Matter Waves

Question 1

de Broglie Hypothesis

Answer 1

-suggested that particles could behave like waves
-this argument is based on symmetry, if light can have both eave and particle nature why cant other particles have wave nature
ρ = h/λ

Question 2

Davisson-Germer Experiment

Equipment

Answer 2

• electrons accelerated through a p.d.
• high speed electrons aimed at a nickel crystal target in a vacuum
• the nickel could be rotated to observe the angular dependence of the electron scattering
• the electron detector was mounted on an arc so that the electron beam could be observed from different angles

Question 3

Davisson-Germer Experiment

Observations

Answer 3

• it was found that at certain angles there was a peak in the scattered electron intensity
• this peak indicated that that the electrons were behaving as waves and being diffracted by the nickel lattice
• this was the first application of wave nature to a particle
• using de Broglie’s Law they were able to estimate the lattice spacing of the nickel crystal

Question 4

The Bragg Condition

Description

Answer 4

• this describes the condition required for x-rays to scatter with constructive interference from the atomic planes in a crystal
• if a wave, II, travels further than a wave, I, then if the path difference is an integer number of wavelengths then the two waves will be in phase an will add constructively/coherently
• the same is true for a particle where, λ = de Broglie wavelength

Question 5

The Bragg Condition

Equation

Answer 5

2dsinθ = nλ

Question 6

G. P. Thomson’s Experiment

Answer 6

• fired x rays and electrons at a target
• behind the target there was a photographic plate
• both xrays and electrons produced the same 2D diffraction pattern
• a series of rings inside each other

Question 7

The Principle of Complementarity

Answer 7

in a measurement, the behaviour will either be a particle or a wave, but not both simultaneously

Question 8

Calculating the Velocity of a Free Electron by treating it as a Wave

Answer 8

Vp = fλ = ω/k
Vp = velocity

E = ℏω = ℏ²k²/2m

ω(k) = ℏk²/2m
this is the dispersion relation for free electrons

Question 9

Dispersion Relation of Free Electrons

Equation

Answer 9

ω(k) = ℏk²/2m

Question 10

Dispersion Relation

Definition

Answer 10

-dispersion relations are functions relating ω and k or f and λ

Question 11

Where does the name dispersion relation come from ?

Answer 11

• when white light enters a prism, the velocity of the light in the prism depends on its wavelength
• the spectrum is revealed because the refraction of the light is dependent on its wavelength, i.e. some wavelengths refract more than others

Question 12

Why do we need wave packets?

Answer 12

Vp = ω/k = ℏk/2m
BUT, ρ = ℏk
SO, Vp = ρ/2m
Vp = mv/2m
Vp = v/22
-this is a problem as it means that the wave will be lagging behind the particle but the wave and particle are supposed to describe the same object
-this isn’t the only problem, a pure sinusoidal wave is also infinite in extent which doesn’t correspond with the idea that it represents a small particle
-hence we need an alternative way do describe particles as waves, we use wavepackets

Question 13

Why are wavepackets a better representation of a particle than a pure sinusoidal wave?

Answer 13

• they peak in a definite place

- does not extend far beyond the peak

Question 14

What is a wave packet?

Answer 14

• constructed from many waves

- it is peaked and rapidly decays to zero on either side of this peak

Question 15

Wavepacket - Wavelength

Answer 15

-the wavelength f a wavepacket is defined as an average and may not be constant over time

Question 16

Wavepacket - Velocity

Answer 16

• there are two velocities associated with a wavepacket
• one, Vp, is the product of the frequency and wavelength
• the other is the velocity with which the wave packet itself moves

Question 17

Electron Wave Function

Answer 17

Ψ = Asin(kx-ωt)

Question 18

Finding Phase and Group Velocity

superposition of two waves with slightly different wave vectors and angular frequencies

Answer 18

Ψ(x,t)=
Asin(k1x-ω1t)+Asin(k2x-ω2t)

[as k1~k2 and ω1~ω2]

=2Acos((k1-k2)x/2-(ω1-ω2)t/2)sin((k1+k2)x/2-(ω1+ω2)t/2)

[if k=k1+k2/2 and Δk=k1-k2/2]

=2Acos(Δkx-Δωt)sin(kx-ωt)
Vp = ω/k
Vg = Δω/Δk

Question 19

Group Velocity of a Free Electron Wavepacket

Answer 19

the dispersion relation for free electrons is:
ω(k) = ℏk²/2m
Vg = Δω/Δk = dω/dk
Vg = d(ℏk²/2m)/dk
Vg = 2ℏk/2m
Vg = ℏk/m, ρ = ℏk
so, Vg = ρ/m = v
-the velocity is the same for the wavepacket as it is when we model the electron as a particle

Question 20

What is the relationship between phase and group velocities for a free electron?

Answer 20

Vg = 2Vp

Question 21

In general, what is the relationship between phase and group velocities?

Answer 21

Vp = ω/k, so ω = kVp

Vg = dω/dk
Vg = d(kVp)/dk
Vg = Vp + k(dVp/dk)

Question 22

What is the relationship between phase and group velocities for photons?

Answer 22

Vp = Vg = c

Question 23

The Born Rule

Answer 23

-the wave function is given by Ψ = Asin(kx-ωt)
-the born rule states that
|Ψ²| = the probability of finding the particle at a certain position
-this probability is not a single number but a range of numbers spread over a distance Δx

Question 24

Heisenberg’s Uncertainty Principle

Answer 24

ΔxΔρ ~ ℏ/2

-we cannot know the position (energy) and momentum (time) of quantum particle simultaneously

Question 25

Particles in a Box

Comparision With Light in a Box

Answer 25

• when we considered light in a box it had quantised energies and boundary conditions at the walls of the box
• if particle behaviour can be modelled as wave behaviour then the same should be true for particles in a box

Question 26

Particles in a Box

Allowed Wavelengths

Answer 26

-for a box of side length L, the particle wavelength is restricted by:
L = Nλ/2, where N
is an integer

Question 27

Particles in a Box

Allowed Energies

Answer 27

-assuming the particle is moving slowly so that special relativity is not needed:
E = ρ²/2m and ρn=h/λn
so, En = ρn²/2m
En = h²/2m*λn²
BUT λn=2L/n
En = n²(h²/8mL²)
En = n²E1
where E1=h²/8mL² is the lowest possible energy, the ground state energy in the box

Question 28

Normalisation Condition

Answer 28

(-∞,∞) ∫ Ψ² dx

• if we a particle at all then the probability of finding it somewhere must be 1
• so the sum of the probabilities over all the possible values of x must equal 1

Question 29

Probability Density

Answer 29

P(x) = Ψ²(x)

Question 30

Ground State Energy for a Particle in a Box

Answer 30

E1=h²/8mL²

Question 31

Standing-Wave Functions for a Particle in a Box

Answer 31

Ψn(x) = √(2/L) * sin(nπ* x/L)