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Quantum Statistical Mechanics Identical Fermions: Fermi-Dirac Distribution Consider the Distribution of a Total Energy of 8E Among 6 Electrons

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Quantum Statistical Mechanics Identical Fermions: Fermi-Dirac Distribution Consider the Distribution of a Total Energy of 8E Among 6 Electrons Quantum Statistical Mechanics Identical fermions: Fermi-Dirac distribution Consider the distribution of a total energy of 8E among 6 electrons: 8E 7E 6E 5E 4E 3E 2E 1E 0 Quantum Statistical Mechanics Identical fermions: Fermi-Dirac distribution Consider the distribution of a total energy of 8E among 6 electrons: 8E 8E 8E 7E 7E 7E 6E 6E 6E 5E 5E 5E 4E 4E 4E 3E 3E 3E 2E 2E 2E 1E 1E 1E 0 0 0 Quantum Statistical Mechanics Identical fermions: Fermi-Dirac distribution Due to the exclusion principle only two electrons can fit in a given state. So there are only 3 arrangements, each with a probability of 1/3 The average number of particles in the energy state E0 is 6 0 3 0 0 0 0 n = n p = 1! + 2 ! + 3 ! + 4 ! + 5 ! + 6 ! = 2.00 0 " i0 i0 i=0 3 3 3 3 3 3 Similarly, n1 = 1.67 n2 = 1.00 n3 = 1.00 n4 = 0.33 n5 = 0.00 n6 = 0.00 n7 = 0.00 n8 = 0.00 Quantum Statistical Mechanics Identical fermions: Fermi-Dirac distribution Due to the exclusion principle only two electrons can fit in a given state. So there are only 3 arrangements, each with a probability of 1/3 The probability of finding a particle with energy E0 is n p(0) = 0 = 0.333 6 Similarly, p(E1 ) = 0.278 p(E2 ) = 0.167 p(E3 ) = 0.167 p(E4 ) = 0.055 p(E5 ) = 0.000 p(E6 ) = 0.000 p(E7 ) = 0.000 p(E8 ) = 0.000 We can make a plot of p(E) vs E Quantum Statistical Mechanics Identical bosons: Bose-Einstein distribution Consider the distribution of a total energy of 8E among 6 identical bosons: 8E 7E 6E 5E 4E 3E 2E 1E 0 Quantum Statistical Mechanics Identical bosons: Bose-Einstein distribution Consider the distribution of a total energy of 8E among 6 identical bosons: Quantum Statistical Mechanics Identical bosons: Bose-Einstein distribution Each of 20 possible arrangements is equally likely. Hence the probability of each arrangement is 1/20. The average number of particles in the energy state E0 is 6 3 5 5 4 1 0 n = n p = 1! + 2 ! + 3 ! + 4 ! + 5 ! + 6 ! = 2.45 0 " i0 i0 i=0 20 20 20 20 20 20 Similarly, n1 = 1.55 n2 = 0.9 n3 = 0.45 n4 = 0.3 n5 = 0.15 n6 = 0.1 n7 = 0.05 n8 = 0.05 Quantum Statistical Mechanics Identical bosons: Bose-Einstein distribution Each of the 20 possible arrangements is equally likely. Hence the probability of each arrangement is 1/20. The probability of finding a particle with energy E0 is n p(0) = 0 = 0.408 6 Similarly, p(E1 ) = 0.258 p(E2 ) = 0.150 p(E3 ) = 0.075 p(E4 ) = 0.050 p(E5 ) = 0.025 p(E6 ) = 0.017 p(E7 ) = 0.008 p(E8 ) = 0.008 We can make a plot of p(E) vs E Quantum Statistical Mechanics For a large number of particles we can derive probability distribution functions for bosons and fermions at an equilibrium temperature T: Identical fermions: Fermi-Dirac distribution " (E ! EF )/kBT $ P(E) = 1 #e + 1% EF: Fermi energy: The probability of finding an electron with energy equal to the Fermi energy is 1/2 at any temperature. Identical bosons: Bose-Einstein distribution E /kBT P(E) = 1 #"e ! 1%$ Distinguishable particles: Maxwell-Boltzmann distribution P(E) = e!E /kBT Derivation of these equations relies on maximizing the number of ways of distributing N particles in k energy states with fixed total energy E. Quantum Statistical Mechanics For a large number of particles we can derive probability distribution functions for bosons and fermions at an equilibrium temperature T: At low temperatures, most bosons occupy the ground energy state. This can cause a change in phase of matter called Bose-Einstein condensation. Quantum Statistical Mechanics Application of Bose-Einstein Statistics: Einstein’s Theory of Specific Heat The molar specific heat of a substance, C is defined as dU C = dT where U is the internal energy added to one mole of the substance. Einstein assumed that the atoms of the solid could be modeled as harmonic oscillators and showed the average energy of the oscillators in 1-D follows the BE distribution: E = !! #e!! /kBT " 1% $ & Therefore the energy of 1mole (Avogadro’s number NA) of atoms oscillating in 3-D is U = 3N E = 3N !! #e!! /kBT " 1% A A $ & Quantum Statistical Mechanics Application of Bose-Einstein Statistics: Einstein’s Theory of Specific Heat The molar specific heat