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1 Nanostructures density of states Faculty of Physics UW

2 Density of states What is the difference between 2D, D i 0D nanostructures? Number of states per unit energy ρ nd (E) depends on the dimension ε n = ħ2 2 k n 2m = ħ2 n 2 π 2 2mL 2 ε 3 = E c + 9ħ2 π 2 2m 0 m L 2 ε 2 = E c + 2ħ2 π 2 m 0 m L 2 ε = E g + ħ2 π 2 2m 0 m L

3 Density of states Density of states If our crystal has a finite size the set of k vectors is finite (though enormous!). For example: we can assume periodic boundary conditions and then: Born- von Karman boundary conditions Finite size of the crystal L x, L y, L z Ψ the Bloch function Ψ(x + L x,y,z) = Ψ(x, y + L y,z) = Ψ(x, y, z + L z ) Number of states per unit energy ρ nd (E) depends on the dimension k y e ik xl x = e ik yl y = e ik zl z = k i = 0, ± 2π L i, ± 4π L i,, ± 2πn i L i k x Number of states in the volume V = 2 2π L 2π x L 2π = y L z 2V 2π 3 2π L y 2π L x 6/5/207 3

4 Density of states Density of states Number of states per unit energy ρ nd (E) depends on the dimension The density of states in k-space of n dimension (and the unite volume) 3D case ρ 3D E de = ρ k 3D dk = 2 2π For a spherical and parabolic band: ρ c 3D E = 2π 2 2m 0 m c ħ 2 3/2 3 4πk 2 dk E E c Fermi sphere T=0 K ρ k nd = 2 2π k y n k x ρ v 3D E = 2π 2 2m 0 m h ħ 2 3/2 E v E 2π L y 2π L x

5 Density of states Density of states Number of states per unit energy ρ nd (E) depends on the dimension The density of states in k-space of n dimension (and the unite volume) 3D case ρ 3D E de = ρ k 3D dk = 2 2π For a spherical and parabolic band: ρ c 3D E = 2π 2 ρ v 3D E = 2π 2 2m 0 m c ħ 2 2m 0 m h ħ 2 3/2 3/2 3 4πk 2 dk E E c E v E Gestosc stanów ρ k nd = 2 2π Energia (ev) n

6 Density of states Inside the well: ψ x, t = 2 L sin k nx e iωt k n = nπ L ε n = ħ2 2 k n 2m = ħ2 n 2 π 2 2mL 2 ε 3 = E c + 9ħ2 π 2 2m 0 m L 2 ε 2 = E c + 2ħ2 π 2 m 0 m L 2 ε = E g + ħ2 π 2 2m 0 m L

7 Density of states 2D density of states ρ 2D E de = ρ k 2D dk = 2 2π ρ a 2D E de = m 0m 2 πħ 2 θ E E a de 2πk dk Step-like θ Heaviside function for a spherical and parabolic band: ρ 2D E de = m 0m πħ 2 a θ E E a de Marc Baldo MIT OpenCourseWare Publication May

8 Density of states 2D ψ kx,k y,n x, y, z = exp ik x x exp ik y y u n z = ψ k,n r, z = exp ik r u n z E n k x, k y = ε n + ħ2 2 k x 2m + ħ2 2 k y 2m E n k = ε n + ħ2 k 2 2m

9 Density of states 2D J. Szczytko, et al. Phys. Rev. Lett. 93, 3740 (2004)

10 Marc Baldo MIT OpenCourseWare Publication May 20 Density of states D D density of states ρ D E de = ρ k D dk = 2 2π for a spherical and parabolic band: 2 dk ρ D E de = 2 π m 0 m 2ħ 2 a x,a y θ E E ax,a y E E ax,a y de

11 Marc Baldo MIT OpenCourseWare Publication May 20 Density of states D 0D density of states For the ISOLATED QD: ΔE 0, Δt ρ 0D E de = g n δ E E n de Suppose that the lifetime of the state energy E is equal to τ, we assume an exponential decay ψ t 2 = A 2 exp t τ, t > 0 ψ t = A exp i E 0t ħ t 2τ, t > 0 Fourrier transform A ψ ω = 2τ + i E 0 ω ħ Lorentz profile ρ 0D E de = 2π ψ ω 2 dω de = 2 π ħ 2τ E E ħ 2τ

12 Marc Baldo MIT OpenCourseWare Publication May 20 Density of states the summary Density of states

13 PL Intensity Harmonic potential 2D CB p Harmonic oscillator model: s-, p-, d-, shells s Allowed interband transition s s p d Wetting layer 0.mW mw 0.5 mw 5mW 0mW VB p f GaAs substrate Energy (mev)

