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Ben Gurion University of the Negev www.bgu.ac.il/atomchip, www.bgu.ac.il/nanocenter Physics 3 for Electrical Engineering Lecturers: Ron Folman, Daniel Rohrlich Teaching Assistants: Ben Yellin, Shai Bartal Week 12. Quantum mechanics – propagation on a crystal lattice • Fermi level/momentum/energy • Kronig-Penney model and band gaps • GHZ and Bell’s theorem • quantum computing and cryptography. Sources: Feynman Lectures III, Chap. 13; Ashcroft and Mermin, Solid State Physics, Chap. 8; Tipler and Llewellyn, Sect. 10.6; selected papers to be distributed. Propagation on a crystal lattice Some solids – including crystals and most metals – are periodic lattices. A periodic lattice obeys the Bragg condition for diffraction of waves from solids: d θ θ 2d sin θ = nλ Propagation on a crystal lattice Some solids – including crystals and most metals – are periodic lattices. A periodic lattice obeys the Bragg condition for diffraction of waves from solids: d θ θ θ 2d sin θ = nλ Electron diffraction X-rays on zirconium oxide electrons Electrons on gold Neutron diffraction Diffraction of X-rays on a single NaCl crystal Diffraction of neutrons on a single NaCl crystal Propagation on a crystal lattice To understand some generic features of electron conduction on a crystal lattice, let’s model a 1D crystal, i.e. a lattice with a periodic potential. V(x) Ion core d x The exact shape of the periodic potential will not matter. Propagation on a crystal lattice To understand some generic features of electron conduction on a crystal lattice, let’s model a 1D crystal, i.e. a lattice with a periodic potential. V(x) x d The exact shape of the periodic potential will not matter. Propagation on a crystal lattice We know how to solve Schrödinger’s equation for each barrier, but how do we connect up the solutions for all the barriers? V(x) e re iKx iKx x teiKx d K 2mE Propagation on a crystal lattice We know how to solve Schrödinger’s equation for each barrier, but how do we connect up the solutions for all the barriers? V(x) te iKx e iKx re iKx d K 2mE x Propagation on a crystal lattice Bloch’s theorem (1D): The eigenstates of 2 d 2 Eψ( x) V ( x) ψ( x) 2 2m dx , where V(x) is periodic with period d, can be written ψ nk ( x) eikx unk ( x) , where unk(x) is periodic with period d, i.e. unk(x+d) = unk(x). That is , every eigenstate has the property ψ( x d ) eikd ψ( x) . Proof of Bloch’s theorem in three steps: ˆ ip d / e 1. The operator shifts any function f(x) by d. Proof: e ipˆ d / f ( x) e ipˆ d / 1 2 ~ f (k )e ikx dk ~ ˆ f (k )e ipd / e ikx dk ~ 1 ikd ikx f ( k ) e e dk f ( x d ) 2 1 2 . ipˆ d / ˆ ipˆ d / [ e , H ] [ e , V ( x)] 0. 2. Therefore 3. Therefore the eigenstates of Ĥ can be written as eigenstates of ˆ e ipd / , which are ψ nk ( x) eikx unk ( x). Fermi level/momentum/energy We will use Bloch’s theorem to find the relation between the momentum k and the energy E of an electron in a crystal lattice. But we ask first, How many momentum states are there in a crystal of length D? Assume that electron wave functions vanish outside the crystal. To be orthogonal, the eigenfunctions of p̂ must be 1, ei 2x / D , ei 4x / D , ei 6x / D , ei8x / D , , in 1D, using “periodic boundary conditions” (x + D ↔ x). If electrons were bosons, we could put them all in the lowestmomentum state. Fermi level/momentum/energy We will use Bloch’s theorem to find the relation between the momentum k and the energy E of an electron in a crystal lattice. But we ask first, How many momentum states are there in a crystal of length D? Assume that electron wave functions vanish outside the crystal. To be orthogonal, the eigenfunctions of p̂ must be 1, ei 2x / D , ei 4x / D , ei 6x / D , ei8x / D , , in 1D, using “periodic boundary conditions” (x + D ↔ x). But electrons are fermions, so we can put just one in each momentum state. Fermi level/momentum/energy In the ground state, all the momentum states are filled up to the state with the highest momentum magnitude kF, which is called the Fermi momentum. The corresponding highest energy EF, which equals E F Fermi energy. k