Photon Statistics 1

1
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).

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

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

5
Density of States 1

• 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.
6
Density of States 2

• 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).

7
Density of States 3

• In dealing with g(k)dk Vk2dk/(2p2), we
  transform from the magnitude of the wave vector k
  to the angular frequency ? i.e.
• g(?)d? g(k)dk g(k)(dk/d?)d?.
• Since c ?/k, g(k) Vk2/(2p2) V?2/(2p2c2),
  and dk/d? 1/c,
• g(?)d? V?2/(2p2c3) d?.
• Thus, the number of photons in the range ? to ?
  d? equals
• the number of photon states in that range g(?)d?
• times
• the occupation of each state ?n(?)? 1/(e ßh?
  1).
• Thus, the energy density u(?) is given by
• u(?)d? (1/V) h? ?n(?)? g(?) d?.

8
Plancks Radiation Law

• The, 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 p2k4/15 h3c3 .
• Note letting x ?h?, u(?,x)d?
  (?4p2c3h3)1?x3dx/(ex 1).

9
Finding the Grand Partition Function ZG
10
Occupation Numbers 1
11
Occupation Numbers 2
12
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.

13
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.

14
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.

15
Comparison of ZG with Z
Bath type Heat bath Heat and particle bath
Probability pr exp ( ?Er)/Z pN,r exp ?(µN EN,r)/Z
Statistical parameter Z ?r1 exp ( ?Er) Partition function ZG ?N0 ?r1 exp ?(µN EN,r) Grand partition function

• 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.

16
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.

17
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.

18
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.

19
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?.

20
Density of States 6

• 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.

21
Mean number of bosons or fermions
If there is a fixed number of particles N,
?ltn(e)gt N
22
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).

23
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?.

24
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).

25
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.

26
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 ?.

27
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, .

28
Free-Electron Theory Calculation of lt?ngt.

• Now U ?0? ??n(?)?f(?) d?.

29
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.

30
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?.
31
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?

32
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.
33
Bose-Einstein Condensation 4
34
Bose-Einstein Condensation 5

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

35
Bose-Einstein Condensation 6
36
Bose-Einstein Condensation 7
37
Bose-Einstein Condensation 8
38
Bose-Einstein Condensation 9
Classical high-temperature value
39
Bose-Einstein Condensation 10
40
AppendixAlternative Approach to Quantum
Statistics

• PHYS 4315
• R. S. Rubins, Fall 2008

41
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.
42
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.
43
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,
.
44
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
45
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
.
46
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.
47
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
.
48
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
49
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
.
50
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.