1 photon statistics 1 a single photon in the state r has energy r = ħω r. the number of photons...

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Photon Statistics 1• A single photon in the state r has energy r = ħωr.• The number of photons in any state r may vary from 0 to .• The total energy of blackbody radiation is ER = ∑r nrr , where nr

is the number of photons in the r’th state, so thatZph(T, V) = ∑R exp (–ER).

• The state R of the complete system may be represented by a set of occupation numbersoccupation numbers (n1, n2, … nr, …).

• We show that ln Zph(T, V) = – ∑r ln [1 – exp(–εr)],

• and <nr> = – (1/) ∂(lnZph)/∂εr,

• which leads to

n(ω) = 1/(e βħω – 1).

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Photon Statistics 2

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Photon Statistics 3

This equation shows how to determine the mean number of systems of energy εr.

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Photon Statistics 4

For a continuous EM (photon) distribution, (ω) = ħω,

so that n(ω) = 1/(e βħω – 1).

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Density of States 1• For a particle in a cube of side L, the wavefunction Ψ is zero at the

walls, so that Ψ(n1,n2,n3) = sink1x sink2y sink3z,

where ki = niπ/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 Δki = π/L , so that the volume per state in k-space is (π/L)3 = π3/V.

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

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Density of States 2• The volume of a spherical shell of radius k is 4πk2dk, which

would contain 4πk2dk/(π3/V) = 4V k2dk/π2, where (π3/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/(2π2).

Density of States 3• In dealing with g’(k)dk = Vk2dk/(2π2), 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/(2π2) = V2/(2π2c2), and dk/d = 1/c,

g()d = V2/(2π2c3) d.

• Thus, the number of photons in the range to + d equalsthe number of photon states in that range g()d

times the occupation of each state n(ω) = 1/(e βħω – 1).

• Thus, the energy density u(ω) is given by

u(ω)dω = (1/V) ħω n(ω) g(ω) dω.

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Planck’s Radiation Law

• The, the energy density u(ω) is given byu(ω)dω = (1/V) ħω n(ω) g(ω) dω.

• Inserting the values g(ω) dω = V ω2 dω/π2c3 , n(ω) = 1/(eħω – 1) ,

we obtainu(ω,) = ħω3 dω / π2c3(eħω – 1) .

• This is Planck’s radiation law, which on integration over all frequencies gives the Stefan-Boltzmann law

u(T) = aT4, where a = π2k4/15 ħ3c3 .

Note: letting x = ħ, u(,x)dω = (4π2c3ħ3)–1∫x3dx/(ex – 1).

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Finding the Grand Partition Function ZG

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Occupation Numbers 1

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Occupation Numbers 2

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Bose Einstein and Fermi-Dirac Statistics• The symmetry requirements placed on a system of identical

quantum particles depends on their spinspin.

• Particles with integer spin (0, 1, 2,…) follow Bose-Einstein Bose-Einstein statistics, statistics, in which the sign of the total wave function is symmetrical with respect to the interchange of any two particles; i.e.

Ψ( ∙ ∙ ∙ Qj ∙ ∙ ∙ Qk∙ ∙ ∙ ) = Ψ( ∙ ∙ ∙ Qk∙ ∙ ∙ Qi ∙ ∙ ∙ ).

• Particles with half-integer spin (1/2, 3/2, …) follow Fermi-Dirac Fermi-Dirac statistics, statistics, in which the sign of the total wave function is antisymmetrical with respect to the interchange of any two particles; i.e.

Ψ( ∙ ∙ ∙ Qj ∙ ∙ ∙ Qk∙ ∙ ∙ ) = – Ψ( ∙ ∙ ∙ Qk∙ ∙ ∙ Qi ∙ ∙ ∙ ).

Thus, two particles cannot be in the same state, since Ψ = 0, when particles j and k are in the same state.

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Two-Particle Systems• Writing the wavefunction for particle j in state A as ψj(qA) etc., we

have the following situations:• Maxwell-Boltzmann statistics: Ψ = ψj(qA)ψk(qB).

• Bose-Einstein (BE) statistics: Ψ = ψj(qA)ψk(qB) + ψj(qB)ψk(qA). In this case, the total wave function Ψ is antisymmetrical.

Examples of bosons bosons are photons and composite particles, such as H1 atoms or He4 nuclei.

• Fermi-Dirac (FD) statistics: Ψ = ψj(qA)ψk(qB) – ψj(qB)ψk(qA). In this case, the total wave function Ψ is symmetrical. Ψ = 0, if both particles are placed in the same state – the Pauli Pauli

exclusion principle.exclusion principle.

