Part 2 Statistical Mechanics

1
Part 2Statistical Mechanics

• PHYS 4315
• R. S. Rubins, Fall 2009

2
Partition Function Z

• Starting from the fundamental postulate of equal
  a priori
• probabilities, the following are obtained
• the results of classical thermodynamics, plus
  their statistical underpinnings
• the means of calculating the thermodynamic
  parameters (U, H, F, G, S ) from a single
  statistical parameter, the partition function Z
  (or Q), which may be obtained from the
  energy-level scheme for a quantum system.
• The partition function for a quantum system in
  contact with a
• heat bath is
• Z ?i exp( ei /kT),
• where ei is the energy of the ith state.

3
Statistical Weight ? (or O)

• The partition function for a quantum system in
  contact
• with a heat bath is Z ?i exp( ei /kT), where
  ei is the
• energy of the ith state.
• The connection to the macroscopic thermodynamic
• function S is through the microscopic parameter O
  (or
• ?), which is known as thermodynamic degeneracy or
• statistical weight, and gives the number of
  microstates
• in a given macrostate.
• The connection between them, known as Boltzmanns
• principle, is S k ln?.
• (S k lnO is carved on Boltzmanns tombstone).

4
The Fundamental Postulate 1

• The foundation of statistical mechanics is the
  Fundamental
• Postulate
• an isolated system is equally likely to be in any
  of its accessible states.
• Quantum case
• Consider a system of N weakly interacting spin ½
  particles of
• magnetic moment m in a strong uniform magnetic
  field B.
• Each particle has a magnetic energy of mB or
  mB.
• If n particles are in the lower state, then the
  statistical weight is
• ? N!/n!(N n)! .

5
The Fundamental Postulate 2

• Classical case
• For non-discrete states, an artificial counting
  method is used.
• Let ?(U,V,N,a) be the number of microstates in
  the macrostate
• (V,N,a), with the energy lying in the range U to
  U ?U, with
• U gtgt ?U gtgt ?U, where ?U is the separation of
  quantum levels.
• a represents one or more variables, for example
  the energy of
• one part of an isolated system separated by a
  partition.
• For each particle of an ideal gas, counting is
  achieved by dividing
• phase space into elemental volumes, given by
  ?px?x ?.

6
Sharpness of Distribution 1

• The sharpness of the peak of the distribution
  ?(a) is given by ?a/ltagt,
• where ?a is the rms deviation ?lt(?a)2gt.
• We show later that ?a/ltagt 1/?N, where N is the
  number of particles.
• For N 1024, ?a/ltagt 1012, so the probability
  of obtaining a value of a other
• than the mean is negligibly small.
• If the particle starts in a state with an
  accessible value of a for which ?(a) is
• negligibly small, the fundamental postulate
  states that it will be found with
• essentially 100 probability of being at the
  value of a corresponding to ?max.
• Thus ? ? ?max , so that ?? ? 0 and ??/?a ? 0.
• Letting S k ln?, we obtain S ? Smax , ?S ? 0
  and ?S/?a ? 0.

7
Thermal Equilibrium

• Equilibrium between two parts of an isolated
  system with a diathermal wall
• U U1 U2, V V1 V2, where U, V, V1, and V2
  are constants.
• The variable parameter a is U1, with ?U1
  ?U2.
• Statistical weights are multiplicative, so that ?
  ?1 ?2.
• Since S k ln?, S S1(U1,V1) S2(U2,V2).
• The equilibrium condition is given by ??/?a 0,
  ? ?S/?a 0, where a U1.
• Thus ?S/?U1 0, so that ?S1/?U1 ?S2/?U1 0, ?
  ?S1/?U1 ?S2/?U2.
• The absolute temperature is defined as 1/T ?
  ?S/?U.
• Note The thermodynamic identity is dU T dS
  P dV so that 1/T ? ?S/?U.

8
Thermal and Mechanical Equilibrium

• Thermal and Mechanical Equilibrium in a two-part
  system
• U U1 U2, V V1 V2, where just U and V are
  constants.
• The variable parameters are U1 and V1, with ?U1
  ?U2 and ?V1 ?V2.
• The equilibrium condition may be written as dS
  0, where
• dS (?S1/?U1 ?S2/?U1)dU1 (?S1/?V1
  ?S2/?V1)dV1 0.
• So that dS 0 only if ?S1/?U1 ?S2/?U2 and
  ?S1/?V1 ?S2/?V2.
• The equilibrium condition gives T1 T2 , where
  1/T ?S/?U.
• The pressure is defined as P ? T ?S/?V, so that
  P1 P2,
• Note P is defined to agree with the standard
  definition of pressure obtained
• from TdS dU PdV.

9
Diffusional Equilibrium 1

• Diffusional Equilibrium in a one-component system
• U U1 U2, V V1 V2, N N1 N2,
• where U, V and N are constants.
• The variables are U1, V1 and N1,
• with ?U1 ?U2, ?V1 ?V2 and ?N1 ?N2.
  
• The equilibrium condition may be written as dS
  0, so that
• (?S1/?U1 ?S2/?U1)dU1 (?S1/?V1 ?S2/?V1)dV1
  (?S1/?N1 ?S2/?N1)dN1 0.
• Thus dS 0 only if
• ?S1/?U1 ?S2/?U2, ?S1/?V1 ?S2/?V2 and ?S1/?N1
  ?S2/?N2.