of a substance, C is defined as dU C = dT U = 3N E = 3N !! #e!! /kBT " 1% A A $ & " !! % e!! /kBT C = 3N Ak $ ' 2 # kT & )e!! /kBT ( 1+ * , Formula agrees well with data Quantum Statistical Mechanics Application of Fermi-Dirac Statistics: Degenerate Fermi Gas " (E ! EF )/kBT $ P(E) = 1 #e + 1% At T=0, P(E) = 1 for E < EF. All energy states with energy below the Fermi energy are occupied. At T=0, P(E) = 0 for E > EF. All energy states with energy above the Fermi energy are empty (not occupied). A system in which electron or fermions completely fill the lowest energy states is called a degenerate system. Quantum Statistical Mechanics Application of Fermi-Dirac Statistics: Degenerate Fermi Gas " (E ! EF )/kBT $ P(E) = 1 #e + 1% Not all fermions are in the lowest energy state at T=0. The energy of the fermions at T=0 is called the zero-point energy. In a metal the valence electrons are loosely bound and can be treated as a Fermi gas. At T=0, for an electron with the cut-off energy EF, 1 mv2 = E 2 F For a typical Fermi energy of 5eV, the velocity at T=0 is 106m/s! Quantum Statistical Mechanics Application of Fermi-Dirac Statistics: Degenerate Fermi Gas " (E ! EF )/kBT $ P(E) = 1 #e + 1% For higher temperatures, the distribution remains close to a step function Quantum Statistical Mechanics Application of Fermi-Dirac Statistics: Degenerate Fermi Gas The Fermi energy is directly related to the total number of fermions. We can model the valence electrons in a metal as being in a 3-D infinite well of dimensions LxLxL. The allowed energies are ! 2!2 E n2 n2 n2 = ( x + y + z ) 2 2mL where nx, ny and nz can be non-zero, positive integers We can represent each combination of nx, ny and nz as a point on a lattice: Quantum Statistical Mechanics Application of Fermi-Dirac Statistics: Degenerate Fermi Gas The Fermi energy is directly related to the total number of fermions. We can model the valence electrons in a metal as being in a 3-D infinite well of dimensions LxLxL. The allowed energies are ! 2!2 E n2 n2 n2 = ( x + y + z ) 2 2mL where nx, ny and nz can be non-zero, positive integers Defining a ‘radius’ 2 2 2 2 R = (nx + ny + nz ) The number of states with energy less than E is the number of points on the lattice lying within a radius R (times 2, for two electrons per state) 3/2 1 4 1 4 # 2mL2 & N = 2 ! " R3 = 2 ! " E 3/2 8 3 8 3 % 2 2 ( $ " ! ' Quantum Statistical Mechanics Application of Fermi-Dirac Statistics: Degenerate Fermi Gas The Fermi energy is directly related to the total number of fermions. We can model the electrons in a metal as being in a 3-D infinite well of dimensions LxLxL. 3/2 1 4 1 4 # 2mL2 & N = 2 ! " R3 = 2 ! " E 3/2 8 3 8 3 % 2 2 ( $ " ! ' At T=0, if we set N to be the total number of electrons, then the energy E in the above expression is the energy of the highest occupied state - i.e. the Fermi Energy: 2 2/3 2 2/3 ! " N % E = 3! ! F ( ) #$ &' 2m V where V is the volume LxLxL. Quantum Statistical Mechanics Application of Fermi-Dirac Statistics: Degenerate Fermi Gas The Fermi energy is directly related to the total number of fermions. We can model the electrons in a metal as being in a 3-D infinite well of dimensions LxLxL. 3/2 1 4 1 4 # 2mL2 & N = 2 ! " R3 = 2 ! " E 3/2 8 3 8 3 % 2 2 ( $ " ! ' The total zero-point energy is EF dN 3 E = E dE = NE total ! F 0 dE 5 The corresponding zero-point pressure (degeneracy pressure) is 2 5/3 dE 2 2 2/3 ! # N & P = ! total = 3" ( ) $% '( dV 5 2m V This pressure helps explain why metals are not compressible. Quantum Statistical Mechanics Application of Fermi-Dirac Statistics: Degenerate Fermi Gas Degeneracy pressure place a crucial role in the equilibrium of white dwarf stars. It balances out the inward pressure due to gravity: 2 5/3 2 2 2 2/3 ! " N % GM 3! " ( ) #$ &' 4 5 2m V R Where the electron density is twice the helium nucleus density N M m ! 2 ! He V 4 R3 3 " For a white dwarf, with mass equal to the mass of the Sun, we can calculate the radius R to be roughly 6000km. Quantum Statistical Mechanics Application of Fermi-Dirac Statistics: Degenerate Fermi Gas Degeneracy pressure place a crucial role in the equilibrium of white dwarf stars. It balances out the inward pressure due to gravity: 2 5/3 2 2 2 2/3 ! " N % GM 3! " ( ) #$ &' 4 5 2m V R Where the electron density is twice the helium nucleus density N M m ! 2 ! He V 4 R3 3 " As the mass of the white dwarf increases, the radius decreases due to higher gravity and the density increases.
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