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Information about Advance Solid State Physics: Electrons in periodic potential

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Energy bands Energy bands What about energies? How do we calculate band structure in real metals? Only describe veryFor each value of p - (infinitely) many solutions of the Schrödinger equation Standard theoretical models: special materials, - Weak periodic potential; but good for illustrative purposes ε l ( p) - Tight-binding model. Properties: Approximate numerical methods: - Orthogonalized plane waves; More powerful, but less - Pseudopotential; illustrative Periodic in the reciprocal lattice: ε l ( p + K ) = ε l ( p) - Augmented plane waves; Even (follows from the time-reversal symmetry): ε l ( p) = ε l (− p ) - Greens’ functions (Korringa – Kohn – Rostoker); -… Each of the functions ε l ( p) must attain a minimum and a maximum in the unit cell of the reciprocal lattice First-principles calculations Very powerful and technically very Energy bands! difficult Electrons in a weak periodic Energy bands in 1D potential 1 ipx / General idea: Unperturbed wave functions: ψ p ( x) = e L Periodic potential is weak Can be treated as perturbation Energies: ε (0) ( p) = p 2 / 2m When can this be good? External potential: Perturbation U ( x) = ∑U n e2π inx / a n Alkali metals (Li, Na, K, Rb, Cs) – “favorite metals of a theorist” Matrix elements: p U p = Un provided p − p = 2π n / a Generally, all s- and p- metals (Examples: Cu, Ag, Al) – reasonably First-order correction: ∆ε1 ( p) = p U p = U 0 good for certain bands Not interesting – renormalization of all energies No good: d- and f- metals (examples: Fe, Co, Cr, Mn) Second-order correction: 2 2 p U p Un Diverges! ∆ε 2 ( p) = ∑ =∑ p ε ( p) − ε ( p ) n ≠ 0 ε ( p ) − ε ( p − 2π n / a) Energy bands in 1D Energy bands Need to use perturbation theory for degenerate states p ≈ p ≈ πn / a Write the wave function as ψ = a pψ p + a p ψ p Insert into Schrödinger equation Solve the resulting secular equation ε p + ε p ⎛ ε −ε ⎞ 2 Solutions: ε= ± ⎜ p p ⎟ + Un 2 2 ⎝ 2 ⎠ The gap opens!ε ε ε 1D: Bands do not overlap 2D and 3D: Bands can overlap p p p 2

Tight-binding model Tight-binding modelUnperturbed states: Electrons localized in isolated atoms Substitute into Schrödinger equation: ⎛ 2 d2 ⎞Perturbation: Overlap of electron shells of neighboring atoms ∑ ⎜ − 2m dx 2 + U ( x − na) ⎟eipna / wn ( x) + ∑ h( x)eipna / wn ( x) V ( x) = ∑U ( x − na) n ⎝ ⎠ n n = ε p ∑ eipna / wn ( x); h( x) ≡ ∑ U ( x − ma)Exact solutions of Schrödinger equation: Bloch waves ψ ( x) = eipx / u ( x) n m≠ n Perturbation: Term with h (only contains overlap between Extended states: inconvenient for the perturbation theory. different atomic states) Define Wannier states wn ( x) = N −1/ 2 ∑ e−ipna / ψ p ( x) in 1D Unperturbed: w0 ( x) = ϕ ( x) Atomic functions p ε p = ε0 Atomic energies wn ( x) is localized on the site x ≈ na wn ( x) = w0 ( x − na)(for instance, without the Bloch modulation, wn ( x) = δ ( x − na) ) Tight-binding model Metals vs insulators ε εFirst-order correction: ∑ h(n) exp(ipna / ) ε p = ε0 + n ∑ I (n) exp(ipna / n ) or ε p = ε 0 + h(0) + 2 [ h(1) − h(0) I (1)] cos pa / p ph(n) = ∫ dxϕ ( x)h( x)ϕ ( x − na) * I (n) = ∫ dxϕ ( x)ϕ ( x − na) * Let us now fill the electron states. ε Insulator ε Metalh(1) h(0), I (1) I (0) = 1 Narrow sinusoidal band3D: Qualitatively the same; details depend on the symmetry of the p p lattice. Always narrow bands. Metals vs insulators Periodic table ε Insulator ε Metal p pMetals can be charged at no energy cost.Insulators can only be charged if energy is supplied (e.g. by temperature). MetalsSemiconductors are insulators with a narrow gap, so that the free carrierscan be easily created (temperature or doping). Also: semimetals, various species of insulators, non-crystalline solids etc. 3

Free electron gas: Fermi energy Free electron gas: Specific heat Finite size system Discrete quasimomentum Need to calculate energy of electron gas at finite temperature. 2V f F (ε ) = (1 + exp((ε − µ ) / kBT )−1 (2π )3 ∫For large systems, the details should not depend on the boundary conditions E= f F ( p 2 / 2m)d 3 p Take a sample Lx × Ly × Lz and periodic boundary conditions: d3 p eipx Lx / = 1 ⇒ px Lx = 2π nx spin ψ ( x) = ψ ( x + Lx ) Step 1: Integrate over angles 2 → g (ε )dε ε = p 2 / 2m (2π )3 d3 p How many states? dnx = Lx dpx / 2π ⇒ dn = 2s V 2 (2π )3 g (ε ) = m3/ 2 ε - density of states h Zero temperature: all states below pF are occupied π2 3 1 4π 3 pF d3 p Vp3 fF # of states: N = 2 V s ∫ (2π )3 = 2V (2π )3 3 pF = 3π 2F 3 0 Step 2: Integrate over energy ε F pF = ( 3π 2 N / V ) 1/ 3 µ - determined only by electron concentration π 2 kBT 2 ∂h 2 εF EF = pF / 2m = ( 3π 2 N / V ) 2 2/3 2 / 2m ∫ dε h(ε ) f F (ε ) = ∫ h(ε )dε + 0 6 ∂ε ε = µ kBT Free electron gas: Specific heat Step 2a: Calculate the number of particles and find the temperature dependence of the chemical potential π 2 kBT 2 g (ε F ) 2 µ (T ) ≈ ε F − 6 g (ε F ) Step 2b: Calculate the energy and the specific heat Only electrons close to the Fermi 1 ⎛ ∂E ⎞ π 2 C= ⎜ ⎟ = kBTg (ε F ) surface are thermally excited V ⎝ ∂T ⎠V 3 Width of the energy strip: kBT Only properties of electrons at the Fermi surface are relevant!!! 4

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