10
Diffusional Equilibrium 2

• Maximizing S requires
• ?S1/?U1 ?S2/?U2, ?S1/?V1 ?S2/?V2 , ?S1/?N1
  ?S2/?N2.
• Thermal equilibrium requires T1 T2 , where
  1/T (?S/?U)V,N.
• Mechanical equilibrium requires P1 P2, where P
  T (?S/?V)U,N.
• Diffusional equilibrium requires µ1 µ2, where
  µ ? T (?S/?N)U,V
• µ is known as the chemical potential.
• The thermodynamic identity becomes dU T dS P
  dV µ dN.
• Likewise dG(T,P,N) S dT V dP µ dN,
• so that µ (?G/?N)T,P g(T,P),
• and dF(T,V,N) S dT P dV µ dN,
• so that µ (?F/?N)T,V f(T,V).

11
The Boltzmann Distribution 1
Single-particle system

• Consider a quantum system with energies Er (r
  1,2 ) in contact with a heat bath of energy Ub
  at temp.T0.
• Assume that the system is in the specific quantum
  state r, so that the total energy U of the system
  bath is
• U Ub Er ? Ub U
  Er .
• Since ?s 1, the total statistical weight is
• ?tot ?s?b 1.?b(Ub) ?b(U Er).

12
The Boltzmann Distribution 2

• The probability of finding the system in the
  state r is proportional to the total statistical
  weight i.e. pr ? ?.
• Now pr ? ?tot ?b(U Er).
• Thus Stot k ln ?tot, pr ? exp(Sb/k), where Sb
  Sb(U Er).
• Since U gtgt Er, a Taylor expansion may be used
  i.e.
• f(x) f(xo) (x xo)f.
• Thus,
• Sb(U Er) ? Sb(U) Er(?S/?Ub) Sb(U) Er/T0,
• since 1/T0 ?S/?Ub

13
The Boltzmann Distribution 3

• Now pr K expSb(U) Er/T0))/k Kexp(
  Er/kT0),
• where K and K are constants.
• If the probabilities are normalized, Srpr 1 .
• The normalized Boltzmann distribution is,
• pr exp( Er/kT0)/Z,
• where the partition function Z is given by
• Z ?r exp ( Er/kT0).
• Although introduced here only to normalize pr,
  the partition function Z may be considered as the
  central concept of statistical mechanics, since
  it links the microscopic and macroscopic
  viewpoints.
• a

14
The Boltzmann Distribution 4

• If the energy states UR of a quantum system is
  known, then the partition function Z may be
  determined.
• For a sufficiently low temperature T, the factor
  exp(UR/kT) becomes negligible, so that often
  only the lowest energy levels need to be
  considered in calculating Z.
• For a system of N very weakly interacting, like
  particles, each with a partition function z, the
  partition function of the total system is given
  by
• Z zN for distinguishable particles (e.g.
  solids)
• Z zN/N? for indistinguishable
  particles (e.g.fluids).
• For very large N, all the thermodynamic
  parameters (U, H, F, G, S ) may be easily
  obtained from Z.

15
Relation of Z to Classical Parameters

• Summary of results to be obtained in this section
  
• ltUgt ?(lnZ)/?ß (1/Z)(?Z/?ß),
• CV lt(?U)2gt/kT2,
• where ß 1/kT, with k Boltzmanns
  constant.
• S kßltUgt kZ ,
• where ltUgt U for a very large system.
• F U TS kT lnZ,
• From dF S dT PdV, we obtain
• S (?F/?T)V and P (?F/?V)T .
• Also,
• G F PV PV kT lnZ.
• H U PV PV ?(lnZ)/?ß.

16
Systems of N Particles of the Same Species

• Z zN for distinguishable particles (e.g.
  solids)
• Z zN/N? for indistinguishable particles
  (e.g.fluids).
• ltugt ?(lnz)/?ß (1/z)(?z/?ß), U Nltugt.
• cV lt(?u)2gt/kT2, CV NcV, CP NcP.
• Distinguishable particles
• F Nf kT ln zN NkT lnz, G Ng.
• Since F U TS, so that S (U F)/T or
  (?F/?T)V,
• Indistinguishable particles
• F kT ln(zN/N?) kT ln(zN) ln N?
  NkT ln(z/N) 1,
• Since for very large N, Stirlings theorem gives
  ln N! N lnN N.
• Also, S (?F/?T)V and P (?F/?V)T as before.

17
Mean Energies and Heat Capacities

• Equations obtained from Z ?r exp ( ?Er),
  where ? 1/kT.
• ?U? ?rprEr/?rpr ?(ln Z)/?? (1/Z) ?Z/??
  .
• ?U2? ?rprEr2/?rpr (1/Z) ?2Z/??2.
• ?Un? ?rprErn/?rpr (1)n(1/Z) ?nZ/??n.
• ?(?U)2? ?U2? (?U?)2 ?2lnZ/??2 or ?
  ?U?/?? .
• CV ? ?U?/?T ? ?U?/?? . d?/dT k?2. ?
  ?U?/??,
• or CV k?2 ?(?U)2? ?(?U)2?/kT2
• i.e. ?(?U)2? kT2CV .
• Notes
• Since ?(?U)2? 0,
• (i) CV 0,
• (ii) ? ?U?/?T 0.

