Part II: Statistical Physics

Transcription

1 Chapter 7: Quantum Statistics SDSMT, Physics 2013 Fall

2 1 Introduction 2 The Gibbs Factor Gibbs Factor Several examples 3 Quantum Statistics From high T to low T From Particle States to Occupation Numbers Bosons and Fermions The Distribution Functions for Bosons and Fermions

3 Systems with a Variable Number of Particles From canonical ensemble to grand canonical ensemble Reservoir: Energy = U R, Number of par8cles: N R, Temperature: T, Chemical poten8al: μ Energy exchange Energy exchange Canonical ensemble E Grand canonical ensemble E, N Par8cle exchange For a system (S) that exchanges particles and energy with a large reservoir (R), both µ and T are dictated by the reservoir (as the reservoirs properties). The equilibrium means the chemical potentials and temperature of a system and its environment become equal to one another.

4 The fundamental assumption of thermodynamics: The fundamental assumption of thermodynamics: (1) In equilibrium, there is no net mass/energy transfer, though the number of particles/energy in a system can fluctuate around their mean values. (2) All the states of the combined (isolated) system are equally probable. By specifying the microstates of the system S, we have reduced Ω S to 1 and S S to 0. Thus, the probability of occurrence of a situation where the system is in state i is proportional to the number of states accessible to the reservoir R. For the reservoir R + open system S we are interested: E = E r + E s = constant, N = N r + N s = constant (1) E(s) E 1, N(s) N 1 (2)

5 The Gibbs Factor Following what we did for a system in canonical ensemble in Chapter 6, we can write down the ratio of probabilities for S to be on two microstates s 1 and s 2 as: P(s 2) P(s = Ω R(s 2) 1) Ω R (s = esr (s2)/k 1) e S R (s 1 )/k = e[s R (s 2 ) S R (s 1 )]/k (3) Since the corresponding change in the reservoir is tiny, so we can describe the change with thermodynamic identities: ds R = 1 T (du R + PdV R µdn R ) Eq. (5.16) (4) ds R = 1 (dus + PdVs µdns), dvs 0 (5) T S R (s 2) S R (s 1) = 1 (E(s2) E(s1) µn(s2) + µn(s1)) T (6) Therefore, (7) P(s 2) P(s 1) = e [E(s2) µn(s2)]/kt e [E(s 1) µn(s 1 )]/kt (8)

6 The Gibbs Factor - cnt. e [E(s) µn(s)]/kt is called a Gibbs Factor When different types of particles are involved, the Gibbs factor is given by e [E(s) i µ i N i (s)]/kt (9) Just like what we did in Chapter 6, the absolute probability of a state is P(s) = 1 Z e [E(s) µn(s)]/kt (10) The grand partition function, or the Gibbs Sum is, (11) Z = s e [E(s) µn(s)]/kt (12)

7 Statistical Ensembles: Summary To summarize: Ensemble Macrostate Probability Thermodynamics micro-canonical U, V, N P n S(U, V, N) = klnω (T fluctuates) = 1 Ω canonical T, V, N P n F (T, V, N) = ktlnz (U fluctuates) = 1 Z e En/kT grand canonical T, V, µ P n Φ(T, V, µ) = ktlnz (U, N fluctuate) = 1 Z e (En µnn)/kt Systems may have different chemical potentials µ. Particles move from system with bigger chemical potential to system with smaller chemical potential. This diffusion cause the chemical potential change in two systems. For Ideal Gas, we obtained: µ = ktln ( VZint Nv Q ), Eq. (6.93).

8 Examples In-class Exercise 07-01: Problem 6.48 Entropy and chemical potential of a diatomic gas at room temperature (a) Show [ that ( the ) entropy ] of a diatomic gas at room temperature is VZ S = Nk ln e Z rot Nv Q + 7, where Z 2 e is the degeneracy of the electronic ground state of the molecule. Use this formula to calculate the entropy of 1 mole of O 2 at room temperature and atmospheric pressure. Z e = 3 for O 2. (b) Calculate the chemical potential of O 2 at room temperature and 1 atmospheric pressure. (c) Calculate the chemical potential of sea level O 2 in the atmosphere at room temperature and atmospheric pressure. Sea level: the partial pressure is about P Oxygen 0.2 atm.

