1. H  E  Schr Ö dinger Equation Hamiltonian H =   (  h 2 /2m      h 2 /2m e )  i  i 2 +     Z  Z  e   r   i     e 2 /r i   i  j  e 2 /r ij Wavefunction Energy

2. Statistical Thermodynamics and Kinetic Theory Driving force of chemical reactions Boltzmann’s distribution Link between quantum mechanics & thermodynamics

3. Contents: Distribution of Molecular States Partition Function Perfect Gas Fundamental Relations Diatomic Molecular Gas Non-ideal Gas Maxwell’s Demon Quantum Statistics Transition State Theory Text Book: Atkins, “Physical Chemistry” (10 th Ed.)

4. Principle of equal a priori probabilities: All possibilities for the distribution of energy are equally probable provided the number of molecules and the total energy are kept the same Democracy among microscopic states !!! a priori: as far as one knows

5. For instance, four molecules in a three-level system: the following two conformations have the same probability. ---------l-l-------- 2  ---------l--------- 2  ---------l----------  ---------1-1-1----  ---------l---------- 0------------------- 0

6. THE DISTRIBUTION OF MOLECULAR STATES Consider a system composed of N molecules, and its total energy E is a constant. These molecules are independent, i.e. no interactions exist among the molecules.

7. Configurations and Weights Imagine that there are total N molecules among which n 0 molecules with energy  0, n 1 with energy  1, n 2 with energy  2, and so on, where  0 <  1 <  2 <.... are the energies of different states. The specific distribution of molecules is called configuration of the system, denoted as { n 0, n 1, n 2,......}

8. {N, 0, 0,......} corresponds that every molecule is in the ground state, there is only one way to achieve this configuration; {N-2, 2, 0,......} corresponds that two molecule is in the first excited state, and the rest in the ground state, and can be achieved in N(N-1)/2 ways. A configuration { n 0, n 1, n 2,......} can be achieved in W different ways, where W is called the weight of the configuration. And W can be evaluated as follows, W = N! / (n 0 ! n 1 ! n 2 !...)

9. 2.Calculate the number of ways of distributing 20 different molecules among six different states with the configuration {1, 0, 3, 5, 10, 1}. Answer: 20! / 1! 0! 3! 5! 10! 1! = 931170240 note: 0! = 1

10. Stirling’s Approximation: When x is large, ln x!  x ln x - x xln x! x ln x - xln A 10.000-1.000 0.081 20.693 -0.614 0.652 4 3.178 1.545 3.157 8 10.605 8.636 10.595 16 30.672 28.361 30.666 20 42.336 39.915 42.332 30 74.658 72.036 74.656 note, A = (2  ) 1/2 (x+1/2) x e -x ln W  ( N ln N - N ) -  ( n i ln n i - n i ) = N ln N -  n i ln n i

11. The Dominating Configuration Imagine that N molecules distribute among two states. {N, 0}, {N-1, 1},..., {N-k, k},..., {1, N-1}, {0, N} are possible configurations, and their weights are 1, N,..., N! / (N-k)! k!,..., N, 1, respectively. For instance, N=8, the weight distribution is then

15. When N is even, the weight is maximum at k = N/2, W k=N/2 = N! / [N/2)!] 2. When N is odd, the maximum is at k = N/2  1 As N increases, the maximum becomes sharper! The weight for k = N/4 isW k=N/4 = N! / [(N/4)! (3N/4)!] | N | 4 8 16 32 256 6.0 x 10 23 |R(N) | 1.5 2.5 7.1 57.1 3.5 x 10 14 2.6 x 10 3e+22 The ratio of the two weights R(N)  W k=N/2 / W k=N/4 is equal to (N/4)! (3N/4)! / [(N/2)!] 2

16. Therefore, for a macroscopic molecular system ( N ~ 10 23 ), there are dominating configurations so that the system is almost always found in the dominating configurations, i.e. Equilibrium Dominating Configuration: Equilibrium Configuration

17. To find the most important configuration, we vary { n i } to seek the maximum value of W. But how? One-Dimensional Function: F(x) = x 2 dF/dx = 0 X

18. Two-Dimensional Case: for instance, finding the minimum point of the surface of a half water melon F(x,y).  F/  x = 0,  F/  y = 0. Multi-Dimensional Function: F(x 1, x 2, …, x n )  F/  x i = 0, i = 1,2,…,n

19. 1. The total energy is a constant, i.e.  n i  i = E = constant 2. The total number of molecules is conserved,  n i = N = constant How to maximize W or lnW ? How to maximize a function ? To find the maximum value of W or lnW,  lnW /  n i = 0,i=1,2,3,... ???