14 Summary Fermi golden rule The probability of transition per unit time: W t = W 0 t τ P mn = w mn τ = 2π ħ m W n 2 δ E m E n Transitions are possible only for states, for which E m = E n W t = w ± e ±iωt 0 t τ P nm = w nm τ = 2π ħ n w± m 2 δ E n E m ± ħω Transitions are possible only for states, for which The perturbation in a form of an electromagnetic wave: E m = E n ± ħω A nm = ω nm 3 e 2 3πε 0 ħc 3 m Ԧr n 2 = 4α 3 ω nm 3 c 2 m Ԧr n 2 P nm = A nm δ E n E m ± ħω

15 Summary Fermi golden rule The transition rate the probability of transition per unit time from the initial state ȁi to final ȁf is given by: Szybkość zmian czyli prawdopodobieństwo przejścia na jednostkę czasu ze stanu początkowego ȁi do końcowego ȁf dane jest wzorem: P mn = 2π ħ f W i 2 ρ E f W- interaction with the field ρ E f - the density of final states Perturbation W does not have to be in the form of an electromagnetic wave

16 Density of states the summary InAs HOMO LUMO NATURE VOL AUGUST

17 The Fermi-Dirac distribution The probability that a state of the energy E will be occupied E F chemical potential Fermions: Bosons: Boltzman distribution: f 0 = E E F e k B T + f 0 = E E F e k B T f 0 = E E F e k B T ± e E E F k B T Electrons Holes Trions (charged excitons) Polaritons Phonons Magnons Excitons, biexcitons Plasmons Anyons eg. composite fermions ȁψ Ψ 2 = e iθ ȁψ 2 Ψ Slave fermions (chargon, holon, spinon) = fermion+bozon with the charge-spin separation E F = F n F = U TS

18 The Fermi-Dirac distribution f 0 = E E F e k B T + Prawdopodobienstwo obsadzenia K 00K 300K Enrico Fermi Energia (ev) Paul Adrian Maurice Dirac

19 The Fermi-Dirac distribution f 0 = E E F e k B T + Prawdopodobienstwo obsadzenia K 00K 300K Energia (ev)

20 The Fermi-Dirac distribution f 0 = E E F e k B T + Prawdopodobienstwo obsadzenia K 00K 300K Energia (ev)

21 The Fermi-Dirac distribution The probability that a state of the energy E will be occupied E F chemical potential f 0 = E E F e k B T + k K 00K 300K Energia (ev) Occupation probability Prawdopodobienstwo obsadzenia

22 The Fermi-Dirac distribution The probability that a state of the energy E will be occupied E F chemical potential f 0 = E E F e k B T + k K 00K 300K Energia (ev) Occupation probability Prawdopodobienstwo obsadzenia

23 The Fermi-Dirac distribution Conduction band Vallence band E g E G lh k hh K 00K 300K Energia (ev) Occupation probability Prawdopodobienstwo obsadzenia

24 Electrons statistics in crystals The case of a semiconductor, in which both the electron gas and hole gas are far from the degeneracy: E c ξ k B T E c E v k B T ξ E v k B T = E g k B T E = E G 2 + E e E e The probability of filling of the electronic states: f e = = E ξ E G ek BT + e2k B T + E e k B T ξ k BT + and of holes: f h = f e f h = = E+ξ e k B T + e E G 2k B T + E h k B T + ξ k B T + E = E G 2 E h E h

25 Electrons statistics in crystals The case of a semiconductor, in which both the electron gas and hole gas are far from the degeneracy: E c ξ k B T E c E v k B T ξ E v k B T = E g k B T E = E G 2 + E e E e The probability of filling of the electronic states: f e = = E ξ E G ek BT + e2k B T + E e k B T ξ e E G 2k B T E e k B T + ξ k B T k BT + and of holes: f h = f e f h = = E+ξ E G e k B T + e 2k B T + E h k B T + ξ k B T + e E G 2k B T E h k B T ξ k B T E = E G 2 E h E h

26 Electrons statistics in crystals The case of a semiconductor, in which both the electron gas and hole gas are far from the degeneracy: the probability of filling of the electronic states: f e e E G 2k B T E e k B T + ξ k B T and of holes f h = f e f h e E G 2k B T E h k B T ξ k B T E c E = E G 2 + E e E e Thus: n ξ = 2 m e k B T 2πħ 2 p ξ = 2 m h k B T 2πħ 2 න 0 xe x dx = 3/2 e E G 3/2 π 2 ξ 2k BT ek BT = N c T e E c ξ k B T e E G 2k B T e ξ ξ E v k B T = N v T e k B T E v E = E G 2 E h E h

