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Astronomy

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Astronomical Distance Units .
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Celestial Coordinates .
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Celestial Navigation .
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Location of North and South Celestial Poles .

Chemistry

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Avogadro's Number
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Balancing Chemical Equations
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The Periodic Table .

Classical Physics

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Archimedes Principle
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Blackbody (Cavity) Radiation and Planck's Hypothesis
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Center of Mass Frame
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Comparison Between Gravitation and Electrostatics
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Compton Effect .
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Doppler Effect
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Double Slit Experiment
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Error Analysis
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Planck Radiation Law .
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Radioactive Decay
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Refractive Index
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Rotational Dynamics
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Specific Heat, Latent Heat and Calorimetry
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The Gas Laws
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Climate Change

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Keeling Curve .

Cosmology

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Baryogenesis
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Cosmic Background Radiation and Decoupling
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Geometries of the Universe
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Hubble's Law
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Inflation Theory
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Introduction to Black Holes .
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Stephen Hawking's Last Paper .
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Stephen Hawking's PhD Thesis .
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The Big Bang Model

Finance and Accounting

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Amortization
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Lecture Notes on International Financial Management
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Game Theory

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The Truel .

General Relativity

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Accelerated Reference Frames - Rindler Coordinates
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Catalog of Spacetimes .
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Tensors
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The Equivalence Principal
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The Metric Tensor
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World Lines Refresher

Lagrangian and Hamiltonian Mechanics

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Classical Field Theory .
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Euler-Lagrange Equation
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Ex: Newtonian, Lagrangian and Hamiltonian Mechanics
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Hamiltonian Formulation .
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Liouville's Theorem
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Symmetry and Conservation Laws - Noether's Theorem

Macroeconomics

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Lecture Notes on International Economics
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Macroeconomic Policy

Mathematics

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Amplitude, Period and Phase
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Basic Group Theory
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Basic Representation Theory
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Building Groups From Other Groups
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Completing the Square
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Complex Numbers
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Composite Functions
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Conformal Transformations .
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Conjugate Pair Theorem
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Contravariant and Covariant Components of a Vector
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Eigenvectors and Eigenvalues
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Euler Formula for Polyhedrons
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Factoring of a3 +/- b3
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Fourier Series and Transforms .
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Fractals
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Gauss's Divergence Theorem
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Grassmann and Clifford Algebras .
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Heron's Formula
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Index Notation (Tensors and Matrices)
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Inequalities
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Integration By Parts
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Introduction to Conformal Field Theory .
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Inverse of a Function
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Law of Sines and Cosines
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Line Integrals, ∮
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Logarithms and Logarithmic Equations
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Matrices and Determinants
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Matrix Exponential
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Mean Value and Rolle's Theorem
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Modulus Equations
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Orthogonal Curvilinear Coordinates .
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Parabolas, Ellipses and Hyperbolas
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Quaternions 1
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Quaternions 2
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Regular Polygons
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Related Rates
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Sets, Groups, Modules, Rings and Vector Spaces
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Similar Matrices and Diagonalization .
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Spherical Trigonometry
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Stirling's Approximation
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Sum and Differences of Squares and Cubes
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Symbolic Logic
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Symmetric Groups
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Tangent and Normal Line
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Taylor and Maclaurin Series .
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The Essential Mathematics of Lie Groups
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The Integers Modulo n Under + and x
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The Limit Definition of the Exponential Function
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Tic-Tac-Toe Factoring
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Trapezoidal Rule
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Unit Vectors
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Vector Calculus
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Volume Integrals

Microeconomics

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Marginal Revenue and Cost

Particle Physics

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Feynman Diagrams and Loops
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Field Dimensions
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Helicity and Chirality
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Klein-Gordon and Dirac Equations
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Regularization and Renormalization
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Scattering - Mandelstam Variables
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Spin 1 Eigenvectors .
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The Vacuum Catastrophe

Probability and Statistics

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Box and Whisker Plots
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Categorical Data - Crosstabs
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Chebyshev's Theorem
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Chi Squared Goodness of Fit
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Conditional Probability
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Confidence Intervals
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Data Types
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Expected Value
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Factor Analysis
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Hypothesis Testing
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Linear Regression
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Monte Carlo Methods
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Non Parametric Tests
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One-Way ANOVA
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Pearson Correlation
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Permutations and Combinations
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Pooled Variance and Standard Error
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Probability Distributions
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Probability Rules
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Sample Size Determination
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Sampling Distributions
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Set Theory - Venn Diagrams
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Stacked and Unstacked Data
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Stem Plots, Histograms and Ogives
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Survey Data - Likert Item and Scale
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Tukey's Test
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Two-Way ANOVA

Programming and Computer Science

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Hashing
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How this site works ...
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More Programming Topics
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MVC Architecture
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Open Systems Interconnection (OSI) Standard - TCP/IP Protocol
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Public Key Encryption

Quantum Computing

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The Qubit .

Quantum Field Theory

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Creation and Annihilation Operators
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Field Operators for Bosons and Fermions
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Lagrangians in Quantum Field Theory
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Path Integral Formulation
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Relativistic Quantum Field Theory

Quantum Mechanics

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Basic Relationships
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Bell's Theorem
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Bohr Atom
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Clebsch-Gordan Coefficients .
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Commutators
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Dyson Series
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Electron Orbital Angular Momentum and Spin
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Entangled States
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Heisenberg Uncertainty Principle
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Ladder Operators .
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Multi Electron Wavefunctions
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Pauli Exclusion Principle
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Pauli Spin Matrices
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Photoelectric Effect
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Position and Momentum States
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Probability Current
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Schrodinger Equation for Hydrogen Atom
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Schrodinger Wave Equation
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Schrodinger Wave Equation (continued)
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Spin 1/2 Eigenvectors
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The Differential Operator
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The Essential Mathematics of Quantum Mechanics
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The Observer Effect
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The Quantum Harmonic Oscillator .
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The Schrodinger, Heisenberg and Dirac Pictures
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The WKB Approximation
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Time Dependent Perturbation Theory
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Time Evolution and Symmetry Operations
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Time Independent Perturbation Theory
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Wavepackets

Semiconductor Reliability

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The Weibull Distribution

Solid State Electronics

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Band Theory of Solids .
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Fermi-Dirac Statistics .
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Intrinsic and Extrinsic Semiconductors
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The MOSFET
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The P-N Junction

Special Relativity

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4-vectors .
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Electromagnetic 4 - Potential
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Energy and Momentum, E = mc2
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Lorentz Invariance
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Lorentz Transform
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Lorentz Transformation of the EM Field
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Newton versus Einstein
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Spinors - Part 1 .
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Spinors - Part 2 .
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The Lorentz Group
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Velocity Addition

Statistical Mechanics

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Black Body Radiation
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Entropy and the Partition Function
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The Harmonic Oscillator
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The Ideal Gas

String Theory

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Bosonic Strings
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Extra Dimensions
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Introduction to String Theory
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Kaluza-Klein Compactification of Closed Strings
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Strings in Curved Spacetime
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Toroidal Compactification

