October 28th, 2010 Class 15 Electrons in a three-dimensional infinite box For the case of electrons in a box, the finite size box may be not the best way to go considering the macroscopic dimensions of a metal that can be regarded, without much approximation, as infinite in size. For a case like that, periodic boundary conditions may be more appropriate. Thus for the three dimensional cases, periodic boundary conditions will be used. By following the same mathematical procedure than the one we followed to determine the number of vibrational modes in a crystal, we found that the solutions for an electron in a three dimensional box with periodic boundary conditions are: 2 2 n 2m 2 2 2 2 2m x 2 y 2 z 2 n Since k r exp ik r d d 2 2 2 ik x expik x x and k x expik x x k x dx dx2 d 2 d 2 2 k y expik y y k y ik y exp ik y y and dy2 dy d d 2 2 2 ik z expik z z and k z expik z z k z 2 dz dz 2 2 2 2 n becomes Thus, 2m x 2 y 2 z 2 2 k x2 k y2 k z2 n 2m or 2 k x2 k y2 k z2 2m 2 2 n k 2m where k 2 k x2 k y2 kz2 n Our relationship between wavevector and wavelength is such that, 2n y 2nx 2nz kx , ky , kz . Lz Ly Lx The quantization comes from the boundary conditions n 0,0,0 n Lx , Ly , Lz Using the quantum mechanical definition of p p k i k k k 1 So the linear momentum of an electron with a wavevector k is ħk and the speed of such an electron is v= ħk/m The energy at the Fermi level (now in 3D called Fermi surface), is given by: F 2 2 kF 2m where kF is the wavevector at the Fermi surface. In the same way we did with the vibrational modes, we can count the number of electronic states and since the solutions are the same, we can see that in the unit of volume there are 3 4 3 L electronic states in the k space. Thus in a sphere of volume k F (volume of the Fermi 3 2 3 V 3 L 4 3 sphere) there are 2 3 k f 2 k F states (counting both spin states). If in the same 3 2 volume there are N atoms and each contributes with 1 electron, then that number above must be equal to N from where 3 2 N k F V 1 2 3 2 N F 2m V 3 2 3 2 k 3 N and F F m m V 1 3 N/V is the electron density (equal to the atomic density assuming each atom contributes with 1 electron). From the above equation it is useful to define the Fermi temperature TF as the temperature for which kBTF=F. This temperature has nothing to do with the thermodynamical temperature, TF~104 is the temperature Tg an electron gas need to be for its thermal energy be kBTg=F. Density of States 2 2 3 2 N 3 , is valid for any value of energy not only for the Fermi energy, The equation 2m V in that case Nε represents the number of electrons up to that energy. So inverting the equation we obtain V N 2 3 2m 2 3 2 where N is the number of one electron states (with or without electrons) considering each energy level as 2 states (one for spin up and the other for spin down) with energy between 0 and for any larger or smaller than the Fermi energy. so the density of states is 2 dN V 2m D d 2 2 2 3 2 1 2 3 3 V 2m 2 3N 2 3 2 2 2 Heat Capacity As said before, only a few electrons near the Fermi level contribute to the heat capacity, electrons with energy much smaller that the Fermi level cannot take advantage of the thermal energy because energy levels above them are occupied and the Pauli principle does not allow them to also occupy those energy levels. Classically, the energy of an electron gas is 3/2kBT per electron, thus each electron contributes to the heat capacity by a quantity 3/2kB (½ per degree of freedom). In the Drudes model (no Pauli), all the N electrons contribute to the heat capacity and thus the electronic contribution to the heat capacity was predicted to be 3/2NkB, or the specific heat capacity 3/2 N/V kB where N/V is the electron concentration. Considering the Pauli principle, electrons with energy within kBT of the Fermi energy can absorb energy, thus only a fraction T/TF can actually contribute to the heat capacity and then qualitatively the total thermal kinetic energy of electrons is Uel≈(NT/TF)kBT leading to a heat capacity of Cel≈NkB(T/TF) which is not the exact value but retains the T dependence found in the experiment. At room temperature, T/TF~0.01 for TF~3x104K which is in the order of a typical value for the Fermi temperature For a more quantitative calculation consider the change in energy of the metal when the temperature increases to a certain temperature T is given by F 0 0 U f , T D d D d Multiplying the identity F 0 0 N f , T D d D d by F F F f , T D d F D d 0 what leads to 0 3 F 0 0 F D d F f , T D d 0 Adding this zero to the equation above and reorganizing we get F 0 0 U F f , T D d F D d F F 0 F 0 F f , T D d F f , T D d F D d or F U f , T D d F F F 1 f , T D d 0 The first term is the product of the energy gained by one electron when it goes from the Fermi level to the energy level , times the number of states at the energy times the probability that the state has been occupied integrated in all the level above the Fermi level. This is equal to the total amount of energy invested to populate energy levels about the Fermi level. The second term is the product of the energy gained by one electron when it goes from the energy level below the Fermi level to the Fermi level times the number of states at the energy times the probability that the state has indeed been vacated integrated in all the level below the Fermi level, that is the total amount of energy invested to de-populate energy levels below the Fermi level. So now the heat capacity can be obtained simply by taken the derivative of the last equation with respect to T. Noticing that only f() depends on T dU df , T Cel F D d dT F dT df , T df , T D d F D d F F dT 0 0 dT Since we are mostly