18
Sharpness of Distribution 2

• Replacing a by U, the sharpness of the
  distribution is given by
• ?U/ltUgt, where ?U ?lt(?U)2gt, so that
• ?U/ltUgt ?(kT2CV)/ltUgt.
• For a system of N identical particles, ltUgt f
  kT, where f 3N is the number
• of degrees of freedom of the system
  (equipartition theorem).
• CV dU/dT f k, so that
• ?U/ltUgt ?(f k2T2)/ltf kTgt f 1/2.
• Thus for a very large system, the peak value of U
  is the only value that a
• measurement could give, allowing the mean value
  ltUgt of statistical mechanics
• to be equated to the internal energy U of
  classical thermodynamics.

19
Relationship between Entropy and Probability

• Consider an ensemble of n replicas of a system.
• If the probability of finding a member in the
  state r is pr, the number of systems that would
  be found in the rth state is
• nr n pr, if n is large.
• The statistical weight of the ensemble On (n1
  systems are in state 1, etc.), is On n?/(n1?
  n2?nr?..),
  so that Sn k ln n? k ?r ln nr?.?
• From Stirlings theorem,
• ln n? n ln n n, ?r ln nr? ?r nr ln
  nr n.
• Thus Sn k n ln n ?r nr ln nr ? k n ln n
  ?r nr ln n ?r nr ln pr,
• so that Sn k ?r nr ln pr kn ?r pr ln pr
  .
• For a single system, S Sn/n i.e. S k
  ?r pr ln pr .

20
Ensembles 1

• A microcanonical ensemble is a large number of
  identical
• isolated systems.
• The thermodynamic degeneracy may be written as
  ?(U, V, N).
• From the fundamental postulate, the probability
  of finding the
• system in the state r is pr 1/?.
• Thus, S k ?r pr ln pr k ?r (1/?) ln ?
  
• (k/?) ln ? ?r1 k ln ?.
• Statistical parameter ?(U, V, N).
• Thermodynamic parameter S(U, V, N) T dS dU
  PdV µdN.
• Connection S k ln ?.
• Equilibrium condition S ? Smax.

21
Ensembles 2

• A canonical ensemble consists of a large number
  of identically
• prepared systems, which are in thermal contact
  with a heat
• reservoir at temperature T.
• The probability pr of finding the system in the
  state r is given by
• the Boltzmann distribution
• pr exp( ?Er)/Z, where Z ?r exp(?Er), and ?
  1/kT.
• Now S k ?r pr ln pr k ?r exp(?Er)/Z
  lnexp(?Er)/Z
• (k/Z) ?r exp(?Er) ln
  exp(?Er) ln Z
• (k?/Z) ?rEr exp(?Er) (k
  lnZ)/Z . ?rexp(?Er),
• so that S k? ?U? k lnZ k lnZ
  k?U.
• Thus, S(T, V, N) k lnZ U/T and F U TS
  kT lnZ.

22
Ensembles 3

• S(T, V, N) k lnZ U/T , F U TS
  kT lnZ.
• Statistical parameter Z(T, V, N).
• Thermodynamic parameter F(T, V, N).
• Connection F kT ln Z.
• Equilibrium condition F ? Fmin.
• A grand canonical ensemble is a large number of
  identical
• systems, which interact diffusively with a
  particle reservoir.
• Each system is described by a grand partition
  function,
• ?G(T, V, µ) ?N?r?(µN EN,r),
• where N refers to the number of particles and r
  to the set of states
• associated with a given value of N.

23
Entropy of an Ideal Monatomic Gas 1

• We wish to find a general expression ?(U,V,N) for
  a system of N
• particles of an ideal monatomic gas, confined to
  a volume V, with
• the total energy in the range U to U ?U.
• Since U p2/2m, the total momentum lies in the
  ranges
• p to (p ?p).
• In order to count microstates, we imagine the
  phase-space to be
• divided into cells of area ?pi?qi h, per
  degree of freedom.
• Note that a state describing N particles moving
  in 3 dimensions is
• a point in 6N-dimensional phase space, while the
  number of
• degrees of freedom of the system is f 3N.

24
Entropy of an Ideal Monatomic Gas 2

• A state describing N particles moving in 3
  dimensions is a point in
• 6N-dimensional phase space,
• The number of degrees of freedom of the system is
  f 3N.
• For f 1 (one particle moving in one dimension),
  the relation
• U p2/2m means that p lies between the values
  p to (p ?p).
• Thus ?(U,V,N) 2 ?p L/h, where is the length of
  the container,
• so that ? ? L V1/3 and ? ? p0 ? U0 .

25
Entropy of an Ideal Monatomic Gas 3

• f 1 (1 atom in 1 dimension)
• ? is the number of squares contained in the two
  rectangles i.e.
• ? 2 ?p L/h ? p0L ? U0V1/3.
• f 2 (1 atom in 2 dimensions or 2 atoms in 1
  dimension)
• ? is the number of squares contained in the area
  between the circles i.e.
• ? 2pp ?p L2/h2 ? p1L2 ? U1/2V2/3.
• f 3 (1 atom in 3 dimensions or 3 atoms in 1
  dimension)
• ? is the number of squares contained in the area
  between the spheres i.e.
• ? 4pp2 ?p L3/h3 ? p2L3 ? U1V.

