Lecture Notes for Statistical Mechanics of Soft Matter

1 Introduction

The purpose of these lectures is to provide important statistical mechanics back- ground, especially for soft condensed matter physics. Of course, this begs the question of what exactly is important and what is less so? This is inevitably a subjective matter, and the choices made here are somewhat personal, and with an eye toward the Amsterdam (VU, UvA, AMOLF) context. These notes have taken shape over the past approximately ten years, beginning with a joint course developed by Daan Frenkel (then, at AMOLF/UvA) and Fred MacK- intosh (VU), as part of the combined UvA-VU masters program. Thus, these lecture notes have evolved over time, and are likely to continue to do so. In preparing these lectures, we chose to focus on both the essential techniques of equilibrium statistical physics, as well as a few special topics in classical statistical physics that are broadly useful, but which all too frequently fall outside the scope of most elementary courses on the subject, i.e., what you are likely to have seen in earlier courses. Most of the topics covered in these lectures can be found in standard texts, such as the book by Chandler [1]. Another very useful book is that by Barrat and Hansen [2]. Finally, a very useful, though more advanced reference is the book by Chaikin and Lubensky [3].

Thermodynamics and Statistical Mechanics The topic of this course is statistical mechanics and its applications, primarily to condensed matter physics. This is a wide ﬁeld of study that focuses on the properties of many-particle systems. However, not all properties of many- body systems are accessible to experiment. To give a speciﬁc example: in an experiment, we could not possibly measure the instantaneous positions and velocities of all molecules in a liquid. Such information could be obtained in a computer simulation. But theoretical knowledge that cannot possibly be tested experimentally, is fairly sterile. Most experiments measure averaged properties. Averaged over a large number of particles and, usually, also averaged over the time of the measurement. If we are going to use a physical theory to describe the properties of condensed-matter systems, we must know what kind of averages we should aim to compute. In order to explain this, we need the language of thermodynamics and statistical mechanics.

1-1 In thermodynamics, we encounter concepts such as energy, entropy and temperature. I have no doubt that the reader has, at one time or another, been introduced to all these concepts. Thermodynamics is often perceived as very abstract and it is easy to forget that it is ﬁrmly rooted in experiment. I shall therefore brieﬂy sketch how “derive” thermodynamics from a small number of experimental observations. Statistical mechanics provides a mechanical interpretation (classical or quantum) of rather abstract thermodynamic quantities such as entropy. Again, the reader is probably familiar with the basis of statistical mechanics. But, for the purpose of the course, I shall re-derive some of the essential results. The aim of this derivation is to show that there is nothing mysterious about concepts such as phase space, temperature and entropy and many of the other statistical mechanical objects that will appear time and again in the remainder of this course.

1.1 Thermodynamics Thermodynamics is a remarkable discipline. It provides us with relations between measurable quantities. Relations such as

dU = q + w (1.1) or dU = T dS P dV + µdN (1.2) − or ∂S ∂P = (1.3) ∂V ∂T T V These relations are valid for any substance. But - precisely for this reason - thermodynamics contains no information whatsoever about the underlying microscopic structure of a substance. Thermodynamics preceded statistical mechanics and quantum mechanics, yet not a single thermodynamic relation had to be modiﬁed in view of these later developments. The reason is simple: thermodynamics is a phenomenological science. It is based on the properties of matter as we observe it, not upon any theoretical ideas that we may have about matter. Thermodynamics is diﬃcult because it seems so abstract. However, we should always bear in mind that thermodynamics is based on experimental observations. For instance, the First Law of thermodynamics expresses the empirical observation that energy is conserved, even though it can be converted in various forms. The internal energy of a system can be changed by performing work w on the system or by transferring an amount of heat q. It is meaningless to speak about the total amount of heat in a system, or the total amount of work. This is not mysterious: it is just as meaningless to speak about the number of train-travelers and the number of pedestrians in a train-station: people enter the station as pedestrians, and exit as train-travelers or vice versa. However if we add the sum of the changes in the number of train-travelers and

1-2 pedestrians, then we obtain the change in the number of people in the station. And this quantity is well deﬁned. Similarly, the sum of q and w is equal to the change in the internal energy U of the system

dU = q + w (1.4)

This is the First Law of Thermodynamics. The Second Law seems more abstract, but it is not. The Second Law is based on the experimental observation that it is impossible to make an engine that works by converting heat from a single heat bath (i.e. a large reservoir in equilibrium) into work. This observation is equivalent to another - equally empirical - observation, namely that heat can never flow spontaneously (i.e. without performing work) form a cold reservoir to a warmer reservoir. This statement is actually a bit more subtle than it seems because, before we have defined temperature, we can only distinguish hotter and colder by looking at the direction of heat flow. What the Second Law says is that it is never possible to make heat flow spontaneously in the “wrong” direction. How do we get from such a seemingly trivial statement to something as abstract as entropy? This is most easily achieved by introducing the concept of a reversible heat engine. A reversible engine is, as the word suggests, an engine that can be operated in reverse. During one cycle (a sequence of steps that is completed when the engine is returned into its original state) this engine takes in an amount of heat q1 from a hot reservoir converts part of it into work w and delivers a remaining amount of heat q2 to a cold reservoir. The reverse process is that, by performing an amount of work w, we can take an amount of heat q2 form the cold reservoir and deliver an amount of heat q1to the hot reservoir. Reversible engines are an idealization because in any real engine there will be additional heat losses. However, the ideal reversible engine can be approximated arbitrarily closely by a real engine if, at every stage, the real engine is sufficiently close to equilibrium. As the engine is returned to its original state at the end of one cycle, its internal energy U has not changed. Hence, the First Law tells us that

dU = q (w + q ) = 0 (1.5) 1 − 2 or q1 = w + q2 (1.6)

Now consider the “efficiency” of the engine η w/q1- i.e. the amount of work delivered per amount of heat taken in. At first,≡ one might think that η depends on the precise design of our reversible engine. However this is not true. η is the same for all reversible engines operating between the same two reservoirs. To demonstrate this, we show that if different engines could have different values for η then we would contradict the Second Law in its form “heat can never spontaneously flow from a cold to a hot reservoir”. Suppose therefore that we ′ have another reversible engine that takes in an amount of heat q1 form the hot reservoir, delivers the same amount of work w, and then delivers an amount of ′ ′ heat q2to the cold reservoir. Let us denote the efficiency of this engine by η .

1-3 Now we use the work generated by the engine with the highest eﬃciency (say η) to drive the second engine in reverse. The amount of heat delivered to the hot reservoir by the second engine is

′ ′ ′ q1 = w/η = q1(η/η ) (1.7)

′ ′ where we have used w = q1η. As, by assumption, η < η it follows that q1 > q1. Hence there is a net heat flow form the cold reservoir into the hot reservoir. But this contradicts the Second Law of thermodynamics. Therefore we must conclude that the efficiency of all reversible heat engines operating between the same reservoirs is identical. The efficiency only depends on the temperatures t1and t2 of the reservoirs (the temperatures t could be measured in any scale, e.g. in Fahrenheit 1 as long as heat flows in the direction of decreasing t). As η(t1,t2) depends only on the temperature in the reservoirs, then so does the ratio q2/q1 = 1 η. Let us call this ratio R(t2,t1). Now suppose that we have a reversible engine− that consists of two stages: one working between reservoir 1 and 2, and the other between 2 and 3. In addition, we have another reversible engine that works directly between 1 and 3. As both engines must be equally

1Daniel Gabriel Fahrenheit (1686 - 1736) was born in Danzig. He was only fifteen when his parents both died from eating poisonous mushrooms. The city council put the four younger Fahrenheit children in foster homes. But they apprenticed Daniel to a merchant, who taught him bookkeeping and took him off to Amsterdam. He settled in Amsterdam in 1701 looking for a trade. He became interested in the manufacture of scientific instruments, specifically me- teorological apparatus. The first thermometers were constructed in around 1600 by Galileo. These were gas thermometers, in which the expansion or contraction of air raised or lowered a column of water, but fluctuations in air pressure caused inaccuracies. The Florentine thermometer showed up as a trade item in Amsterdam, and it caught young Fahrenheit’s fancy. So he skipped out on his apprenticeship and borrowed against his inheritance to take up thermometer making. When the city fathers of Gdansk found out, they arranged to have the 20-year-old Fahrenheit arrested and shipped off to the East India Company. So he had to dodge Dutch police until he became a legal adult at the age of 24. At first he had simply been on the run; but he kept traveling – through Denmark, Germany, Holland, Sweden, Poland, meeting scientists and other instrument makers before returning to Amsterdam to begin trading in 1717. In 1695 Guillaume Amontons had improved the gas thermometer by using mercury in a closed column, and in 1701 Olaus Roemer devised the alcohol thermometer and developed a scale with boiling water at 60 and an ice/salt mixture at 0. Fahrenheit met with Roemer and took up his idea, constructing the first mercury-in-glass thermometer in 1714. This was more accurate because mercury possesses a more constant rate of expansion than alcohol and could be used over a wider range of temperatures. In order to reflect his greater sensitivity, Fahrenheit expanded Roemer’s scale using body temperature (90F) and ice/salt (0F) as fixed reference points, with the freezing point of water awarded a value of 30 (later revised to 32). He used the freezing point of water for 32 degrees. Body temperature he called 96 degrees. Why the funny numbers? Fahrenheit originally used a twelve-point scale with zero, four, and twelve for those three benchmarks. Then he put eight gradations in each large division. That is body temperature in the Fahrenheit scale is 96 (actually, it is a bit higher) – it is simply eight times twelve. Using his thermometer, Fahrenheit was able to determine the boiling points of liquids and show that liquids other than water also had fixed boiling points that varied with atmospheric pressure. He also researched the phenomenon of supercooling of water: the cooling of water below freezing without solidification, and dis- covered that the boiling point of liquids varied with atmospheric pressure. He published his method for making thermometers in the Philosophical Transactions in 1724 and was admitted to the Royal Society the same year.

