigt into te beavior. In tis approximation, we neglect te correlations between neigboring spins, and assume tat tey are eac fluctuating independently wit te same statistical distribution. In te following two sections, I will explain tis approximation to you in two different ways..3.1 Mean-Field Teory in Terms of Energy and Entropy For a first approac to mean-field teory for te Ising model, let us continue to work in terms of energy and entropy. By analogy wit te non-interacting model, we will classify te microstates into macrostates according to teir order parameter M. We already worked out te entropy as a function of M in Eq. (.18), and te field energy as a function of M in Eq. (.19), and we will continue to use tose results. Hence, we just need to work out an expression for te interaction energy as a function of M. For any particular microstate wit specific spins σ 1, σ,... σ N, te interaction energy is given by E int = J i, j σ i σ j. (.6) Clearly it does not just depend on M; it depends on all te spins. Can we at least calculate te expectation value of tat energy? Te expectation value is given by

9 .3 Interacting Spins 15 E int = J i, j σ i σ j. (.7) It involves correlations between neigboring spins σ i and σ j, wic we do not know. In mean-field teory, we now make a HUGE approximation. We neglect te correlations between neigboring spins, and write σ i σ j σ i σ j. (.8) If you ave any experience wit statistics, you will not like tat line; you will complain to me tat it is not true. You are rigt; it is not true; it is an approximation! We cannot yet tell weter it is a good approximation or a bad approximation. As you will see, it actually works surprisingly well, muc better tan we ave any rigt to expect. If you believe te mean-field approximation, ten we can write E int J i, j σ i σ j. (.9) Because all te spins are identical, tey all ave te same expectation value: σ i = σ j =M for all sites i and j. Hence, eac term in te sum reduces to M.How many terms are in te sum? Well, let us suppose tat eac site on te lattice as q nearest neigbors; q is called te coordination number of te lattice. In general, q depends on te lattice type and dimensionality. For te D square lattice sown in Fig..1, weaveq = 4. If tere are N sites on te lattice, and eac site interacts wit q neigbors, you migt expect tat tere are Nq pairs of nearest neigbors in te sum. Unfortunately, tis argument double-counts te pairs, i. e. it counts σ 1 as a neigbor of σ, and σ as a neigbor of σ 1. Wen we eliminate te double-counting, te number of nearest-neigbor pairs in te sum is 1 Nq. Hence, our mean-field approximation for te interaction energy is E int 1 NJqM. (.30) We now construct te free energy F = E TStat includes te mean-field approximation for te interaction energy, along wit te field energy and te entropy, ( ) ( ) F(M) Jq Nk B T = M M (.31) k B T k B T ( ) ( ) ( ) ( ) 1 + M 1 + M 1 M 1 M + log + log. Tis free energy depends on two parameters, Jq/k B T and /k B T. To understand its beavior, we need to consider various cases.

10 16 Ising Model for Ferromagnetism (a) F 0.45 (b) F 0.50 (c) M F M (d) M F M Fig..4 Free energy of te interacting Ising model, wit /k B T = 0, as a function of M. a For Jq/k B T = 0.9. b For Jq/k B T = 1. c For Jq/k B T = d For Jq/k B T = 1.5 (Interactive version at ttp:// Selinger+Interactive+Figures.zip?SGWID= p ) Let us first consider te case were tere is no applied field, = 0. We begin at ig temperature, so tat Jq/k B T is small. In tat case, te free energy as a function of M as te form sown in Fig..4a. It is approximately a parabola pointing upward, wit te minimum at M = 0. Now we gradually reduce te temperature, i.e., increase te value of Jq/k B T. As we reduce te temperature, te sape of te free energy around M = 0 gradually gets flatter and flatter. (I strongly recommend tat you try te interactive figure to see tis trend for yourself.) At a critical value Jq/k B T = 1, sown in Fig..4b, te minimum becomes so flat tat te second derivative goes to zero. At tis point, te curve no longer looks like a parabola; instead, it looks like a fourt-order function. As we reduce te temperature furter, te single minimum at M = 0 splits up into two minima at small positive and negative values of M, as sown in Fig..4c. As te temperature continues to decrease, te minima move outward and eventually approac ±1, as sown in Fig..4d. (Te cases wit nonzero field in Fig..5 will be discussed later.) Notice wat is appening ere: At ig temperature, te system goes to te state wit no magnetic order, M = 0, as favored by entropy. By comparison, at low temperature, te system goes to a state wit some magnetic order. Because tere is no applied field, te system as no preference about weter te magnetic order sould point up or down. Hence, it randomly cooses one of te minima, wit positive or negative M. Tis low-temperature state wit spontaneous (not induced by field) magnetic order is called a ferromagnetic pase. By contrast, te ig-temperature

