Quantum statistics Notes of the lecture from Professor Dr. Jaroslav Fabian in the summer term 2010

Transcription

1 Quantum statistics Notes of the lecture from Professor Dr. Jaroslav Fabian in the summer term 200 Florian Rappl July 20, 200

2 Contents Concepts of probability 5. Random numbers, probabilities, simple rules Mean values, standard deviations Uncertainty, disorder, entropy Maximum uncertainty distribution Useful probability distributions and central limit theorem Binomical distribution Poisson distribution Normal (Gaussian) distribution Central limit theorem Classical and quantum microstates 7 2. Configuration, momentum and phase space Phase space volume, density of states The ergodic hypothesis Phase-space invariance and Liouville s theorem Microstates in quantum mechanics, the density matrix/operator Quantum Liouvilles theorem Counting quantum states Classical Ensembles Microcanonical ensemble Microcanonical ideal gas, Gibbs paradox, indistinguished particles Canonical ensemble (T, V, N) Grand canonical ensemble (T, V, µ)

3 Contents 4 Ideal classical, bose and fermi gas Ideal classical gas Maxwell-Boltzmann probability distribution Ideal bose and fermi gases; occupation numbers Quantum gases at high temperatures - the classical limit Ideal quantum gases and the maaximum uncertainty principle Blackbody radiation Bose-Einstein condensation (BEC) Vibrations in solids (phonon gas) Equipartition theorem, virial theorem, thermodynamics of diatomic gases Degenerate Fermi Gas Non-ideal gas 8 5. Virial expansion Van-der-Waals equation Thermodynamics Work and pressure The first law of thermodynamics Enthalpy and heat capacities Second law of thermodynamics Third law of thermodynamics Free energies Maxwell s relations Gibbs-Duhem relation Conditions for thermodynamic equilibrium and stability Phase transitions and critical phenomena Phase equilibria Abrupt phase transitions - Clausius-Clapeyron equation Nucleation Continuous phase transitions - Landau s theory The Ising model: One dimensional One dimensional Ising model: renormalization group (RG)

4 Contents 7.7 Ising model in two dimensions: Renormalization Group Appendix 9 8. Volume of a sphere in n dimensions Stirling s formula Bose-Einstein and Fermi-Dirac Integrals Equipartition theorem Sommerfeld s expansion Pressure and entropy in statistics

5 Concepts of probability In the theory of common sense we have a fair coin with 2 possible outcomes head (H) and tail (T) and a fair die (pl. dice) with outcomes from to 6. In this context fair means unbiased.. Random numbers, probabilities, simple rules Definition.. A sequence x, x 2,..., x N of numbers is called random, if the probability for x N+ to have a value of x does not depend on the previous numbers. An example: Sequences like HTHTHTHT or HHHHHHH let us think that the next event must be H in both cases. This is common sense and not a good random sequence. When there s a pattern it s not a random sequence. In reality we can never be sure if the sequence is random. A truly random sequence is infinite. In computers we have quasi random numbers. This means they are random FAPP (for all pratical purposes). Assigning probabiliy We have to distinguish between Trial, process: Tossing (flipping) a coin or throwing a dice. Outcome of a trial: 5

6 Chapter. Concepts of probability H, T,, 2, 3, 4, 5, 6 or all so called microstates. Sample space (which is the set of all outcomes): {H, T}, {, 2, 3, 4, 5, 6} or all so called phase space. For a fair coin we obtain p(h) the probability of outcome H, p(t) the probability of outcome T, p(h) = p(t) =. 2 Similar for a fair die: In general for a probability we require that p() = p(2) = p(3) = p(4) = p(5) = p(6) = 6. p(outcome i) = p i 0. As boundary condition we have p i = (something always happens). outcomes i For continuous outcomes x R we find that p i p(x)dx is the possibility to find the outcome in [x, x + dx]. While p i is dimensionless, p(x) is a probability density (think of ψ(x) 2 ). So we have to normalize it. The normalization of the continuous distribution is 6

7 Chapter. Concepts of probability The simplest example for this would be p(x)dx =. p(x) =, b a 0, else. x [a, b],. Addition rule The probability that two independent outcomes i or j occur is p(i or j) = p(i) + p(j). Example: The probability that a die gives or 3 is p( or 3) = p() + p(3) = = 3. Multiplication rule The probability that two independent outcomes i and j occur is p(i and j) = p(i) p(j). Example: The probability that a die gives and (then) 3 is p( and 3) = p() p(3) = 6 6 = 36. Example: Suppose a DNA evidence is likely with probability 0 6. The chances are 0 6 : that the defendant is guilty. The systematic error in collecting, analysing, handling, reporting DNA evidence is %. Question: What is the probability that the defendant is guilty?.2 Mean values, standard deviations The mean value or average of x is defined as 7

8 Chapter. Concepts of probability x = x = x i p(i) = i dxxp(x). The mean square of x is defined as x 2 = x 2 = x 2 i p(i) = dxx 2 p(x). i The variance or dispersion of x is defined as σ 2 x = x 2 x 2 = (x x) 2. The standard deviation or r.m.s. (root mean square) of x is defined as σ x + σ 2 x. As an example we consider a coin with H = and T = 0. We obtain x = p(h) + 0 p(t) = 2, x 2 = 2 p(h) p(t) = 2, σ 2 x = x 2 x 2 = 2 4 = 4, σ x = σ 2 x = 2. This means that the actual values of the outcome are likely in the interval [x σ x, x+σ x ]. For the coin we found 2 ± 2..3 Uncertainty, disorder, entropy Shannon did some research in the 940 s. He found that that a system is in order when not much (the least) information (knowledge) about a system is required in order to describe it, while a system is in disorder when a lot of (more than the least) information is required in order to reconstruct it. 8

9 Chapter. Concepts of probability Example: A child in a house with n rooms. We bring the child in the morning to the house. We then come back in the evening to see in which room the child is. Task: Calculcate the probability (p(i)) for the child being in room i. We already know that there are two main cases in this scenario.. A certain (well-behaved) child: We have full knowledge of it. So the probability will be p(i) = δ i,i0, where i 0 is the room in which the child stays all the time. 2. An uncertain child (this one is running around): We have the least knowledge of where he or she is. We probability will be p(i) = n. Our task is now to find a function of p(i) which cover our main cases. In the case of minimum uncertainty we have to obtain a minimum of p(i) while in the case of maximum uncertainity we have to obtain a maximum. In addition we require that the function is additive. Example: Two children and two houses. We then get for our function S that S = S + S 2, where S is the function for the first child in the first house and S 2 is the function for the second child in the second house. So we have to obtain p(, 2) = p() p(2). Satz.3. (Shannon) There is only one such function: S = k p i ln p i = kln p, (.) i where p i p(i) and k is some constant. This function is called Shannon s (information) entropy. Example: If p(i) = δ i,i0 we get S = k ln = 0. If p(i) = n we obtain that S = k ln n. 9

10 .4 Maximum uncertainty distribution Chapter. Concepts of probability Task: Find p i such that S has a maximum and the average value of a quantity A i, A = p i A i, i is fixed. This is an optimization problem with Lagrange multipliers. The function to be maximized is with the constraints that S = k p i ln p i, i Calculating we obtain p i = i and p i A i = A. i 0! = δ S λ ( p i ) λ 2 ( p i A i A), ( ) δs = k δp i ln p i + p i δp i p i i = k (ln p i + )δp i. i i 0 = k (ln p i + )δp i λ δp i λ 2 A i δp i = i = k (ln p i + ) λ λ 2 A i = = Finally we get individually i i i i i = ( ln p i + + λ k + λ ) 2 k A i. i i which gives us ln p i + + λ k + λ 2 k A i = 0, 0

11 Chapter. Concepts of probability ( ( p i = exp + λ k + λ )) 2 k A i. The coefficients λ and λ 2 can be determined by the constraints above. We can simplify the solution for p i in order to get p i = ( Z exp λ ) 2 k A i, Z = i ( exp λ ) 2 k A i. (.2).5 Useful probability distributions and central limit theorem.5. Binomical distribution Suppose an event A can occur with probability p. Then, out of N trials, the event A will be found exactly k times with the probability p (N) k = N k pk ( p) N k. (.3) In this formula we have N k the combinatorial factor, read N choose k which is N!. (N k)!k! number of ways to choose k objects out of N. That is the p k the probability to find k events A. ( p) N k the probability that the remaining N k events are not A. Example: Suppose I flip a coin 00 times. What is the probability that H comes 20 times? p (00) 20 = = Example: Suppose you play darts (L little squares). You can shoot (throw) N times. What is the probability to hit the same (pre-determined) little square 2 times when p = L?

12 Chapter. Concepts of probability p (N) 2 = N 2 ( ) N 2. L 2 L Question: What is the probability that in the first two throws I hit the same little square? It is L 2. Another question: What is the probability that in the first two throws only I hit the ( same little square? In all other throws I do not hit it. It is L N 2. L) 2 Let us calculate the average value of k and variance σ 2 k. For the mean value we calculate: k = = N k=0 kp (N) k = N N k k pk ( p ) N k = }{{} k=0 N N! k (N k)!k! pk q N k = p p k=0 q N k=0 N! (N k)!k! pk q N k = = p p (p + q)n = pn( p + q ) N = pn, }{{} which is the result we expected. Now we do the same for k 2 : k 2 = N k=0 So we get for the variance = ( k 2 P (N) = p ) 2 (p + q) N = p ( ) k Np(p + q) N = p p = Np ( (p + q) N + (N )p(p + q) N 2) = Np( + pn p). σ 2 k = k2 k 2 = Np( + pn p) p 2 N 2 = Np( p). Finally we have the standard deviation as σ k = Np( p), σ k k = p p N. 2

13 Chapter. Concepts of probability Figure.: Plot of a binomical distribution with FWHM 2 Np( p) for p = 2 and N = Poisson distribution The Poisson distribution is given as P k = λk k! exp( λ), k = 0,, 2,...,, (.4) where λ is the parameter of the distribution. Let us now calculate the average k and variance σ 2 k. For the average value we get By exactly the same we get k = ( ) λ k k exp( λ) = exp( λ) k! k=0 = λ exp( λ) k= k= λ k (k )! = λ k (k )! = λ exp( λ) λ k k! k=0 } {{ } exp(λ) = λ. 3

14 Chapter. Concepts of probability k 2 = k=0 = exp( λ)λ = λ 2 + λ. ( ) λ k 2 k exp( λ) = exp( λ) k! λ k k (k )! = k= (k + ) λk k! = exp( λ)λ k=0 k=0 k λk k! + k=0 λ k k! = For the variance we get σ 2 k = k2 k 2 = λ 2 λ 2 + λ = λ. So we found the standard deviation which is λ. The relative error is σ k k = λ. We will show that the poisson distribution is a limiting case of the binomical one as N (many experiments) and p 0 (small probability). p (N) k = = N! (N k)!k! pk ( p) N k λ=pn = N (N ) (N k + ) N k } {{ } λ k k! exp( ) λ =exp( λ) ( λ N ) k ( λ N ) N k = N! (N k)!k! ( λ ) N λ ( λ N λ ) k = N } {{ }} {{ } = λk k! exp( λ)..5.3 Normal (Gaussian) distribution It is a continuous distribution for a variable x: p(x) = ( ) exp (x a)2. (.5) 2πσ 2 2σ 2 Let us calculate the average and variance. 4

15 Chapter. Concepts of probability Figure.2: Plot of a Gaussian distribution with FWHM 2σ x for a = 5 and σ x = 2. For the mean value we calculate, that ( x = dxxp(x) = dxx exp ) (x a) 2 = 2πσ 2 2 σ 2 ( = dx(x a) exp ) (x a) 2 + 2πσ 2 2 σ 2 } {{ } + In the next step we get ( a dx exp ) (x a) 2 = a. 2πσ 2 2 σ 2 } {{ } x 2 = 2πσ 2 = σ 2 + a 2. =a =0 ( dxx 2 exp ) (x a) 2 = 2 σ 2 So we get what we expected: σ 2 x = x 2 x 2 = σ 2 + a 2 a 2 = σ 2, σ x = σ. 5

16 Chapter. Concepts of probability The normal distribution also follows from the binomical one, in the limit of N and taking k x, centered at k a, and considering expansion around k. See Gold and Tobachnik for a proof..5.4 Central limit theorem Suppose x, x 2, x 3,...x N are distributed according to some distribution p(x). The average of x is x and variance of the distrubtion, σ 2 x, exists (not infite). Then the average are distributed according to y N = N N x i, i= P(y N ) = ( exp ) (y N x) 2, 2πσ 2 2 σ 2 y y and σ y = σ x N. For a proof see Gold and Tobachnik. Alternativly for the sums the distribution is S N = N i= x i So the standard deviation is p(s N ) = ( exp 2πσ 2 2 S ) (S N Nx). σ 2 S The relative error is again σ x N xn N. σ S = σ x N. Example: Shopping in a supermarket. Typical price is approximently x = 2 e. Buy N 20 items. Guess the total price. What is the total error when σ x = 0.5 e. The average sum is Nx which is 40 e. The total error is e. So we have (40 ± 2) e. 6

17 2 Classical and quantum microstates 2. Configuration, momentum and phase space The from the classical probability theory known systems coin or dice now go over to more complicated systems like gas, solids, etc. We need to find the sample space for the generic physical system. We first discuss the classical systems. Consider a collection of N particles. Assume that the particles cannot escape from the box. We say they are confined. Assume also, that the total energy is preserved, i.e. particles do not give away energy to the walls of the box. (N, V, E) are fixed. The collection of the positions of particles is Q = ( r, r 2,..., r N ) = (x, y, z,..., x N, y N, z N ), which forms the configuration space. This space is 3N dimensional. The collection of the momenta of the particles is P = ( p, p 2,..., p N ), which forms the momentum space. This space is again 3N dimensional. The state of the system of N particles is uniquely defined / determined by the collection (Q, P). We call this collection the phase space. It has 3N + 3N = 6N dimensions. It is huge! The important question is: How big is the phase space? We will argue that the phase space is the sample space. A leap in thinking brought us to the conclusion that... 7

