Lecture Frontier of complexity more is different Think of a spin - a multitude gives all sorts of magnetism due to interactions

Transcription

1 Lecture 1 Motivation for course The title of this course is condensed atter physics which includes solids and liquids (and occasionally gases). There are also interediate fors of atter, e.g., glasses, rubber, polyers, and soe biophysical systes. Basically this is the branch of physics that covers the things we see and touch in everyday life, i.e., real stuff. Most of the aterials we eet in every day life are aorphous, but since we understand crystalline aterials so uch better, that is what we will spend ost of our tie talking about. Why should we study condensed atter physics? 1. Because it s there. 2. Real-life physics 3. Frontier of coplexity ore is different Think of a spin - a ultitude gives all sorts of agnetis due to interactions 4. Analogies with eleentary particle physics, e.g., Higgs echanis, topological winding nubers, broken syetry, etc. 5. Practical applications, e.g., transistors. Drude Theory of Metals (a) Phenoenology of etals high electrical conductivity, shiny (reflecting), ductile alleable, high theral conductivity, etc. Found generally in coluns 1A and 2A of the periodic table, aong heavier III-VI colun eleents, and in transition etals and rare earths. In general, they have 1-2 extra electrons above a closed shell. Typically ρ etal few µω-c versus ρ insulator Ω-c for insulators like polystyrene. (b) Basic concepts - The extra electrons are called conduction electrons and they are free to ove within the volue. Core electrons stay hoe. The nuber of conduction electrons n e = ZnAvagadro electrons/c 3 1.1

2 where Z is the cheical valence (see table 1.1 of AM). Electronic density is often defined in ters of r s = radius of sphere whose volue is equal to the volue per conduction electron: V N = 1 ( ) 1/3 n = 4πr3 s 3 3 ; r s = 4πn Typically r s 1 3Å. Natural unit is Bohr radius a 0 = 2 /e 2 = c. r s a For coparison, note that a typical atoic (ionic) radius is 0.3 2Å. So conduction electrons occupy a larger sphere than ions. (c) Electrical Conductivity (resistivity) σ = 1 ρ j = σ E E = ρ j Let A = cross sectional area of wire, L = length, j = I A and V = EL V = ρjl = ρ I A L = IR R = ρ L A or ρ = RA L Longer wires have ore resistance. Larger A eans ore anuverability for electrons and less resistance. As we said before, ρ(300 K) 1µΩ-c. At not too low T, ρ(t ) T (phonon scattering). As T 0, ρ(t ) ρ 0 = residual resistivity due to scattering of ipurities. This yields Matthiessen s Rule: ρ(t ) = ρ 0 ρ T (1.1) ρ ρ~ T ρ o T 1.2

3 Assuptions of the Drude odel: (i) Electrons ove independently under the influence of local electric field between collisions. (ii) Collisions are instantaneous, with soe unspecified but energy-nonconserving echanis. (iii) Collisions are rando, with probability dt/τ per unit tie (no history dependence). (iv) Electrons totally theralized to local teperature by inelastic collisions. DC Conductivity ( B = 0, E = const) Electric current j = ne v (1.2) The inus sign is due to the negative charge of the electrons. There are two contributions to d j dt : total d j dt = d j total dt d j collision dt Field: Force = F = d p dt d v = ee dt d j dt = d(ne v ) = ne2 E field dt Collisions: Collisions knock electrons out of the current flow. So we expect d j dt < 0: coll degrade current field δ v = v (prob of collision fraction of particles affected) = v dt (τ is relaxation tie) τ d v dt = v coll τ d j dt = j coll τ 1.3

4 So d j dt = ne2 E total j τ In a steady state with E = const, j ust be constant: d j dt = 0 total j = ne2 τ E = σ 0 E where the DC (E = constant) conductivity is given by σ 0 = ne2 τ sign of charge doesn t atter (When E = E(t), σ = σ(ω), i.e., the conductivity has frequency dependence.) Fro experiental values of σ 0 and n, we can work out τ (see AM, table 1.2). Typically, τ sec. at roo teperature (τ 1 T ). At low T, τ 10 9 sec and is liited by ipurity scattering. Matthiessen s rule: τ 1 τ 1 0 (τ 1 ) T. We can define a ean free path l vτ. How do we estiate v? Drude used kinetic theory of gases and said 1 2 v2 = 3 2 kt v rs 10 7 c s at T = 300K l 1 10Å lattice spacing or distance between ions But this is isleading. (Should use v F 10 8 c/s) Conduction in a Magnetic Field In the presence of a agnetic field B, an additional Lorentz force acts on the electrons. F = e( E v c B) This leads to the Hall Effect. Consider a etal bar with current flowing in it carried by electrons with average velocity v. 1.4

5 x z y v d J E Now suppose we apply a agnetic field in the ˆx direction. This initially causes a downward deflection of the oving electrons. B E Negative charge builds up at the botto; positive charge at the top. The transverse electric field E t counters the agnetic force so that the electrons again flow in the ŷ direction. B E t J E Notice that if the charge carriers had been positively charged, Et would point in the opposite direction ( j is in the sae direction as before). Thus if we easure the voltage difference between top and botto, the sign should tell us the sign of the carriers. (We expect negative, but soeties it s positive. More on this later.) It is easy to deterine the agnitude of E t by balancing the electric force with the agnetic force in the z- direction. (Let s use q rather than ( e).) q E t = q v c B E t = v c B We know j = nq v v = j/nq E t = ( ) 1 jb = R H jb nqc 1.5

