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Schedule for Rest of Quarter • Review: 3D Free Electron Gas • Heat Capacity

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Schedule for Rest of Quarter • Review: 3D Free Electron Gas • Heat Capacity Lecture 16 Housekeeping: schedule for rest of quarter Review: 3D free electron gas Heat capacity of free electron gas Semi classical treatment of electric conductivity Review: 3D fee electron gas This model ignores the lattice and only treats the electrons in a metal. It assumes that electrons do not interact with each other except for pauli exclusion. 풌∙풓 Wavelike-electrons defined by their momentum eigenstate k: 휓풌(풓) = 푒 Treat electrons like quantum particle in a box: confined to a 3D box of dimensions L in every direction Periodic boundary conditions on wavefunction: 휓(푥 + 퐿, 푦, 푧) = 휓(푥, 푦, 푧) 휓(푥, 푦 + 퐿, 푧) = 휓(푥, 푦, 푧) 휓(푥, 푦, 푧 + 퐿) = 휓(푥, 푦, 푧) Permissible values of k of the form: 2휋 4휋 푘 = 0, ± , ± , … 푥 퐿 퐿 Free electron schrodinger’s eqn in 3D + boundary conditions gives energy eigenvalues: ℏ2 휕2 휕2 휕2 − ( + + ) 휓 (푟) = 휖 휓 (푟) 2푚 휕푥2 휕푦2 휕푧2 푘 푘 푘 ℏ2 ℏ2푘2 (푘2 + 푘2 + 푘2) = 휖 = 2푚 푥 푦 푧 푘 2푚 There are N electrons and we put them into available eigenstates following these rules: o Every state is defined by a unique quantized value of (푘푥, 푘푦, 푘푧) o Every state can hold one spin up and one spin down electron o Assume that single-electron solution above applies to collection of electrons (i.e. electrons do not interact and change the eigenstates) o Fill low energy states first. In 3D, this corresponds to filling up a sphere in k space, one 2 2 2 2 ‘shell’ at a time. Each shell is defined by a radius k, where 푘 = 푘푥 + 푘푦 + 푘푧 , and every state in the shell has the same energy, although different combinations of 푘푥, 푘푦, 푘푧 Important concepts: Fermi energy Fermi temperature Fermi velocity Density of states Fermi surface Fermi energy: When we have used up all our electrons, we are left with a filled sphere in k space with radius 푘퐹 (called the Fermi momentum) such that ℏ2 휖 = 푘2 퐹 2푚 퐹 The boundary of this sphere is called the fermi surface and it demarcates the separation between occupied and unoccupied states. 4 2휋 3 This sphere in k-space has a volume 휋푘3 and it is divided into voxels of volume ( ) 3 퐹 퐿 If we divide the total volume of the sphere by the volume of each ‘box’ and account for the fact that each box holds 2 electrons, we get back how many electrons we put in: 4 3 휋푘퐹 2 ∗ 3 = 푁 = 푉푘3/3휋2 2휋 3 퐹 ( ) 퐿 Here, 푉 = 퐿3 is the volume of the solid. We can use this relationship to solve for k_F and show that it depends on electron density (N/V) 1/3 3휋2푁 푘 = ( ) 퐹 푉 Plugging this back into the expression for 휖퐹 we get: 2/3 ℏ2 3휋2푁 휖 = ( ) 퐹 2푚 푉 Fermi temperature is the temperature equivalent of the Fermi energy: 푇퐹 = 휖퐹/푘퐵푇. Physically, the fermi temperature represents the temperature when a free electron gas starts to act like a classical gas instead of a quantum gas. As we calculated in the last lecture, a typical Fermi temperature is >30,000K, well above the melting point of any metal. Fermi velocity describes the typical velocity of electrons at the Fermi energy, and it is given by 푣퐹 = ℏ푘퐹/푚. Typical values are ~10^6 m/s. The density of states describes the number of eigenstates at a given energy. 