18 Chapter 2. Classical and quantum microstates the only preserved quantity, constraining the phase space (in addition to V, N) is the energy! Large systems forget their initial conditions, i.e. all other integrals of motions are forgotten. 2.2 Phase space volume, density of states We introduce 3 measures for the phase-space volume, Γ, g and Ω. First of all we introduce with the step function dqdp Γ(E) = Θ(E H(Q, P)) = h3n Θ(x) =, x 0, 0, x < 0. H(Q,P) E dqdp, (2.) h3n H(Q, P) is the Hamiltonian of the system. So Γ(E) tells us the volume of the phase space up to the energy E. This brings us to h which is the yet to be determined constant (which will be the Planck constant), of dimension [xp] =kgm 2 s, introduced to make Γ dimensionless. Second we introduce the density of state The dimension of g(e) is [ E ]=J. Finally we introduce the number of states g(e) = dγ(e) dqdp de = δ(e H(Q, P)). (2.2) h3n Ω(E) = g(e)δe. (2.3) This is again dimensionless and gives us the number of states in the interval [E, E + δe]. 8

19 Chapter 2. Classical and quantum microstates Example: -dimensional linear harmonic oscillator We set our Q = x and P = p. We already know that H = 2m p2 + 2 mω2 x 2. The surface of the equal energy is H(Q, P) = E 2m p2 + 2 mω2 x 2 = E p2 2mE + x2 =. 2E mω 2 To calculate the volume we use So we get Γ(E) = area of ellipse h = πab h π 2mE = h 2E mω 2 = E ω. g(e) = dω de = δe and Ω(E) = g(e)δe = ω ω. Important example: Phase space volume of classical ideal gas We have N particles in an ideal gas (no/weak interacting particles) with no potential energy, i.e. P H = 2 N 2m = p 2 i 2m. Our task is now to calculate Γ(E), which is an integral i= Γ(E) = We already know that H(Q,P) E dqdp h = 3N h 3N dq d 3 p N p 2 d 3 p 2 d 3 p N. i i= 2m E dq = V N. The remaining momentum integral is the volume of a sphere in 3N dimensions of radius 2mE. Generally we have 9

20 Chapter 2. Classical and quantum microstates p p2 N = p2 x + p2 y + p2 z p2 Nx + p2 Ny + p2 Nz ( 2mE) 2. This is a mathematical problem. The solution is discussed in appendix 8.. We now know that R = 2mE, n = 3N, dp = P 2 2mE The measures π 3N 2 Γ ( 3N 2 3N 3N 2 E 2 = (2πmE) 3N 2 + )(2m) ( 3N)!. 2 Γ(E) = VN 3N (2πmE) 2 h 3N ( 3N)! E 3N 2, 2 g(e) = dγ de = VN 3N (2πmE) 2 3N h 3N ( 3N)! 2E = Γ(E) N Ω(E) = Γ(E) δe 2 3 N All thermodynamics of classical gases follows from here! Important remark The phase-space volume change very very fast with E. Suppose there are N = 0 23 particles. If you change E by δe ( E one particle, the volume of Γ increases twice: N) 2 3, which is the energy of just Ω(E) Γ(E)! 2.3 The ergodic hypothesis We consider a coin with H = 0 and T =. We can now calculate the average value:. By flipping the coin many times: HHTHTT and so on we get 20

21 Chapter 2. Classical and quantum microstates avg = flip(t = ) + flip(t = 2) flip(t = 6) = 6 number of flips [ ] flip(t = ) + flip(t = 2) flip(t = 6) t = = number of flips t T i flip(t i ) t = time average, i.e. average over time, dtflip(t) T = where t is the difference in time between the flips and T is the total time. 2. Guess the answer by finding the sample space and the probabilities for the events. So we have a guess, like p H=0 = 2, p T= = 2. The average is then avg = 0 p H + p T = i p i dxxp(x) = 2. This is the ensemble average - an ensemble is the sample space. In this case it would be (H, T). The ergodic hypthesis then states that time average (T ) is equal to the ensemble average. Flipping a coin is an ergodic process as long as the flipper explores the whole ensemble. Example for a non-ergodic-process: An ant cannot flip the coin due to the energy barrier. So the time-average is H = 0. 2 Important: H, T need to have the same (gravitational) energy - but in this scenario there is an energy barrier. Example: Glass (amorpheous material) has also energy barriers which prevent the material to be in a crystal formation. Also in an experiment we measure the time averages i=h,t A time = T T 0 dta(t). 2

22 Chapter 2. Classical and quantum microstates In theory due to the high number of particles (0 23 ) we have A ensemble = dqdpa(q, P)ϱ(Q, P), phase space where ϱ(q, P) is a probability density in phase space. The ergodic hypothesis now states, that A time T = A ensemble. (2.4) 2.4 Phase-space invariance and Liouville s theorem We already know that phase-space trajectories cannot cross because if they would cross, the system would have two futures. The question now is (for a phase-space volume): how much does the volume of a phase-space region change in time? The answer is that it does not change! We are now going to proof this. Proof Denote q α as coordinates and p α as momenta with α =, 2,..., 3N. So we have (q, q 2,..., q 3N ) = (x, y, z, x 2, y 2, z 2,..., x N, y N, z N ), (p, p 2,..., p 3N ) = (p x, P y, p z, p x2, p y2, p z2,..., p xn, p yn, p zn ). A general point (Q, P) evolves in time to (Q, P ) according to q α = q α + q α δt = q α(q α, q α ), p α = p α + ṗ α δt = p α(p α, ṗ α ). So we can say that it is (Q, P) (Q, P ), Q Q (Q, P), P P (Q, P). 22

23 Chapter 2. Classical and quantum microstates Remark q α and ṗ α are functions of q α, p α. So the volume at a time t > 0 is equal to the trajectory times the volume at time t = 0, dq dp (Q = JdQdP, J =, P ) (Q, P) = q q q p 3N q... So to get a feeling for this formula we first consider a simple one-dimensional case with particle, i.e. p 3N p 3N p 3N. q = q + qδt, p = p + ṗδt. Now we calculate the Jacobian: q = + q q q δt, p p = + ṗ p δt, q = q p p δt, p q = ṗ q δt, q q q p J = p p = + q δt q δt q p ṗ δt + ṗ δt = q p q p ( = + q ) ( q δt + ṗ ) p δt q p δt ṗ q δt = ( q = + q + ṗ ) δt + O(δt 2 ). p δt 2 and higher can be neglected because our taylor expansion was also just to order δt 2. We already know the Hamilton equations, Inserting them we obtain q = H p, ṗ = H q. ( q q + ṗ ) ( ) 2 H = p p q 2 H = 0. q p Therefore we get that J =! The same proof can be done for more particles in more dimensions, then 23

24 Chapter 2. Classical and quantum microstates J = + 3N α= ( qα + ṗ ) α δt + O(δt 2 ). q α p α } {{ } =0 Thus it is proofen. We now go one step further. Since the volume does not change it is like an incompressable liquid, where the shape changes but the area is preserved. We are now going to introduce the phase-space density of states ϱ(q, P), dqdpϱ(q, P) =. (2.5) all phase-space Consider a volume V in the phase-space (not in the real-space - V is not the box volume!). Suppose we have n ensemble members (points) total. The number of ensemble points in volume V is then n v = n dqdpϱ(q, P). V We consider time evolution. The net increase of the number of points inside V is then given through dn v dt = d dt n V ϱ(q, P) dqdpϱ(q, P) = n dqdp. V t The net decrease is accompanied by the flow through the surface of the volume, dn v dt = V d S j, j = nϱ v, where d S is an infinitesimal surface element, v the velocity (6N dimensional vector, ( Q, Ṗ)) and nϱ the number of points (density) at (Q, P). By using Gauss theorem we obtain dn v dt = n dqdp ϱ V t = ds j = dqdp j = n dqdp (ϱ v). V V V So we found another continuity equation in form of ϱ t + (ϱ v) = 0. (2.6) 24

25 Chapter 2. Classical and quantum microstates By using div(ϱ v) = we find Liouville s equation, 3N α= ( ϱ qα q α + ϱṗ ) α = p α α ( ϱ q α + ϱ ) ṗ α q α p α ϱ ( ϱ t + q α + ϱ ) ṗ α dϱ q α p α dt α = 0. (2.7) Conclusions We can directly make two statements:. We see that dϱ = 0, where dϱ is a full derivative, i.e. the observer moves with the dt dt flow, while ϱ is the stationary observer case. t 2. If ϱ t = 0 then ϱ = ϱ(e) only. So the energy is conserved. 2.5 Microstates in quantum mechanics, the density matrix/operator We already know the Schrödinger equation: where i ψ t = 2 2m ( N ), ψ = ψ( r,..., r N ) : wave function N : Number of particles ψ 2 d 3 r : probability to find particle(s) in a small neighorhood of volume d 3 r arround r, ψ( r,... r N ) : defines the microsopie state = microstate. (2.8) Stationary states φ n have Any wavefunction can be expanded as Ĥφ n = E n φ n. 25

26 Chapter 2. Classical and quantum microstates ψ = c n φ n. n Normalization requires that c n 2 =. n The average value of a physical observable A, to which there is a hermitian operator Â, is given by A = dqψ Âψ = ψ Â ψ, where A is the expectation value of A. We continue, A = dq c nc nφ nâφ n = nn c nc n nn dq φ nâφ n } {{ } A nn = c nc na nn. nn We now introduce ϱ nn }{{} density matrix c nc n, A = ϱ nn A nn = tr ˆϱÂ. In operator notation we have ˆϱ = ψ ψ. φ n ˆϱ φ n = c n { }} { φ n ψ ψ φ n } {{ } c n nn = c n c n = ϱ nn. Diagonal elements, ϱ nn = c n 2, are called probabilities. Off-diagonal elements, are called coherences. trϱ = n c n 2 = ˆϱ 2 = ˆϱ ˆϱ = ψ ψ ψ ψ = ψ ψ = ˆϱ ˆϱ : Complete description of the quantum mechanical state. Up to now, considering only pure states, it is a redundant description. 26

27 Chapter 2. Classical and quantum microstates The motion of mixed states (Schrödinger s cat). Mixed state: there is a probability p α, that the system is in state ψ α. I cannot write a wave function for such a mixed state. Average Value of A is Ā = α statistical average {}}{ p α A α = dqψ αâψ a. A α }{{} QM average Let us rewrite: Ā = p α ψ α Â ψ α = p α c α n c α n α α nn = p α ϱ nn A nn = tr( ˆϱÂ) α nn dqφ nâφ n = α nn p α c α n c α n A nn = ϱ nn = α p α c α n c α n, ˆϱ = α p α ψ α ψ α : describes mixed states tr( ˆϱ) = α p α n c n 2 = α p α =, Summary: ˆϱ 2 ϱ! Signature of mixed states! pure states mixed states ˆϱ ψ ψ α p α ψ α ψ α, α p α = tr ˆϱ ˆϱ 2 = ˆϱ ˆϱ A tr [ ˆϱÂ ] tr [ ˆϱÂ ] 27

28 2.6 Quantum Liouvilles theorem Chapter 2. Classical and quantum microstates t = = α t ( pα ψ α >< ψ α ) = ( p α i Ĥ ψ α ψ α α ( ψα > p α ψ α + ψ α ψ ) α = t t }{{} α SGL ) i ψ α ψ α Ĥ = (Ĥ ) ˆϱ ˆϱĤ = i = i [Ĥ, ˆϱ ]. The Von Neumann s equation is ϱ t = [Ĥ, ] ˆϱ. i In QM we learned that the time evolution of an operator  is By using this we obtain d dtâ =  t + [Â, ] Ĥ. i For time independent ˆϱ: d dt ˆϱ = i [Ĥ, ˆϱ] + [ ˆϱ, Ĥ] = 0, i d ˆϱ = 0 Quantum Liouvilles theorem. dt ϱ t ˆϱ: integral of motion, conserved quantity. = 0 [ ˆϱ, Ĥ] = 0. Definition 2.6. Energy is the only integral of motion left, (quantum chaos) ˆϱ ˆϱ(E). It is then natural to take φ n as stationary states, Ĥφ n = E n φ n, and ϱ nn = ϱ(e n )δ nn, Diagonal in the basis of stationary states, ϱ(e n ) = c n 2 = p n. 28

29 Chapter 2. Classical and quantum microstates Is the probaility to find the system in a microstate (state) of energy E n. ( the probability that the system has energy E n ). What happens if there are degenerate states (having the same energy E n )? ϱ nn = p n (E n )δ nn = For degenerate states, the coherences ρ n n phase assumption or a priori equal probabilities. p n (E n ) p n (E n ) p n (E n ). are also zero. This is called the random We see that Ā = p n (E n )A n = p n A n, n n A n = φ n  φ n. The task of statistical physics is to find p(e n ) or p n. 2.7 Counting quantum states In analogy with classical physics, define Number of states below E, The density of states, Γ(E) = Θ(E E n ). n g(e) = dγ de = δ(e E n ). n And the number of states in [E, E + de], Ω(E) = g(e)de. Important example: Single-particle states. Consider one particle in a box of linear dimension L, with periodic boundary conditions (see QM lecture). We calculate 29

30 Chapter 2. Classical and quantum microstates Ĥ = 2 2m 2, ψ( r + L m) = ψ( r) with an integer vector m. Therefore we can calculate the spectrum (eigenstates and eigenenergies), ψ k = V e i k r, ε k = 2 k 2 2m, Trick: from discrete sums to integrals: 2π k = n n: integer vector. L k f ( k) = = f ( k) = f ( k) n x n y n z = }{{} n z }{{}}{{} n x n y n x n y n z f ( ( ) 2π 3 ( ) 2π 3 k) k x k y k z = L L k 3, n x n y n z kx ky kz ( ) 2π 3 d 3 k f ( k). }{{} L continuum Back to our example: Γ(E) = Θ(E E states ) = Θ(E E k ), states k ( ) 2π 3 d }{{} 3 2 k 2 ( 2π kθ(e L 2m ) = L continuum ) 3 k 2mE 2 d 3 k = ( ) 3 V 4π 2mE 2. (2π) Recall the classical result (N = ), which was Γ(E) cl = V h 3 4π 3 (2mE)