1-4 eﬃcient, it follows that

R(t3,t1)= R(t3,t2)R(t2,t1) (1.8)

This can only be true in general if R(t1,t2) is of the form

f(t2) R(t2,t1)= (1.9) f(t1) where f(t) is an, as yet unknown function of our measured temperature. What we do now is to introduce an “absolute” or thermodynamic temperature T given by T = f(t) (1.10) Then it immediately follows that

q2 T2 = R(t2,t1)= (1.11) q1 T1 Note that the thermodynamic temperature could just as well have been defined as c f(t). In practice, c has been fixed such that, around room temperature, 1 degree× in the absolute (Kelvin) scale is equal to 1 degree Celsius. But that choice is of course purely historical and - as it will turn out later - a bit unfortunate. Now, why do we need all this? We need it to introduce entropy, this most mysterious of all thermodynamic quantities. To do so, note that Eqn.1.11 can be written as q q 1 = 2 (1.12) T1 T2 where q1 is the heat that flows in reversibly at the high temperature T1, and q2 is the heat that flows out reversibly at the low temperature T2. We see therefore that, during a complete cycle, the difference between q1/T1 and q2/T2 is zero. Recall that, at the end of a cycle, the internal energy of the system has not changed. Now Eqn.1.12 tells us that there is also another quantity that we call “entropy” and that we denote by S that is unchanged when we restore the system to its original state. In the language of thermodynamics, we call S a state function. We do not know what S is, but we do know how to compute its change. In the above example, the change in S was given by ∆S = (q /T ) (q /T )=0. In general, the change in entropy of a system due to the 1 1 − 2 2 reversible addition of an infinitesimal amount of heat δqrev from a reservoir at temperature T is δq dS = rev (1.13) T We also note that S is extensive. That means that the total entropy of two non-interacting systems, is equal to the sum of the entropy of the individual systems. Consider a system with a fixed number of particles N and a fixed volume V . If we transfer an infinitesimal amount of heat δq to this system,then the change in the internal energy of the system, dU is equal to δq.Hence, ∂S 1 = (1.14) ∂U T V,N

1-5 The most famous (though not most intuitively obvious) statement of the Second Law of Thermodynamics is that Any spontaneous change in a closed system (i.e. a system that exchanges neither heat nor particles with its environment) can never lead to a decrease of the entropy. Hence, in equilibrium, the entropy of a closed system is at a maximum. Can we understand this? Well, let us first consider a system with an energy U, volume V and number of particles N that is in equilibrium. Let us denote the entropy of this system by S0(U,V,N). In equilibrium, all spontaneous changes that can happen, have happened. Now suppose that we want to change something in this system - for instance, we increase the density in one half and decrease it in the other. As the system was in equilibrium, this change does not occur spontaneously. Hence, in order to realize this change, we must perform a certain amount of work, w (for instance, my placing a piston in the system and moving it). Let us perform this work reversibly in such a way that E, the total energy of the system stays constant (and also V and N). The First Law tells us that we can only keep E constant if, while we do the work, we allow an amount of heat q, flow out of the system, such that q = w. Eqn.1.13 tells us that when an amount of heat q flows out of the system, the entropy S of the system must decrease. Let us denote the entropy of this constrained state by S1(E,V,N) < S0(E,V,N). Having completed the change in the system, we insulate the system thermally from the rest of the world, and we remove the constraint that kept the system in its special state (taking the example of the piston: we make an opening in the piston). Now the system goes back spontaneously (and irreversibly) to equilibrium. However, no work is done and no heat is transferred. Hence the finally energy E is equal to the original energy (and V and N) are also constant. This means that the system is now back in its original equilibrium state and its entropy is once more equal to S0(E,V,N). The entropy change during this spontaneous change is equal to ∆S = S0 S1.But, as S1 < S0, it follows that ∆S > 0. As this argument is quite general,− we have indeed shown that any spontaneous change in a closed system leads to an increase in the entropy. Hence, in equilibrium, the entropy of a closed system is at a maximum. From this point on, we can derive all of thermodynamics, except one “law” - the so-called Third Law of Thermodynamics. The Third law states that, at T = 0, the entropy of the equilibrium state of pure substance equals zero. Actually, the Third Law is not nearly as “basic” as the First and the Second. And, anyway, we shall soon get a more direct interpretation of its meaning.

1.2 Elementary Statistical Mechanics Statistical thermodynamics is a theoretical framework that allows us to compute the macroscopic properties of systems containing many particles. Any rigorous description of a system of many atoms or molecules, should be based on quantum mechanics. Unfortunately, the number of exactly solvable, non-trivial quantum- mechanical models of many-body systems is almost negative. For this reason, it is often assumed that classical mechanics can be used to describe the motions of atoms and molecules. This assumption leads to a great simpliﬁcation in almost

1-6 all calculations and it is therefore most fortunate that it is indeed justified in many cases of practical interest (although the fraction of non-trivial models that can be solved exactly still remains depressingly small). It is therefore somewhat surprising that, in order to ‘derive’ the basic laws of statistical mechanics, it easier to use the language of quantum mechanics. I will follow this route of ‘least resistance’. In fact, for our derivation, we need only little quantum mechanics. Specifically, we need the fact that a quantum mechanical system can be found in different states. For the time being, I limit myself to quantum states that are eigenvectors of the Hamiltonian H of the system (i.e. energy eigenstates). For any such state i >, we have that H i > = Ei i >, where Ei is the energy of state i >. Most| examples discussed| in quantum-mechanics| textbooks concern systems| with only few degrees of freedom (e.g. the one-dimensional harmonic oscillator or a particle in a box). For such systems, the degeneracy of energy levels will be small. However, for the systems that are of interest to statistical mechanics (i.e. systems with (1023) particles), the degeneracy of energy levels is astronomically large - actually,O the word “astronomical” is misplaced - the numbers involved are so large that, by comparison the total number of particles in the universe is utterly negligible. In what follows, we denote by Ω(U,V,N) the number of energy levels with energy U of a system of N particles in a volume V . We now express the basic assumption of statistical thermodynamics as follows: A system with fixed N, V and U is equally likely to be found in any of its Ω(U) energy levels. Much of statistical thermodynamics follows from this simple (but highly non-trivial) assumption. To see this, let us first consider a system with total energy U that consists of two weakly interaction sub-systems. In this context, ‘weakly interacting’ means that the sub-systems can exchange energy but that we can write the total energy of the system as the sum of the energies U1 and U2 of the sub-systems. There are many ways in which we can distribute the total energy over the two sub- systems, such that U1+U2 = U. For a given choice of U1, the total number of degenerate states of the system is Ω1(U1) Ω2(U2). Note that the total number of states is not the sum but the product of× the number of states in the individual systems. In what follows, it is convenient to have a measure of the degeneracy of the sub-systems that is extensive (i.e. additive). A logical choice is to take the (natural) logarithm of the degeneracy. Hence:

ln Ω(U ,U U ) = ln Ω (U ) + ln Ω (U U ) (1.15) 1 − 1 1 1 2 − 1 We assume that sub-systems 1 and 2 can exchange energy. In fact, in thermodynamics we often consider such a process: it is simply heat transfer (because the two systems do not exchange particles, nor does one system perform work on the other). What is the most likely distribution of the energy? We know that every energy state of the total system is equally likely. But the number of energy levels that correspond to a given distribution of the energy over the sub-systems, depends very strongly on the value of U1. We wish to know the most likely value of U , i.e. the one that maximizes ln Ω(U ,U U ). The 1 1 − 1

1-7 condition for this maximum is that ∂ ln Ω(U ,U U ) 1 − 1 = 0 (1.16) ∂U 1 N,V,U or, in other words, ∂ ln Ω (U ) ∂ ln Ω (U ) 1 1 = 2 2 . (1.17) ∂U ∂U 1 N1,V1 2 N2,V2 We introduce the shorthand notation ∂ ln Ω(U,V,N) β(U,V,N) . (1.18) ≡ ∂U N,V With this deﬁnition, we can write Eqn. 1.17 as

β(U1, V1,N1)= β(U2, V2,N2). (1.19)

Clearly, if initially we put all energy in system 1 (say), there will be energy transfer from system 1 to system 2 until Eqn. 1.19 is satisﬁed. From that moment on, there is no net energy ﬂow from one sub-system to the other, and we say that the two sub-systems are in thermal equilibrium. This implies that the condition β(U1, V1,N1)= β(U2, V2,N2) must be equivalent to the statement that the two sub-systems have the same temperature. ln Ω is a state function (of U, V and N), just like S. Moreover, when thermal equilibrium is reached, ln Ω of the total system is at a maximum, again just like S. This suggests that ln Ω is closely related to S. We note that both S and ln Ω are extensive. This suggests that S is simply proportional to ln Ω:

S(N,V,U) k ln Ω(N,V,U) (1.20) ≡ B where kB is Boltzmann’s constant which, in S.I. units, has the value 1.3806503 10−23J/K. This constant of proportionality cannot be derived - as we shall see later, it follows from the comparison with experiment. With this identiﬁcation, we see that our assumption that all degenerate energy levels of a quantum system are equally likely immediately implies that, in thermal equilibrium, the entropy of a composite system is at a maximum. Hence, in the statistical picture, the Second Law of thermodynamics is not at all mysterious, it simply states that, in thermal equilibrium, the system is most likely to be found in the state that has the largest number of degenerate energy levels. The next thing to note is that thermal equilibrium between sub-systems 1 and 2 implies that β1 = β2. In thermodynamics, we have another way to express the same thing: we say that two bodies that are brought in thermal contact are in equilibrium if their temperatures are the same. This suggests that β must be related to the absolute temperature. The thermodynamic deﬁnition of temperature is ∂S 1/T = (1.21) ∂U V,N

1-8 If we use the same deﬁnition here, we ﬁnd that

β =1/(kBT ) . (1.22)

It is of course a bit unfortunate that we cannot simply say that S = ln Ω. The reason is that, historically, thermodynamics preceded statistical thermodynamics. In particular, the absolute thermodynamic temperature scale (see below Eqn.1.11) contained an arbitrary constant that, in the previous century was chosen such that one degree Kelvin matched one degree Celsius. If we could have introduced an absolute temperature scale now, we could have chosen it such that S = ln Ω and the new absolute temperature T ′ would be equal to kBT . However, this would create many practical problems because, in those units, room temperature would be of the order of 5 10−21Joule (that is: entropy would be dimensionless, but temperature would have the dimension of energy). Few people would be happy with such a temperature scale. So we leave things as they are. The Third Law of thermodynamics is also easy to understand from Eqn1.20: it simply states that, at T = 0, the number of accessible states of a pure substance, (Ω) is equal to one. In other words: at absolute zero, the system is in its ground state - and this ground state is non-degenerate. One ﬁnal comment: we mentioned that Ω is usually “super-astronomically” large. Let me quantify this: at room temperature the entropy of one mol of argon 25 is 307.2 Joule/Kelvin and hence Ω = 1010 . That is a large number.... Now we can also understand the Second Law of thermodynamics - in its form “the entropy of a closed system cannot decrease spontaneously”. Strictly speaking, this is not true. The probability that a system spontaneously transform from a state with an entropy SA into a state with a lower entropy SB is simply equal to ΩB = exp([SB SA]/kB) (1.23) ΩA − For instance, the probability that the entropy of one mol of argon at room temperature spontaneously decreases by as little as 0.00000001% would still be 16 10−10 which is -for all practical purposes - equal to zero. This illustrates the fact that Second Law of thermodynamics is an empirical law (i.e. it states what is or is not observed) and not a mathematical law - mathematicians would only say that the entropy of a closed system is unlikely to decrease.