Superconductivity

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BCS Theory
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Introduction to Superconductors
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Superconductivity (Lectures 1 - 10)
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Superconductivity (Lectures 11 - 20)

Supersymmetry (SUSY) and Grand Unified Theory (GUT)

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Chiral Superfields
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Generators of a Supergroup
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Grassmann Numbers
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Introduction to Supersymmetry
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The Gauge Hierarchy Problem

The Standard Model

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Electroweak Unification (Glashow-Weinberg-Salam)
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Gauge Theories (Yang-Mills)
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Gravitational Force and the Planck Scale
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Introduction to the Standard Model
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Isospin, Hypercharge, Weak Isospin and Weak Hypercharge
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Quantum Flavordynamics and Quantum Chromodynamics
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Special Unitary Groups and the Standard Model - Part 1 .
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Special Unitary Groups and the Standard Model - Part 2
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Special Unitary Groups and the Standard Model - Part 3 .
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Standard Model Lagrangian
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The Higgs Mechanism
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The Nature of the Weak Interaction

Topology

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Units, Constants and Useful Formulas

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Constants
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Formulas
Last modified: October 13, 2020

Special Unitary Groups and the Standard Model - Part 1 Special Unitary Groups and the Standard Model - Part 3 Special Unitary Groups and the Standard Model - Part 2 ------------------------------------------------------ The Adjoint Representation -------------------------- In Special Unitary Groups and the Standard Model - Part 1 we talked about the fundamental (defining) representation in which the matrices produced by the generators are the elements of the group. For SU(3) this is the set of complex 3 x 3 unitary matrices with determinant = 1. These matrices act on the 3-component column vectors with complex components that represent the quark fields that form a basis. However, there is another representation of the group called the ADJOINT representation. Adjoint Representation from Structure Constants ----------------------------------------------- For the fundamental representation we had the following definition of the Lie algebra: [Ta,Tb] = ifabcTc The matrix elements of the adjoint representation are given by the structure constants obtained from the fundamental representation. (Ta(A))bc := -ifabc Where (Ta(A))bc is a particular element of the matrix Ta(A). The dimension of (Ta(A))bc is given by: (N2 - 1) x (N2 - 1) And there would be (N2 - 1) of them. SU(2) ----- For SU(2) we write [Ta(A),Tb(A)] = εabcTc(A) and, (Ta(A))bc = εabc The generators are constructed from: (T1(A))23 = ε123 = 1 (T1(A))32 = ε132 = -1 (T1(A))11 = ε111 = 0 etc. etc. This eventually gives: - - - - - -   | 0 0 0 |  | 0 0 -1 |  | 0 -1 0 | T1(A) = | 0 0 1 | T2(A) = | 0 0 0 | T3(A) = | 1 0 0 |   | 0 -1 0 |  | 1 0 0 |  | 0 0 0 | - - - - - - Note: These are also the generators of SO(3) that represents rotations around the x, y and z axes. SU(3) ----- Consider T3(A). The structure constants are: f312 = 1 f321 = -1 f345 = 1/2 f354 = -1/2 f367 = -1/2 f376 = 1/2 Therefore, T3(A) looks like: - -   | 0 -i 0 0 0 0 0 0 |   | i 0 0 0 0 0 0 0 |   | 0 0 0 0 0 0 0 0 | T3(A) = | 0 0 0 0 -i/2 0 0 0 |   | 0 0 0 i/2 0 0 0 0 |   | 0 0 0 0 0 0 i/2 0 |   | 0 0 0 0 0 -i/2 0 0 |   | 0 0 0 0 0 0 0 0 | - - For T8(A). The structure constants are: f845 = √3/2 f854 = -√3/2 f678 = √3/2 f687 = -√3/2 f867 = √3/2 f876 = -√3/2 Therefore, T8(A) looks like - -   | 0 0 0 0 0 0 0 0 |   | 0 0 0 0 0 0 0 0 |   | 0 0 0 0 0 0 0 0 | T2(A) = | 0 0 0 0 -i√3/2 0 0 0 |   | 0 0 0 i√3/2 0 0 0 0 |   | 0 0 0 0 0 0 -i√3/2 0 |   | 0 0 0 0 0 i√3/2 0 0 |   | 0 0 0 0 0 0 0 0 | - - We will now use the JACOBI IDENTITY to show that the Ta(A)'s also form a representation of the Lie algebra. For a compact Lie algebra, the Lie bracket [A,B] satisfies: [A,[B,C]] + [B,[C,A]] + [C,[A,B]] = 0 This is the JACOBI IDENTITY. Proof: [A,(BC - CB)] + [B,(CA - AC)] + [C,(AB - BA)] = ABC - ACB - (BCA - CBA) + BCA - BAC - (CAB - ACB) + CAB - CBA - (ABC - BAC) = ABC - ACB - BCA + CBA + BCA - BAC - CAB + ACB + CAB - CBA - ABC + BAC = 0 In terms of the generators, the first