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Quantum statistics Contents 1 Boltzmann distribution 1.1 Detailed balance for the Boltzmann distribution . . . . . . . . 1 4 2 Fermi-Dirac distribution 5 3 Bose-Einstein distribution 9 4 Density of states 9 4.1 DOS in a 3D uniform solid . . . . . . . . . . . . . . . . . . . . 11 4.2 DOS for a 2D solid . . . . . . . . . . . . . . . . . . . . . . . . 13 4.3 DOS for a 1D solid . . . . . . . . . . . . . . . . . . . . . . . . 14 1 Boltzmann distribution Boltzmann distribution represents the simplest model of the energy distribution for distinguishable, identical, and non-interacting particles. The particles are called distinguishable because they are distinct and their wavefunctions do not overlap. They are called identical since two particles with the same energy are taken to be similar. They are non-interacting, because the presence of a particle in a energy state does not affect the probability of another particle occupying the same energy state. Consider a finite number of these particles in an isolated system with a certain total energy. The Boltzmann distribution predicts the energy distribution of these particles in the system. In terms of probability, the Boltzmann distribution gives the probability of a particle having a certain energy at a given temperature. To derive the expression for the Boltzmann distribution consider an isolated system with five particles. The system has quantized energy levels with the lowest energy level equal to zero. For simplicity, the other energy levels are equally distributed with energy difference equal to ∆E. Thus, the energy 1 MM3010: Physics of materials levels for this system are 0, ∆E, 2∆E, 3∆E, 4∆E .... and so on. A general system can have continuous energy states. This is similar to a quantized system, with ∆E → 0. The isolated system is at some finite temperature T. Consider a scenario when the total energy of the five particles is zero. This is only possible, when all five particles are in the lowest energy state, and all the other energy states are empty. Such a situation can be considered to be the ground state of the isolated system. Now, consider a system with total energy equal to ∆E. This is possible, if there is one particle in the ∆E level, and the remaining four particles in the zero energy level. Out of the five particles, any one can be in the energy level ∆E. Thus, there are a total of five possibilities for this to happen, since any of the five particles can be chosen and they are all identical. The remaining four particles go to energy level zero. If n(E) is defined as the number of particles with energy E, then for this scenario n(∆E) = 1 and n(0) = 4. All the other energy levels are unoccupied. If p(E) is the probability of occupation of energy level E, then p(∆E) = 15 and p(0) = 45 , since one out of five particles occupy the energy level, ∆E, and four out of five particles occupy zero energy level. Thus, higher the energy level lower is the occupation probability. Consider, a system with total energy 2∆E. There are two ways of achieving this Scenario 1: 1 particle with energy 2∆E and 4 particles with zero energy. Scenario 2: 2 particles with energy ∆E and 3 particles with zero energy. For scenario 1, there are 5 possibilities and for scenario 2 there are 10 possibilities (5 × 4 divided by 2, since we do not distinguish the order of choosing two particles). There are a total of 15 possibilities, with 5 particles in each possibility. The product works out to be 75. We can calculate n(E) for this system, for each of the three energy states. The other states, above 2∆E, have zero occupancy. State zero: n(0) = 5 × 4 + 10 × 3 = 50 State ∆E: n(∆E) = 10 × 2 = 20 State 2∆E: n(2∆E) = 5 × 1 = 5 For each state, we have multiplied the number of possibilities with the number of particles. This can be converted into a probability, p(E). Both n(E) and p(E) are listed in table 1. The number of particles and the corresponding 2 MM3010: Physics of materials Table 1: n(E) and p(E) for 5 particles in a system with total energy 2∆E. Energy state n(E) p(E) 0 50 50/75 ∆E 20 20/75 2∆E 5 5/75 Table 2: Different scenarios for arranging 5 particles with total energy 4∆E and the number of ways of doing this. Energy Scenario 1 4∆E 1 3∆E 2∆E 1∆E 0 4 Possibilities 5 Scenario 2 Scenario 3 Scenario 4 Scenario 5 1 2 1 3 20 3 10 1 2 2 30 occupation probability decreases with increasing energy. Consider a more general case for this system, where the total energy is 4∆E. There are more possibilities now for distributing the five particles and different values for n(E) and p(E). Working with the same logic as the previous example, the different scenarios are given in table 2. The total number of