11 .3 Interacting Spins 17 (a) F (b) F M M Fig..5 Free energy of te interacting Ising model, wit /k B T = 0.0, as a function of M. a For Jq/k B T = 0.9. b For Jq/k B T = 1.5 (Interactive version at ttp:// content/document/cda_downloaddocument/selinger+interactive+figures.zip?sgwid= p ) state wit no spontaneous magnetic order is called a paramagnetic pase. Te cange from paramagnetic to ferromagnetic at a specific temperature is a pase transition. Notice also a point of symmetry: Te ig-temperature paramagnetic pase as a symmetry between up and down, wit no preference for eiter direction. In te lowtemperature ferromagnetic pase, tis symmetry is broken and te system randomly goes one way or te oter. Tis random selection is called spontaneous symmetry breaking. I sould empasize tat te ig-temperature state as more symmetry and it is disordered. By contrast, te low-temperature state as less symmetry and it is more ordered. In tis sense, order means te opposite of symmetry; it means broken symmetry. (Students sometimes get confused about tis point. Tey tink tat order is good and symmetry is good, so order must be te same ting as symmetry. No, it is not a question of good and bad!) We migt want to know te pase transition temperature precisely (not just by playing wit te graps). One feature of te pase transition tat we already noticed is tat te second derivative F/ M = 0atM = 0. Hence, we calculate te second derivative: ( ) F M = Jq Nk B T k B T + 1 (1 + M) + 1 (1 M). (.3) At M = 0, it becomes Hence, te transition occurs at te temperature ( ) F M = 1 Nk B T M=0 Jq k B T. (.33) T C = Jq k B. (.34)

12 18 Ising Model for Ferromagnetism (a) M T T C (b) M T T C Fig..6 Ising order parameter M as a function of temperature T. a For /k B T = 0. b For /k B T = 0.0 (Interactive version at ttp:// downloaddocument/selinger+interactive+figures.zip?sgwid= p ) Tis temperature is called te critical temperature, and ence is given te notation T C. Note tat T C is proportional to te interaction strengt J and te coordination number q. If tose quantities increase, ten it is easier to get an ordered pase, so te pase transition occurs at a iger temperature. We migt also want to know ow te equilibrium value of M varies wit temperature, as te system cools from T = T C down to T = 0. One approac is to minimize te free energy numerically at eac temperature. A plot of te numerical result is sowninfig..6a. Alternatively, we can try to find te minimum by setting te first derivative equal to zero: ( ) ( ) F Jq = M M Nk B T k B T k B T + tan 1 M = 0. (.35) We cannot solve tis equation analytically in general, but we can make a good approximation wen M is small (wic is valid for T sligtly below T C ). In tis case, we approximate te inverse yperbolic tangent by its Taylor series to obtain ( ) F M Nk B T ( TC T ) M (M k B T ) M3 = 0. (.36) (In te first term of tis expression, T C is substituted in place of Jq/k B.) For = 0, te solution is 3(TC T ) 3(T C T ) M = 0orM ± ±. (.37) T T C Note tat M = 0 is te free energy minimum for T > T C, and te free energy maximum for T < T C. Te nonzero solution is only defined for T < T C,wereitis te free energy minimum.