27 Electrons statistics in crystals What is the concentration of carriers for T>0? In the thermodynamic equilibrium for an intrinsic semiconductors (półprzewodniki samoistne), the concentration of electrons in the conduction band is equal to the concnetration of holes in the valence band (because they appear only as a result of excitation from the valence band). n = p = n i (an intrinsic case) ln(n) n p = n i 2 = 4 n = p = n i = 2 k BT 2πħ 2 k BT 2πħ 2 3 m e m h 3 2 e E g k BT = N c N v e E g k B T 3 2 me m h 3 4 e E g 2k B T = N c N v e E g 2k B T e E g 2k B T /T

28 Intrinsic carrier concentration What is the concentration of carriers for T>0? In the thermodynamic equilibrium for an intrinsic semiconductors (półprzewodniki samoistne), the concentration of electrons in the conduction band is equal to the concnetration of holes in the valence band (because they appear only as a result of excitation from the valence band). n p = n i 2 = 4 k BT 2πħ 2 3 m e m h 3 2 e E g k BT = N c N v e E g k B T (general formula) filled band empty band n = p = n i (an intrinsic case) n = p = n i = 2 k BT 2πħ me m h 3 4 e E g 2k BT = N c N v e E g 2k B T E g 2ξ E N g c = e k BT ξ = N v 2 E c + E v k BT ln m h m e T in our notation the middle of the band is

29 R. Stępniewski Intrinsic carrier concentration What is the concentration of carriers for T>0? The intrinsic semiconductors (półprzewodniki samoistne) in thermal equilibrium, the concentration of electrons in the conduction band is equal to the concnetration of holes in the valence band. The concentration values less than 0 0 cm 3 do not make sense because the concentration of impurities, and thus the concentration resulting from unintentional doping, is greater n = p = N c N v e Eg 2k B T E F E c n = N c e k B T p = N v e E F E v k B T

30 Domieszkowanie półprzewodników Carriers: holes + electrons - Dopants: Acceptors (p-type) Donors (n-type) Semiconductors II III IV V VI Be B C N O Mg Al Si P S Zn Ga Ge As Se Cd In Sn Sb Te Group IV: diamond, Si, Ge Group III-V: GaAs, AlAs, InSb, InAs... Group II-VI: ZnSe, CdTe, ZnO, SdS

31 Dopants, impurities and defects How to control the concentration of carriers? In semiconductors we can find several deviations from the ideal crystal structure: Defects of the crystal structure, vacancies, the atoms in interstitial positions, dislocations created eg. during the growth process. Foreign atoms (dopants) introduced intentionally or by adding impurities As a result of their presence: allowed states in the forbidden gap due to deviations from the ideal crystalline potential space charges in insulators screening by free carriers Dopant states can be classified into: Deep - the short-range potential located mainly in the area of one unit cell - eg. a vacancy, izoelectronics dopant (the same valency as atom of the base material e.g. N InP). Shallow - mainly long-range Coulomb potential

32 Dopants, impurities and defects Hydrogen-like model Atom of a valency higher than the base material atom becomes a source of Coulomb potential (modified by the dielectric constant of the crystal ε r ) caused by an extra proton in the nucleus. Extra electron in the conduction band feels this potential. The states are described by the effective mass equation: T = ħ2 2m Δ U = 4πε 0 e 2 ε r r II III IV V VI Be B C N O Mg Al Si P S Zn Ga Ge As Se Cd In Sn Sb Te T + U φ Ԧr = Eφ Ԧr Group IV: diamond, Si, Ge Group III-V: GaAs, AlAs, InSb, InAs... Group II-VI: ZnSe, CdTe, ZnO, SdS

33 Dopants, impurities and defects Hydrogen-like model Finally, the problem reduces to the problem of hydrogen atpom with a free carrier of mass m in the medium of dielectric constant ε and with a small "perturbation" potential. II III IV V VI Be B C N O E n = m m 2 0 ε Ry r n 2 Mg Al Si P S For typical semiconductors m e * 0.m e e s 0 a B = 4πε rε 0 ħ 2 m 0 m 0 m 0 e 2 m = a B ε r m For Hydrogen Ry = 3.6 ev and a B = nm For GaAs semiconductor Ry 5 mev and a B 0 nm Zn Ga Ge As Se Cd In Sn Sb Te Group IV: diamond, Si, Ge Group III-V: GaAs, AlAs, InSb, InAs... Group II-VI: ZnSe, CdTe, ZnO, SdS

34 Far infrared spectroscopy E D 0 2p+ 2p c 2p0 2p- B s s Silicon R y =30.28(5) mev

35 Far infrared spectroscopy E D 0 2p+ 2p c 2p0 2p- B s s Oxygen R y =30.28(5) mev

36 Electron energy Dopants, impurities and defects Hydrogen-like model ionization of the dopant n-type p-type conduction band conduction band donor level acceptor level vallence band vallence band x x

