Fermi-Dirac and Bose-Einstein

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1 Fermi-Dirac and Bose-Einstein Beg. chap. 6 and chap. 7 of K &K Quantum Gases Fermions, Bosons Partition functions and distributions Density of states Non relativistic Relativistic Classical Limit Fermi Dirac Distribution Fermi Energy Electrons in solids Nuclear matter White Dwarf Bose Einstein Distribution Bose-Einstein Condensation Liquid Helium, Superconductivity Modern experiments with Bose Einstein Condensates

2 Quantum Gases Bosons, Fermions Consider quantum system and its quantum states Quantum Field Theory => Integer spin= Boson = number of particles in a given state is arbitrary Half Integer spin=fermion= at most one on each state: Pauli exclusion principle Partition Functions and mean occupation numbers Thermodynamical system=state of energy ε Use Gibbs method and grand partition function Fermion At most one Z = + exp µ ε s( ε) =τ logz µ = Boson Sum on all integers Z = s( ε) = τ logz µ = exp µ ε s( ε ) = + exp µ ε exp s µ ε = s = exp µ ε exp µ ε τ s( ε ) = exp µ ε 2 exp ε µ + exp ε µ

3 Ideal quantum gases Ideal gas approximation States are not modified by presence of other particles No correlation between states Density of states multiplicity x density in phase space g i d 3 x d 3 p h 3 change of variable to energy p ε g p 2 dp dω i = D(ε)dε where D(ε) is the density of states per unit volume Ω Non relativistic h 3 dp = mdε p = m 2 Ultra-Relativistic ε = p2 2m p2 = 2mε dε ε D ε ( )dε = 4πg i ε = pc p D( 2 dp ε)dε = 4πg i h 3 = Note: 2 notations either with or without volume included. These notes do not include the volume, and use the density of states per unit volume. 3 p 2 dp h 3 = g i 4π 2 g i 2π 2 3 c 3 ε2 dε 2m εdε

4 Behavior Sign is critical for (ε-µ)/τ small Fermi-Dirac Bose-Einstein Bose condensation For (ε-µ)/τ large, classical limit s( ε ) = s( ε ) = Occupation number << s ε exp ε µ = + 2 for ε = µ Same old results of Boltzmann Prob ε ( ) s( ε) N < N >= V s( ε) = for ε << µ exp ε µ for ε µ ( ) exp µ ε ( ) s ε N τ exp ε exp ε = = Vn Q D ( ε )dε = V exp µ Z exp ε d 3 p = V exp µ h 3 n Q µ = log 4 <s(ε)> µ ε = λ exp ε independent of F.D. or B.E. n Q = mτ 2π 3/2 n n Q

5 Thermo functions for ideal quantum Number of Particles D( ε)dε N = V s( ε) D( ε)dε = V exp ε µ ± µ(τ) set by requirement that N=total number of particles Energy Entropy U = ε = V σ ε case of Black Body N γ α τ 3 εd( ε)dε exp ε µ ± ( ) = ( τ log Z) - Bose- Einstein + Fermi-Dirac ( ) τ = ε µ s + log Z log τ prob s =< s > = ε µ s ± log ( ± s ) τ + Bose- Einstein = ± ( ± s )log( ± s ) s log s - Fermi-Dirac 5

6 Fermi Gas Ground State (Non relativistic) Fermi Energy Calculation : d 3 xd 3 p density of states g i and ε = p 2 h 3 2m N = V 2m 2 ε F 2π 2 2 εdε Watch out: here n=density Ex: Electrons in metal Energy U = V Free Energy s( ε) Pressure ε F Spin 2 g i = 2 ε τ << µ s( ε) = ε F = µ ( τ = ) ε F = 5 ev v F = 8 cm/s = 3 3 c exp ε µ for ε < µ + for ε > µ N = V s ( ε ) D( ε)dε = V D( ε)dε ε F εd( ε)dε = 3 5 Nε F F = V ε F F( ε)d( ε)dε F = U = 3 5 Nε F p = F V τ = 2 5 nε F V 3 ε F ε F = 2 2m 3π2 n Kittel+Kroemer do this calculation with integer n density Spin /2 ( ) 2 3 with n = N V F( ε ) =U at τ = σ = for s( ε ) = or Repulsive

7 Fermi Gas Ground State (Relativistic) Note that even at zero temperature very large kinetic energies: in some case ultra-relativistic Fermi Energy Energy Pressure ε F = µ ( τ = ) ε F N = V D ( ε )dε = U = F = V V 3π 2 3 c 3 ε F 3 ε F = ( 3π 2 n) 3 c ε F εd( ε)dε = 3 4 Nε F τ << µ p = F = V τ 4 nε F V 4 3 Note: same general relation P = / 3u (ultra relativistic) 7

