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Classical Physics

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Special Relativity

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String Theory

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Superconductivity

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Supersymmetry (SUSY) and Grand Unified Theory (GUT)

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Introduction to Supersymmetry
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test

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test

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: December 2, 2019

Special Unitary Groups and the Standard Model - Part 2 Special Unitary Groups and the Standard Model - Part 3 Special Unitary Groups and the Standard Model - Part 1 ------------------------------------------------------ The Fundamental (Defining) Representation ------------------------------------------ The special unitary group of degree N, SU(N), is the LIE GROUP where the elements are N x N unitary matrices with determinant = +1. The group operation is that of matrix multiplication. Lie groups are best studied in terms of their Lie algebra involving the generators that create the group elements. Consider a general complex transformation in 2 dimensions, x' = Ux which, in matrix form, reads: - - - - | x' | = U| x | | y' | | y | - - - - - - - - - - | x' | = | a b || x | | y' | | c d || y | - - - - - - This has 2N2 = 8 free parameters. To satisfy the condition |x|2 + |y|2 = |x'|2 + |y'|2 requires that the matrix be unitary. Thus, UU = I. - - - - - - - - | a* c* || a b | = | aa* + cc* a*b + c*d | = | 1 0 | | b* d* || c d | | b*a + d*c b*b + d*d | | 0 1 | - - - - - - - - Thus, aa* + cc* = 1 a*b + c*d = 0 b*a + d*c = 0 b*b + d*d = 1 These four conditions means that the original 8 free parameters are reduced to 4. In other words, if the matrix is unitary it is characterized by N2 = 4 parameters. If, in addition to these conditions, we require that the determinant of the matrix = 1, the matrix must have the form: - - - - - - | x' | = | a b || x | | y' | | -b* a* || y | - - - - - - It is easy to see that UU = I. - - - - - - - - | a* -b || a b | = | aa* + bb* a*b - ba* | = | 1 0 | | b* a || -b* a* | | b*a - ab* b*b + aa* | | 0 1 | - - - - - - - - So, aa* + bb* = 1 ... 1. a*b - ba* = 0 ... 2. b*a - ab* = 0 ... 3. b*b + aa* = 1 ... 4. 1. and 4. are the same so now there are only 3 free parameters which we can generalize to N2 - 1. This formula holds for all unitary groups. The implication of this is that there are only 3 matrices that we can be constructed that meet these requirements. These are the 3 SKEW HERMITIAN (anti-Hermitian) Pauli matrices iσ1, iσ2 and iσ3. Aside: A skew Hermitian (anti-Hermitian) matrix is one that has the property that A = -A. This is accomplished by multiplying a Hermitian matrix by i. Thus, for example: - - iσ1 = | 0 i |   | i 0 | - - - - - - | 0 i | = | 0 -i | = -iσ1 | i 0 |  | -i 0 | - - - - - - | 0 i | a = 0, a* = 0, b = i, b* = -i | i 0 | - - - - | 0 1 | a = 0, a* = 0, b = 1, b* = 1 | -1 0 | - - - - | i 0 | a = i, a* = -i, b = 0, b* = 0 | 0 -i | - - If we break U down into its real and imaginary components we get: - - U = | aR + iaI bR + ibI | | -bR + ibI aR - iaI | - -   - -   - -   - -   - - = aR| 1 0 | + iaI| 1 0 | + bR| 0 1 | + ibI| 0 1 |   | 0 1 |  | 0 -1 |  | -1 0 |  | 1 0 |   - -   - -   - -   - -   - -   - -   - -   - - = aR| 1 0 | + iaI| 1 0 | + ibR| 0 -i | + ibI| 0 1 |   | 0 1 |  | 0 -1 |  | i 0 |  | 1 0 |   - -   - -   - -   - - = aRI + aI3 + bR2 + bI1 So any element can be represented as a linear combination of I and the Pauli matrices. It is worthwhile pointing out that this is just the Lie algebra of SU(2). We are in a vector space where σ0, iσ1, iσ2 and iσ3 are the bases which produce, through exponentiation, the special unitary matrices that are the elements of the group SU(2). In terms of infnitesimal generators we can write: U = exp(iθiσi/2) The generators satisfy the following commutation rule: [Ti,Tj] = iεijkTk Where, Ti = σi/2 and εijk is the LEVI-CIVITA symbol. Hermitian Generators -------------------- The requirement UU = I requires that the generators be Hermitian. To see this we write: U = exp(iθiσi/2) = I + iθT (dropping the factor of 2) and, U = I - iθT Thus, UU = (I + iθT)(I - iθT) - i2θ2TT = I = I + iθ(T - T) = I This implies that (T - T) = 0. For this to be true, T has to be Hermitian. For example, - - - - σ2σ2 = | 0 -i || 0 -i | = I     | i 0 || i 0 | - - - - σ2 - σ2 = 0 What about the determinant of T? We can write for U: - - - - - - | 1 0 | + | iθT11 iθT12 | = | 1 + iθT11 iθT12 | | 0 1 | | iθT21 iθT22 | | iθT21 1 + iθT22 | - - - - - - det of RHS = (1 + iθT11)(1 + iθT22) - (iθT12iθT21) = 1 + iθ(T11 + T22) Therefore, to get det = 1, T11 = -T22. In other words, the trace of the generators has be 0. Equivalently, we can also see the from the Jacobi Formula: 1 = detU = det(exp(iθG)) = exp(iθTr(T)) ∴ Tr(T) = 0. This is clearly the case for the σ's. Note: SU(1) ≡ U(1) since T is I (unit matrix) and therefore is just a phase rotation. Therefore, U = exp(iθ) For SU(3) we replace the σ's with the 8 GELL-MANN matrices. Thus, U = exp(iθiλi/2) - -   | 0 1 0 | λ1 = | 1 0 0 |   | 0 0 0 | - - - -   | 0 -i 0 | λ2 = | i 0 0 |   | 0 0 0 | - - - -   | 1 0 0 | λ3 = | 0 -1 0 |   | 0 0 0 | - - - -   | 0 0 1 | λ4 = | 0 0 0 |   | 1 0 0 | - - - -   | 0 0 -i | λ5 = | 0 0 0 |   | i 0 0 | - - - -   | 0 0 0 | λ6 = | 0 0 1 |   | 0 1 0 | - - - -   | 0 0 0 | λ7 = | 0 0 -i |   | 0 i 0 | - - - -   | 1 0 0 | λ8 = 1/√3 | 0 1 0 |   | 0 0 -2 | - - The Gell-Mann matrices generalize the Pauli matrices for SU(2) that are indicated in orange. Like, the σ's the matrices are both traceless and Hermitian. The generators satisfy the following commutation rule: [Ti,Tj] = ifijkTk Where, Ti = λi/2 and fijk are the structure constants. Structure Constants ------------------- The structure constants are found from the fundamental representation as: fijk = -2iTr([Ti,Tj]Tk) Proof: Consider: [Ta,Tb]Td This is equivalent to: ifabcTcTd Therefore, [Ta,Tb]Td = ifabcTcTd If we now take the trace of this we get: Tr([Ta,Tb]Td) = ifabcTr(TcTd) But from before, Tr(TcTd) = (1/2)δbc. Therefore, Tr([Ta,Tb]Td) = ifabc(1/2)δbc Or, fabc = -2iTr([Ta,Tb]Td) Example: f458 f458 = -2iTr[[T4,T5]T8]     = -2iTr(T4T5T8 - T5T4T8) - - - -     | i/8√3 0 0 | | -i/8√3 0 0 |     = -2iTr(| 0 0 0 | - | 0 0 0 |     | 0 0 i/4√3 | | 0 0 -i/4√3 | - - - - - -     | i/4√3 0 0 |     = -2iTr(| 0 0 0 |)     | 0 0 i/2√3| - -     = (-2i)(3i/4√3)     = √3/2 f ijk ---- ------------------ 1 123 1/2 147, 246, 257, 345 -1/2 156, 367 (√3)/2 458, 678 All others permutations are 0. The structure constants obey the following rules: fabc = fbca = fcab = -fbac = -facb = -fcba Thus, f123 = -f312 etc. Structure Constants as a Tensor ------------------------------- Consider the easier case of the Levi-Civita symbol in 3D, εijk.     { +1 if ijk is (1,2,3), (2,3,1) or (3,1,2) εijk = { -1 if ijk is (3,2,1), (1,3,2) or (2,1,3)     { 0 if i = j or j = k or k = i We can represent this as a 3D matrix as follows. - - | 0 1 0 | | -1 0 0 | ij3 | 0 0 0 | - - - - | 0 0 -1 | | 0 0 0 | ij2 | 1 0 0 | - - - - | 0 0 0 | | 0 0 1 | ij1 | 0 -1 0 | - - Thus, ε123 = 1 and ε321 = -1 In general, the dimension of the f tensor is: (N2 - 1) x (N2 - 1) And there would be (N2 - 1) of them. Trace Orthonormality -------------------- The G-M matrices also obey the trace orthonormality relation: Tr(λaλb) = 2δab Consider Tr(λ1λ1) and Tr(λ8λ8) - - - - - - | 0 1 0 || 0 1 0 | | 1 0 0 | Tr(| 1 0 0 || 1 0 0 |) = Tr(| 0 1 0 |) = 2 | 0 0 0 || 0 0 0 | | 0 0 0 | - - - - - - - - - - - - | 1 0 0 || 1 0 0 | | 1 0 0 | Tr((1/3)| 0 1 0 || 0 1 0 |) = Tr((1/3)| 0 1 0 |) = 2 | 0 0 -2 || 0 0 -2 | | -2 0 4 | - - - - - - Or, in terms of Ti: Tr(TaTb) = (1/2)δab - - - - - - | 0 1/2 0 || 0 1/2 0 | | 1/4 0 0 | Tr(| 1/2 0 0 || 1/2 0 0 |) = Tr(| 0 1/4 0 |) = 1/2 | 0 0 0 || 0 0 0 | | 0 0 0 | - - - - - - NOTE: WE USE BOTH λ AND T IN THIS DISCUSSION SO BE MINDFUL OF FACTORS OF 2 IN SOME OF THE RESULTS. SU(2) Subgroups --------------- The following form SU(2) subgroups of SU(3): T1, T2 and T3 T4, T5 and (1/2)(√3T8 + T3) T6, T7 and (1/2)(√3T8 - T3) Proof: [T4,T5] = if453T3 + if458T8 [T4,T5] = (i/2)λ3/2 + (i√3/2)λ8/2 - - - -    | i/4 0 0 | | i/4 0 0 |    = | 0 -1/4 0 | + | 0 i/4 0 |    | 0 0 0 | | 0 0 -i/2 | - - - - - -    | i/2 0 0 |    = | 0 0 0 |    | 0 0 -i/2 | - - [T6,T7] = (-i/2)λ3/2 + (i√3/2)λ8/2 - - - -    | -i/4 0 0 | | i/4 0 0 |    = | 0 i/4 0 | + | 0 i/4 0 |    | 0 0 0 | | 0 0 -i/2 | - - - - - -    | 0 0 0 |    = | 0 i/2 0 |    | 0 0 -i/2 | - - Casimir Operators ----------------- The Casimir operator is an operator that commutes all other generators. As such it is the center of the Lie algebra. Thus, _ [C2(R),Ta(R)] = 0 _ The QUADRATIC CASIMIR OPERATOR, C2(R), operator is: Where (R) denotes the particular representation. _ C2(R) = ΣTa(R)Ta(R) n Note: For SU(3) there is also a cubic CO given by: _ C2(R) = Σijkfijkλiλjλk/8 However, we can ignore it for our purposes. Casimir Invariant ----------------- _ Since C2(R) commutes with all operators we can use SCHUR'S LEMMA that says: If D is an irreducible representations of a group G, A is a matrix and [D(g),A] = 0 then A = kI where k is some number and I is the identity matrix. In our case we can write: - C2(R) = C2(R)I Where C2 is the QUADRATIC CASIMIR INVARIANT. The invariant also follows the relationship: dim R * C2(R) = dim T * T(R) Where, dim R is the dimension of the representation. dim T is the number of generators. T(R) is the INDEX of the representation given by: T(R)δab = Tr(TaTb) (δaa = δbb = ... = 1, 0 otherwise) For the fundamental representation of SU(2), and using σ2 we get: Index: - - - - - - | 0 -i/2 || 0 -i/2 | = | 1/4 0 | ∴ T(F) = 1/2 | i/2 0 || i/2 0 | | 0 1/4 | - - - - - - Commutator: - - - - - - - - | 1/4 0 || 0 -i/2 | - | 0 -i/2 || 1/4 0 | = 0 | 0 1/4 || i/2 0 | | i/2 0 || 0 1/4 | - - - - - - - - Casimir Invariant: 2 * C2(F) = 3 * 1/2 ∴ C2(F) = 3/4 Casimir Operator: _ C2(F) = Σii/2)2   = (σ1/2)2 + (σ2/2)2 + (σ3/2)2   - -   = | 3/4 0 |   | 0 3/4 |   - -   - -   = 3/4| 1 0 |   | 0 1 |   - - _ Which is in agreement with C2(F) = C2(F)I For the fundamental representation of SU(3) and using λ1 we get: Index: - - - - - - | 0 1/2 0 || 0 1/2 0 | | 1/4 0 0 | | 1/2 0 0 || 1/2 0 0 | = | 0 1/4 0 | ∴ T(F) = 1/2 | 0 0 0 || 0 0 0 | | 0 0 0 | - - - - - - Commutator: - - - - - - - - | 1/4 0 0 || 0 1/2 0 | | 0 1/2 0 || 1/4 0 0 | | 0 1/4 0 || 1/2 0 0 | - | 1/2 0 0 || 0 1/4 0 | = 0 | 0 0 0 || 0 0 0 | | 0 0 0 || 0 0 0 | - - - - - - - - Caimir Invariant: 3 * C2(F) = 8 * 1/2 ∴ C2(F) = 4/3 Casimir Operator: _ C2(F) = Σii/2)2   = (λ1/2)2 + ... (λ8/2)2   - -   | 4/3 0 0 |   = | 0 4/3 0 |   | 0 0 4/3 |   - -   - -   | 1 0 0 |   = (4/3)| 0 1 0 |   | 0 0 1 |   - - _ Which is in agreement with C2(F) = C2(F)I Raising and Lowering Operators ------------------------------ For SU(3) we can define the operators: I± = (1/2)(λ1 ± iλ2) gives d <-> u V± = (1/2)(λ4 ± iλ5) gives s <-> u U± = (1/2)(λ6 ± iλ7) gives s <-> d Therefore, - - - -   | 0 1 0 |  | 0 0 0 | I+ = | 0 0 0 | I- = | 1 0 0 |   | 0 0 0 |  | 0 0 0 | - - - - - - - -   | 0 0 1 |  | 0 0 0 | V+ = | 0 0 0 | V- = | 0 0 0 |   | 0 0 0 |  | 1 0 0 | - - - - - - - -   | 0 0 0 |  | 0 0 0 | U+ = | 0 0 1 | U- = | 0 0 0 |   | 0 0 0 |  | 0 1 0 | - - - - The operators follow the commutation relations: [I+,I-] = λ3 [V+,V-] = (√3/2)λ8 + (1/2)λ3 [U+,U-] = (√3/2)λ8 - (1/2)λ3 [I+,V-] = -U- [I+,U+] = V+ [U+,V-] = I- [I-,U-] = -V- [I-,V+] = U+ [U-,V+] = -I+ and every other commutation of I,V and U = 0 These relationships are emphasized in the following diagrams. The left hand case is not allowed and so the commutator is 0. The right hand case is [I+,U+] producing the root vector, V+. These operators act on the column vectors: - - - - - - | 1 | | 0 | | 0 | u = | 0 | d = | 1 | s = | 0 | | 0 | | 0 | | 1 | - - - - - - Proof for d -> u: - - - - - -   | 0 1 0 || 0 | | 1 | I+ = | 0 0 0 || 1 | = | 0 |   | 0 0 0 || 0 | | 0 | - - - - - - Because I+ and I- are associated with d -> u and u -> d associated with neutrons and protons, they are referred to as the isospin or I-spin operators. ----------------------------------------------------- Digression: Isospin ------- The proton and neutron have very similar masses and the nuclear force is found to be approximately charge-independent. In 1932, Heisenberg proposed that if you could switch off the electric charge of the proton there would be no way to distinguish between a proton and neutron. He proposed that the neutron and proton should be