Chapter 6 Free Electron Fermi Gas Chapter 6 Free Electron Fermi Gas Chapter G Free Electron Fermi Gas Introduction � Free electron model extremely powerful way of looking at metals (eg especially for[.]
Chapter 6: Free Electron Fermi Gas Introduction • • • Free electron model: extremely powerful way of looking at metals (eg especially for simple metals such as Li, Na, K) In this case, we assume that the conduction electrons can freely move throughout the lattice (without feeling the postively charged ions in the lattice) What does this look like? Have we seen this idea in modern physics before? Particles in a box • • • • • U(x) = (0 ≤ x ≤ L), ∞ otherwise Let’s assume that we have N electrons (1 for each ion) that can roam around freely in a lattice of size L x L x L Let’s assume that they not feel the positive cores, or interact with one another (these ideas will be tested later on) What would the possible energy levels look like? We call this model the free electron gas – the electrons act like a “gas” of non-interacting particles confined to a box L Let’s look at this in 1D first Particle in a box : 1D • To find the possible energy levels for particles which can exist in this box, we must solve the Schrödinger equation: energy of electron in orbital n HΨn ( x) = ε n Ψn ( x) Hamiltonian = Kinetic Energy + Potential Energy • Wavefunction for orbital n For particles confined to this box, there is only a kinetic energy term to the Hamiltonian H p2 d d H= = (−ih )(−ih ) 2m 2m dx dx h2 d H =− 2m dx so we have to solve Quantum mechanical operator h2 d Hψ n = − ψ n = ε nψ n 2m dx Particle in a box: 1D • • • Before we find solutions for the possible energy levels εn, we need to consider boundary conditions At x = 0, and x = L, the wavefunction must be zero (because the potential is infinite, or another way of saying this is that the electron is confined to the box) The solution to this wavefunction is sinelike: 2π ψ n = A sin λn λn = L / n x U(x) = (0 ≤ x ≤ L), ∞ otherwise Ψ(x) λ=(2/3)L λ=L λ=2L L Particle in a box: 1D • • Notice that this solution satisfies the boundary conditions (the wavefunction is zero at x = and x = L) To show that this is a solution: dΨn nπ nπ = A cos dx L L x d Ψn nπ nπ = − A x sin dx L L • So, the energy for these levels are: h d ψn h nπ − = ε nψ n ⇒ ε n = 2m dx 2m L 2 2 Particle in a box: 1D • • Another way of looking at this: the energy looks like a free particle of wavelength λn = 2L/n Therefore, the energy is: p h k h 2π = = εn = 2m 2m 2m λn 2 2 h 2π h nπ εn = = 2m L / n 2m L • The energy goes like (and it is quantized) n2 n=3 ε1=9(ħπ)2/(2mL2) = Ψ(x) ε1=4(ħπ)2/(2mL2) n=2 ε1=(ħπ)2/(2mL2) n=1 L