26
Entropy of an Ideal Monatomic Gas 4

• General result for f degrees of freedom
• ? ? p(f 1) Lf or p(f 1) Vf/3 .
• ? ? (mU)(f 1)/2 Vf/3 (mU)(3N 1)/2 VN.
• For large N, 3(N 1)/2 3N/2.
• Omitting the dependence on m, we obtain
• ?(U,V,N) B(N) U3N/2 VN,
• so that
• S(U,V,N) k C(N) N lnV (3N/2) lnU.
• Assuming that S is an extensive function, we have
• S(U,V,N) N s(U,V) Nk K lnv (3/2)
  ln u.
• Using the results u (3/2)kT and v V/N, we
  obtain
• S(U,V,N) Nk D (3/2)lnT ln(V/N).

27
Mixing of gases 1 Gibbs Paradox

• If N molecules of an ideal monatomic gas make a
  free expansion
• from a V to 2V, ?U 0, so that
• ?S Sf Si Nk ln(2V/V) Nk ln 2.
• If the container of volume 2V is now redivided
  into equal parts,
• so that (N/2) molecules are in each half, then
• ?S Sf Si 2(N/2)k ln(V/2V) Nk ln 2.
• This is Gibbs paradox, which is removed by
  assuming that the
• molecules are indistiguishable i.e. that
• ?(U,V,N)
  B(N) U3N/2 VN/N!.
• The term Nk lnV in the expression for S is
  replaced by Nk ln(V/N!),
• where (by Stirlings theorem),
• Nk ln(V/N!) Nk ln(V/N) 1.

28
Entropy of an Ideal Monatomic Gas 5

• If S is assumed to be an extensive quantity,
  Gibbs paradox may be avoided.
• Letting S Ns, with s(U,V) k lnC lnv (3/2)
  lnu, where C
  B(N)/N, v V/N, and u U/N.
• Thus, S(U,V,N) Nk lnC (3/2)ln(U/N)
  ln(V/N).
• Temperature is defined as 1/T ? (?S/?U)V,N
  (3/2)Nk/U, since U
  (3/2)NkT, so that
• S(U,V,N) Nk D (3/2)lnT ln(V/N).
• Note that P/T (?S/?V)T,N Nk/V, so that PV
  NkT.
• Mixing gases
• Gas A in the left compartment, a different gas B
  in the right compartment.
• ?Stot ?S1 ?S2.
• b. Gas A in the left compartment, the same gas
  B in the right compartment.
• ?Stot Sfinal S1initial S2initial.

29
Some Useful Integrals

• In polar coordinates, r varies from 0 to ?.
• In Cartesian coordinates x, y and z vary from
  ? to ?.

30
Maxwell Velocity Distribution 1

• One molecule of a monatomic ideal gas has
  internal energy ? ½ mv2, so that the
  probability of a molecule lying in the range v to
  v dv is
• P(v) d3v exp( ?/kT) d3v /?d3v exp( ?/kT),
• since its energy does not involve its
  position coordinates.
• The Maxwell velocity distribution function
  f(v)d3v is the number of molecules per unit
  volume with velocities from v to v dv.
• The distribution function g(vx)dvx for the
  velocity component vx is the number of molecules
  per unit volume with components in the range vx
  to vx dvx.
• The Maxwell Speed Distribution F(v) dv is the
  number of molecules per unit volume with speeds
  between v and v dv, regardless of direction.

31
Maxwell Velocity Distribution 2

• For an ideal monatomic gas in uniform
  surroundings, the Maxwell velocity distribution
  function f(v) d3v is the number of molecules per
  unit volume with velocities between v and
• v dv.
• Now f(v)d3v (N/V) P(v)d3v, where
• P(v)d3v exp( mv2/2kT) d3v / ?d3v exp(
  mv2/2kT).
• Thus
• f(v)d3v (N/V) exp( mv2/2kT) d3v / ?d3v exp(
  mv2/2kT)
• Since d3v dvxdvydvz,
• f(v)d3v (N/V)exp( mv2/2kT) d3v /??? dvx
  exp( mvx2/2kT)3.
• Now I0(a) ½ (p/a)1/2, so that
• f(v)d3v (N/V)(m/2pkT)3/2 exp(mv2/2kT) d3v.

32
Distributions of Velocity Components 1

• The distribution function g(vx)dvx for the
  velocity component vx is the number of molecules
  per unit volume with components in the range vx
  to vx dvx.
• To obtain g(vx)dvx, the Maxwell velocity
  distribution f(v)d3v is integrated over all
  values of vy and vz.

33
Distributions of Velocity Components 2

• The distribution of velocity components is,
• g(vx)dvx (N/V)(m/2pkT)1/2 exp(mvx2/2kT) dvx.
• This is a Gaussian distribution, symmetrical
  about the mean value, ltvxgt 0, with a
  root-mean-square (rms) deviation,
• v(?vx2) v(vx2) v(kT/m).

34
Mean Value Calculations 1
35
Maxwell Speed Distribution 1

• The Maxwell Speed Distribution F(v) dv is the
  number of molecules per unit volume with speeds
  between v and v dv, regardless of direction.
• F(v) dv is obtained by working in polar
  coordinates and integrating f(v) d3v over all
  orientations (?, F), so that the volume element
  d3v is replaced by v2sin? dF d? dv.