Examples of bosons bosons are electrons, protons, neutrons, and composite particles, such as H2 atoms or He3 nuclei.

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Grand Partition Function ZG

• To obtain the grand partition functiongrand 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 ensemblegrand 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.

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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].

• FermionsFermions (particles with half-integer spin): ni = 0 or 1.

• BosonsBosons (particles with integer spin): ni = 0,1, 2,…… ∞.

Bath type Heat bath Heat and particle bath

Probability pr = exp (– Er)/Z pN,r = exp (μN – EN,r)/Z

Statistical parameter

Z = r=1 exp (– Er)

Partition function

ZG= N=0 r=1 exp (μN – EN,r)

Grand partition function

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Fermi-Dirac Statistics

ni is the mean no of spin-1/2 fermions in the i’th state.

All values of μ, positive or negative are allowable, sinceni always lies in the range 0 ≤ ni ≤ 1.

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Bose-Einstein Statistics

∑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.

ni is the mean no of bosons in the i’th state.

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Density of States 4

• For a set of spin-zero bosonsbosons, , g’(k)dk = V k2dk /(2π2).

• For a set of spin-½ fermions, fermions, g’(k)dk = 2V k2dk /(2π2),

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 byf()d = g’(k)dk = g’(k)(dk/d)d.

• Now = p2/2m = (kħ)2/2m, so that

k = √(2m)/ħ , dk/d = (1/2ħ)(2m/)1/2.

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Density of States 5Bose-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/(2π2), k = √(2m)/ħ , dk/d = (1/2ħ)(2m/)1/2.

• Hence, f()d = Vk2(dk/d)d/2π2 = V[4πm/ħ3](2m/)1/2d;

i.e. f()d = (2πV/h3)(2m)3/21/2 d.

Free electron theory (spin ½ system)

f()d = (4πV/h3)(2m)3/21/2 d.

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Density of States 6

• Suppose that for an N-particle system with continuous ε,

i. the number of states in the range ε to ε + dε is f()d; ii. the mean number of particles of energy is <n()>.

• The number of particles with energies in the range ε to ε + dε is

dN(ε) = <n()> f()d.

• Thus, the total no. of particles is given by N = ∫dN(ε), and the total energy is given by U = ∫ε dN(ε) = ∫ε <n()> f()d, where the integration limits are 0 and ∞.

• The values of <n()> for quantum systems are given by<n()> = 1/{exp[β(ε – μ)] ± 1}.

• The distribution f()d is called the density of statesdensity of states.

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Mean number of bosons or fermions

If there is a fixed number of particles N, ∑<n(ε)> = N

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Classical limit• Quantum statistics gives

<nr> = 1/{exp[β(εr – μ)] ± 1}.

• In the classical limit, the energy states r are infinitesimally close, so that

<nr> → 0, <nr>–1 → ∞, exp[β(εr – μ)] » 1

and<nr> → exp[β(μ – εr)] = exp(βμ). exp(– εr) ,

where z is the single-particle partition function.

• The single-particle Boltzmann distribution ispr = <nr>/N = exp(– εr).z .

• Thus,

N = z exp(βμ).

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 ε = p2/2m.

• The number of states in the range ε to ε + dε is given by f(ε) dε = C ε1/2 dε,

where C = (2πV/h3)(2m)3/2 for spin 0 bosons, and (4πV/h3)(2m)3/2 for spin ½ fermions.

• The occupation numbers of each state are given by the BE and FD distribution functions

<n(ε)> = 1/{exp[β(ε – μ)] – 1} and 1/{exp[β(ε – μ)] +1} respectively.

The distribution of particles is given by dN(ε) = <n()> f()d.

LOW TEMPERATURE FD DISTRIBUTION• As β → 0, exp[β(ε – μ)] behaves as follows:• If ε < μ, exp[β(ε – μ)] → exp(-∞) = 0, so that <n(ε)> = 1; if ε > μ, exp[β(ε – μ)] → exp(∞) = ∞, so that <n(ε)> = 0.

• Thus dN(ε) = f(ε)dε for ε < μ, and dN(ε) = 0 for ε > μ. The The Fermi energyFermi energy F is defined as F ≡ μ(T → 0).

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Free-Electron Theory: Fermi Energy 1

The Fermi energy F ≡ μ(T=0) At T=0, the system is in the state of lowest energy, so that the N lowestsingle-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, andstates 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 withone conduction electron per atom. Below room temperature, T/TF < 0.03, and μ ≈ F.

