Introduction

Question 1

What is quantum mechanics?

Answer 1

-the description of the behaviour of matter and light in all its details and, in particular, of the happenings of an atomic scale

Question 2

Waves or Particles?

Answer 2

• Newton thought that light was made up of particles
• later it was discovered that it behaved like a wave
• now we say that it is like neither (or both)

Question 3

Quantum Behaviour of Atomic Objects

Answer 3

• electrons behave like light
• the quantum behaviour of atomic objects (electrons, protons, neutrons, photons etc.) is the same for all
• they are all ‘particle waves’ or whatever you want to call them

Question 4

Quantum Mechanics Timeline

Answer 4

1901 - Planck: Blackbody radiation
1905 - Einstein: Photoelectric effect
1913 - Bohr: Atomic spectra
1922 - Compton: Scattering photons off electrons
1924 - Pauli: Exclusion principle
1925 - de Broglie: Matter waves
1927 - Heisenberg: Uncertainty principle
1927 - Born: Interpretation of the wave function

Question 5

Double Slit Experiment - With Bullets

Answer 5

• bullets fired in all directions
• pass through slits
• hit a detector screen
• the probability when one slit is closed gives a peak directly behind the slit
• when both are open the pattern is the sum of the one from each slit individually

Question 6

Double Slit Experiment - With Water Waves

Answer 6

-for each slit individually the same pattern for the bullets is produced
-but when both slits are open, an interference pattern is produced
I12 = I1 + I2 + 2√(I1I2) cos𝛿

Question 7

Double Slit Experiment - With Electrons

Answer 7

• for each slit individually the result is the same as for the bullets
• when both slits are open an interference pattern is produced
• HOWEVER
• if the electrons are observed as they pass through the slit then the same result as for the bullets with both slits open is observed

Question 8

Why is the curve smooth for bullets but not for electrons (unobserved)?

Answer 8

• they are not actually different

- but for macroscopic objects like bullets the interference pattern is so narrow that the curve appears smooth

Question 9

probability amplitude

definition

Answer 9

-the probability of an event in an ideal experiment is given by the absolute value of a complex number ϕ:
P = |ϕ|² = ϕ ϕ*
-where ϕ* is the complex conjugate

Question 10

Interference Present - Summary

Answer 10

-when an event can occur in several alternative ways, the probability amplitude for the event is the sum of the probability amplitudes for each way considered separately, there is interference:
P = | h1 + h2 + … |²

Question 11

Interference Lost - Summary

Answer 11

-if an experiment is performed which is capable of determining whether one or another alternative is actually taken, the probability of the event is the sum of the probabilities for each alternative:
P = |h1|² + |h2|² + …

Question 12

Double Slit - Actual Experiment

Answer 12

• two possible paths, same collection point
• one trajectory passes through a quantum dot (the dot serves as an electron weigh station , so knows if the electron has gone that way)
  1. when the detector sensitivity is low, an interference pattern is seen
  2. when the detector is sensitive enough to detect an electron passing, interference is lost

Question 13

Classical Wave Equations

Answer 13

v² * ∂²y(x,t)/∂x² = ∂²y(x,t)/∂t²
-this is solved by:
y(x,t) = A sin(kx - ωt)
ω/k = v

Question 14

Time Dependent Schrodinger Equation

Answer 14

-ħ²/2m * ∂²Ψ(x,t)/∂x² = iħ * ∂Ψ(x,t)/∂t

Question 15

Solution to the Time Dependent Schrodinger Equation

Answer 15

-starting with the time dependent form, solve using separation of variables to obtain:
Ψ = Ψ(x) * e^(-iEt/ħ)

Question 16

Time Independent Schrodinger Equation

Answer 16

-starting with the time dependent Schrodinger equation, sub in the solution:
Ψ = Ψ(x) * e^(-iEt/ħ)
-cancel out the exponential function on both sides to obtain the time independent form:
-ħ²/2m * ∂²Ψ/∂x² = E * Ψ(x)

Question 17

Schrodinger Equation Modified for an External Potential

Answer 17

(-ħ²/2m ∂²/∂x² + U(x)) * Ψ(x) = E * Ψ(x)

• where -ħ²/2m ∂²/∂x² + U(x) is the Hamiltonian
• and -ħ²/2m ∂²/∂x² is kinetic energy
• and U(x) is the potential energy

Question 18

Problems with the Schrodinger Equation

Answer 18

1) we cannot derive it from first principles as it itself is a postulate of quantum mechanics
2) it is a complex equation (contains i) but how can we measure a complex wave function?

