Subject PHYSICS Paper VI Paper Code PHY522 Topic Statistical Physics
Dr. Mahesh Chandra Mishra
Associate professor of physics
Millat ollege, Darbhanga
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M.Sc Physics:Second Semester STATISTICAL PHYSICS Paper: VI Paper Code: PHY 522 UNIT-I Objective of statistical macrostates, microstates, phase space and ensembles. Ergodic hypothesis Postulates of equal a priori probability and equality of ensemble average and time average Boltzman’s postulates of entropy Counting the number of microstates in phase space Entropy of ideal gas SackurTetrode equation and Gibb’s paradox. Liouville’s theorem. UNIT-II System in contact with a heat reservoir Expression of entropy Canonical partition function Helmholtz free energy Fluctuation of internal energy. System in contact with a particle reservoir Chemical potential Grand canonical partition function and grand potential fluctuation of particle number Chemical potential of ideal gas. UNIT-III Mean field theory and Vanderwaal’s equation of state Density matrix Quantum Liouville theorem
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Density matrices for microcanonical canonical and grand canonical systems Simple examples of density matrices- one electron in a magnetic field,, particle in a box; Identical particles- B-E and F-D distributions
UNIT-IV Equation of state Bose condensation Equation of state of ideal Fermi gas Fermi gas at finite T Ising model Partition function for one dimensional case Chemical equilibrium and Saha ionisation formula.
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UNIT - I
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Statistical Mechanics Statistical Physics: It is the study of macroscopic parameters of a system in equilibrium with the help of microscopic properties of its constituent particles using the laws of mechanics. It is different from thermodynamical approach as in thermodynamics, macroscopic system (bulk matter) in equilibrium is studied with the help of macroscopic properties. The macroscopic behavior is related to bulk properties i.e. large scale whereas microscopic behavior is related to individual particles Classification of Statistical Physics The study of statistical physics is mainly classified into two categories (I) Classical Statistics or Maxwell-Boltzmann (M.B.) Statistics (II) Quantum Statistics or Bose-Einstein (B.E.) and Fermi-Dirac (F.D.) Statistics. (I) Classical Statistics: It is based on the classical results of Maxwell’s laws of distribution of molecular velocities and Boltzmann theorem relating entropy and probability. So it is also known as Maxwell-Boltzmann (M.B.) Statistics. M.B. Statistics deals with the distinguishable identical particles of any spin. (II) Quantum Statistics ; It was developed by Bose, Einstein, Fermi and Dirac. So it is also known as Bose-Einstein (B.E.) and Fermi-Dirac (F.D.) Statistics. The B.E. statistics deals with the indistinguishable identical particles having zero or integral spin.The F.D. Statistics deals with the indistinguishable identical particles having half-integral spin.The particles obeying B.E. Statistics are called bosons whereas those obeying F.D. Statistics are called fermions and obey Pauli exclusion principle.Classical statistics is only a limiting case of quantum statistics.
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Objectives of Statistical Mechanics: The main objective of statistical mechanics is to establish the relation between the macroscopic behaviour of the substance in terms of microscopic behavior of particles.
Macrostates and Microstates: Macrostates – A macrostate of the ensemble may be defined by the specification of phase points in each cell. Microstates – A microstate of the ensemble may be defined by the specification of the individual position of phase points for each system or molecule of the ensemble There may be many different microstates which may correspond to the same macrostate. Understanding macrostate with example To clearly understand macrostate, we consider an example. Let four distinguishable particles a, b, c & d to be distributed into two exactly similar boxes. When any particle is thrown, there is ½ probability of going it into either of the two boxes. There are five different ways of distribution like (0,4), (1,3), (2,2), (3,1) & (4,0) i.e. zero particle in first box and 4 particles in second box as one way, one particle in first box and three particles in second box as second way and so on. So the total no. of macrostates is five. For n no. of particles the total no. of macrostates is (n+1)
Understanding microstate with example Each distinct arrangement of particle is known as microstate of the system. In the same example, first arrangement is (0,4) has zero particle in first box and all the four particles a, b, c, d in second box, so the possible arrangement is only one.In (1,3) macrostate, one particle in first box and three particles in second box.
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So arrangements can be (a, bcd), (b, cda), (c, dab), (d, abc). Thus possible arrangements are four. In (2,2) macrostate, 2 particles are in first box and the two particles in second. So arrangements may be (ab, cd), (ac, bd), (ad, bc), (bc, ad), (bd, ac), (cd, ab). Hence possible arrangements are six. Similar to (1,3), (3,1) macrostate can be arranged as (bcd, a) (cda,b) (dab,c), (abc,d) resulting in four possible arrangements. Finally (4,0) macrostate is similar to (0,4) i.e (abcd,0) with possible arrangement only one. Thus in the above example of four particles the total number of microstates is equal to total number of arrangements i.e 1+4+6+4+1=16=24. Hence in general, for a system of n particles, total number of microstates are 2n.
Phase space, Ensemble & Ensemble average Phase space:- It is defined as the combination of position space and momentum space. The phase space has six dimensions i.e. three position co-ordinates (x, y, z) and three momentum co-ordinates (px, py, pz). Thus the position of a particle in phase space is specified by a point with six co-ordinates x, y, z, px, py, pz.By these six co-ordinates the complete information like position and momentum of any particle can be obtained.
Phase space and Phase cell A small element in phase space is denoted by d and is given by
d = (dx dy dz)(dpx dpy dpz) Where, dx, dy, dz, dpx, dpy, dpz are the sides of six dimensional cells. Such cells are called phase cells. According to uncertainty principle –
dxdpx ≥ h
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similarly dydpy h & dz dpz h d h3 So, a point in the phase space is like a cell whose minimum volume is ̴ h3. Hence the particle in the phase space can not be considered exactly located at point x, y, z, px, py, pz but within a phase cell centered at that point. The phase space is denoted by space. Ensemble:- It is defined as a collection of large number of macroscopically identical but essentially independent systems. Here macroscopically identical means each of the system constituting an ensemble satisfies the same macroscopic conditions like volume, energy , pressure, temperature, total no. of particles etc. Independent system means the system consisting an ensemble are mutually non-interacting. A system is the collection of a no. of particles. Ensemble Average It is the average at a fixed time over all the elements in an ensemble. This average closely agrees with time average provided (1) The system is a macroscopic system consisting of a large no. of molecules so that the microscopic variables can be truly randomise. (2) The no. of elements forming the ensemble at one time is large so that they can truly represent the range of states accessible to the system over a long period of time. It is to be noted that all members of an ensemble are identical in features like no. of particles N, volume V, energy E etc and are called elements. Kinds of ensembles According to Gibbs, there are three types of ensembles. They are 1. Microcanonical ensemble 2. Canonical ensemble 3. Grand canonical ensemble
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1. Microcanonical ensemble It is the collection of a large no. of independent systems having the same energy
E, Same volume V and same no. of particles N. The individual systems of a microcanonical ensemble are separated by rigid, impermeable and well insulated walls such that the values of E, V and N for a particular system are not affected by the presence of other systems.
2. Canonical ensemble It is the collection of a large no. of independent systems having the same temperature T, same volume V and same no. of identical particles N. The individual systems of a canonical ensemble are separated by rigid, impermeable but conducting walls. The equality of temperature of all the systems can be achieved by bringing each in thermal contact with large heat reservoir at constant temperature T or bringing all the systems in thermal contact with one another
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3. Grand canonical ensemble It is the collection of a large no. of essentially independent systems having the same temperature T, same volume V and the same chemical potential . The individual systems are separated by rigid, permeable and conducting walls. As the separating walls are conducting and permeable, the exchange of heat energy as well as of that of particles between the systems takes place in such a way that all the systems arrive at common temperature T and chemical potential .
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Comparison of Ensembles
Property Microcanonical Canonical Grand Canonical 1. Contact with System is isolated Systems are in System can the Energy and no. of thermal contact exchange both environment particles of the with heat reservoir. energy and matter system are So system can i.e. particles constant exchange energy but no particles Temperature(T), 2.Same Energy (E), Volume Volume (V) & no. of Temperature(T), parameters (V) & no. of particles (N) same Volume (V) &
particles (N) same. chemical potential
3.Probability u same distribution Probability density Probability density (E)= e[( /k] E (E)= constant (E) = Ae kT In range E to E+dE 휌(퐸) = 0 4. Fluctuation Outside this range Fluctuation in Fluctuation in energy only energy and no. of No fluctuation particles 5.Partition Partition function function Z= Partition function Partition function E Z= e d Z= en / Zn n0
The individual The individual The individual 6. Separation systems are systems are systems are separated by rigid, separated by rigid, separated by rigid, impermeable and impermeable and permeable and well insulated conducting walls walls conducting walls
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Entropy and Probability
Entropy: It is defined as the thermal property of a substance which remains constant during an adiabatic change. It depends on state and does not depend on the process by which the substance is brought from previous state to present state. When an isolated system undergoes an irreversible process, there is a net increase of entropy of the system and an increase of randomness.