term, [A,[B,C]], becomes: [Ta(A),[Tb(A),Tc(A)]] Now [Tb(A),Tc(A)] = -ifbcdTd(A). Therefore, [Ta(A),[Tb(A),Tc(A)]] = [Ta(A),-ifbcdTd(A)] The RHS can be rewritten as: = -ifbcd[Ta(A),Td(A)] = -ifbcd(ifadeTe(A)) = fbcdfadeTe(A) The second and third terms are cyclic permutations as follows: [Tb(A),[Tc(A),Ta(A)]] = fcadfbdeTe(A) And, [Tc(A),[Ta(A),Tb(A)]] = fabdfcdeTe(A) Adding together we get: (fbcdfade + fcadfbde + fabdfcde)Te(A) = 0 Therefore (since Te(A) ≠ 0): fbcdfade + fcadfbde + fabdfcde = 0 We now make use of the following: (Tx)yz = -ifxyz ∴ fxyz = i(Tx)yz fabc = fbca = fcab = -fbac = -facb = -fcba Therefore, -fabdfcde = fbcdfade + fcadfbde = fbcdfade - facdfbde = i(Tb(A))cdi(Ta(A))de - i(Ta(A))cdi(Tb(A))de = -(Tb(A))cd(Ta(A))de + (Ta(A))cd(Tb(A))de = -(Tb(A)Ta(A))ce + (Ta(A)Tb(A))ce = -[Tb(A),Ta(A)]ce -fabdfcde = [Ta(A),Tb(A)]ce -ifabd(Tc(A))de = [Ta(A),Tb(A)]ce ifabd(Td(A))ce = [Ta(A),Tb(A)]ce This is the familiar commutator associated with the Lie algebra. In the adjoint representation, the T(A)'s act on a basis that is comprised of the generators themselves and not the u, d and s column vectors associated with the defining representation. From the discussion of Lie groups and algebras, the adjoint action of a Lie algebra on itself is given by: adXy ≡ ad(x)y = [x,y] Where x and y are group generators. Therefore, ifabd(Td(A))ce = [Ta(A),Tb(A)]ce = ad(Ta(A))Tb(A) Which can be written in simpler notation as: [ea,eb] = ifabdTd = ad(ea)eb SU(2) ----- The bases are constructed from the 3 Pauli matrices: - - e1 = iσ1/2 = | 0 i/2 |    | i/2 0 | - - - - e2 = iσ2/2 = | 0 1/2 |    | -1/2 0 | - - - - e3 = iσ3/2 = | i/2 0 |    | 0 -i/2 | - - Therefore, ad(e1)e1 = 0 ad(e1)e2 = [e1,e2] = ε123e3 = -e3 ad(e1)e3 = [e1,e3] = ε132e2 = e2 The Lie algebra lives in a tangent (vector) space around the identity element of the group. The vector space is equipped with the operations of matrix multiplication together with addition and scalar multiplication. Therefore, ad(e1) is expressed as the linear combination: - - - -     | 0 || e1 | 0e1 - e2 + e3 = | -1 || e2 |     | 1 || e1 | - - - - In matrix form this looks like: - -   | 0 0 0 | e1 ad(e1) = | 0 0 1 | e2   | 0 -1 0 | e3 - -   e1 e2 e3 Likewise, ad(e2)e1 = [e2,e1] = ε213e3 = e3 ad(e2)e2 = [e2,e2] = 0 ad(e2)e3 = [e2,e3] = ε231e1 = -e1 - - - -     | -1 || e1 | -e1 - 0e2 + e3 = | 0 || e2 |     | 1 || e1 | - - - - - -   | 0 0 -1 | e1 ad(e2) = | 0 0 0 | e2   | 1 0 0 | e3 - -   e1 e2 e3 Finally, ad(e3)e1 = [e3,e1] = ε312e2 = e2 ad(e3)e2 = [e3,e2] = ε321e1 = -e1 ad(e3)e3 = [e3,e3] = 0 - - - -     | -1 || e1 | -e1 + e2 + 0e3 = | 1 || e2 |     | 0 || e1 | - - - - - -   | 0 -1 0 | e1 ad(e3) = | 1 0 0 | e2   | 0 0 0 | e3 - -   e1 e2 e3 SU(3) ----- The bases can be constructed from ei = iλei. If we go through the same procedure as before we get: - -   | 0 -i 0 0 0 0 0 0 | e1   | i 0 0 0 0 0 0 0 | e2   | 0 0 0 0 0 0 0 0 | e3 ad(e3) = | 0 0 0 0 -i/2 0 0 0 | ee   | 0 0 0 i/2 0 0 0 0 | e5   | 0 0 0 0 0 0 i/2 0 | e6   | 0 0 0 0 0 -i/2 0 0 | e7   | 0 0 0 0 0 0 0 0 | e8 - -   e1 e2 e3 e4 e5 e6 e7 e8 And, - -   | 0 0 0 0 0 0 0 0 | e1   | 0 0 0 0 0 0 0 0 | e2   | 0 0 0 0 0 0 0 0 | e3 ad(e8) = | 0 0 0 0 -i√3/2 0 0 0 | e4   | 0 0 0 i√3/2 0 0 0 0 | e5   | 0 0 0 0 0 0 -i√3/2 0 | e6   | 0 0 0 0 0 i√3/2 0 0 | e7   | 0 0 0 0 0 0 0 0 | e8 - -   e1 e2 e3 e4 e5 e6 e7 e8 Which agrees exactly with the result using the structure constants determined from the fundamental representation. The Adjoint Representation by Conjugation, Ad(e1) ------------------------------------------------- Consider SU(2): - - | 1 0 0 | exp(ad(e1)) = | 0 cosθ sinθ | | 0 -sinθ cosθ | - - - - - - exp(| 0 i/2 |) = | cos(θ/2) isin(θ/2) | | i/2 0 | | isin(θ/2) cos(θ/2) | - - - - - - [exp(e1)]-1 = | cos(θ/2) -isin(θ/2) |     | -isin(θ/2) cos(θ/2) | - - exp(e1)e1[exp(e1)]-1 gives: - - - - - - | cos(θ/2) isin(θ/2) || 0 i/2 || cos(θ/2) -isin(θ/2) | | isin(θ/2) cos(θ/2) || i/2 0 || -isin(θ/2) cos(θ/2) | - - - - - - - - = | 0 i/2 | = e1 + 0e2 + 0e3 | i/2 0 | - - - - - - | 1 ? ? || e1 | | 0 ? ? || e2 | | 0 ? ? || e3 | - - - - exp(e1)e2[exp(e1)]-1 gives: - - - - - - | cos(θ/2) isin(θ/2) || 0 1/2 || cos(θ/2) -isin(θ/2) | | isin(θ/2) cos(θ/2) || -1/2 0 || -isin(θ/2) cos(θ/2) | - - - - - - - - | -isin(θ/2)cos(θ/2) (-sin2(θ/2) + cos2(θ/2))/2 | | (-cos2(θ/2) + sin2(θ/2))/2 isin(θ/2)cos(θ/2) | - - - - | (-isinθ)/2 (cosθ)/2 | | (-cosθ)/2 (isinθ)/2 | - - We can break this down as follows: - - - - | 0 (cosθ)/2 | + | -(isinθ)/2 0 | | (-cosθ)/2 0 | | 0 (isinθ)/2 | - - - - - - - - | 0 1/2 |cosθ + | -i/2 0 |sinθ | -1/2 0 | | 0 i/2 | - - - - 0e1 + e2cosθ - e3sinθ - - - - | 1 0 ? || e1 | | 0 cosθ ? || e2 | | 0 -sinθ ? || e3 | - - - - exp(e1)e3[exp(e1)]-1 gives: - - - - - - | cos(θ/2) isin(θ/2) || i/2 0 || cos(θ/2) -isin(θ/2) | | isin(θ/2) cos(θ/2) || 0 -i/2 || -isin(θ/2) cos(θ/2) | - - - - - - - - | i(cos2(θ/2) - sin2(θ/2))/2 sin(θ/2)cos(θ/2) | | -sin(θ/2)cos(θ/2) i(sin2(θ/2) - cos2(θ/2))/2 | - - - - | (icosθ)/2 (sinθ)/2 | | (-sinθ)/2 (-icosθ)/2 | - - We can break this down as follows: - - - - | 0 (sinθ)/2 | + | (icosθ)/2 0 | | (-sinθ)/2 0 | | 0 (-icosθ)/2 | - - - - - - - - | 0 1/2 |sinθ + | i/2 0 |cosθ | -1/2 0 | | 0 -i/2 | - - - - = 0e1 + e2sinθ + e3cosθ The full matrix becomes: - - | 1 0 0 | | 0 cosθ sinθ | = Ad(e1) | 0 -sinθ cosθ | - - Which is the same as exp(ad(e1)). The derivative of Ad(e1) is: ad(e1) = dAd(e1)/dθ|θ=0 - - | 0 0 0 | = | 0 -sinθ cosθ | | 0 -cosθ -sinθ | - - - - | 0 0 0 | = | 0 0 1 | at θ = 0 | 0 -1 0 | - - = ad(e1) The Cartan-Killing Metric (aka the Killing Form) ------------------------------------------------ The metric tensor takes as its input a pair of tangent vectors at a point on a manifold and produces a real number scalar in a way that generalizes the dot product of vectors in Euclidean space. This is discussed in detail in the note, 'The Essential Mathematics of General Relativity'. In the same way we can associate a metric with the inner product on a finite dimensional Lie algebra. In this case the metric is defined by: gab = B(a,b) = Tr(Ta(A)Tb(A)) = -facdfbdc The Killing form is the scalar product of the algebra, defined in terms of the adjoint representation. Example SU(2): - - - -    | 0 0 0 || 0 0 0 | Tr(ad(e1)ad(e1)) = Tr(| 0 0 1 || 0 0 1 |)    | 0 -1 0 || 0 -1 0 | - - - - - - | 0 0 0 | = Tr(| 0 -1 0 |) | 0 0 -1 | - - = -2 g(1,1) = Tr(ad(e1)ad(e1)) = -2 g(1,2) = Tr(ad(e1)ad(e2)) = 0 g(1,3) = Tr(ad(e1)ad(e3)) = 0 g(2,1) = Tr(ad(e2)ad(e1)) = 0 g(2,2) = Tr(ad(e2)ad(e2)) = -2 g(2,3) = Tr(ad(e2)ad(e3)) = 0 g(3,1) = Tr(ad(e3)ad(e1)) = 0 g(3,2) = Tr(ad(e3)ad(e2)) = 0 g(3,3) = Tr(ad(e3)ad(e3)) = -2 - -    | -2 0 0 | gab = | 0 -2 0 |    | 0 0 -2 | - - Using the structure constants: For SU(2) -facd is replaced by εacd to get: gab = Tr(ea(A)eb(A)) = εacdεbdc      = Σεacdεbdc      cd For Tr(e1(A)e1(A)) we get: gab = Tr(e1(A)e1(A)) = ε1cdε1dc      = Σε1cdε1dc      cd      = ε111ε111 = 0 x 0 = 0      +      ε121ε112 = 0 x 0 = 0      +      ε131ε113 = 0 x 0 = 0      +      ε112ε121 = 0 x 0 = 0      +      ε122ε122 = 0 x 0 = 0      +      ε132ε123 = -1 x 1 = -1      +      ε113ε131 = 0 x 0 = 0      +      ε123ε132 = 1 x -1 = -1      +      ε133ε133 = 0 x 0 = 0      = -2 as we obtained before. The Cartan-Killing metric can be used as a metric tensor to raise and lower indeces. For example, fabc is antisymmetric in a and b but lacks any symmetries for c. This is where the Killing metric comes into play. It can be used to lower the unwelcome index and create a completely anti- symmetric fabd as follows: fabcgcd = fabd Which is totally antisymmetric in the indeces. The use of the Cartan-Killing metric allows us to be loose with regard to defining upper and lower indices on the Levi-Civita tensor and the structure constants tensor and these indeces are frequently not distinguished. Therefore, fabc ≡ fabc etc. The Cartan-Killing form can be used to find the quadratic Casimir operator as follows: _ C2(A) = gabTa(A)Tb(A) = Ta(A)Ta(A) = T1(A)2 + T2(A)2 + T3(A)2 SU(2): - - _  | -2 0 0 | C2(A) = | 0 -2 0 |   | 0 0 -2 | - - The Cartan Subalgebra Revisited ------------------------------- We deliberately showed the calculations for ad(e3) and ad(e8) which correspond to the adjoint form of the 2 Cartan generators, H1 and H2, for SU(3). However, these 2 matrices do not have a basis of common eigenvectors that are linearly independent as they do in the fundamental representation. While they can be individually diagonalized, they cannot be simultaneously diagonalized. We can overcome this by using the raising and lowering operators I±, U± and V± as a basis instead of the iλi's. This is called the CARTAN-WEYL BASIS. In this basis the H's now look like: - -   | 1 0 0 0 0 0 0 0 | I+   | 0 -1 0 0 0 0 0 0 | I-   | 0 0 -1/2 0 0 0 0 0 | U+ ad(H1) = | 0 0 0 1/2 0 0 0 0 | U-   | 0 0 0 0 1/2 0 0 0 | V+   | 0 0 0 0 0 -1/2 0 0 | V-   | 0 0 0 0 0 0 0 0 | I3   | 0 0 0 0 0 0 0 0 | Y - -   I+ I- U+ U- V+ V- I3 Y Therefore, for example, ad(H1)U+ = [H1,U+] - - - - - - - - | 1/2 0 0 || 0 0 0 | | 0 0 0 || 1/2 0 0 | = | 0 -1/2 0 || 0 0 1 | - | 0 0 1 || 0 -1/2 0 | | 0 0 0 || 0 0 0 | | 0 0 0 || 0 0 0 | - - - - - - - - - - | 0 0 0 | = -1/2| 0 0 1 | | 0 0 0 | - - = (-1/2)U+ Which leads to: 0I+ + 0I- + (-1/2)U+ + 0U- + 1V+ + 0V- + 0I3 + 0Y -> - - | 0 | | 0 | | -1/2 | | 0 | the orange column in the matrix. | 0 | | 0 | | 0 | | 0 | - - Likewise, - -   | 0 0 0 0 0 0 0 0 | I+   | 0 0 0 0 0 0 0 0 | I-   | 0 0 √3/2 0 0 0 0 0 | U+ ad(H2) = | 0 0 0 -√3/2 0 0 0 0 | U-   | 0 0 0 0 √3/2 0 0 0 | V+   | 0 0 0 0 0 -√3/2 0 0 | V-   | 0 0 0 0 0 0 0 0 | I3   | 0 0 0 0 0 0 0 0 | Y - -   I+ I- U+ U- V+ V- I3 Y Therefore, for example, ad(H2)U+ = [H2,U+] - - - - - - - - | 1/2√3 0 0 || 0 0 0 | | 0 0 0 || 1/2√3 0 0 | = | 0 1/2√3 0 || 0 0 1 | - | 0 0 1 || 0 1/2√3 0 | | 0 0 -1/√3 || 0 0 0 | | 0 0 0 || 0 0 -1/√3 | - - - - - - - - - - | 0 0 0 | = √3/2| 0 0 1 | | 0 0 0 | - - = (√3/2)U+ Which leads to: 0I+ + 0I- + (√3/2)U+ + 0U- + 1V+ + 0V- + 0I3 + 0Y -> - - | 0 | | 0 | | √3/2 | | 0 | the green column in the matrix. | 0 | | 0 | | 0 | | 0 | - - And, just like in the fundamental representation: [ad(H1),ad(H2)] = 0 Root Diagrams ------------- The Roots, or Roots Vectors, of a Lie algebra, are the weights of the adjoint representation. The number of Roots is equal to the dimension of the Lie