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:
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~mo at non-relativistic speed. Suggested in De B. PhD thesis & implied by Bohr model of atom
Direct Experimental evidence??
N.B. Significance is for SMALL objects…
or λ =hp
�
λ = hp
= hmv
p =ωc
= k =hλ
h = 2π = 6.625 ×10−34 Js
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…
(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˚
crystal planes
θ θ d
Analysis of D & G Expt. cont.
2dsinθ = nλ λ = h/p θ = 65˚ find p from KE KE = 54 eV (<< 0.51 MeV (= moc2) so non-relativistic)
KE = mov2/2 = (mov)2/2mo so p = mov = √(2moKE)
(n=1) λ= 2dsin 65˚ = (2)(0.091 nm)(0.906) = 0.165 nm
crystal planes
θ θ
�
λ = hmov
= h2moKE
λ = 6.63x10−34 J s2( ) 9.1x10−31 kg( ) 54eV( ) 1.6x10−19C( )
= 0.166nm
d
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)
OK for particle, just bounces between walls.
What if it’s a wave? Means standing waves in box – nodes at walls
c.f. Guitar strings, Waves in cup, etc…
L
�
E = KE = 12mov2 =
mov( )2mo
2
=hλ( )
2mo
2
= h2
2moλ2
Particle in a Box (cont.) λ1 = 2L/1 λ2 = 2L/2 λ3 = 2L/3 λ4 = 2L/4 Where n = 1, 2, 3, ….
En: “quantised” energy level. n is the “quantum number”
E ≠ 0 (since v=0 implies infinite λ)
So lowest energy level is E1 called “the ground state”
L
En =h2
2moλn2 =
h2
2mo2L
n( )2=
n2h2
8moL2
λn = 2L n
E1
E3
E2
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 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
Also written Velocity v = f λ
Is v the same as the ‘particle velocity’ vp? E=hf p=mvp=h/λ
!?
Clearly not! ⇒ Wave ‘group’ Superposition of many waves
y = Acos2π ft = Acosωt
y = Acos ωt − kx( )
�
ω = 2πf k = 2π λ
y = Acos2π f t − x v( ) = Acos2π ft − fx v( ) = Acos2π ft − x λ( )
v = fλ = hfλh=Ep=mc2
mvp=c2
vp
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 = y1 + y2 (using cos a + cos b = 2cos((a+b)/2).cos((a-b)/2)
For Δω and Δk small:
i.e. the basic wave ‘modulated’ by ‘beat’ frequency Δω/2 and wavenumber Δk/2
Velocity of group (beat) Group of many waves:
y1 = Acos ωt − kx( )y2 = A cos ω + Δω( )t − k + Δk( )x[ ]
y = 2Acos 12 2ω + Δω( )t − 2k + Δk( )x[ ]cos 12 Δωt − Δkx[ ]
y = 2Acos ωt − kx[ ]cos Δω 2t − Δk2 x[ ]
�
vg =Δω
2Δk2
= ΔωΔk
vg =dωdk
Group velocity and particle velocity
v g =dωdk
=dω
dvdk
dv
= vAre they the same? ??
So…
ω =E=mc2
=
m0c2
1− v2 c2
dωdv
=m0v
1− v2 c2( )32
k =p=mv
=m0v
1− v2 c2
dkdv
=m0
1− v2 c2( )32
Now
And So…
⇒ v g =dω
dvdk
dv
= v
Group velocity equals particle velocity
Group velocity and “dispersion”
Can equate particle velocity (v) with group velocity (vg), not with phase velocity (vp)!
Wave group built from many individual waves; each wave has a
phase velocity (vp). If vp is independent of ω or λ (as for light in a vacuum),
then cannot represent a particle!
Corollary: it is a further requirement of de Broglie wave group that the phase velocity varies with wavelength: This “dispersion” has implications (later)
�
vp = ωk
⎛ ⎝ ⎜
⎞ ⎠ ⎟
�
vg = dωdk
= v
�
vg = dωdk
= ωk
= vp
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