Imperfect (non-ideal) Gases

Transcription

1 Chapter 2 Imperfect non-ideal) Gases Before getting to a completely quantum treatment of particles, however, let s step back for a moment and consider real gases. Virtually all textbooks on statistical mechanics focus on ideal gases, because the math is verysimple. But the real worldis made up ofrealgases, which interact with one another. At high temperatures, these interactions consist mostly of the billiard-ball type, i.e. where the atoms or molecules can be considered as hard spheres of some radius that simply scatter off each other elastically. It turns out that the math for interacting gases isn t really hard, though, and I think that it s worth your while to work out the statistical properties of the gases you re likely to encounter in real life like the ones you re breathing in right now). 2.1 Virial Coefficients Now consider the possibility that the particles can interact. For very dilute gases i.e. where the particle density is very low), one wouldn t expect particles to find each other to collide very frequently. But for dense gases, we might expect collisions to happen frequently, in which case the presence of interactions would become more important. It seems reasonable to expect that the equation of state for a non-ideal gas would be a Taylor series expansion in density: P = Nk BT V [ 1+ ) N B 2 T)+ V ) 2 N B 3 T)+...], 2.1) V whereb 2 andb 3 arecalledthesecondandthirdvirialcoefficientsthefirstisjustone!). Thequestion is: how to calculate these coefficients if you know the nature of the particle-particle interactions? In general, the Hamiltonian for an N-particle system is H = N i=1 ) p 2 i 2m +V i + U ij, i<j where V i is the external potential felt by each particle, and U ij = Ur ij ) = U r i r j ) is the particle-particle interaction. Note that the sum for the particle interactions is over both i and j, but if all of the atoms were summed over twice, it would count each interaction twice i.e. particle i 39

2 PHYS Course Notes interacts with j but also j with i); ordering i < j guarantees no double-counting. In the absence of any external potential V i, the N-particle partition function is given by the equipartition expression: Z N = 1 ) ) ) N! h 3 d 3 r 1 d 3 p 1 h 3 d 3 r 2 d 3 p 2 h 3 d 3 r N d 3 p N N exp β p 2 i 2m + U ij i=1 i<j = 1 ) ) ) 1 N! h 3 d 3 p 1 exp β p m h 3 d 3 p 2 exp β p m h 3 d 3 p N exp β p2 N 2m d 3 r 1 d 3 r 2 d 3 r N exp U ij β i<j = Z N ideal 1 V N d 3 r 1 d 3 r 2 d 3 r N exp β i<j U ij. 2.2) The Hamiltonian is a sum of two contributions: the kinetic energy plus the external potential energy, which is the same for the ideal gas; and the interaction energy, which is new to the imperfect gas. Because all observable measurable) quantities depend on lnz), then we can write lnz N ) = lnz ideal )+lnz imperfect ) = lnz ideal )+δlnz). Because F = lnz N ), we immediately have 1 δf = ln V N d 3 r 1 d 3 r 2 d 3 r N i<j exp U ) ij, where the sum in the exponential has turned into a product of exponentials. I can add and subtract a one for no apparent reason: 1 {[ δf = ln V N d 3 r 1 d 3 r 2 d 3 r N exp U ) ] } ij 1 +1, but notice that if I do this, the factor of +1 integrates to V N which cancels the denominator, and I obtain: δf = ln 1+ 1 [ V N d 3 r 1 d 3 r 2 d 3 r N exp U ) ] ij 1, Now suppose that the only interactions are binary; that is, the only interactions are between two particles, and we never have three-body or four-body interactions. There are NN 1)/2 N 2 /2 pairs of particles, so we can simplify the above expression δf = ln {1+ N2 2V 2 d 3 r i d 3 r j [exp U ) ]} ij 1. Now also we can assume that the interaction potential only depends on the distance between the particles, and not on any angles, so that δf = ln {1+ N2 2V 24πV drr [exp 2 Ur) ) ]} 1. 0 i<j i<j

3 PHYS Course Notes Now, because this is an expression for the correction to the free energy, we expect it to be small. In this case, we can assume that the term in square brackets must also be small, because ln1 = 0. Accordingly, we make use of the Taylor series ln1+x) = x x 2 /2)+... in which case δf N2 V 2π drr 2 [exp Ur) The change in the pressure caused by the interactions is therefore δp = δf ) 2 N V k BT 2π drr [exp 2 V ) 1 ]. Ur) ) ] 1. Comparison with Eq. 2.1) immediately gives the expression for the second virial coefficient: B 2 T) = 2π drr [exp 2 Ur) ) ] 1. What can the second virial coefficient tell us? Most interaction potentials Ur) have a generic shape as a function of r: the particles strongly repel each other at short distances because of the electron-electron repulsion; this is associated with large values of Ur) near r = 0. The particles attract each other weakly at longer distances mostly because the outermost electrons in atoms and molecules act as dipoles, and there is a long-ranged magnetic dipole-dipole interactions sometimes called the dispersive force) Ur 0) 1/r 6 ; in general, the potential has an attractive bowl shape over some length scale r 0, so that Ur r 0 ) < 0. For r, Ur) = 0. See the attached figure. The attractive part of the interaction would make the particles stick together, which should lower the pressure relative to the ideal gas; likewise the short-rangle repulsive part should increase the pressure. We know from ideal gases that the mean energy U will increase with temperature though perhaps not linearly in the imperfect gas). So, we can superimpose a constant energy line corresponding to U onto the potential curve. It is clear that at high temperatures, the particles mostly experience the short-range repulsion, while at low temperatures the particles spend most of their time in the attractive part of the potential. It is likely, therefore, that the pressure will be higher than that of the ideal gas at high temperatures and lower at low temperatures. Does the functional form of B 2 T) bear this out? The simplest possible model for particle interactions is the hard-sphere potential: { r < d Ur) = 0 r > d where a is the diameter of the sphere. Let s calculate B 2 T). For the non-interacting region r > d, exp0/) = 1 for all temperatures; clearly B 2 T) = 0 in this regime, as expected. For the hard core region r < d, exp /) = 0 for all temperatures. So we need only consider the repulsive region, and where R = 2d is the sphere radius. B 2 T) = 2π d 0 drr 2 = 2πd3 3 = 16πR3, 3 A slightly better model is to assume that the interaction potential has an attractive part in addition to the short-range repulsion, so that we can define the second virial coefficient as 2r0 B 2 T) = 2π drr [exp 2 Ur) ) ] [ 1 2π drr 2 exp Ur) ) ] r 0