algebra, which is also equal to the dimension of the adjoint representation. Therefore, we can associate a root to every element of the algebra. The states in the adjoint representation correspond to the generators (roots). The action of a generator on a state is given by the Eigenvalue equation: H|x> = α|x> where x = I±, V±, U± and Hi The LHS can be evaluated as follows: Ta|Tb> = |Tc><Tc|Ta|Tb> From Quantum Mechanics, Ta is interpreted as a matrix element so we can replace it by [Ta]bc. Thus, Ta|Tb> = |Tc>[Tc]ab    = ifabc|Tc>    = |ifabcTc>    = |[Ta,Tb]> Does this look familiar? It is the equivalent of ad(Ta)Tb = [Ta,Tb]. Therefore, by comparison: ad(Ta)Tb ≡ Ta|Tb> ----------------------------------------------------- Digression: We can see this another way. Now that we are in a basis where the Cartan generators are in diagonal form we can write: - -     | γ(1) .... |a (Ti(A))ab = -| .......... |     | .... γ(8) | b - -     = -γi(a)δab Where i goes from 1 to the rank and a,b go from 1 to 8. Now, (Ti(A))ab = -ifiab By comparison: ifiab = γi(a)δab = γi(a)δba since δab = δba fiab in this case will form the diagonal elements of Ti(A) and are not 0 as they would be in the case of the fundamental representation. Now, we also know: [Ti(A),Ta(A)] = ifiabTb(A)     = γi(a)δbaTb(A)     = γi(a)Ta(A) Therefore, γi(a) is just the ith eigenvalue of Ti(A) and so we can write: [Ti,Ta] ≡ Ti|Ta> = γi(a)|Ta> ----------------------------------------------------- Therefore, the eigenvalue equation becomes: |[Ta,Tb]> = αi|Tb> With this in mind we now find the eigenvalues, αi, for the following commutators: [H1,I+] = I+ [H2,I+] = 0 Thus α1 = ±(1,0) [H1,V+] = (1/2)V+ [H2,V+] = (√3/2)V+ Thus α2 = ±(1/2,√3/2) Proof: [H1,U+] = (-1/2)U+ [H2,U+] = (√3/2)U+ Thus α3 = ±(-1/2,√3/2) [H1,T3] = [H2,T3] = 0 [H1,T8] = [H2,T8] = 0 Diagramatically, root space looks like: The roots are the differences of the weights from the fundamental representation. ----------------------------------------------------- Historical Digression: The root diagram corresponds to the EIGHTFOLD WAY coined by Gell-Mann and Ne'eman in the early 60s that describes the meson and baryon octets. The eightfold way explained a "strange" feature of certain particles produced in accelerator reactions that decayed slowly as if something was hindering the process. The particles exhibited a property termed "strangeness" which would be conserved in production but could be changed during decays. Plots of electric charege, Q, versus strangeness for particles most of which were known in the late 50s, are shown below. We will see more of this in the discussion of Tensor decomposition towards the end of this document. ----------------------------------------------------- w1 = (1/2,1/2√3) w2 = (-1/2,1/2√3) w3 = (0,-1/√3) V+: w1 - w3 (1/2,1/2√3) - (0,-1/√3) = (1/2,√3/2) V-: w3 - w1 (0,-1/√3) - (1/2,1/2√3) = (-1/2,-√3/2) I+: w1 - w2 (1/2,1/2√3) - (-1/2,1/2√3) = (1,0) I-: w2 - w1 (-1/2,1/2√3) - (1/2,1/2√3) = (-1,0) U+: w2 - w3 (-1/2,1/2√3) - (0,-1/√3) = (-1/2,√3/2) Ui: w3 - w2 (0,-1/√3) - (-1/2,1/2√3) = (1/2,-√3/2) Raising and Lowering Operators ------------------------------ For a given state, |i3,y>, the raising and lowering operators produce: I+|i3,i8> = |i3 + 1, y + 0> I-|i3,i8> = |i3 - 1, y + 0> U+|i3,i8> = |i3 - 1/2, y + √3/2> U-|i3,i8> = |i3 + 1/2, y - √3/2> V+|i3,i8> = |i3 + 1/2, y + √3/2> V-|i3,i8> = |i3 - 1/2, y - √3/2> Ladder Operators ---------------- A ladder operator is an operator that increases or decreases the eigenvalue of another operator. Suppose that two operators X and N have the commutation relation: [X,N] = cX If n is an eigenvector of N such that: N|n> = α|n> then NX|n> = (XN + [N,X]|n> = XN|n> + [N,X]|n> = Xα|n> + cX|n> = (α + c)X|n> In other words, if |n> is an eigenvector of N with eigenvalue α then X|n> is an eigenvalue of N with eigenvalue (α + c). The operator X is a raising operator for N if c is real and positive, and a lowering operator for N if c is real and negative. Let's apply this to I3 (≡ H1)and Y (≡ H2) and a state |i3,y>. For example, we let N = I3 and X = U±. - -   | 1/2 0 0 | I3 = | 0 -1/2 0 |   | 0 0 0 | - - I3U±|i3,y> (U±I3 + [I3,U±])|i3,y> (U±I3 -/+ (1/2)U±)|i3,y> (i3 -/+ 1/2)U±|i3,y> Likewise, - - | 1/2√3 0 0 | Y = | 0 1/2√3 0 | | 0 0 -1/√3 | - - YU±|i3,y> = (y +/- √3/2)U±|i3,y> Therefore U+ decreases i3 by 1/2 and increases y by √3/2. Note: [U-,V+] ≡ U-|V+> is an illegal operation. Change of Notation ------------------ We now rename the 6 generators I±, U± and V± as follows. Eα ≡ E1/2,√3/2 = V+ E ≡ E-1/2,√-3/2 = V- etc. etc. We can now write the eigenvalue equation in this new notation as: [Hi,Eα] ≡ Hi|Eα> = αiEα where αi are the root vectors. We can also write: -[Hi,Eα] = αiEα From which we conclude: Eα = E Example: [H1,E1/2,√3/2] ≡ [H1,V+] gives: - - - - - - - - - - | 1/2 0 0 || 0 0 0 | | 0 0 0 || 1/2 0 0 | | 0 0 0 | | 0 -1/2 0 || 0 0 0 | - | 0 0 0 || 0 -1/2 0 | = (-1/2)| 0 0 0 | | 0 0 0 || 1 0 0 | | 1 0 0 || 0 0 0 | | 1 0 0 | - - - - - - - - - - Likewise [H2,E1/2,√3/2] ≡ [H2,V+] gives: - - | 0 0 0 | -√3/2| 0 0 0 | | 1 0 0 | - - So (1/2,√3/2) is a root and (-1/2,-√3/2) is a root. Now