1.3 Classical Statistical Mechanics Thus far, we have formulated statistical mechanics in purely quantum-mechanical terms. The entropy is related to the density of states of a system with energy E, volume V and number of particles N. And similarly, the Helmholtz free energy is related to the partition function Q, a sum over all quantum states i of the Boltzmann factor exp( Ei/kBT ). To be speciﬁc, let us consider the average value of some observable−A. We know the probability that a system at temperature T will be found in an energy eigenstate with energy Ei and we can

1-9 therefore compute the thermal average of A as exp( E /k T ) i A i A = i − i B h | | i . (1.24) h i exp( E /k T ) P i − i B This equation suggests how we shouldP go about computing thermal averages: first we solve the Schrödinger equation for the (many-body) system of interest, and next we compute the expectation value of the operator A for all those quantum states that have a non-negligible statistical weight. Unfortunately, this approach is doomed for all but the simplest systems. First of all, we cannot hope to solve the Schrödinger equation for an arbitrary many-body. And secondly, even if we could, the number of quantum states that contribute to the average 25 in Eqn. 1.24 would be so astronomically large ( (1010 )) that a numerical evaluation of all expectation values would be unfeasible.O Fortunately, Eqn. 1.24 can be simplified to a more workable expression in the classical limit (h 0). To this end, we first rewrite Eqn. 1.24 in a form that is independent of→the specific basis set. We note that exp( Ei/kBT ) = i exp( H/kBT ) i , where H is the Hamiltonian of the system. Using− this relation,h | we− can write| i i exp( H/k T )A i A = ih | − B | i h i i exp( H/k T ) i P ih | − B | i T r exp( H/k T )A = P − B (1.25) T r exp( H/k T ) − B where T r denotes the trace of the operator. As the value of the trace of an operator does not depend on the choice of the basis set, we can compute thermal averages using any basis set we like. Preferably, we use simple basis sets, such as the set of eigenfunctions of the position or the momentum operator. The reason why use of these basis sets may be advantageous can be understood as follows. We recall that the Hamiltonian H is the sum of a kinetic part K and a potential part U. The kinetic-energy operator is a quadratic function of the momenta of all particles. As a consequence, momentum eigenstates are also eigenfunctions of the kinetic energy operator. Similarly, the potential energy operator is a function of the particle coordinates. Matrix elements of U are therefore most conveniently computed in a basis set of position eigenfunctions. However, H = K +U itself is not diagonal in either basis set, nor is exp( β(K +U)). However, if we could replace exp( βH) by exp( βK) exp( βU),− then we could simplify equation 1.25 considerably.− In general,− we cannot make− this replacement because

exp( βK) exp( βU) = exp( β ((K + U)+ ([K,U]))) − − − O where [K,U] is the commutator of the kinetic and potential energy opera- tors while ([K,U]) is meant to denote all terms containing commutators and higher-orderO commutators of K and U. It is easy to verify that the commutator [K,U] is of order ¯h. Hence, in the limit ¯h 0, we may ignore the terms of order ([K,U]). In that case, we can write → O T r exp( βH) T r exp( βU) exp( βK) (1.26) − ≈ − −

1-10 If we use the notation r for eigenvectors of the position operator and k for eigenvectors of the momentum| i operator, we can express Eqn. 1.26 as | i

T r exp( βH)= r e−βU r r k k e−βK k k r (1.27) − h | | ih | ih | | ih | i Xr,k All matrix elements can be evaluated directly:

r exp( βU) r = exp( βU(rN )) h | − | i − where U(rN ) on the right hand side is no longer an operator, but a function of the coordinates of all N particles. Similarly,

N k exp( βK) k = exp( β p2/(2m )) h | − | i − i i i=1 X where pi =¯hki. And ﬁnally,

r k k r =1/V N h | ih | i where V is the volume of the system and N the number of particles. Finally, we can replace the sum over states by an integration over all coordinates and momenta. The ﬁnal result is 1 T r exp( βH) dpN drN exp( β( p2/(2m )+ U(rN )) (1.28) − ≈ hdN N! − i i i Z X where d is the dimensionality of the system. The factor 1/N! is a direct result of the indistinguishability of identical particles. In almost the same way, we can derive the classical limit for T r exp( βH)A and ﬁnally, we can write the classical expression for the thermal average− of the observable A as

dpN drN exp( β( p2/(2m )+ U(rN ))A(pN , qN ) A = − i i i (1.29) h i dpN drN exp( β( p2/(2m )+ U(rN )) R P− i i i Equations 1.28 andR 1.29 are the startingP point for virtually all classical simulations of many-body systems.

1.3.1 System at constant temperature Now that we have defined temperature, we can consider what happens if we have a system (denoted by S) that is in thermal equilibrium with a large “heat- bath”(B). The total system is closed, i.e. the total energy U=UB+US is fixed (we assume that the system and the bath are weakly coupled, so that we may ignore their interaction energy). Now suppose that the system S is prepared in a specific state i with energy ǫi (note that there may be many states with the same energy – but here we consider one specific state i out of this large number). The bath then has an energy U = U ǫ and the degeneracy of the B − i

1-11 bath is given by Ω (U ǫ ). Clearly, the degeneracy of the bath determines the B − i probability P (ǫi) to ﬁnd system S in state i. Ω (U ǫ ) P (ǫ )= B − i . (1.30) i Ω (U ǫ ) i B − i To compute Ω (U ǫ ), we expandP ln Ω (U ǫ ) around ǫ =0. B − i B − i i ∂ ln Ω (U) ln Ω (U ǫ ) = ln Ω (U) ǫ B + (1/U) (1.31) B − i B − i ∂U O or, using Eqns. 1.21 and 1.22,

ln Ω (U ǫ ) = ln Ω (U) ǫ /k T + (1/U) (1.32) B − i B − i B O If we insert this result in Eqn. 1.30, we get exp( ǫ /k T ) P (ǫ )= − i B (1.33) i exp( ǫ /k T ) i − i B This is the well-known BoltzmannP distribution for a system at temperature T . Knowledge of the energy distribution allows us to compute the average energy U = ǫ of the system at the given temperature T S h i ǫ exp( ǫ /k T ) ǫ = ǫ P (ǫ )= i i − i B h i i i exp( ǫ /k T ) i i i B X P − ∂ ln ( exp( ǫ /k T )) = i −P i B − ∂1/k T P B ∂ ln Q = , (1.34) −∂1/kBT where, in the last line, we have deﬁned the partition function Q.

Q exp( ǫ /k T ) (1.35) ≡ − i B i X If we compare Eqn. 1.34 with the thermodynamic relation ∂F/T U = , (1.36) ∂1/T we see that the Helmholtz free energy F is related to the partition function Q:

F = k T ln Q. (1.37) − B Strictly speaking, F is only ﬁxed up to a constant. Or, what amounts to the same thing, the reference point of the energy can be chosen arbitrarily. In what follows, we can use Eqn. 1.37 without loss of generality. The relation between the Helmholtz free energy and the partition function is often more convenient to use than the relation between ln Ω and the entropy. As a consequence, Eqn. 1.37 is the “workhorse” of equilibrium statistical thermodynamics.

1-12 Further links with thermodynamics Let us go back to the formulation of the second law of thermodynamics that states that, in a closed system at equilibrium, the entropy of the system is at a maximum. As argued above, we have a simple understanding of this law: the system is most likely to be found in the state that has the largest degeneracy. Now consider again a thermal reservoir in contact with a (macroscopic) system, under the condition that system plus reservoir are isolated. Then we know that the entropy of this combined system should be a maximum at equilibrium. As before, we can write the total degeneracy of system plus bath as

Ω =Ω (U )Ω (U U ) (1.38) total system S bath total − S where US is the internal energy of the system, and Utotal the total internal energy of system plus bath. The condition for equilibrium is that the derivative with respect to U of ln Ω vanishes. As in Eqn.1.32, we expand ln Ω (U S total bath total − US) up to linear order in US and we ﬁnd ln Ω = ln Ω (U ) + ln Ω (U ) βU (1.39) total system S bath total − S

Note that ln Ωbath(Utotal) does not depend on US. Hence, to ﬁnd the maximum of ln Ωtotal, we have to locate the maximum of ln Ωsystem(US) βUS. Hence, we arrive at the statement that, for a system in contact with a− heat bath, the condition for equilibrium is that ln Ωsystem(US) βUS is at a maximum or, what amounts to the same thing βU ln Ω −(U ) is at a minimum. Now S − system S we make use of the fact that we have identiﬁed kB ln Ωsystem with the entropy S of the system. Hence, we conclude that, at constant temperature, the condition for equilibrium is β(U TS) is at a minimum (1.40) S − But from thermodynamics we know that U TS is nothing else than the Helmholtz free energy F . Hence, we immediately− recover the well-known statement that - at constant temperature and volume - F is at a minimum in equilibrium.

