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). DOS does not depend on temperature (except due to change of material parameters), while Fermi level depends on temperature. Goal for today: find DOS and Fermi energy (at 𝑈 = 0 ) for quantum wires (1D), quantum wells (2D), and solids (3D)

Quantum wire: large 1D well (not in textbook) 𝐹 𝑜 = 𝑜 2 𝜌 2 ℏ 2 ( 𝑜 ≫ 1 , 𝑏 is large) 2𝑛𝑏 2 2 𝑏 sin 𝑜𝜌 2 𝑏 sin 𝑙 𝑜 𝑦 𝜔 𝑜 𝑦 = 𝑏 𝑦 = 𝑙 𝑜 = 𝑜𝜌 𝑏 ≫ . . . 𝜀𝑙 3𝜀𝑙 5𝜀𝑙 7𝜀𝑙 𝑏 𝜀𝑙 = 𝜌 One state per 𝑙 𝑏 0 2𝜀𝑙 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 −3𝜀𝑙 −2𝜀𝑙 𝑙 −𝜀𝑙 0 2𝜀𝑙 3𝜀𝑙 4𝜀𝑙 𝜀𝑙 Equivalently, one state per 𝜀𝑞 = ℏ 𝜀𝑙 = 2𝜌ℏ 𝑏 𝑞 −3𝜀𝑞 2𝜀𝑞 3𝜀𝑞 4𝜀𝑞 −2𝜀𝑞 −𝜀𝑞 𝜀𝑞 0

Quantum wire (cont.) 𝜀𝑞 = 2𝜌ℏ Rule: one state (quantum level) per 𝑏 Δ𝑂 = Δ𝑞 𝑏 length in space × length in 𝑞 space So, number of states 2𝜌ℏ 2𝜌ℏ 𝐸 𝐹 ≡ Δ𝑂 Density of states (usually measured in 1/eV) Δ𝐹 𝐹 𝑜 = 𝑜 2 𝜌 2 ℏ 2 Δ𝐹 = 2𝑜 𝜌 2 ℏ 2 Δ𝐹 = 𝑛𝑏 2 Δ𝑜 𝑜𝜌 2 ℏ 2 = 𝑏 𝑛 ⟹ Δ𝑜 ⟹ 2𝑛𝑏 2 2𝑛𝑏 2 𝜌ℏ 2𝐹 𝐸 𝐹 ∝ 1 𝐸 𝐹 = 𝑏 𝑛 × 2 (spin) decreases 𝜌ℏ 2𝐹 𝐹 with energy (absent for high magnetic field) We have 𝑂 ≫ 1 electrons, what is the maximum occupied energy? Fermi energy 𝑂 2 𝑂 2 2 𝜌 2 ℏ 2 = 𝜌 2 ℏ 2 𝑏 : (linear) density of electrons 𝑂 𝐹 𝐺 = 𝐹 𝐺 is usually measured in eV 2𝑛𝑏 2 8𝑛 𝑏 (sometimes in Kelvin) spin (absent for high B-field)

Quantum well, 2D electron gas, 2DEG (large 2D well) (not in textbook) 𝐹 𝑜,𝑚 = 𝜌 2 ℏ 2 𝑏 2 + 𝑚 2 𝑜 2 𝑐 𝑜, 𝑚 ≫ 1 𝑐 2 2𝑛 𝑏 2 𝑏 sin 𝑜𝜌 2 𝑐 sin 𝑚𝜌 𝜔 𝑦, 𝑧 = 𝑏 𝑦 𝑐 𝑧 𝑙 𝑦 𝐹 = ℏ 2 𝑙 𝑧 2 + 𝑙 𝑧 2 ) ⟹ 2𝑛 (𝑙 𝑦 equal 𝑙 𝑧 𝑙 𝑧 𝑞 𝑧 𝑚 equal energy energy line line 𝑜 𝑙 𝑦 𝑙 𝑦 𝑞 𝑦 Δ𝑂 = (Δ𝑞 𝑦 𝑏)(Δ𝑞 𝑧 𝑐) Again the rule 2𝜌ℏ 2 (so far no spin) area in space × area in 𝑞 space 2𝜌ℏ 2

𝑞 𝑧 2D electron gas (cont.) 𝑐 Δ𝑂 = (Δ𝑞 𝑦 𝑏)(Δ𝑞 𝑧 𝑐) 𝑞 𝑦 2𝜌ℏ 2 𝑏 2 + 𝑞 𝑧 2 = 𝑞 2 Density of states 𝐹 = 𝑞 𝑦 Δ𝐹 = 2𝑞Δ𝑞 = 𝑞Δ𝑞 𝑞 𝑧 2𝑛 2𝑛 2𝑛 𝑛 Δ𝑞 𝑞 𝑞 Δ𝑂 = 𝑏𝑐 2𝜌𝑞 Δ𝑞 𝐸 𝐹 = Δ𝑂 Δ𝐹 = 𝑏𝑐 𝑛 𝑞 𝑦 2𝜌ℏ 2 2𝜌ℏ 2 𝐸 𝐹 𝑛 𝐸(𝐹) does not depend = × 2 (spin) 𝐵 = 𝑏𝑐 (area) on energy 𝐹 2𝜌ℏ 2 𝐵 (absent for high magnetic field) Fermi energy 2 2 2𝑛 = 𝜌ℏ 2 2 𝑂 = 𝐵 𝜌𝑞 𝐺 2𝜌ℏ 2 × 2 spin = 𝐵𝑞 𝐺 𝐹 𝐺 = 𝑞 𝐺 𝑂 𝑞 𝑧 𝑞 𝐺 2𝜌ℏ 2 𝑛 𝐵 𝑞 𝑦 (twice larger 𝐹 𝐺 in high B-field) 2D density in space