ways of arranging this is 70 and there are 5 particles, so the product is 350. Once again, it is possible to calculate n(E) and p(E) for the different energy levels, table 3. The n(E) calculated in table 3 is obtained by multiplying the number in the scenario column in table 2 with the number of possibilities and adding all the scenarios for a given energy. The occupation probability decreases with increasing energy. For states above 4∆E, the occupation probability is zero. It is possible to plot the probability as a function of energy, as shown in figure 1. It is possible to fit the data with an exponential Table 3: No of particles and occupation probabilities for the various energy levels Energy 4∆E 3∆E 2∆E 1∆E 0 n(E) 5 20 50 100 175 3 p(E) 5/350 20/350 50/350 100/350 175/350 4 1 5 MM3010: Physics of materials Figure 1: The Boltzmann distribution as a function of energy. The dots represent the data for p(E) from table 3 and the line is an exponential fit to the data. function of the form p(E) = A exp(−βE) (1) where A and β are constants. Equation 1 shows that as the energy level increases then the occupation probability drops exponentially. So for a system of identical non-interacting particles, most of the particle occupy the lowest energy states. For atomic systems, like atoms in a solid that is a finite temperature T the constant β is equal to kB T , which is the average thermal energy of an atom that is treated as a simple harmonic oscillator. Also, the system is defined such that the probability of occupation of the ground state (E = 0) is 1, so that A = 1. Thus, the Boltzmann distribution can be written as (E − E0 ) ] (2) p(E) = exp[− kB T where E0 is the ground state energy. For electrons in a metal, when the Boltzmann approximation is used, E0 is taken to be equal to the Fermi energy, EF . 1.1 Detailed balance for the Boltzmann distribution Consider a system where the particles satisfy the Boltzmann distribution. Consider two energy states, E1 and E2 , with n1 and n2 particles. Let R12 be the rate of transition per particle from state 1 → 2. Similarly, R21 is the rate of transition from 2 → 1. To get the total rate this needs to be multiplied with the number of particles. For a system in dynamic equilibrium the rate 4 MM3010: Physics of materials of particles going from 1 → 2 equal to 2 → 1. This can be written as n1 R12 = n2 R21 n2 R12 = R21 n1 (3) This dynamic equilibrium between individual energy states is called detailed balance. Since the number of particles is directly proportional to the probability of occupation, p(E), using equation 2, the detailed balance, equation 3, can be written as p2 (E2 − E1 ) ∆E R12 = = exp[− ] = exp[− ] R21 p1 kB T kB T (4) Thus, the rates for the transitions are dependent on the energy difference between the two levels. If E2 > E1 then ∆E in equation 4 is greater than zero. This means, that R12 < R21 . Thus, it is easier for a particle to drop down to a lower energy level than for a particle to occupy a higher energy level. An external stimulus is needed for this transition to happen. Higher the value of ∆E, lower is the rate for this transition. 2 Fermi-Dirac distribution The Boltzmann distribution is used to describe the occupation statistics for a system consisting of non-interacting particles. Electrons in a solid, on the other hand, can interact with each other due to electrostatic repulsion. This has an impact on the occupation of energy levels by electrons. Electrons are also considered indistinguishable since when two electrons come close to each other their wavefunctions can overlap. The consequence of electron interaction is the Pauli exclusion principle, which states that no two electrons can have the same set of quantum numbers. A more general way of stating the exclusion principle is that two electrons cannot occupy the same space. We can use the Boltzmann statistics along with the exclusion principle to derive a new statistics for electron distribution as a function of temperature and energy, called the Fermi-Dirac distribution, Consider a metal with N atoms. Corresponding, to the N atoms there are N energy levels. These energy levels are formed by overlap of the electron wavefunctions. Each of these energy levels can take two electrons, of opposite spin, without violating the exclusion principle, so that there are a total of 2N energy states. If each atom contributes one electron then there are N electrons and these electrons occupy energy states from 0 to N . Thus, half 5 MM3010: Physics of materials of the energy levels are full and half are empty. If the lowest energy level is taken as zero (reference) the energy level separating the filled and empty states is called the Fermi level and the corresponding energy