20. Let’s investigate the water melon’s surface: cutting the watermelon how to find the minimum or maximum of F(x, y) under a constraint x = a ? L = F(x, y) - x  L /  x = 0  L /  y = 0 x = a

21. Generally, to minimize or maximize a function F(x 1, x 2, …, x n ) under constraints, C 1 (x 1, x 2, …, x n ) = Constant 1 C 2 (x 1, x 2, …, x n ) = Constant 2. C m (x 1, x 2, …, x n ) = Constant m L = F(x 1, x 2, …, x n ) -  i i C i (x 1, x 2, …, x n )  L/  x i = 0,i=1,2,..., n

22. JUSTIFICATION dL = dF -  i i dC i under the constraints, dC i = 0, thus dF = 0 i.e., F is at its maximum or minimum.

23. This method is called the method of Lagrangian Multiplier, or Method of undetermined multipliers Procedure Construct a new function L, L = lnW +   i n i -   i n i  i Finding the maximum of L by varying { n i },  and  is equivalent to finding the maximum of W under the two constraints, i.e.,  L/  n i =  lnW/  n i +  -  i = 0

24. Since ln W  ( N ln N - N ) -  i ( n i ln n i - n i ) = N ln N -  i n i ln n i  lnW/  n i =  (N ln N)/  n i -  (n i ln n i )/  n i = - ln (n i /N) Therefore, ln (n i / N) +  -  i = 0 n i / N = exp(  -  i )

25. The Boltzmann Distribution P i = exp (  -  i ) Interpretation of Boltzmann Distribution 1 =  i n i / N =  i exp(  -  i ) exp(  ) = 1 /  i exp(-  i )  = - ln [  i exp(-  i )]  > 0, more molecules occupying the low energy states.

26. Physical Meaning of  Imagine a perfect gas of N molecules. Its total energy E =  i n i  i =  i N exp(-  i )  i /  i exp(-  i ) = -Nq -1 dq /d  = -N dlnq / d  where q =  i exp(-  i ), n i and  i are the population and energy of a state i, respectively. It will be shown later that q = V /  3 and  = h (  /2  m) 1/2 [  is called the thermal wavelength] dlnq/d  = -3dln  /d  = -3/2 

27. Therefore, E = N = 3N/2 , where is the average kinetic energy of a molecule. Therefore, = = 3/2 . On the other hand, according to the Maxwell distribution of speed, the average kinetic energy of a molecule at an equilibrium,

28.  : the reciprocal temperature 1 /  = kT where k is the Boltzmann constant. Physical Meaning of 

29. Example 1: Consider a molecular whose ground state energy is -10.0 eV, the first excited state energy -9.5 eV, the second excited state energy -1.0 eV, and etc. Calculate the probability of finding the molecule in its first excited state T = 300, 1000, and 5000 K.

30. Answer The energies of second, third and higher excited states are much higher than those of ground and first excited states, for instance, at T = 5,000 K, the probability of finding the first excited state is, p 2 = exp(  -  2 ) with  2 = -9.5 eV And, the probability of finding the second excited state is, p 3 = exp(  -  3 ) with  3 = -1.0 eV

31. the ratio of probabilities between second & first excited states is, exp[-(-1.0+9.5) x 11600/5000] = exp(-20) [ 1 eV = 1.60 x 10 -19 J; k = 1.38 x 10 -23 J K -1 ; 1 eV  11600 k K -1 ] i.e., compared to the first excited state, the chance that the molecule is in second excited state is exceedingly slim. Thus, we consider only the ground and first excited states, a two-state problem.

32. The probability of finding the molecule in the first excited state is p = exp(-  2 ) / [exp(-  1 ) + exp(-  2 )] where,  1 = -10.0 eV, and  2 = -9.5 eV (1) T = 300 K

33. (2) T = 1000 K (3) T = 5000 K

34. Molecular Partition Function

56. Fundmental Thermodynamic Relationships

63. Example 7 Calculate the translational partition function of an H 2 molecule confined to a 100-cm 3 container at 25 o C Example 8 Calculate the entropy of a collection of N independent harmonic oscillators, and evaluate the molar vibraitional partition function of I 2 at 25 o C. The vibrational wavenumber of I 2 is 214.6 cm -1 Example 9 What are the relative populations of the states of a two-level system when the temperature is infinite? Example 10 Evaluate the entropy of N two-level systems. What is the entropy when the two states are equally thermally accessible?