Example 𝑜 = 𝑂 1 𝐵 = 10 12 cm 2 . 2DEG in GaAs, Find Fermi energy. 𝑛 eff = 0.067 𝑛 0 𝐹 𝐺 = 𝜌ℏ 2 𝑜 = 𝜌 1.05 ∙ 10 −34 Js 2 0.067 ∙ 9.1 ∙ 10 −31 kg ∙ 10 16 1 m 2 = 𝑛 = 5.68 ∙ 10 −21 J = 35 meV = 410 K eV = 1.6 ∙ 10 −19 J 𝑙 𝐶 = 1.38 ∙ 10 −23 J K

Large 3D well: theory of metals 𝑊 = 𝑏𝑐𝑑 (slightly different in textbook) 𝑑 Same rule Δ𝑂 = (Δ𝑞 𝑦 𝑏)(Δ𝑞 𝑧 𝑐)(Δ𝑞 𝑨 𝑑) = 𝑊 Δ𝑞 𝑦 Δ𝑞 𝑧 Δ𝑞 𝑨 𝑐 2𝜌ℏ 3 2𝜌ℏ 3 𝑏 𝐸 𝐹 = Δ𝑂 𝑞 𝑨 Density of states Δ𝐹 Δ𝑞 𝑞 Δ𝑂 = 𝑊 4𝜌𝑞 2 Δ𝑞 𝑞 2 = 𝑞 Δ𝐹 = Δ 𝑛 Δ𝑞 𝑞 𝑧 2𝜌ℏ 3 2𝑛 𝑞 𝑦 = 𝑊 𝑛 3/2 𝐹 Δ𝑂 Δ𝐹 = 𝑊 4𝜌𝑞𝑛 2𝜌ℏ 3 = 𝑊 𝑛 2𝑛𝐹 2𝜌 2 ℏ 3 2 𝜌 2 ℏ 3 = 𝑛 3/2 𝐹 𝐸(𝐹) 3D: 𝐸 𝐹 ∝ 𝐹 × 2 (spin) 𝑊 2 𝜌 2 ℏ 3 2D: 𝐸 𝐹 ∝ 𝐹 0 1D: 𝐸 𝐹 ∝ 1 𝐹 Some people call 𝐸(𝐹)/𝑊 density of states

Theory of metals (cont.) 𝑊 = 𝑏𝑐𝑑 𝑑 Δ𝑂 = 𝑊 Δ𝑞 𝑦 Δ𝑞 𝑧 Δ𝑞 𝑨 2𝜌ℏ 3 𝑐 𝑏 Fermi energy 𝑞 𝑨 𝑊 4 3 3 𝜌𝑞 𝐺 3 2𝜌ℏ 3 ∙ 2 spin = 𝑊 𝑞 𝐺 𝑞 𝐺 𝑂 = 3𝜌 2 ℏ 3 𝑞 𝑧 𝑞 𝑦 1/3 1/3 𝑞 𝐺 = ℏ 3𝜌 2 𝑂 𝑙 𝐺 = 𝑞 𝐺 3𝜌 2 𝑂 ℏ = 𝑊 𝑊 2/3 2 2𝑛 = ℏ 2 𝐹 𝐺 = 𝑞 𝐺 3𝜌 2 𝑂 2𝑛 𝑊

Example 1 mass density 𝜍 = 8.96 g g atomic mass 𝑁 = 63.5 Copper (Cu) cm 3 mole 2/3 𝐹 𝐺 = ℏ 2 Find 𝐹 𝐺 3𝜌 2 𝑂 2𝑛 𝑊 6.02 ∙ 10 23 𝑂 𝑊 = 1 ∙ atoms = 𝜍 𝑂 𝑁 = 8.96 ∙ 10 3 kg kg mole = 8.5 ∙ 10 28 1 1 mole 𝐵 m 3 m 3 63.5 ∙ 10 −3 m 3 2/3 𝐹 𝐺 = ℏ 2 = 1.05 ∙ 10 −34 2 3𝜌 2 𝑂 3 ∙ 3.14 2 ∙ 8.5 ∙ 10 28 2/3 = 2 ∗ 9.1 ∙ 10 −31 2𝑛 𝑊 eV = 1.6 ∙ 10 −19 J = 1.1 ∙ 10 −18 J = 7.0 eV = 8.1 ∙ 10 4 K 𝑙 𝐶 = 1.38 ∙ 10 −23 J K 𝑈 𝐺 ≫ 300 K ! degenerate electron gas 2𝐹 𝐺 𝑛 = 1.6 ∙ 10 6 m Fermi velocity 𝑤 𝐺 = (very high but still nonrelativistic) s

Example 2 1 𝑛 eff = 0.067 𝑛 0 10 18 GaAs (bulk), doping of cm 3 Find 𝐹 𝐺 2/3 𝐹 𝐺 = ℏ 2 1.05 ∙ 10 −34 2 3𝜌 2 𝑂 2 ∗ 0.067 ∙ 9.1 ∙ 10 −31 3𝜌 2 ∙ 10 24 2/3 = = 2𝑛 𝑊 = 8.65 ∙ 10 −21 J = 54 meV = 630 K still > 300 K, behaves almost as a metal (degenerate semiconductor) However, if doping of 10 17 cm −3 , then only 140 K.