8 Pauli exclusion principle Interpretation 2 fermions cannot be in the same quantum state In particular not at the same position=> fermion does not have the full volume available but only Heisenberg uncertainty principle => large random momenta Fermi energy: =>pressure Δp x Δx Non relativistic Relativistic V N = n Δp x n / 3 ε F E Δp2 2m 2 2m ε F E cδp cn /3 n2 /3 8

9 Symmetry of Fermi Dirac distribution Basic symmetry (except for lower bound at δ =-µ ). The distribution can be described either as the presence of electrons or the absence of electrons (=holes). Decomposition into hole-like and electron-like excitations f ( ε,τ ) N D ε dε = ε F D ε dε Hole and Electron Excitations f ( ε,τ ) = exp ( ε µ ) + Writing ε = µ +δ f ( ε,τ ) = exp δ + ε f ( ε,τ ) = describes Fermions of energy -ε and chem.pot. - µ exp δ + V = ( ) f ( ε,τ ) ( ) D( ε ) f ( ε,τ ) dε = ε F D( ε ) f ( ε,τ ) dε + D ε ε F D( ε ) f ( ε,τ ε ) dε = ε F D( ε )( f ( ε,τ )) dε # F ##" ### $ #### "##### $ B ="electrons" A ="holes" 9 f ( ε,τ) = s( ε) τ ( ) f ( ε,τ ) dε

10 Electron and hole energy Change of energy with temperature ( ) f ε,τ ε F u( τ ) u dε ε F εd( ε ) dε dε + εd ε dε ε F εd ε dε ( ) = εd( ε ) f ( ε,τ ) = ε F εd( ε ) f ( ε,τ ) ( ) f ( ε,τ ) Note that as referenced to the Fermi energy, the energy of holes are opposite to that of the corresponding missing electrons and are positive ε ε F ε ε F +ε ε F F D( ε ) f ( ε,τ ) dε + D( ε ) f ( ε,τ ε ) dε ε F D( ε ) dε F #########" ######### $ = = ( ε ε F )D( ε ) f ( ε,τ ) dε + ε F ( ε F ε )D( ε ) f ( ε,τ ) dε ε F ##### "##### $ "electrons" Electron like excitation: increasing energy k Hole like excitation: increasing energy "holes" ( ) ( ) ######" ###### $

11 Electrons in crystals: Quantum States Energy Band structure ε v ε c E Gap energy Chap. 7)) conduction band valence band k Periodicity of the lattice (e.g., spacing a) Periodic zones in momentum space k k + G with G = 2π a Resonant tunneling: free propagation of specific modes (Bloch Waves) Discrete E( k) These relationships do not necessarily overlap in => gap (cf. Kittel, introduction to Solid States Physics ev Seen in projection on the energy axis: energy bands Valence band Conduction band Metal: Fermi level = chemical potential in conduction band => conduction can be described by free Fermi gas Insulator: Fermi level in gap

12 Order of magnitude Phys 2 (F24) 7 Fermi Dirac/Bose τeinstein Electrons in metals τ=o : very good approximation at room temperature and below Common notation: f ( ε,τ) = s( ε) τ VεD ( ε )f ( ε,τ ) Heat Capacity but conservation of number of electrons Then use: df /dt only large close to µ does not vary fast with τ ε F large negative τ Change variable ( )τ C el k B VD ε F C k VD ε el B F ε F ε F 5 ev T F 5 4 K >> T lab du C el = k B dτ = k d B dε = k B V εd ε dτ ε x = ε ε F τ ε = ε F µ ε F 2 d dτ ( ) d ( ) dτ f ε,τ D( ε) f ( ε,τ ) dε = ε F D ε dτ f ( ε,τ ) dε = C el = k B V ε ε df F dε k VD ε dτ B F x 2 e x dx k ( e x +) 2 B VD( ε F )τ df dτ ( ε ε F )D ε ( ε ε F ) τ 2 ( ) 2 ε ε F x 2 e x τ 2 ( ) 2 e x + ( ) df ( ) d dε dτ ( ) exp ε ε F τ ( ( ) + ) 2 exp ε ε F τ exp ε ε F τ exp ε ε F τ + 2 dε dx = π2 3 k BVD( ε F )τ dε

13 Heat capacity of metals Finally Hence taking into account phonons C tot = C el + C ϕ = γt + AT 3 f ( ε,τ ) ε F ε Important puzzle historically Very few electrons affected C el << 3 2 Nk B 3