Pressure Up to this point we have been considering a system that exchanges energy with a bath. Now let us consider a system that can exchange volume with a reservoir. As before, the total energy and the total volume of system plus reservoir are ﬁxed. Let us denote this total volume by Vtot and the volume of the system of interest by V. As before, the condition for equilibrium is that the total entropy is at a maximum. Hence we have to determine the maximum with respect to V of

ln Ω(V, V V ) = ln Ω (V ) + ln Ω (V V ) (1.41) tot − sys bath tot − or, using the identiﬁcation between entropy and kB ln Ω ∂S ∂S sys + bath = 0 (1.42) ∂V ∂V U,N U,N

1-13 However, from thermodynamics we know that dU P µ dS = + dV dN (1.43) T T − T and hence ∂S P = (1.44) ∂V T U,N This expresses the fact that, if a system and a bath can exchange both energy and volume, then the conditions for equilibrium are

Tsys = Tbath (1.45) and Psys = Pbath (1.46) In practice, it is often more convenient to use the relation between the change in volume and the change in Helmholtz free energy F dF = P dV SdT + µdN (1.47) − − and the corresponding expression for the pressure ∂F P = (1.48) − ∂V T,N to obtain ∂ ln Q P = k T (1.49) B ∂V T,N Later on, we shall use this expression to compute the pressure of gases from our knowledge of the partition function. In order to obtain an explicit expression for the pressure, it is convenient to write the partition function Q in a slightly diﬀerent way. We assume that the system is contained in a cubic box with diameter L = V 1/3. We now deﬁne scaled coordinates sN by r = Ls for i =1, 2, ,N. (1.50) i i ··· If we now insert these scaled coordinates in the expression for the partition function, we obtain V N 1 1 Q(N,V,T )= dsN exp[ βU(sN ; L)]. (1.51) Λ3N N! ··· − Z0 Z0 In equation (1.51), we have written U(sN ; L) to indicate that U depends on the real rather than the scaled distances between the particles. If we now insert this expression for Q in Eqn. 1.49, we obtain ∂ ln V N ∂ 1 1 P = k T + k T ln dsN exp[ βU(sN ; L)] B ∂V B ∂V ··· − T,N Z0 Z0 T,N (1.52)

1-14 The ﬁrst part on the right-hand side simply yields the ideal-gas pressure NkB T/V . To evaluate the second term on the right-hand side, we note that the only quantity that depends on V is the potential energy U(sN ; L). We use the chain rule to compute the derivative

N ∂ ∂L ∂r ∂ = i ∂V ∂V ∂L ∂r i=1 i X N 1 ∂ = s 3L2 i ∂r i=1 i X N 1 ∂ = r . (1.53) 3V i ∂r i=1 i X We can now perform the diﬀerentiation in the second term on the right-hand side of Eqn.1.52 to obtain

N Nk T 1 ∂U P = B r . V − 3V i ∂r *i=1 i + X N Nk T 1 = B + r .f (1.54) V 3V i i *i=1 + X where fi denotes the force on particle i due to all other particles. In the special case that the intermolecular forces are pairwise additive, we can write

fi = fij Xj6=i where fij is the force on particle i due to particle j. We can now write

N Nk T 1 P = B + r .f (1.55) V 3V i ij *i=1 + X Xj6=i If we permute the dummy indices i and j, we get

N Nk T 1 P = B + r .f (1.56) V 3V j ji *j=1 + X Xi6=j We now use Newton’s third law (f = f ) to get ij − ji N Nk T 1 P = B + r .f (1.57) V 6V ij ij *j=1 + X Xi6=j where rij ri rj . Eqn. 1.57 shows how we can compute the pressure of the system from≡ knowledge− of the particle positions and the intermolecular forces.

1-15 1.3.2 Other ensembles In the previous section, we considered a system that could exchange energy with a large thermal bath. This allowed us to derive the relevant statistical mechanical expressions that describe the behavior of a system of N particles in a volume V at a temperature T . As in the case in the case of a system at constant N,V and U, a system at constant NVT can be found in any one of a very large number of quantum states. Such a collection of states is usually referred to as an ensemble. The probability to find the system in any one of these states depends on the external conditions (constant NVU or constant NVT ). The choice of ensemble depends on the “experimental” conditions that one aims to model: an isolated system will be described with the NVU-ensemble (which, for historical reasons, is often called the micro-canonical ensemble). A system at fixed volume and temperature will be described by the NVT (or canonical) ensemble. But often it is convenient to consider systems at constant pressure, or systems that can exchange particles with a reservoir. For every condition, one can introduce the appropriate ensemble. The procedure is very similar to the “derivation” of the canonical ensemble described above. For instance, consider the situation that a system of N particles can exchange energy and volume with a large reservoir. The probability to find the system in a given state with energy Ei and volume Vi is, as before, determined by the number of realizations Ω (U E , V V ) of the “bath”. As before, we expand lnΩ to first order B tot − i tot − i B in Ei, and now also in Vi. Using ∂S 1 = ∂U T and ∂S P = ∂V T we obtain

Ei + P Vi ln ΩB(Utot Ei, Vtot Vi) = ln ΩB(Utot, Vtot) − − − kBT And the probability to ﬁnd the system with volume V is given by

Q exp( βP V ) (V )= NV T − . (1.58) P ∞ dV Q exp( βP V ) 0 NV T − Another ensemble of great practicalR importance is the constant µV T or grand- canonical ensemble. As before, we consider a system of volume V in contact with a large reservoir. The system can exchange energy and particles with the reservoir. The probability to ﬁnd N particles in the system in a state with energy Ei is determined by the number of realizations of the bath Ω(Utot Ei,Ntot N). Using − − ∂S µ = − ∂N T

1-16 and following the same procedure as before, we ﬁnd that the probability to ﬁnd N particles in the system in volume V at temperature T , is given by

QNV T exp(βµN) (N)= ∞ . (1.59) P N=0 QNV T exp(βµN) For systems at constant N, V andP T , the canonical partition function provides the link between statistical mechanics and thermodynamics (through the relation F = kBT ln Q(N,V,T ). Similarly, we can deﬁne appropriate partition function for− systems at constant N, P and T or µ,V,T . The partition function for a system at constant µ, V and T is called the Grand-Canonical partition function (Ξ). It is deﬁned as follows:

∞ Ξ Q(N,V,T ) exp(βµN) ≡ NX=0 The relation to thermodynamics is given by

P V = kBT ln Ξ

It is easy to derive that the average number of particles in a system at constant µ,V,T is given by ∂ ln Ξ N = h i ∂βµ It is possible to generate a large number of diﬀerent ensembles. However, the NVT , NPT and µV T ensembles are by far the most important.

1.3.3 Langmuir adsorption Why do we need all these different ensembles? Surely, one should be enough? After all, if we can compute the Helmholtz free energy of a system at constant N,V and T , using F = kBT ln Q, then all other measurable quantities can be derived by using thermodynamic− relations. This is true. But some calculations may be more difficult in one ensemble than in another. The choice of ensemble is therefore primarily a matter of convenience. As an illustration, let us consider the problem of adsorption on a surface. We consider a highly simplified model: the surface is represented as a lattice with M sites. A site can either be empty, or it is singly occupied by an adsorbed atom. If an atom is adsorbed, its energy is lowered by an amount ǫ. We assume that different adsorbed atoms do not interact. Let us first follow the canonical (NVT ) route and, after that, the Grand-Canonical (µV T ) route. The answer is the same, but the amount of effort involved to get there, will turn out to be different. Suppose that we have N particles adsorbed over M lattice sites (in this case, M plays the role of the “volume” of the system). Then the potential energy of the system is U(N)= Nǫ . −

1-17 The number of ways in which N atoms can be distributed over M lattice sites, is M! Ω(N)= N!(M N)! − The canonical partition function of this system is therefore M! Q(N,M,T )= exp(βNǫ) (1.60) N!(M N)! − The chemical potential is ∂ ln Q µ = k T − B ∂N = ǫ + k T (ln N ln(M N)) (1.61) − B − − where, in the last line, we have used the Stirling approximation ln x! x ln x x + ln √2πx. In computing the derivatives of ln x!, we have retained≈ only the− leading terms ∂ ln x! ln x ∂x ≈ Let us deﬁne the density of adsorbed atoms N ρ ≡ M Then we can write ρ µ + ǫ = k T ln (1.62) B 1 ρ − or ρ = exp(β(µ + ǫ)) (1.63) 1 ρ − We can rewrite this as an equation for ρ: exp(β(µ + ǫ)) ρ = 1 + exp(β(µ + ǫ)) 1 = (1.64) 1 + exp( β(µ + ǫ)) − Now, let us derive the same result using the Grand-Canonical ensemble. From Eqn. 1.60 it follows that the Grand-Canonical partition sum Ξ is equal to

M Ξ = Q(N,M,T ) exp(Nβµ) NX=0 M M! = exp(Nβ(µ + ǫ)) N!(M N)! NX=0 − = (1 + exp(β(µ + ǫ)))M (1.65)

1-18 The average number of particles in the system is ∂ ln Ξ N = h i ∂βµ M exp(β(µ + ǫ)) = (1.66) 1 + exp(β(µ + ǫ)) It then immediately follows that the average density ρ is equal to h i 1 ρ = (1.67) h i 1 + exp( β(µ + ǫ)) − The answer is the same as the one obtained in Eqn.1.64, but the eﬀort involved was less. In more complex cases, a judicious choice of ensemble becomes even more important.

1.4 Fluctuations We started our discussion of statistical mechanical systems by considering a system of N particles in a volume V with total energy E. We noted that such a system can be found in a very large number (Ω(N,V,E)) eigenstates. Subsequently, we considered a system of N particles in a volume V that could exchange energy with a large thermal “bath”. In that case, the probability to find the system in any of the Ω(N,V,E) states with energy E was given by Eqn. 1.33 Ω(N,V,E) exp( E/k T ) (E)= − B P exp( E /k T ) i − i B using Eqn. 1.21, we can express ΩP in terms of the entropy S and we find (E) exp( E/k T ) exp(S(N,V,E)/k ) P ∼ − B B The most likely energy of the system, E∗, is the one for which ∂S =1/T. ∂E ∗ E=E Expanding S in a Taylor series in ∆E E E∗, we get ≡ − 1 ∂2S ln (∆E)= c + (∆E)2 + ((∆E)3); . (1.68) P 2k ∂E2 O B In the limit of large N, we can ignore terms of order ∆E3 and we find

1 ∂2S (∆E) = constant exp (∆E)2 (1.69) P × 2k ∂E2 B and, recalling that ∂2S 1 = − , ∂E2 C T 2 V

1-19 we obtain 2 1 (∆E) 2 − 2 (∆E)=(2πkB CV T ) exp 2 . (1.70) P −2kBCV T From Eqn. 1.70 we immediately see that the mean-square ﬂuctuation in the energy of a system a constant N,V,T is directly related to the heat capacity CV : (∆E)2 = k T 2C . (1.71) h i B V Using Eqn. 1.70 we can relate any average in the NV E ensemble to the corresponding average in the NVT ensemble by Taylor expansion: ∂A 1 ∂2A A = A + ∆E + (∆E)2 + ( ∆E3 ) (1.72) h iNV T h iNVE ∂E h i 2 ∂E2 h i O h i A well-known application of this conversion is the expression derived by Lebowitz, Percus and Verlet for the relation between kinetic energy ﬂuctuations in the NV E and NVT ensembles [4]:

3Nk2 T 2 3Nk (∆K)2 = B 1 B (1.73) h iNVE 2 − 2C V Of course, one can use a similar approach to relate averages in other ensembles. For instance, it is possible to derive an expression for the ﬁnite-size correction on the excess chemical potential by comparing the averages in the canonical (NVT ) and grand-canonical (µV T ) ensembles [5]:

1 ∂P ∂ρ (∂2P/∂ρ2) ∆µ (N)= 1 k T ρk T . (1.74) ex 2N ∂ρ − B ∂P − B (∂P/∂ρ)2 1.4.1 Fluctuations and Landau Free energies Let us look once more at the probability to ﬁnd a system with volume V in an NPT ensemble. According to Eqn. 1.58 this probability is given by

(V )= cQ(N,V,T ) exp( βP V ) P − 0 where c is a normalization constant. We use the notation P0 to distinguish that applied pressure (i.e. the pressure of the reservoir) from P (N,V,T ), the pressure of the system. Using the relation between the canonical partition function Q(N,V,T ) and the Helmholtz free energy F (N,V,T ), we can rewrite Eqn. 1.58 as (V )= c exp( β(F (N,V,T )+ P V ) . (1.75) P − 0 From this equation we see that the probability to ﬁnd the system in volume V is determined by the behavior of F + P0V . But this is nothing other than the Gibbs free energy G. The most likely volume is the one for which G is at its minimum. This happens when ∂F (N,V,T ) P (N,V,T )= P ∂V ≡− − 0

1-20 i.e. when the thermodynamic pressure of the system is equal to the applied pressure. In fact, if we could measure the histogram (V ), we could directly measure the variation of the Helmholtz free energy withP volume. The higher the free energy of a given volume fluctuation, the less likely we are to observe this fluctuation. In principle, we could determine the complete equation of state of the system from knowledge of (V ). To see this, consider P ∂ ln (V ) P = β(P (N,V,T ) P ) (1.76) ∂V − 0 One amusing consequence of this expression is the relation between two-phase coexistence and van der Waals loops in the equation of state [6]. Suppose that a system at a given pressure P0 undergoes a first-orde phase transition from a state with volume V1 to a state with volume V2. At coexistence, the system is equally likely to be found in either state, but it is unlikely to be in a state with intermediate density. Hence, the histogram of volumes, (V ) will be double- peaked. Eqn. 1.76 the immediately shows that the pressureP as a function of volume should exhibit an oscillation around P0. The relation between probabilities of fluctuations and free energies is, in fact, quite general. To see this, let us consider a system with N particles in volume V in contact with a reservoir at constant T . Now let us assume that we are not interested in the fluctuations of the energy of the system, but in fluctuations of some other observable property, e.g. the total magnetic moment M. We wish to know the probability that the system is found in a state with magnetic moment M. To do so, we should sum the probabilities of all states i that satisfy the constraint Mi=M. exp( βE )δ (M)= i − i Mi,M P exp( βE ) P i − i where the Kronecker delta constrainsP the sum to those terms that have the required magnetic moment. We see that the restricted sum in the numerator has the form of a partition function and we denote it by Q(N,V,T,M). We also define an associated free energy F (N,V,T,M) k T ln(Q(N,V,T,M)) (1.77) ≡− B We refer to F (N,V,T,M) as the Landau free energy associated with the variable M. Clearly, there is a close connection between (Landau) free energies and constraints. We define a subset of all possible states of the system by the constraint Mi=M. The Landau free energy then determines the probability that the system will spontaneously be found in a state that satisfies this constraint: (M)= c exp( βF (N,V,T,M)) (1.78) P − 1.5 Free energy and phase behavior The Greeks distinguished four elements: earth, water, air and fire. Three of these elements are prototypical for what we call phases of matter: solid, liquid and gaseous. The existence of different states of aggregation follows from

1-21 the properties of individual atoms or molecules, but not in a trivial way. In fact, knowledge of the properties of individual atoms and molecules is only the starting point on the way towards understanding matter. In this sense, Dirac’s famous quote that, once the laws of Quantum Mechanics are understood ”the rest is chemistry”, places a heavy burden on “chemistry”. Dirac’s “chemistry” includes virtually all of condensed matter physics (not to mention biology). The number of distinct phases of matter is far greater than the three listed above: in addition to crystalline solid phases (of which there very many), there exist dozens of distinct liquid-crystalline phases with order between that of a crystalline solid and of an isotropic liquid. Then there are materials with magnetic or ferro-electric ordering, superconductors, superﬂuids, and the list is still far from complete. One of the aims of statistical mechanics is to provide an explanation for the existence of this bewildering variety of phases, and to explain how, and under what conditions, a transition from one phase to another takes place. In this course, I shall treat only a small number of very simple examples. However, before I begin, it may be useful to point out that one of the phases that we know best, viz. the liquid phase, is actually one of the least obvious. We are so used to the occurrence of phenomena such as boiling and freezing that we rarely pause to ask ourselves if things could have been diﬀerent. Yet the fact that liquids must exist is not obvious a priori. This point is eloquently made in an essay by V. F. Weisskopf [7]: ...The existence and general properties of solids and gases are relatively easy to understand once it is realized that atoms or molecules have certain typical properties and interactions that follow from quantum mechanics. Liquids are harder to understand. Assume that a group of intelligent theoretical physicists had lived in closed buildings from birth such that they never had occasion to see any natural structures. Let us forget that it may be impossible to prevent them to see their own bodies and their inputs and outputs. What would they be able to predict from a fundamental knowledge of quantum mechanics? They probably would predict the existence of atoms, of molecules, of solid crystals, both metals and insulators, of gases, but most likely not the existence of liquids.

Weisskopf’s statement may seem a bit bold. Surely, the liquid-vapor transition could have been predicted a priori. This is a hypothetical question that can never be answered. In his 1873 thesis, van der Waals gave the correct explanation for a well known, yet puzzling feature of liquids and gases, namely that there is no essential distinction between the two: above a critical temperature Tc, a vapor can be compressed continuously all the way to the freezing point. Yet below Tc, a first-order phase transition separates the dilute fluid (vapor) from the dense fluid (liquid). In the Van der Waals picture, the liquid- vapor transition is due to a the competition between short-ranged repulsion and longer-ranged attraction. We shall discuss the Van der Waals theory in more detail in section 1.6. Here we focus on the question: why does the liquid phase exist at all? To discuss this question, we must go beyond the original Van der Waals model because it does not allow for the possibility of crystallization. Half

1-22 a century after Van der Waals death, Longuet-Higgins and Widom [8] showed that the van der Waals model 2 is richer than expected: it exhibits not only the liquid-vapor transition but also crystallization. The liquid-vapor transition is possible between the critical point and the triple point. In the Van der Waals model, the temperature of the critical point is about a factor two higher than that of the triple point. There is, however, no fundamental reason why this transition should occur in every atomic or molecular substance, nor is there any rule that forbids the existence of more than one fluid-fluid transition. It turns out that the possibility of the existence of a liquid phase, depends sensitively on the range of attraction of the intermolecular potential: as this range is decreased, the critical temperature approaches the triple-point temperature. When Tc drops below the latter, only a single stable fluid phase remains. For instance, in mixtures of spherical colloidal particles and non-adsorbing poly- mer, the range of the attractive part of the effective colloid-colloid interaction can be varied by changing the size of the polymers. Experiment, theory and simulation all suggest that when the width of the attractive well becomes less than approximately one third of the diameter of the colloidal spheres, the colloidal ‘liquid’ phase disappears. Fortunately for Van der Waals and Kamerlingh Onnes, the range of attraction of all noble gases and other simple molecules, is longer than this critical value. That is why liquids exist in everyday life 3. But let us consider another (for the time being, hypothetical) limit. We take a model system of spherical particles and decrease the range of the attractive forces. As the range of attraction decreases, the liquid-vapor curve moves into the meta-stable regime. But that is not the end of the story. For very short- ranged attraction (less than 5% of the hard-core diameter), a first-order iso- structural solid-solid transition appears in the solid phase [10]. This transition is similar to the liquid-vapor transition, as it takes place between two phases that have the same structure and only differ in density. Such iso-structural transitions have, thus far, not been observed in colloidal systems. Nor had they been predicted before the simulations appeared. This suggests that Weisskopf may have been right.

1.5.1 First-order phase transitions The condition for coexistence of two or more phases I,II, is that the pressure of all coexisting phases must be equal (P = P = ···= P ), as must be the I II ··· temperature (TI = TII = = T ) and the chemical potentials of all species α α α ··· (µI = µII = = µ ). These conditions follow directly from the Second Law of thermodynamics.··· This can be seen by considering a closed system of N particles in a volume V with a total energy E. Suppose that this system consists of two distinct phases I and II that are in equilibrium. Phase I consists of NI particles in a volume V , with a total energy E . Phase II consists of N = N N particles I I II − I 2In the Van der Waals model, molecules are described as hard spheres with an inﬁnitely weak, inﬁnitely long-ranged attraction[9]. 3In fact, everyday life is hard to imagine without the existence of the liquid-vapor transition

1-23 in volume VII = V VI and energy EII = E EI . Note that we ignore any contributions of the− interface between I and II −to the extensive properties N, V and E. This is allowed in the thermodynamic limit. The second law of thermodynamics states that, at equilibrium, the total entropy (S = SI + SII ) is at a maximum. Hence:

dS = dSI + dSII =0

But 1 P µ dS = dE + dV dN T T − T Hence:

1 1 PI PII µI µII dS = dEI + dEII + dVI + dVII dNI dNII TI TII TI TII − TI − TII

As dEI + dEII = 0, dVI + dVII = 0 and dNI + dNII = 0, we can write 1 1 P P µ µ 0= dE + I II dV I II dN T − T I T − T I − T − T I I II I II I II

As dEI , dVI and dNI are independent variations, all three terms should vanish individually. The ﬁrst term yields: 1 1 = TI TII or TI = TII . Let us call this temperature T . Then the vanishing of the second term implies that: P P I = II T T or PI = PII . Finally, from the fact that the third term is also zero, we get: µ µ I = II T T or µI = µII .