Entropy and Probability: Generally all spontaneous processes represent changes from a less probable to a more probable state and since in such process the entropy increases, there should be relation between entropy (s) of the system and probability of that state (w). Entropy is the measure of degree of randomness in an assembly. The entropy of a physical assembly in a definite state depends solely on the probability of that state.
Boltzmann postulates of entropy According to Boltzmann, the probability of the system in equilibrium state is maximum. But the equilibrium state of a system is the state of its maximum entropy. Thus in equilibrium state, both the entropy and probability are maximum.
Boltzmann’s relation between entropy & probability Following his assumptions that the entropy and probability are maximum in equilibrium state, Boltzmann gave a relation between entropy and probability.
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Boltzmann assumed that entropy(s) of a system in any state is a function of maximum probability w of that state. i.e. S = f(w) ….…..(1)
Consider two independent systems A and B having entropies S1 and S2. The entropy of the combined system. S = S1 + S2 ………(2) Since entropy is additive whereas probability is multiplicative physical quantity, so
S = f(w) = f(w1 × w2) S1 = f(w1) ………………………..(3) S2 = f(w2) w1 & w2 are probability of A & B f(w1 × w2) = f(w1 + f(w2) ……………………..(4) n Differentiating eq (4) w.r. to w1, we get w2 f (w1 × w2)= (w1) ………………….……..(5) n Differentiating eq (4) w.r. to w2, we get
w1 (w1 × w2)= (w2) ………(6) Dividing eqn (5) by (6), we have f (w ) w 1 = 2 ….…..(7) f (w2 ) w1
W1 (w1)= W2 (w2)= constant= K (let)
k So (w1)= w1 k & (w2)= ………..(8) w2 Integrating eqn (8), we get
f(w1) = K log(w1) + C1 ……….(9) & f(w2) = K log(w2) + C2
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In general F(w) = S = K log (w) + C ……..(10) Where C is constant of integration In case of perfect gas, at absolute zero( T = 0) Entropy S = 0 & W = 1 Putting these values in eqn (10) C = 0 Hence eqn (10) becomes S = K log (w)
This is the required relation between entropy and thermodynamic probability. This is known as Boltzmann’s entropy relation. Here K is universal constant known as Boltzmann’s constant. Thus we see that the entropy of a system is proportional to the logarithm of thermodynamic probability of that system.
Entropy of ideal gas: Sackur – Tetrode equation & Gibb’s Paradox From Boltzmann’s entropy relation, we know S = K log W ………(1) Where S is entropy, K is Boltzmann’s constant and W is probability From Maxwell – Boltzman statistics K Log W = log N! + (푛푖 푙표푔 푔푖 − 푙표푔푛푖!) ………..(2) i
Where N is the total no. of particles in the system, ni is the no. of th particles in i energy level and 푔푖 is the quantum states. K log w = log N! + ni log gi ni logni ni (Using Sterling’s formula) i K = log N! + ni (loggi logni 1) i
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k g i log W = ni log 1 log N! ……………(3) i ni For maximum value of W or log W,
e E ni = e gi i KT E g i Or i e KT ....………(4) ni Taking logarithm on both sides g E log i i ………….(5) ni KT
Substituting this value in eqn (3) k Ei log W = ni 1 logN! i KT 1 log W = ni Ei ni ni logN! .………..(6) kT i i i 3 ni Ei = total energy E NKT i 2
& ni N i
So, eqn (6) can be written as E log W = N N logN! KT 1 3 = NKT N N logN! KT 2 3 = N N N logN! 2 5 log W = N N logN! ………(7) 2 3 2 N h 2 But log V 2mkT
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3 2 5 N h 2 logW N N log log N! 2 V 2mkT Putting this value in eqn (1), we get 3 2 5 N h 2 S = Nk Nk log k log N! 2 V 2mkT 3 5 V 2mKT 2 S = Nk Nk log k log N! ……….(8) 2 N h2
To resolve the paradox, Maxwell – Boltzmann exprn in eqn – (2) should be divided by N! i.e. subtract the term K log N from eqn (8)
3 . V 2mKT 2 5 . S NK log 2 Nk . N h 2
This is Sackur-Tetrode formula for entropy of a perfect gas. Gibb’s Paradox
In eqn (8) , Klog N! is the additional term which can not be accounted for identical gases.
When two different ideal gases (distinguishable) with no. of particles N1 in box A & N2 in box B are at same pressure & temp and partition is removed then volm is added at the same temp. Since the process is irreversible, the entropy of the mixture should be greater than the sum of the entropies before mixing.
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If the gases are of the same kind (indistinguishable) then change in entropy should be zero. So, eqn (8) gives the correct result for two different gases But for two identical gases eqn (8) gives increase in entropy . This is in contradiction. This contradiction is known as Gibb’s Paradox.
A B
N N 1 2
Fig : Mixing of two gases.
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UNIT- II
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Probability density for canonical ensemble Let us consider a macrocanonical ensemble representing a very large isolated system A. Let each system (t) is made of a large no. of subsystems which are in mutual thermal contact and can exchange energy but not the particles with each other. Here subsystem is denoted by S and rest of the system or heat reservoir by r. As the total energy of the system is
Et = Er + Es = constant------(1) Where Es = Energy of the subsystem
& Er = Energy of the rest of the system.
Let d t represent the probability that the total system t is in an
element of volume dt of the phase space, then for a micro conical ensemble, we have
d t (E)=C for Et =0, outside energy range Where C is constant We can split into two elements one for subsystem ds and other for reservoir dr . Then, = ds . dr ------(3) So, equation (2) can be written as, d t(E) = Cd s . d r for Et PG SEMESTER:- II, PAPER: - VI PHYSICS CHE522: STATISTICAL PHYSICS If d S is the probability that the sub system is in the volume d S, provided the total energy remains between Et and Et+ Et then d s= Cd S . r ------(5) where r is the volume of the phase space of the reservoir From definition of entropy in classical statistical mechanics, the entropy of the reservoir r= log r ------(6) exp ( r)= ------(7) But Er = Et - ES [From equation (1)] r(Er) = r[Et – ES] r(E ) = r(Et) - t .Es+ ------(Expanding by Taylor’s theorem) Et r(E ) r(Er) = r(Et) - t .Es ------(8) Et (Neglecting higher terms) Putting this value in equation (7) r(E ) r exp ( r) = exp r(Et) - t .Es ------(9) Et 1 = Where = 𝓀푇; T= abs. temperature. Also Et=Er So, equation (9) becomes [ Es ] exp [ r(Et)].exp Substituting this value in equation (5), we get E s r(E )] [ s ] d흎s= Cd exp[ t . PG SEMESTER:- II, PAPER: - VI PHYSICS CHE522: STATISTICAL PHYSICS E [ s ] s d흎s=C exp [ r(Et)].exp d d흎s= A exp d s Where A= C exp [ r(Et)] The normalization factor A is to be determined such that d 흎s= 1 = A exp d ------(10) But the normalization condition is 1= d ------(11) Comparing equation (10) & (11) we get (E) = A exp [ E ] ------(12) Where E is the energy of the entire system. [ E ] (E) = A exp kT ( kT ) An ensemble characterized by the probability density (E) is called Gibb’s canonical ensemble or macro canonical ensemble. Perfect monatomic gas in canonical ensemble and partition function In thermodynamics, the entropy is determined by the energy independently of whether the system is isolated (micro canonical) or in thermal contact with the heat reservoir (canonical). While calculating the entropy of a perfect gas isolated system, the exponential term is neglected. This clearly mean that canonical distribution can serve the purpose equally well and the given energy E may be chosen as equal to the mean energy E of the system. PG SEMESTER:- II, PAPER: - VI PHYSICS CHE522: STATISTICAL PHYSICS (A) Partition Function The classical partition function for canonical ensemble is given by 1 Z = ∫ 푒−퐸/휏 푑휏------(1) 푁!