64. Example 11 Calculate the ratio of the translational partition functions of D 2 and H 2 at the same temperature and volume. Example 12 A sample consisting of five molecules has a total energy 5 . Each molecule is able to occupy states of energy j  with j = 0, 1, 2, …. (a) Calculate the weight of the configuration in which the molecules share the energy equally. (b) Draw up a table with columns headed by the energy of the states and write beneath then all configurations that are consistent with the total energy. Calculate the weight of each configuration and identify the most probable configuration. Example 13 Given that a typical value of the vibrational partition function of one normal mode is about 1.1, estimate the overall vibrational partition function of a NH 3.

65. Diatomic Gas Consider a diatomic gas with N identical molecules. A molecule is made of two atoms A and B. A and B may be the same or different. When A and B are he same, the molecule is a homonuclear diatomic molecule; when A and B are different, the molecule is a heteronuclear diatomic molecule. The mass of a diatomic molecule is M. These molecules are indistinguishable. Thus, the partition function of the gas Q may be expressed in terms of the molecular partition function q, The molecular partition q where,  i is the energy of a molecular state I, β=1/kT, and  ì is the summation over all the molecular states.

67. Factorization of Molecular Partition Function The energy of a molecule j is the sum of contributions from its different modes of motion: where T denotes translation, R rotation, V vibration, and E the electronic contribution. Translation is decoupled from other modes. The separation of the electronic and vibrational motions is justified by different time scales of electronic and atomic dynamics. The separation of the vibrational and rotational modes is valid to the extent that the molecule can be treated as a rigid rotor.

68. The translational partition function of a molecule  ì sums over all the translational states of a molecule. The rotational partition function of a molecule  ì sums over all the rotational states of a molecule. The vibrational partition function of a molecule  ì sums over all the vibrational states of a molecule. The electronic partition function of a molecule  ì sums over all the electronic states of a molecule. where

69. Vibrational Partition Function Two atoms vibrate along an axis connecting the two atoms. The vibrational energy levels: If we set the ground state energy to zero or measure energy from the ground state energy level, the relative energy levels can be expressed as 5--------------5hv 4--------------4hv 3--------------3hv 2--------------2hv 1--------------hv 0--------------0 n= 0, 1, 2, ……. kT 

70. Then the molecular partition function can be evaluated Therefore, Consider the high temperature situation where kT >>hv, i.e., Vibrational temperature  v High temperature means that T>>  v    F 2 HCl H 2  v /K 309 1280 4300 6330 v/cm -1 215 892 2990 4400 where

71. Rotational Partition Function If we may treat a heteronulcear diatomic molecule as a rigid rod, besides its vibration the two atoms rotates. The rotational energy where B is the rotational constant. J =0, 1, 2, 3,… where g J is the degeneracy of rotational energy level ε J R Usually hcB is much less than kT, =kT/hcB  h/8cI  2 c: speed of light I: moment of Inertia  hcB<<1 Note: kT>>hcB

72. For a homonuclear diatomic molecule Generally, the rotational contribution to the molecular partition function, Where  is the symmetry number. Rotational temperature  R 

73. Electronic Partition Function =g 0 =g E where, g E = g 0 is the degeneracy of the electronic ground state, and the ground state energy  0 E is set to zero. If there is only one electronic ground state q E = 1, the partition function of a diatomic gas, At room temperature, the molecule is always in its ground state

74. Mean Energy and Heat Capacity The internal energy of a diatomic gas (with N molecules) (T>>1) Contribution of a molecular to the total energy Translational contribution (1/2)kT x 3 = (3/2)kT Rotational contribution (1/2)kT x 2 = kT Vibrational contribution (1/2)kT + (1/2)kT = kT kinetic potential the total contribution is (7/2)kT q V = kT/hv q R = kT/hcB The rule: at high temperature, the contribution of one degree of freedom to the kinetic energy of a molecule (1/2)kT

75. the constant-volume heat capacity (T>>1) Contribution of a molecular to the heat capacity Translational contribution (1/2) k x 3 = (3/2) k Rotational contribution (1/2) k x 3 = k Vibrational contribution (1/2) k + (1/2) k = k kinetic potential Thus, the total contribution of a molecule to the heat capacity is (7/2) k