Stability We can use thermodynamic arguments to ﬁnd conditions for the stability of a single phase. To this end, we ﬁrst consider a homogeneous phase in closed system (constant N,V,E) that is divided in two (equal) parts. In equilibrium, the entropy of the total system is at a maximum. Let us now consider what happens if we transfer a small amount of energy ∆E from one half of the system to the other. In equilibrium, the temperature in the two halves of the system is the same, and hence to linear order in ∆E, the entropy does not change. When we consider the second-order variation we get: 1 ∂1/T 1 ∂1/T ∆S = (∆E)2 + (∆E)2 2 ∂E 2 ∂E

1-24 As S is a maximum. ∆S 0. Hence: ≤ ∂1/T 0 ∂E ≤ or ∂E 0 ∂1/T ≤ which implies that: ∂E T 2 = T 2C 0 − ∂T − V ≤ and hence the heat capacity CV 0. In other words, thermodynamic stability implies that the heat capacity is≥ never negative. A similar argument can be used to show that the compressibility of a system is non-negative. To this end, we consider the condition for equilibrium at constant N, V and T . Under these conditions, the Helmholtz free energy F must be a minimum. As before, we divide the system in to equal parts (both containing N/2 particles in a volume V/2 at temperature T ). We now vary the volume of one half of the system by an amount ∆V (and that of the other half by an amount ∆V . To first order in ∆V , the variation of the free energy of the system vanishes− (because the pressures in the two halves are equal). To second order we have: ∂2F ∆F = (∆V )2 ∂V 2 but this can be written as ∂P κ ∆F = (∆V )2 = (∆V )2 −∂V V where κ is the isothermal compressibility of the system. As F is a minimum in equilibrium, this implies that κ 0. ≥ 1.5.2 Coexistence How do we compute the point where two phases coexist? Let us consider the case where we have an analytical approximation of the Helmholtz free energy of both phases. For simplicity, we assume that we are dealing with a one- component system. The question is then: given F (N,V,T ) for both phases, how do we find the point where the two phases coexist. It turns out that there is a nice graphical method to do this (see fig. 1). For a given temperature, we plot F1(N,V,T ) and F2(N,V,T ) as a function of V (while keeping N and T constant). We now have to check if it is possible to draw a line that is tangent to both curves. If this common tangent exists, then this line touches F1 at a volume V1 and it touches F2 at a volume V2. As the curves F1 and F2 have a common tangent, the derivatives of F1 and F2 at V1 and V2 respectively are the same. However as ∂F = P ∂V −

1-25 m1= m2

V1 V2

Figure 1: The ﬁgure shows a free-energy curve that is locally non-convex. It is then possible to draw a line that is tangent to the free-energy curve at two distinct points (V1 and V2). The systems at V1 and V2 have the same pressures (because they have the same slope of F versus V ) and they have the same chemical potentials (because they have the same intercepts).

This implies that the pressure of the phase 1 at V1 is the same as that of phase 2 at V2. But as the two curves have a common tangent, the tangents have the same intercepts at V = 0. The intercept is ∂F F 1 V = F + P V = Nµ 1 − ∂V 1 1 1 1 and this is equal to ∂F F 2 V = F + P V = Nµ 2 − ∂V 2 2 2 2

Hence, the volumes V1 and V2 are the volumes of the coexisting phases. A completely analogous analysis can be performed if we plot F/V versus N/V . But then the slope is related to the chemical potential and the intercept is related to the pressure. In the previous section it was shown that the free energy must be a convex function of V . Hence the non-convex part does not correspond to an equilibrium situation. This is sketched in figure 2. Between volumes V1 and a, and also between b and V2, the system is “meta-stable”, i.e. it is stable with respect to small fluctuations but will eventually phase separate in the thermodynamically stable phases 1 and 2. Between a and b, the free energy curve corresponds to a situation that is absolutely unstable (negative compressibility). The system will phase separate spontaneously under the influence of infinitesimal fluctuations. The boundary between the meta-stable and the unstable region is called the “spinodal”. It should be stressed that the spinodal, although a useful qualitative concept, is not well defined. It appears naturally when we use approximate

1-26 expressions for the free energy. However, an exact expression for the equilibrium free energy is necessarily convex and has therefore no spinodal.

F(V)

a b V1 V2

Figure 2: Between V1 and V2, the free-energy curve does not correspond to the lowest free energy of the system. Between V2 and a, and between b and V2, the free-energy curve describes a meta-stable phase. Between a and b, the curve corresponds to a state that is absolutely unstable (negative compressibility).

1.5.3 Thermodynamic Integration The second law of thermodynamics states that for a closed system with energy E, volume V and number of particles N, the entropy S is at a maximum when the system is in equilibrium. From this formulation of the second law it is simple to derive the corresponding equilibrium conditions for systems that can exchange heat, particles or volume with a reservoir. In particular, if a system is in contact with a heat bath, such that its temperature T , volume V and number of particles N are ﬁxed, then the Helmholtz free energy F E TS is at a minimum in equilibrium. Analogously, for a system of N particles≡ − at constant pressure P and temperature T , the Gibbs free energy G F + P V is at a minimum. ≡ If we wish to know which of two phases (denoted by α and β) is stable at a given temperature and density, we should simply compare the Helmholtz free energies Fα and Fβ of these phases. It would seem that the obvious thing to do is simply to measure Fα and Fβ. Unfortunately, it is not possible to measure the free energy (or entropy) directly. What we can measure are averages of mechanical quantities, i.e. averages of functions of the coordinates and momenta of the molecules in the system, such as the pressure or the dielectric constant. If we denote such a mechanical quantity by A(pN , qN ), then the average of A

1-27 that can be measured in an experiment at constant N, V and T is

dpN dqN A(pN , qN )exp( βH(pN , qN )) A = − (1.79) h iNV T dpN dqN exp( βH(pN , qN )) R − where H is the Hamiltonian ofR the system expressed as a function of the mo- N N menta p and coordinates q . kB is Boltzmann’s constant and β = 1/(kBT ). However, entropy, free energy and related quantities are not simply averages of functions of the phase-space coordinates of the system. Rather they are directly related to the volume in phase space that is accessible to a system. For instance, in classical statistical mechanics, the Helmholtz free energy F is directly related to the canonical partition function QNV T :

dpN dqN exp( βH(pN , qN )) F = k T ln Q k T ln − (1.80) − B NV T ≡− B hdN N! R where h is Planck’s constant and d is the dimensionality of the system. Unlike quantities such as the internal energy, the pressure or polarization, QNV T itself is a canonical average over phase space. Rather, it is a measure for the accessible volume of phase space. This is why F or, for that matter, S or G, cannot be measured directly. We call quantities that depend directly on the available volume in phase thermal quantities. In experiments, one cannot measure the free energy directly. However, derivatives of the free energy with respect to volume V or temperature T can be measured:

∂F = P (1.81) ∂V − NT and ∂F/T = E. (1.82) ∂1/T V T As the pressure P and the energy E are mechanical quantities, they can be measured. In order to compute the free energy of a system at given temperature and density, we should find a reversible path in the V T plane, that links the state under consideration to a state of known free energy.− The change in F along that path can then simply be evaluated by integration of Eqns. 1.81 and 1.82. There are only very few thermodynamic states where the free energy of a substance is known. One state is the ideal gas phase, another may be the perfectly ordered ground state at T =0K. In order to compute the free energy of a dense liquid, one may construct a reversible path to the very dilute gas phase. It is not really necessary to go all the way to the ideal gas. But at least one should reach a state that is sufficiently dilute that the free energy can be computed accurately, either from knowledge of the first few terms in the virial expansion of the compressibility factor PV/(NkT ), or that the chemical potential can be computed by other means. The obvious reference state for solids is the harmonic lattice. Computing

1-28 the absolute free energy of a harmonic solid is relatively straightforward, at least for atomic and simple molecular solids. However, not all solid phases can be reached by a reversible route from a harmonic reference state. For instance, in molecular systems it is quite common to ﬁnd a strongly anharmonic plastic phase just below the melting line. This plastic phase is not (meta-) stable at low temperatures.

Artificial paths Fortunately, in theoretical and numerical studies we do not have to rely on the presence of a ‘natural’ reversible path between the phase under study and a reference state of known free energy. If such a path does not exist, we can construct an artificial path. This is in fact a standard trick in statistical mechanics (see e.g. [11]). It works as follows: Consider a case where we need to know the free energy F (V,T ) of a system with a potential energy function U1, where U1 is such that no ‘natural’ reversible path exists to a state of known free energy. Suppose now that we can find another model system with a potential energy function U0 for which the free energy can be computed exactly. We define a generalized potential energy function U(λ), such that U(λ = 0) = U0 and U(λ = 1) = U1. The free energy of a system with this generalized potential is denoted by F (λ). Although F (λ) itself cannot be measured directly in a simulation, we can measure its derivative with respect to λ: ∂F ∂U(λ) = (1.83) ∂λ ∂λ NVTλ NVTλ If the path from λ = 0 to λ = 1 is reversible, we can use eq. 1.84 to compute the desired F (V,T ). We simply measure ∂U/∂λ for a number of values of λ between 0 and 1. h i

1.6 Perturbation Theory The aim of thermodynamic perturbation theory is to arrive at an estimate of the free energy (and all derived properties) of a many-body system, using as input information about the free energy and structure of a simpler reference system. We assume that the potential energy function of this reference system is denoted by U0 while the potential energy function of the system of interest is denoted by U1. In order to compute the free energy diﬀerence between the known reference system and the system of interest, we use the linear parametrization of the potential energy function The free energy of a system with this generalized potential is denoted by F (λ). The derivative of F (λ) with respect to λ is ∂F ∂U(λ) = (1.84) ∂λ ∂λ NVTλ NVTλ and hence we can express the free energy diﬀerence F F as 1 − 0 1 F F = dλ U U (1.85) 1 − 0 h 1 − 0iNVTλ Z0

1-29 This expression (ﬁrst derived by Kirkwood), allows us to put bounds on the free energy F1. Consider ∂2F ∂λ2 NVTλ For the linear parametrization considered here, it is straightforward to show that

∂2F = β (U U )2 (U U ) 2 (1.86) ∂λ2 − 1 − 0 NVTλ − h 1 − 0 iNVTλ NVTλ The important thing to note is that the second derivative of F with respect to λ is always negative (or zero). This implies that

∂F ∂F ∂λ ≥ ∂λ NVTλ=0 NVTλ and hence F F + U U . (1.87) 1 ≤ 0 h 1 − 0iNVTλ=0 This variational principle for the free energy is known as the Gibbs-Bogoliubov

F(l )

0 l 1

Figure 3: The perturbation-theory estimate of the free energy of a system is an upper bound to the true free energy. In the ﬁgure, the perturbation theory estimate of F1 is given by the intersection of the dashed line with the line λ = 1. It clearly lies above the true F1. The reason is that the true free energy is always a convex function of λ. inequality. It implies that we can compute an upper bound to the free energy of

1-30 the system of interest, from knowledge of the average of U1 U0 evaluated for the reference system. Note that the present “derivation” was− based on classical statistical mechanics. However, the same inequality also holds in quantum statistical mechanics [12](a quick-and-dirty derivation is given in appendix A). Of course, the usefulness of Eqn. 1.87 depends crucially on the quality of the choice of reference system. A good reference system is not necessarily one that is close in free energy to the system of interest, but one for which the fluctuations in de potential energy difference U1 U0 are small. Thermodynamic perturbation theory for simple liquids has been− very successful, precisely because the structure of the reference system (hard-sphere fluid) and the liquid under consideration (e.g. Lennard-Jones) is very similar. As a result, U1 U0 λ hardly depends on λ and, as a consequence, its derivative (Eqn. 1.86)h is− veryi small.

