Solids (free electron gas) Simplest model : non-interacting - - PowerPoint PPT Presentation

EE201/MSE207 Lecture 13

Solids (free electron gas)

Simplest model : non-interacting electrons (no Coulomb interaction, no exchange correlation) Idea: electrons occupy energy states from the lowest energy up (fermions) At zero temperature, the highest occupied energy is called Fermi energy. So, we need to count available states. Assume many electrons, so that even though states are discrete, we always consider “thick slices” in energy, and are interested only in density of states. Density of states (DOS): number of available states per unit of energy (eV). Goal for today: find DOS and Fermi energy (at 𝑈 = 0) for quantum wires (1D), quantum wells (2D), and solids (3D) DOS does not depend on temperature (except due to change of material parameters), while Fermi level depends on temperature.

Quantum wire: large 1D well

𝑏 ≫ . . .

𝐹𝑜 = 𝑜2𝜌2ℏ2 2𝑛𝑏2

(𝑜 ≫ 1, 𝑏 is large)

𝜔𝑜 𝑦 = 2 𝑏 sin 𝑜𝜌 𝑏 𝑦 = 2 𝑏 sin 𝑙𝑜𝑦 𝑙𝑜 = 𝑜𝜌 𝑏

One state per

𝜀𝑙 = 𝜌 𝑏

4𝜀𝑙 6𝜀𝑙 8𝜀𝑙 However, it is more convenient to use both positive and negative 𝑙 (physically corresponds to two directions of momentum) Then one state per 𝜀𝑙 = 2𝜌

𝑏

𝑙 𝜀𝑙 2𝜀𝑙 3𝜀𝑙 5𝜀𝑙 7𝜀𝑙 𝑙 𝜀𝑙 2𝜀𝑙 3𝜀𝑙 4𝜀𝑙 −𝜀𝑙 −2𝜀𝑙 −3𝜀𝑙 Equivalently, one state per 𝜀𝑞 = ℏ 𝜀𝑙 = 2𝜌ℏ

𝑏

𝑞 𝜀𝑞 2𝜀𝑞 3𝜀𝑞 4𝜀𝑞 −𝜀𝑞 −2𝜀𝑞 −3𝜀𝑞 (not in textbook)

Quantum wire (cont.)

Rule:

• ne state (quantum level) per

𝜀𝑞 = 2𝜌ℏ 𝑏

So, number of states

Δ𝑂 = Δ𝑞 𝑏 2𝜌ℏ

length in space × length in 𝑞 space 2𝜌ℏ Density of states

𝐸 𝐹 ≡ Δ𝑂 Δ𝐹 𝐹𝑜 = 𝑜2𝜌2ℏ2 2𝑛𝑏2 ⟹ Δ𝐹 = 2𝑜 𝜌2ℏ2 2𝑛𝑏2 Δ𝑜 ⟹ Δ𝑜 Δ𝐹 = 𝑛𝑏2 𝑜𝜌2ℏ2 = 𝑏 𝜌ℏ 𝑛 2𝐹 𝐸 𝐹 = 𝑏 𝜌ℏ 𝑛 2𝐹 × 2 (spin) 𝐸 𝐹 ∝ 1 𝐹

decreases with energy Fermi energy We have 𝑂 ≫ 1 electrons, what is the maximum occupied energy?

𝐹𝐺 = 𝑂 2 2𝜌2ℏ2 2𝑛𝑏2 = 𝜌2ℏ2 8𝑛 𝑂 𝑏

2

spin (absent for high B-field) 𝑂 𝑏 : (linear) density of electrons (usually measured in 1/eV) 𝐹𝐺 is usually measured in eV (sometimes in Kelvin)

(absent for high magnetic field)

𝜔 𝑦, 𝑧 = 2 𝑏 sin 𝑜𝜌 𝑏 𝑦 2 𝑐 sin 𝑚𝜌 𝑐 𝑧

Quantum well, 2D electron gas, 2DEG (large 2D well)

(not in textbook)

𝑏 𝑐 𝐹𝑜,𝑚 = 𝜌2ℏ2 2𝑛 𝑜2 𝑏2 + 𝑚2 𝑐2 𝑜, 𝑚 ≫ 1

𝑙𝑦 𝑙𝑧

⟹ 𝐹 = ℏ2 2𝑛 (𝑙𝑦

2 + 𝑙𝑧 2)

𝑜 𝑚 𝑙𝑦 𝑙𝑧

equal energy line

𝑙𝑦 𝑙𝑧 𝑞𝑦 𝑞𝑧 Again the rule

Δ𝑂 = (Δ𝑞𝑦 𝑏)(Δ𝑞𝑧 𝑐) 2𝜌ℏ 2

equal energy line

area in space × area in 𝑞 space 2𝜌ℏ 2 (so far no spin)

2D electron gas (cont.)

𝑏

𝑞

𝑐

𝑞𝑦 𝑞𝑧

Δ𝑂 = (Δ𝑞𝑦 𝑏)(Δ𝑞𝑧 𝑐) 2𝜌ℏ 2

Density of states 𝑞𝑦 𝑞𝑧 Δ𝑞

𝐹 = 𝑞𝑦

2 + 𝑞𝑧 2

2𝑛 = 𝑞2 2𝑛 Δ𝐹 = 2𝑞Δ𝑞 2𝑛 = 𝑞Δ𝑞 𝑛 Δ𝑂 = 𝑏𝑐 2𝜌𝑞 Δ𝑞 2𝜌ℏ 2 𝐸 𝐹 = Δ𝑂 Δ𝐹 = 𝑏𝑐 𝑛 2𝜌ℏ2 𝐸 𝐹 𝐵 = 𝑛 2𝜌ℏ2 × 2 (spin)

(absent for high magnetic field)

𝐵 = 𝑏𝑐 (area)

𝐸(𝐹) does not depend

• n energy 𝐹

Fermi energy 𝑞𝑦 𝑞𝑧

𝑂 = 𝐵 𝜌𝑞𝐺

2

2𝜌ℏ 2 × 2 spin = 𝐵𝑞𝐺

2

2𝜌ℏ2 𝑞𝐺 𝐹𝐺 = 𝑞𝐺

2

2𝑛 = 𝜌ℏ2 𝑛 𝑂 𝐵

2D density in space

(twice larger 𝐹𝐺 in high B-field) 𝑞

Example

2DEG in GaAs,

𝑜 = 𝑂 𝐵 = 1012 1 cm2 .