consider the commutator of 2 generators. [Eα,Eβ] = Nα,βEα+β Example: [I+,U+] ≡ [E1,0,E-1/2,√3/2] = E1/2,√3/2 ≡ V+ (i.e. Nα,β = 1) [I+,V+] ≡ [E1,0,E1/2,√3/2] = 0 (i.e. Nα,β = 0) Consider the commutator, [Hi,[Eα,Eβ]] We can use the Jacobi identity to get: [Hi,[Eα,Eβ]] = -[Eα[Eβ,Hi]] - [Eβ[Hi,Eα,]] = -Eα(-βiEβ) - EβiEα) = (αi + βi)[Eα,Eβ] This shows that root vectors add. We can now let β = -α to get: [Hi,[Eα,E]] = -[Eα[E,Hi]] - [E[Hi,Eα,]] = -Eα(-αiE) - EiEα) = (αi - αi)[Eα,E] = 0 Now, the implication of this is that [Eα,E] must be some linear combination of Hi's since [Hi,Hj] = 0 (i.e. diagonal matrices commute). Therefore, we can write: [Eα,E] = αiHi Where αi are coefficients. Note that αi should not be confused with the roots, αi, at this point. the subscript i is deliberate and indicates that αiHi implies sum over i (Eimstein notation). Therefore, the commutator becomes a linear combination of Hi's Computation of αi ----------------- [Eα,E] = αiHi To get αi we take the dot product of the state with Hj. Hj.[Eα,E] = αiHi.Hj In Bra-Ket notation the dot product is: <Hj|[Eα,E]> = αi<Hj|Hi> We now make use of the following: <Hi|Hj> = Tr(HiHj) = kDδij This is the dot product of the 2 diagonal matrices. Therefore, - -   | 1/2 0 0 | H1 = | 0 -1/2 0 |   | 0 0 0 | - - - -   | 1/2√3 0 0 | H2 = | 0 1/2√3 0 |   | 0 0 -1/√3 | - - - -    | 1/4 0 0 | H1H1 = | 0 1/4 0 | with Tr = 1/2    | 0 0 0 | - - - -    | 1/12 0 0 | H2H2 = | 0 1/12 0 | with Tr = 1/2    | 0 0 1/3 | - - - -      | 1/4√3 0 0 | H1H2 = H2H1 = | 0 -1/4√3 0 | with Tr = 0      | 0 0 0 | - - Tr(Hj[Eα,E])/kD = αiTr(HjHi)/kD Tr(HjEαE) - Tr(HjEEα) = αiTr(HjHi) We can now make use of the cyclic property of traces, Tr(ABC) = Tr(BCA) = Tr(CAB), on the second term to write (CAB): Tr(HjEαE) - Tr(EαHjE) = αiTr(HjHi) Tr([Hj,Eα]E) = αiTr(HjHi) Tr(αiEαE) = αiTr(HjHi) αiTr(EαE) = αiTr(HjHi) Now, from before, E = Eα so, αiTr(EαEα) = αiTr(HjHi) αj<Eα|Eα> = αiTr(HjHi) Therefore, αj = αiTr(HjHi) But Tr(HjHi) is the Cartan-Killing metric, gij so we get: αj = αigij Consequently, αj ≡ αi More About Roots ---------------- The roots are the weights of the generators that are not Cartan. All of the states in the adjoint representation with zero weight vectors are Cartan generators (and they are orthonormal). - All root vectors are of equal length = 1, i.e. √[(0 + 1/2)2 + (1/2√3 + 1/√3)2] = 1 - Angles between root vectors are < 90° (60° in this case). - The sum of 2 roots sometimes equals another root, sometimes not, i.e. I+ + U+ = V+ I- + U+ ≠ root. - The sum of a root with itself does not exist. - A root is POSITIVE if its 1st non zero component is positive. For example, in the fundamental representation the weight (0,-1/√3) is negative because its first component is 0 so the sign is determined by the the sign of the second component. In the adjoint representation the positive roots are (1/2,-√3/2), (1,0) and (1/2,√3/2)). - If we know the positive roots it easy to determine all of the roots in the representation (i.e. reflect about the y axis). - SIMPLE ROOTS are the positive roots that cannot be written as a sum of other positive roots. - The number of simple roots is equal to the rank of the group, r, and so equals the number of Cartan generators. - If αi and αj are simple, then αi - αj is not a root. Why? If say, αj > αi then αj - αi is positive. We can then write αj = αi + (αj - αi). But αj is then the sum of 2 positive roots which violates the first bullet item. - The angle between the simple roots is given by: cosθαβ = ±√(mn)/2 = √((-1)(-1))/2 = ±1/2 Now, the dot product α12 = |α1||α2|cosθαβ = -1/2 dictates that θαβ is in the 2nd quadrant. Therefore, θαβ = 120° - Any POSITIVE, φk, root can be decomposed as a sum of simple roots with integer coefficients ≥ 0. Therefore, φk = Σkαα α Where k = Σkα α In this sense the simple roots form a basis. Example: φ3 = α1 + α2 - A simple root raises weights by a minimal amount. For SU(3) the simple roots are: V+: α1 = (1/2,√3/2) and U-: α2 = (1/2,√-3/2) Therefore, α1 + α2 = (1,0) = I+ (not a simple root) The Highest Weight, ν --------------------- One can define an ordering of weights and roots such that if μ ≥ ν then μ - ν is positive, and from this find the highest weight of the representation, i.e., For the 3 representation: (1/2,1/2√3) - (0,-1/√3) = (1/2,√3/2) (1/2,1/2√3) - (-1/2,1/2√3) = (1/2,0) Therefore, the highest weight is (1/2,1/2√3). For the adjoint representation: (1,0) - (1/2,√3/2) = (1/2,-√3/2) (1,0) - (1/2,-√3/2) = (1/2,√3/2) Therefore, the highest weight is (1,0). The state with the highest weight, |ν>, satisfies: I+|ν> = V+|ν> = U+|ν> = 0. Or in the 'new' notation: Eα|ν> = 0 In other words, it it is the state that vanishes upon application of any of the raising operators. In the above case, the highest weight intuitively is (1/2,1/2√3) corresponding to the state u. Note: There must be a highest weight state, because otherwise one could construct infinitely many eigenstates by repeated application of the raising operators. Once the highest weight has been found it is possible to construct all of the remaining states by application of the lowering operators until