Integration over all angles gives F(v)dv
4pv2f(v)dv.
The factor v2 present in F(v) dv, makes
the distribution unsymmetrical.
F(v)dv 4p(N/V)(m/2pkT)3/2 v2 exp( mv2/2kT) dv.
36
Maxwell Speed Distribution 2

• vpeak 1.41 v(kT/m), ltvgt 1.60 v(kT/m), vrms
  1.73 v(kT/m).

(V/N)
37
Mean Value Calculations 2
38
Partition Function of an Ideal Monatomic Gas 1

• z ?i exp( ??i) ? ???d3x??? d3p exp( ??)/h3
  (classical system).

39
Partition Function of an Ideal Monatomic Gas 2
40
Entropy of an Ideal Monatomic Gas 6

• Z zN/N!, where z V(2pmkT/h2)3/2,
• and
• U (3/2)N kT.
• Also
• S k(lnZ ?U) k ln zN lnN! (3/2)N,
• Nk lnz lnN 1 3/2.
• Thus
• S Nk K ln(V/N) (3/2) ln T,
• with
• K ln (2pmk/h2)3/2 5/2.

41
Equipartition Theorem 1
42
Equipartition Theorem 2
43
Equipartition Theorem Examples

• Harmonic oscillator in one dimension
• ? p2/2m ½ k x2, where k is the spring
  constant.
• ??? ½ kT ½ kT kT.
• Ideal diatomic gas
• 50 K ?vib ,?rot ?? T ?? ?trans, so that ???
  (3/2)kT and CV (3/2)nR.
• 500 K ?vib ?? T ?? ?trans ,?rot , so that ???
  (5/2)kT and CV (5/2)nR.
• CP CV nR (7/2)nR (Mayers
  equation), so that ? 7/5 1.2.
• 5000 K (assuming no dissociation) T ?? ?trans
  ,?rot ,?vib ,
• 
  so that CV (7/2)nR.

44
Harmonic Oscillator in One Dimension 1

• For a 1D harmonic oscillator, ?n (n
  ½) h?, where ? v(k/m).

45
Harmonic Oscillator in One Dimension 2
46
Einstein Model of Specific Heats 1

• The Einstein model represents a solid of N
  identical atoms by 3N
• one-dimensional oscillators, each vibrating at
  the Einstein
• frequency ?E.
• This is an isotropic model, which neglects the
  coupling of
• neighboring atoms.
• The Einstein temperature ?E is defined by the
  relationship
• k?E h?E.
• The heat capacity CV is 3NcV, where cV k
  x2ex/(ex 1)2, where
• x ?h?E h?E/kT ?E/T.
• Thus,
• CV 3Nk exp(?E/T)2/ exp(?E/T) 12.

47
Einstein Model of Specific Heats 2

• CV Nk exp(?E/T)2/ exp(?E/T) 12.
• High temperature limit
• CV ? 3Nk (classical equipartition theorem).
• Low temperature limit
• CV ? 3Nk(?E/T)2exp(?E/T) ? 0 as T ? 0 (3rd
  Law).
• The low temperature drop in CV is steeper than
  measured in most materials.
• This model applies to a solid with a phonon
  spectrum dominated by a narrow band of
  frequencies.
• For example, Al10V fits the Einstein model with
  ?E 22 K, since one Al atom per molecule
  vibrates in a relatively large cavity
• Phys.Rev.Lett. 30, 1138 (1973).

48
Debye Model of Specific Heats 1

• The Debye model is a long wavelength
  approximation (? ?? a, the lattice
• spacing), which takes account of interactions
  between neighboring molecules.
• The Debye frequency ?D is an artificial upper
  limit to the frequency distribution,
• chosen so that the number of modes of oscillation
  is 3N.
• The Debye temperature ?D is defined by the
  relationship k?D h?D.
• ?0?D g(?) d? 3N, where g(?) is the density of
  states.
• Insulator CV BT3, where B(2.4)p4Nk/?D3.
• Conductor CV AT BT3.

49
Photon Statistics 1

• A single photon in the state r has energy ?r
  h?r.
• The number of photons in any state r may vary
  from 0 to ?.
• The total energy of blackbody radiation is ER
  ?r nr?r , where nr is the number of photons in
  the rth state, so that
• Zph(T, V) ?R exp (?ER).
• The state R of the complete system may be
  represented by a set of occupation numbers (n1,
  n2, nr, ).
• We show that ln Zph(T, V) ?r ln 1
  exp(?er),
• and ltnrgt (1/?) ?(lnZph)/?er,
• which leads to
• ?n(?)? 1/(e ßh? 1).

50
Photon Statistics 2
51
Photon Statistics 3
This equation shows how to determine the mean
number of systems of energy er.
52
Photon Statistics 4

• For a continuous EM (photon) distribution, ?(?)
  h?,
• so that ?n(?)? 1/(e ßh? 1).

