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

Particle versus photon General relations apply to both, some specifics do not:

photon particle mo = 0 mo ≠ 0

�

E 2 = mo2c 4 + p2c 2

ω( )2 = mo2c 4 + k( )2c 2

ω( )2 = k( )2 c2ω k = fλ = c

ω k ≠ c