13 .3 Interacting Spins 19 From Eq. (.37), or just from looking at Fig..6a, we see tat M increases as te square root of te temperature difference wen T drops sligtly below T C.Tis beavior is an example of a scaling relation. In general, people say tat M (T C T ) β, (.38) were te exponent β is called a critical exponent. Here, we see tat β = 1 in meanfield teory for te Ising model. Tis mean-field prediction is not exactly correct: More precise teories (and experiments!) sow tat β = 1 8 for a D Ising system and β 0.31 for a 3D Ising system. Te mean-field prediction for te exponent is only exact for 4D or iger. However, mean-field teory is still teacing us a lot: It sows tat tere is a pase transition and tat te order parameter as singular beavior just below te transition, wic can be described by a power law. So far we ave only considered te interacting system wen tere is no applied field, but we can also consider te case wit a field. If we apply a magnetic field, it breaks te symmetry between M > 0 and M < 0, so te free energy is no longer an even function of M. ForT > T C, te free energy as te form sown in Fig..5a. It as only a single minimum, and tat minimum is not exactly at M = 0; it is displaced from zero by te applied field. By contrast, for T < T C, te free energy as te form sown in Fig..5b. It as two minima, and tese minima are not equally deep. Te applied field breaks te symmetry between te minima, and favors te state wit magnetic order aligned wit te field. Under an applied field, tere is a smoot crossover from T > T C to T < T C. Figure.6b sows te equilibrium value of M as a function of T for a small but nonzero field. Note tat tere is no singularity in M, i.e., no pase transition! Rater, M increases rapidly but smootly around T T C. We can understand tis smoot beavior by saying tat te symmetry between M > 0 and M < 0 is already pre-broken by te field, so tere is no need for a symmetry-breaking transition. For an alternative view of te same pysical beavior, imagine an experiment tat varies te field at fixed temperature. For T > T C, te beavior is sown in Fig..7a. At tis ig temperature, tere is no pase transition; rater M scales linearly wit for small fields, and ten saturates for large fields. As T approaces T C, te linear response to small field becomes sarper and sarper. Wen T = T C, te linear response to small field becomes infinitely sarp, i.e., te plot of M() as an infinite slope rigt at = 0, as sown in Fig..7b. Wen T < T C, te response actually as a discontinuity at = 0, as in Fig..7c. Tink about te free energy curve at tis low temperature, wic as two minima. As passes troug 0, te order parameter jumps from te minimum at M < 0 to te minimum at M > 0. Te magnitude of tis discontinuity increases as T decreases furter below T C,asinFig..7d. (You sould compare te interactive Figs..5 and.7 to see tis beavior for yourself.) Figure.8 sows a 3D plot of M as a function of bot temperature T and field. It provides te same information as in Figs..6 and.7, but in a somewat more beautiful visualization. Tis 3D plot looks like a partially torn seet of paper. Te

14 0 Ising Model for Ferromagnetism (a) M k B T M (c) k B T (b) M k B T M (d) k B T 0.5 Fig..7 Ising order parameter M as a function of field. a For Jq/k B T = 0.9. b For Jq/k B T = 1. c For Jq/k B T = d For Jq/k B T = 1.5 (Interactive version at ttp:// cda/content/document/cda_downloaddocument/selinger+interactive+figures.zip?sgwid= p ) Fig..8 Ising order parameter M as a function of temperature T and field, in a 3D plot (Interactive version at ttp:// Selinger+Interactive+Figures.zip?SGWID= p )

15 .3 Interacting Spins 1 Ferromagnetic (spin up) Critical point Paramagnetic Ferromagnetic (spin down) Fig..9 Pase diagram of te Ising model as a function of temperature T and field torn part on te left represents te discontinuous response to a field for T < T C. Te non-torn part on te rigt represents te linear response to a field for T > T C. If we project tis 3D plot down into te (T, ) plane, we obtain te pase diagram sown in Fig..9. Tis pase diagram sows tree types of beavior: For T < T C, te system is ferromagnetic. Across te line = 0, tere is a first-order pase transition between spin-up and spin-down, wit a discontinuous cange in te order parameter M. Tis discontinuity corresponds to te torn part of te 3D plot. We can recognize te first-order transition in te free energy plots because tere are two competing minima, and te system jumps from one minimum to te oter. Te point T = T C and = 0, were te first-order transition terminates, is a very special point called te critical point, were te system as a second-order pase transition. At te second-order pase transition, tere is no discontinuity in te order parameter, but tere is a discontinuity in te derivative of te order parameter M/ T. We can recognize te second-order transition in te free energy plots because te single minimum at ig temperature becomes flat and breaks up into two minima at low temperature. At te critical point, te system as interesting singular beavior, wic is caracterized by critical exponents see te discussion of te exponent β in Eq. (.38), as well as te problems below. For T > T C, te system is paramagnetic. As varies between positive and negative values, tere is no pase transition! Tere is just a disordered pase tat responds smootly to te applied field. Problem: Calculate te response of te system to a small applied field in te paramagnetic pase, for T > T C.