53
Plancks Radiation Law

• Let g(?)d? be the number of quantum states in the
  range ? to ? d?, and ?n(?)? the no. of photon
  of frequency ?.
• The number of photons in the range ? to ? d? is
• n(?)d? ?n(?)? g(?) d?.
• Thus, the energy density u(?) is given by
• u(?)d? (1/V) h? ?n(?)? g(?) d?.
• Inserting the values
• g(?) d? V ?2 d?/p2c3 , ?n(?)? 1/(e?h?
  1) ,
• we obtain
• u(?,?) h?3 d? / p2c3(e?h? 1) .
• This is Plancks radiation law, which on
  integration over all frequencies gives the
  Stefan-Boltzmann law
• u(T) aT4,
• where a p2 k4 / 15 h2c3 .

54
Finding the Grand Partition Function ZG
55
Occupation Numbers 1
56
Occupation Numbers 2
57
Bose Einstein and Fermi-Dirac Statistics

• The symmetry requirements placed on a system of
  identical quantum particles depends on their
  spin.
• Particles with integer spin (0, 1, 2,) follow
  Bose-Einstein statistics, in which the sign of
  the total wave function is symmetrical with
  respect to the interchange of any two particles
  i.e.
• ?( Qj Qk ) ?( Qk
  Qi ).
• Particles with half-integer spin (1/2, 3/2, )
  follow Fermi-Dirac statistics, in which the sign
  of the total wave function is antisymmetrical
  with respect to the interchange of any two
  particles i.e.
• ?( Qj Qk ) ?( Qk
  Qi ).
• Thus, two particles cannot be in the same
  state, since ? 0, when particles j and k are in
  the same state.

58
Two-Particle Systems

• Writing the wavefunction for particle j in state
  A as ?j(qA) etc., we have the following
  situations
• Maxwell-Boltzmann statistics ? ?j(qA)?k(qB).
• Bose-Einstein (BE) statistics ? ?j(qA)?k(qB)
  ?j(qB)?k(qA).
• In this case, the total wave function ? is
  antisymmetrical.
• Examples of bosons are photons and composite
  particles, such as H1 atoms or He4 nuclei.
• Fermi-Dirac (FD) statistics ? ?j(qA)?k(qB)
  ?j(qB)?k(qA).
• In this case, the total wave function ? is
  symmetrical.
• ? 0, if both particles are placed in the
  same state the Pauli exclusion principle.
• Examples of bosons are electrons, protons,
  neutrons, and composite particles, such as H2
  atoms or He3 nuclei.

59
Grand Partition Function ZG

• To obtain the grand partition function ZG , we
  consider a system in which the number of
  particles N can vary, which is in contact with a
  heat reservoir.
• The system is a member of a grand canonical
  ensemble, in which T, V and µ (the chemical
  potential) are constants.
• Assume that there are any number of particles in
  the system, so that 0 N No ? ?, and an
  energy sequence for each value of N,
• UN1 UN2 UNr
• in which,
• Vo V Vb, Uo U Ub, No N Nb,
• where Vb etc. refer to the reservoir and Vo
  etc. to the total.

60
Comparison of ZG with Z

• The state N,r of the system may be written as a
  set of particle occupation-numbers (n1, n2,, nr,
  ), with
• ?ni? (1/?)?(lnzGi)/?µ.
• Fermions (particles with half-integer spin) ni
  0 or 1.
• Bosons (particles with integer spin) ni 0,1,
  2, 8.

61
Fermi-Dirac Statistics?ni? is the mean no of
spin-1/2 fermions in the ith state.

• All values of µ, positive or negative are
  allowable, since
• ?ni? always lies in the range 0 ?ni? 1.

62
Bose-Einstein Statistics
?ni? is the mean no of bosons in the ith state.

• ?ni N, the total number of particles in the
  system of like particles.
• ?ni? must be positive and finite, ?i ? µ for all
  i.
• For an ideal gas, ?i p2/2m ? ?min 0, so that
  µ must be negative.
• For a photon gas, µ 0.

63
Mean number of bosons or fermions
If there is a fixed number of particles N,
?ltn(e)gt N
64
Classical limit

• Quantum statistics gives
• ltnrgt 1/expß(er µ) 1.
• In the classical limit, the energy states r are
  infinitesimally close, so that
• ltnrgt ? 0, ltnrgt1 ? 8, expß(er µ) 1
• and
• ltnrgt ? expß(µ er) exp(ßµ). exp( ?er) ,
• where z is the single-particle partition
  function.
• The single-particle Boltzmann distribution is
• pr ltnrgt/N exp( ?er).z .
• Thus,
• N z exp(ßµ).

65
Density of States 1

• Suppose that for an N-particle system with
  continuous e,
• i. the number of states in the range e to
  e de is f(?)d?
• ii. the mean number of particles of energy
  ? is ltn(?)gt.
• The number of particles with energies in the
  range e to e de is
• dN(e) ltn(?)gt f(?)d?.
• Thus, the total no. of particles is given by N
  ?dN(e), and the total energy is given by U ?e
  dN(e) ?e ltn(?)gt f(?)d?, where the integration
  limits are 0 and 8.
• The values of ltn(?)gt for quantum systems are
  given by
• ltn(?)gt 1/expß(e µ) 1.
• The distribution f(?)d? is called the density of
  states.

66
Density of States 2

• For a particle in a cube of side L, the
  wavefunction ? is zero at the walls, so that
• ?(n1,n2,n3) sink1x sink2y sink3z,
• where ki nip/L (i 1,2,3), and each
  k-state is characterized by the set of positive
  integers (n1, n2, n3).
• Neighboring states are separated by ?ki p/L ,
  so that the volume per state in k-space is (p/L)3
  p3/V.