1.7 Van der Waals Equation In some cases, the fluctuation term can be made to vanish altogether. Consider, for instance, the free energy difference between a hard sphere reference system and a model fluid of particles that have a hard-sphere repulsion and a very weak, attractive interaction ǫ that extends over a very large distance Rc. In particular, we consider the limit− ǫ 0, R , such that → c → ∞ 4π R3ǫ =2a 3 c where a is a constant. In the limit R , the potential energy of the per- c → ∞ turbed fluid can be computed directly. Within a shell of radius Rc, there will be, on average, Nc other particles, all of which contribute ǫ/2 to the potential energy of the fluid. For R , the number of neighbors− N also tends to c → ∞ c infinity and hence the relative fluctuation in Nc becomes negligible (as it is of order 1/√Nc). It then follows that the fluctuations in the perturbation (i.e. U1 U0) also become negligible and, as a consequence, the Gibbs-Bogoliubov inequality− becomes and identity: in this limit of weak, long ranged forces, perturbation theory becomes exact [9]! I refer to this limit as the “van der Waals” limit, for reason that will become obvious. The free energy per particle of the van der Waals fluid is

f (ρ,T )= f (ρ,T ) ρa (1.88) vdW HS − and the pressure is ∂f P = ρ2 vdW = P (ρ,T ) aρ2 (1.89) vdW ∂ρ HS − where PHS denotes the pressure of the hard sphere reference system. Of course, van der Waals did not know the equation of state of hard spheres, so he had to make an approximation [13]:

P = ρk T/(1 ρb) (1.90) HS B −

1-31 where b is the second virial coeﬃcient of hard spheres. If we insert Eqn. 1.90 in Eqn. 1.89, the well-known van der Waals equation results 4. If, on the other hand, we use the “exact” equation of state of hard spheres (as deduced from computer simulations), then we can compute the “exact” equation of state of the van der Waals model (i.e. the equation of state that van der Waals would have given an arm an a leg for). Using this approach, Longuet-Higgins and Widom [8] were the ﬁrst to compute the true phase diagram of the van der Waals model. As an illustration (Figure 4), I have recomputed the Longuet-Higgins-Widom phase diagram, using more complete data about the hard-sphere equation of state than were available to Longuet-Higgins and Widom.

0.2

τ 0.1

0.0 0.0 0.5 1.0 1.5 ρ

Figure 4: Phase diagram for the van der Waals model. This phase diagram is computed by using the hard-sphere system as reference system and adding a weak, long-ranged attractive interaction. Density ρ in units σ−3, where σ is the diameter of the hard spheres. The ‘temperature’ τ is defined in terms of the HS HS second virial coefficient: B2/B2 1 1/(4τ), where B2 is the second virial coefficient of the hard-sphere reference≡ − system.

1.8 Mean-ﬁeld theory The Gibbs-Bogoliubov inequality can be used as a starting point to derive mean- ﬁeld theory, which can be considered as a systematic approximation for the free

4Van der Waals explained the liquid-vapor transition by assuming that atoms were impen- etrable, spherical objects that exert attractive forces on each other. It should be stressed that Van der Waals proposed his theory some 40 years before the atomic hypothesis was generally accepted. The short-ranged repulsion between atoms was only explained in the 1920’s on basis of the Pauli exclusion principle. And the theory for the attractive (”Van der Waals”) forces between atoms was only formulated by London in 1930.

1-32 energy of a many-body system. For the sake of simplicity, we consider a many- body Hamiltonian of interacting spins (Ising model)

J N U = s s . 1 − 2 i j i=1 jnni X X The Ising model is not just relevant for magnetic system. In appendix C we show that a lattice-gas model that can be used to describe the liquid-vapor transition is, in fact, equivalent to the Ising model. We wish to approximate this model system, using a reference system with a much simpler Hamiltonian, namely one that consists of a sum of one-particle contributions, e.g. N U = hs 0 − i i=1 X where h denotes the eﬀective “ﬁeld” that replaces the interaction with the other particles. The free energy per particle of the reference system is given by

f (h)= k T ln ds exp(βhs) 0 − B Z If we only have two spin states (+1 and -1), this becomes

f (h)= kT ln(exp(βh) + exp( βh)) 0 − − We can easily compute the average value of s in the reference system

∂f (h) s = 0 (1.91) h i0 − ∂h In the case of spins 1: ± s = tanh(βh) (1.92) h i0 Now, we consider the Gibbs-Bogoliubov inequality (Eqn. 1.87

f f + u u 1 ≤ 0 h 1 − 0i0 J = f + s s + hs 0 h− 2 i j ii0 jnni X J = f z s 2 + h s (1.93) 0 − 2 h i0 h i0 where, in the last line, we have introduced z, the coordination number of particle i. Moreover, we have used the fact that, in the reference system, diﬀerent spins are uncorrelated. We now look for the optimum value of h, i.e. the one that minimizes our estimate of f1. Carrying out the diﬀerentiation with respect to

1-33 h, we ﬁnd

J 2 ∂f0 z s + h s 0 0 = − 2 h i0 h i ∂h ∂ s = s (Jz s h) h i0 + s −h i0 − h i0 − ∂h h i0 ∂ s = (Jz s h) h i0 (1.94) − h i0 − ∂h And hence, h = Jz s (1.95) h i0 If we insert this expression for h in Eqn. 1.91, we obtain an implicit equation for s 0, that can be solved to yield s 0 as a function of T . For spins equal to 1,h i h i ± s = tanh(βJz s ) (1.96) h i0 h i0 The free energy estimate that we obtain when inserting this value of s 0 in Eqn. 1.93 is h i J f = f + z s 2 (1.97) MF 0 2 h i0 The subscript MF in this expression stands for the “mean-field” approximation. It is very important to note that the free energy that results from the mean-field approximation is not simply the free energy of the reference system with the effective field.

1.9 Landau Theory Matter can occur in many different phases. We can induce transitions from one phase to another, by changing the temperature or pressure of the system, or by varying some other thermodynamic parameter, such as an external electric or magnetic field. At the phase transition, the state of the system changes. But the way in which this change takes place is not the same for all phase transitions. At a phase transition, the temperatures, pressures and chemical potentials of the coexisting phases are the same. This implies that the Gibbs free energies of coexisting phases are equal. But,as was first pointed out by Ehrenfest [14], phase transitions can be classified according to the behavior of the derivatives of the Gibbs free energy. If the first derivatives of the Gibbs free energy (i.e. V and S) are discontinuous, then the transition is called a ”first- order” phase transition. If the first derivatives are continuous, but the second are not, then the transition is called a “second-order” phase transition, and so on... In the light of the modern theory of critical phenomena, this classification cannot strictly be maintained. However, the Ehrenfest classification provides, at the very least, a useful language to discuss the difference between first-order and continuous phase transitions. Can we predict whether a give phase transition will be of first order or not? This is the question that Landau addressed in 1937 [15, 16]. Surprisingly, the

1-34 answer is yes (for the purists ”mostly, yes”). And, even more surprisingly, the answer can be given without much detailed knowledge of the intermolecular interactions. What is crucial, though, is the symmetry of the system. As Lan- dau already pointed out in his first paper [15], a symmetry is either there or not there. Unlike the density, the symmetry of a system cannot change continuously. In this respect, Landau says, a so-called continuous phase transformation that involves a symmetry change, is less “continuous” then the first-order transformation from liquid to vapor, that can also be carried out continuously by following a path around a critical point. To formulate the Landau theory of phase transitions, we need to quantify the symmetry properties of the free energy of the system. To be more precise, we must define a quantity that provides a measure for the degree of ordering of a system. In order to formulate the theory, we need to introduce a new concept: the order parameter. When a phase transition involves a change in symmetry of the system (e.g. the transition from an isotropic liquid to a crystalline solid, or the transition from a para-magnetic to a ferromagnetic phase) then it is useful to use an order parameter that is zero in the disordered phase, and becomes non- zero in the ordered phase. In the case of magnetic systems, the magnetization is such an order parameter. In section 1.4.1, we showed that it is possible to relate a free energy to the probability to observe a particular spontaneous fluctuation in the value of an observable quantity (see Eqn. 1.78. For an observable M, we have: F (N,V,T,M = x)= k T ln (M = x) + constant (1.98) − B P where (M = x) denotes the probability that, due to a spontaneous fluctuation, the observableP M equals x. Clearly, the most likely situation corresponds to the maximum of , and hence to the minimum of F . We now considerP the situation that M is the order parameter that distin- guishes the ordered phase (usually at low temperatures) from the disordered phase (usually at high temperatures). In the disordered phase, the most likely value of M is zero. In the ordered phase, the minimum of F corresponds to a non-zero value of M. We now make the assumption that we can expand the free energy of the disordered phase in powers of the order-parameter M.