the lowest weight state is encountered. su(2) Subalgebra ---------------- Note again that isospin follows exactly the same mathematics as spin. For SU(2) we can define the operators: J± = J1 ± iJ2 = (1/2)(σ1 ± iσ2) - - J+ = (1/2)(σ1 + iσ2) = | 0 1 | d -> u     | 0 0 | - - - - J- = (1/2)(σ1 - iσ2) = | 0 0 | u -> d     | 1 0 | - - These operators act on the column vectors: - - - - u = | 1 | d = | 0 | | 0 | | 1 | - - - - Proof for d -> u: - - - - - - I+ = | 0 1 || 0 | = | 1 |   | 0 0 || 1 | | 0 | - - - - - - The commutators are: [J3,J±] = ±J± and, [J+,J-] = J3 Where, J3 = σ3/2 J+ = 1/2(σ1 + iσ2) J- = 1/2(σ1 - iσ2) The Cartan generator is: - - H = σ3/2 = | 1/2 0 |   | 0 -1/2 | - - The weights are found from the eigenvalue equations. - - - - H| 1 | = 1/2| 1 | ∴ μ = 1/2 for u | 0 | | 0 | - - - - - - - - H| 0 | = -1/2| 0 | ∴ μ = -1/2 for d | 1 | | 1 | - - - - This looks like: Here we have neglected the hypercharge (Y coordinate) since for our purposes we are only interested in the isospin, I3 component. Previously we said that the roots are the differences of the weights from this fundamental epresentation. Therefore, α = w1 - w2 = (1/2,0) - (-1/2,0) = (1,0) We now want to use our new notation in place of the J's. Let, E3 replace J3 E+ replace J+ E- replace J- We get: [J3,J±] -> [E3,E±] = ±E± Define E+ = (1/|α|)Eα and E- = (1/|α|)E [E+,E-] = (1/|α|2)[Eα,E] = (1/|α|2)αH If we compare this to [J+,J-] = J3 we can immediately see that, E3 = (1/|α|2)αH = ±E± Let's double check this: [E3,E±] -> [(1/|α|2)αH,(1/|α|)E±α] = (1/|α|3)[αH,E±α] = (1/|α|3)α(±α)E±α = ±(1/|α|)E±α = ±E± Therefore, [E3,E±] = ±E± Which is what we had before! In other words, the operators acting horizontally in SU(3) behave just the raising and lowering operators in SU(2). As such they form a su(2) subalgebra. The Master Formula ------------------ The master formula puts constraints on the allowed roots. It is: 2αiw/(αi)2 = -(p - q) Where μ is a weight in some representation, αi is a root, p is the mumber of times w can be raised and q is the number of times it can be lowered. For the special case of the adjoint representation this becomes: 2αiαj/(αi)2 = -(p - q) Proof: We make use of the su(2) subalgebra. For any state, μ, of a representation the E3 eigenvalue is: E3 = (1/|αi|2iH Therefore, E3|μ> = (1/|αi|2iH|μ> But, H|μ> = αj|μ> Therefore, E3|μ> = (1/|αi|2ij|μ> E3 produces either integer or 1/2 integer eigenvalues in the x direction, i.e., (-1/2,√3/2) -> (1/2,√3/2) ... integer 1 (-1,0) -> (1,0) ... integer 2 (1/2,√3/2) -> (1,0) ... integer 1/2 This means that: 2αiαj/|αi|2 is always an integer. We can write: (E+)p|μ> = αij + pαi)/|αi|2|μ> If we let this correspond to the heighest weight, ν, such that (E+)p+1|μ> = 0, then the E3 value is: E3 = αij + pαi)/|αi|2 = αiαj/|αi|2 + p = ν Likewise, we can do the same thing for the lowest state: (E-)q|μ> = αij - pαi)/|αi|2|μ> with E3 = αij - qαi)/|αi|2 = αiαj/|αi|2 - q = -ν If we add the highest and lowest state E3 values together we get: 2αiαj/|α|2 + p - q = 0 or, 2αiαj/|α|2 = -(p - q) The MASTER FORMULA is interpreted as the root αi adding to or subtracting from the weight/root, αj, to produce another valid weight/root. αj adding to or subtracting from αi would be 2αjαi/|αj|2. The numbers -(p - q) are called the DYNKIN COEFFICIENTS. For the fundamental representation, 3: 2(1/2,√3/2)(1/2,1/2√3) = 1 ∴ p = 0, q = 1 We can lower (1/2,√3/2) with (1/2,1/2√3) to get: (1/2,1/2√3) - (1/2,√3/2) = (0,-1/√3) However, if we try and lower (0,-1/√3) again with (1/2,√3/2) the master formula tells us: 2(0,-1/√3)(1/2,√3/2) = -1 ∴ p = 1, q = 0 So we cannot go any further and have to pick another root. We can use this procedure to construct the following weights/roots: 2(1/2,-√3/2)(0,-1/√3) = 1 ∴ p = 0, q = 1 (0,-1/√3)) - (1/2,-√3/2) = (-1/2,1/2√3) 2(1,0)(-1/2,1/2√3) = -1 ∴ p = 1, q = 0 (-1/2,1/2√3) + (1,0) = (1/2,1/2√3) _ For the anti-fundamental representation, 3: 2(1/2,-√3/2)(1/2,-1/2√3) = 1 ∴ p = 0, q = 1 (1/2,-1/2√3) - (1/2,√-3/2) = (0,1/√3) 2(1/2,√3/2)(0,1/√3) = 1 ∴ p = 0, q = 1 (0,1/√3) - (1/2,√3/2) = (-1/2,-1/2√3) 2(1/2,√3/2)(-1/2,-1/2√3) = 1 ∴ p = 1, q = 0 (-1/2,-1/2√3) + (1/2,√3/2) = (0,1/√3) For the adjoint representation: 2(1,0)(-1/2,√3/2) = -1 ∴ p = 1, q = 0 (-1/2,√3/2) + (1,0) = (1/2,√3/2) 2(1/2,√3/2)(1/2,-√3/2) = -1 ∴ p = 1, q = 0 (1/2,-√3/2) + (1/2,√3/2) = (1,0) 2(1/2,√3/2)(1,0) = 1 ∴ p = 0, q = 1 (1,0) - (1/2,√3/2) = (1/2,-√3/2) 2(1/2,√3/2)(1/2,√3/2) = 1 ∴ p = 0, q = 2 (1/2,√3/2) - (1/2,√3/2) = (0,0) (0,0) - (1/2,√3/2) = (-1/2,-√3/2) 2(-1/2,√3/2)(-1/2,-√3/2) = -1 ∴ p = 1, q = 0 (-1/2,-√3/2) + (-1/2,√3/2) = (-1,0) 2(1/2,-√3/2)(1/2,-√3/2) = 2 ∴ p = 0, q = 2 (1/2,-√3/2) - (1/2,-√3/2) = (0,0) (0,0) - (1/2,-√3/2) = (-1/2,√3/2) For the fundamental representation, 2: 2(1,0)(1/2,0)/(1)2 = 1 ∴ q = 1 (1/2,0) - (1,0) = (-1/2,0) We can go one step further with the master formula. We can also write: 2(α.β)/(β.β) = -(p' - q') = n If we multiply both together we get: 4|α||β||β||α|cos2θ = mn Therefore, cos2θαβ = (α.β)2/(α.α)(β.β) = mn/4 If we divide the two we get: (α.α)/(β.