4 PHYS Course Notes The first term is the hard-core part, and it always satisfies Ur) for some sufficiently small r 2r 0. So B 2 core) b = 16πr0/3 3 where I have defined the quantity b. For most temperatures, the attractive part always satisfies Ur) for sufficiently large r > 2r 0. In this regime, exp Ur)/) 1 Ur)/ so that B 2 attract) 2π 2r 0 drr 2 Ur) a, where I have defined another parameter a. So, B 2 T) b a/). It doesn t seem that I have actually done anything, though! But consider the free energy F = F 0 + N2 b a ). V Because P = F/ V, we obtain the van der Waals equation of state [ PV = N 1+ N b a )]. V Note that B 2 is negative at low temperatures, and positive at high temperatures, as expected from general considerations. So, if the model is a relatively good description of the interacting gas around room temperature, a series of measurements of the pressure as the volume and temperature are varied will uniquely determine the parameters a and b. These in turn will give insights into the nature of the microscopic interactions between particles, because b uniquely determines r 0 and a places constraints on the form of Ur) for r > 2r 0. Slightly more sophisticated model potentials can allow you to infer the real inter-particle interactions surprisingly well. 2.2 Cluster Expansion It is obvious from the general expression 2.1) that there are higher-order density-dependent corrections to the equation of state than the one calculated above. The way one goes about systematically finding these correction terms is called the cluster expansion. It is widely used in physics and chemistry for its calculational simplicity. Perhaps its nicest feature is that the perturbation series can be calculated almost entirely in terms of diagrams, much like the Feynman series. In fact, the two are quite closely related. The goal is to calculate the contribution to the total partition function coming from the particle interactions, everything not in the Z ideal of Eq. 2.2). The configuration integral is defined as Z c 1 V N d 3 r 1 d 3 r N exp β U ij = 1 V N d 3 r 1 d 3 r N exp βu ij ). i<j Because we expect the corrections to be small when the particle density is low, we can replace the term e βuij 1+f ij with the Mayer function f ij. Then f ij + f ij f kl +... pairse βuij = 1+ pairs distinct pairs pairs

5 PHYS Course Notes Inserting this expansion into the definition of Z c, one obtains the first and second virial coefficients from the first and second terms, respectively the first virial coefficient is just unity). The second virial coefficient is represented by the following figure: 1 N 2 2 V 2 d 3 r 1 d 3 r 2 f 12. Pictorially, the second virial coefficient is represented by a two-site graph, with each vertex representing a particle an atom or molecule), and an edge representing an interaction between them. The next correction corresponds to two two-particle interactions. This could either be something like particle 1 interacting with 2, then 2 interacting with 3, in which case it would be represented by = 1 NN 1)N 2) 2 V 3 d 3 r 1 d 3 r 2 d 3 r 3 f 12 f 23 1 N 3 2 V 3 d 3 r 1 d 3 r 2 d 3 r 3 f 12 f 23, or it would correspond to particle 1 interacting with 2 and particle 3 interacting with 4: = 1 NN 1)N 2)N 3) 8 V 4 d 3 r 1 d 3 r 2 d 3 r 3 d 3 r 4 f 12 f 34 1 N 4 8 V 4 d 3 r 1 d 3 r 2 d 3 r 3 d 3 r 4 f 12 f 34. To construct any diagram, it is sufficient to follow these rules, akin to the Feynman rules: 1. Number the dots starting with 1, and for each dot i, write 1 V first dot, N 1 for the second dot, etc. d 3 r i. Multiply by N for the 2. For a line connecting the dots i and j, write down a factor f ij. 3. Divide by the symmetry factor, which is the number of ways of numbering the dots without changing the corresponding product of f-functions. Alternatively, the symmetry factor is the number of permutations of the graph vertices that you can perform without changing the graph pattern. It s clear that we could devise a graph for every possible combination of interactions, but this is tedious to do! What is often done in practice is to group certain similar graphs together, for example = ! + 1 2! = exp. So all the disconnected diagrams can be summed simultaneously to give a single exponential. Though it is not that easy to prove, it can be shown that in the thermodynamic limit N and V but with N/V a constant, the configuration integral is: Z c = exp Each of these diagrams is called a cluster, and the formula is the cluster expansion. The perturbation series is well-behaved for a low-density gas: a cluster diagram with more dots always gives a smaller contribution than one with fewer.