푛3푁 Where N = Total number of independent identical spinless molecules, For a perfect gas, there are no mutual interaction between the molecules, so we have, 2 pi E = 2m i Also, d di dqi dpi i i So, equation (1) becomes 3N pi2 1 Z = [ e 2m dq i dpi ------(2) N!h3N i1 3N pi2 1 V N e 2m dp Z = 3N i ------(3) N!h 3푁 N Here ∫푖=1 휋 dqi=V Using standard definite integral p 2 1 i e 2m dp (2m ) 2 i = ------(4) 2 1 ay 2 e dy a Substituting this value of equation (4) in equation (3), we get 3N 1 1 N 2 Z = 3N V (2m ) N!h PG SEMESTER:- II, PAPER: - VI PHYSICS CHE522: STATISTICAL PHYSICS 3N/2 푉푁 2휋푚kT = ( kT) ------(5) 푁 ! ℎ2 Thermal de Broglie wavelength 1 h h2 2 = 1 2m kT 2m kT 2 So, Equation (5) becomes 3N N N V N 1 V N 1 1 V Z N! N! 3 N! 3 Partition fraction 1 Z f N N! 3 V 2mkT 2 Where f V 3 h2 This is the formula for partition fraction for perfect monatomic gas. (B) Helmholtz Free energy (F) We know the relation F= - kT log Z f n F kT log N! N N Using Sterling approximation N! e N ef ef F= kT log NkT log N N f e = NkT log NkT loge N PG SEMESTER:- II, PAPER: - VI PHYSICS CHE522: STATISTICAL PHYSICS f F = NkT log NkT N (C) Entropy of monatomic gas F Entropy S= V T f But F NkT log NkT N f S NkT log NkT T N f f = Nk log NkT log Nk N T N 3 2mkT 2 Substituting the value of f V , we get – h2 3 3 V 2mkT 2 V 2mkT 2 S Nk log NkT log Nk N h2 T N h2 3 3 1 V 2mkT 2 1 V 3 2mk 2 S Nk log NkT T 2 Nk N h2 3 N 2 h2 V 2mkT 2 N h2 3 V 2mkT 2 5 S Nk log Nk N h2 2 This is the required expression for entropy which is same as Sackur – Tetrode equation for perfect monatomic gas in microcanonical ensemble. Thus, it justifies the statement that entropy is determined by energy independently of whether the system is isolated or in thermal contact with the reservoir. PG SEMESTER:- II, PAPER: - VI PHYSICS CHE522: STATISTICAL PHYSICS (D) Internal energy Mean energy is defined as – E i Eie i th E E ------(i) Where Ei = Energy of the system in (i) i e i state & kT Partition function E i Z = e i log Z = log Differentiating both sides with respect to , we have E i logZ log e i E 1 i . e = E i e i i E 1 i 1 . e . E = E i 2 i e i i E i Eie 1 i E = 2 E 2 [From above equation (i)] i e i E 2 logZ ------(ii) T From Helmholtz free energy F = logZ F log Z = PG SEMESTER:- II, PAPER: - VI PHYSICS CHE522: STATISTICAL PHYSICS Substituting is in equation (ii), we get F F E 2 2 ------(iii) f N Since, F = - logZ log N! F N log f logN! Using sterling’s approximation F = N log f N logN N = Nlog f logN N f F N log N N Substituting this value of F in equation (iii), we get 2 1 f E N log N N 2 f = N log N N 3 3 V 2m 2 2m 2 = N 2 log N 0 & putting value of f V 2 2 N h h 3 1 1 V 2m 2 3 2 2 N 3 . 2 . = N h 2 V 2m 2 N h2 1 3 1 = . . 2 3 2 2 3 = N 2 3 E NkT ( kT ) 2 PG SEMESTER:- II, PAPER: - VI PHYSICS CHE522: STATISTICAL PHYSICS Internal energy of system 3 U E N kT 2 (E) Specific heat at constant volume (CV) Specific heat at constant volume – E 3 3 CV NkT Nk T V T 2 V 2 Since in the case of perfect monatomic gas 3 E E N kT 2 3 CV Nk 2 This is the expression for specific heat at constant volume of monatomic perfect gas. PG SEMESTER:- II, PAPER: - VI PHYSICS CHE522: STATISTICAL PHYSICS UNIT – III PG SEMESTER:- II, PAPER: - VI PHYSICS CHE522: STATISTICAL PHYSICS Dr. Abhay Kumar Yadav Associate Professor of Physics University Department of Physics L.N.M.U. Darbhanga, PG SEMESTER:- II, PAPER: - VI PHYSICS CHE522: STATISTICAL PHYSICS Density matrix A double average is taken in obtaining the value for a thermodynamic variable in quantum statistical mechanics. When the system is described as being in the quantum state n , the first average of the expectation value of the appropriate quantum mechanical operator R , is given by Rn = < | | > ------(1) The second average the usual statistical mechanical ensemble is given by 1 M < >A = Rn ------(2) M n1 Where the sum is over the ensemble. It is written as 1 * < >A = a nj .ank j | R | R ------(3) M n j k This can be expressed as a product of two matrices. If define the elements of a matrix , the density matrix , by M 1 * kj a nj .ank ------(4) M n1 Then equation (3) may be written as < >A = kj = j | R | j = tr( R ) ------(5) j k j Here < >A = trace ( ) yields a compact way of expression the averaging process required. Choosing the unit operator for , we see that Trace( 1 ) = trace ⋌ ( ) = 1 ------(6) Which serves as a normalization condition for the density matrix and related it to a probability density, the diagonal elements of the density matrix give the probability in those systems do in that quantum state. As the density matrix is defined as it is Hermitean and its diagonal elements are real. PG SEMESTER:- II, PAPER: - VI PHYSICS CHE522: STATISTICAL PHYSICS In the context of micro canonical ensemble, the density matrix applies to any ensembles, provided density matrix is selected to conform to the particular constraint imposed by the ensemble. The density matrix can be applied to the dynamics of a quantum mechanical system also. If the state describing the system are not energy eigen states or if the system Hamiltonian in changing in time. The density matrix can be time dependent. The development in time of a quantum state is governed by the time dependent Schrodinger equation. ih H ------(7) t n n Which becomes, ihanj H nk akj ------(8) R With Hnk = < n | H | > Taking the derivative of equation (4) with respect to time and using equation (8), we get time dependence of density matrix ih jk H jhhk jhH nk ------(9) Since in quantum mechanics the commutator bracket of a pair of operators A and B is defined as [ , ] = - , then using this bracket notation , equation (9) is written as ih H ------(10) When a system is in equilibrium, the commutator bracket of with H vanishes. For the micro canonical ensemble, all the system states lying within the small energy