31 Chapter 2. Classical and quantum microstates This becomes the quantum result, if we identify h = 2π. Classical-quantum corespondence fixes the value of h to be Planck s constant. 3

32 3 Classical Ensembles Classical physics {Q, P} microstate ϱ(q, P) probability density dqdpϱ(q, P) = A = dqdpa(q, P)ϱ(Q, P). Quantum physics n ϕ n microstate p n probability to be in n ˆϱ density operator p n = n A = p n A n = tr [ ˆϱ ], n A n = n  n. 3. Microcanonical ensemble We consider a system which is totally closed. We have only 3 parameters, the total energy E, the volume V and the number of particles N. These parameters are fixed and describe a macrostate. 32

33 Chapter 3. Classical Ensembles We need to find the probability that, given the macrostate (E, V, N), the system is in a specified microstate. In equilibrium (things do not change with time) we can guess: a priori equal probabilities,. classical physics p microstate = nr. of microstates. E,V,N ϱ(q, P) = δ [E H(Q, P)]. h 3N g(e) 2. quantum physics p n = Ω(E), if E E n E + δe = 0, else. ˆϱ = ϕ n ϕ n. Ω(E) n,e E n E+δE 3. entropy We know S = S(E, V, N). From the first chapter we still know So we identify S = k B p n ln(p n ) = k B ln Ω(E). n S(E, V, N) = k B ln Ω(E, V, N), (3.) with the Boltzmann constant k B = J/K. This constant connects statistics with thermodynamics. Important remark: As N we see that up to negligable constants. S = k B ln Ω = k B ln Γ = k B ln(gδe), 33

34 Chapter 3. Classical Ensembles 4. temperature For this we consider a system which we devide into two parts. This system can exchange energy. In this system we see that E = E A + E B, V = V A + V B, N = N A + N B, are fixed and E A = E A + E, E B = E B E, E A + E B = E A + E B = E. In equilibrium the enrgy is distributed (on average) such, that S = S A + S B will have a maximum S = 0, S = S A E S B E = 0. E A E B So we conclude for every division, that S A E A = S B E B. (3.2) It is useful to introduce a name for this derivative, the thermodynamic temperature T or T = ( ) S. (3.3) E V,N This is the partial derivative of S with respect to E, keeping V and N fixed. 5. pressure and chemical portential By doing the same as before we obtain (by changing the volume) the pressure over the temperature 34

35 Chapter 3. Classical Ensembles Summary P T = ( ) S. (3.4) V E,N And if we allow particles N to be exchanged we get the negative chemical potential over the temperature µ ( ) S T =. (3.5) N E,V macrostate p n thermodynamic potential thermodynamics E, V, N S = k Ω(E,V,N) B ln Ω = ( ) S T E V,N ds = de + PdV µ dn P = ( ) S T T T T V E,N µ = ( ) S T N E,V 3.2 Microcanonical ideal gas, Gibbs paradox, indistinguished particles Thermodynamics of an ideal gas (weakly interacting particles) using microcanonical ensemble. We will use We calculated Γ in chapter 2.2, S = k B ln Γ. Therefore we get Γ(E, V, N) = VN 3N (2πmE) 2. h 3N (3N/2)! S = k B ln Γ = k B ( N ln V + 3N 2 ln(2πme) ln 3N 2! 3N ln h ). We can easily see that for the temperature we obtain 35

36 Chapter 3. Classical Ensembles T = This leads to the well known equation ( ) S = 3N E 2 V,N E k B. E = 3 2 Nk BT. (3.6) We say that k B T is the thermal energy. The equipartitian theorem says that for every (quadratic) degree of freedom there is, in equilibrium, the energy k 2 BT for a particle. This is classical physics. In an ideal gas we have three degrees of freedom. They are quadratic since H p2 x 2m + p2 y 2m + p2 z 2m. We can also calculate the pressure over the temperature, P T = ( ) S = k B N V V PV = Nk BT. E,N This is also well known as the equation of state of an ideal gas.. Gibbs paradox Striling s formula (Appendix) is given by Therefore we see that lim ln N! N ln N N. N ( 3N ln 2 )! 3N 2 ln 3N 2 3N 2. The entropy then changes to ( ) 3 4πmE S k B N ln V 2 3Nh k BN. This is a problem, cause V should be V. We will see this by considering 2 gases, N where T = T 2 and P = P 2. Therefore we get E N = E 2 N 2, N V = N 2 V 2. 36

37 Chapter 3. Classical Ensembles Suppose we remove the partition between the two gases and let the gases mix. We can calculate As we see we find S = S (S + S 2 ) } {{ } mixing entropy. S = k B N ln V V + k B N 2 ln V V 2. Example: N = N 2 = N, V 2 = V 2 = V. So we get 2 S = k B N ln 2 > 0. This is a paradox, because if we take the same gas left and right we see that S is still greater than zero. So the entropy increases - that makes it ill defined. The formula must be wrong since S expect = 0. The solution by Gibbs was to devide the phase-space volume(s) by N!. So we get Γ(E) Γ(E) N! Then, we see that the V problem in S is fixed, Now we get S = 0 for the same gases. 2. Undistinguished particles = VN (2πmE) 3N 2 h 3N N! (3N/2)! S = k B N ln V ( ) 3 4πmE 2 N 3Nh k BN. In quantum mechanics two particles are indistinguishable in principle. Example: two particles in three states.. We have 3! = 6 possible states of the system. These are called 6 microstates. If the particles are undistinguished we have only 37

38 Chapter 3. Classical Ensembles states, where N! 3 = 3! 2! removes the degeneracy due to undistinguished particles. 3. Correct classical phase space measures dqdp Γ(E) = Θ(E H(Q, P)), h 3N N! g(e) = dγ de, Ω(E) = g(e)δe. If there are two types of particles we have to use N! N! N 2!, Canonical ensemble (T, V, N) Canonical ensemble in a set of microstates of a system (A) in contact with thermal bath (B). We say that a thermal bath is a reservoir. The system itself can be very small (even particle). The reservoir is to be very large and is not changing due to the contact with the system. We exchange energy between the reservoir and the system. We propose E = E A + E B = const., where E is the total system energy. Thus the total system is isolated. A and B can exchange energy. Question What is the probability to find system A in microstate n of energy E n? Answer We will show that p n = Z exp( βe n), β = k B T. 38

39 Chapter 3. Classical Ensembles The number of microstates corresponding to a fixed energy E A is Ω A (E A ) Ω B (E E A ), } {{ } =E B with the number of microstates of A of energy E A, Ω A, and the number of microstates of B of energy E B, Ω B. The number of microstates available is E A Ω A (E A ) Ω B (E E A ). Question What is the probability to find A with energy E A? Answer We know that this is given by p(e A ) = Ω A(E A ) Ω B (E E A ) E A Ω A (E A ) Ω B (E E A ). Question What is the probability to find A in a microstate n of energy E n = E A? p n = p n (E n = E A ) = Ω B (E E A ) E A Ω A (E A ) Ω B (E E A ). This probability is given mainly by Ω B (E E A ), the number of microstates of the reservoir! We know that ( ) Ω B (E B ) = exp S B (E B ), kb with the entropy S B of B. So Ω B (E B ) is a steep function of E B. Therefore it is the best to deal with ln Ω = k B S, when we do a Taylor expansion. In equilibrium, A will have average (most probable) energy E A. Ω B (E E A ) around E A. We calculate We expand 39

40 Chapter 3. Classical Ensembles S B (E E A ) S B (E E A ) + S B E A (E A E A ) +..., EA =E A S B (E E A ) S B E B (E A E A ) +... = EB =E E A = S B (E E A ) T (E A E A ) +... Neglecting higher order terms ( 2 S/ E 2 ) means that we assume T = const. So we calculate ( ) ( ) Ω B (E E A ) exp S(E E A ) exp kb k B T (E A E A ), Ω B (E E A ) p n = E A Ω A (E A ) Ω B (E E A ) = ( ) E Ω B (E E A ) exp A k B exp ( ) E A T k B T = ( ) E Ω B (E E A ) exp A Ω k B T EA A(E A ) exp ( ) = E A k B T exp( E A /k B T) = E A Ω A (E A ) exp( E A /k B T) = E A = E n, Ω = E A n = exp( E n /k B T) = n exp( E n /k B T). Thus we found the Boltzmann probability with the partition function Z, p n = Z exp( βe n), (3.7). Classical physics β = /k B T, Z = exp( βe n ). n We see that ϱ(q, P) = exp( βh(q, P)), Z N! h3n dqdp Z = exp( βh(q, P)). h 3N N! 40

41 Chapter 3. Classical Ensembles 2. Quantum physics Again we observe that ϱ nn = p n = Z exp( βe n), Z = exp( βe n ), n ˆϱ = Z exp( βĥ). 3. Maximum uncertainty principle In chapter.4 we found that if we fix the average of an observable A, A = p n A n, then the probability that maximum uncertainity is n p n = Z exp( λ 2A n ). Thus the canonical ensemble is such that the uncertainty is maximized and E = λ 2 = p n E n = const., n k B T = β. 4. Internal energy E (or just E) and specific heat C We know that E = p n E n = E n exp( βe n ) = ln Z Z β. n n This is called the internal energy. For the specific heat or heat capacity we calculate 4

42 Chapter 3. Classical Ensembles c v = = = ( ) E T Here we must know that = β E T β = V ( ) z k B T 2 β Z β ( E k B T 2 E 2). 2 = k B T 2 Z 2 ( Z β ) 2 + Z 2 Z β = 2 E 2 = c v = p n E 2 n = exp( βe n )E 2 n = Z n 2 = exp( βe Z β 2 n ) = Z n [ E k B T 2 E 2] = σ2 E 2 k B T. 2 This forumla is important, since it shows that a) c v is proportional to the energy fluctuation (σ 2 E ). b) c v is greater than 0! n 2 Z β 2, c) c v is proportional to N due to ( E/ T) V. Therefore we calculate σ 2 E N σ E N. The relative error in measuring the energy is thus given by for N. σ E E N N N 0, In the thermodynamic limit, i.e. N, V and thus N/V = const the average energy will be E E fixed! Thus it is very important to see that in the thermodynamic limit the canonical ensemble is equivalent to the microcanonical ensemble! 42

43 Chapter 3. Classical Ensembles 5. Entropy and free energy F We see that [ S = k B p n ln p n = k B Z exp( βe n) ln n)] Z exp( βe = n = k B Z exp( βe n )( ln Z βe n ) = n n = k B Z ln Z exp( βe n ) + k β B E n exp( βe n ) = Z n = k B ln Z + T E. n We call F = k B T ln Z = E TS, (3.8) the (Helmholz) free energy. It is the thermodynamic potential, F F(T, V, N) for the canonical ensemble. 6. Thermodynamics We will now calculate some of the thermodynamic quantities. We see that ( ) F T V,N Therefore we have = k B ln Z k B T dβ ln Z T β S = = F T + k BT ( E) = S. k B T2 ( ) F. (3.9) T V,N Similarly we introduce two other thermodynamic quantities. We have the pressure and the chemical potential P = ( ) F, V T,N 43

44 Chapter 3. Classical Ensembles µ = ( ) F. N T,V Summary macrostate p n thermodynamic potential thermodynamics T, V, N exp( βe Z n) F F(T, V, N) S = ( ) F T V,N Z = n exp( βe n ) F = k B T ln Z P = ( ) F V T,N df = SdT PdV + µdn µ = ( ) F N T,V 3.4 Grand canonical ensemble (T, V, µ) Again our system A can be small, even particle. The heat reservoir B is very very large and E A + E B = E, N A + N B = N, as well as V A and V B are fixed. The system A can exchange energy and particles with the reservoir. Question What is the probability to find system A in a (one) microstate n corresponding to the energy E n and number of particles N n? Answer We will show that p n = exp( β(e n µn n )), Z G Z G = exp( β(e n µn n )). n Z G is called the grand partition function (sometimes also written as Z). The number of microstates of the total isolated system A + B corresponding to a fixed energy E A and a fixed number of particles N A is Ω A (E A, N A ) Ω B (E E A, N N A ), with the number of microstates of A of energy E A and particles N A, Ω A, and the number of microstates of B of energy E B and particles N B, Ω B. 44

45 Chapter 3. Classical Ensembles The number of microstates available is Ω A (E A, N A ) Ω B (E E A, N N A ). E A,N A Question What is the probability to find A with energy E A and particles N A? Answer We know that this is given by p(e A, N A ) = Ω A (E A, N A ) Ω B (E E A, N N A ) E A,N A Ω A (E A, N A ) Ω B (E E A, N N A ). Question What is the probability to find A in a (one) microstate n of energy E n = E A and number of particles N n = N A? p n = p n (E n = E A, N n = N A ) = Ω B (E E A, N N A ) E A,N A Ω A (E A, N A ) Ω B (E E A, N N A ). This probability is given mainly by Ω B (E E A, N N A ), the number of microstates of the reservoir! Assuming that E B E A, N B N A we can do the mathematics of large numbers. In equilibrium, A will have average (most probable) energy E A and number of particles N A that corresponds to the maximum of S A + S B = S A+B. Then we can expand Ω B (E E A, N N A ) around E A in E A and N A in N A. We calculate S B (E E A, N N A ) S B (E E A, N N A ) + S B E A (E A E A ) + EA =E A + S B N A (N A N A ) +..., NA =N A S B (E E A, N N A ) S B E B (E A E A ) EB =E E A S B N B (N A N A ) +... = NB =N N A = S B E B = EB =E E A T, S B N B = µ NB =N N A T = = S B (E E A, N N A ) T (E A E A ) + µ T (N A N A )