6 PHYS Course Notes Ising Model of the Ferromagnet Recell fromlast termthat in the Pauliparamagnet,there weren spins that couldeither orientthemselves parallel or antiparallel with an external magnetic field. Different orientations corresponded to different energies, and from this principle we obtained the total energy and magnetization of the system as a function of temperature and magnetic field. Suppose now that these spins were fixed to the positions of a regular lattice, and were now allowed to interact with each other. In the simplest model, we suppose that only nearest neighbouring spins interact. If the two spins are pointing in the same direction i.e. i j or i j ) then there is an additional energy J ij ; if the two spins are pointing in opposite directions i.e. i j or i j ) then the energy contribution is J ij J ij > 0). Here i and j label neighbouring sites; on a one-dimensional line, j = i±1. In other words, a given spin lowers its total energy by having its neighbours aligned parallel, and raises its energy if its neighbours are antiparallel. At low temperatures, in the absence of an external magnetic field which complicates things), the system will want to form a ferromagnet, where all of the spins are pointing in the same direction. The ground state will clearly be doubly degenerate, because the energy of a system with all spins is the same as that with all spins. The system somehow has to choose which direction the spins will point. With a small magnetic field, one direction will be biased versus the other. In the absence of any field, the choice is arbitrary but irreversible as temperature is lowered. This is might be the first example you have seen of a system that exhibits spontaneous symmetry breaking, where the system spontaneously chooses one macroscopic configuration over another equally likely) possibility. A similar thing happened in the early universe, when the forces the strong, weak, electromagnetic, and gravitational) decoupled from each other as the temperature cooled. The spontaneously broken symmetry is manifested as the different forces, but the total symmetry of the universe is preserved by the existence of special particles called Goldstone bosons of which the proposed Higgs particle is the most widely known). Based on these general considerations, we can then write the total energy of the system as E = ij J ij S i S j H i S i, where ij represents a sum over nearest-neighbours only, H is proportional to the magnetic field, and S i = ±1 is the value of the spin at site i. If the interaction energy is the same for all neighbours in the system, then the energy corresponds to the Ising model of a ferromagnet: E = J ij S i S j H i S i. 2.3) The Ising model is very simple to write down, but beyond the one-dimensional case no one knows how to solve it exactly. In larger dimensions it can be solved approximately using the mean-field approximation, however. It is a very interesting model to study; in more than one dimension it is one of the simplest models where the system undergoes a phase transition, in this case from a disordered state with no total spin to an ordered state with a total spin. Even in one dimension, many of these interesting features are displayed.

7 PHYS Course Notes D case: No external magnetic field In one dimension, this model can be solved exactly. Let s consider the situation H = 0 first. The energy can then be written as E = J S 1 S 2 +S 2 S S N 1 S N ), so the canonical) partition function is Z = e βjs1s2 e βs2s3 e βjsn 1SN. S 1 S 2 S N Consider first the final sum over S N : S N e βjsn 1SN = e βjsn 1 +e βjsn 1 = e βj +e βj = 2coshβJ), independent of the value of S N 1. We can now do the same sum for the S N 1, and so on, until we reach S 1 : Z = 2 N 1 [coshβj)] N 1 S 1 1) = 2 N [coshβj)] N 1 [2coshβJ)] N in the limit N 1. Now that we have the partition function, we can calculate the total energy U: U = lnz) = N {ln2)+ln[coshβj)]} = NJ tanhβj). β β In the limit of low temperatures β 1, the energy becomes U NJ, corresponding to all spins aligned with each other though their direction is not known). For high temperatures β 0 the mean energy is zero, corresponding to randomly-oriented spins. This result should look familiar! It is indeed the same as for the original Pauli paramagnet, except that now the coupling energy J is taking the role of B in the Pauli case. Using this analogy, we can define a magnetization as M J lnz) = Nk BTβtanhβJ) = N tanhβj), which corresponds to M J at low temperatures and M 0 at high temperatures, as we expected D case: With external magnetic field Suppose now that we keep the term in the Ising model 2.3) proportional to H. The partition function is then Z = e βjs1s2 e βs2s3 e βjsn 1SN e βhs1 e βhs2 e βhsn S 1 S 2 S N = { [ exp β S 1 S 2 + H ]} { [ 2 S 1 +S 2 ) exp β S N 1 S N + H ]} 2 S N 1 +S N ). S 1 S 2 S N There are four possible outcomes for the two spins S i and S i+1, so it is convenient to represent these in terms of a transfer matrix T, whose matrix elements correspond to { [ T i,i+1 exp β S i S i+1 + H ]} 2 S i +S i+1 ).

8 PHYS Course Notes The transfer matrix is then e βj+h) e T = βj e βj e βj H) Now we can rewrite the partition function in terms of the elements of transfer matrices: Z = ) T 12 T 23 T N 1,N = T 12 T 23 T N 2,N 1 T N 1,N. S 1 S 2 S N S 2 S N 1 S 1 S N In the last line I have explicitly paired up terms with a common S j factor. But from elementary linear algebra we know that S j T j 1,j T j,j+1 = T 2 j 1,j+1, so we can rewrite the partition functon as Z = S 1 S N S 3 T 2 13 T2 35 ) ). S N 2 T 2 N 4,N 2 T2 N 2,N Using the same trick successively and ignoring the boundary conditions N 1) gives Z S 1 T N 11 Tr T N),. where Tr stands for the trace i.e. the summation over the diagonals). Well, this is much prettier than what we started with! To evaluate the trace, we need to use some more linear algebra. If we diagonalize the matrix T, i.e. find its eigenvalues λ 1 and λ 2 and orthonormal eigenvectors u 1 and u 2, then the matrix of eigenvectors U = u 1 u 2 ) diagonalizes T: ) UTU 1 λ1 0 = Λ. 0 λ 2 This is handy, because the Trace is defined as the sum of the eigenvalues. The partition function is Z = Tr T N) = Tr [ U 1 ΛU ) U 1 ΛU ) U 1 ΛU )] = Tr [ U 1 Λ N U ] = Tr [ UU 1 Λ N] = Tr [ Λ N] = λ N 1 +λn 2. So, the evaluation of the partition function reduces to the problem of obtaining the eigenvalues of T. We need to solve the secular equation DetT λi) = 0, or eβj+h) λ e βj e βj e βj H) λ = 0. After lots of algebra, one obtains the eigenvalues ) λ ± = e coshβh)± sinh βj 2 βh)+e 4βJ