is called the Fermi energy. This is the scenario at T = 0 K. We are interested in what happens to the electrons as temperature is increased. Also, we can take N to be sufficiently large, so that the energy levels are very close to each other i.e. continuous energy states. To give an idea of the magnitude of N , 1 gram of Li contains around 1023 atoms! Consider the detailed balance expression from equation 3. In this n1 and n2 are the average number of particle (electrons) in the two energy states given by E1 and E2 . Because of the exclusion principle, the instantaneous values of n1 and n2 can be either 0 or 1. Also, because of the exclusion principle the presence of an electron in a energy states prevents another electron from occupying this energy state. This is given by an inhibition factor, (1 − n), so that when n = 0 there is no inhibition, while when n = 1 there is complete inhibition. Hence, for electrons, the detailed balance equation 3 is modified as (5) n1 (1 − n2 ) R12 = n2 (1 − n1 ) R21 The transition rate R1 2 is multiplied by the inhibition factor, (1 − n2 ), since the rate is affected by the presence of electrons in energy state 2. Rearranging, and using equation 4 gives exp(− kEB1T ) R21 n1 (1 − n2 ) = = n2 (1 − n1 ) R12 exp(− kEB2T ) (6) Writing the terms corresponding to energy state 1 on one side and energy state 2 on the other side gives n1 E1 n2 E2 α exp( ) = exp( ) = exp( ) 1 − n1 kB T 1 − n2 kB T kB T (7) Here, α is a constant. Considering only the energy state 1 in equation 7, it can be rearranged to write n1 as 1 n1 = 1 + exp( E1 − α ) kB T (8) n1 is equivalent to the probability of finding the electron in energy state E1 and is written using the symbol f (E1 ). The constant α has to be evaluated. For electrons in a metal, the Fermi level, EF , is taken to have a probability of one-half (1/2) at all temperatures. This is possible only when α = EF . 6 MM3010: Physics of materials Figure 2: (a) The Fermi-Diract distribution as a function of energy. The temperature is taken at 300 K and EF = 5 eV . (b) A close up of the region around EF . Within 0.2 eV or 200 meV the Fermi function becomes close to 0 (for E > EF ) or 1 (for E < EF ). Plots were generated in MATLAB. Thus, the Fermi-Dirac distribution for finding the probability of occupation of energy level, E, is written as f (E) = 1 (E − EF ) 1 + exp kB T (9) The distribution satisfies the following conditions for electron distribution in a metal Condition 1: At T = 0, f (E) = 1, when E < EF . All energy levels below Fermi level are fully occupied. Condition 2: At T = 0, f (E) = 0, when E > EF . All energy levels above Fermi level are empty. Condition 3: At any T , f (E) = 1/2, when E = EF . The Fermi level occupation probability is always one-half. The Fermi-Dirac distribution is plotted in figure 2. Since, the function has a exponential term the occupation probability decreases rapidly when energy is higher than the Fermi energy. In figure 2(b), it can be seen that above 0.2 eV above EF the probability is close to zero. The behavior of the Fermi function also depends on the temperature. f (E) at three different temperatures are plotted in figure 3. At room temperature the probability decreases rapidly 7 MM3010: Physics of materials Fermi function 1 T = 30 K T = 300 K T = 3000 K f(E) 0.8 0.6 0.4 0.2 0 4.7 4.8 4.9 5 Energy (eV) 5.1 5.2 5.3 Figure 3: (a) The Fermi-Diract distribution as a function of energy for different temperatures. EF = 5 eV and the temperatures are 30 K, 300 K, and 3000 K. The first represents cryogenic temperatures while the last is higher than the melting points of most metals. Plot was generated in MATLAB. Table 4: p(E) vs. f (E) for different values of (E − EF )/kB T . Relative energy (E − EF )/kB T 0.1 0.2 0.5 1 2 5 10 p(E) f (E) 0.91 0.82 0.61 0.37 0.14 6.7 × 10−3 4.5 × 10−5 0.48 0.45 0.38 0.27 0.12 6.7 × 10−3 4.5 × 10−5 above EF . At all temperatures f (EF ) = 0.5. When, E − EF kB T , the Fermi function can be approximated by the Boltzmann distribution. Table 4 lists the two functions at different values of (E − EF )/kB T . For small values of (E − EF ) the Boltzmann distribution overestimates the occupation probability compared to the Fermi Dirac function. But as (E − EF ) becomes ≈ 2kB T , both functions start to converge. To put this in energy units, at room temperature, kB T is approximately, 0.025 eV or 25 meV . Thus, at energies of 0.05 eV above EF the Fermi function can be replaced by the Boltzmann approximation. Electrons belong to a general class of particles called fermions. These are particles, with a spin of one-half, and include electrons, positrons, protons, neutrons, and muons. They obey the exclusion principle and are described 8 MM3010: Physics of materials by the Fermi-Dirac distribution. 