46 Chapter 3. Classical Ensembles Neglecting higher order terms ( 2 S/ E 2 ) means that we assume T = const, because B is very very big, so that So we calculate T E A, µ N A 0. p n = = Ω B (E E A, N N A ) E A,N A Ω A (E A, N A ) Ω B (E E A, N N A ) = ( ) E Ω B (E E A, N N A ) exp A k B T ( Ω B (E E A, N N A ) exp = exp( β(e NA µn A )) n exp( β(e n µn n )). E A µn A k B T exp ( ) ( E A k B T exp µ N A k B T ) exp ( ) µ N A k B T ) EA,NA Ω A(E A, N A ) exp ( E A µn A k B T ) = We call the Z = exp(βµ) fugacity, which means tendency to escape. Overall we found. classical physics We see that Z N = exp( βe n (N)), Z G = Z n Z N, (3.0) n Z = exp(βµ). N=0 2. Quantum physics Here it is Z G = Z N N=0 ϱ(q, P) = Z G N=0 dqdp exp( βh(q, P)), h 3N N! Z N exp( βh(q, P, N)). h 3N N! 46

47 Chapter 3. Classical Ensembles p n = Z G exp( β(e n (N n ) µn n )), ˆϱ = Z G exp( β(ĥ µ ˆN)), Z G = tr [ exp( β(ĥ µ ˆN)) ]. Note: The trace means N=0 n(n). ˆN n = N n, 3. Information theory point of view The grand canonical ensemble is the one which maximizes uncertainty given the constraints E = p n E n, N = p n N n, where T, µ are the Lagrange multipliers in that case. 4. Particle number fluctuations Let us calculate N/ µ. n n N µ = N n exp( β(e n µn n )) = µ Z G n = Z G N Z 2 n exp( β(e n µn n )) + µ Z β Nn 2 exp( β(e n µn n )) = G n n } {{ }} {{ } = N2 N 2 k B T = σ2 N k B T. = N 2 β This result is important since a) This means that N/ µ > 0! b) The error in N is σ N /N which is proportional to / when N goes to infinity. We also see =N 2 β N. This goes to zero σ 2 N N µ N. So the average number of particles N is very well defined (small σ N ) in the thermodynamic limit. This can be considered as fixed. Therefore in the 47

48 Chapter 3. Classical Ensembles thernodynamic limit with N, V and N/V = const. the microcanonical, canonical and grandcanonical ensembles are equivalent! The reason is that E fixed 5. Grand free energy Φ(T, V, µ) ( ) σe E 0, N fixed ( ) σn N 0. We now introduce another potential, Φ(T, V, µ) = k B T ln Z G. Remark The notation in the literature is Ω, L, Ψ, G,... Let us calculate the entropy, S = k B p n ln p n = N [ ] = k B exp( βe n + βµn n ) ln exp( βe n + βµn n ) = Z G Z G n = k B Z G ( ln Z G ) Therefore we see that exp( βe n + µβn n ) + n β + k B (E n µn n ) exp( ηe n + µβn n ) = Z G n = k B ln Z G + T (E µn) = Φ T + E µn. T In thermodynamics we see that E E and N N. 6. Thermodynamics We now calculate some relations. Φ = E ST µn. (3.) 48

49 Chapter 3. Classical Ensembles We will see later that Φ µ = k BT µ ln Z G = k B T Z G Z G µ = = k B T βn n exp( β(e n µn n )) = N, Z G n Φ T = k dβ ln Z G B ln Z G k B = dt β = Φ T + (E n µn n ) exp( β(e n µn n )) = T n = Φ T (E µn) = S. T Summary Φ V = P. macrostate p n thermodynamic potential thermodynamics T, V, µ Z G exp( β(e n µn n )) Φ Φ(T, V, µ) S = ( ) Φ T V,µ Z G = n exp( β(e n µn n )) Φ = k B T ln Z G P = ( ) Φ V T,µ dφ = SdT PdV Ndµ N = ( ) Φ µ T,V 49

50 4 Ideal classical, bose and fermi gas classical gas non-interacting classical particles, low occupation probabilities. bose gas non-interacting bosons (integer spin), photons, phonons, integer-spin atoms, cooper pairs,... with no restriction on the occupation of states. fermi gas non-interacting fermions (half-integer spin), electrons, protons, neutrons, up to one particle in state (Pauli exclusion principle). In the limit of T all the gases are equivalent. 4. Ideal classical gas In chapter 3.2 we found that S(E, V, N) = k B N ln in the microcanonical ensemble.. Canonical treatment We know that [ ( ) V 4πmE 3/2] + 5 N 3Nh 2 2 k BN, We need to calculate H(Q, P) = P 2 2m = p2 2m +... p2 N 2m. 50

51 Chapter 4. Ideal classical, bose and fermi gas Z Z(T, V, N) = We need to calculate the partition function dqdp exp( βh(q, P)). h 3N N! Z = = h 3N N! [ V N h 3N N! dq d 3 p d 3 p 2 d 3 p N exp( β p 2 /2m) exp( β p2 N /2m) = )] N d 3 p exp (β p2 = ZN 2m N!. In the factorization we used the one particle partition function which is Z = V [ ] dp h 3 x exp( βp 2 x/2m) dp y exp( βp 2 y/2m) dp z exp( βp 2 z/2m) = = V ( π2mkb T ) 3 V = h 3 λ = h 2 2πmk B T = ( h 2πmkB T 2π 2 mk B T. ) 3 = V λ 3, This thermal-de-broglie-wavelength is equivalent to the de-broglie-wavelength (h/momentum) of a particle of energy k B T. Question Why can Z be factorized into the product of Z? Answer Each particle interacts (exchanges energy) independently with the reservoir. This leads to V N V N Z = N! h (2πmk BT) 3N/2 = 3N N! λ. 3N The free energy is F = k B T ln Z. Therefore we calculate [ ( ) V N ] ( ) V F = k B T ln = k N! λ 3 B T ln N! k B TN ln = λ 3 [ ] N = ln N! N ln N N = k B TN ln V λ3 k B TN = ( [ ] ) N = k B TN ln V λ3 N. 5

52 Chapter 4. Ideal classical, bose and fermi gas The entropy is S = ( F/ T) V,N. So we calculate Remark We see that ( [ ] N S = k B N ln V λ3 ( ) N = k B N ln V λ3 ) k B TN T ln λ3 = k BN. T T ln λ3 = T T 3 ln λ = 3T dλ λ dt = 3 2. Therefore we have found the Sackur-Tetrole equation, [ ] N S S(T, V, N) = k B N ln V λ k BN. (4.) The internal energy E = F + TS, E = 3 ( )3/2 2 k V 4πmE BTN S = k B N ln N 3Nh k BN, as in the microcanonical treatment (E E). 2. Grandcanonical treatment We calculate z = exp(βµ), Z G = z N Z N = N=0 = exp exp(βµn) ( ) V N = N! λ 3 N=0 (exp(βµ) V λ 3 ). The grand canonical potential Φ = k B T ln Z G is N=0 ( ) V Φ = k B T exp(βµ). λ 3 ( exp(βµ) V ) N = N! λ 3 52

53 Chapter 4. Ideal classical, bose and fermi gas The average particle number N = ( Φ/ µ) T,V is Chemical potential ( ) V N = k B T β exp(βµ) = βφ Φ = k λ 3 B TN. µ = k B T ln ( ) N V λ3. Remark If N V/λ 3 we see that µ. This is the case when the distance between the particles is much greater than λ or when there are much more boxes than particles. The entropy is S = ( Φ/ T) V,µ. We see ( ) ( V S = k B exp(βµ) + k λ 3 B TV 3 ) dλ λ 4 dt = Φ T 3 Φ 2 T + µ k B T [ ] N = k B N ln V λ k BN. Φ T = 5 2 k BN N µ T = ( ) V dβ exp(βµ) + k B T λ 3 dt µ exp(βµ) = This is the same as in the canonical treatment in the thermodynamic limit (N N). The internal energy is E = Φ + TS + µn. We see E = k B TN TN µ T k BTN + µn = 3 2 k BTN. Again in agreement with the equipartition theorem. 4.2 Maxwell-Boltzmann probability distribution Question What is the probability to find a given one particle in a state of momentum p? Answer We know that 53

54 Chapter 4. Ideal classical, bose and fermi gas We now want to calculate ϱ ( p) = (2πmk B T) 3/2 exp ϱ ( r, p ) = Zh exp( βh( r, p 3 )), Z = ( p2 2mk B T This is the canonical ensemble for one particle! So we see that ). d 3 r d 3 p h 3 exp( βh( r, p )). ϱ ( r, p ) = exp ( ) p2 2mk B T h 3 d 3 r h 3 d 3 p exp = exp( p 2/2mk BT), V (2πmk B T) 3/2 ϱ ( p ) = d 3 r ϱ ( p, r ) = Vϱ ( r, p ), ϱ ( p ) = ( ) p 2 (2πmk B T) 3/2 exp. 2mk B T ( ) = p2 2mk B T Question Suppose the particle is in an external potential U( r). What is ϱ ( p)? Answer It is the same because we also devide through the potential term. Thus we do not have any space dependence! This is very important. The velocity distribution ϱ ( p)d 3 p = ϱ ( p)4πp 2 dp ϱ (p = p )dp, ϱ (p) = 4πp 2 (2mk B T) 3/2 exp( p 2 /2mk B T), ϱ (p)dp = ϱ (v)dv = ϱ (v) m dp, ( ) ϱ (v) = 4π(vm) 2 m(2πmk B T) 3/2 exp mv2. 2k B T So we found the Maxwell distribution We see that we get ( ) ϱ (v) = 4πv 2 m 3/2 ) exp ( mv2. (4.2) 2πk B T 2k B T 54

55 Chapter 4. Ideal classical, bose and fermi gas 2k B T v max = m, v = The distribution is shown in figure (4.). 8k B T πm. Figure 4.: Plot of the Maxwell distribution with k B = m = at T = Ideal bose and fermi gases; occupation numbers Consider N identical (nondistinguishable) particles. wavefunction The system is described by a ψ = ψ( r,..., r N ). bosons The wavefunction is symmetric with respect to the exchange (swap) of particles ψ( r, r 2,..., r N ) = ψ( r 2, r,..., r N ) = ψ( r p,..., r pn ), with p i the permuations of particle i, e.g. (, 2, 3) P (2, 3, ) = (p, p 2, p 3 ) σ(p) = +. Here σ(p) is the sign of the permutation P, defined 55

56 Chapter 4. Ideal classical, bose and fermi gas σ(p) = ( ) number of permutations. If σ(p) = + we have an even number of permutations, while with σ(p) = we have an odd number of permutations. fermions The wavefunction is antisymmetric with respect to the exchange of two fermions, algorithm ψ( r,..., r N ) = ψ( r 2, r,..., r N ) = σ(p)ψ( r p,..., r pn ). Write down the Hamiltonian for N particles, Ĥ = 2 2m ( N ) + u( r,..., r N ). } {{ }} {{ } kinetic Solve to get eigenstates ψ N and energies E n, potential Ĥψ n = E n ψ n. Symmetrize (antisymmetrize) to get bosonic (fermionic) states. ψ n,bosons = c b ψ n ( r p,..., r pn ), P ψ n,fermions = c f σ(p)ψ n ( r p,..., r pn ). The bosonic and fermionic states n have energy E n. It can happen that ψ n,fermions = 0. Then the corresponding state n does not exist and E n is absent in the spectrum. Now consider ideal bose and fermi gases, in which the particles do not interact, u = 0, P Ĥ = 2 2m ( N ). 56

57 Chapter 4. Ideal classical, bose and fermi gas Let ψ k ( r) be the single-particle wave function, 2 2m 2 ψ k ( r) = E k ψ k ( r), obtained by solving the single-particle hamiltonian. The single-particle energy is ε k. The quantum numbers of the single particle states are k. A generic N-particle wavefunction (eigenstate of H) can be written as The corresponding energy is ψ( r,..., r N ) = ψ ki ( r i ). i The boson wavefunction is E = ε ki. i The fermion wavefunction is ψ boson ( r, r 2,..., r N ) = c b P N ψ ki ( r pi ). i= N ψ fermion ( r, r 2,..., r N ) = c f σ(p) ψ ki ( r pi ). P i= Therefore we can rewrite this with the help of the so called Slater-Determinate, ψ fermion ( r, r 2,..., r N ) = ψ k ( r )... ψ k ( r N ). N!..... ψ kn ( r )... ψ kn ( r N ) Example The Pauli-exclusion principle follows directly from this. single-particle state k and two particles. Consider a ψ k,boson = ψ k ( r )ψ k ( r 2 ), E = 2ε k, ( ψ k,fermion = ψk (r )ψ k (r 2 ) ψ k (r 2 )ψ k (r ) ) = 0. 2! 57

58 Chapter 4. Ideal classical, bose and fermi gas The state in which two fermions (or more) occupy a single state, does not exist. This is the Pauli-exclusion principle. Remark The distinction between bosons and fermions is important as long as the single-particle wavefunction overlap. For non-interacting particles, the state is uniquely given by the set of occupation numbers n k (number of particles occupying the single particle state k), E = n k ε k. We have the constraint N = k n k. Grand canonical ensemble treatment. We have k bosons n k = 0,, 2,..., N, fermions n k = 0,. The grand partition function is then Z G = exp( βe state (N) + βµn). N=0 states With E state (N) = k n k ε k we can rewrite this to Z G = exp β n k ε k + βµ n k. Therefore we find that N=0 {n k }, k n k =N k k 58