9 PHYS Course Notes and therefore the partition function [ ) N ) ] N Z = e βnj coshβh)+ sinh 2 βh)+e 4βJ + coshβh) sinh 2 βh)+e 4βJ. Let s test this result against our expectations. First, let s set H = 0 to see if it reproduces the results of the case in the previous section: [ ZH = 0) = e βnj 1+e 2βJ ) N + 1 e 2βJ ) ] [ N e = βj +e βj) N + e βj e βj) ] N = 2 N [coshβj)] N { 1+[tanhβJ)] N} [2coshβJ)] N, in the limit N 1. This checks out. Now, let s set J = 0 which would correspond to the Pauli paramagnet. In this case we have [ N ) ] N ZJ = 0) = coshβh)+ 1+sinh βh)) 2 + coshβh) sinh 2 βh)+1 = [2coshβH)] N because of the identity 1+sinh 2 x) = cosh 2 x). So everything reduces to what we expect. All interesting quantities can be obtained from the Helmholtz) free energy, which is F = lnz) NJ N ) coshβh)+ sinh β ln 2 βh)+e 4βJ, where the term ln [ λ /λ + ) N] 0 in the limit of large N. In the limit of large temperature β 0), the free energy becomes FT 1) NJ N ln2), reflecting the two possible configurations the spins can take. At low temperature β 1), then coshβh),sinhβh) e βh /2 so that FT 0) NJ. To get a feeling for what the results mean, consider the spin susceptibility, defined as ) M χ =, H where M = F/ H) T is the magnetization. The susceptibility corresponds to the sensitivity of the magnetization to changes in the eternal field at fixed temperature; high susceptibility means that the magnetization will change a lot with a small change in magnetic field, and vice versa. The magnetization is shown in Fig For constant J, the susceptibility as a function of external magnetic field is shown to become increasingly strongly peaked near zero as temperature is lowered, reflecting the fact that at low temperatures and low magnetic fields the system becomes increasingly ferromagnetically ordered. When H = 0, the susceptibility follows a universal curve as a function of /J, diverging as temperature decreases, or alternatively at high temperatures as the inter-spin interaction increases. The plot of the heat capacity at constant volume is shown in Fig Notice that it is zero both at high temperature because the mean energy goesto H 2 +J 2 )/ in this limit) when the thermal T

10 PHYS Course Notes Figure 2.1: The susceptibility χ for the one-dimensional Ising model is plotted as a function of the magnetic field H in a) and as a function of the dimensionless parameter /J in b). In a), the results are plotted for J = 1 and various temperatures β = 0.5, 0.2, and 0.1, with the function becoming increasingly peaked at lower temperatures larger β). In b) I have assumed that H = 0. energy is much larger than that of the external field, and low temperature when the external field contributes much more than the thermal energy. It reaches a peak near /J 1, which suggests something interesting going on, but what that is exactly will only be revealed when we consider the two-dimensional case, investigated in the next section. The divergence of the thermodynamic quantities, such as the susceptibility near T, H) = 0, 0), is one signature of a phase transition. Consider what really happens to a chain of spins as the temperature is lowered. At high temperatures, all the spins are random, but at low temperatures, adjacent spins want to be aligned. One could imagine that at some intermediate temperature, one would have regions consisting of spins, and other regions with spins. The size of these spin domains would presumably grow as the temperature decreases. To get a feeling for what is going on, consider the free energy for a single spin domain, F 1 dom = NJ N ln2). Suppose now that there are two adjacent domains of arbitrary size,. There are 2N 1) possible ways of having two adjacent domains like this, and the one domain wall separating them costs energy 2J. The free energy is then F 2 dom = NJ +2J N ln[2n 1)]. The free energy difference is then ) 2N 1) F = F 2 dom F 1 dom = 2J N ln 2J N lnn) 2 when N 1. Clearly, the system will prefer to form these domain walls and will therefore not be able to settle down to a nice ferromagnet) if F < 0, or 2J < N lnn). As N, it is always possible to satisfy this inequality for any choice of temperature. This means that the system can only become a true ferromagnet at exactly T = 0: the phase transition occurs only at T = 0. This is very reminiscent of the two-dimensional BEC case.

11 PHYS Course Notes Figure 2.2: The specific heat divided by k B ) of the Ising model is calculated as a function of the dimensionless parameter /J Mean Field Solution No exact solution is known to exist for the general Ising model for two or three dimensions, though the 2D case with no magnetic field was solved exactly by Lars Onsager in 1944 using an extension of the the tranfer matrix approach described above. There is nevertheless a powerful approximate solutionmethodthatworksforanydimensionandforanykindoflattice. Thisiscalledthemean-field approximation. With this approach, you solve the problem by considering only a single spin, which can be aligned either up or down, interacting with fixed neighbours whose average configuration you guess. The solution is found by ensuring that the guess is valid for any choice of spin you consider. Mathematically, we have Si S i +S i ) Sj S j +S j ) H E = J S i S j H S i = J ij i ij i = J [ ) ) Si S i Sj +S i Sj S j +Si S j + ) )] S i S i Sj S j H S i, ij i where I have added and substracted the terms S i and S j for no apparent reason. These terms correspond to the mean values of the spin at site i or j. If we have a uniform system then we can set S i = S j m, where m is the site-independent mean magnetization. Specifically, m = 1 S i = S k k. N i The lastterm, J ) ) ij Si S i Sj S j, is relatedtothe fluctuationsofthe meanmagnetization about its average, so as long as these remain small we can ignore it we ll address this issue below). The mean-field Ising energy then becomes E MF J [ Si +S j )m m 2] H S i ij i S i