3 Bose-Einstein distribution 1 Similar to the fermions, there are a set of particles called bosons, which have integral spin i.e. spin of 0 or 1. Examples of bosons include the α particle, ground state of the He atom, π meson, photons, phonons, and deuterons. Their behavior is opposite to that of fermions, in that the presence of a particle in an energy state increases the transition rate for another particle to move to that energy state. Thus, instead of an inhibition factor, there is an enhancement factor for these particles. Thus, the detailed balance for bosons can be written as n1 (1 + n2 ) R12 = n2 (1 + n1 ) R21 (10) Rearranging equation 10 and using equation 4 gives n2 α E1 E2 n1 ) = ) = exp( ) exp( exp( 1 + n1 kB T 1 + n2 kB T kB T (11) where α is a constant. n1 represents the probability of occupation of a given energy level by bosons, as a function of temperature and this is given by n(E) = 1 (E − α) exp − 1 kB T (12) This distribution is called the Bose-Einstein distribution. This is usually not applicable to electrons, since they obey the exclusion principle. But, when materials exhibit superconductivity, electrons in them become paired. The process is mediated through the lattice vibrations (phonons). These are called Cooper pairs. While an electron has a spin of one-half, the Cooper pairs have integral spin, ‘0’ or ‘1’. Hence, these are bosons and obey the BoseEinstein statistics. This, pairing is responsible for the negligible resistance in superconductors. 4 Density of states In a solid with N atoms there are a total of 2N energy states. These are discrete states but for large values of N the spacing between them are so 1 Section 3 can be skipped without loss of continuity 9 MM3010: Physics of materials Figure 4: Calculated DOS of Cu around the Fermi level. The filled states lie below EF while the empty states lie above EF . The broad peak below EF corresponds to the filled 3d states. Adapted from http://www.personal.psu.edu/ams751/VASP-Cu/ small that they can be considered to be continuous. Density of states (DOS) is defined as the number of available energy states per unit energy per unit volume. The units are J −1 m−3 or eV −1 cm−3 and it provides information on how the energy states are distributed in a given solid. It is typically denoted as g(E). The experimental density of states of a material can be measured by photo electron spectroscopy or Scanning Tunneling Microscopy (STM) or Electron Energy Loss spectroscopy (EELS). These techniques probe the density of empty or filled states around the Fermi energy, called Local DOS (LDOS). Density functional calculations can also be used to calculate the density of states. These model the electron density distribution in a solid and also model the atomic potential within the solid. Figure 4 shows the calculated DOS for Cu around the Fermi energy. The density of states can be used to calculate the total number of electrons. If g(E) is the DOS, then the total number of electrons, S(E) is given by Z S(E) = g(E)dE (13) E where the limits of the integration is usually from 0 to the Fermi energy (at 0 K). This equation assumes that the probability of occupation of the energy 10 MM3010: Physics of materials state is 1. This is not always true, since at any temperature, the occupation probability is defined by a Fermi function, f (E). An energy expression for the density of states, g(E), will be useful for evaluation of the integral in equation 13. 4.1 DOS in a 3D uniform solid To simplify the calculation for g(E) consider a 3D box with uniform potential. This is a simplification of a real solid but is a good approximation of the valence band of metals where the electrons are loosely bound to the atom and are delocalized. We will also use this approximation for electrons and holes near the edge of the band. For simplicity the uniform potential can be taken to be zero. For this solid, the electron is defined by 3 quantum numbers (n1 ,n2 , n3 ) and its energy is given by h2 (n21 + n22 + n23 ) E = 2 8me L (14) Equation 14 is for a cubic solid of length L with the 3 quantum numbers for the 3 axes. me is the free electron mass. We can replace the 3 quantum numbers by a single value n so that equation 14 is modified into E = h2 n2 8me L2 (15) For small values of n these energy levels are quantized, but for large values of n the spacing between them are close so that the energy levels can be considered to be continuous. So n represents the radius of a sphere, where the total number of states within the sphere is given by its volume. This is shown schematically in figure 5. Since the quantum numbers can only be positive (the quantum numbers represent the electron wavefunction and it can be shown