59 Chapter 4. Ideal classical, bose and fermi gas Z G = = = = Z k = exp β (ε k µ)n k = N=0 {n k }, k n k =N exp β (ε k µ)n k δ k n k,n = N=0 {n k } exp( β (ε k µ)n k ) {n k } k k k N=0 δ k n k,n } {{ } n n 2 k exp( β(ε µ)n ) exp( β(ε 2 µ)n 2 ) = Z k, n k exp( β(ε k µ)n k ). = = We call Z k the single-state k grand canonical function. Bose-Einstein-statistics Since we have n k = 0,, 2,..., we can see by using the geometric series for µ < ε k that we have So we obtain Z k = exp( β(ε k µ)n k ) = n k =0 exp( β(ε k µ)). The grand free energy is Z G = Z k = exp( β(ε k µ)). k k Φ = k B T ln Z G = k B T ln( exp( β(ε k µ))). The average number of particles is N = ( ) Φ = µ T,V k Therefore we got the Bose-Einstein statistics k exp(β(ε k µ)) = n k. k 59

60 Chapter 4. Ideal classical, bose and fermi gas n k = The average occupation of single-particle state k. exp(β(ε k µ)) n(ε k). (4.3) Fermi-Dirac-statistics Here we have n k = 0, which gives us So the grand canonical function is The grand free energy is Z k = + exp( β(ε k µ)). Z G = Z k = ( + exp( β(ε k µ))). k k Φ = k B T ln Z G = kbt ln( + exp( β(ε k µ))). The average number of particles is N = ( ) Φ = µ T,V k Therefore we got the Fermi-Dirac statistics n k = k exp(β(ε k µ)) + = n k. exp(β(ε k µ)) + f (ε k). (4.4) The average number of fermions in a single-particle state k is described by this. k 4.4 Quantum gases at high temperatures - the classical limit From the last section we know that n k = exp(β(ε k µ)), 60

61 Chapter 4. Ideal classical, bose and fermi gas Figure 4.2: Comparison of fermions and bosons at low and high temperatures with for bosons and + for fermions. This results in figure 4.2. We get n k if exp(β(ε k µ)). In this limit we have n k exp( β(ε k µ)). (4.5) This is called the Maxwell-Boltzmann statistics. This is used to describe classical gases. For the average particle number we get N = n k = exp(βµ) exp( βε k ) = exp(βµ)z exp(βµ) = N. Z Overall we found that k k n k = N Z exp( βε k ). Thus the probability that a single-particle energy level ε k is occupied is given by Remark Compare this with p k = n k N = Z exp( βε k ). (4.6) p n = Z exp( βe n). The difference is that the one above is always valid and refers to the probability to find the (whole) system in many particle state of energy E n while the one we found only describes one particle states and is only valid in the classical limit. Question At which physical conditions is the classical limit realized? What is in T? 6

62 Chapter 4. Ideal classical, bose and fermi gas Answer For n k ε k we get that µ = βµ. k B T In chapter 4.2 we found for a classical gas µ = k B T ln ( ) N V λ3. We see that µ/k B T if λ 3 N/V. Therefore we see that ( ) /3 V λ. N This can be interpreted when we see that V/N is the volume available for one particle. The classical limit is reduced when the distance between particles is much greater than the thermal de-broglie wavelength λ. We should note that λ = λ(t) / T. The summary is shown in a graph, figure 4.3. Figure 4.3: Comparison of the most important statistics 4.5 Ideal quantum gases and the maaximum uncertainty principle Consider levels i of energy ε i, each g i degenerate. Let n i be the occupation number of (all degenerate) levels of energy ε i. We will now investigate using 62

63 Chapter 4. Ideal classical, bose and fermi gas. Bosons N = n i, E = ε i n i. ε i ε i Question In how many ways can we distribute n i Bosons in g i states? We see that we have n Bosons and g seperating walls. Thus we have total n + g circles. This results in n i + g i Ω i =. This is the number of ways to distribute n i Bosons in g i states. For all energy levels we have n i Ω = Ω i = ε i ε i n i + g i n i. 2. Fermions Here we have Ω i = Therefore we find for all energy levels g i n i. Ω = ε i g i n i. 3. Maxwell-Boltzmann particles Classical, indistinguished particles. There is no correlation between particle occupation. Distinguished particles, Ω i = g n i i. 63

64 Chapter 4. Ideal classical, bose and fermi gas Undistinguished particles, Ω i = gn i i n i! n g i i Ω = n i!. ε i 4. Maximum uncertainty principle We will now do the basic calculation for the Bose-Einstein statistics. The calculation for the Fermi-Dirac and Maxwell-Boltzmann statistics is nearly the same. Question What is n i if the entropy should be a maximum, given the two constraints E = ε i n i, N = n i? ε i ε i Answer We can do the calculation with the method of Lagrange-multipliers and S = k ln Ω. A variation of this results in δ S λ n i λ 2 ε i n i = 0. ε i By using δg i = 0 (g i is a constant) and Stirling s approximation we calculate ε i n i + g i S = k ln Ω = k ln ln ε i n = k n i + g i ln i ε i n = i [ = k ln(ni + g i )! ln n i! ln(g i )! ], ε i δs ( = δ ln(ni + g i )! δ ln n i! ) = k ε i ( = δ((ni + g i ) ln(n i + g i ) ln(n i + g i )) δ(n i ln(n i ) n i ) ) = ε i ( = δn i ln(n i + g i ) + (n i + g i ) n i + g ) ( ) i n i δn i δn i ln n i + n i δn i δn i = δ n ε i i ( ) ( ni + g ) i = δn i ln(ni + g i ) ln n i = δn i ln. n ε i i ε i 64

65 Chapter 4. Ideal classical, bose and fermi gas The extremum problem 0! = δs λ δn i λ 2 ε i δn i = ε i ε i ( ( ni + g ) i = k δn i ln λ n i k λ ) 2 k ε i, ε i can then be solved with the knowledge of λ 2 /k = β, λ /k = βµ. Overall we get n i = g i exp(ε i λ 2 /k + λ /k) = g i exp(β(ε µ)). To get the occupation of one of the states g i divide by g i. Therefore we found the Bose-Einstein statistics, n state of energy εi = exp(β(ε i µ)). 4.6 Blackbody radiation We already know that photons have a wave vector k and a polarization λ. We also know that there are two independent polarization states. The photon energy is given by ω = ω k = kc, with the speed of light c. Photons are bosons (with spin ), so they obay the Bose-Einstein statistics. The chemical potential of photons is zero (µ γ = 0). The reason for this is the creation and annihilation of photons. Their number is not fixed. The average number of photons is determined by T, V. Then we see that 65

66 Chapter 4. Ideal classical, bose and fermi gas F F(T, V) µ = ( ) F = 0. N Another (alternative) explenation will be given in the recitation class. Photon-statistics: n k = exp(β ω k ) = 2 exp(β ω k ). λ Question How many photons there are at given frequency ω? Answer The photon spectral density is n(ω) = = = n k δ(ω ck) = k 2V (2π) 4π 3 V π 2 c 3 ω 2 exp(β ω). V (2π) 3 d 3 2 k δ(ω ck) = exp(β ω k ) ( ) dkk 2 ω exp(β ω k ) c δ c k = So we get the number of photons of frequencies in [ω, ω + dω] in a box of volume V, n(ω)dω = V π 2 c 3 ω 2 exp(β ω) dω. Historically important is the spectral energy density, u(ω) per unit volume, defined as Vu(ω)dω = ωn(ω)dω. This is the equation known as Planck s law, u(ω) = π 2 c 3 ω 3 exp( ω/k B T). (4.7) Therefore the shape of u(ω) depends only on T! This is a way to measure T very precise. The general shape is shown in figure

67 Chapter 4. Ideal classical, bose and fermi gas Figure 4.4: Planck s law for c = = with ω in units for k B T A black body absorbs all radiation and emits according to Planck s law. The number of photons is calculated by N = = Total radiated energy is then n(ω)dω = 0 x = β ω dx = β dω = V π 2 c 3 0 V π 2 c 3 (β ) 3 = V ( ) 3 kb T T 3. π 2 c ω 2 exp(β ω) dω = x 2 dx 0 exp(x) } {{ } = E = 0 = V π 2 c 3 (β ) 4 Vu(ω)dω = = 4 c σvt4 T 4, 0 V π 2 c ω 3 dω = x = β ω = 3 0 exp(β ω) x 3 dx = V (k B T) 4 π 4 exp(x) π 2 c = } {{ } =π 4 /4 σ = c π π 2 c 3 k4 B 5 = π 2 k 4 B 60 3 c. 2 We call σ the Stefan-Boltzmann constant. 67

68 4.7 Bose-Einstein condensation (BEC) Chapter 4. Ideal classical, bose and fermi gas Consider bosons of mass m (real atoms) of energy ε k = 2 k 2 /2m. The number of bosons is given by Remark We can calculate N = exp(β(ε k µ)). k f (ε k ) So we see that this is k V (2π) 3 d 3 k f (ε k ) = V (2π) 3 dε d 3 kδ(ε ε k ). dεg(ε) f (ε), g(ε) = V (2π) 3 d 3 kδ(ε ε k ) = Vm3/2 2π2 3 ε. Then we get N = n k = k 0 dεn(ε)g(ε) = Vm3/2 2π2 3 0 ε dε exp(β(ε µ)). We get µ < 0 because µ has to be smaller than the smallest ε, which is ε = 0. The maximum number of bosons corresponds to µ = 0. We calculate Nmax cont = Vm3/2 ε dε = βε = x = 2π2 3 0 exp(βε) ( ) 3 2πmkB T 2 x = V h π 0 exp(x) dx = ( ) V 2 ( ) x V = λ 3 π 0 exp(x) λ 3 } {{ } =ζ(3/2)γ(3/2) Here we have a problem by using the integral instead of the sum. We forget about the first (ground) state. We get g(0) = 0 - but we know that there is state. We can fix this problem by using 68

69 Chapter 4. Ideal classical, bose and fermi gas N = N ε=0 + N cont. So this one is fixed at T, µ. N ε=0 is the number of bosons at the lowest energy state, k = 0 and ε = 0. We see that N cont max V = 2.62 particles thermal de-broglie value. The temperature at which the lowest state is populated is called T C, critical temperature. We see that N cont max(t C ) = N, ( ) ( ) V 2.62 = N k λ 3 B T C = 2 N 2/3. 2πm 2.62V At T = T C we speak of the BEC. Question What is the number of atoms in ε = 0 level? Answer We see that Therefore we can calculate that Nmax cont V = N ε>0 V = N V ( T TC ) 3/2. N ε=0 V = N ( ( ) T 3/2 ). V TC The relative number of particles is illustrated in figure Vibrations in solids (phonon gas). Einstein s model Atoms are connected to a spring of frequency ω. The energy of the spring is given by 69

70 Chapter 4. Ideal classical, bose and fermi gas Figure 4.5: The relative number of particles for N ε=0 (blue) and N ε>0 (red) ( ε = ω n + ). 2 The average energy for one such oscillator is ( ε = ω n + ), n = 2 exp(β ω), given by the Bose-Einstein statistics for quantized vibrations. Atomic vibrations are bosons with µ = 0. There are N atoms. There are then 3N independent oscillators. The internal energy of Einstein s solid is The heat capacity is c V = [ E = 3N ω exp(β ω) + ]. 2 ( ) E = β ( ) 2 E ω T T β = 3N exp(β ω) k V B T (exp(β ω) ) k B. 2 For large T with k B T ω we get that c V = 3Nk B = 6 2 k BN. 70

71 Chapter 4. Ideal classical, bose and fermi gas Therefore 6N times equipartition theorem value. So 6 quadratic degrees of freedom. This is the classical result. For low T with k B T ω we get that ( ) 2 ω c V = 3Nk B exp( β ω) T 0 0. k B T Figure 4.6: Einstein s model (red) in comparison to experiment to a.u. (blue) Remark If there are conduction electrons present in a solid at a very low T, we have c V T. For insulators we have c V T 3 for low T. 2. Debye model The atoms are connected by springs. The dynamics decomposes to that of 3N normal modes (phonons), plane waves. These are characterized by momentum k and polarization λ. There are 3 polarizations: 2 transversal and longitudinal. The spectrum is given by ω kλ = v λ k ω kt = v t k ω kl = v l k. Typically we have v l > v t with v m/s. The density of states is 7

72 Chapter 4. Ideal classical, bose and fermi gas g(ω) = = δ(ω ω kλ ) = k,λ V (2π) 3 = V 2π 2 λ λ ω 2 V (2π) 3 λ 4π dkk 2 δ 0 v λ ( v 3 λ = Vω2 2π 2 ( ω vλ d 3 kδ(ω vλ k) = ) k = ) + + v 3 t v 3 t v 3 l } {{ } = Vω2 3. 2π 2 v 3 s So we found that 3/v 3 s The total number of phonons is given by 3N = ωd 0 dωg(ω) = V 2π 2 3 v 3 s g(ω) ω 2. (4.8) ωd 0 ω 2 dω, ω D = ( 6π 2 N V v3 s) /3, with the Debye frequency ω D, which is the maximum allowed frequency. The internal energy is ωd ω 3 E = dω 0 exp(β ω) = β ω = x dω = dx β = = V ( ) 4 3 β ωd x 3 dx 2π 2 v 3 s β 0 exp(x) = = v3 s = V ( ) 3 3N 2π 2 ω3 D = 9Nk kb T ω D k B T x 3 BT dx ω D 0 exp(x) = = T D = ω ( ) D T 3 TD /T = 9Nk x 3 BT dx TD exp(x). k B We call T D the Debye temperature. For high temperatures T T D we get 0 TD /T 0 x 3 dx exp(x) TD /T 0 dx x3 x = 3 ( TD Therefore we calculate that E 3Nk B T and thus c V 3Nk B. T ) 3. 72

73 Chapter 4. Ideal classical, bose and fermi gas For low temperatures T T D we calculate TD /T Therefore we calculate that 0 x 3 dx exp(x) 0 x 3 dx exp(x) = π4 5. ( ) T 3 π 4 E(T T D ) 3Nk B T TD 5 T4, ( ) c V (T T D ) 2π4 T 3 5 Nk B T 3. TD This is in agreement with the experiment. 4.9 Equipartition theorem, virial theorem, thermodynamics of diatomic gases. Equipartition theorem Let q i be generalized coordinates and p i be generalized momenta. Thermal averages (classical physics only) Proof on the web of the lecture. 2. Virial theorem q i H q i = p i H p i = k B T. This follows directly from the equipartition theorem, q i = H q i 3N i= q i ṗ i = 3Nk B T. 3. Thermodynamics of ideal diatomic gases (H 2, HCl, N 2 ) at high temperatures This is an application of the equipartition theorem. We calculate 73