interested in temperatures around room temperature, then T/TF= kBT/F~0.01 where the density of states can be approximated by the density of states at the Fermi level and the chemical potential can be considered independent of temperature and ~F Then Cel D F F 0 f , T df , T d dT 1 F exp 1 k BT F exp k BT df , T F k BT 2 F 2 dT exp 1 k T B 4 F exp k BT F ex 2 2 Cel k B D F d k TD x dx B F 2 2 x k T e 1 B 0 F exp F 1 k BT k BT x=(-F)/ kBT 2 In order to integrate, we can replace the lower limit with -, which is approximately true for kB<<F. Thus Cel kB2TD F x 2 e ex x d kB2TD F 2 1 2 1 2 D F k B2T 3 3 Using that D F 3N 3N 2 F 2k B TF 1 3N 1 T C el 2 k B2 T 2 Nk B 3 2 k B TF 2 TF Experimentally Determined Heat Capacity For very low temperatures, the total heat capacity can be considered the sum of the electron and phonon contributions, thus C=T+AT3 Where the electronic contribution is proportional to T and the phonon contribution is proportional to T3. Dividing by T, we obtain a linear relationship between C/T and T2 with intercept and slope A. (see figure 9 chapter 6, in page 145) The observed value of the intercept does not agree very well with the predicted value and the reason is that the electron seems not to have the classical mass when in the solid but an effective mass. Usually an effective mass is defined as mth measured m calculated The reason for which the apparent mass is not the free electron mass involves three separate effects 1) The interaction of electrons with positive ions. The resulting mass is called the band effective mass 5 2) The interaction between electrons and phonons what distorts the lattice that in turn interact with other electrons increasing their effective mass 3) Electron-electron interaction what also increases the effective mass Electrical Conductivity The momentum of a charge is given by mv=ħk, thus its equation of motion in an electric field can be written as: dv dk F m qE (where q=-e for electrons) dt dt qEt the same for ALL electrons, thus the entire Fermi sphere is moving! What is consistent with a free charge in an electric field. What leads to k In a real system, electrons are not really free, they interact with the ions that are hopping in place (phonon) and they interact with imperfections in the crystal. A more appropriate model consists in assuming that these electrons are free to accelerate just during a time t= after which they just stop and start over. The overall effect is that these electrons find a steady state at a higher speed than before the electric field is applied (sort of like a terminal velocity). So the shift in the position of the Fermi sphere is qE/ħ while the speed increases an amount v=ħk/m= qE/m. At equilibrium, the average speed of the electron gas is zero, thus v is the drift velocity. Consider n electrons per unit of volume moving at a speed v, the number of electrons that will cross a surface of unit area perpendicular to the direction of motion in a time to is equal to the number of electrons in a cylinder of unit cross section area and length l=vto, thus, in the unit of time, the total charge crossing a unit are is: j=nqv=nq2E/m j is the current density and the above is known as the Ohm’s law where = nq2/m is the conductivity and the inverse = m/nq2 is the resistivity. Another important property of charge carriers is their mobility which is defined as μq=v/E where E is the electric field. Carriers with high mobility will pick up speed at small values of electric field thus leading to larger currents. With this definition j=nqv=nq μqE and = nq μq or μq=q/m The two main sources for resistivity are the electron-phonon interaction (interaction with the ions that are hopping in their place) and the interaction with impurities and crystal defects. The former is more relevant at high temperature while the later is relevant at all temperature. If the 6 electric field is then removed, collisions will make the electrons go back to their unbiased speed. The rate of these collisions is independent from each other, thus f=fL+fi where f is the total number of collisions per unit of time fL is the number of collisions with the lattice (charge-phonon collisions) per unit of time, and fi is the number of collisions with impurities per unit of times thus the characteristic time can be obtained as: 1 1 1 L i The equation above then indicates that the total number of collisions per unit of time is the sum of the total number of collisions with the lattice (phonons) per unit of time plus the total number of collisions with impurities per unit of time) or multiplying by m/ne2 =L+i Hall Effect If a magnetic field is applied perpendicular to a current, the electrons will suffer a force perpendicular to both, the drift velocity and B, of a magnitude F=qvxB This force produces an accumulation of charge and an electric field that generates an electrical force that opposes to the magnetic force. In equilibrium -eEH=evB or EH=-vB By defining the Hall coefficient RH as EH=-RH JB vB= RH JB or RH=v/J (notice that RH can be measured by measuring the electric field perpendicular to the current and it is a property of the material). Using J=nqv RH=1/nq (where q is negative for electrons) For Li, monovalent, the measured Hall coefficient at room temperature is -1.7x10-10 m3/C and the calculated value -1.4x10-10 m3/C. For Al (trivalent) both values are ~ -0.3x10-10 m3/C. But for Zinc, the calculated value is -0.5x10-10 m3/C while the measured one is +0.3x10-10 m3/C. The reason is that the carriers are positive. Although electrons are always the ones moving, some material, especially divalent, behave as if the carriers were positive. We’ll talk about that later. 7