In this 2-dimensional figure, each point
represents an allowed k-state, associated with an
area in k-space of (p/L)2.
67
Density of States 3

• The volume of a spherical shell of radius k is
  4pk2dk, which would contain 4pk2dk/(p3/V) 4V
  k2dk/p2, where (p3/V) is the volume in k-space
  associated with each state (n1, n2, n3).
• However, since only positive values of ni
  represent physical situations, the number of
  k-states in range k to k dk is (1/8)th of that
  for the total shell i.e. g(k)dk Vk2dk/(2p2).

68
Density of States 4

• For a set of spin-zero bosons,
• g(k)dk V k2dk /(2p2).
• For a set of spin-½ fermions,
• g(k)dk 2V k2dk /(2p2),
• since each set of quantum numbers (n1, n2,
  n3), has two possible spin states.
• The number of states in the range ? to ? d? is
  given by
• f(?)d? g(k)dk g(k)(dk/d?)d?.
• Now ? p2/2m (kh)2/2m,
• so that
• k v(2m?)/h , dk/d? (1/2h)(2m/?)1/2.

69
Density of States 5

• Bose-Einstein condensation (spin 0 system)
• The number of states in the range ? to ? d? is
  given by
• f(?)d? g(k)dk g(k)(dk/d?)d?,
• with
• g(k) Vk2/(2p2),
• k v(2m?)/h , dk/d? (1/2h)(2m/?)1/2.
• Hence,
• f(?)d? Vk2(dk/d?)d?/2p2
  V4pm?/h3(2m/?)1/2d?
• i.e. f(?)d? (2pV/h3)(2m)3/2?1/2
  d?.
• Free electron theory (spin ½ system)
• f(?)d? (4pV/h3)(2m)3/2?1/2 d?.

70
Summary

• Consider N particles of an ideal quantum gas,
  with closely spaced states, which may be taken as
  a continuum.
• Such is the case for the energy of a molecule of
  an ideal monatomic gas e p2/2m.
• The number of states in the range e to e de is
  given by
• f(e) de C e1/2 de,
• where C (2pV/h3)(2m)3/2 for spin 0 bosons,
• and (4pV/h3)(2m)3/2 for spin ½ fermions.
• The occupation numbers of each state are given by
  the BE and FD distribution functions
• ltn(e)gt 1/expß(e µ) 1 and 1/expß(e
  µ) 1
• respectively.
• The distribution of particles is given by
  dN(e) ltn(?)gt f(?)d?.

71
LOW TEMPERATURE FD DISTRIBUTION

• As ß ? 0, expß(e µ) behaves as follows
• If e lt µ, expß(e µ) ? exp(-8) 0, so that
  ltn(e)gt 1
• if e gt µ, expß(e µ) ? exp(8) 8, so that
  ltn(e)gt 0.
• Thus dN(e) f(e)de for e lt µ, and dN(e) 0 for
  e gt µ.
• The Fermi energy ?F is defined as ?F µ(T ?
  0).

72
Free-Electron Theory Fermi Energy 1

• The Fermi energy ?F µ(T0)
• At T0, the system is in the state of lowest
  energy, so that the N lowest
• single-particle states are filled, giving a sharp
  cut-off in ?n(?)? at T TF.
• At low non-zero temperatures, the occupancies are
  less than unity, and
• states with energies greater than µ are partially
  occupied.
• Electrons with energies close to µ are the ones
  primarily excited.
• The Fermi temperature TF ?F/k lies in the range
  104 105 K for metals with
• one conduction electron per atom.
• Below room temperature, T/TF lt 0.03, and µ ?F.

73
Free-Electron Theory Heat capacity

• Simplified calculation for T ltlt TF
• Assume that only those particles within an
• energy kT of ?F can be excited and have
• mean energies given by equipartition i.e.
• Neff ? N(kT/?F) NT/TF.
• Thus, U ? Neff(3/2)kT (3/2)NkT2/TF,
• so that
• CV dU/dT ? 3NkT/TF.
• In a better calculation, 4.9 replaces 3.
• Thus, for a conductor at low temperatures,
• with the Debye term included,
• CV AT3 ?T,
• so that
• CV/T AT2 ?.

74
Free-Electron Theory Fermi Energy 2

• The number of electrons with energies between ?
  and ? d? is given by
• dN(?) ?n(?)? f(?) d?, where ?n(?)? 1/exp?(?
  µ) 1, .

75
Free-Electron Theory Calculation of lt?ngt.

• Now U ?0? ??n(?)?f(?) d?.

76
LOW TEMPERATURE B-E DISTRIBUTION

• The distribution of particles dN(e) ltn(?)gt
  f(?)d? cannot work for BE particles at low
  temperatures, since all the particles enter the
  ground-state, while the theoretical result
  indicates
• that the density of states f(e) C e1/2 is
  zero at e 0.
• The thermodynamic approach to Bose-Einstein
  Condensation shows the strengths and weaknesses
  of the statistical method
• mathematical expressions for the phenomena
  are obtained quite simply, but a physical picture
  is totally lacking.
• The expression dN(e) ltn(?)gt f(?)d? works down
  to a phase transition, which occurs at the Bose
  or condensation temperature TB, above which the
  total number of particles is given by N ?
  dN(e).
• Below TB , appreciable numbers of particles are
  in the ground state.