Question 19

Solution to the Problems with the Wave Equation

Answer 19

-instead of measuring a complex wave function we measure the probability density:
|Ψ(x,t)|²

Question 20

Probability Density

Answer 20

-the probability of finding a particle between x and x+dx
-given by:
|Ψ(x,t)|²

Question 21

Normalising Probability Density

Answer 21

-as the particle must be somewhere in space, integrating the probability density over all space must give 1:
∫ |Ψ(x,t)|² dx = 1
-where the integral is taken from -∞ to +∞

Question 22

How to Normalise a Wave Function

Answer 22

-given an un-normalised wave function Ψ(x) :

Ψ(x) / √[ ∫ |Ψ(x)|² dx] = normalised wave function

Question 23

Example - The Plane Wave

Answer 23

Ψ(x,t) = A*e^(i(kx-ωt)
-sub into the Schrodinger Equation to obtain:
ħ²k²/2m = ħω
-which is correct as both sides = E, so a plane wave satisfies the Schrodinger Equation
-probability density:
|Ψ(x,t)|² = A²
-a particle that can be found any where in space with equal probability i.e. a wave

Question 24

Does an ordinary wave f(x) = Asin(kx-wt) solve the Schrodinger Equation?

Answer 24

NO

Question 25

Principle of Linear Superposition

Answer 25

• the Schrodinger equation is a linear differential equation therefore:
• if Ψ1(x,t) and Ψ2(x,t) are solutions of the Schrodinger equation, then c1*Ψ1(x,t) + c2Ψ2(x,t) is also a solution
• this can be shown simply by subbing into the schrodinger equation

Question 26

Schrodinger’s Cat

Description

Answer 26

• cat is sealed in a box with a flask of poison and a radioactive source
• if an internal monitor detects radioactivity (e.g. single atom decay) the flask is shattered releasing the poison

Question 27

Schrodinger’s Cat

Copenhagen Interpretation

Answer 27

• after a while the cat is simultaneously dead AND alive
• when the box is opened and the cat is observed the wave function collapses into one state or the other i.e. the cat is dead OR alive

Question 28

Schrodinger’s Cat

Many Worlds Interpretation

Answer 28

• every event is a branch point
• the cat is both alive and dead regardless of observation
• BUT the ‘alive’ and ‘dead’ cats are in different branches of the universe that are equally real but cannot interact with each other

Question 29

Copenhagen Interpretation

Answer 29

-a system stops being a superposition of states and becomes either one or the other when an observation takes place

Question 30

What is uncertainty?

Answer 30

• measuring a quantity n times we get may get different measurements
• we can calculate a mean and a standard deviation ΔQ
• if ΔQ=0 we always measure the same values of Q
• if ΔQ≠0 results fluctuate around some value and we cannot be certain of a measurement

Question 31

Uncertainty

Double Slit

Answer 31

• in the double slit experiment we can measure two things:
  1) momentum of electrons, from trace on screen
  2) position of electrons exiting the slit
• if the width of the slit is large then Δpx is small as the electrons are less spread out on the screen
• if the width of the slit is small then Δpx is larger as the electrons are diffracted more and spread out on the screen

Question 32

Heisenberg’s Uncertainty Principle

Answer 32

Δx Δpx ≥ ħ/2

-where px is momentum in the x direction

Question 33

What does Heisenberg’s uncertainty principle tell us?

Answer 33

• because of a particles wave nature, it is theoretically impossible to measure both the particles position and its momentum along the same axis
• measurement always perturbs the system, usually this has non observable consequences for macroscopic object but, at an atomic level this is very important

Question 34

Heisenberg Uncertainty Principle and Gaussian Wave Packets

Answer 34

-for most states the uncertainty principle is strictly greater than
-but there are special states for which the equals sign hold
-one such instance is the Gaussian wave packet:
Ψ(x) = C*e^(-(x/2e)²)

Question 35

Heisenberg Uncertainty Principle and Zero Point Motion

Answer 35

-from the uncertainty principle we can show that if a particle is enclosed within a box, it can never be at rest:
-consider a box of width L
-since the particle is definitely in the box:
Δx ≤ L
-rearrange the uncertainty principle:
Δx Δp ≥ ħ/2
Δp ≥ ħ/2Δx
-sub in Δx ≤ L
Δp ≥ ħ/2Δx ≥ ħ/2L
-calculate average (to account for fluctuations) energy:
= <p>/2m ≥ (ħ/2L)²/2m > 0
-the particle constantly ‘wiggles’ with some kinetic energy, even at absolute zero, this is referred to as zero point motion</p>

Question 36

Heisenberg Uncertainty Principle in Three Dimensions

Answer 36

Δx Δpx ≥ ħ/2
Δy Δpy ≥ ħ/2
Δz Δpz ≥ ħ/2

Question 37

Heisenberg Uncertainty Principle - Energy and Time

Answer 37

ΔE Δt ≥ ħ/2

Question 38

What is a Fourier Transform used for?

Answer 38

• to analyse more complicated waves we can decompose them into harmonics
• applying the Fourier transform to a function gives the amplitude of different wave vectors each corresponding to a harmonic that makes up the function

Question 39

Fourier Transform Equation

Answer 39

A(k) = ∫ dx f(x) e^(-ikx)

• where the integral is taken from -∞ to +∞
• and f(x) is the function of the wave of interest

Question 40

Fourier Transform Inverse Equation

Answer 40

f(x) = ∫ dk/2π A(k) e^(ikx)

-where the integral is taken from -∞ to +∞

Question 41

What do particles look like as waves?

Answer 41

• a particle can be visualised as a packet of waves created by adding together a continuum of harmonic waves
• this creates a perfectly localised wave packet or ‘particle’