9 Examples - cnt. Carbon Monoxide Poisoning Illustrate the use of Gibbs factors for adsorption of oxygen by hemoglobin molecules (C 34H 32FeN 4O 4). A single heme site can be Occupied by O 2 Unoccupied Occupied by CO E = ev E=0 ev E= ev

10 Examples - cnt. Carbon Monoxide Poisoning A single heme site can have three different states: open, occupied by O 2, occupied by CO. The energy of these three states are 0 ev, 0.7 ev, 0.85 ev. When we consider Oxygen only, the grand partition function for the single-site system is: Z = 1 + e (ɛ µ)/kt In lung, the blood is proximately in diffusive equilibrium with the atmosphere. Approximate the air by ideal gas, we can calculate the chemical potential: µ = ktln ( VZint Nv Q ) 0.6 ev (See previous in-class exercise.) e (ɛ µ)/kt e (0.1 ev )/kt = e (0.1 ev )/( ev /K (37+273)) = 42 P(occupied by O 2) = 42 = 97.7% 1+42 When there is some carbon monoxide present, there will be three microstates. The grand partition function becomes: Z = 1 + e (ɛ µ)/kt + e (ɛ µ )/kt

11 Examples - cnt. Carbon Monoxide Poisoning - cnt. Assume the abundance of carbon monoxide is α times less than oxygen, the chemical potential of carbon monoxide (also use ideal gas approximation) would be: ( ) ( ) µ CO = ktln VZint N CO v Q ktln VZint N O2 v Q /α : because, for example, Z rot,o2 kt, Z 2ɛ rot,co ( ) kt ɛ µ CO = ktln VZint N O2 v Q ktlnα = µ O2 ktlnα Some numbers: α = 10, ktlnα = , µ CO = µ = = ev ; α = 100, ktlnα = 0.118, µ CO = µ = = ev ; α = 1000, ktlnα = 0.177, µ CO = µ = = ev ; When α = 100, we have e (ɛ µ )/kt = e ( 0.85 ev ev )/kt = e (0.13 ev )/kt = 130 The probability of the heme site to be occupied by O 2 is P(occupied by O 2) = 42 = 24% only!

12 Example - cnt. In-class Exercise When a system is in thermal and diffusive equilibrium with a reservoir, show that (a) The average number of particles in the system is N = kt Z partial derivative is taken at fixed temperature and volume. (b) The mean square number of the particles is < N 2 >= (c) Show the standard deviation of N is σ N = kt ( N/ µ). (d) Calculate σ N in terms of N for ideal gas. Z, where the µ (kt )2 Z 2 Z µ 2

13 High T vs Low T : The difference at the microscopic level In dense systems, two or more identical particles may have significant probability to occupy the same single-particle state. Two par3cles into 5 single par3cle states 2 par6cles S 1, E(s 1 ) S 2, E(s 2 ) S 3, E(s 3 ) S 4, E(s 4 ) S 5, E(s 5 ) 5 single par6cle states Each digit for each energy level/single par6cle state: 1: occupised; 0: empty All allowed system states For dis3nguishable par3cles: 2*(5+5) + 5 = 25 system states For iden3cal par3cles: (5+5) + 5 = 15 system states 25/2!

14 Occupation Numbers Let s introduce a new way of characterizing the quantum system consisting of identical particles. Single- par+cle states ε 4 1 N (1) Systems with a fixed number Ztotal = Z1 ε 3 N! of particles in contact with the ε U reservoir, occupancy n i = ln Z 1 β ε 1 U = nu 1 (2) Systems which can exchange both energy and particles with a reservoir, arbitrary occupancy n i. ε 4 ε 3 ε 2 ε 1 Occupa+on number n( E 4 ) N = n i i E = n E i i i

15 System and reservoir in two cases (Top) Systems with a fixed number of particles in contact with the reservoir The energy in the system fluctuates; Total number of particles is fixed; The role of the thermal reservoir is to fix the mean energy of each particle (i.e., each system). The identical systems in contact with the reservoir constitute the canonical ensemble. This approach works well for the high-temperature (classical) case, which corresponds to the occupation numbers << 1.

16 System and reservoir in two cases (Bottom) Systems which can exchange both energy and particles with a reservoir, arbitrary occupancy n i When the occupation numbers are 1, it is more convenient to choose a single quantum level as the system (instead of particles), with all particles that might occupy this state. Each energy level is considered as a sub-system in equilibrium with the reservoir; Each energy level is populated from a reservoir independently of the other levels.

17 In terms of Occupation Numbers Consider a system of identical non-interacting particles at the temperature T, ɛ i is the energy of a single-particle state i, n i is the occupation number (the occupancy) for this state. We can do the transformation from (s) to {n i } = {n 1, n 2,...}: E(s) = n 1ɛ 1 + n 2ɛ = i Z = ( exp n iɛ i µn ) i kt i,{n i } n i ɛ i (13) (14) The sum is taken over (1) all possible occupancies {n i } AND (2) all states for each occupancy. The Gibbs sum depends on the single-particle spectrum ɛ i, the chemical potential µ, the temperature T, and the occupancy {n i }. {n i } depends on the nature of particles that compose a system (fermions or bosons).