12 PHYS Course Notes = 2Jm S i +Jm 2 1) H ij ij i = 2Jm z S i +Jm 2Nz 2 2 H i i S i S i = Jzm+H) i S i + NJz 2 m2, 2.4) where z is the coordination number, i.e. the number of spins bordering a given spin. In one dimension, z = 2; a two-dimensional square lattice has z = 4, a three-dimensional regular cubic lattice has z = 6, etc. The obvious advantage of the mean-field energy is that it depends now only on single spin terms, so it can be solved for any geometry. To calculate various physical properties, one first needs to obtain the partition function as usual: Z = e βnjzm2 /2 e βjzm+h)s1 e βjzm+h)s2 e βjzm+h)sn S 1 S 2 S N The free energy is then = e βnjzm2 /2 {2cosh[βJzm+H)]} N. F = lnz) = NJzm2 2 The total magnetization is then ) F M = H T = NJzm m H N N ln{2cosh[βjzm+h)]}. Jz m ) H +1 tanh[βjzm+h)]. But m M/N so this can be re-written as the following condition for m: ) Jzχ N +1 {tanh[βjzm+h)] m} = 0, where I have made use of the fact that the spin susceptibility is χ = M/ H = N m/ H. Because the first term in parentheses is always positive see for example Fig. 2.1), this means that tanh[βjzm+h)] = m. 2.5) There was a simpler way to get the same result. In the mean-field approximation, the energy contribution for each site is E MF S i ) = ǫ MF S i, where ǫ MF = Jzm+H. Here we neglect the overall constant i.e. S i -independent) contribution to the energy. The single-spin Boltzmann probability is ps i ) = e βemfsi) Z = e βemfsi) e βǫmf +e βǫmf = e βǫmfsi e βǫmf +e βǫmf, In fact the more rigorous reason is that χ is closely related to the standard deviation of the magnetization, which is always a positive number.

13 PHYS Course Notes where S i {±1}. So the mean magnetization of a given site is m = S = S i=±1 ps i )S i = eβǫmf e βǫmf e βǫmf +e βǫmf = tanhβǫ MF) = tanh[βjzm+h)]. In general, Eq. 2.5) is impossible to solve analytically. Let s consider the low and high-temperature results first to build some insight. At high temperatures β 0) we have βjzm + H) = m or m = 0, as we expected: at high temperature the spins are totally randomized yielding no net magnetization. At low temperatures β 1) we have m = ±1, depending on the sign of H or the spontaneous direction of the spins as they cool. So we have a magnet at low temperature. Actually m = 0 is still a solution, so if the magnet is cooled very carefully, one can remain in a metastable unmagnetized state. I say metastable because at low temperatures this state will have a higher energy that one with finite magnetization. The total magnetizations of these three states correspond to i S i = {±N,0} which yield mean-field energies 2.4) E MF = { NJz/2±NH,0}. Note that the sign of m depends on the sign of Jz + H and so in fact we only have two energies E MF = { NJz/2 NH,0}. The zero-magnetization state is therefore only metastable. Because the system transforms from a zero-magnetization state to a magnetized state at low temperature, there must be some finite temperature at which the magnetization changes spontaneously. This is reminiscent of the BEC transition. The solutions with m > 0 appear when the slope of the tanh function at the origin is greater than one, because then the tanh curve and the m line as a function of m can touch at a point other than m = 0. We know that tanhx) x x 3 /3 near x = 0, so that at zero field we have m βjzm 1 3 βjzm)3. If m > 0 then m > βjzm or < Jz. The critical temperature known as the Curie temperature, after Pierre Curie) in the mean-field approximation is therefore T MF) c = Jz k B. This is a pretty remarkable result, stating that the transition temperature only depends on the coordination number of the lattice, but otherwise is totally independent of the details of the physical system. For a two-dimensional square lattice, T c MF) = 4J/k B. As we ll discover below, the result is also pretty far from the correct one found by Onsager, T c J/k B. Even worse, we have already proved that in one dimension there is no finite-temperature phase transition! So mean-field theory is something that gives tractable results, but it can be quite misleading. In spite of the fact that mean-field theory can be both inaccurate and misleading, it is nevertheless instructive. In fact under certain circumstances it can give very accurate results. For example, for four or more dimensions, the mean-field approximation for the Ising model is exact! Let s calculate the magnetization per site as a function of temperature. Using the expression for the transition temperature, we have to solve the non-linear equation m = tanhmt c /T). I have solved this numerically using mathematica, and plotted it in Fig Notice that there are two

14 PHYS Course Notes m T Tc Figure 2.3: The magnetization per site is plotted as a function of the temperature for the Ising model in the mean-field approximation. solutions below T = T c, one positive and one negative. It is useful to calculate the behaviour of the magnetization right near the transition temperature. If m 0 and T T c then m m T ) 3 c T m3 Tc m 2 = 3 1 T ) ) 3 c T. 3 T T Let s define the reduced temperature t T c T T c = 1 T T c, valid for T T c. So t 0 and is smallest near T c. In terms of t, we obtain T/T c = 1 t and 1 T c /T = T c /T)t = t/1 t) = t/t 1). The equation for the magnetization is then m = 3t t 1 1 t)3 = 3t1 t) 2. At small t, the magnetization goes like m 3t) 1/2. In other words, the leading behaviour of the magnetization near the transition temperature has a characteristic exponent of 1/2. Let s also calculate the susceptibility. To do this we need to go back to the field-dependent expression for the magnetization, Eq.2.5). If the field is weak then the transition temperature won t be strongly affected, and we can again expand the tanh: m m T c T + H m3 3 ) 3 Tc, T where I have dropped the cubic contribution from the field because it is very small. Taking the derivative of both sides with respect to the field, we have: χ = χ T c T + 1 ) 3 Tc χm2. T T c