shell are accorded equal weight and if there are N energy eigen state forming this set, the density matrix formed with energy eigen states is diagonal and has the value 1/N for each of those states whose eigen values lie within the energy shell. For an ensemble appropriate to a system in contact with a heat bath at a fixed temperature the canonical ensemble in the appropriate choice and the functional form of the density matrix is PG SEMESTER:- II, PAPER: - VI PHYSICS CHE522: STATISTICAL PHYSICS eH / Z ------(11) Again for a system which can exchanged particles and energy with its surroundings, the appropriate ensemble is the grand canonical ensemble and the functional from density matrix for the grand canonical ensemble is e H N ------(12) Z In the pure case, density matrix holds 2 . The eigen value of density matrix will be positive number and less than or equal to one 1. They satisfy the normalization condition, where their sum will be equal to 1. In the mixed case, at least two of the eigen value will be non-zero. The statistical state can be described by a non – negative density f4 f (q1…qf , p1…pf ,t) The probability of finding the system in interval dq1,…dqf,…dp1,…dpf at time t is ( dq1,…dqf,…dp1,…dpf ) The quantum analogue of classical density function is known as operator. Since operators can be expressed by Matrix and hence the density operator expressed as Matrix is known as Density matrix. We consider ensemble of N system in normalized states i i = 1, 2, 3,…, N Let i be the probability for assemble to be in , we then define the density matrix in { } represented by ij iij ------(1) PG SEMESTER:- II, PAPER: - VI PHYSICS CHE522: STATISTICAL PHYSICS If a measurement is made on an oberservable whose operator is A , having eigen function i , the result will be eigen value an corresponding to n . If i be the state of assembly, we get = Cij j ------(2) j as a linear combination of i * The probability that will give an is then simplify C inCin . But probability that assembly is in state is i e. where A will give anj the robability is just * * iCin Cin i ijCin Cin i i j * ijCinCin i j * * C C dq in in i j i j * C C * dq in jn j i * dq n n ------(3) Probability that will give an is just * dq n n nn ------(4) To calculate the average value of , we simply write A = an nn * a dq = n n n n * A dq = n n ------(5) = Trace [ A ] PG SEMESTER:- II, PAPER: - VI PHYSICS CHE522: STATISTICAL PHYSICS Here A is a double, quantum much and state much average. The ideal gases and microcanonical ensembles. The various volume elements in the allowed region of the phase space in generally known as various possible microstates. The resulting ensemble is referred to as the microcanonical ensemble. In this ensemble the microstate of a system is given as the number of molecules N, the volume V and the energy E. The Hamiltonian of N identical particles having N non interacting member is given as N p 2 H= i ------(1) i1 2m 2 Where pi = pi . and is the momentum operator of the i-th particle. Hamiltonian is independent of the positions of the particle such as spin. For non-interacting identical particles, we have three cases: a. ideal Bose gas, b. ideal Fermi gas, and c. ideal Boltzmann gas. Thermodynamics of these ideal gases in the formalism of microcanonical ensemble. We shall obtain the number of state (E) of the system having the energy eigen value which lies between E and E+ . Any energy eigen value of an ideal system is the sum of single particle energies and are called levels and are given by 2 E p = p / 2m ------(2) Where p= |p| and p is the momentum eigen value of the single particle, 2h p= .n ------(3) L Where n is a vector whose components are 0 or ±1,2 etc. L is the cube root of the volume of the system i.e. L= V 1/ 3 In the limit V 0, the possible values of p form a continuum. Then v d 3 p ------(4) p h3 PG SEMESTER:- II, PAPER: - VI PHYSICS CHE522: STATISTICAL PHYSICS Where ħ = 2 h and is Planck’s constant. If there are n p particle having momentum –p, the total energy E and total number of particles N of the state are given by E np ∑ np E= p p and N= 푝 ------(5) The allowed value for any are = 0,1,2………. For spinless boson ------(6) 0,1 for spinless fermions For a Boltzmann gas = 0,1,2……… but ( ) specifies N!∏ ( !) states of the N- particle system P. When V α the levels (eqn.2) form a continuum, If we devide spectrum as g1, g2 , g3 ……….. levels , each group is called a cell and has an average energy i . The occupation number ni of the i th cell is the sum of n p over all the levels is the i-th cell. Let W{ ni} = no. of states of the system corresponding to the sets of occupation numbers { ni }, then (E) Wni ------(7) ni When the sum extend ni over all sets of integers { }, satisfying the conditions E= Eini and N= ni ------(8) i i Since interchanging particle is different cells lends to a new state of the system. If we consider all N particles together. In the case of interchanging particles in different cells leads to a new state of the system. We have W{ } = wj Ni PG SEMESTER:- II, PAPER: - VI PHYSICS CHE522: STATISTICAL PHYSICS Bose gas : Since each level can be occupied by any number of particles, (ni gi 1)! Wi= nil(gi 1)! (ni gi 1)! Hence, W{ ni} = wi ------(9) i i ni!(gi 1)! Fermi gas: The number of particles in each of the gi subcells of the i- th cell is either 0 or 1. Therefore wi is equal to no. of ways in which ni things can be chosen out of gi things g ! w i i gi = ------(10) h ni!(gi ni )! n Hence, w{ i } = wi = ------(11) i i Boltzmann gas: The N particles are placed into cells where i-th cell have hi particles. There are N!/ i ni ! ways to do this. Within the i-th cell, there are gi levels. Among in an i-particles in the i-th cell, the first can occupy the level in ways . The second and all subsequent ones hi can also occupy the level in ways. Therefore there are (gi ) ways in which ni particles can occupy the levels. The total no. of ways to obtain { ni } is given as ni gi ni i ni! ni gi Hence, w{ ni }= ------(12) i ni! The entropy of the system is given as S= k log w{ ni } ------(13) g Where = i (Bose Fermi) Z 1e ti 1 ti gi ze Boltzmann ------(14) PG SEMESTER:- II, PAPER: - VI PHYSICS CHE522: STATISTICAL PHYSICS Using Stirling’s approximation and neglecting 1 in comparision to gi, we have from (12) & (13) g n i i ni log1 gi log1 forBose i ni gi S g n logw{n } n log i 1 g log1 i forFermi i i i ------(15a) R i ni gi gi ni log forBoltzmann` i ni The validity of these equations depend on the assumption that 2 2 hi hi 2 1 ------(16) hi The condition E = hiti requires that 2 2 p t ZV p E Z g t eti Z t e p dp4p2 e 2m i i p 3 i p h 0 2m 3 = N K T ------(17) 2 Therefore T is the absolute temperature. The entropy is given by (15b) S 2etp (tp logz) E N logz R p 3 2 3 N 2h 2 N N log ------(18) 2 V mkT