74 Chapter 4. Ideal classical, bose and fermi gas Therefore we see that H = 3N i= a i p 2 i + 3N i= b i q 2 i. So overall we have 3N i= p i H p i + q i H q i = 2 3N i= a i p 2 i + 2 3N i= b i q 2 i = 2H. 2H = 3Nk B T + 3Nk B T H = (3N + 3N) 2 k BT. For the heat capacity we derive c V = (3N + 3N) 2 k B. Per quadratic degree of freedom the heat capacity is 2 k B. a) Translation: 3, b) Rotation: 2, c v,trans N = 3 2 k B. c v,rot N = 2 2 k B. The energy of the rotation is given by ε = 2 2I l(l + ), H = 2I L 2, with the moment of inertia I and the orbital momentum quantum number l. Only two rotation freedoms, because in the 3rd one, I is very small. Therefore 2 /2I the disocciation energy. Effectivly two rotational degrees of freedom. c) Vibrations: 2 (momentum and position), c v,vib N = 2 2 k B. 74

75 Chapter 4. Ideal classical, bose and fermi gas The energy of the vibrations are given by At very very large T we have ( ε = ω n + ). 2 c V = ( ) N 2 k B = 7 2 k B. Typically vibrations are excited only at very large T, so usually we observe only translation and rotation, c V N = 5 2 k B. (4.9) Figure 4.7: The addition of degrees of freedom with increasing temperatures 4.0 Degenerate Fermi Gas Electrons in metals, white dwarfs, 3 He,... We know that ε F is the largest occupied level at zero temperature and is called Fermi energy. We thus can define T F, the Fermi temperature through 75

76 Chapter 4. Ideal classical, bose and fermi gas k B T F = ε F. (4.0) We can analyse three special limits.. For T = 0 we get f (ε) = Θ(µ ε). 2. For T T F with T > 0 we get f (ε) = exp(β(ε µ)) For T T F we see that µ which leads to f (ε) exp( β(ε µ)), β(ε µ). We can see the plots to these limits in figure 4.7. The case of T T F is the case of degenerate Fermi gas. Figure 4.8: T = 0 (blue) vs. T T F (red) vs. T T F (gold) with β = Example Electrons in metal have E F 0 ev. Therefore we calculate that T F = E F /k B = 0 5 K. 76

77 Chapter 4. Ideal classical, bose and fermi gas So electrons in metals are degenerate fermions. In this section wie deal with degenerate electron gases.. The density of states g(ε) We have ε = 2 k 2 /2m with a spin degeneracy of g = 2. We know that g(ε) = 2 V (2π) 3 d 3 kδ(ε ε k ) = 2 Vm3/2 2π2 3 ε ε. 2. The chemical potential First we calculate the number of particles with a factor 2 due to the spin degeneracy, N = 2 n k = 2 V (2π) 3 k d 3 k f (ε k ) = 0 dεg(ε) f (ε). This equation defines our µ! So we just have to calculate it - which gives us N = 2 Vm3/2 dε ε 2π2 3 exp(β(ε µ)) + = 2 Vm3/2 2π2 I(µ). 3 So we can determine the chemical potential using 0 I(µ) = 0 dε ε exp(β(ε µ)) +, and Sommerfeld s expansion (see appendix). We get I(µ) 2 3 µ3/2 + π2 6 (k BT) 2 2 µ +... By inserting this into the number of particles equation we get For the special case T = 0 we get [ N = 2 Vm3/2 2 2π2 3 3 µ3/2 + π2 2 (k BT) 2 ] (4.) µ N = 2 Vm3/2 2 2π2 3 3 ( }{{} µ(0) ) 3/2 = 2 Vm3/2 2 2π2 3 =ε F 3 ε3/2 F. (4.2) 77

78 Chapter 4. Ideal classical, bose and fermi gas By dividing equation 4. through equation 4.2 we obtain = ( µ εf ) 3/2 + π2 8 ( kb T ε F ) (µ/e F ) /2 We can solve this for µ = µ(t) by iteration. The lowest order os µ ε F. This solution is inserted into the second term on the right. So we get = ( µ εf ) 3/2 + π2 8 ( ) 2 kb T +... By using ( + x) n + nx for small x we get that ε F ( µ εf ) 3/2 = π2 8 ( ) 2 kb T. ε F µ ε F π2 8 ( ) 2 2/3 kb T ε F Two other important identites are ε F π2 2 ( ) 2 kb T < ε F. ε F 3. Internal energy E We see that E = 0 Γ(ε) = g(ε) = dγ dε = dεεg(ε) f (ε) 3 5 V 2(π) π ( 2mE 2 ) 3/2, V (2π) π ( 2m 2 ) 3/2 ε. ( µ εf ) 3/2 Nµ + 5π2 8 By substituting µ = µ(t) from the last subsection we get E = 3 5 Nε F At T = 0 we see that E T=0 = 3 5 Nε F. 4. Heat capacity We calculate + 5π2 2 ( ) 2 kb T ε F ( ) 2 kb T +... µ. 78

79 Chapter 4. Ideal classical, bose and fermi gas c V = Therefore we found that ( ) de = 3 dt 5 Nε 5π 2 2k 2T B F 2 V ε 2 F = π2 2 Nk B ( ) kb T. ε F This is important since phonons are T 3 for instance. st Remark The Fermi momentum k F c V T. (4.3) We can calculate the radius of the Fermi sphere in momentum space and obtain ( k F = 3π 2 N ) /3, ε F = 2 k 2 F V 2m, where k F is the highest momentum of an electron at T = 0. We found that Typically k F /Å. Therefore λ F Å. 2nd Remark Physics behind c V T ( N 2π 2 3 )/3 2 3 ε F = µ(t = 0) = V 2 m 3/2 2. Many energy levels. With T T F we have all levels occupied till ε F. Since k B T is the thermal energy per particle and the Pauli-exclusion principle does not allow lower electrons to go up with their energy because the upper states are occupied, only electrons of energies k B T below ε F can be thermally excited. So N excited electrons T. The energy of the excited electrons is k B T/electron. energy (compared to T = 0) is Thus the total change in E T T = T 2, because the number of electrons is also proportional to T. So overall we found that 79

80 Chapter 4. Ideal classical, bose and fermi gas c V = d E dt T. 80

81 5 Non-ideal gas 5. Virial expansion We consider a non-ideal, which is an interacting gas. Assume pairwise interaction, u ij ( r) = u( r i r j ), with the distance between i and j being r i r j. Figure 5.: Typical potential with a strong repulsion at the beginning and weak attraction in r 0 We consider weakly interacting systems, dilute gases (low density), so that u(typical distance) is much smaller than k B T. Note At room temperature k B T = 25 mev. The total potential energy is 8

82 Chapter 5. Non-ideal gas The total energy is then U = u ij. i<j H = p2 2m + U, p2 = p 2 + p p2 N. Task Calculate (as well as possible) Z. We see that dqdp Z = h 3N N! exp( βh) = = V N dp h 3N N! exp( βp2 /2m) } {{ } dqdp h 3N N! exp( βp2 /2m) exp( βu) = dq V exp( βu) = Z N id dq V N exp( βu), Z id Z id = N! ( V λ 3 ) N. We need to calculate the exponential function, exp( βu) = exp( β u ij ) = exp( βu ij ) = = i<j i<j ( + exp( βu ij ) ) = } {{ } ( + f ij ). i<j We call f ij the Mayer f -function. We expand f ij i<j exp( βu) = ( + f 2 )( + f 3 ) ( + f N,N ) = + f ij + f ij f kl +..., and continue evalulating i<j i<j k<l Z = Z id Z = Z (0) + Z () + Z (2)

83 Chapter 5. Non-ideal gas We see directly that Z (0) = Z id. For the first order term we obtain that Z () = Z id dq V N i<j N 2 = N = Z id V 2 2 = r = r r 2 = Zid N 2 = Z id V 2 4π N(N ) f ij = Z id V N 2 d 3 r d 3 r 2 f ( r r 2 ) = N 2 d 3 r V 2 d 3 r f (r) = 2 drr 2 f (r). If we neglect higher-order terms ( f 2, f 3,...) we get d 3 r d 3 r 2 d 3 r N f ( r r 2 ) = or Z Z (0) + Z () N = Z id + Z id V 2πN drr 2 f (r), Z Z id NnB 2, B 2 = 2π drr 2 f (r), with n = N/V and the second virial coefficiant B 2. We call Z = Z (0) + Z () +... the virial expansion.. Cluster expansion Remark We see that Z/Z id is NnB 2. We expected something to be to the power of N - not linear in N. This is a problem, because we want something to be F ln Z N. Let us go back: dq V N i<j f ij = N(N ) 2 V 2 d 3 r d 3 r 2 f 2. Denote this as symbol. The diagrammatic rules to calculate Z/Z id are Number the dots from to n, 2. For each dot i write 83

84 Chapter 5. Non-ideal gas V d 3 r i. Give a factor of N, N, N 2,... for dots, 2, 3,... corresponding. Finally divide by the symmetry factor, which is the number of permutations of the diagram leaving the product f ij f kl (or the diagram) unchanged. Let us look at O( f 2 ) terms, dq V N i<j k<l f ij f kl. There are 2 ways to make the diagrams with 2 links: connected or disconnected. a) connected Figure 5.2: The diagram for two connected links The symmetry factor is 2 because we can only change 3 in order to get the same diagram again. The integral is then N(N )(N 2) 2V 3 d 3 r d 3 r 2 d 3 r 3 f ( r r 2 ) f ( r 2 r 3 ). b) disconnected The symmetry factor is 8 because we can change 2, 3 4 and The integral is then N(N )(N 2)(N 3) 8 V 4 d 3 r d 3 r 2 d 3 r 3 d 3 r 4 f 2 f 34 2 ( ) 2. 84

85 Chapter 5. Non-ideal gas Figure 5.3: The diagram for two disconnected links If we continued further we would get Z Z id = The is called cluster expansion. We make an approximation ( f ) and take a subject of connected diagrams (ireducible diagrams). Then connect them and form higher-order terms, so that they can be cut. We take the simplest, S. So we get Z + + Z id 2 This is called Dyson equation. 2. The free energy and pressure ( ) 2 ( ) 3 ( ) = exp. (5.) 3! We know that S = NnB 2. First we calculate the free energy F = k B T ln Z + F id F id k B T ln(exp(s )) F id k B T = Z id = F id + k B TNnB 2, B 2 = 2π drr 2 f (r). Now we calculate the pressure, ( ) Fid P = k B TN ( N ) B V V T,N 2 2 = Nk BT V + Nk BTn B 2 V. Therefore we get a correction to the equation of state for gases, PV Nk B T = + nb 2. (5.2) 85

86 Chapter 5. Non-ideal gas 5.2 Van-der-Waals equation Let the potential be u(r) = We calculate the second virial coefficiant,, r < r 0. c, r > r r m 0 Then B 2 = 2π 0 r0 drr 2 f (r) = 2π 0 = 2π drr 2 (exp( ) ) 2π 0 = 2π 3 r3 0 + ( k }{{} B T 2π drr 2 c ) = r 0 r m } {{ } b = b a k B T. a drr 2 (exp( βu) ) = r 0 drr 2 ( βu ) = PV Nk B T So we obtain the Van-der-Waals equation, na = + nb k B T nb n a k B T, P Nk B T ( ) Nk B T N V nb V V a k B T. (P + an 2 )(V Nb) = Nk B T. (5.3) We call Nb the exclusion volume and say that an 2 is the pressure from the attrative long distance force. 86

87 6 Thermodynamics 6. Work and pressure We consider a box with a volume V which can be compressed by a pisten moving dx with the area A. So we get V V + dv, The work done by the pisten on the gas is dv = Adx. Remark The work by the gas is +PdV. dw = Fdx = PAdx = PdV. In going from thermodynamic state to 2 the work done on the system is 2 W = PdV. Assumption: Such (and similar) formula hold for quasistatic processes only. Quasistatic means very very slow. That is slower than the relaxation in the system. The system is (close to) in equilibrium. Important: dw is the infinitesimal work - this is not an exact differential. Sometimes dw is written asdw. Exact differential d f performs like 2 d f = f (2) f () d f = 0. 87

88 Chapter 6. Thermodynamics Imperfect differential dg performs like dg 0. The integration of 2 dg depends on the path from to 2. Example: Let f (x, y) = xy. So we see that d f = ydx + xdy is an exact differential. If we now just take dg = ydx we get this imperfect differential. (,) (0,0) dg = if the first way is x = y (straight line) and the second way is from the x-axis to the y-axis (y is zero for the first part). Functions f whose infinitesimal d f is an exact differential are called state functions. Popular examples are E, S, F. State functions depend on the state, not on the path. /2, Example: Ideal gas at constant T. If we change the volume from V V 2, 0, 2 V2 W = PdV = PdV = V PV = Nk B T, P = Nk BT V = Nk BT ln [ V2 ]. V 6.2 The first law of thermodynamics Energy conservation in form of de = dq + dw. In this case de is the increase of internal energy (E, E, U) of the system, dq is the amount of heat added to the system and dw is the amount of work done on the system. de = dq PdV is an exact differential. Therefore dq must be an imperfect differential (since PdV is an imperfect differential). The dimension of energy is Joule [J]. The dimension of thermal energy on the move (heat) is also Joule, but more often it is given in calories, 88