F (N,V,T,M)= F (N,V,T, 0)+ aM + bM 2 + cM 3 + dM 4 + (1.99) ··· As this is an expansion of the free energy of the disordered phase, every term must be invariant under all symmetry operations of the disordered phase. This implies that the term linear in M must vanish (because if M itself had the full symmetry of the disordered phase, it would not be a good measure for the change in symmetry at the phase transition). In general the quadratic term in Eqn. 1.99 will be non-zero. The cubic term may or may not vanish, depending on nature of the order parameter. To give an example: if we consider spontaneous magnetization, then the ﬂuctuations with magnetization +M and M are equally likely, and hence the Landau free energy (Eqn. 1.99) must be− an even function of M. But in other cases (e.g. the ﬂuid-solid transition or the isotropic-nematic transition), the cubic term in the expansion is not symmetry

1-35 forbidden. This is easy to see in the case of the isotropic-nematic transition. In the isotropic phase, the distribution of molecular orientations has spherical symmetry. In the nematic phase, the distribution has cylindrical symmetry. A positive nematic order parameter corresponds to the case that the molecules tend to align parallel to the symmetry axis: the distribution function of molecular orientations resembles a prolate ellipsoid. If the nematic order parameter is negative, then the molecules have a propensity to align perpendicular to the symmetry axis. In that case, the orientational distribution function is more like an oblate ellipsoid. These two distributions are physically distinct, and hence there is no symmetry between states with positive and negative order parameter. We assume that the coefficients b, c, d etc. in Eqn. 1.99 all are continuous functions of temperature. For the study of phase transitions, we focus on the temperature regime where b changes sign at a temperature Tc. Close to Tc we can ignore the temperature dependence of the other coefficients. Then we can write F (N,V,T,M) F (N,V,T, 0)+ α(T T )M 2 + cM 3 + dM 4 + (1.100) ≈ − c ··· Now we can distinguish several situations. First, we consider the case that the Landau free energy is an even function of M and that the coefficient d is positive: F (N,V,T,M) F (N,V,T, 0)+ α(T T )M 2 + dM 4 (1.101) ≈ − c Above Tc, the most minimum of the free energy corresponds to the state M = 0. In other words, above Tc, the disordered phase is stable. Below Tc, the point M = 0 changes form a minimum in the free energy to a local maximum. The minimum of the free energy is easily found to be

α(T T ) M = c − ±r 2d The free energy below Tc is α2(T T )2 F (N,V,T,M) F (N,V,T, 0) − c (1.102) ≈ − 4d

Note that, in this case, ordering sets in at Tc and that the free energy and its first derivative are continuous at the transition. However, the heat capacity exhibits a jump at the phase transition. This is characteristic for a second- order phase transition in the Ehrenfest classification. An interesting alternative situation is that the coefficient d is negative. Then we have to consider the next non-vanishing term in the series. Let us assume that the coefficient f of M 6 is positive. Then it is easy to show that at high temperatures, the absolute minimum of the free energy corresponds to the point M = 0. However, when d2 T T = − c 4fα The point M = 0 is still a local minimum, but no longer the global minimum. The global minimum of the free energy jumps to the point M = d/2 . Of ± | | p 1-36 course, the free energy itself is continuous at the transition. But now there is a discontinuity in its first derivative: in the Ehrenfest language, we have a first-order phase-transition. Finally, let us consider the very important case that F contains odd powers of M: F (N,V,T,M) F (N,V,T, 0)+ α(T T )M 2 + cM 3 + dM 4 + (1.103) ≈ − c ··· We assume that d is positive. At high temperatures, the global minimum of the free energy is a M = 0. But at the point c2 T T = − c 4dα the minimum jumps to M = c/2 and a first-order phase transition takes place. Hence, according to Landau− theory, both the freezing transition and the isotropic-nematic transition must be first order. The Landau theory is a very useful way to analyze phase transitions. But it is not rigorous. In particular, it makes the assumption that the state of a system is determined by the minimum of Landau free energy. Fluctuations around that minimum are ignored. In practice, fluctuations may have a large effect on the behavior of a system near a phase transition.

A Quantum Gibbs-Bogoliubov inequality

We start with the expansion of the exponential operator to second order in the perturbation5: 1 −β(H0+λH1) −β(H0+λH1 ) −β(1−s)H0 −βsH0 e = e o λβ ds e H1e (A.1) − 0 1 s Z +λ2β2 ds du e−β(1−s)H0 H e−β(s−u)H0 H e−βuH0 + (λ3) 1 1 O Z0 Z0 and take the trace and divide by Q(0) = Tr e−β(H0+λH1)o: Q(λ) 1 s Tr e−β(1−s+u)H0 H e−β(s−u)H0 H =1 λβ H + λ2β2 ds du 1 1 1 H0 −β(H0+λH1) Q(0) − h i 0 0 Tr e o Z Z (A.2) where we have used (and will use repeatedly below) the fact that Tr AB = Tr BA, even if A and B do not commute, and that for any operator: exp(λA)exp(µA)= exp(λ + µ)A. Take the opposite of the logarithm, and keep terms up to second order in λ:

1 s −β(1−α)H0 −βαH0 2 2 1 2 Tr e H1e H1 βF (λ) βF (0) = λβ H1 +λ β H1 ds dα − h iH0 2 h iH0 − Tr e−β(H0+λH1)o 0 0 Z Z (A.3)

5 (A+B)t At t A(t−s) (A+B)s Obtained from recursive use of the equality: e = e + R0 ds e Be for t = 1, A = −βH0, and B = −βλH1. The derivation of this result is straightforward when eAt d e−Ate(A+B)t considering dt .

1-37 where we have also made the change of variable α = s u. − To prove Gibbs-Bogoliubov inequality, we only need to show that F is con- cave at λ = 0 (by shifting the origin, we can then conclude that it is con- cave everywhere), i.e. that the term between brackets is negative. To show this, we note that for a hermitian operator H such as the Hamiltonian, and for any real scalar λ, (exp λH)† = exp λH. The idea is then to consider the scalar product (A, B) Tr A†B for which the Cauchy-Schwartz inequality is: Tr A†B 2 (Tr A†A)(Tr7→ B†B). We can now rewrite: | | ≤ 1−α α α 1−α −β(1−α)H0 −βαH0 −β 2 H0 −β 2 H0 −β 2 H0 −β 2 H0 † Tr e H1e H1 = Tr e H1e e H1e = Tr A A (A.4) with A = exp( β α H )H exp( β 1−α H ). The trick is to apply the Cauchy- − 2 0 1 − 2 0 Schwartz inequality by introducing B = exp( β H ): − 2 0

1−α α β 2 −β H0 −β H0 − H0 2 Tr e 2 H1e 2 e 2 −βH0 −β(1−α)H0 −βαH0 Tr e H1 2 −βH0 Tr e H1e H1 β β = = H1 H0 Tr e h − H0 − H0 i −βH0 ≥ Tr e 2 e 2 Tr e h i (A.5) Thus a lower bound for the double integral in (A.3) is:

1 s Tr e−β(1−α)H0 H e−βαH0 H 1 s 1 ds dα 1 1 H 2 ds dα = H 2 −β(H0+λH1) 1 H0 1 H0 0 0 Tr e o ≥ h i 0 0 2 h i Z Z Z Z (A.6) which proves that F ′′(λ = 0) 0, and hence the Gibbs-Bogoliubov inequality. This elegant derivation is≤ due to Dr Benjamin Rotenberg.

B Virial pressure

Clausius proposed the following, very general, method to derive an expression for the pressure of a many body system. Consider a system of N-particles in a volume V at total energy E. As the volume and the total energy of the system are ﬁnite, the following quantity

N V r .p ≡ i i i=1 X must be bounded. This means that, for a system in equilibrium, the average value of V˙ must vanish V˙ =0 where the bar over V˙ denotes the time average. It the follows that

N N ˙ri.pi + ri. ˙pi =0 i=1 i=1 X X

2-38 Using ˙ri = pi/mi and ˙pi = fi , where fi is the force on particle i, we can write

N N 2 pi /mi + ri.fi =0 i=1 i=1 X X The ﬁrst term on the left is simply twice the kinetic energy of the system. Using the equipartition theorem, we can write this as 3NkBT . We then get

N 3NkBT + ri.fi =0 i=1 X Next, we write fi as the sum of the force due to intermolecular interactions, and the force due to interactions with the wall of the container. Let us consider the contribution to virial due to the interaction with a surface element dS of the wall of the container. We denote the position of this surface element by rS. Whenever a particle collides with this surface element, its coordinate is given by rS. Hence, collisions with this surface element make the following contribution to the force virial

N (dS) N (dS) ri.fi = rS.fi i=1 ! i=1 X X N (dS) = rS. fi i=1 X (dS) = rS.Ftot = r .( P )dS , (B.1) S − where the superscript (dS) indicates that we only consider forces due to the element dS of the container wall. In the last line of the above equation, we have used the fact that the average of the total force exerted on an area dS equals P dS and hence the force exerted by that area on the particles is P dS. We can now integrate over the entire surface of the container to obtain−

N r .f = P r .dS (B.2) i i − S i=1 ! S X wall I Using Gauss’s theorem, this can be rewritten as

P r .dS = P ( .r) V − S − ∇ IS

2-39 In 3D, .r = 3, and hence ∇ P r.dS = 3P V − − IS With this result, we can write

N 3Nk T + r .f inter 3P V =0 B i i − i=1 X or N 1 P V = Nk T + r .f inter B 3 i i i=1 X which should be compared to the result that we obtained in Eqn. 1.54

N Nk T 1 P = B + r .f (B.3) V 3V i i *i=1 + X C Ising model and Lattice gas

The Ising model for a lattice of interacting magnetic spins is among the most widely used models in physics because many other models can be mapped onto it. As an example, we consider the relation between the Ising model and a lattice gas. The lattice gas model is a simple model to describe the liquid- vapor transition. Let us ﬁrst consider the Ising model. In the Ising model, neighboring parallel spins have an interaction energy J, anti-parallel spins have an interaction energy +J. We denote the value of− the spin on lattice site i by si. si can take on the values +1 and -1. The “Ising Hamiltonian” is given by N J H = s s . I − 2 i j i=1 jnni X X In the presence of a magnetic ﬁeld B, the Hamiltonian becomes

J N N H = s s B s . I − 2 i j − i i=1 jnni i=1 X X X The partition function of the Ising model can be written as:

N N J Q = exp( β( s s B s )) I − − 2 i j − i i=1 jnni i=1 X{s} X X X where β =1/kT and s denotes all possible sets of spin values, i.e (s1 =0, 1), (s =0, 1) (s =0{, 1).} 2 ··· N

3-40 Now consider a gas on a lattice with M sites. There can be at most one particle per lattice site. Neighboring particles have an interaction energy ǫ. The Hamiltonian is − ǫ N H = n n . LG −2 i j i=1 jnni X X where ni denotes the number of particles (0 or 1) on site i. The partition function is N ǫ Q (M)= exp( β( n n )) LG − −2 i j ′ i=1 jnni {Xn } X X ′ where n denotes the set of occupation numbers that satisfy i ni = M. It turns{ out} to be more convenient to consider the Grand-canonical partition function Ξ. P N N ǫ Ξ= exp( β( n n ))eβµM − −2 i j i=1 jnni MX=0 X{n} X X Now we can change the order of the summation to convert the sum in an unre- stricted sum over n ,n , ,n : 1 2 ··· N ǫ N Ξ= = exp( β( n n ) µ n ) − −2 i j − i i=1 jnni X{n} X X X It is immediately obvious that the grand-canonical partition function of this lattice gas is identical in form to the canonical partition function of an Ising model in a magnetic ﬁeld, with the identiﬁcation s =2n 1, J = ǫ and B = µ. −

3-41 References

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