푑푁 퐷(휖) ≡ 푑휖 We can find it by expressing N in terms of 휖 and taking a derivative. We begin by considering a sphere in k-space with an arbitrary radius k and asking how many electrons that will hold 푁(푘) = 푉푘3/3휋2 The relationship between energy and momentum in a free electron gas is pretty straightforward too (unlike with phonons): ℏ2푘2 휖 = 2푚 Solving for k, and plugging in above we get 푉 2푚휖 3/2 푁(휖) = ( ) 3휋2 ℏ2 Now we can just take the derivative with respect to energy and get: 3 푑푁 푉 2푚 2 퐷(휖) ≡ = ( ) 휖1/2 푑휖 2휋2 ℏ2 Effect of temperature Temperature introduces a ‘cutoff’ by the Fermi-dirac function 1 푓(휖) = 푒(휖−휇)/푘퐵푇 + 1 Such that some states with 휖 > 휖퐹~휇 can be occupied and some states with 휖 < 휖퐹~휇. Temperature only affects states roughly within 푘퐵푇 of the Fermi energy. Another way to think of the effect of temperature is the fuzzing out of the boundary of the Fermi surface. Heat Capacity of free electron gas In chapter 5, we learned about lattice heat capacity—how inputting energy into a solid raises the temperature by exciting more vibrational modes. However, in metals, particularly at low temperature, this is not the whole story, because electrons can absorb heat as well. Qualitative derivation In a free electron gas, only electrons with energy within ~푘퐵푇 of the Fermi level do anything. This represents a small fraction of the total electrons N, given by 푁푇/푇퐹 where 푇퐹 is the Fermi temperature which is usually ~104퐾, well above the melting point of metals. Thus, the total electronic thermal kinetic energy when electrons are heated from 0 to temperature T is 푁푇 푈푒푙 ≈ ( ) 푘퐵푇 푇퐹 The heat capacity is found from the temperature derivative: 휕푈 퐶 = ≈ 푁푘 푇/푇 푒푙 휕푇 퐵 퐹 This sketch of a derivation is intended only to achieve the proper temperature dependence: 퐶푒푙 ∝ 푇, which we will show more rigorously in the next section Quantitative derivation This derivation of electron heat capacity is applicable to the regime when a Fermi gas does not behave like a classical gas—when 푘퐵푇 ≪ 휖퐹 The change in internal energy when electrons are heated up to temperature T from 0K is given by: ∞ 휖퐹 Δ푈 = 푈(푇) − 푈(0) = ∫ 푑휖 휖퐷(휖)푓(휖) − ∫ 푑휖 휖퐷(휖) 0 0 1 Where 푓(휖) is the Fermi function, 푓(휖) = , which describes the occupation probability of a 푒(휖−휇)/푘퐵푇+1 given energy level. It is equal to 1 for 휖 ≪ 휇 and 0 for 휖 ≫ 휇 and something in between 0 and 1 for |휖 − 휇|~푘퐵푇. The parameter 휇 is called the “chemical potential”, and it’s value is temperature dependent and close to 휖퐹 for most temperatures one might realistically encounter. 