89 Chapter 6. Thermodynamics cal = 4.86 J. One calory is the amount of energy needed to raise the temperature of g of water from 4.5 C to 5.5 C. The food industry also introduced Cal = 486 J = 000 cal = kcal. Some terminology is needed for many processes. adiabatic process: dq = 0. isentropic system - adiabatic and quasistatic - ds = 0. isothermal system has dt = 0. isobaric system has dp = 0. isochoric system has dv = 0. Extensive quantities depend on the size of the system ( N) - like E, S, F, V, N,..., while intensive quantities donot depend on the size like T, P, µ, n = N/V, Enthalpy and heat capacities We know that For the heat capacity we calculated We also saw that c V = dq = de + PdV. c = Experimentally it is easier to measure ( ) dq. dt process ( ) dq = dt V ( ) de. dt V 89

90 Chapter 6. Thermodynamics ( ) dq c P =. dt P We get to this point by doing a simple Legendre-Transformation, dq = de + PdV = de + d(pv) VdP = d(e + PV) VdP = dh VdP. We call H the enthalpy, H = E + PV, c P = Example: Ideal gas with E = 3 2 Nk BT, PV = Nk B T gives us ( ) dh. (6.) dt P H = 5 2 Nk BT, c V = 3 2 Nk B, c P = 5 2 Nk B. So we do see that c V > c P. This relation is always given. 6.4 Second law of thermodynamics Ireversibility: There exists a state function, called entropy S, such that in an isolated system S 0. In a quasistatic process, ds = dq/t, or S(2) S() = 2 dq/t, with ds being an exact differential. Therefore dq/t is an exact differential, which is quite remarkable. Since S is a state function we see that S S(E, V, N) like in the microcanonical ensemble (in the thermodynamic limit all ensembles are the same). So we get ds = ( ) S de + E V ( ) S dv, V E with N being constant and not considering variations with N for simplicity. The first law then states that de = dq PdV ( ) ds S = dq dq = dq ( S/ E) V E T. V 90

91 Chapter 6. Thermodynamics The entropy is extensive: S A+B = S A + S B. The heat flows from a hotter to a cooler object. connected. T A > T B and Q A = Q B. So we calculate S A+B = S A + S B = Q A T A + Q B T B = Q B ( T B T A ) } {{ } Consider two systems A, B Therefore we see that Q B must be greater than zero - which means that the heat does indeed only flow from a hotter to a cooler object Removing internal constraints lead in general to entropy increase. This also causes some irritations because we can show that from the SEQ we can see that in the microscopic context ds/dt = 0. But according to the second law ds/dt 0. Therefore from microscopic to macroscopic something happens (the reason for this is still unknown - this is why it is called a law). Things are ireversible. 6.5 Third law of thermodynamics At the limit T 0 we must have c V (T) 0. Remark As T 0, S goes to zero or a finite number so that S/N 0. Classical ideal gas does not follow the third law of thermodynamics, because c V is a constant. This is fixed for Fermi and Bose gases. 6.6 Free energies st law: de = TdS PdV + µdn, with the new term µdn. There µ is the energy the system gains when a particle is added to it, keeping V and S constant. Remark Since for bosons µ < 0 (if ε k > 0), if bosons can be created, µ 0 otherwise the energy would be decreased to (by producing bosons). 9

92 Chapter 6. Thermodynamics 2nd law: S 0. uncertainty). Therefore in equilibrium S has its maximum (maximum. Helmholtz free energy F Consider a system in contact with a reservoir. We have V, N fixed, but the energy can be exchanged, E + E R = 0, because the whole system is isolated. In this notation E is for the system and E R for the reservoir. But the 2nd law states that S + S R 0. Since the reservoir is not affected we have E R = T S R (from the st law). This brings us to Then we have S R = T E R = T E. As T > 0 we find that S + S R = (T S E) 0. T We found the Helmholtz free energy F, E T S = F 0. F(T, V, N) = E TS. (6.2) From the 2nd law we see that for a system in contact with reservoir, fixed T, V, N F 0. Therefore in equilibrium F has a minimum. 2. G, H, Φ In the same way we can show that if the system is in contact with a reservoir for fixed T, P, N we get the Gibbs free energy G, 92

93 Chapter 6. Thermodynamics G(T, P, N) = E TS + PV. (6.3) We see directly that G reaches minimum, G 0. For S, P, N we get the Enthalpy H, H(S, P, N) = E + PV. (6.4) Again H reaches the minimum with H 0. Finally we get the Grand Potential Φ which reaches the minimum as well ( Φ 0) for T, V, µ, Legendre transformation Take f (x,..., x n ) with d f = i u i dx i, u i = ( f x i )x j. Now consider Φ(T, V, µ) = F µn. (6.5) This gives us n g = f u i x i, dg = i=r+ n n n u i dx i du i x i u i dx i. i= i=r+ i=r+ dg = r u i dx i i= n x i du i g g(x,..., x r, u r+,..., u n ). i=r+ We say that (u, x) are conjugate variables. Different free energies are connected by Legendre transformation. An example would be: S = S(E, V, N), T = ( ) S. E V,N Thus (T, E) are conjugate variables. We perform the Legendre transformation and obtain S T E = T (E TS) = F T. 93

94 Chapter 6. Thermodynamics 6.7 Maxwell s relations Take ( ) S for instance. First find out which free energy (potential) has T, V, N as V T,N natural variables? We see that this is the Helmholtz free energy F with the differential df = SdT PdV + µdn. Now we use the permutation of the second derivative and obtain ( ) ( ) ( ) S 2 F 2 F = = = V T V V T T,N ( ) P T This is a Maxwell s relation. Another example would be ( S P would be dg = SdT + VdP + µdn. We obtain ( ) ( ) S V =. P T T,N P,N V,N. ) T,N. Here the potential We can use this to investigate the relation between c V and c P. For this we fix N. We get c P = T By taking the derivate we obtain ( ) S, ds = T P ( ) S dt + T V ( ) ( ) ( ) ( ) S S S V = +. T T V T P V T P }{{}} {{ }} {{ } T c P By using Maxwell s relations we get T c V T c P = T c V We now define two important variables, ( ) P V T Maxwell [( ) ] 2 V. T P ( ) S dv. V T κ T ( ) V V P α ( ) V V T P T (compressibility), (expansibility). 94

95 Chapter 6. Thermodynamics We can now write c P c V = V T κ T α 2 > Gibbs-Duhem relation First we need to know something about homogeneous functions. Let f = f (x) be such that f (λx) = λ k f (x). Then f is called homogeneous of k-order. We now consider k =, thus f (λx) = λ f (x). By taking the derivative d/dλ we see that f (λx) d(λx) d(λx) dλ! = f (x) f (x) = x d f (x) dx. For n variables we have the famous Euler relation, ( ) f f (x,..., x n ) = x i. (6.6) x i x j Back to thermodynamics. This is useful for extensive functions, like E = E(S, V, N), e.g. E(2S, 2V, 2N) = 2E(S, V, N). We see that E = ( ) E S + S V,N i ( ) E V + V S,N ( ) E N. N S,V From the st law we already know that E = TS PV + µn. But since the differential of the from the Euler s equation obtained function is we see directly that de = dts + TdS dpv PdV + dµn + µdn! = TdS PdV + µdn, SdT VdP + Ndµ = 0. (6.7) This is the Gibbs-Duhem relation. Applications of this equation would be G = E T S + PV = TS PV + µn TS + PV = µn, Φ = E TS µn = PV. 95

96 Chapter 6. Thermodynamics So overall we can see that G = µn, Φ = PV. (6.8) 6.9 Conditions for thermodynamic equilibrium and stability From the 2nd law we know that when E, V, N are constant we have S 0. Therefore S has a maximum in equilibrium.. Principal of minimal E at S, V, N, constant Removing the constraint at fixed E leads to S(E) 0. Put the system in contact with reservoir, such that E can change, but S is kept constant. So S(E + E). A Taylor expansion gives us S(E) + ( ) S E = 0, E = T S(E) 0. E V,N So we found that E must be smaller than 0 when we have fixed S, V, N. Remark We already know that E = i ε i p i. This gave us δe = δε i p i + ε i δp i, i meaning that we have a work and an entropy part. Therefore the average E at fixed S, V, N has a minimum in equilibrium. This gives us a contrast: Fixed E, V, N: S has maximum. Fixed S, V, N: E has minimum. One analogy to this would be a circle with a perimeter. The circle is an object which has for a fixed area a minimum perimeter and for a fixed perimeter a maximum area (Area entropy, perimeter energy). 2. Conditions for equilibrium We consider S, V, N constant and i 96

97 Chapter 6. Thermodynamics S = S A + S B δs A = δs B, V = V A + V B δv A = δv B, N = N A + N B δn A = δn B. Removal of an internal constaint in equilibrium gives δe, change in energy: δe δe () = (T A T B )δs A (P A P B )δv A + (µ A µ B )δn A = 0, because E = E(S, V, N) has a minimum. In equilibrium we have T A = T B the thermal equilibrium, P A = P B the mechanical equilibrium and µ A = µ B the chemical equilibrium. We call T = const., P = const. and µ = const. the thermodynamic equilibrium. Example We consider two connected boxes with µ A > µ B with particles can be exchanged. We do want to know about the particle flow. We know that V and E are fixed and N A = N B. Since E is fixed (energy conservation) we see that So overall we have E = 0 = T S + µ N S = µ T N. S = S A + S B = µ A T N A µ B T N B = T (µ A µ B ) N A 0. So we know that N A must be lower than 0 which gives us the particle flow from A to B. The particles flow from higher to lower µ. 3. Conditions for stability E = E(S, V, N) should have a minimum in equilibrium. In equilibrium we have (δe) S,V,N = 0, (δ 2 E) S,V,N 0. Remark This could be a confusion, since the 2nd law of thermodynamics states E 0. But this is only in equilibrium (going into the equilibrium by removing 97

98 Chapter 6. Thermodynamics internal constraints). In this case we are looking at spontaneous changes δe 0, thus going away from equilibrium. Suppose a gas in equilibrium and a pushable wall between two boxes (spontaneous fluctuation). We say that δ 2 E > 0 is stable equilibrium. δ 2 E = 0 is a state which cannot decide. variations. We do have to look at higher δ 2 E < 0 is unstable equilibrium. Now consider δn A = δn B = 0 and δv A = δv B = 0. We also have δs A = δs B. We calculate δ 2 E = δ 2 E A + δ 2 E B = ( ) 2 E A δs 2 2 S 2 A + ( ) 2 E B δs 2 2 S 2 B = A V A,N A B V B,N B = ( ) ( ) 2 2 S 2 2 A + S 2 B E A V A,N A E δs2 A = [ TA + T ] B δs 2 A B V B,N B 2 c V,A c = V,B = T A = T B = ( 2 T A + ) δs 2 A c V,A c > 0. V,B This follow because the partition ration A to B is arbitrary. Therefore we found that c V > 0! We did already know this from chapter??. Consider the Helmholtz free energy F = F(T, V, N). We see that δf = 0 and δ 2 F > 0 with fixed T, V, N. By looking at T A = T B with N A, N B fixed and δv A = δv B we see that δ 2 F = ( ) PA 2 V A T A,N A + ( ) PB V B So we conclude that P/ V < 0 for fixed T, N. c P > c V > 0. T B,N B δv2 A > 0. We see also that κ > 0 and Remark We can partition extensive variables only! For example we cannot say that T A + T B is fixed. 98

99 7 Phase transitions and critical phenomena 7. Phase equilibria From the Gibbs-Duhem relation SdT VdP + Ndµ = 0 we get that dµ = S N dt + V dp = sdt + vdp, s = S/N, v = V/N. N Therefore we conclude that µ is a function of the two variables T, P only, µ µ(t, P). Suppose now that we have ν different phases. A phase can be a gas, liquid or solid etc. Question Can different phases coexist (be together) in equilibrium? Answer T, P are equal for all phases (µ must be equal) so that µ (T, P) = µ 2 (T, P) =... = µ ν (T, P). So we have ν equations. Define f, the number of degrees of freedom, which is the number of unknown variables minus the number of equations, so f = 2 (ν ) = 3 ν. (7.) This is the so called Gibbs phase rule. Discussion For ν = 3 we get f = 0 which is a unique solution (T 0, P 0 ) (triple point or TP). Three phases can coexist in a point in the (P, T) plane. 99

100 Chapter 7. Phase transitions and critical phenomena For ν = 2 we get f = which gives us P P(T) - that specifies a line. Two phases can coexist along a coexistence line (curve). Remark The strategy to solve µ (T, P) = µ 2 (T, P) is to fix T first and then to get P(T). For ν = we get f = 2 and thus the whole (P, T) plane. One phase exists in a plane. For ν 4 we find that four or more different phases cannot coexist together in a one component system. If the system has r components (types of atoms / molecules) then we have f = 2 + r ν. (7.2) Therefore in a two component system for example we can have 4 phases in a point. A generic phase diagram is shown in figure 7.. Figure 7.: A valid phase diagram with three phases, two triple points and a critical point We can build an invalid phase diagram by having another phase in it, with a point where all phases are combined but having only a one component system. Therefore we do have to look for the intersection of such coexistence lines. 00

101 Chapter 7. Phase transitions and critical phenomena At the critical point is a phase transition of the 2nd type (continuous). Above the critical point (CP) we cannot distinguish between phases. So by going through a line of coexistence we are able to see the change (e.g. in entropy) but if we go from one phase to the other above the critical point we see not difference between the phases. So for example we would say that water and vapor are the same! 7.2 Abrupt phase transitions - Clausius-Clapeyron equation We set G = µn and consider two phases with G and G 2. We say that if G < G 2 then phase wins, if G > G 2 then phase 2 wins and if G = G 2 then we have µ = µ 2 and therefore coexistence. We can draw phase diagram like the graphs in figure 7.2. Figure 7.2: Possible graphs of heat capacity (C) against temperature (T) at a phase transition We derive that V = ( ) G = N P T,N In the same way we obtain that S = ( ) G = N T P,N ( ) µ P T ( ) µ T, v =, s = P ( ) µ P T. ( ) µ T. P 0