77
Bose-Einstein Condensation 1

• The number of particles with energies in the
  range e to e de is
• dN(e) ltn(?)gt f(?)d?,
• with
• ltn(?)gt 1/expß(e µ) 1
• and
• Thus,
• dN(e) (2pV/h3)(2m)3/2?1/2 d? /expß(e µ)
  1,
• and the total number of particles N is given
  by
• N (2pV/h3)(2m)3/2 ??1/2 d? /expß(e µ)
  1,
• which is integrated from ? 0 to 8.

f(?)d? (2pV/h3)(2m)3/2?1/2 d?.
78
Bose-Einstein Condensation 2

• Since the number of particles are fixed, with
• N (2pV/h3)(2m)3/2 ??1/2 d? /expß(e µ)
  1,
• the integral
• ??1/2 d? /expß(e µ) 1
• must be positive and independent of
  temperature.
• Since ?min 0 for an ideal gas, the chemical
  potential µ 0, the factor
• exp(ß µ) 1 exp(µ/kT) 1,
• must be constant, so that (µ/T) is
  independent of temperature.
• This can happen only down to the Bose Temperature
  TB, at which µ becomes 0.
• What happens below TB?

79
Bose-Einstein Condensation 3

• At the Bose (or condensation) temperature TB, µ
  0, so that
• N (2pV/h3)(2m)3/2 ??1/2 d? /exp(e/kTB) 1,
• which yields on integration, TB
  (h2/2pmk)(N/2.612V)2/3.
• The expression f(?)d? K?1/2 d?, with K
  (2pV/h3)(2m)3/2,
• indicates that for T gt TB, the number of
  particles in the ground state (? 0) is
  negligible, since f(?) K?1/2 ? 0.

Behavior of µ above TB.
80
Bose-Einstein Condensation 4
81
Bose-Einstein Condensation 5

• For T gtTB, the number of particles in the ground
  state (N) is zero.

82
Bose-Einstein Condensation 6
83
Bose-Einstein Condensation 7
84
Bose-Einstein Condensation 8
85
Bose-Einstein Condensation 9
Classical high-temperature value
86
Bose-Einstein Condensation 10
87
AppendixAlternative Approach to Quantum
Statistics

• PHYS 4315
• R. S. Rubins, Fall 2008

88
Lagrange Method of Undetermined Multipliers 1

• Simple example
• How to find an extremum for a function f(x,y),
  subject to the constraint f(x,y) constant.
• Suppose f(x,y) x3 y3, and f(x,y) xy 4.
• Method 1
• Eliminating y, f(x,y) x3 (4/x)3, so that
  df/dx 3x2 - 3(4/x)4
• When df/dx 0, x6 64, ? x 2, y 2.
• Method 2 (Lagrange method)

and
a, so that
In this example, a is a Lagrange undermined
multiplier.
89
Lagrange Method of Undetermined Multipliers 2
Suppose that the function of
is needed.
This occurs when df (?f/?x1)dx1
(?f/?xn)dxn 0.
Let there be two constraints
N,
U,
where
in the calculations of the
mean number of particles in the state j.
Lagranges method of undetermined multipliers
gives the following set of equations
In the calculations that follow, the function f
equals ln(?), where ? is the thermodynamic
degeneracy.
90
Alternate Fermi-Dirac Calculation 1
If the jth state has degeneracy gj, and contains
Nj particles, Nj gj for all j, since the limit
is one particle per state e.g.
The number of ways of dividing N
indistinguishable particles into two groups is
In the Fermi-Dirac case,
.
91
Alternate Fermi-Dirac Calculation 2
The total no. of microstates is obtained by
summing over all j i.e.
Therefore,
Using Stirlings theorem, ln N! N ln N N, we
obtain
92
Alternate Fermi-Dirac Calculation 3
Since the constraints are
,
we let f(N1Nj) N, and ?(N1 Nj) U, so that
,
where a and ß are Lagrange multipliers. Inserting
the expression for ln ?FD and remembering that
we obtain
.
93
Alternate Fermi-Dirac Calculation 4
reduces to
.
Thus ltnjgt
The constraint a has been replaced by µ/kT, where
µ is the chemical potential, and ß by 1/kT.
94
Alternate Bose-Einstein Calculation 1
The jth energy level has gj quantum states, and
contains a total of Nj identical particles, with
up to Nj particles in each state. All possible
microstates can be obtained by rearranging (gj
1) partitions and Nj dots, in a diagram like
that shown below.
The number of microscopes for a given Nj and gj is
.
95
Alternate Bose-Einstein Calculation 2
The total no. of microstates is obtained by
summing over all j i.e.
Therefore,
Using Stirlings theorem, ln N! N ln N N, we
obtain
96
Alternate Bose-Einstein Calculation 3
Using the method of Lagrange multipliers as
before,
where a and ß are Lagrange multipliers. Inserting
the expression for ln ?FD, we obtain
hence
.
97
Alternate Bose-Einstein Calculation 4
reduces to
.
Thus ltnjgt
The constraint a has been replaced by µ/kT, where
µ is the chemical potential, and ß by 1/kT.