76. Summary Principle of equal a priori probabilities: All possibilities for the distribution of energy are equally probable provided the number of molecules and the total energy are kept the same. A configuration { n 0, n 1, n 2,......} can be achieved in W different ways or the weight of the configuration W = N! / (n 0 ! n 1 ! n 2 !...) Dominating Configuration vs Equilibrium The Boltzmann Distribution P i = exp (-  i ) / q

77. Partition Function q =  i exp(-  i ) =  j g j exp(-  j ) Q =  i exp(-  E i ) Energy E= N  i p i  i = U - U(0) = - (  lnQ/  ) V Entropy S = k lnW = - Nk  i p i ln p i = k lnQ + E / T A= A(0) - kT lnQ H = H(0) - (  lnQ/  ) V + kTV (  lnQ/  V) T Q = q N or (1/N!)q N

78. Non-Ideal Gas Now let’s derive the equation of state for real gases. Consider a real gas with N monatomic molecules in a volume V. Assuming the temperature is T, and the mass of each molecule is m. So the canonical partition function Q can be expressed as Q =  i exp(-E i / kT) where the sum is over all possible state i, and E i is the energy of state i.

79. In the classical limit, Q may be expressed as Q =(1/N!h 3N )  …  exp(-H / kT) dp 1 … dp N dr 1 … dr N where, H = (1/2m)  i p i 2 +  i>j V(r i,r j ) Q = (1/N!) (2  mkT / h 2 ) 3N/2 Z N Z N =  …  exp(-  i>j V(r i,r j ) / kT) dr 1 … dr N [ note: for ideal gas, Z N = V N, and Q = (1/N!) (2  m kT / h 2 ) 3N/2 V N ] Z N = V N Q = (1/N!) (2  m kT / h 2 ) 3N/2 V N

80. The equation of state may be obtained via p = kT(  lnQ/  V) T We have thus, p / kT = (  lnQ/  V) T = (  lnZ N /  V) T (  lnZ N /  V) T Z N =  …  { 1 + [ exp(-  i>j V(r i,r j ) / kT) - 1 ] } dr 1 … dr N = V N +  …  [ exp(-  i>j V(r i,r j ) / kT) - 1 ] dr 1 … dr N  V N + (1/2) V N-2 N(N-1)  [ exp(- V(r 1,r 2 ) / kT) - 1 ] dr 1 dr 2  V N { 1 - (1/2V 2 ) N 2  [ 1 - exp(- V(r 1,r 2 ) / kT) ] dr 1 dr 2 } = V N { 1 - B N 2 / V } where, B = (1/2V)  [ 1 - exp(- V(r 1,r 2 ) / kT) ] dr 1 dr 2 B = (1/2V)  [ 1 - exp(- V(r 1,r 2 ) / kT) ] dr 1 dr 2

81. Therefore, the equation of state for our gas: p / kT = N / V + (N / V) 2 B = n + B n 2 Comparison to the Virial Equation of State The equation of state for a real gas P / kT = n + B 2 (T) n 2 + B 3 (T) n 3 + … This is the virial equation of state, and the quantities B 2 (T), B 3 (T), … are called the second, third, … virial coefficients.

82. Thus,

83. A. HARD-SPHERE POTENTIAL  r 12 <  U(r 12 ) = 0r 12 >  B 2 (T) = (1/2)  0  4  r 2 dr = 2  3 /3 = (1/2)  0  4  r 2 dr = 2  3 /3

84. B. SQUARE-WELL POTENTIAL  r 12 <  U(r 12 ) = -  < r 12 <  0r 12 >  B 2 (T) = (1/2)  0  4  r 2 dr = (2  3 /3) [1 - ( 3 -1) ( e  - 1 )] = (1/2)  0  4  r 2 dr

85. C. LENNARD-JONES POTENTIAL U(r) = 4  [ (  /r) 12 - (  /r) 6 ] Withx =  /r, T * = kT /  =  0  { 1 - exp[(-4/T * ) ( x 12 - x 6 )] } x 2 dx B 2 (T) = ( )  0  { 1 - exp{( ) [ ]} } 4  r 2 dr

86. Maxwell ’ s Demon

87. Second Law of Thermodynamics ΔS > 0 ΔS = ΔQ / T S = k B logW

89. http://www.youtube.com/watch?v=k3N5vtauD QU

90. Maxwell ’ s Demon (1867)

91. Thermal Fluctuation (Smolochowski, 1912) In his talk “Experimentally Verifiable Molecular Phenomena that Contradict Ordinary Thermodynamics”,… Smoluchowski showed That one could observe violations of almost all the usual statements Of the second law by dealing with sufficiently small systems. … the increase of entropy… The one statement that could be upheld… was the impossibility of perpetual motion of the second kind. No device could be ever made that would use the existing fluctuations to convert heat completely into work on a macroscopic scale … subject to the same chance fluctuations…. -----H.S. Leff & A.F. Rex, “Maxwell’s Demon”

99. Szilard ’ s one-molecule gas model (1929) To save the second law, a measure of where-about of the molecule produces at least entropy > k ln2

100. Measurement via light signals (L. Brillouin, 1951) h  k T A Temporary Resolution !!!

101. Mechanical Detection of the Molecule (Rothstein, 1974) Counter-clockwise rotation always !!! A Perpetual Machine of second kind?

