(quantum statistics) Classical statistical mechanics 1. - - PowerPoint PPT Presentation

EE201/MSE207 Lecture 14

Particle distributions at 𝑈 ≠ 0 (quantum statistics)

• 1. Microcanonical ensemble

Classical statistical mechanics 𝐹 = const 𝑂 = const

(gas of particles distributed among energy/velocity levels) All microstates are equally probable (postulate)

• 2. Canonical ensemble

big reservoir

𝑂 = const 𝐹 ≠ const

(exchange of heat) Probability of a state

𝑄 𝐹 ∝ 𝑓−𝐹/𝑈

(𝑙𝐶 = 1, 𝑙𝐶𝑈 → 𝑈) Follows from the postulate for microcaconical ensemble (this is how temperature is introduced)

Classical statistical mechanics (cont.)

𝐹 ≠ const

(particles can penetrate) Probability of a state 𝑄 𝐹, 𝑂 ∝ 𝑓−(𝐹−𝜈𝑂)/𝑈 (two parameters: temperature and chemical potential)

𝑂 ≠ const

• 3. Grand canonical ensemble

Chemical potential 𝜈: average energy cost of bringing an extra particle from big reservoir The formula for 𝑄(𝐹, 𝑂) also follows from equiprobability in microcanonical ens. From 𝑄(𝐹, 𝑂) we can derive 𝑜(𝜁): average number of particles with energy 𝜁

𝐹 = 𝑗 𝑜 𝜁𝑗 𝜁𝑗 𝑜 𝜁 = exp − 𝜁 − 𝜈 𝑈

Maxwell-Boltzmann distribution Derivation 𝑄𝑙: probability to have 𝑙 particles with quantized (binned) energy 𝜁 𝑄𝑙 = 𝑄0 𝑓−𝑙 𝜁−𝜈 /𝑈 𝑙! (𝑙! comes from number of combinations) 𝑙=0

∞

𝑄𝑙 = 1  𝑄0 = 1 exp[exp(−(𝜁 − 𝜈)/𝑈)]  𝑙 = exp[− (𝜁 − 𝜈) 𝑈]

Quantum statistics

Main difference: indistinguishable particles (instead of a question “which one” we are only allowed to ask “how many”) Example: two particles

• n two levels

classical

12 12 1 2 1 2

equal probabilities, 1/4 each quantum

II I I

equal probabilities, 1/3 each

II

Fermions

Either 0 or 1 particle on a level with energy 𝜁 (spin increases number of levels, still no 2 particles on the same level)

𝑄

1

𝑄 = 𝑓− 𝜁−𝜈 /𝑈

Still use classical relation 𝑄 𝐹, 𝑂 ∝ 𝑓− 𝐹−𝜈𝑂 /𝑈

𝑄

0 + 𝑄 1 = 1

⇒

𝑄

0 =

1 1 + 𝑓− 𝜁−𝜈 /𝑈 , 𝑄

1 =

𝑓− 𝜁−𝜈 /𝑈 1 + 𝑓− 𝜁−𝜈 /𝑈 ,

⇒

𝑜 = 𝑄

1 =

1 1 + 𝑓 𝜁−𝜈 /𝑈

Fermi-Dirac distribution (Fermi statistics) 𝜈 is Fermi level (chemical vs. electrochemical)

Quantum statistics (cont.)

