Lecture 7Phys 446 Solid State Physics Lecture 7 (Ch. 4.1 – 4.3, 4.6.)Last time: Finished with phonons, optical and thermal properties. Today: Start with electronic properties of metals.Free electron model. Fermi energy.Density of states. Electronic heat capacityElectrons in metals: free electron model•Simplest way to represent the electronic structure of metals •Although great simplification, works pretty well in many cases, describes many important properties of metals•In this model, the valence electrons of free atoms become conduction electrons in crystal and travel freely •Neglect the interaction of conduction electrons with ions of the lattice and the interaction between the conduction electrons – a free electron gas•Fundamental difference between the free electron gas and ordinary gas of molecules: 1) electrons are charged particles ⇒ to maintain the charge neutrality of the whole crystal, we need to include positive ions. This is done within the jelly model : the positive charge of ions is smeared out uniformly throughout the crystal - charge neutrality is maintained, no field on the electrons exerted2) Free electron gas must satisfy the Pauli exclusion principle, which leads to important consequences.Free electron gas in one dimensionAssume an electron of mass m is confined to a length L by infinite barriersSchrödinger equation for electron wave function ψn(x): En- the energy of electron orbitalassume the potential lies at zero ⇒ H includes only the kinetic energy ⇒Note: this is a one-electron equation –neglected electron-electron interactionsGeneral solution: Asin qnx+ Bcos qnxboundary conditions for the wave function: ⇒ B = 0; qn= πn/L ; n -integerSubstitute, obtain the eigenvalues: What is Hamiltonian?First three energy levels and wave-functions of a free electron of mass m confined to a line of length L: picture from KittelWe need to accommodate N valence electrons in these quantum states.Pauli principle: no two electrons can have identical quantum numbers. Electronic state in a 1D solid is characterized by quantum numbers n and ms, where n describes the orbital ψn(x), and ms- the projection of the spin: ms= ±½.⇒ each orbital labeled by the quantum number n can accommodate two electrons, one with spin up and one with spin down orientation.Let nF- the highest filled energy level. Start filling the levels from the bottom (n = 1) and continue until all N electrons are accommodated. Condition 2nF= N determines nFFermi energyThe energy of the highest occupied level is called the Fermi energy EF For the one-dimensional system of N electronsFinite temperature: the Fermi - Dirac distributionThe ground state of the N electron system at zero temperature: all the electronic levels are filled up to the Fermi energy. All the levels above are empty.What happens if the temperature is increased? The kinetic energy of the electron gas increases with temperature ⇒ some energy levels become occupied which were vacant at 0 K; some levels become vacant which were occupied at 0 K. The distribution of electrons among the levels is described by the distribution function, f(E) - the probability that the level E is occupied11)()(+=− TkEBeEfµFermi - Dirac distribution: µ− the chemical potential. It can be determined in a way that the total number of electrons in the system is equal to N. At T = 0 K µ = EF11)()(+=− TkEBeEfµf(E) at T = 0 K and T> 0 KAt any T if f(E) = 1/2 when E = µHigh energy tail of f(E), when E - µ >> kBT:called Maxwell – Boltzmann distributionTkEBeEf)()(−=µEffect of temperature on Fermi-Dirac distributionFree electron gas in three dimensionsThe Schrödinger equation in the three dimensions:If the electrons are confined to a cube of edge L, the solution isintroduce periodic boundary conditions, as we did for lattice vibrations– assume that our crystal is infinite and disregard the influence of the outer boundaries of the crystal on the solution – require that our wave function is periodic in x, y, and z directions with period L, so that and similarly for the y and z coordinates. The solution of the Schrödinger equation satisfying these boundary conditions has the form of the traveling plane wave:rkkr⋅=iAe)(ψprovided that the component of the wave vector k satisfywhere nx, ny, and nz- integerssubstitute this to the Schrödinger equation, obtain the energy of the orbital with the wavevector k:Wave functions ψk– the eigenfunctions of the momentum operator The eigenvalue of the momentum is ħk.∇−= =ipThe velocity of the electron is defined by v = p/m= ħk/m/FBEkFermi energy and Fermi momentumIn the ground state a system of N electrons occupies states with lowest possible energies ⇒ all the occupied states lie inside the sphere of radius kF. The energy at the surface of this sphere is the Fermi energy EF.The magnitude of the Fermi wave vector kFand the Fermi energy are related by the equation:The Fermi energy and the Fermi momentum are determined by the number of valence electrons in the system N.We need to count the total number of energy orbitals in a sphere of radius kFwhich should be equal to N.The volume element in the k space (volume per single set of kx, ky, and kz) is equal to 32⎟⎠⎞⎜⎝⎛LπThus in the sphere of the total number of states is343FFkVπ=where does the factor 2 come from ?⇒3123⎟⎟⎠⎞⎜⎜⎝⎛=VNkFπ- depends only of the electron concentrationObtain then for the Fermi energy:and the Fermi velocity: 322232⎟⎟⎠⎞⎜⎜⎝⎛=VNmEFπ=3123⎟⎟⎠⎞⎜⎜⎝⎛=VNmvFπ=density of statesEFkBTDensity of statesDefined as the number of electronic states per unit energy range –an important characteristic of electronic properties of a solid To find it, write the total number of orbitals of energy < E. We had 3222)(32⎟⎟⎠⎞⎜⎜⎝⎛=VENmEπ=⇒232223)(⎟⎠⎞⎜⎝⎛==mEVENπSo, the density of states D(E) is32122223()()22dN V m N EDE EdE Eπ⎛⎞== =⎜⎟⎝⎠=the integral- total number of electrons in system (at 0K)At T ≠ 0 should take into account the Fermi distribution:Heat capacity of the electron gas• Classical statistical mechanics - a free particle should have 3kB/2 ;N atoms each give one valence electron and the electrons are freely mobile ⇒ the heat capacity of the electron gas should be 3NkB/2• Observed electronic contribution at room T is usually < 0.01 of this value• The discrepancy is resolved by taking into