This is Sackur Terode equation. It is to be noted that (18) does not satisfy the third law of thermodynamics. This shows that the third law of thermodynamics is not an automatic consequence of the general principles of quantum mechanics but depends on the nature of the density of states hear the ground state. PG SEMESTER:- II, PAPER: - VI PHYSICS CHE522: STATISTICAL PHYSICS More explicitly ti log z ti gi 1 t log(1 ze )Bose z e i 1 i S ti log z ti gi log(1 ze )Fermi 1 ti R i z e 1 g eti (t log z)Boltzmann` (15b) i i i THE PROBABLITY OF A SYSTEM The probability of the system having an energy E, when in equilibrium with its surrounding. The probability P(E) of the system S having an energy E is consequently proportional to the number of state, (E) available to it, which is equal to the number of states e ( Et0l - E) available to the environment i.e. (E) s(E) e(Et0l E) (1) The environment is much bigger than the system and hence the environment energy Ee Et0l and expanded in a Taylor series with only the lowest term related. Then equation (1) becomes Te 21 loge 휌(E)= e (Ee E) e (Ee ) E 0(E e (Ee ){1 E ------(2) fe Ee This leads to loge loge 2 Log 휌(E)log e (Ee )+log {1 E }= log (Ee ) -E E +0(E ) Ee Ee loge =Log e (Ee ) - E× ……… (3) Ee We assume that the entropy Se of the environment is k in e ,then E Se log(E) loge(Ee ) . …………. (4) R Ee PG SEMESTER:- II, PAPER: - VI PHYSICS CHE522: STATISTICAL PHYSICS S 1 Since, e , the temperature of environment and in equilibrium the Ee Te temperature of the system T = Te, yields E log (E) log (E ) e e RT E E or (E) e(Ee )e RT A.e RT ……………(5) where A is constant, we thus find that if the system S is in thermal equilibrium with its environment and is characterized by a temperature T, then the probability of its having an energy E is given by equation (5). The distribution should be normalized to unity, which implies that E 1 RT 1 A = e z …………….(6) allmicrostates E RT Where, z e …………..(7) is called the Canonical partition function of the system. It is also called ‘Sum – over – states’. It shows that how to gas molecules of an assembly are distributed or partitioned among the various energy levels. Linear Harmonic Oscillator We consider a system of harmonic oscillators, having each of mass m in thermal equilibrium. If they execute small oscillation with the same frequency ω. The Hamiltonian of the system of N such oscillators are given by N 2 1 pi 2 2 H(q, p) m qi …………… (1) 2 i1 m Where qi, pi are respectively the generalized coordinate and generalized moment in classical mechanics, but in quantum mechanics, the energy level of the system are given by PG SEMESTER:- II, PAPER: - VI PHYSICS CHE522: STATISTICAL PHYSICS 1 n h E i …………… (2) n i 2 Where ni is an integer, 1, 2, 3, … The partition function of the system is given by N QN Q1 …………. (3) Where Q1 is the single particle partition function of an oscillator, using equation (1), we have, 2 pi mqi 1 2m = exp dq dp 2 kT i i 1 p 2 m 2q 2 exp i dp exp i dq = i i 2h 2mkT 2kT Integrating over the momentum and co-ordinate, we have 1 1 1 2kT 2 2 = (2 mkt) . 2 2h m = (kT / h ) N …………… (4) N And QN= [kT/ ] ………… (5) Therefore energy of the system is then A= -kT log QN = -NkT log (kT/ ) ………….. (6) Then energy is given by (6) as S=nk[log(kT/ +1)] and energy is ………….. (7) U= NkT ………….. (8) The sp. Heat at constant volume is given by Cv= Nk ………….. (9) From equation (8), we find that the mean energy per oscillator is kT. PG SEMESTER:- II, PAPER: - VI PHYSICS CHE522: STATISTICAL PHYSICS The single particle partition function is written as h 1 Q1 exp ni ni 1 kT 2 h = exp 1 exp(h kT 2kT Hence the total partition function is given as N exp (h 2kT) QN ………….(11) 1 exp(h kT) Using equation (11), the free energy is 1 A N h kT log1 exp( h kT) 2 = N kT log [2sin hh 2kT)] …………… (12) Other thermodynamical properties can be calculated as h kT h S =Nk log 1 exp h kT exp. 1 kT 1 h h = Nk cot log2sinhh 2kT) …………… (13) 2 kT 2kT 1 h U = N 2 exp(h kT) 1 1 h = Nh coth ………….. (14) 2 2kT 2 2 exp ℏ휔/푘푇 Cv = Nk (ℏ휔/푘푇) exp(ℏ휔/푘푇)−1 2 1 h = Nk cosec2hh 2kT ------(15) 4 RT ` h At low temperature << 1, and (15) reduces to kT PG SEMESTER:- II, PAPER: - VI PHYSICS CHE522: STATISTICAL PHYSICS C =Nk 1 h kT .... v 1 (1 h kT ....2 2 = Nk, as T . Thus the classical result is valid at high temperature only A free particle in a box We shall apply ensemble theory to quantum mechanical system composed of indistinguishable entities. It deal in the language of operators and the wave function for studying quantum system. We consider a free particle of mass m having in the cubical box of side L. The Hamiltonian of the particle is given by h2 h2 2 2 2 2 H= - 2 2 2 ………. (1) 2m 2m x y z The eigen function of the Hamiltonian satisfying boundary conditions (x L, y, z) (x, y L, z) (x, y, z L) (x, y, z) ………. (2) are given by 1 E (r) 3 exp(iK.r ) ……….. (3) L2 2 2 The corresponding eigen value E = h K …………. (4) 2m 2 Where K= (ku, ky, kz)= (nx, ny, nz) ………… (5) L Where nx, ny, and nz are positive , negative or zero quantum number. Since the wave vector k is written as 2 k .n L PG SEMESTER:- II, PAPER: - VI PHYSICS CHE522: STATISTICAL PHYSICS Where n is a vector with integral components 0, 1,2...... then the density matrix ( ˆ ) of this system in the canonical ensemble may be evaluated by co-ordinate representation. Since. 휌푛= C.exp( Ex ) , where C is a constant. Applying normalization conditions nn 1 and Tr( () 1 , value of constant C will be. n 1 1 1 C = H ………. (7) exp(En ) QN () Tr e and |n n | is the identically the unit operator where Qn ( ) is N called the partition function of the system. The energy E is determined by the Boltzmann factor exp( En ) , 1/ rT, hence we have r | eH | r r / E eE E / r E E * = e E r E (r ) ………… (8) E Substituting from (3) and making use of (4) and (6) we obtain. 1 훽퐻² PG SEMESTER:- II, PAPER: - VI PHYSICS CHE522: STATISTICAL PHYSICS 2휋푣 g(E)= (2m)3/2 E3/2, ħ³ which gives single particle partition function, 훼 -βE 푉 2휋푚 /2 Q1(B) =∫ 푒 g(E) d E = ( )³ ...... (11). ° ℎ³ 훽 Combining equation (12) and (13), we obtain for the density matrix in the coordinate representation 1 푚 ℎ² 2 푚 This may be written as : 푇푟.(퐻̂푒ˉ훽퐻̂) ə PG SEMESTER:- II, PAPER: - VI PHYSICS CHE522: STATISTICAL PHYSICS Let us suppose that the indistinguishable element that constitute the system under consideration are divided into a series of quantum group or level with the no. of elements n1, n2...... , ni in each group the energy is constant viz E1, E2, E3...... , Ei respectively. Then the total energy Ʃi (ni = n = constant, and the total energy = ∑ 휖푖ni = E = constant If the de-generacy of the level is equal to gi, the total no. of eigen states for the group of ni elements is equal to the number of ways in which the ni elements supposed to be distributed among the gi section. Then ni particles will be distributed among the gi Let us imagine a box divided by(gi-1) particles into gi, sections, the ni particles then among the gi compartment will be (푛𝑖 + 푔푖 − 1)! 