14 Example Ge,Si,GaAs from Sze Physics of Semiconductor devices p3 Wiley-InterScience 98 4

15 Insulators: Density of states cf. K&K chap 3 In particular intrinsic semiconductors (no role of impurities) Statistical distribution Still good approximation to consider free electrons as quantum ideal gas => occupation number Density of states ε conduction band valence band f ( ε) = exp ( ε µ )/τ We can then get the total number of electrons n et = ε c Gap ε v ( ) + f ( ε )D( ε)dε = f ( ε)d h ( ε )dε + f ( ε)d e ( ε)dε ε v ε v = D exp ( h ε µ ε )/ ( )dε + ε D (ε) ( ) + D e (ε)dε = 2 4π 2 D h (ε)dε = 2 4π 2 ε c ε c 2m* e 2 2m* h 2 exp ( ε µ )/τ 2 spin states Electron state density below the gap (ε ε c )dε (ε v ε)dε ( ) + D e ε Parabolic at gap edge ( )dε 5

16 This can be rewritten as Electrons and Holes Therefore we can describe the electron population by two non relativistic gases : holes and electrons (cf. what we did with metals). The equality of the number of holes and electrons fixes the chemical potential: Charge neutrality Fermi level: in the middle of the gap if m h *=m e * Measuring from the edge of the valence and conduction band respectively n et = ε v n vt exp (( µ ε )/τ ) + D h ( ε )dε + exp( ( ε µ )/τ ) + D e ( ε )dε ε " $ $ $ $ # $ $ $ $ $ % " c $ $ $ $ # $ $ $ $ $ % total in valence band at zero temperature =total number of elctrons exp (( µ ε )/τ ) + D h ( ε )dε = " $ $ $ $ # $ $ $ $ $ % n e = 2 4π 2 for τ << ε v = n h = 2 4π 2 * 2m e 2 holes * 2m h 2 3/2 3/2 ε v holes exp( ( ε µ )/τ ) + D e ( ε )dε ε " c $ $ $ $ # $ $ $ $ $ % electrons electrons exp( ( ε ' ( µ ε c )) / τ ) + ε 'dε ' n Qe exp ε µ c with n Qe = 2 n Qh exp µ ε v exp ε ' ε v µ with n Qh = 2 (( ( )) / τ ) + ε 'dε ' m * eτ 2π 2 m * hτ 2π

17 from Sze Physics of Semiconductor devices p85 Wiley-InterScience 98 Example Ge,Si,GaAs 7

18 Determining the chemical potential f No impurities: intrinsic semiconductors /2 occupation number ε v µ ε ε c Yes do intersect at /2 but does not fix position logn e (µ) logn h µ ( ) ε v µ µ ε c charge neutrality does 8

19 Semiconductors ε v ε c Large role of impurities: localized states (Not band ) in gap If they are shallow ( 4meV (Si) mev (Ge)), such states can be excited at room temperature. This modifies totally the behavior k ε D ε A Donors Acceptors d o d + + e a a o + e note: 2 A state because a bond is missing and the missing electron can be spin up or down, A- bond established (pair of electrons of antiparallel spins) : state The number of free electrons(holes) is no more constant Number of electrons can be increased by donors and decreased by acceptors But we need to keep charge neutrality = method to compute the Fermi level For large enough impurities concentration, the Fermi level can move close to the edge of the gap (Thermally generated) conductivity either dominated by electron like excitation: negative carriers (n type) hole like excitation: positive carriers (p type) n d = n d + + n d o n a = n a + n a o 9

20 Donors negative carriers (n type) Acceptors Semiconductors cf. K&K fig 3.6 d d + + e exp ε + n d + = n d exp ε exp ε µ = n d + 2exp µ ε d.4 ev Si with ε d ε ε + ε c. ev Ge a + e d + exp ε µ τ n a = n a 2exp ε + exp ε µ = n a + 2 exp ε µ a.4 ev Si with ε a ε ε ε v +. ev Ge positive carriers (p type) logn e (µ) logn h µ ( ) ( ) logn d + µ logn e (µ) logn h µ 2 ( ) ( ) logn d + µ Electric neutrality n e = n d + + n h n d + + µ ε v ε c µ µ ε d Electric neutrality n e + n a = n h n a ε v ε c µ ε a -

21 Other examples of degenerate Fermi gas 3 He Spin /2 cf. Problem set Very different behavior from 4 He: phase separation Basis for dilution refrigerators (K&K Chapter 2: evaporation of 3 He into superfluid 4 He which acts as an excellent pump: Works down down to mk) Nuclear Matter n p n n 5 37 Fermi momentum N p N n A 2 R.3 3 A3 cm cm -3 ε F = 4 2 J 3 MeV T F 3 K 948 sub threshold production LBL α( 38 MeV)+ 2 C π