18 Bosons and Fermions One of the fundamental results of quantum mechanics is that all particles can be classified into two groups: Bosons and Fermions: Bosons: particles with zero or integer spin (in units of?). Examples: photons, all nuclei with even mass numbers. The wave function of a system of bosons is symmetric under the exchange of any pair of particles: Φ(..., q j,...q i,..) = Φ(..., q i,...q j,..). The number of bosons in a given state is unlimited. Fermions: particles with half-integer spin (e.g., electrons, all nuclei with odd mass numbers); the wave function of a system of fermions is anti-symmetric under the exchange of any pair of particles: Φ(..., q j,...q i,..) = Φ(..., q i,...q j,..). The number of fermions in a given state is zero or one (the Pauli exclusion principle).

19 Bosons and Fermions - cnt. A few more words: (1) The Bose or Fermi character of composite objects: the composite objects that have even number of fermions are bosons; those containing an odd number of fermions are themselves fermions. For example: 3 He atom: 3 He = 2 electrons + 2 protons + 1 neutron, hence 3 He atom is a fermion. (2) In general, if a neutral atom contains an odd number of neutrons then it is a fermion, and if it contains an even number of neutrons then it is a boson. (3) In order to treat systems with quantum particles, we need the explicit formulae for µ and n i for bosons and fermions.

20 Distribution Functions for Bosons Lets consider a system that consists of just one single state of energy ɛ i. The total energy of this state: n i ɛ i. The probability of this state to be occupied by n i particles is: P(ɛ i, n i ) = 1 exp ( Z(ɛ i n i ɛ i µn i ) ) kt, The grand partition function for all particles in the i th single-particle state (the sum is taken over all possible values of n i ): Z(ɛ i ) = n i exp ( n i ɛ i µn i ) kt, If we put all energy levels together, assuming single-particle states are independent from each other, the probability is P(n i ) = ɛ i P(ɛ i, n i ) The corresponding full partition function will be Z FPF = ɛ i Z(ɛ i )

21 Distribution Functions for Bosons - cnt. Let s look at one single-particle state of a system which will have energy ɛ when it is occupied by one particle. For Bosons, the number of particles that can be on this state can be 0, 1, 2, 3,... Therefore, Z = n i = n i =0 e n i (ɛ i µ)/kt (15) = 1 + e (ɛ µ)/kt + e 2(ɛ µ)/kt +... (16) = 1 + [e (ɛ µ)/kt ] 1 + [e (ɛ µ)/kt ] (17) Z = 1 1 e (ɛ µ)/kt (18)

22 Distribution Functions for Bosons - cnt. The average number of particles in the state can be calculated: n = n np(n) = 0 P(0) + 1 P(1) + 2 P(2) + (19) make the variable change: x = (ɛ µ)/kt n = n e nx Z = 1 Z x e nx = 1 Z Z x n n 1 n = n BE = e (ɛ µ)/kt 1 (20) (21) This is called the Bose-Einstein Distribution. n BE approches when ɛ approaches µ + δ

23 Distribution Functions for Fermions For Fermions, the number of particles that can be on a single-particle state can only be 0, 1. Therefore, Z = 1 + e (ɛ i µ)/kt (22) (23) The average number of particles in the state can be calculated: n = 0 P(0) + 1 P(1) = n FD = 1 e (ɛ µ)/kt + 1 This is called the Fermi-Dirac distribution. e (ɛ i µ)/kt 1 + e (ɛ i µ)/kt (24) (25)

24 Distribution Functions for Fermions - cnt. The distribution for Fermions looks like this: 1 Low T High T ~ k B T 0 T =0 (with respect to µ) ε = µ

25 Boltzmann distribution P(s) = 1 Z 1 e ɛ/kt (26) n = NP(s) = N Z 1 e ɛ/kt (27) Since (HW-6.44, p.251) µ = ktln(z 1/N) e µ/kt = Z1 N n = e µ/kt e ɛ/kt = e (ɛ µ)/kt. (28)

26 To summarize: Three distributions Comparison among three Distributions Boltzmann Bose-Einstein Fermi-Dirac 1 nk = ε µ exp kbt indistinguishable Z=(Z 1 ) N /N! (n K <<1) spin doesn t matter localized particles Ψ don t overlap gas molecules at low densities unlimited number of particles per state (n K <<1) 1 nk = ε µ exp 1 kbt indistinguishable integer spin 0,1,2 bosons Wave functions overlap total Ψ symmetric photons 4 He atoms unlimited number of particles per state 1 nk = ε µ exp + 1 kbt indistinguishable half-integer spin 1/2,3/2,5/2 fermions Wave functions overlap total Ψ anti-symmetric free electrons in metals electrons in white dwarfs never more than 1 particle per state

27 To summarize: Three distributions Three distributions as function as energy 2 1 FD MB BE ε = µ 0