Bosons

𝑄

1

𝑄 = 𝑓− 𝜁−𝜈 /𝑈, 𝑄

2

𝑄 = 𝑓−2 𝜁−𝜈 /𝑈, 𝑄

3

𝑄 = 𝑓−3 𝜁−𝜈 /𝑈, . . . 𝑄

𝑜 = 1

⇒

𝑄

0 =

1 1 + 𝑓− 𝜁−𝜈 /𝑈 + 𝑓−2 𝜁−𝜈 /𝑈+ . . . = 1 − 𝑓− 𝜁−𝜈 /𝑈

𝑜 = 0 ∙ 𝑄0 + 1 ∙ 𝑄

1 + 2 ∙ 𝑄2 + . . . = 1 − 𝑓−𝜁−𝜈 𝑈

(1 ∙ 𝑓−𝜁−𝜈

𝑈 + 2 ∙ 𝑓−2𝜁−𝜈 𝑈 + ⋯ )

= 1 − 𝑓−𝜁−𝜈

𝑈

𝑓−𝜁−𝜈

𝑈

1 − 𝑓−𝜁−𝜈

𝑈

+ 𝑓−2𝜁−𝜈

𝑈

1 − 𝑓−𝜁−𝜈

𝑈

+ ⋯ = 𝑓−𝜁−𝜈

𝑈

1 − 𝑓−𝜁−𝜈

𝑈

𝑜 = 1 𝑓 𝜁−𝜈 /𝑈 − 1

Bose-Einstein distribution (Bose statistics) 𝜈 ≤ 0 (if energy starts from 0),

• therwise infinity at 𝜁 = 𝜈

Particle distributions: summary

𝑜 𝜁 =

𝑓− 𝜁−𝜈 /𝑈 1 𝑓 𝜁−𝜈 /𝑈 + 1 1 𝑓 𝜁−𝜈 /𝑈 − 1

Bose-Einstein Fermi-Dirac Maxwell-Boltzmann

 

 

𝜁

    

𝜁 𝑈 = 0 𝜈

    

𝜁 𝜈

𝜈 is Fermi level

To find 𝜈: 𝑂 = 𝑜 𝜁 𝐸 𝜁 𝑒𝜁

density of states

𝑜(𝜁) depends on temperature

⇒ 𝜈 depends on temperature

Remark 1. Often notation 𝑔 𝜁 instead of 𝑜(𝜁), especially for Fermi distribution Remark 2. Large-𝜁 tails of Fermi-Dirac and Bose-Einstein distributions coincide with Maxwell-Boltzmann distribution

(Boltzmann) (Fermi) (Bose)

2D case

(not in textbook)

𝐸 𝜁 𝐵 = 𝑛 2𝜌ℏ2 2𝑡 + 1

𝑡 is spin, in general 2𝑡 + 1 is degeneracy

𝑂 𝐵 =

∞

𝑛 2𝜌ℏ2 2 1 𝑓(𝜁−𝜈)/𝑈 + 1 𝑒𝜁 = 𝑛 2𝜌ℏ2 2 𝑈 ln(1 + 𝑓𝜈/𝑈)

Electrons (Fermi, 𝑡 = 1 2) spin spin No spin factor of 2 in high magnetic field

𝑂 𝐵 =

∞

𝑛 2𝜌ℏ2 1 𝑓(𝜁−𝜈)/𝑈 − 1 𝑒𝜁 = 𝑛 2𝜌ℏ2 𝑈 ln(1 − 𝑓𝜈/𝑈)

Bosons with 𝑡 = 0 𝐸 𝜁 is density of states, 𝐵 is area

𝐹 𝑊 =

∞

𝜁 𝑛3/2𝜁1/2 2 𝜌2ℏ3 1 𝑓(𝜁−𝜈)/𝑈 ± 1 (2𝑡 + 1) 𝑒𝜁

3D case

𝐸 𝜁 𝑊 = 𝑛3/2𝜁1/2 2 𝜌2ℏ3 2𝑡 + 1

𝑡 is spin, in general 2𝑡 + 1 is degeneracy (including valleys, etc.)

𝑂 𝑊 =

∞ 𝑛3/2𝜁1/2

2 𝜌2ℏ3 1 𝑓(𝜁−𝜈)/𝑈 ± 1 (2𝑡 + 1) 𝑒𝜁

degeneracy 𝐸 𝜁 is density of states, 𝑊 is volume Fermi: “+”, Bose: “−” Unfortunately, these integrals cannot be calculated analytically. Simplification if −𝜈 ≫ 𝑈, then F-D and B-E distributions reduce to M-B.