22 White Dwarfs and Neutron Stars hite dwarf stars (and core of massive stars) => Degenerate Fermi gas Fermi pressure balances gravity => Condition for equilibrium Look at total energy of system: Assume constant density Non relativistic Ultra Relativistic ρ 6 g/cm 3 n 3 cm -3 ε F.5 3 J 3 5 ev T F 3 9 K >> T star U T = ε F n / 3 M/3 UFD has the same R dependence R as gravitational energy ( U T = dm 4 /3 bm 2 ) R Degeneracy pressure cannot balance gravity if M too big Chandrasekhar limit.4 M eutron stars ε Same story for neutrons (uncertainty of equation of state) F 3 MeV Similar Chandrasekhar limit 3 M if larger => black hole Gravitational potential energy U G GM2 Kinetic energy (neglect angular momentum) R U FD nvε F Mε F ε F n 2 / 3 M2 / 3 5/3 am bm 2 R 2 R 2 R minimum of total energy: stable 22 ρ = U T Rel M 4 /3πR 3 n M R 3 U U FD Non Relativistic U T NR U U G U FD Relativistic U G R R

23 White Dwarf Explosions: SN Ia An accelerating universe? Supernovae Type Ia at high redshift (2 groups) Ω m -Ω Λ Distant supernovae appear dimmer than expected in a flat universe Potential problems Are supernova properties really constant? Dust? The Uninvited Guest: Dark Energy Large negative energy a = G u acceleration r 2 c p V energy density pressure " $$ $ # $$$$ % GR gravitational mass Luminosity Fainter 23 Ω m = Ω Λ = time Gravity becomes repulsive Distance

24 Bose-Einstein Condensation Calculation of chemical potential Let us take the origin of energy at the ground state As usual the chemical potential can be computed by imposing the average number of particles in the system Separating between the ground state s= and the other states exp µ + s > exp ε µ = N s At low enough temperature, we can solve the equation by having exp µ and put nearly all the particles in the ground state, with the occupation of the higher states given by exp ε s µ exp ε << N s or exp ε s >> + 24 N τ <<ε N s Bose Einstein condensation For large densities not a very low temp. phenomenon

25 B-E Condensation: Quantitative Approach Chemical potential at low temperature With the energy origin at ground orbital. The occupation number of the ground state is f (,τ ) = s( ) = exp µ τ for µ << τ µ for 4 He (µ< because <ε o =) Comparison with 2nd state Cannot use continuous approximation For 4 He but f,τ ( ) = N µ τ N N N = 22 cm -3 µ =.4 45 J at T =K and V = cm 3 ε( n x,n y,n z ) = 2 ( π ) 2 2m L ( n x + ny + nz ) ε(,, ) = 3 2 ( π ) 2 ε( 2,,) = 6 2 π 2m L 2m L ( π ) 2 L = J for L = cm>> µ Δε = ε( 2,,) ε(,, ) = 3 2 2m ( ),τ => the occupation number of 2nd state is much smaller NΔε = J >> τ = at T =K f ε 2,, ( ) = exp Δε µ τ ( ) 2 ( ) << ( exp µ ) τ = 25

26 Bose-Einstein Condensation(2) Temperature dependence Calculate separately condensed phase and normal phase N = f ( ",τ # ) + f ( ε,τ ) condensed ε( 2,,) $ " $ # excited Replace sum by integral exp µ, Defining the Einstein condensation temperature τ E = 2π 2 2/3 N N m 2.62 V exc = N τ N x /2 dx e x For large densities, τ E is not very small exp µ + V 4π 2 2m.36 π N exc = /2 ε /2 dε exp µ exp ε mτ 2π 2 3/2 V = 2.62n Q V does not depend on N if µ τ τ E 3/2 26

27 Properties of 4 He Liquid Helium 4 Loose coupling =>liquid (4.2K at Atm) ideal gas 4 He has spin => boson 3 He spin /2 C 4 He => expect condensation at 3.K Experimentally lambda point 2.7K (Landau temp.) Phase transition => peculiar properties Macroscopic quantum state cf energy of electromagnetic field => Quantization phenomena =>Superfluidity Ψ = n /2 e iϕ( t, x ) e.g. Vortex Equivalent of Josephson effect No resistance to flow <= impossibility to transfer energy to a quantum liquid E n"ω ˆ E e ( ) i" ωt k x Δϕ = 2nπ where n is integer 2.7K T 27