F 2 2me in the simplest cases, is called the Fermi level/momentum/energy In 3D, the momentum states are vectors with components kx, ky, kz taking values 1, ei 2x / D , ei 4x / D , ei6x / D , ei8x / D , . We can think of triplets (kx,ky,kz), e.g. (4π/D,–16π/D,6π/D ), as points on a “reciprocal lattice”. When states are occupied up to the Fermi momentum, the triplets lie inside a ball of radius kF. The surface of this ball is called the Fermi surface. kz kF [From here.] ky kx Compare Maxwell-Boltzmann and Fermi-Dirac statistics to see the effect of the Pauli exclusion principle: TF = 500 K PMB ( E ) e E / k BT PFD ( E ) 1 e E / k BT 1 Compare Maxwell-Boltzmann and Fermi-Dirac statistics to see the effect of the Pauli exclusion principle: PFD PMB Average occupancy vs. average kinetic energy PMB PFD PMB ( E ) e E / k BT PFD ( E ) 1 e E / k BT 1 Exercise: What is the minimum energy of a population of N fermions? Exercise: What is the minimum energy of a population of N fermions? Solution: N (2 spin states)(4 /3) (k F ) 3 (2 /D) 3 (k F D) 3 3 2 2 (k F ) 2 2 (3 2 N ) 2/3 (3 2 N )1/3 . so k F and E F 2me D 2me D 2 Exercise: What is the minimum energy of a population of N fermions? Solution: N (2 spin states)(4 /3) (k F ) 3 (2 /D) 3 (k F D) 3 3 2 2 (k F ) 2 2 (3 2 N ) 2/3 (3 2 N )1/3 . so k F and E F 2me D 2me D 2 (2me ) 3/2 D 3 dN The “density of states” dN/dEF is dEF 2 2 3 EF . Exercise: Find the 2D versions of EF and dN/dEF. Exercise: Given that electron densities in metals are of order 1022/cm3, find the speed of the fastest electrons. Kronig-Penney model and band gaps So far, we have not discussed E(p), the relation between the energy E and the momentum p = ħk of electrons in crystals. But E(p) has dramatic consequences. E Allowed band E E Forbidden band Allowed band k k Free electron Electron in crystal Kronig-Penney model and band gaps So far, we have not discussed E(p), the relation between the energy E and the momentum p = ħk of electrons in crystals. But E(p) is remarkable, and has dramatic consequences. E Allowed band E Forbidden band Allowed band k Free electron EF conductor Electron in crystal k Kronig-Penney model and band gaps So far, we have not discussed E(p), the relation between the energy E and the momentum p = ħk of electrons in crystals. But E(p) is remarkable, and has dramatic consequences. E Allowed band Forbidden band E EF Allowed band k Free electron insulator Electron in crystal k Kronig-Penney model and band gaps So far, we have not discussed E(p), the relation between the energy E and the momentum p = ħk of electrons in crystals. But E(p) is remarkable, and has dramatic consequences. E Allowed band Forbidden band E EF Allowed band k Free electron Where does the “band gap” come from? insulator Electron in crystal k Kronig-Penney model and band gaps The “Kronig-Penney” model is a 1D lattice of square potential wells with lattice spacing is d. V(x) x -d d 2d 3d Kronig-Penney model and band gaps The “Kronig-Penney” model is a 1D lattice of square potential wells with lattice spacing is d. The energy E of an electron as a function of its wave number k is shown below for an electron in this lattice. Energy gaps appear when k = nπ/d. 0 π/d 2π/d 3π/d k The gap arises from standing waves that resonate through the lattice at k = nπ/d. For n = 1, for example, the waves are ψ ( x) (e ix / d e ix / d ) / 2 cos(x / d ) , ψ ( x) (e ix / d e ix / d ) / 2i sin(x / d ) . The figure shows the spatial probability distribution of both waves. |ψ–|2 |ψ+|2 traveling wave x d Prob. 8.1 in Solid State Physics by Ashcroft and Mermin is a mathematical analysis of band gaps in 1D. Assume a potential V(x) such that V(–x) = V(x) and V(±d/2) = 0 and V(x+d) = V(x). There are two scattering solutions, ψl(x) and ψr(x), with the following behaviors at x = ±d/2: ψl ( x) eiKx re iKx , x d / 2 , teiKx , x d / 2; V(x) e iKx re iKx teiKx K 2mE x d Prob. 8.1 in Solid State Physics by Ashcroft and Mermin is a mathematical analysis of band gaps in 1D. Assume a potential V(x) such that V(–x) = V(x) and V(±d/2) = 0 and V(x+d) = V(x). There are two scattering solutions, ψl(x) and