that to avoid exponential increasing functions the quantum numbers have to be non-zero positive integers) we can only take the first quadrant of the sphere. Hence the total number of ‘orbitals’ (energy states), Sorb (n), is given by the volume of the sphere in the first quadrant Sorb (n) = 1 1 4 3 ( πn ) = πn3 8 3 6 (16) Since each orbital can take two electrons of opposite spin, the total number of energy states (including spin), S(n), is given by S(n) = 2Sorb (n) = 11 1 3 πn 3 (17) MM3010: Physics of materials Figure 5: Energy states in a solid with uniform potential. The portion of the sphere corresponds to the constant energy surface. Only the first quadrant is to be considered since the quantum numbers are non-zero positive integers. Adapted from Principles of Electronic Materials - S.O. Kasap. We can relate n (quantum number of the electron in the solid) to the energy E using equation 14 to write the total number of states in terms of energy, S(E). This is given by S(E) = 3 3 L 1 π (8me E) 2 3 3 h (18) Dividing by the volume of the cube will give total number of states per unit volume, Sv (E). The density of states is the differential of the total number of states, so that g(E) is given by √ me 3 √ g(E) = 8π 2 ( 2 ) 2 E h (19) Equation 19 gives the DOS in a solid with a uniform potential. At the reference point where energy is set to 0, DOS is zero. As the energy increases g(E) also increases. The functional form is shown schematically in figure 6. g(E) represents the density of available states. It does not provide information whether those states are occupied or not. The occupation is given by the Fermi function and is usually a function of temperature. 12 MM3010: Physics of materials 3D DOS 46 x 10 10 9 g(E) (J−1m−3) 8 7 6 5 4 3 2 1 0 0 1 2 3 Energy (eV) 4 5 Figure 6: Plot of DOS vs. E for a 3D solid using free electron mass. This is a plot of equation 19 using MATLAB. 4.2 DOS for a 2D solid The density of states function will change upon changing the dimensionality of the solid. Consider the case of a 2D solid with a uniform potential. There are 2 quantum numbers, n1 and n2 , which are related to the energy, similar to equation 14. h2 n2 h2 2 2 (n + n ) = (20) E = 1 2 8me L2 8me L2 For a 2D case, n, represents the radius of a circle, shown in figure 7, and only the first quadrant can be considered since the quantum numbers should be positive. It is possible to calculate the density of states per unit area, Sarea (n), including spin πn2 Sarea (n) = (21) 2 Using equation 20 it is possible to calculate the DOS in terms of energy, Sarea (E) 4πme Sarea (E) = E (22) h2 Differentiating equation 22 gives the density of states in two dimensions g(E) = 4πme h2 (23) The density of states function is independent of energy, unlike 3D where g(E) increases with energy. It is represented as a step function at different energy values. 13 MM3010: Physics of materials Figure 7: Energy states in a 2D solid with uniform potential. Instead of a sphere states of constant energy form a circle and only the first quadrant is relevant. Adapted from Principles of Electronic Materials - S.O. Kasap. 4.3 DOS for a 1D solid The calculation for a 1D solid is similar to the earlier calculations except that there is only one quantum number and spatially it is represented on a line (instead of circle in 2D and sphere in 3D). The total number of states, per unit length, S(n), is just 2n and this is related to energy by equation 24. r 8me E (24) n = h2 The density of states is given by r g(E) = 8me 1 √ h2 E (25) In a 1D solid the density of states decreases with energy. For a zero dimensional solid, energy states are only discrete. Solids with two, one, and zero dimensionality can be obtained by reducing the length in one or more dimensions. A thin film is an example of a two dimensional solid (or a single or bi layer of graphene ), while a quantum wire is a one dimensional solid. A quantum dot is a zero dimensional solid. The density of states and hence the electronic properties of these materials are different from a bulk solid. The DOS for solids of different dimensionalities are shown in figure 8. 14 MM3010: Physics of materials Figure 8: DOS for bulk, 2D, 1D, and 0D solids. The y-axis is a qualitative representation for the different dimensions. Adapted from http://britneyspears.ac/physics/dos/dos.htm References The following source materials were used for preparing this handout 1. Quantum Physics of Atoms, Molecules, Solids, Nuclei, and Particles by R. Eisberg and R. Resnick, Wiley Student edition, 2011 2. Principles of electronic materials and devices by S.O. Kasap, McGrawHill India, 2007 The PDF was prepared in LATEX, using Texmaker and MiKTeX compiler. The plots were generated in MATLAB. 15