15 PHYS Course Notes Χ t Figure 2.4: The magnetic susceptibility per site is plotted as a function of the reduced temperature for the Ising model in the mean-field approximation. Solving for χ: 1 χ = [1 T c /T)+m 2 T c /T) 3]. For T > T c the magnetization is zero and χ = 1 c 1 1 t)[1 1/1 t)] 1 c 1 1 t)[1 1+t)] 1 c t t 1. Note that t < 0 for T > T c which explains the negative sign above). This is called the Curie-Weiss law, and is usually expressed as χ = C/T T c ) with C a material-specific quantity. In the above derivation, the constant is inversely proportional to the Curie temperature T c. For T < T c one obtains χ = 1 1 c 1 t)[1 1/1 t)+3t1 t) 4 /1 t) 3 ] 1 1 c 1 t)[ t+3t1 t)] 1 2 c t t 1. Notice that while the scaling with t is the same on both sides, there is a factor of two difference, which means that the susceptibility is not a smooth function of temperature across the transition. Of course, it also clearly diverges right at T c, because t 0 at this point. I evaluated χ numerically in mathematica, and it is plotted in Fig The most important and obvious feature is the divergence at T = T c or t = 1. Such a divergence is a hallmark of a phase transition, and is analogous to the divergence of the heat capacity of superfluid Helium at the so-called lambda point, so named because the shape of the heat capacity curve looks like the Greek letter λ.

16 PHYS Course Notes As discussed above, near the phase transition the order parameter and the susceptibility behave as t a, where a is some critical exponent. In the mean-field model, we have m m ± t β with β = 1/2, and χ c ± t γ with γ = 1. Note that β and γ are always associated to the order parameter and generalized) susceptibility for any system, respectively. For example, the exponent α is reserved for the heat capacity, and ν is reserved for correlation lengths more on these below). In the mean-field model these come out to α = 0 and ν = 1/2, respectively; note that there is no divergence of the heat capacity in the mean-field model. While all phase transitions are associated with critical exponents, their actual values are generally difficult to predict, for reasons that will be discussed further below. The values found in the mean-field model above are in fact quite wrong. The exact values in two dimensions are {α,β,γ,ν} = {0,1/8,7/4,1},while in three dimensions they are approximately {0.11, , , }. It turns out that many quantities are related, and in fact only β, γ and ν are needed to specify any phase transition. These critical exponents are universal, in that they don t depend on the details of the system; furthermore, totally different systems can have the same critical exponents. The magnetization per site m can be considered as an order parameter, which characterizes the order of a system. Above T c, there is no average magnetic order, so m = 0; below T c magnetic order begins to manifest itself and so m > 0. This idea of finding an order parameter to describe phase transitions applies very generally, though the choice of specific parameter depends on the system. We ll be coming back to order parameters throughout this course. The phase transition for the Ising model described by mean-field theory is called a continuous phase transition, because the order parameter varies continuously from zero to non-zero through the phase transition albeit piecewise continuously). It is also called a second-order phase transition. The order parameter can be defined as the first derivative of the free energy with respect to the external field in this case the magnetic field). The susceptibility corresponds to taking a second derivative of the free energy here the derivative of the order parameter with respect to the external field). Because it is discontinuous through the phase transition, the transition is called second-order. To see how the order parameter is the first derivative of the free energy with respect to the field, consider the Helmholtz) free energy: F lnz N ) ln [ Tr e βe)] = ln Tr exp βh S i +βj S i S j. i ij Now, [ F Tr S i exp βh H = i S i +βj )] ij S is j [ Tr exp βh i S i +βj )] NS = M. 2.6) ij S is j Improved Mean-Field Theory: Bethe Approximation In the regular mean-field approximation, we assumed that every site was exactly the same as any other; a given spin only interacted with the mean magnetization of any neighbor. We can systematically improve this by considering ever larger clusters of spins exactly. The first improvement is to consider a cluster with one central spin with index 0, and z neighbours with indices i = 1,2,...,z. The energy of the cluster is then z z E = JS 0 S i HS 0 H S i. i=1 i=1

17 PHYS Course Notes Now, the ordinary field H acting on the neighboring spins has been replaced by an effective field H, much like in the previous mean-field approximation the spin operator was replaced by an effective value, its mean. The partition function of the cluster is readily evaluated: Z = e βe = e βh {2cosh[βJ +H )]} z +e βh {2cosh[βJ H )]} z. S i=±1 The mean values are for the central spin and S 0 = eβh {2cosh[βJ +H )]} z e βh {2cosh[βJ H )]} z e βh {2cosh[βJ +H )]} z +e βh {2cosh[βJ H )]} z S i = eβh 2sinh[βJ +H )]{2cosh[βJ +H )]} z 1 e βh 2sinh[βJ +H )]{2cosh[βJ H )]} z 1 e βh {2cosh[βJ +H )]} z +e βh {2cosh[βJ H )]} z for the z spins neighboring the central one. Because the system is translationally invariant, all of the sites mean values must be the same, i.e. S 0 = S i i, yielding {2cosh[βJ +H )]} z {2cosh[βJ H )]} z = 2sinh[βJ +H )]{2cosh[βJ +H )]} z 1 2sinh[βJ +H )]{2cosh[βJ H )]} z 1, where I have assumed that the real external magnetic field is zero to make my life easier! Let s simplify: {2cosh[βJ +H )]} z 1 {2cosh[βJ +H )] 2sinh[βJ +H )]} = {2cosh[βJ H )]} z 1 {2cosh[βJ H )] 2sinh[βJ H )]} {2cosh[βJ +H )]} z 1 e βj+h ) = {2cosh[βJ H )]} z 1 e βj H ) cosh[βj +H ) z 1 )] cosh[βj H = e 2βH. )] Nowwe need to solvethis equationforthe effective field H. It s prettyobviousthat H = 0is always a solution; this corresponds to the m = 0 high-temperature case or to the metastable solution when T < T c ). So we can assume that the ordered phase corresponds to H 0, and the phase transition occurs at the moment when H changes from zero. So we can take a Taylor series around H = 0 of both sides: 1+2βH z 1)tanhβJ) 1+2βH, equivalent to tanhβ c J) = 1/z 1) or cothβ c J) = z 1. We can now solve for the transition temperature: J T c = k B coth 1 1 z). We can use the identity coth 1 x) = ln[x+1)/x 1)]/2 to obtain 2J T c = k B ln[z 2)/z]. For two dimensions z = 4 and T c J/k B, which is much closer to the exact solution of T c 2.269J/k B than the naïve mean-field approximation. Even better, this improved mean-field