Find Fermi energy.

𝑛eff = 0.067 𝑛0

𝐹𝐺 = 𝜌ℏ2 𝑛 𝑜 = 𝜌 1.05 ∙ 10−34Js 2 0.067 ∙ 9.1 ∙ 10−31kg ∙ 1016 1 m2 = = 5.68 ∙ 10−21J = 35 meV = 410 K

𝑙𝐶 = 1.38 ∙ 10−23 J K eV = 1.6 ∙ 10−19 J

Large 3D well: theory of metals

𝑏 𝑑 𝑐

Same rule

Δ𝑂 = (Δ𝑞𝑦 𝑏)(Δ𝑞𝑧 𝑐)(Δ𝑞𝑨 𝑑) 2𝜌ℏ 3 = 𝑊 Δ𝑞𝑦Δ𝑞𝑧Δ𝑞𝑨 2𝜌ℏ 3

𝑊 = 𝑏𝑐𝑑 (slightly different in textbook) 𝑞𝑧 𝑞𝑨 𝑞𝑦 𝑞 Δ𝑞 Density of states 𝐸 𝐹 = Δ𝑂 Δ𝐹

Δ𝑂 = 𝑊 4𝜌𝑞2Δ𝑞 2𝜌ℏ 3 Δ𝐹 = Δ 𝑞2 2𝑛 = 𝑞 𝑛 Δ𝑞 Δ𝑂 Δ𝐹 = 𝑊 4𝜌𝑞𝑛 2𝜌ℏ 3 = 𝑊 𝑛 2𝑛𝐹 2𝜌2ℏ3 = 𝑊 𝑛3/2 𝐹 2 𝜌2ℏ3 𝐸(𝐹) 𝑊 = 𝑛3/2 𝐹 2 𝜌2ℏ3 × 2 (spin)

3D: 𝐸 𝐹 ∝ 𝐹 2D: 𝐸 𝐹 ∝ 𝐹0 1D: 𝐸 𝐹 ∝ 1 𝐹 Some people call 𝐸(𝐹)/𝑊 density of states

Theory of metals (cont.)

𝑏 𝑑 𝑐 Δ𝑂 = 𝑊 Δ𝑞𝑦Δ𝑞𝑧Δ𝑞𝑨 2𝜌ℏ 3

𝑊 = 𝑏𝑐𝑑 𝑞𝑧 𝑞𝑨 𝑞𝑦 𝑞𝐺

Fermi energy 𝑂 = 𝑊 4 3 𝜌𝑞𝐺

3

2𝜌ℏ 3 ∙ 2 spin = 𝑊 𝑞𝐺

3

3𝜌2ℏ3 𝑞𝐺 = ℏ 3𝜌2 𝑂 𝑊

1/3

𝐹𝐺 = 𝑞𝐺

2

2𝑛 = ℏ2 2𝑛 3𝜌2 𝑂 𝑊

2/3

𝑙𝐺 = 𝑞𝐺 ℏ = 3𝜌2 𝑂 𝑊

1/3

Example 1

Copper (Cu) mass density 𝜍 = 8.96 g cm3 atomic mass 𝑁 = 63.5 g mole

Find 𝐹𝐺

𝑂 𝑊 = 1 ∙ atoms m3 = 𝜍 𝑂

𝐵

𝑁 = 8.96 ∙ 103 kg m3 6.02 ∙ 1023 1 mole 63.5 ∙ 10−3 kg mole = 8.5 ∙ 1028 1 m3

𝐹𝐺 = ℏ2 2𝑛 3𝜌2 𝑂 𝑊

2/3

𝐹𝐺 = ℏ2 2𝑛 3𝜌2 𝑂 𝑊

2/3

= 1.05 ∙ 10−34 2 2 ∗ 9.1 ∙ 10−31 3 ∙ 3.142 ∙ 8.5 ∙ 1028 2/3 = = 1.1 ∙ 10−18J = 7.0 eV = 8.1 ∙ 104K

𝑙𝐶 = 1.38 ∙ 10−23 J K eV = 1.6 ∙ 10−19 J 𝑈

𝐺 ≫ 300 K ! degenerate electron gas

Fermi velocity 𝑤𝐺 =

2𝐹𝐺 𝑛 = 1.6 ∙ 106 m s

(very high but still nonrelativistic)

Example 2

GaAs (bulk), doping of 1018 1 cm3 𝑛eff = 0.067 𝑛0

Find 𝐹𝐺 𝐹𝐺 = ℏ2 2𝑛 3𝜌2 𝑂 𝑊

2/3

= 1.05 ∙ 10−34 2 2 ∗ 0.067 ∙ 9.1 ∙ 10−31 3𝜌2 ∙ 1024 2/3 = = 8.65 ∙ 10−21J = 54 meV = 630 K

still > 300 K, behaves almost as a metal (degenerate semiconductor) However, if doping of 1017cm−3, then only 140 K.