102 Chapter 7. Phase transitions and critical phenomena At the coexistence curve we have a densities change v v 2 and latent heat s s 2. This is called abrupt transition. At the critical point we have v = v 2 and s = s 2. This is why this is called continuous transition (higher derivatives of G are not continuous). At the coexistence line we have µ (T, P) = µ 2 (T, P). Along the coexistence line we find dµ = dµ 2. By using the Gibbs-Duhem relation dµ = sdt + vdp we obtain that dµ = s dt + v dp! = s 2 dt + v 2 dp = dµ 2. This gives us dp(v v 2 ) = dt(s s 2 ) which leads to the Clausius-Clapeyron equation, dp dt = s s 2 = l v v 2 T v. (7.3) We introduced the latent heat of the transition l, with s s 2 = s = l/t, by using the temperature of the transition T. The solution of this equation is P P(T), the coexistence curve. Typical phase diagrams are shown in figure 7.3. Figure 7.3: The phase diagrams for a generic (st picture) and water (2nd picture) substance 02

103 Chapter 7. Phase transitions and critical phenomena Figure 7.4: A P V diagram for the phase transition of a generic (not an anomaly) substance 7.3 Nucleation Vapor to liquid transition at temperature T tr. Condensation into water droplets. Droplets growth is irupeded by surface tension σ. It costs energy 4πR 2 σ to grow a droplet of radius R. At T = T tr we have G gas = G liquid. At T = T tr + T we can make a Taylor expansion and get G gas G liquid T (G gas G liquid ) T = (S gas S liquid ) Ttr T = N s T. T=Ttr We can rewrite this and obtain G liquid = G gas + Vn l T tr T. If T > 0 we get that G liquid > G gas and therefore we are in the gas phase. If T < 0 we get that G liquid < G gas and therefore we are in the liquid phase. Consider the energy required to make one droplet of radius R. Cost of energy compared to having just gas is Therefore we obtain that E Droplet = G liquid + 4πR 2 σ G gas. 03

104 Chapter 7. Phase transitions and critical phenomena E Droplet = 4 3 πr3 }{{} =V Droplet l n liquid T tr We get a plot like the one shown in figure 7.5. T +4πR }{{} 2 σ = αr 3 + βr 2. <0 Figure 7.5: The cost of energy E to make one droplet of radius R The cost of energy to go from gas to water is given by a barrier called B. We see a maximum at R C which gives us this barrier B. We can calculate that For the barrier B we obtain that E (R = R C ) = 0 R C = 2σT tr n liquid l T σ/ T. B = E(R = R C ) = 6πσ3 T 2 tr 3n 2 liquid The droplet creation (nucleation) rate is given by l 2 T 2 σ3 / T 2. exp( B/k B T tr ). An application of this would be how to boil water in a microwave. By using a clean glass (washed by a dishwasher) we have a large σ and thus a large R C. Therefore we obtain large bubbles which could damage the glass. If we use a cloth to wipe out the glass we have a reduced σ and thus the boiling is less violent. 04

105 Chapter 7. Phase transitions and critical phenomena 7.4 Continuous phase transitions - Landau s theory st order phase transitions (abrupt) are claled nucleation (last section). 2nd order phase transitions (continuous) perform spontaneous symmetry breaking.. mechanical analog (rod-spring model) The question is in which direction does the rod fall? The Hamiltonian ist symmetric with respect to the possible directions. But the physical realization (ground state) is not! Thus we say that the symmetry is broken. We now support the rod with springs. We have to find two potential energies: The potential energy of the two springs. The potential energy of the rod. We find that u spring = 2 2 k(displacement/streching)2 = kr 2 ϕ 2, u rod = mg l 2 cos ϕ G l ( 2 2 ϕ2 + ) 24 ϕ4. The total energy is then given by u = u spring + u rod = 2 Gl + 2 We have to dinstinguish now two cases: (2kR 2 2 Gl ) ϕ 2 + Gl 48 ϕ4. 2kR 2 > Gl. We see in figure 7.6 (Graph ) that we have a minimum at ϕ = 0. 2 Therefore we are in equilibrium and are always stable! 2kR 2 < Gl. We see in figure 7.6 (Graph 2) that we have two minima - 2 therefore we have two equal choices. This is called spontaneous symmetry breaking. The minima occur at ±ϕ 0 with ϕ 0 = 2 ( 2 2kR ) = Gl 6 G G c G (G G c ) /2, G c 4kR 2 /l. 05

106 Chapter 7. Phase transitions and critical phenomena Figure 7.6: The first graph is always stable in ϕ = 0 while in the second graph the symmetry is broken for the second case with two minima - symmetric in ϕ Figure 7.7: We see no jump in ϕ G G c - therefore continuous We say that /2 is the critical exponent. We see in figure 7.7 that ϕ G G c for G > G c. we call ϕ the order-parameter with two special variants: ϕ = 0: disordered phase ϕ 0: ordered phase Going through G c is a phase transition. The order parameter ϕ changes continuously at G = G c. Thus we have a continuous phase transition. The three most important graphs are shown in figure 7.8. The system is symmetric when we have u(ϕ) = u( ϕ). 06

107 Chapter 7. Phase transitions and critical phenomena Figure 7.8: The graphs for G < G c, G = G c and G > G c (phase transition) 2. Landau theory of paramagnet-ferromagnet transitions Experimental we get a graph like the one shown in figure 7.9. comes the concept of order parameter - here: magnetization M. From Landau Figure 7.9: Experimental picture with the temperature T in units of the critical temperature T C The Landau free energy is defined as F F(T, V, M) = F 0 (T, V) + 2 am2 + 4 bm4, a a(t), b b(t) > 0. This is valid close to T C, where M is small. We set b(t) > 0 due to stability. If it would be negative we would have to look at higher orders. Question Why there are no odd powers of M? Answer F(M) = F( M) - so symmetry is the reason! 07

108 Chapter 7. Phase transitions and critical phenomena In equilibrium we see that F will have a minimum, so For a > 0 we only have M = 0. For a < 0 we have F M = 0 am + bm3 = 0. M = ± The big idea from Landau was that a a(t) and that the most reasonable a b. (sometimes called biblical) ansatz would be a (T T C ). At T < T C we have M = ± α b (T T C) (T C T) /2, with the critical exponent β being /2. So we say that Magnetization at T < T C, but close to T C, goes as (T C T) /2. 3. Critical exponents Close to T C we have: M (T C T) β, T < T C, M B /δ, T = T C, χ (T T C ) γ, T > T C, χ (T C T) γ, T < T C, ( ) S c B = T (T T C ) α, T > T C, T B c B (T C T) α, T < T C. So we have critical exponents α, β, γ, δ,...! 08

109 Chapter 7. Phase transitions and critical phenomena 4. Critical exponents in Landau-theory In a magnetic field B we have From our minimum condition we get F = F am2 + 4 bm4 BM. F M = am + bm3 B = 0 M(a + bm 2 ) = B. Therefore at T = T C with a = 0 we get So we found out that δ = 3. bm 3 = B M B /3. We can do the same for χ. We know that ( ) M χ = µ 0, am + bm 3 = B a M M + 3bM2 B B B =. So we see that at T > T C with M = 0 we have χ = µ 0 a T T C = (T T C ), which is γ =. At T < T C we haev M 2 = a/b and thus χ = µ 0 a 3a = µ 0/2a (T C T). Therefore γ = as well. This gives us a graph like the one in figure 7.0. We set B = 0 and get F. At T > T C we see that S = S 0. At T < T 0 we obtain that S = S 0 + α 2 2 b (T T C). So F and S are continuous at T = T C. We now calculate c B which is (close to T C ) ( ) S c B = T C. T B 09

110 Chapter 7. Phase transitions and critical phenomena Figure 7.0: χ diverges at T C as we calculated with the temperature in units of T C We obtain that at T > T C we have c B0 and at T < T C we have c B0 + α 2 T 2 b C. So we say that the critical exponent is 0 in Landau s theory. We also see from picture 7. that c B has a jump at T C. So this is discontinuous / abrupt! Figure 7.: c B has a jump at T = T C with c B0 being a function of T in units of the critical temperature The gap is 2 T Cα 2 /b. 7.5 The Ising model: One dimensional Let the Hamilton operator be 0

111 Chapter 7. Phase transitions and critical phenomena H = J s i s j, with the nearest neighbors i, j, s i = ± and J the exchange intergral (exchange coupling). We know that J > 0 for ferromagnetic ground state ( ) and J < 0 for antiferromagnetic ground state ( ). Note In quantum mechanics we learned that Coulomb interaction and Pauli principle lead to the Heisenberg Hamiltonian i,j H = J s s 2 for two electrons. The Ising model is a simplified classical version. Let us solve the one dimensional Ising model. We need to calculate the partition sum, Z Z(β) = exp( βh(s)) = exp(βjs s 2 + βjs 2 s βjs N s N ). {s} s =± s2=± s N =± Take first the last sum exp(βjs N s N ) = exp(βs N ) + exp( βs N ). s N The good thing is that for s N = ± we have exp(βjs N s N ) = 2 cosh(βj), which is independent of s N! We get s N s N = : exp(βj) + exp( βj) = 2 cosh(βj), s N = : exp( βj) + exp(βj) = 2 cosh(βj).

112 Chapter 7. Phase transitions and critical phenomena This leads to Z = exp(βj(s s s N 2 s N ))2 cosh(βj) = s s 2 s N = exp(βj(s s s N 3 s N 2 ))(2 cosh(βj)) 2 = s s N 2 = ( 2 cosh(βj) ) N = 2 N cosh N (βj) N = ( 2 cosh(βj) )N, s =± }{{} =2 which is exact in the thermodynamic limit! We will now calculate the thermodynamic quantities. The free energy is The internal energy is F = k B T ln Z = k B TN ln(2 cosh(βj)). E = d ln Z = NJ tanh(βj). dβ For the limit T 0 with β we obtain that E NJ. Otherwise for T with β 0 we see that NJ 2 β 0. Since F is analytic (no divergences in its derivation) at all T > 0, there are no phase transitions in the one dimensional Ising model! In one dimensional Ising model the phase is a paramagnet at all T > 0! In T = 0 it is a ferromagnet - but since T = 0 cannot be reached it is obsolete. 7.6 One dimensional Ising model: renormalization group (RG) K. Wilson calculated this in the 970 s starting with Z = exp βj s i s j = exp K s i s j, {s} i,j {s} i,j 2

113 Chapter 7. Phase transitions and critical phenomena with K = βj = J/k B T. We consider a one dimensional Ising model with periodic boundary conditions, meaning s N+ s. We calculate N Z = exp K s i s i+ = exp(k(s s 2 + s 2 s )) = {s} i= s s 2 s N = exp(ks 2 (s + s 3 ) + Ks 4 (s 3 + s 5 ) +...). s s N Let us sum over all the even spins only: [ Z = e K(s +s 3 ) + e ] [ K(s +s 3 ) e K(s 3+s 5 ) + e ] K(s 3+s 5 ) [...]. s 3 s We now want to rewrite this new (called decimated) system to look like the old one. We see that the effective distance has increased from a 2a and the coupling constant has changed from K K. This is called resclaing or coarse graining. Mathematically this is exp(k(s + s 3 )) + exp( K(s + s 3 ))? = A(K) exp(k s s 3 ). We know that when s = s 3 = ± we have exp(2k) + exp( 2K)! = A(K) exp(k ). We also know that for s s 3 = ± we obtain that +! = A(K) exp(k ). So we have two equations with two unknown variables. This can be solved. We get A(K) = 2 exp(k ), 2 cosh(2k) = A(K) exp(k ) = 2 exp(2k ). Therefore afterall we have K = 2 ln cosh(2k), A(K) = 2 cosh(2k). (7.4) 3

114 Chapter 7. Phase transitions and critical phenomena Then we take Let us write Z = [A(K)] N/2 exp(k (s s 3 + s 3 s )). odd spins Z Z(K, N) = [A(K)] N/2 Z(K, N/2). We introduce now f = (ln Z)/N = F/k B TN with our free potential F = k B T ln Z. We calculate f = f (K) = N ln Z(K, N) = N ln ( [A(K)] N/2 Z(K, N/2) ) = So we see that = N N 2 ln A(K) + N ln Z(K, N/2) = 2 ln A(K) + 2 N 2 ln Z(K, N/2) = } {{ } = [ ln A(K) + f (K ) ]. 2 = 2 f (K ) f (K ) = 2 f (K) ln A(K) = 2 f (K) ln [ 2 cosh(2k) ]. We found the renormalization group equations, which give us the renormalization group flow, K = 2 ln [cosh(2k)], f (K ) = 2 f (K) ln [ 2 cosh(2k) ]. (7.5) Let s take for example K =. Then K is. By calculating K we receive 0.66 then 0.35 then 0. and so on... Therefore we see that At K = 0 which is T = we have a stable renormalization group fixed point, because K = 0 gives us K = 0. At K = which is T = 0 we have an unstable renormalization group fixed point. 4

115 Chapter 7. Phase transitions and critical phenomena So the flow is from K = to K = 0. We say that K = is ordered while everything else is disordered. We also see a phase transition at T = 0. A fixed point is where all length scales are equivalent. So fractal, universality, power laws. Therefore we see that a critical exponent is the same for all ferromagnets. Universality means that α, β, γ, δ,... depend on the dimensionality and geometry. If d 4 all systems behave as in Landau s theory. The reason for this is that Landau neglected fluctuations, which are very important for systems with d < Ising model in two dimensions: Renormalization Group By rotating a square lattice by π/4 (decimation), and therefore summing over every other spin. We are doing the in figure 7.2 shown process on the lattice. Figure 7.2: The rough sketch of the setup used for our calculation Let us look at the structure of the partition function, Z = = exp(k(s s 5 + s 2 s )) = exp(ks 5 (s + s )) exp(ks 6 (s 2 + s )) = {s} {s} [exp(k(s + s 2 + s 3 + s 4 )) + exp( K(s + s 2 + s 3 + s 4 )) ] [exp(k(s 2 + s 3 + s 7 + s 8 )) + exp( K(s 2 + s 3 + s 7 + s 8 )) ]. We would like to find K and A(K) such that 5