28 Liquid Helium Superfluidity Consider an object of mass M going through helium at velocity V Because coherent quantum state, cannot transfer energy to single atom: only possibility is transfer to excitations phonons v = Single atom (no superfluidity) Conservation energy and momentum in collision => subtracting 2M m + M Energy transfer If the mass of an atom is m / 2MV 2 = / 2MV ' 2 + / 2mv 2 / 2MV 2 = / 2 MV' 2 +ω k M V = M V ' +m v M V = M V ' +" k 2mM v 2 mv2 = 2mM 2 m + M ( M V m v ) 2 = ( M V ') 2 ( M ) 2 = ( M ) 2 28 V " k phonons M 2 V 2 mmv 2 = M 2 V' 2 M 2 V 2 2ω k M = M 2 V' 2 V + m( m + M)v 2 = V cosθ θ = recoil angle in the lab ( ) 2 V 2 cos 2 θ m only possible if V is large enough (>velocity of sound) V ' 2M " k " V + 2 k 2 + 2ω k M = 2 ω k 2 ω k = 2Mc s ( V cosθ c s ) > c s 2 ω k V cosθ 2M ω k c s V cosθ + 2ω k M = k

Much cleaner system: Alcali Vapors BE condensation for atoms demonstrated in 995 => 2 Nobel Prize in Physics awarded jointly to Eric A. Cornell of NIST / JILA; Wolfgang Ketterle of MIT; and Carl E.

29 Much cleaner system: Alcali Vapors BE condensation for atoms demonstrated in 995 => 2 Nobel Prize in Physics awarded jointly to Eric A. Cornell of NIST / JILA; Wolfgang Ketterle of MIT; and Carl E. Wieman of CU / JILA. Time sequence of images showing one cycle of the ringing of a Bose- Einstein condensate (BEC) in the JILA TOP (time-averaged orbiting potential) trap after being driven by strong oscillations of trap potential. 29

30 Cooling in a trap See: BEC Atoms Images of the velocity distributions of the trapped atoms Left: just before the appearance of the Bose-Einstein condensate Center: just after the appearance of the condensate Right: after further evaporation. 3

31 Superconductivity 3 He Pairing of Fermions Low Tc: Pairing of electrons s= (Cooper pairs) <= phonon interaction But condensation theory bad approximation (not free) τ superconductivity << τ condensation Similar effects Spin /2 Zero resistance Quantization of flux : Vortices 2 phases of pairing s= similar to superconductivity but magnetic properties τ condensation =.95mK and 2.5mK 3

32 Energy Density of Ultra Relativistic Gases Generalization Important for behavior of early universe (energy density =>expansion) Density of energy g 2 x 3 dx e x = π 4 5 Bosons π c 3 k BT Suppose that particles are non degenerate (µ<<τ) f ( ε ) exp ε ± u = εd ( ε ) f( ε)dε with ( )dε = g ε 2 dε 2 π 2 c 3 3 D ε ( ) 4 = g 2 a BT 4 u = where g is the spin multiplicity u = g τ 4 x 3 x 3 dx dx e 2 π 2 c 3 3 x = e x ± g bosons g fermions Fermions a B 2 T 4 7 g 8 2 Effective number of degrees of freedom for a relativistic π 4 5 π 2 ( 5 3 c 3 k BT) 4 = 7 g 8 2 a BT 4 32

33 Is the pressure the force per unit area? A last task: Show that the pressure we compute is indeed the average force per unit area cf. Kittel and Kroemer Chapter 4 p. 39 Describe the particles by their individual density in momentum space (ideal gas) density in d 3 p = n( p) p 2 dpdω θ vδt If the particles have specular reflection by the wall, 2 pcosθ the momentum transfer Δp for a particle arriving at angle θ is P = Force d da da = Δt da = 2 pcosθ daδt " # $# vδtdacosθ #" # $ n( p) p2 dpdω ## "## $ 2 Integration on angles gives 2π pv n( p) p 2 dp 3 that we would like to compare with the energy density u = ε n( p)d 3 p = 4π ε n( p) p 2 dp non relativistic pv = 2ε pressure P = 2 u (energy density) 3 u = 3 N 2 V τ P = N τ σ τ = same pressure as thermodynamic definition = V V U,N ultra relativistic pv = ε P = 3 u 33 Momentum transfer Volume of cylinder = dϕ d cosθ 2 pvcos 2 θ n p 2π ( ) p 2 dp density in cylinder

5. Systems in contact with a thermal bath

5. Systems in contact with a thermal bath So far, isolated systems (micro-canonical methods) 5.1 Constant number of particles:kittel&kroemer Chap. 3 Boltzmann factor Partition function (canonical methods)