ψr(x), with the following behaviors at x = ±d/2: ψl ( x) eiKx re iKx , teiKx , and ψ r ( x) teiKx , e iKx re iKx , x d / 2 , x d / 2; x d / 2, x d / 2. We now define a general solution ψ(x) = Aψl(x) + Bψr(x) and impose two conditions: ψ( x d ) e ikd ψ( x) [from Bloch' s theorem] , and ψ' ( x d ) eikd ψ' ( x) [by differenti ating] . Imposing these conditions on ψ(x) we can derive the following: t 2 r 2 iKd 1 iKd cos kd e e , 2t 2t and since in general rt* is pure imaginary and |r|2+|t|2 = 1, we can also derive cos(Kd ) cos kd , |t | where t = |t|eiδ . Now if |t| < 1, there are no solutions for k when Kd+δ ≈ nπ. This is the origin of the band gaps. Exercise: For weak scattering (|t| ≈ 1, δ ≈ 0), derive the width of the band gap. Solution: The gap occurs for cos Kd > |t|. Since |r|2+|t|2 = 1 and |t| ≈ 1, we can approximate |t| ≈ 1–|r|2 /2. The relevant values of K are those for which Kd ≈ nπ. For those values, we can approximate cos Kd as 1 – |Kd–nπ |2/2. Thus, the limits of the gap occur when |Kd±nπ| = |r|, i.e. when K = (|r|±nπ)/d. Since the energy is we have 2 E 2me 2K 2 E , 2me | r | n 2 | r | n 2 2 2 n | r | d d me d 2 as the width of the band gap. Electrons perform miracles! In superconductivity (which occurs at temperatures as high as 160 K), currents encounter no resistance. Electrical resistance suddenly vanishes as a superconductor is cooled below its critical temperature Tc. normal metal superconductor Electrons perform miracles! In superconductivity (which occurs at temperatures as high as 160 K), currents encounter no resistance. Superconductors expel all magnetic flux (Meissner effect) Electrons perform miracles! In superconductivity (which occurs at temperatures as high as 160 K), currents encounter no resistance. Superconductors expel all magnetic flux (Meissner effect) The explanation for superconductivity put forward by J. Bardeen, L. Cooper and J.R. Schrieffer [BCS] in the Physical Review 106 (1957) 162 is based on “bosonic” pairs of electrons having k (with spin up) and –k (with spin down) that are far apart but interact via the lattice. GHZ and Bell’s theorem Is the uncertainty principle a fundamental limit on what we can measure? Or can we evade it? Einstein and Bohr debated this question for years, and never agreed. GHZ and Bell’s theorem Is the uncertainty principle a fundamental limit on what we can measure? Or can we evade it? Einstein and Bohr debated this question for years, and never agreed. Today we are certain that uncertainty will not go away. Quantum uncertainty is even the basis for new technologies such as quantum cryptology. It may be that the universe is not only stranger than we imagine, but also stranger than we can imagine. GHZ and Bell’s theorem In 1935, after failing for years to defeat the uncertainty principle, Einstein argued that quantum mechanics is incomplete. The famous “EPR” paper GHZ and Bell’s theorem In 1935, after failing for years to defeat the uncertainty principle, Einstein argued that quantum mechanics is incomplete. Note that [x, ˆp] ≠ 0, but [x2–x1, pˆ 2+pˆ 1] = [x2, pˆ 2] – [x1, pˆ1] = 0. That means we can measure the distance between two particles and their total momentum, to arbitrary precision. So we can measure either x2 or p2 without affecting Particle 2 in any way, via a measurement on Particle 1 (and vice versa). That means both x2 and p2 are simultaneously real. Quantum mechanics has no place for simultaneous x2 and p2 , so quantum mechanics is incomplete. GHZ and Bell’s theorem Some reactions to the EPR argument: Bohr (1935) The uncertainty principle still applies. Pauli (1954) “One should no more rack one's brain about the problem of whether something one cannot know anything about exists all the same, than about the ancient question of how many angels are able to sit on the point of a needle. But it seems to me that Einstein's questions are ultimately always of this kind.” Bell (1964) Quantum mechanics contradicts EPR! GHZ and Bell’s theorem We will consider a theorem similar to Bell’s, proved in 1988 by D. Greenberger, M. Horne, and A. Zeilinger (GHZ). It describes three spin-½ particles prepared