WAVE-PARTICLE DUALITY Of MATTER

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5.61 Fall 2007Lecture #6page 1WAVE-PARTICLE DUALITY of MATTERConsequences (II)ΔxΔpx Heisenberg Uncertainty Principle!2Consider diffraction through a single slitxDspeak-nulldistancellight, λD λ ls D λlsNow consider a beam of electrons with de Broglie wavelength λ. The slitrestricts the possible positions of the electrons in the x direction: at the slit,the uncertainty in the electron x-position isxsΔ xComing out of the slit, the electronsspread out to form a diffraction patternwith width D.DThis means the electrons must go through the slit with some range ofvelocity components vxxv!vΔxv!

5.61 Fall 2007Lecture #6Δvxpage 2D λ λ v!s ΔxΔxΔvx λ v or ΔxΔpx λ pBut from deBroglieλ hp ΔxΔpx hSo the position and momentum of a particle cannot both be determined witharbitrary position! Knowing one quantity with high precision means that theother must necessarily be imprecise!The conventional statement of the Heisenberg Uncertainty Principle isΔxΔpx !2(depends on how “uncertainty” is defined: 1/e half-width, FWHM, etc.)Uncertainty can always be larger than ! 2 , but not smaller.Note that this sort of uncertainty is standard in classical wave mechanics. Ifyou focus a light beam or a water wavelet to a small spot size, at the focusthere is a wide range of propagation directions. What is new is the idea thatparticles inherently show wavelike behavior, with similar consequences.Implications for atomic structureApply Uncertainty Principle to e- in H atomme 9.11 x 10-31 kgΔx 10-10 m (1 Å)

5.61 Fall 2007Lecture #6Δpx !2Δx page 3!1 x10 6 m s2mΔx 2Δvx Basically, if we know the e- is in the atom, then we can’t know its velocity at all!Bohr had assumed the electron was a particle with a known position and velocity.To complete the picture of atomic structure, the wavelike properties of theelectron had to be included.So how do we properly represent where the particle is?Schrödinger(1933 Nobel Prize)A particle in a “stable” or time-independent state can be representedmathematically as a wave, by a “wavefunction” ψ(x ) (in 1-D) which is a solution tothe differential equation!2 2ψ V x ψ x Eψ x2m x 2() ()potential energy()Time-independentSchrödingerequationtotal energyWe cannot prove the Schrödinger equation. But we can motivate why it might bereasonable.Av( )()φ1 x,t Asin kx ω t is a right-traveling wave.k 2πωω 2πνλν vSimilarly, a left-traveling wave can be represented as( )()φ2 x,t Asin kx ω t .Both are solutions to the wave equation

5.61 Fall 2007Lecture #6( ) 2φ x,t x 2( )( )page 4( )21 φ x,tv 2 t 2( )Further, the sum Ψ x,t φ1 x,t φ2 x,tof left and right traveling waves isalso a solution.( )()()( ) ( )Ψ x,t A sin kx ω t sin kx ω t 2 Asin kx cos ω tThis is a stationary wave or standing wave. Its peaks and nulls remain stationary.At various times during a full cycle (2π/ω):( )( )Ψ x,0 2 Asin kxt 0 2 1 2π Ψ x, 2 A sin kx t π/4ω 8 ω 2 ( ) 1 2π Ψ x, 0 4 ω t π/2ωt 3π/4ω 2 3 2π Ψ x, 2 A sin kx 8 ω 2 ( ) 1 2π Ψ x, 2 A 1 sin kx 2 ω ( ) ( )nodest π/ωAs in a vibrating violin string, the node positions are independent of time. Onlythe amplitude of the fixed waveform oscillates with time.More generally, we can write wave equation solutions in the form( )() ( )()( )Ψ x,t ψ x cos ω tIn the particular case above, ψ x 2 Asin kx .

5.61 Fall 2007Lecture #6page 5For the general case,( ) ( ) 2 Ψ x,t21 Ψ x,t ω 2 2 Ψ x,t k 2 Ψ x,t22v tv x 2( )( )( )(ωk v)() ( )Plugging in Ψ x,t ψ x cos ω t ( ) k ψ 2ψ x x22hλ pde Broglie relation !()( ) pψ2()2() ! ψ x x 2 x( ) 2( x)(assuming t-independent potential)()!2 ψ x V x ψ x Eψ x2m x 22 ( ) p 2ψ x2But p 2m K.E. 2m E V x () 2ψ x 22 2π x ψ x λ () ()()time-independent Schrödinger equation in one dimensionWe now have the outline of: a physical picture involving wave and particle duality of light and matter ! a quantitative theory allowing calculations of stable states and theirproperties !

The conventional statement of the Heisenberg Uncertainty Principle is ΔxΔp x 2 (depends on how “uncertainty” is defined: 1/e half-width, FWHM, etc.) Uncertainty can always be larger than ! 2 , but not smaller. Note that this sort of uncertainty is

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This wave will be moving with a phase velocity given by vphase ! k. Phase velocity is the speed of the crests of the wave. Using E ! 1 2 mv2 and p k where v is the velocity of the particle we get: vphase ! k E p 1 2 mv2 mv 1 2 v That is, the wave is moving with half the speed of its associated particle. Semester 1 2009 PHYS201 Wave .