β) = m/n mn θ -- --------- 0 90° 1 60°, 120° 2 45°, 135° 3 30°, 150° 4 180° This means that there are only 4 possibilities for angles between roots. Cartan Matrix ------------- We can construct a matrix using the above master formula and the simple roots. Aij = 2(αij)/(αii) Therefore, for the simple roots of SU(3): A11 = 2(1/2,√3/2)(1/2,√3/2)/(1/2,√3/2)(1/2,√3/2) = 2 A12 = 2(1/2,√3/2)(1/2,-√3/2)/(1/2,√3/2)(1/2,√3/2) = -1 A21 = 2(1/2,-√3/2)(1/2,√3/2)/(1/2,√-3/2)(1/2,-√3/2) = -1 A22 = 2(1/2,-√3/2)(1/2,-√3/2)/(1/2,-√3/2)(1/2,-√3/2) = 2 This leads to the following Cartan matrix: - - A = | 2 -1 | | -1 2 | - - The rows correspond to the DYNKIN COEFFICIENTS, -(p - q). The first row is associated with the action of αi and the second row is asociated with the action of αj. The invidual rows of the matrix can be written as: -------- | 2 | -1 | -------- ^ ^ | | | \ \ q = 0, p = 1 q = 2, p = 0 and, -------- | -1 | 2 | -------- ^ ^ | | | \ \ q = 2, p = 0 q = 0, p = 1 These are called DYNKIN LABELS. Fundamental Weights ------------------- SU(3) has 2 fundamental weights that form a basis which are orthogonal to the simple roots of the adjoint representation. Since SU(3) has 2 simple roots it also has 2 fundamental weights, w1 and w2. The fundamental weights of SU(3) are the highest weight of the 3, (1/2,1/2√3) and the highest weight _ of the 3, (1/2,-1/2√3). Proof: 2wjαk/(αk)2 = δjkjk = 0 if i ≠ j)) α1 = (1/2,√3/2) and w1 = (1/2,1/2√3) α2 = (1/2,-√3/2) and w2 = (1/2,-1/2√3) α1w1: 2(1/2,√3/2)(1/2,1/2√3)/(1/2,√3/2) = 1 α1w2: 2(1/2,√3/2)(1/2,-1/2√3)/(1/2,√3/2) = 0 (orthogonal) α2w1: 2(1/2,-√3/2)(1/2,1/2√3)/(1/2,√3/2) = 0 (orthogonal) α2w2: 2(1/2,-√3/2)(1/2,-1/2√3)/(1/2,-√3/2) = 1 Which can be written as the matrix: - - | 1 0 | = δjk | 0 1 | - - We can expand any weight of any representation in terms of the fundamantal weights as: r w = Σaiωi i Where ai are the DYNKIN COEFFICIENTS obtained from the Cartan matrix. The 3 weights of the fundamental representation of SU(3): p = 0 0 q = 1 0 --------- | 1 | 0 | --------- -> 1(1/2,1/2√3) + 0(1/2,-1/2√3) = (1/2,1/2√3) p = 0 1 q = 0 0 --------- | 0 | -1 | --------- -> 0(1/2,1/2√3) - 1(1/2,-1/2√3) = (-1/2,1/2√3) p = 0 0 q = 1 1 --------- | -1 | 1 | --------- -> -1(1/2,1/2√3) + 1(1/2,-1/2√3) = (0,-1/√3) The 3 weights of the anti-fundamental representation of SU(3): p = 0 0 q = 1 0 --------- | 0 | 1 | --------- -> 0(1/2,1/2√3) + 1(1/2,-1/2√3) = (1/2,-1/2√3) p = 0 1 q = 1 0 --------- | 1 | -1 | --------- -> 1(1/2,1/2√3) - 1(1/2,-1/2√3) = (0,1/√3) p = 1 0 q = 0 0 --------- | -1 | 0 | --------- -> -1(1/2,1/2√3) + 0(1/2,-1/2√3) = (-1/2,-1/2√3) The 6 roots of the adjoint representation of SU(3): p = 0 0 q = 1 1 --------- | 1 | 1 | --------- -> 1(1/2,1/2√3) + 1(1/2,-1/2√3) = (1,0) p = 0 1 q = 2 0 --------- | 2 | -1 | --------- -> 2(1/2,1/2√3) - 1(1/2,-1/2√3) = (1/2,√3/2) = α1 p = 1 0 q = 0 2 --------- | -1 | 2 | --------- -> -1(1/2,1/2√3) + 2(1/2,-1/2√3) = (1/2,-√3/2) = α2 p = 1 2 q = 0 0 --------- | 1 | -2 | --------- -> 1(1/2,1/2√3) - 2(1/2,-1/2√3) = (-1/2,√3/2) = -α2 p = 2 0 q = 0 1 --------- | -2 | 1 | --------- -> -2(1/2,1/2√3) + 1(1/2,-1/2√3) = (-1/2,-√3/2) = -α1 p = 1 1 q = 0 0 --------- | -1 | -1 | --------- -> -1(1/2,1/2√3) - 1(1/2,-1/2√3) = (-1,0) We can use the Dynkin labels to construct diagrams showing all the weight/roots of a representation. The 3 Representation: p = 0 0 q = 1 0 ------- ν: | 1 | 0 | (1/2,1/2√3) ------- / p = 0 1 q = 2 0 ------- | 2 |-1 | Subtract (1/2,√3/2) ------- / p = 1 0 q = 0 1 ------- ν - α1: |-1 | 1 | to get (0,-1/√3) ------- \ p = 1 0 q = 0 2 ------- | 1 |-2 | Subtract (-1/2,√3/2) ------- \ p = 0 1 q = 0 0 ------- ν - α1 - α2: | 0 |-1 | to get (-1/2,1/2√3) ------- Therefore, ν = (1/2, 1/2√3) ν - α1 = (1/2,1/2√3) - (1/2,√3/2) = (0,-1/√3) ν - α1 - α2 = (1/2,1/2√3) - (1/2,√3/2) - (1/2,-√3/2) = (-1/2,1/2√3) _ The 3 Representation: _ We can do the same thing for the 3 representation. ------- ν: | 0 | 1 | (1/2,-1/2√3) ------- \ ------- |-1 | 2 | Subtract (1/2,-√3/2) ------- \ ------- ν - α2: | 1 |-1 | to get (0,1/√3) ------- / ------- | 2 |-1 | Subtract (1/2,√3/2) ------- / ------- ν - α2 - α1: |-1 | 0 | to get (-1/2,-1/2√3) ------- Therefore, ν = (1/2,-1/2√3) ν - α2 = (1/2,-1/2√3) - (1/2,-√3/2) = (0,1/√3) ν - α2 - α1 = (1/2,-1/2√3) - (1/2,-√3/2) - (1/2,√3/2) = (-1/2,-1/2√3) The adjoint representation: Here we start by adding the 2 simple roots to get the highest weight state and adding the negatives of the simple roots to get the lowest weight state. p = 0 0 q = 1 1 --------- α1 + α2: | 1 | 1 | k = 2 --------- \ / / / \ / \ p = 0 1 1 0 q = 2 0 0 2 --------- --------- α12: | 2 | -1 | | -1 | 2 | k = 1 --------- --------- \ / \ / \ / \ / --------- Cartan | 0 | 0 | k = 0 generators: --------- \ / / / \ / \ p = 0 2 2 0 q = 1 0 0 1 --------- --------- -α1,-α2: | 1 | -2 | | -2 | 1 | k = -1 --------- --------- \ / \ / \ / \ / p = 1 1 q = 0 0 --------- -α1 - α2: | -1 | -1 | k = -2 --------- Dynkin Diagrams --------------- A Dynkin diagram depicts the Cartan matrix by a graph with simple roots as vertices connected by one or more 'edges'. So for our Cartan matrix we can draw the following Dynkin diagram: (1/2,√3/2) o----o (1/2,-√3/2) meaning that we have 2 simple roota and angle between them is 120°.