102. Bennett ’ s solution (1982) Demon’s memory To complete thermodynamic circle, Demon has to erase its memory !!! Memory eraser needs minimal Entropy production of k ln2 (R. Landauer, 1961)

103.  rate Feynman ’ s Ratchet and Pawl System (1961) T 1 =T 2, no net rotation T 1 > T 2, counter-clockwise rotation T 1 < T 2, clockwise rotation Mechanical Rectifier Feynman’s Lecture Notes

104. A honeybee stinger potential coordinate

106. A two-chamber design: an analogy to Feynman’s Ratchet and Pawl Feynman’s Lecture Notes Our two-chamber design string

107. Potential of the pawl string radian potential

109. Movie of Feynman ’ s ratchet-pawl system

110. July 1998GunaHua CHEN @copyright

112. Determination of temperature at equilibrium

113. Simulation results The ratchet (B) moves when the pawl (L) is cooled down

117. Micro-reversibility Pawl a transition state

118. Density of distribution in the phase space  q 1 …q f,p 1 …p f )  q 1  q f  p 1  p f Liouville’s Theorem: d  /dt = 0 Coarse-grained density over  q 1  q f  p 1  p f at  q 1 …q f,p 1 …p f ) : P =  …   q 1 …q f,p 1 …p f ) dq 1  dq f dp 1... dp f /  q 1  q f  p 1  p f Boltzmann’s H: H =  …  P log P dq 1  dq f dp 1... dp f where  q 1 …q f,p 1 …p f ) is fine-grained density at  q 1 …q f,p 1 …p f )

119. d(  …   log  dq 1  dq f dp 1...dp f )/dt = 0 Q =  log  -  log P -  + P  0 At t 1,  1  P 1 H 1 =  …   1 log  1 dq 1  dq f dp 1... dp f At t 2,  2  P 2 H 2 =  …  P 2 log P 2 dq 1  dq f dp 1... dp f H 1 - H 2  0 Boltzmann’s H-Theorem

120. Double-walled Nano-Oscillators MD simulaton of DWNT Oscillator: Zhao, Ma, Chen & Jiang, PRL, 2003

121. OscillationHibernationRevival

122. Entropy [Q: Partition Function] S = k lnW = - Nk  i p i ln p i = k lnQ - (  lnQ/  ) V / T Revival Hibernation

124. Fluctuations Theorems (FT) Evans-Searles FT – systems evolving from equilibrium toward an NSS: Entropy production function

126. Quantum Statistics Reference: Grasser & Richards, “An Introduction to Statistical Thermodynamics”

127. Ensemble An ensemble is a collection of systems. A Thought Experiment to construct an ensemble To set up an ensemble, we take a closed system of specific volume, composition, and temperature, and then, replicate it A times. We have A such systems. The collection of these systems is an ensemble. The systems in an ensemble may or may not exchange energy, molecules or atoms.

131. Microcanonical Ensemble: N, V, E are common; Canonical Ensemble: N, V, T are common; Grand Canonical Ensemble: , V, T are common. Microcanonical System: N, E are fixed; Canonical System: N is fixed, but E varies; Grand Canonical System: N, E vary.

132. Example: What kind of system is each of the following systems: (1) an isolated molecular system; (2) an equilibrium system enclosed by a heat conducting wall; (3) a river; (4) a system surrounded by a rigid and insulating material.