푉 2푚 3/2 And 퐷(휖) is the density of states, where for a 3D free electron gas, 퐷(휖) = ( ) 휖1/2 2휋2 ℏ2 The integral terms above take each energy slice, multiply it by how many electrons have that energy (via the density of states multiplied by the Fermi function), and sum up over all the energies available to an electron. The second integral truncates at 휖 = 휖퐹 at its upper bound because at T=0, 휇 = 휖퐹, and the Fermi function is a step function which is equal to zero for 휖 > 휖퐹. The Fermi energy is determined by the number of electrons and there are two ways to express this similarly to the integrals above: ∞ 휖퐹 푁 = ∫ 푑휖 퐷(휖)푓(휖) = ∫ 푑휖 퐷(휖) 0 0 The right-most integral is the total number of electrons at zero temperature, and the other integral is the total number of electrons at finite temperature. They must be equal since electrons (unlike phonons) cannot be spontaneously created. We now multiply both integrals by 휖퐹, which is a constant. This is just a mathematical trick. ∞ 휖퐹 ∫ 푑휖 휖퐹퐷(휖)푓(휖) = ∫ 푑휖 휖퐹퐷(휖) 0 0 And split up the first integral: 휖퐹 ∞ 휖퐹 ∫ 푑휖 휖퐹퐷(휖)푓(휖) + ∫ 푑휖 휖퐹퐷(휖)푓(휖) = ∫ 푑휖 휖퐹퐷(휖) 0 휖퐹 0 휖퐹 ∞ ∫ 푑휖 휖퐹퐷(휖) (푓(휖) − 1) + ∫ 푑휖 휖퐹퐷(휖)푓(휖) = 0 0 휖퐹 Use this to rewrite the expression for Δ푈 ∞ 휖퐹 Δ푈 = ∫ 푑휖 휖퐷(휖)푓(휖) − ∫ 푑휖 휖퐷(휖) 0 0 ∞ 휖퐹 휖퐹 휖퐹 Δ푈 = ∫ 푑휖 휖퐷(휖)푓(휖) + ∫ 푑휖 휖퐷(휖)푓(휖) − ∫ 푑휖 휖퐷(휖) + ∫ 푑휖 휖퐹퐷(휖) (푓(휖) − 1) 휖퐹 0 0 0 ∞ + ∫ 푑휖 휖퐹퐷(휖)푓(휖) 휖퐹 ∞ 휖퐹 Δ푈 = ∫ 푑휖 (휖 − 휖퐹)퐷(휖)푓(휖) − ∫ 푑휖 (휖퐹 − 휖)퐷(휖)[1 − 푓(휖)] 휖퐹 0 The first integral describes the energy needed to take electrons from the Fermi level to higher energy levels, and the second integral describes the energy needed to excite electrons from lower energy levels up to the Fermi level. The heat capacity is found by differentiating Δ푈 with respect to temperature, and the only terms in the integrals which have temperature dependence are 푓(휖) 휕Δ푈 ∞ 휕푓(휖, 푇) 퐶푒푙 = = ∫ 푑휖 (휖 − 휖퐹)퐷(휖) 휕푇 0 휕푇 Where we have spliced the integrals back together after the temperature derivative produced the same integrand At low temperature, 휇~휖퐹 And the temperature derivative of 푓(휖, 푇) is peaked close to 휖퐹, so the density of states can come out of the integral (this is another way of saying that only electrons very close to the Fermi level matter) ∞ 휕푓(휖, 푇) 퐶푒푙 ≈ 퐷(휖퐹) ∫ 푑휖 (휖 − 휖퐹) 0 휕푇 To solve this integral, first set 휇 = 휖퐹. This is a decent approximation for most ordinary metals at 푘 푇 휏 temperatures we might realistically encounter (remember that 퐵 = ~0.01 at room temperature ) 휖퐹 휖퐹 휖−휖퐹 휖 − 휖퐹 푘 푇 ( 2 )푒 퐵 휕푓 푘퐵푇 = 2 휕푇 휖−휖퐹 (푒 푘퐵푇 + 1) Define a new variable x and plug back into integral 휖 − 휖퐹 푥 ≡ 푘퐵푇 ∞ 푥2푒푥 퐶 ≈ 퐷(휖 )푘2푇 ∫ 푑푥 푒푙 퐹 퐵 (푒푥 + 1)2 −휖퐹/푘퐵푇 Since we are working at low temperature, we can replace the lower bound of the integral by −∞ because 푘퐵푇 ≪ 휖퐹 (our starting assumption) ∞ 2 푥 2 2 푥 푒 2 휋 퐶푒푙 ≈ 퐷(휖퐹)푘퐵푇 ∫ 푑푥 푥 2 = 퐷(휖퐹)푘퐵푇 −∞ (푒 + 1) 3 We can further express the Density of states at the Fermi energy in another way: 3푁 퐷(휖퐹) = = 3푁/2푘퐵푇퐹 2휖퐹 1 This gives 퐶 = 휋2푁푘 푇/푇 푒푙 2 퐵 퐹 This is very similar to our ‘qualitative derivation’ from earlier, except the prefactors are exact.
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