18 PHYS Course Notes theory correctly predicts that there is no finite-temperature phase transition for one-dimensional systems z = 2). The calculation gets better and better as the size of the cluster increases, as expected. That said, the critical exponents don t change, a result that will be left as an exercise to the reader. This implies that mean-field theory even this improved version and in fact any improved version) is fundamentally unable to capture the essential physics of phase transitions. Why this is the case will be the topic of the next section Fluctuations In the mean-field approximation to the Ising model discussed in the sections above, at least one spin operator was singled out as special. The remaining spins are treated in some average way. In the most naïve formulation, we obtained Eq. 2.4) from the exact Hamiltonian 2.3) by neglecting the J ) ) Si S i Sj S j ij term in the mean-field Hamiltonian. Let s define the spin correlation function g ij S i S i ) Sj S j ) = Si S j S i S j. Omitting the term above in the mean-field Hamiltonian therefore corresponds to setting g ij = 0. In other words, any averages involving spins on different sites i and j are replaced by products of average spin values on single sites. This is equivalent to neglecting any correlations between spins on different sites. It turns out that correlations are important, particularly near phase transitions, which is why the mean-field theory doesn t do a very good job predicting the Curie temperature or critical exponents. Recall that we can obtain the magnetization directly from the partition function, Eq. 2.6). I mentioned earlier that the Ising phase transition was second-order because the second derivative is discontinuous at T c, so let s consider this now: 2 F 2 H 2 = H 2 [ k BT lnz N )] = N [ ] 1 Z H Z H [ 1 2 Z = N Z H 2 1 ) ] 2 Z Z ) H We already know that so that F H = Nk BT Z Z H 1 Z Z H = m. = NS = Nm = M, The second term in Eq. 2.7) must be m 2 /. We also know that 1 2 Z Z H 2 = β2 S 2, so that the first term in Eq. 2.7) is S 2 /. Putting things together gives 2 F H 2 = Nk BT) 2 2 lnz) H 2 = N S 2 S 2) = N S) 2.

19 PHYS Course Notes The spin susceptibility is therefore χ = M H = N m H = F 2 H 2 = N S)2. In other words, the spin susceptibility is proportional to the standard deviation of the magnetization, i.e. to the fluctuations of the magnetization. Alternatively, we can write the susceptibility in terms of the spin correlation function: χ = 1 ) 2 S i i i S i )2 = 1 ij Si S j S i S j ) = 1 g ij. 2.8) Eq.2.8) is called the fluctuation-susceptibility equation, and is a special case of the fluctuationdissipation theorem, which will be discussed in the chapters on non-equilibrium statistical mechanics. In the case studied here where the system is translationally invariant J and H are independent of the site), one can choose any value of i as a reference position. The susceptibility is then often written as χ = N g 0j = N gr 0,r j ),= N gr j r 0 ), 2.9) j j where i = 0 is a convenient reference position, and in the last line I have explicitly made use of the translational invariance of the system. The important point of all of this is that in the mean-field approximation we have χ = N2 1 m 2 ). At zero temperature this makes sense; m = 1 and χ = 0 which means that the fluctuations are all zero, which is correct if all the spins are identically pointing in the same direction. But it is really wrong at higher temperatures, because it clearly doesn t diverge at T c. In fact, the behavior of this value of χ isn t even the same as in the derivation above, so the mean-field calculation hasn t been consistent with respect to the definition of the spin susceptibility. Recall that the mean-field Hamiltonian 2.4) is j ij E MF = Jzm+H) i S i + NJz 2 m2 + E, where E is the fluctuation contribution E = J ij Si S i ) Sj S j ) = Jz 2 k BTχ. Forthe mean-field theoryto be valid, wemust havethe magnetic field is setto zeroforconvenience): Jz 2 k BTχ Jzm S i + NJz 2 m2 = NJzm2 + NJz 2 m2 = Jz 2 Nm2. i