133. Distribution of a Microcanonical Ensemble State123…k… EnergyEEE…E… Occupationa 1 a 2 a 3 …a k … W = A! / a 1 !a 2 ! a 3 !… Constraint  i a i = A

134. To maximize lnW under the constraint, we construct a Lagrangian L = lnW +   i a i Thus, 0 =  L/  a i =  lnW/  a i + 

135. the probability of a system being found in state i, p i = a i /A = exp(  ) = constant or, in another word, the probabilities of all states with the same energy are equal. Utilizing the Stirling’s approximation, ln x! = x ln x - x  lnW/  a i = - ln a i /A = - ,

136. Distribution of a Canonical Ensemble State123…k… EnergyE 1 E 2 E 3 …E k … Occupationa 1 a 2 a 3 …a k … Constraints:  i a i = A  i a i E i =  where,  is the total energy in the ensemble. W = A! / a 1 !a 2 ! a 3 !…

137. To maximize lnW under the above constraints, construct a Lagrangian L = lnW +   i a i -   i a i E i 0 =  L/  a i =  lnW/  a i +  -  E i ln a i /A =  -  E i the probability of a system being found in state i with the energy E i, p i = a i /A = exp(  -  E i )

138. The above formula is the canonical distribution of a system. Different from the Boltzmann distribution of independent molecules, the canonical distribution applies to an entire system as well as individual molecule. The molecules in this system can be independent of each other, or interact among themselves. Thus, the canonical distribution is more general than the Boltzmann distribution. (note, in the literature the canonical distribution and the Boltzmann distribution are sometimes interchangeable).

139. Distribution of a Grand Canonical Ensemble State123…k… EnergyE 1 E 2 E 3 …E k … Mol. No.N 1 N 2 N 3 …N k … Occupationa 1 a 2 a 3 …a k …

140. Constraints:  i a i = A  i a i E i =   i a i N i = N where,  and N are the total energy and total number of molecules in the ensemble, respectively. W = A! / a 1 !a 2 ! a 3 !…

141. To maximize lnW under the above constraints, construct a Lagrangian L = lnW +   i a i -   i a i E i -   i a i N i 0 =  L/  a i =  lnW/  a i +  -  E i -  N i ln a i /A =  -  E i -  N i the probability of a system being found in state i with the energy E i and the number of particles N i, p i = a i /A = exp(  -  E i -  N i )

142. The above formula describes the distribution of a grand canonical system, and is called the grand canonical distribution. When N i is fixed, the above distribution becomes the canonical distribution. Thus, the grand canonical distribution is most general.

143. Quantum Statistics Quantum Particle: Fermion(S = 1/2, 3/2, 5/2,...) e.g. electron, proton, neutron, 3 He nuclei Boson(S = 0, 1, 2,...) e.g.deuteron, photon, phonon, 4 He nuclei Pauli Exclusion Principle: Two identical fermions can not occupy the same state at the same time. Question: what is the average number particles or occupation of a quantum state?

144. Fermi-Dirac Statistics System: a fermion’s state with an energy  (   -  /  ) ------------------ --------l--------- occupation n = 0 n = 1 energy 0  probability exp(0) exp[-  (  -  )]

145. There are only two states because of the Pauli exclusion principle. Thus, the average occupation of the quantum state , 1 / {exp[  (  -  )] + 1}

146. Therefore, the average occupation number n(  ) of a fermion state whose energy is , n(  ) = 1 / {exp[  (  -  )] + 1}  is the chemical potential. When  = , n = 1/2 For instance, distribution of electrons

147. Bose-Einstein Statistics System: a boson’s state with an energy  Occupation of the system may be 0, 1, 2, 3, …, and correspondingly, the energy may be 0, , 2 , 3 , …. Therefore, the average occupation of the boson’s state,

148. 1 / {exp[  (  -  )] - 1} = Therefore, the average occupation number n(  ) of a boson state whose energy is , n(  ) = 1 / {exp[  (  -  )] - 1}

149. the chemical potential  must less than or equal to the ground state energy of a boson, i.e.    0, where  0 is the ground state energy of a boson. This is because that otherwise there is a negative occupation which is not physical. When  =  0, n(  )  , i.e., the occupation number is a macroscopic number. This phenomena is called Bose-Einstein Condensation! 4 He superfluid: when T  T c = 2.17K, 4 He fluid flows with no viscosity.

150. Classical or Chemical Statistics When the temperature T is high enough or the density is very dilute, n(  ) becomes very small, i.e. n(  ) > 1. Neglecting +1 or -1 in the denominators, both Fermi-Dirac and Bose-Einstein Statistics become n(  ) = exp[-  (  -  )] The Boltzmann distribution!

151. July 1998GunaHua CHEN @copyright

152. July 1998GunaHua CHEN @copyright

153. July 1998GunaHua CHEN @copyright

154. July 1998GunaHua CHEN @copyright

155. July 1998GunaHua CHEN @copyright

156. July 1998GunaHua CHEN @copyright