푛! (푔푖 − 1)! Since the element are in distinguishable there is only one way in which this subdivision can be made. In each group the no. of eigen states is given below by expression of the same type. Thus the total number of eigen states for the whole system is (푛₁+푔₁−1)! (푛₂+푔₂−1)! (푛₁+푔₁−1)! (푛 +푔 −1)! G= ...... = πᵢ 푖 푖 푛 푖!(푔ᵢ−1)! 푛2!(푔₂−1)! 푛푖!(푔₁−1)! 푛푖!(푔푖−1)! Where Π denotes the product of the similar term. Now the probability w of the system having the particular distribution specified is proportional to the total no. of given eigen states. Hence (푛₁+푔₁−1) W= πᵢ x constant...... (2) ∟푛₁(푔₁−1)! Log w = Ʃᵢ {log (nᵢ + gᵢ - 1) – log nᵢ! - log ∟(gᵢ- 1) +e + constant (c) = Ʃᵢ{log∟ (nᵢ + gᵢ) - log∟nᵢ -log∟gᵢ } +c as nᵢ >> 1, so i can be neglected. Making the use of strilling formulae. PG SEMESTER:- II, PAPER: - VI PHYSICS CHE522: STATISTICAL PHYSICS Log∟n= n log n-n log w = Ʃᵢ [ (nᵢ + gᵢ) log (nᵢ + gᵢ) - (nᵢ + gᵢ)]- nᵢ log nᵢ+ nᵢ - gᵢ. Log gᵢ+gᵢ}+c Log w= Ʃᵢ [(nᵢ + gᵢ) log (nᵢ + gᵢ) - nᵢ log nᵢ - gᵢ log gᵢ]+for maximum probability δ log w=0 ˌ δ log w= Ʃᵢ[ log (nᵢ + gᵢ) δ nᵢ + δ nᵢ - log nᵢ δ nᵢ- δ nᵢ Since gi is constant δ gi = 0 (nᵢ + gᵢ) δ log w = Ʃᵢ { log [ ]} δ ni...... (3) nᵢ Now Ʃ nᵢ = n = constant so that Ʃ δ nᵢ = δn= 0 and Ʃ Eᵢ nᵢ = E = constant ----(A) ∴ δ Ʃ 휀푖 δni = δ E = 0 ------(B) We cannot conclude from above equation that the co efficient of dnᵢ is equal to zero because nᵢ are dependent on n and E. We get by lagengiam method of undetermined multipliers i.e. multiplying equation (A) by α and (13) by B and subtracting the resulting equation to equation (3) (nᵢ + gᵢ) Ʃᵢ [log - α - β휀i ] δni =0 nᵢ Since δni are independent of one another or, (nᵢ + gᵢ) 푔ᵢ Log = log (1+ ) = α + βεi nᵢ 푛ᵢ 푔ᵢ α Or, 1+ = e + βεi ------(4) 푛ᵢ 푔푖 푔푖 Or, ni = = 휀−휇 ------(5) (훼+훽휀 )−1 − 푒 푖 푒 휃 1 When β = 1/θ, 휇/휃 = −훼 This is the mathematical representation of the B-E statistics for the most probable distribution of elements among energy levels. PG SEMESTER:- II, PAPER: - VI PHYSICS CHE522: STATISTICAL PHYSICS UNIT- IV Bose Einstein distribution PG SEMESTER:- II, PAPER: - VI PHYSICS CHE522: STATISTICAL PHYSICS For the ideal gas of N molecules in a volume v , the most probable number of particle with energy εi is 푔ᵢ 푔ᵢ 푛̅i (ɛᵢ) = = ...... (1) 푒푥푝 (훼+훽ɛ)−1 푒푥푝 [(ɛᵢ−휇)푘퐵푇]−1 where β = 1/푘퐵푇, 훼 = −휇/푘퐵푇 and gi is the degeneracy of i-th the level. The parameters α or µ is determined as a function of N and T by the condition. ∝ 푔0 푔₁ N= ∑ 푛̅ = + + ...... 푖=0 푖 푒푥푝 [ 훽( 휀−휇 )]−1 푒푥푝 [훽 (ɛᵢ−휇)]−1 =n0 + n1 +...... (2) Where the sum is over the energy level as we have included gi in (1). Equivalently we replace gi by 1 is (1) and then sun over the quantum states in (2). We have ni ≥ 0, because the number of particle in a level cannot be negative. For a boson gas at all temperature (휀ᵢ - ) must be greater than zero for all ɛᵢ. Therefore, exp (-휇/kβT) ≥ 1 or μ ≤0 ...... (3) We know replace the sum by an integral in (2) by using in place of gi , then density of state become 2휋푉 g(휀) dε = (2m)3/2 ×E½ . dε ...... (4) ℎ³ 훼 푔 (ɛ)푑ɛ using it, equation (2) becomes N = ∫ 0 푒−훽푒훽휀−1 1 ⁄2 2휋푣 3/2 훼 휀 푑휀 푣 = (2m) ∫ 1 = F3/2 (na ) = µna F3/2 (na)...... (5) ℎ³ 0 푒훽휀−1 ƛ³ 푛푎 Where na is the absolute activity for a gas, na = 1/ƛ³ is the quantum concentration associated with one atom in a cube of side equal to ƛ. ℎ βμ Since e = na ≤ 1 , 휆 = 3/2...... (6) (2휋𝓂푘훽푇) And x=β휀 = 휀/kβT, we have PG SEMESTER:- II, PAPER: - VI PHYSICS CHE522: STATISTICAL PHYSICS 2 훼 푥²푑푥 2 훼 1/2 -x -x 2 -2x F3/2 (푛a) = ∫ 1 = ∫ 푑푥x 푛a e (1+푛a e + n a e +...... ) 휋1/2 0 푒푥−1 휋1/2 0 푛ₐ 푛ₐ² 푛ₐ² 훼 푛ₐ² =na + + +...... =∑ ...... (7) 푍3/2 33/2 푛=1 푛3/2 훼 푛ₐᴺ The F3/2 (na)= ∑ ...... (8) 푛=1 푛⁵ Is special case of the general class of function and are shown in figure 1a and 1b. 2.612 Z A F3/2(na) 1.341 ↑ B 1 Fs(α) 0 →α 2 0 na 1 Fig. 1a Fig. 1b At the limiting value na = 1 or μ = 0. The derivative of F3/2 (na) 훼 1 diverges but the series (7) converges to F3/2 (na=1) = ∑ 푛=1 푛3/2 1 1 3 = 1+ + +...... = G ( ) = 2. 612 ...... (9) 23/2 33/2 2 Where G is the Riemann zeta function. This is the maximum possible value of F3/2 (na) due to (3) The maximum temperature To is called the critical temperature at which n0 has the maximum value 1. (equ.5) PG SEMESTER:- II, PAPER: - VI PHYSICS CHE522: STATISTICAL PHYSICS 푉 2휋푚푘 푇 N= 2. 612 = V ( 퐵 0)3/2 x 2.612...... (10) ƛ³ ℎ² ℎ2 휌2/ And T0 = ( 3 ) 휌= n/v ...... (11) 2휋푚푘퐵 2.612 If we take one mole of gas, so that N is Avogadro number, 115 T0 2/3 K ……..(12) 푀푉 푚 Where M is the molecular weight and Vm is the molar volume is in cm³ mol-1. BOSE EINSTEIN CONDENSATION When A finite fraction of the particles constituting a Bose- Einstein assembly occupy the level with momentum p=0, the phenomenon is known as the B.E Condensation of the temperature in an ideal Bose gas is lowered to zero, the particle will beginning to crowd into a few level. 푛ℎ³ 푒−2훼 The lowest energy level is zero i.e then = e-α + + 푣푔(2휋푚푘푡) 3/2 23/2 푒−3훼 + ...... (6) 33/2 We introduced the fugacity, defined by Z= e-α = eμ/KT = eβH...... (7) Equation (6) can be written as, 푛ℎ³ 1 3/2 = ∑ ( ) 2푙...... (8) 푣푔(2휋푚푘푡)3/2 푙=1 푙 If T is decreased equation (8) requires that 1/kT should become smaller and the series cannot be equated to its first term. We lower the temperature T or increase n/v to the point where in 휇 equation (8) = 0 and Z is nearly equal to 1. 푘푇 PG SEMESTER:- II, PAPER: - VI PHYSICS CHE522: STATISTICAL PHYSICS Therefore, if T=T0 and Z =1,the series in equation (8) becomes 1 3/2 I3/2 (Z) = ∑ ( ) Zl 푙=1 푙 1 3/2 When I3/2 (1)= ∑ ( ) =2.612 ------(9) 푙=1 푙 The derivative of I3/2 (Z) at Z=1 diverges. But its value is finite. A graph for I3/2 (Z) is shown below. (Z) I3/2 2.612 Z Z = 1 Number of particles in lowest state is given as 3/2 2휋푚푘푇 3/2 1 N=n0 + v ( ) ∑ ( ) zl...... (10) ℎ² 푙=1 푙 And on substituting µ =0 or z=1, we get for T≤T0 푇 3⁄2 N0=n[1-( )) ] ...... (12) 푇0 This n0 is the number of particles in the lowest state with p=0 or energy 휀=0, and for assume of particle obeying B.E. statistics, we can write 푤0푧 푤0 N0 = = … … … … (13) 1−푧 푒−휇/푘푇− 1 PG SEMESTER:- II, PAPER: - VI PHYSICS CHE522: STATISTICAL PHYSICS Combining equation (12) and (13), we get -1 −휇 휔 푇 3/2 = log [1 + 0 {1 − ( ) } ] …………14 푘푇 푁 푇0 For T At temperature much below To , n0 is very nearly equal to the total number of particles in the gas. The number of particles in the excited states changes as, (for T 푇 3/2 Nn = n-n0 =n ( ) ...... (16) 푇0 As T approaches absolute zero, the whole of the gas condenses into the lowest state. PARTITION FUNCTION The term partition function is related to distribution function for an canonical enable. It indicates that how the particle are partitioned amongst the various energy states. It is given by ∑ −퐸1⁄휏 Z= 퐸1 Ω (E1) 푒 Where Ω (EI is the number of states between energies El + 훿El and 휏 is the temperature. Characterizing every part of the system because the system is in the thermal contact. PG SEMESTER:- II, PAPER: - VI PHYSICS CHE522: STATISTICAL PHYSICS Changing the summation over to integration as the distribution of energy is continuous and writing E1 = E, partition function comes out to be for an assembly of in distinguishable particles. 