20 PHYS Course Notes Therefore, Nχ Nm) 2 = N 2 S 2. But we already know that χ = N criterion for the validity of the mean-field approximation is S 2 S 2 S 2 S S S 2 S 2 S 2), so that This is the so-called Ginzburg criterion; for mean-field theory to be valid, the relative fluctuations must be small. But we alreadyknowfromthe same mean-fieldtheorythat χ t 1 fort T c from either above or below T c. Together with the fact that S 0 near T c, this means that the Ginzburg criterion χ NS 2 will definitely break down for some temperature region in the vicinity of the transition., i.e. mean-field theory will only be valid for T < T c T and T c + T < T, where 2 T is the width of the region around T c where mean-field theory definitely breaks down Correlation function and correlation length Recall in Sec that near T c, various parameters are characterized by power-laws of the reduced temperature t; examples included the magnetization, spin susceptibility, the heat capacity, and the correlation length. While the first two of these were discussed at length the heat capacity will be left as exercises for you!), it is important to talk about correlation functions and correlation lengths in detail. Correlation functions are widely discussed in physics, and so the sooner you see one analyzed the better! The discussion in the previous section made it clear that the spin-spin correlation function is closely tied to the spin susceptibility, but it isn t yet clear what is the spatial-dependence of the function, if any, or what it might reveal about the properties of the system. Recall the solution to the mean-field Hamiltonian, Eq. 2.5): m = tanh[βjzm+h)]. Instead of a constant magnetic field, suppose that it is actually has site-dependent values {H i }. Then we could write the magnetization in terms of an effective magnetic field S i = tanh[βh eff ], where H eff = H i +J S j, j where the sum runs only over neighbors of i. If the mean magnetization varies slowly enough in space differences in H i from site to site are gradual enough), then I can take a Taylor series to obtain: H eff H i +JzS i +cja 2 2 S i, where a is the lattice constant spacing between sites) and c is a constant that depends on the lattice you need to take the appropriate continuum limit of a finite-difference approximation; the stencil applied depends on the lattice geometry). Note that the linear odd) term vanishes by because of the assumed reflection symmetry of the sites j around the central site i. We also know that tanhβh eff ) βh eff 1 3 βh eff) 3 when βh eff 1, valid at large temperatures. One therefore obtains S i β H i +JzS i +cja 2 2 S i ) 1 3 βjz)3 S 3 i,

21 PHYS Course Notes keeping only terms linear in H i and lowest-order in derivatives. Rearranging gives βh i [1 βjz)+ 13 ) βjz)3 S 2i βcja2 2 S i. In the mean-field approximation, the critical temperature is defined by c = Jz, so we have [ βh i 1 T ) c + 1 ) 3 Tc S 2 i T 3 T T ) c ca 2) 2 S i. T z Using the relations for the reduced temperature t 1 T/T c ) dicussed above gives 1 T c /T) = t/1 t) t and T c /T = 1/1 t) 1 because t 0. Then we obtain βh i [ t+ 13 ) S2i c a 2 2 S i, where I have defined a new constant of order unity, c = c/z. When T > T c then S i = 0 and t < 0; then the expression simplifies to βh i T > T c ) t c a 2 2) S i, where t t so that t > 0. We already know that χ = M/ H = N S i / H = N/) j g ij, so S i / H j = βg ij. Thus, βδ ij t c a 2 2) βg ij δ ij t c a 2 2) g ij. In the continuum limit we can express this equation as t c a 2 2) gr) = δr), which can be Fourier tranformed to give t+c a 2 k 2) gk) = 1. The correlation function is therefore gr) d d e ik r ka) t+c a 2 k 2, where I have dropped the V/2π) 3d prefactor to put more emphasis on the scaling. I ll continue to do this below. It s not difficult to put in all the factors if you are interested, i.e. to calculate the susceptibility using Eq. 2.9). Integrating over generalized angles gives gr) 0 a d ) d 2)/2 1 a t+c a 2 k 2 J d 2)/2kr)k d 1 2 dk kr ξr ) d 2)/2 K d 2)/2 ) r, ξ where J µ x) is a Bessel function and K µ x) is a modified Bessel function, and ξ = a c /t is the correlation length. The most important thing to notice is that ξ diverges as T T + c as t 1/2. The argument of the modified Bessel function is therefore small, and in this limit K µ x) x µ as long as µ > 0; one then obtains ) a 2 d 2)/2 ) d 2)/2 ξ a ) d 2, gr) ξr r r valid for r ξ and d > 2. When d = 2, the modified Bessel function at large argument goes like K 0 x) lnx/2) so the correlation function becomes ) ) r ξ gr) ln ln. 2ξ r

22 PHYS Course Notes Finally, when d = 3 the modified Bessel function is exactly K 1/2 x) = π/2xe x for all x, we obtain πa 2 gr) 2r 2 e r/ξ a r e r/ξ. The correlation function is telling us about the spatial-dependence of the fluctuations in the spin, and the correlation length is therefore telling us about the length over which these fluctuations occur. For temperatures far from T c, the correlation length is small, ξ a c a, of order the lattice spacing. But because ξ as T T + c or t 0, this is saying that as the critical temperature is approached, the fluctuations over a length that eventually encompasses the entire system in twodimensional and three-dimensional systems! This is a signature of the incipient establishment of order over the whole system, but also tells us that fluctuations can no longer be neglected, as they have for the mean-field model in the first place. Of course, this is paradoxical just as it was for the susceptibility to which the correlation function is closely related as have already seen), since the mean-field model was used to derive this result. But the conclusion is that mean-field theory necessarily breaks down as the critical temperature is approached, and the correlations/fluctuations become more and more important. A couple of last notes. First, the correlation length diverges near T c as ξ t 1/2 t ν ; this is the same critical exponent ν = 1/2 mentioned in the section above, but not derived. Second, note that the correlation function for d > 3 appears to not depend on the correlation length at all. This seems to suggest that the Ginzburg criterion is satisfied for all temperatures, and that mean-field theory is always valid. In fact, this has been proven to be true in the Ising model: d = 4 is the critical dimension for the validity of the mean-field model. This will be discussed further in the next chapter. The divergence of the correlation length also tells us that right at the phase transition, there no longer any length scale governing the fluctuations. The absence of any length scale in the problem other than the lattice spacing) implies that the phase transition is somehow universal, in that other systems at criticality should behave the same way. Note furthermore that the critical exponents don t depend in any way on the details of the lattice, such as the coordination number z or the strength of the coupling constant J or magnetic field H. Much more on this in the next chapter!