Quantum Physics Lecture 3 If light (waves) are particle-like, are particles wave-like? Electron diffraction - Davisson & Germer Experiment Particle in a box -Quantisation of energy Wave Particle?? Wave groups and velocity
The Story so far.. Electromagnetic radiation interacts with matter in 3 ways: Photoelectric, Compton, Pair production Relative strength depends on energy, in order as shown If light (waves) particle-like are particles wave-like?
Wave Properties of Particles Recall, photon has momentum: p = ω c = k = h λ or λ = h p De Broglie (1924) suggested that this relation is general: particles which have momentum have an associated (or de Broglie) wavelength: where m is relativistic mass. m~m o at non-relativistic speed. Suggested in De B. PhD thesis & implied by Bohr model of atom Direct Experimental evidence?? λ = h p = h mv N.B. h = 2π = 6.625 10 34 Js Significance is for SMALL objects
Davisson & Germer Experiment Diffraction of Electrons by a Crystal of Nickel The Physical Review 1927 The investigation reported in this paper was begun as the result of an accident which occurred in this laboratory in April 1925.. D & G had been working on electron scattering (1921) from polycrystalline nickel: during the accident the target became oxidised. After removing the oxide by heating, the scattering was dramatically altered : The target was now more crystalline: electrons were diffracted by the crystal, just like x-rays!
Analysis of D & G Experiment just like x-rays θ θ d crystal planes (planes drawn in blue) 2dsinθ = nλ electron energy KE = 54 ev normal incidence θ : angle with planes enhanced reflectivity at 50 d : plane separation to find θ bisect 50 ( 25 ) (Bragg) θ = 90-25 = 65
Analysis of D & G Expt. cont. θ 2dsinθ = nλ λ = h/p θ θ = 65 crystal planes find p from KE KE = 54 ev (<< 0.51 MeV (= m o c 2 ) so non-relativistic) d KE = m o v 2 /2 = (m o v) 2 /2m o so p = m o v = (2m o KE) λ = λ = h m o v = h 2m o KE 6.63x10 34 J s ( 2) ( 9.1x10 31 kg) ( 54 ev )( 1.6x10 19 C) = 0.166nm (n=1) λ= 2dsin 65 = (2)(0.091 nm)(0.906) = 0.165 nm
Conclude: Particles can act as if they are also waves Diffraction becomes significant if λ (= h/p) is similar in size to aperture/spacing What if the particle (wave) is confined? e.g. In a box
Particle in a Box confinement of moving particle implies energy quantisation Particle in a box, making elastic collisions with rigid walls. Cannot go outside box Has a kinetic energy KE (& zero potential energy change in box) Total energy (non-relativistic) E = KE = 1 2 m ( ov 2 = m v o ) 2m o OK for particle, just bounces between walls. 2 = ( h λ ) 2 2m o = h 2 2m o λ 2 L What if it s a wave? Means standing waves in box nodes at walls c.f. Guitar strings, Waves in cup, etc
Particle in a Box (cont.) λ 1 = 2L/1 L λ 2 = 2L/2 λ 3 = 2L/3 λ n = 2L λ 4 = 2L/4 n Where n = 1, 2, 3,. E n = h 2 2m o λ n 2 = 2m o h 2 ( ) 2 = n 2 h 2 2L 8m n o L 2 E n : quantised energy level. n is the quantum number E 0 (since v=0 implies infinite λ) E 3 E 2 E 1 So lowest energy level is E 1 called the ground state
Conclude: Confining a wave restricts possible wavelengths only certain λ and hence Energies allowed! Wider implication: Confining a wave (particle) restricts possible states e.g. Diffraction possible directions affected by spacing d The atom (Lecture 6) and much else.
Waves of what? everything in the future is a wave, everything in the past is a particle Light: wave of E & M fields Matter: wave of existence But: a particle is at a point, whereas a wave is extended So where is the particle? And how does a point particle interact with more than one atom at the same time to give diffraction pattern? Consider waves, groups and packets
General fomula for Waves 1-D wave: Simple harmonic function of time t and distance x y = Acos2π ft = Acosωt At x = 0: (amplitude A, frequency f) At general x? - travelling wave speed v wave travels a distance x in time t = x/v Amplitude at any x, at any time t, is amplitude at x = 0 but at the earlier time t - x/v y = Acos2π f t x v ( ) = Acos2π ( ft fx v) = Acos2π ( ft x λ) y = Acos ωt kx ( ) ω = 2πf k = 2π λ Also written Velocity v = f λ Is v the same as the particle velocity v p? E=hf p=mv p =h/λ v = f λ = hf λ h = E p = mc2 = c2!? mv p v p Clearly not! Wave group Superposition of many waves
Wave Groups The wave y = A cos (ωt - kx) cannot reasonably be associated with a particle because of its (infinite) extent. Instead, consider a wave group Simplest example of wave group is beats : 2 waves of slightly different frequencies: y 1 = Acos( ωt kx) y 2 = A cos ( ω + Δω )t ( k + Δk)x [ ] y = y 1 + y 2 (using cos a + cos b = 2cos((a+b)/2).cos((a-b)/2) y = 2Acos 1 2[ ( 2ω + Δω)t ( 2k + Δk )x ]cos 1 2 [ Δωt Δkx ] For Δω and Δk small: y = 2Acos[ ωt kx]cos[ Δω 2 t Δk 2 x ] i.e. the basic wave modulated by beat frequency Δω/2 and wavenumber Δk/2 Δω Velocity of group (beat) v g = 2 = Δω Group of many waves: Δk Δk 2 v g = dω dk
Group velocity and particle velocity v g = dω dk = dω dv dk dv = v Are they the same??? Now ω = E = mc2 = m 0 c 2 dω So 1 v 2 c 2 dv = m 0 v ( 1 v 2 c 2 ) 3 2 And k = p = mv = m 0 v dk So 1 v 2 c 2 dv = m 0 1 v 2 c 2 ( ) 3 2 v g = dω dv dk dv = v Group velocity equals particle velocity
Group velocity and dispersion v g = dω dk = v Can equate particle velocity (v) with group velocity (v g ), not with phase velocity (v p )! v p = ω k Wave group built from many individual waves; each wave has a phase velocity (v p ). If v p is independent of ω or λ (as for light in a vacuum), then cannot represent a particle! v g = dω dk = ω k = v p Corollary: it is a further requirement of de Broglie wave group that the phase velocity varies with wavelength: This dispersion has implications (later)
Particle versus photon General relations apply to both, some specifics do not: E 2 = m o 2 c 4 + p 2 c 2 ( ω ) 2 = m 2 o c 4 + ( k) 2 c 2 photon particle m o = 0 m o 0 ( ω ) 2 = ( k) 2 c 2 ω k = f λ = c ω k c