1 ʃ e-E (q,p) / kT ℎ³ᴺ 1 Z= ∫(푒−퐸(푞,푝) /푘푇).d휏 ℎ3푁 This is called classical partition function the summation is changed into integration as the assembly has a continuous energy distribution. Where h3N is the volume occupied by N particle. If we account for an assembly of in distinguishable particle n! is to be introduced. △휏 So that number of states becomes Ω(E)= ℎ3푁ͪ.푛! This is called semi-classical partition function. The quantum partition function for canonical ensemble is given by βE Z=∑푟 푒⁻ Where r signifies the state of the system and β= 1/kT. THE PARTITION OF A SYSTEM OF FREE PARTICLE Consider a system of N identical particles of mass contained in a cube of volume V. Position of N particles are denoted by (r1, r2 ...... ,rn). If the system is the sum of the kinetic energy operator K and the potential energy operator Ω now for the ideal gas, Ω(1...... ,N) = 0. The eigen function of the Hamiltonian are the free particles wave function ɸp (1,...... ,N) and are labeled by a set of N momenta P=(P,...... ,PN) …….(1), which satisfy the eigen value equation. K ɸP (1,...... ,N)= Kp ɸp (1,...... ,N) ...... (2) PG SEMESTER:- II, PAPER: - VI PHYSICS CHE522: STATISTICAL PHYSICS 1 Where Kp ≅ (p21 + ...... +Pn2) ...... (3) 2푚 When the periodic boundary condition with respect to the volume v is imposed, pi has the allowed values. 2휋ℎ푛 Pi = ...... (4) where N is a vector whose component may 푣1/3 be 0,±1, ±2, … … … … … then ɸp(i,...... ,n) is given by 1 ɸ1 (1,...... ,N) = ∑푝 훿p [ up (p1),...... upn(PN)] √푛푖 1 = ∑푝 훿p [upp1 (1),...... ,uppn(N)] ...... (5) √푛푖 1 Where up (푟̇) = eip.r/ħ ...... (6) √푣 Where p is the permutation of N object that sends the ordered set (1,2,...... N) to the ordered set (p1,p2,...... pn) and where δp = ±1 In the limit v=α, a momentum sum may be replaced by an 푣 integral.∑ ʃ d³p ...... (7) 푝→ ℎ³ 푣푁 3푁 3푁 Therefore, tr e-βk = ʃ 푑푝 푑푟 | ɸp (1,...... ,N)2| e-βkP ...... (8) 푁!ℎ³푁 Using, equation (5), we may write. 1 ∗ ∗ | ɸp (1,...... ,N) |2 = ∑ ∑ ˈ 훿p훿 ˈ [𝑢 p1 (P1)uPP1 (1) ...... [ 𝑢 p1 푁! 푝 푝 푝 (PN)𝑢푃1푃푁(N)] ……….(9) Every term is in P1 sum will give the same contribution to the integral in equation (8). This we replace the above by N1 times, any one term in the P1 sum : 2 ∗ |ɸ̄ (1,...... ,11)| – ∑푝 훿p [𝑢 p1 (P1) up1 (1)] .... [𝑢푃푁푃(11)𝑢푃푁(푁)] 1 푖 = ∑ ʃp exp [pi. (ri-pri)+...... +pN. (r1-pr1) ………… (10) 푣ᴺ 푝 ℎ PG SEMESTER:- II, PAPER: - VI PHYSICS CHE522: STATISTICAL PHYSICS After substituting in equation(8) each momentum integral be expressed in term of the function ʃ푑푝3 푒−훽(푝2⁄2푚)+푝−푟⁄ℎ 2 2 f(r) = =푒−휋푟 /λ ……………(11) ʃ푑푝3푒−훽(푝2⁄2푚) Where r≃|r| and 휆=√2휋ħ²/푚푘푇, the thermal wavelength then the result is given as 1 -βk β 2 Tr e = 푛! ʃ dp³ᴺdr³ᴺe⁻ (p1²+...... +pn )/2m*∑푝 ʃp [f(rn- ℎ3ᴺ pr)……..f(rn-prn ) ] ……(12) For high temperature 2 ∑푝 훿푝[f(r1-pr1)….f(rN-pN)]=1± ∑푖<푗 푓푖푗 +∑푖푗푘 푓푖푗푓푗푘푓푘푗 ± ……(13) Where fij ⋍ 푓(ri-rj) and where the plus sign applies to bosons and the minus sign to ferimions. At high temperatures |ri-rj| >>휆 , then since inter particle distance becomes very greater than thermal wavelength and we have 2 1± ∑푖<푗 푓 ij = exp (-β∑푗<푖 𝑣̄ ij)...... (15) Where v ij = -kT log (1±푓²ij)= -kT log 2 −2휋(푟 −푟 ) [1±푒푥푝( 푖 푗 )]...... (16) ƛ2 +sign is for bosons and –sign is for fermions. Hence, applying equation (16), over (14), the improvement equation is: 1 푝푖² Tr e-βk = ʃ dp³ᴺ dr³ᴺ exp [ -β(∑ + ∑ 𝑣̄ ij) ] ...... (17) 푛!ℎ³ᴺ 푖 2푚 푖<푗 PG SEMESTER:- II, PAPER: - VI PHYSICS CHE522: STATISTICAL PHYSICS The above equation shows that the first quantum correction to the partition function of an ideal gas has the same effect, v (r). The potential v (r) is attractive for bosons and respulsive for fermions, an shown below: U(r) ← for fermions h/ √mkT 0 ↑ for bosons Again v (r) depends on the temperature and thus can not be -kTregarded log2 as a true inter particle potential. ∑ 푝 contains N! term, permutation P=1 and [f(0)]ᴺ=1. The term interchanges ri and rj is |(ri-rj)|2 1 = ʃ dp³ᴺ dr3N푒−훽(훽2 +………,푝 2)/2푚 푛!ℎ³ᴺ 1 푁 Neglecting higher terms in equation (13) we have, ∑ 2 1+ 푖<푗 푓² ij = Πi PG SEMESTER:- II, PAPER: - VI PHYSICS CHE522: STATISTICAL PHYSICS 1 3푁 푝푖² Tr e-βk = ʃ푑푝 푑푟³ᴺ푒푥푝 − 훽 (∑ + ∑ 𝑣̄̅ij) 푛!ℎ³ᴺ 푖 2푚 푖<푗 Ensemble density, 휌(p,q)=e-H(p,q)/kT Where H (p,q) = E The volume in Γspace occupied by the canonical ensemble is called the partition function. 푑푝³ᴺ 푑푞³ᴺ Qn (V,T) = ʃ . 푒⁻βH(p,q) 푁!ℎ³ᴺ 1 Where β= , and we have introduced a constant h, to make QN 푘푇 1 dimension less. Has been used for correct Boltzmann counting. 푁! Fermi Energy The Fermi level lies just middle of the band gap energy at absolute zero temperature in case of a semiconductor or in an insulator at all temperatures. It is an unoccupied state of electron. The Fermi energy is an energy required for an electron in the valence band to escape in the conduction band. At T=0, all single particle states up to the energy equivalent to the Fermi energy and are completely filled with one particle in each state according to Paulis exclusion principle. Thus all single particle states with greater the Fermi energy are empty. Thus Fermi energy is the energy of the top most occupied level T=0 such a behaviour can not be seen in bose gas. We considered an ideal Fermi gas, which obey F.D. statistics. For such a gas the number of quantum states with energy E and E+dE is given by 푔(퐸)푑퐸 g(E) dE= ...... (1) 푍−1푒훽퐸−1 1 Where β= and Z is the fugacity of the gas related to the chemical 푘푇 potential μ through the formula Z=eβE. ……….(2) PG SEMESTER:- II, PAPER: - VI PHYSICS CHE522: STATISTICAL PHYSICS The energy of the particle is the kinetic energy of its translation 푃² motion, which is given by E= ………..(3) 2푀 The number of quantum states, considering the spins of the particle causing g=(2s+1)-fold degeneracy, we have 4훱푔푉 g(E) dE = . p³dp...... (4) ℎ³ Hence the number of quantum states with the energy range E and E+dE,; wih equation (3) gives 2훱푔푣 g(E) dE = (2m)3/2 E1/2 dE...... (5) ℎ³ The mean occupation number of the single particle state E is 1 PG SEMESTER:- II, PAPER: - VI PHYSICS CHE522: STATISTICAL PHYSICS The limiting chemical potential μ0 is called the Fermi energy EF of the system. Using equation (10/at T=0, equation (8) reduces to 1 2휋푔푉 ⁄ 퐸퐹 N= .(2푚)3 2. ∫ 퐸2dE ℎ3 0 (4푔푉) = × (2푚)3⁄2 × 퐸 3⁄2 ……….