1 𝑓(𝜁−𝜈)/𝑈 ± 1 ≈ 𝑓− 𝜁−𝜈 /𝑈 𝜁 − 𝜈 ≫ 𝑈

when

(e.g., for heat capacity)

Nondegenerate semiconductor

conduction band Assume n-type (p-type similar), −𝜈 ≫ 𝑈, 2𝑡 + 1 = 2

𝑂 𝑊 ≈

∞ 𝑛3/2𝜁1/2

2 𝜌2ℏ3 𝑓−(𝜁−𝜈)/𝑈2 𝑒𝜁 = . . . 𝜈

(Fermi level) valence band (neglect) > 3𝑈 Room temperature: 𝑈 = 26 meV

= 2 𝑓𝜈/𝑈 𝑛𝑈 2𝜌ℏ2

3/2

𝜈 = 𝑈 ln 𝑂 𝑊 1 2 2𝜌ℏ2 𝑛𝑈

3/2

degeneracy; can be larger, Si: 26

𝐹 𝑊 ≈

∞

𝜁 𝑛

3 2𝜁 1 2

2 𝜌2ℏ3 𝑓−

𝜁−𝜈 𝑈2 𝑒𝜁 = . . . = 3

2 𝑈 𝑂 𝑊 𝐹 = 3 2 𝑈𝑂

Bose-Einstein condensation

For Bose-Einstein distribution usually 𝜈 < 0 (cannot be 𝜈 > 0). However, at small enough 𝑈, it becomes 𝜈 = 0, then

𝑂 𝑊 =

∞ 𝑛3/2𝜁1/2

2 𝜌2ℏ3 1 𝑓−𝜈/𝑈 − 1 𝑒𝜁 = 2.61 𝑛𝑈 2𝜌ℏ2

3

(𝑡 = 0) Therefore critical temperature

𝑈

𝑑 = 2𝜌ℏ2

𝑛 𝑂 2.61 𝑊

2/3

Below 𝑈

𝑑 particles crowd into the ground state

(finite fraction of all particles occupy ground state) Different calculation:

𝑂 = 𝑂 0 + 𝑜 𝜁 𝐸 𝜁 𝑒𝜁

Examples: superconductivity, superfluidity, B-E condensation of atoms

Massless particles (photons, phonons)

𝜁 = ℏ𝜕 𝑙 = 2𝜌 𝜇 = 𝜕 𝑑

speed of light or sound velocity

Number of particles is not conserved ⇒

𝜈 = 0

(creation of extra particle does not cost extra energy)

𝑜(𝜕) = 1 𝑓ℏ𝜕/𝑈 − 1

(bosons) DOS:

𝑒𝑂 = 𝑒𝑦 𝑒𝑙𝑦 𝑒𝑧 𝑒𝑙𝑧 𝑒𝑨 𝑒𝑙𝑨 2𝜌 3 ⇒ 𝑒𝑂 𝑊 = 𝑒𝑙𝑦 𝑒𝑙𝑧 𝑒𝑙𝑨 2𝜌 3 𝑒𝑂 𝑊 𝑒𝜕 = 4𝜌𝑙2 2𝜌 3 𝑒𝑙 𝑒𝜕 = 𝜕2 2𝜌2𝑑3 × 2 for photons (two polarizations) × 3 for phonons, better

2 𝑑⊥

3 + 1

𝑑∥

3

Average energy per d𝜕 (for photons)

𝑒𝐹 𝑊 𝑒𝜕 = ℏ𝜕 2𝜕2 2𝜌2𝑑3 1 𝑓ℏ𝜕/𝑈 − 1 = 2ℏ𝜕3 2𝜌2𝑑3(𝑓ℏ𝜕/𝑈 − 1)

(Planck’s formula)