in one laboratory and sent to three different laboratories, operated by Alice, Bob, and Claire. For simplicity, we assume the particles are not identical. GHZ and Bell’s theorem Setting for Greenberger-Horne-Zeilinger (GHZ) paradox: Alice A Bob B time O space Claire C GHZ and Bell’s theorem We will consider a theorem similar to Bell’s, proved in 1988 by D. Greenberger, M. Horne, and A. Zeilinger (GHZ). It describes three spin-½ particles prepared in one laboratory and sent to three different laboratories, operated by Alice, Bob, and Claire. For simplicity, we assume the particles are not identical. The spin state of the three particles is GHZ 2 1 A B 1 0 where and . 0 1 C A B C , GHZ and Bell’s theorem Alice can measure x(A) or (yA) on her system; Bob can measure x(B) or (yB) on his system; Claire can measure x(C) or (yC) on her system; 0 1 0 i 1 0 , y , z . where x 1 0 i 0 0 1 0 1 ˆ 0 i ˆ 1 0 ˆ , S y , S z ). (Recall S x 1 0 2 i 0 2 0 1 2 The result of each x or y measurement can be 1 or –1. Note x , x , y i , y i . GHZ and Bell’s theorem To prove: ( A ) ( B) ( C ) 1. x x x | GHZ | GHZ ( A ) ( B) ( C ) | 2. GHZ x y y | GHZ (yA ) x( B) (yC) | GHZ (yA ) (yB) x(C) | GHZ That is, Alice, Bob and Claire discover two rules: ( A) ( B) (C) 1. x x x 1 (A) (B) (C) (A) (B) (C) (A) (B) (C) 2. x y y 1 y x y y y x GHZ and Bell’s theorem But now 1 x( A ) (yB) (yC) (yA ) x( B) (yC) (yA ) (yB) x(C) x( A ) x( B) x(C) 1 , ( A) 2 because x ( A) 2 y 1, etc. CONTRADICTION! GHZ and Bell’s theorem But now 1 x( A ) (yB) (yC) (yA ) x( B) (yC) (yA ) (yB) x(C) x( A ) x( B) x(C) 1 , ( A) 2 because x ( A) 2 y 1, etc. CONTRADICTION! This contradiction arose because we tacitly assumed, with EPR, that observables have values whether we measure them or not. Quantum computing and cryptography. Introducing…the qubit (quantum bit). Like an ordinary “classical” bit, a qubit has two states: , . But unlike an ordinary bit, a qubit can be in any superposition of these two states! Quantum “CNOT” gate: Quantum computing and cryptography. Introducing…the qubit (quantum bit). Like an ordinary “classical” bit, a qubit has two states: , . But unlike an ordinary bit, a qubit can be in any superposition of these two states! In principle, a quantum computer, given a superposition of inputs, computes the final result for each input simultaneously. Therefore there is a great speed-up of the computation. (The problem is that the final results, as well, are superposed.) In 1994, P. Shor devised an algorithm for factoring large numbers into primes, using a quantum computer. It is exponentially faster than any classical computer. Quantum computing and cryptography. Shor’s algorithm puts in jeopardy all codes based on the RSA encoding method, which is based on the difficulty of factoring large numbers into primes. But have no fear! !תורת הקוונטים הקדימה תרופה למכה There are quantum codes, and they are unbreakable! Example of quantum cryptography: “the one-time pad”. 100011111111110110 –10010101011010100 “10001010100100010” Bob “10001010100100010” +10010101011010100 100011111111110110 Alice How Alice, Bob and Claire can share a one-time pad: 1. They share many GHZ triplets (indexed), on which they measure σz . 2. From time to time they measure σx and σy to check that no one is trying to interfere with their key-sharing. Alice A time 10010101011010100 space Bob B 10010101011010100 O Claire C 10010101011010100 The Wall Street Journal (European edition), 5 March, 2004, p. A7 “…new technologies such as quantum computers...could perform calculations quickly enough to crack today’s best encryption systems. For that reason, military agencies in both the U.S. and Britain are funding quantum weirdness studies; they probably wouldn’t mind knowing how to put their troops in two places at once, either.” A Quantum Century “In the 19th century, life was transformed by the conscious application of classical mechanics….. In this [20th] century, a similar transformation has been wrought by electromagnetism…. It is easy to predict that in the 21st century it will be quantum mechanics that influences all our lives.” - Sir Michael Berry