(11A) 3ℎ3 퐹 Hence Fermi Energy EF is given as 3푁 ℎ² 3푝 ℎ² EF=( )2/3. ( ) = ( )2/3×( )...... (11) 4휋푔푉 2푚 4휋푔 2푚 Where 휌=N/V, is the particle density. Thus the Fermi energy EF depend on the particle density 휌 and on the mass m of the particle. The Femi temperature TF is defined by TF= kTF...... (12) Since the electrons have spine half, we say these electrons as an ideal Fermi gas obeying the F.D statistics. Then the Fermi energy EF of the electron gas, with g=2 is given by equation (11) as , 3휋²푛 ℎ² ℎ² EF=( )2/3, = (3²휌)2/3 ...... (13) 푉 2푚 2푚 Substituting equation (10) in equation (9), the total ground state energy of the gas is given by: 2훱푔푉 퐸퐹 3⁄2 4휋푔푣 U0 = (2m)3/2 . ∫ 퐸 dE = . (2m)3/2 EF5/2...... (14) ħ³ 0 5ℎ³ From, equation (11a) and equation (14), we get. 푉0 3 = . EF...... (15) 푁 5 Thus the average energy per particle of Fermi Gas at T=0 is 3/5 times the Fermi energy EF. Hence the ground state equation of state of the gas at T=0 is given as: PG SEMESTER:- II, PAPER: - VI PHYSICS CHE522: STATISTICAL PHYSICS 2 푈표 2 푁퐸퐹 2 Po= ∙ = = ∙ 휌 ∙ 퐸 ………….(16) 3 푉 5 푉 5 퐹 Now substituting the value of EF ,we have 2 3 P0 =( ). ( 휋푔)2/3 . (h2/2m) . 휌5/3...... (17) 5 4 Hence:- P0=α 휌5/3...... (18) Thus the pressure of a Fermi gas at T=0 is proportional to the 5/3 power of the density. Thermodynamical behaviour of an ideal Fermi gas Thermodynamical behaviour of an ideal Fermi gas we shall study the different thermo. Dynamical behaviour of the ideal Fermi gas for an Fermi gas. For an ideal Fermi gas 푃푉 1 =∑ 푙표푔(1+Ze-βe) and N=∑ ...... (1) 푘푇 푒 푒 푍−1푒훽퐸+1 휇ˈ Where Z=exp and β = 1/kT. 푘푇 Replacing summation over E by corresponding integration equation (11) becomes. 푝 푔 푁 푔 = f5/2 (Z), and = f3/2(Z)...... (2) 푘푇 ƛ³ 푉 휆3 Where g is a weight factor arising from the internal structure of the particle (such as spin) , ƛ is the mean thermal wavelength of the particle ℎ ƛ= ...... (3) (2휋푚푘푇)1/2 F휈 (Z) are Fermi Dirac function and given by PG SEMESTER:- II, PAPER: - VI PHYSICS CHE522: STATISTICAL PHYSICS 1 ∞ (푥휈−1)푑푥 푧² 푧³ F휈 (Z) = ∫ = = 푧 − + ...... (4) Γ(푥) 0 푍−1푒푥+1 푧푣 3푣 The internal energy U of the Fermi Gas is given by 3 푔휈 U= 푘푇 푓5 (푧) 2 휆3 ⁄2 3 푓5⁄ (푧) = 푁푘푇 2 …………(5) 2 푓3 (푧) ⁄2 2 푈 This system satisfies the relationship , p= ……….(6) 3 푉 The specific heat Cv of the gas can be obtained by differentiating (5) w.r.t. to T, keeping N and V Constant and using relation 푓3 (푧) 1 훿푍 3 ⁄2 ( )v = - ...... (7) 푍 훿푇 2푇 푓1 (푧) ⁄2 퐶푣 15 푓5/2(푍) 9 푓3/2(푍) The final result is = , − , ...... (8) 푁푘 4 푓3/2(푍) 4 푓1/2(푍) For the helmhotz free energy of the gas, we get 푓5/2(푍) A=Nμ-Pv = NkT {log z – }...... (9) 푓3/2(푍) And for the entropy, 푈−퐴 5푓5/2(푍) S= =Nk { -logz}...... (10) 푇 2푓3/2(푍) In the case of very low density of gas and or very high temperature .Thus we have highly non-degenerate, 푛ℎ³ F3/2 (Z) = <<1...... (11) 푔(2훱푀푘푇)3/2 When 2<<1, fu(Z) = Z, 푁퐾푇 3 3 P= , U= NKT, Cv= NK...... (12) 푉 32 2 푛ƛ 5 ℎ휆3 A= NKT {log -1} and S=NK { - log ( )}...... (13) 푔 2 푔 PG SEMESTER:- II, PAPER: - VI PHYSICS CHE522: STATISTICAL PHYSICS The equation of state takes the form of the virial expn 푃푉 ∞ ƛ3 =∑ (−1)l-1al( )l-1...... (14) 푁푘푇 푙=1 푔푣 Where 휈=1/n, is the volume per particle, the coefficient al are referred to as virial co-efficient of the system, we obtain sp.heat 3 ∞ 푙−1 (5−3푙) 3 푙−1 Cv= 푁푘 ∑ (−1) 푎 (휆 /푔휈) 2 푙=1 2 푙 ƛ³ ƛ³ ƛ³ Cv=3/2 Nk[1-0.0884( )+0.0066( )2-0.0004 ( )+...... ]...... (15) 푔푣 푔푣 푔푣 Thus at finite temperature, the sp. heat of gas is smaller than its limiting values 3/2 NK. ℎƛ³ If >>i, we have degenerate gas, where nƛ3/푔−훼 , then in the 푔 limit T→0, the mean occupation hambers of the single particle state E(p) becomes, 1 Where μ0 is the chemical potential of the system at T=0, the function 1.0 ↑ (ex-ɛ + 1)-1 0 X→ ɛ PG SEMESTER:- II, PAPER: - VI PHYSICS CHE522: STATISTICAL PHYSICS Thus at T=0, all single particle states upto E=μ0 are completely filled with one particle per state, where all single particle states with E>μ0 are empty. The limiting energy μ0 is referred to as the Fermi energy of the system and is denoted by Ef. The density of states of the system, a(e) is given as: 푔푣 푑푝 A(E) = . 4πp² ...... (17) ℎ³ 푑퐸 4훱푔푣 3푁 Where, N= ×P3f, Pf = ( )1/3...... (18) 3ℎ³ 4훱푔푣 In the non-relativistic case, 3푁 ℎ² 6훱²ℎ ℎ² Ef=( )2/3. =( )...... (19) 4훱푔푣 2푚 푔 2푚 The ground state or zero point energy of the system is given by 4훱푔푣 푝푓 푝² 2훱푔푣 E0= ∫ ( ) p²dp = . P5f ...... (20) ℎ³ 0 2푚 5푚ℎ³ 2 퐸0 3푝 퐹 3 Where, = = EF ...... (21) 푁 10푚 5 The ground state pressure of the system is given by 퐸0 2 P0 = 2/3( ) = nEF ...... (22). 푉 5 Substituting value of EF, we get 6훱² ℎ² p0 = ( )2/3 . n5/3 ...... (23) 푔 5푚 At-T = oK, the particle constituting the system can not settle down into a single energy state. From the fig. 1 we conclude that the thermal excitation of the particle occurs only in a narrow energy range which is located around the energy value E=μo and has a width o(kT) accordingly, the sp. PG SEMESTER:- II, PAPER: - VI PHYSICS CHE522: STATISTICAL PHYSICS 푁퐾,퐾푇 Heat of the system will be 0( ). Thus the low temperature sp. 퐸퐹 heat of a Fermi system differs from the classical value 3/2 NK. For the value of v, we have to the first approximation. 8 5훱² F5/2 (Z) = (logZ)5/2 [ 1+ (logZ)⁻²+...... ] 5훱ᴵᴵ² 8 4 훱² F3/2 (Z) = (logZ)3/2 [1+ (logZ)⁻²+...... ] 3훱ᴵᴵ² 8 2 훱² F1/2 (Z) = (logZ)1/2 [1- (logZ)-2 +...... ] 훱ᴵᴵ² 24 Substituting above equation into (2), we obtain :- 푁 4훱푔 2푚 훱² = ( )3/2 . (KT log z)3/2 [1+ (log2)⁻²+...... ] 푉 3 ℎ² 8 In the zeroth approximation, this gives 3푁 ℎ² KT log z = μ =( )2/3...... (25) 4훱푔푣 2푚 Which is identical with the ground state result µ0=EF. In the next approximation, we obtain 훱² 퐾푇 KT log z= μ = EF . [ 1- ( )²] ...... (26) 12 퐸퐹 Finally we have 푣 3 5훱² 퐾푇 = 퐸F [ 1+ ( )²+...... ] ...... (27) 푁 5 12 퐸퐹 The pressure of the gas is then given by 2 푢 2 5훱² 퐾푇 P= = = he [ 1+ ( )²+...... ] ...... (28) 3 푣 5 12 퐸푝 From the temperature dependent part of (27), we have temperature spiheat of the gas. PG SEMESTER:- II, PAPER: - VI PHYSICS CHE522: STATISTICAL PHYSICS 푐 훱²퐾푇 = + ...... (29) 푁퐾 2퐸푓 Is the Fermi temperature of the system, the sp. Heat varies as the first power of temperature, the overall variation of cv 1.5 cv/nk ↑ 0 T/TF 2 → 3 With T is shown in fig:-2 the helmholty free energy of the system follows directly from equation (19) and (22) 퐴 푃푣 =μ- 푁 푁 With the help equation (20) 푈 3 5훱² 퐾푇 = Ef [ 1+ ( )²+...... ] 푛 5 12 퐸푝 The pressure of μe gas in the given by 2 푢 2 5훱2 퐾푇 P= = ℎ 푒푓 [1 + ( )2 + ...... ] ...... (28) 3 푣 5 12 퐸푓 From the temperature dependent part of (27), we have low temperature specific heat of the gas. 퐶 훱²푘푇 푣 = + ...... (29) 푁푘 2퐸퐹 PG SEMESTER:- II, PAPER: - VI PHYSICS CHE522: STATISTICAL PHYSICS 퐸퐹 Thus for T< PG SEMESTER:- II, PAPER: - VI PHYSICS CHE522: STATISTICAL PHYSICS