37 Electron energy Dopants, impurities and defects Doping conduction band donor level E D E g /2 0 E F The carrier concentration in extrinsic semiconductor (niesamoistny) Consider a semiconductor, in which: N A concentration of acceptors N D concentration of donors p A concentration of neutral acceptors n D concentration of neutral donors n c concentration of electrons in conduction band p v concentration of holes in valence band acceptor level vallence band E A -E g /2 From the charge neutrality of the crystal: x

38 Electron energy Dopants, impurities and defects Doping conduction band donor level E D E g /2 0 E F The carrier concentration in extrinsic semiconductor (niesamoistny) Consider a semiconductor, in which: N A concentration of acceptors N D concentration of donors p A concentration of neutral acceptors n D concentration of neutral donors n c concentration of electrons in conduction band p v concentration of holes in valence band acceptor level vallence band E A -E g /2 From the charge neutrality of the crystal: n c +(N A - p A )= p v + (N D - n D ) n c + n D = (N D - N A )+ p v + p A x

39 The occupation of impurity levels Occupation of impurity / defect levels in the thermodynamic equilibrium Occupation of localized or band states means the exchange of particles (electrons) between the reservoir and considered subsystem (microstate). The grand canonical ensemble (subsystem exchange particles and energy with the environment) Thermodynamic probability (unnormalized) of finding subsystem in a state j,in which there are n j particles (electrons) and which subsystem energy is E j (the total energy of all n j particles): P j = e β E j n j ξ, β = k B T ξ- chemical potential. Statistical sum: Z = P j = j j e β E j n j ξ

40 The occupation of impurity levels Statistical sum : Statistical means: Z = P j = j j e β E j n j ξ A = σ j A j e β E j n j ξ σ j e β E j n j ξ Examples: free electron occupying (or not) the quantum state of the k-vector and spin: 2 possible states of the subsystem (microstate): n 0 = 0; E 0 = 0 n = ; E = E (the occupation only for one spin state) the average number of particles of the subsystem: n = 0 e0 + e β E ξ + e β E ξ = = f E, T eβ E ξ + (Fermi-Dirac distribution)

41 The occupation of impurity levels Examples : free electron occupying (or not) the quantum state of the k-vector and ANY spin: 4 possible states of the subsystem (microstate): n 0 = 0; E 0 = 0 n = ; E = E (spin ) n 2 = ; E 2 = E (spin ) n 3 = 2; E 3 = 2E (spin ) the average number of particles of the subsystem: n = 0 e0 + e β E ξ + e β E ξ + 2 e β 2E 2ξ + e β E ξ + e β E ξ + e β 2E 2ξ = 2 e β E ξ + e β E ξ e β E ξ + 2 = = 2f E, T (Fermi-Dirac distribution x2)

42 The occupation of impurity levels The ratio of the probability of finding dopant / defect of n + electrons and of n electrons: p n+ = N n+/n total σ j:nj =n+ e β E j n+ ξ = p n N n /N total σ k;nk =n e β E = g n+ e β E n+ E n ξ k nξ g n σ n N n = N impurity (dopants) concentration E n+ i E n the lowest of all subsystem energies E j with n + and n electrons respectively Successive impurity energy levels are filled with the increase of the Fermi level. E n+/n so-called energy level of the impurity/ defect numbered by charge states n + and n g n+, g n so-called degeneration of states of subsystem of n + and n electrons 6/5/207 42

43 The occupation of impurity levels g n+, g n so-called degeneration of states of subsystem of n + and n electrons. The degeneracy g n+ and g n takes into account the possibility of the existence of different subsystem states corresponding to a same number of particles (including the excited states): g n = α n,0 + i=,2, α n,i e βε n,i α n,0 i α n,i they are respectively: the degeneracies of the n-electronic ground state and its excited states of energies higher than the ground state by ε n,i (excitation energies) Such defined degeneration g n generally depends on temperature 6/5/207 43

44 The occupation of impurity levels Example donor (we omit the excited states) charge state (+) implemented in way : g + = charge state(0) implemented in 2 ways (spin or ): g 0 = 2 the energy of the donor state Ε 0/+ = E D Doping concentration of donors N D p n+ p n = p 0 p + = g 0 g + e β E D ξ, p + + p 0 = The occupation probability of the donor state: p 0 = n = + g + g0 e = β E D ξ + 2 eβ E D ξ The concentration of occupied donor states (neutral donors are N D 0 ) N D N 0 D = + 2 eβ E D ξ

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