16 Ising Model for Ferromagnetism Solution: We return to Eq..36 for te Ising order parameter M. If te applied field is small, ten te induced M must also be small. In tat case, we can neglect te M 3 term in comparison wit te M terms in te equation, and we obtain Te solution of tis equation is ( TC T ) M k B T M = Hence, te susceptibility to an applied field is χ M + M = 0. (.39) k B (T T C ). (.40) = =0 1 k B (T T C ). (.41) Tis result sows tat te susceptibility diverges as T T C, i.e., te system becomes more and more sensitive to an applied field as it approaces te critical point. Te beavior is caracterized by te scaling relation χ (T T C ) γ, (.4) were γ is a critical exponent. Tus, we see tat γ = 1 in mean-field teory. (Remember tat Eq. (.1) was derived for te non-interacting Ising model; ere, we are generalizing it to te Ising model wit interactions.) Problem: Calculate te response of te system to a small applied field at te critical point, for T = T C. Solution: Again, we return to Eq..36 for M. Rigt at T = T C,teM terms in te equation cancel eac oter, so we cannot neglect te M 3 term. In tat case, te equation becomes k B T M3 = 0, (.43) and ence M = ( ) 3 1/3. (.44) k B T Tis result sows tat te system as an infinite susceptibility at T = T C, i.e., a nonlinear response to a small applied field, as sown in Fig..7b. Tis beavior is caracterized by te scaling relation M 1/δ, (.45) were δ is anoter critical exponent. Tus, we see tat δ = 3 in mean-field teory.

17 .3 Interacting Spins 3.3. Mean-Field Teory in Terms of Average Field At tis point, dear students, you ave learned a lot about mean-field teory. You migt be wondering, Wy is it called mean-field teory? I don t see anyting about mean fields in te teory! In tis brief section, I will sow you an alternative derivation of te teory, wic sows te significance of te name. Some of you may ave already seen tis derivation in classes on magnetism, solid-state pysics, or termal pysics ere I will sow tat it is equivalent to our previous derivation. Let us go back to te energy of te Ising model defined in Eq. (.1), E = J σ i σ j i, j i σ i. (.46) We will consider a particular spin σ i as our test spin. How does te energy depend on σ i? Well, let us rewrite te energy as E = + J j=neigbor of i σ j σ i + (terms tat do not depend on σ i ). (.47) Tis expression sows tat our test spin σ i experiences an effective field consisting of te actual field plus an interaction wit eac of te neigboring spins. Of course, we do not know tese neigboring spins; tey may be correlated wit te test spin in a complicated way. As our approximation, we neglect tese correlations and replace eac of tese spins by its average value. Because te spins are all identical, te average value of eac neigboring spin is σ j =M. Recall tat te coordination number of te lattice is q, so tat tere are q neigbors in te sum. Hence, our approximation for te energy is E = mean σ i + (terms tat do not depend on σ i ), (.48) were mean = + JqM. (.49) In tis expression, mean can be considered te effective field or average field or mean field acting on σ i. (Now you see te terminology!) From our discussion of te non-interacting Ising model, we derived Eq. (.11)for te response to a field. Now we assume tat σ i as te same response to te effective field, ( ) mean σ i =tan. (.50) k B T

18 4 Ising Model for Ferromagnetism Of course, our test spin σ i is really just te same as all te oter spins, so it must ave te same expectation value σ j =M. Hence, we can combine Eqs. (.49) and (.50) to obtain ( ) + JqM M = tan. (.51) k B T Tis equation represents te condition for self-consistency: M must satisfy tis equation so tat te effective field of te neigboring spins is consistent wit te magnetic order of te test spin. In te previous section, Eq. (.35), we minimized te free energy by setting its first derivative equal to zero: ( ) ( ) F Jq = M M Nk B T k B T k B T + tan 1 M = 0. (.5) Note tat tis equation is exactly equivalent to te self-consistency equation! Hence, minimizing te free energy means te same ting as solving te self-consistency equation. All of our results from te previous section carry over uncanged. In general, I prefer te approac based on free energy because it provides some extra information, compared wit te approac based on self-consistency. At low temperature, te self-consistency equation as tree solutions. By looking at te free energy plot, as in Figs..4c, d and.5b, we can see tat tese solutions correspond to tree places were F/ M = 0: first a minimum, ten a maximum, and ten anoter minimum. We can disregard te maximum, and identify wic of te local minima is te lowest, i.e., te absolute minimum of te free energy. If we did not ave te free energy, we would ave to fall back onto oter pysical arguments to select wic solution of te self-consistency equation is correct. Furter Reading Te Ising model is discussed in many textbooks on magnetism, solid-state pysics, and termal pysics, suc as te books listed at te end of Cap. 1. For a discussion more advanced tan te current capter, I recommend: 1. P.M. Caikin, T.C. Lubensky, Principles of Condensed Matter Pysics (Cambridge, 1995) If you are interested in te istory of te Ising model, you migt want to read Ising s original article:. E. Ising, Beitrag zur Teorie des Ferromagnetismus. Z. Pys. 31, (195). doi: %fbf as well as te istorical review article: 3. S.G. Brus, History of te Lenz-Ising Model. Rev. Mod. Pys. 39, (1967). doi: /revmodpys

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