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Free-Electron Theory: Heat capacity

Simplified calculation for T << TF

Assume that only those particles within anenergy kT of F can be excited and havemean 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 + γT,so that

CV/T = AT2 + γ.

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Free-Electron Theory: Fermi Energy 2The number of electrons with energies between and + d is given by

dN() = n() f() d, where n() = 1/{[exp( – μ)] + 1}, .

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Free-Electron Theory: Calculation of <n>.

Now U = 0 n()f() d.

LOW TEMPERATURE B-E DISTRIBUTION

• The distribution of particles dN(ε) = <n()> 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(ε) = C ε1/2] is zero at ε = 0.• The thermodynamic approach to Bose-Einstein Bose-Einstein

Condensation 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(ε) = <n()> f()d works down to a phase transition, which occurs at the Bose or condensation Bose or condensation temperaturetemperature TB, above which the total number of particles is given by N = ∫ dN(ε).

• Below TB , appreciable numbers of particles are in the ground state.

Bose-Einstein Condensation 1• The number of particles with energies in the range ε to ε + dε is

dN(ε) = <n()> f()d, with

<n()> = 1/{exp[β(ε – μ)] – 1}

and

• Thus,dN(ε) = (2πV/h3)(2m)3/21/2 d /{exp[β(ε – μ)] – 1},

and the total number of particles N is given by

N = (2πV/h3)(2m)3/2 ∫1/2 d /{exp[β(ε – μ)] – 1},

which is integrated from = 0 to ∞.30

f()d = (2πV/h3)(2m)3/21/2 d.

Bose-Einstein Condensation 2• Since the number of particles are fixed, with

N = (2πV/h3)(2m)3/2 ∫1/2 d /{exp[β(ε – μ)] – 1},

the integral• ∫1/2 d /{exp[β(ε – μ)] – 1}

must be positive and independent of temperature.

• Since min= 0 for an ideal gas, the chemical potential μ ≤ 0, the factor

exp(β |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 Bose Temperature TB, at which μ becomes 0.

What happens below TB?31

Bose-Einstein Condensation 3• At the Bose (or condensation) temperature TB, μ ≈ 0, so that

N = (2πV/h3)(2m)3/2 ∫1/2 d /{exp(ε/kTB) – 1},

which yields on integration, TB = (h2/2πmk)(N/2.612V)2/3.

• The expression f()d = K1/2 d, with K = (2πV/h3)(2m)3/2, indicates that for T > TB, the number of particles in the

ground state ( = 0) is negligible, since f() = K1/2 → 0.

Behavior of μ above TB.

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Bose-Einstein Condensation 4

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

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Classical high-temperature value

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Appendix

Alternative Approach to Quantum Statistics

PHYS 4315R. S. Rubins, Fall 2008

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Lagrange Method of Undetermined Multipliers 1Simple example• How to find an extremum for a function f(x,y), subject to the

constraint φ(x,y) = constant.• Suppose f(x,y) = x3 +y3, and φ(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

= α, so that

In this example, α is a Lagrange undermined multiplier.

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Lagrange Method of Undetermined Multipliers 2Suppose 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

Lagrange’s method of undetermined multipliers gives the following set of equations:

mean number of particles in the state j.

In the calculations that follow, the function f equals ln(ω), where ω is the thermodynamic degeneracy.

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Alternate Fermi-Dirac Calculation 1

If the j’th 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,

.

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Alternate Fermi-Dirac Calculation 2The total no. of microstates is obtained by summing over all j; i.e.

Therefore,

Using Stirling’s theorem, ln N! ≈ N ln N – N, we obtain

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Since the constraints are ,

we let φ(N1…Nj…) = N, and ψ(N1 …Nj…) = U, so that

where α and β are Lagrange multipliers.Inserting the expression for ln ωFD and remembering that

,

we obtain

.

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reduces to

Thus <nj> = .

The constraint α has been replaced by μ/kT, where μ is the chemical potential, and β by –1/kT.

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Alternate Bose-Einstein Calculation 1

The j’th energy level has gj quantum states, and contains atotal 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

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Alternate Bose-Einstein Calculation 2The total no. of microstates is obtained by summing over all j; i.e.

Therefore,

Using Stirling’s theorem, ln N! ≈ N ln N – N, we obtain

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where α and β are Lagrange multipliers.Inserting the expression for ln ωFD, we obtain

hence

.

Using the method of Lagrange multipliers as before,

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reduces to

Thus <nj> = .

The constraint α has been replaced by μ/kT, where μ is the chemical potential, and β by –1/kT.

1 photon statistics 1 a single photon in the state r has energy r = ħω r. the number of photons...

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