6 where R H = nqc 1 is called the Hall coefficient. For q = e, R H = nec 1 Experientally, R H = E t jb E t j B. Note that because E t cancels the effect of the agnetic field, we still have j y = σ 0 E y (different coords than AM). You can check this by looking at d j dt. Experientally, total this isn t always true. Drude odel is too siple. ( B = 0, E(t)) σ(ω) AC Conductivity Consider an electric field that is varying in tie: E(t) = E 0 cos ωt = Re ( E 0 e iωt) The response of the electrons as well as the current will also vary in tie. This leads to a frequency dependent conductivity. j(t) = Re ( j 0 e iωt) where j 0 = σ(ω) E 0 In general σ(ω) will be coplex, indicating that j is out of phase with E. Start with Calculate σ(ω) d j dt = ne2 E(t) total j τ Plug in j(t) = j 0 e iωt and E(t) = E 0 e iωt to get iω j 0 = ne2 E0 j 0 τ ( iω 1 τ ) j 0 = ne2 E0 ( ) j 0 = ne2 1 E iω 1 0 τ σ(ω) = ne2 τ ( 1 ) 1 iωτ = σ 0 1 iωτ 1.6

7 where tan δ = ωτ. If E = E 0 cos ωt, then j = σ 0 E 0 cos(ωt δ) 1 ω2 τ 2 We can relate σ(ω) to the frequency dependent dielectric constant ε(ω). Consider a piece of etal that is free-standing. Suppose we irradiate it with electroagnetic radiation. There will be no free current j f but there will be a polarization current because the electrons slosh back and forth: j = P t P = j iω = σ E iω = iσ E ω D = ε E = E 4π P ε = 1 4π P E = 1 4πiσ ω At high frequencies (ωτ 1) where ω 2 p = 4πne2 ε(ω) = 1 4πiσ(ω) ω Plasa Frequency (ωτ 1) σ 0 σ(ω) = 1 iωτ iσ 0 ωτ ε(ω) = 1 4πσ 0 ω 2 τ = 1 4π ne 2 τ ω 2 τ. This is called the plasa frequency. = 1 ω2 p ω 2 What does this ean physically? ω p is the characteristic frequency for the electrons to slosh back and forth. These are called plasa oscillations, or plasons. AM give a siple odel of this. Iagine displacing the entire electron gas, as a whole, through a distance d with respect to the fixed positive background of ions. 1.7

8 surface charge = ned N / Z ions d N electrons surface charge = ned Plason: longitudinal excitation without Eext The resulting surface charge gives rise to an electric field of agnitude 4πσ, where σ is the charge per unit area (recall Gauss Law). The electron gas obeys the equation of otion d F = Ne E ( E ext = 0. This E is internally generated.) N d = Ne E = Ne 4πσ = Ne(4πnde) N d = 4πne 2 Nd F = ẍ = kx N = total nuber of electrons ω 2 p = k = 4πne2 n = N V There is yet another way to derive the plasa frequency: go back to j t = ne2 E j τ At high frequencies (ωτ 1), ω 1 τ think of sloshing electrons as producing a polarization current. we can neglect the last ter. This leaves j t = ne2 E 1.8

9 ρ Recall the continuity equation: = j. So t [ ] j t = ne2 E 2 ρ t = ne2 2 E = 4πne2 ρ or 2 ρ t = 2 ω2 pρ where again ωp 2 = 4πne2 (ẍ = kx for) Transverse EM Waves If we shine EM radiation on a etal, it will not penetrate very far (and in fact, it will be reflected) for low frequencies because the electrons respond quickly enough to screen it. At high frequencies, however, (ω ω p ) the electrons aren t fast enough to respond to E(t) and the radiation gets through. Thus the etals becoe transparent to ultraviolet light. To see this atheatically, go back to Maxwell s eqns. and derive the wave equation. set µ = ε = 1 ρ = 0 E = 0 B = 0 B = 4π c j 1 c E = 1 B c t E t ( j = σ E) ( E) = 1 c t ( B) ( E) 2 E 1 = c t (4π c σ E 1 c E t ) 1.9

10 Fourier Transfor w.r.t. tie using E E o e iωt : ( 2 E = iω 1 ) c 4πσ E ω2 E 2 c ( ) 2 = ω2 1 4πiσ E c 2 }{{ ω } ε(ω) At low frequencies, ωτ 1, σ(ω) σ 0 and ω2 c 2 2 E i ω c 2 4πσ E ter is negligible. Hence For E = E0 e i k r, 2 E = k 2 E ω = i c 4πσ E 2 k 2 = i 4πσω k = c 2 4πσω c ( 1 i 2 ) = k ik E = E 0 e ik r e k r EM wave decays as it enters the etal Skin depth = 1 k = c 2πσω (ωτ 1) High Frequencies For ωτ 1, σ(ω) σ 0 iωτ = i ne2 ω ( recall σ(ω) = σ ) 0 1 iωτ so 2 E = 4π c 2 ne 2 E ω2 c 2 E ω 2 p = 4πne2 = ω2 p ω 2 E k 2 = ω2 p ω 2 c 2 c 2 For ω < ω p, this leads to exponential decay with decay length c ω 2 p ω 2. For ω > ω p, we get propagation and the etal becoes transparent at a frequency ν p sec