considered as two states of a single entity - the nucleon. - - - - p = | 1 | n = | 0 | | 0 | | 1 | - - - - The 2 states are analogous to the spin up and spin down states of a spin 1/2 particle. He referred to this as isospin. Isospin has nothing to do with spin but shares the same mathematics. The neutron and proton form an isospin doublet with total isospin I = 1/2 and third component I3 = ±1/2 We can extend this idea to the u and d quarks since they have similar masses: - - - - u = | 1 | d = | 0 | | 0 | | 1 | - - - - ----------------------------------------------------- These operators are the operators that make the transitions between the weights. Cartan Generators ----------------- The Cartan subalgebra of a Lie algebra is the largest subset of elements of the algebra that are traceless, Hermitian and can be diagonalized simultaneously. In other words, it is the largest subset of elements of the algebra that all commute with one another. The RANK of a Lie algebra, r, is the number of elements contained in its Cartan subalgebra. It is conventional to denote these elements as Hi. For SU(3) the Cartan generators are: - -   | 1 0 0 | H1 = (1/2)| 0 -1 0 | = λ3/2 = I3   | 0 0 0 | - - - -   | 1 0 0 | H2 = (1/2√3)| 0 1 0 | = λ8/2 = Y/2 = I8   | 0 0 -2 | - - It is easy to show that [H1,H2] = 0 (orthogonal) and Hi = (Hi). Hence SU(3) has a rank of 2. These matrices share the same set of linearly independent eigenvectors but not necessarily the same eigenvalues. As such they can be simultaneously diagonalized. Proof: - - - - - - - - | 1/2 0 0 | | 1 | | 0 | | 0 | | 0 -1/2 0 | -----------> 1/2| 0 |; 0| 0 |; -1/2| 1 | | 0 0 0 | | 0 | | 1 | | 0 | - - - - - - - - - - - - - - - - | 1/2√3 0 0 | | 1 | | 0 | | 0 | | 0 1/2√3 0 | -> 1/2√3| 0 |; -1/√3| 0 |; 1/2√3| 1 | | 0 0 -1/√3 | | 0 | | 1 | | 0 | - - - - - - - - SU(2) has 1 Cartan generator so it has a rank of 1. The Cartan generator is σ3/2. - - H = | 1/2 0 | | 0 -1/2 | - - ----------------------------------------------------- Digression: Simultaneous Diagonalization ---------------------------- 2 n x n matrices A and B are simultaneusy diagonalizable if they commute with one another and both can be diagonalized using the same invertible matrix S such that: S-1AS = M and S-1BS = P Where M and P are both diagonal matrices. The use of the same invertible matrix S implies that simultaneously diagonalizable matrices share a common basis of linearly independent eigenvectors. Example: - - - - A = | 2 2 | and B = | 1 2 | | 2 2 | | 2 1 | - - - - - - - - - - eigenvectors/eigenvalues of | 2 2 | = 0| 1 | and 4| 1 | | 2 2 | | -1 | | 1 | - - - - - - - - - - - - eigenvectors of | 1 2 | = -1| 1 | and 3| 1 | | 2 1 | | -1 | | 1 | - - - - - - - - - - | 1 1 |-1 = | 1/2 1/2 | | 1 -1 |   | 1/2 -1/2 | - - - - - - - - - - - - | 1/2 1/2 || 2 2 || 1 1 | = | 4 0 | = C | 1/2 -1/2 || 2 2 || 1 -1 | | 0 0 | - - - - - - - - - - - - - - - - | 1/2 1/2 || 1 2 || 1 1 | = | 3 0 | = D | 1/2 -1/2 || 2 1 || 1 -1 | | 0 -1 | - - - - - - - - [A,B] = [C,D] = 0 ----------------------------------------------------- Weight Diagrams --------------- The Weights of a representation contain the eigenvalues of the elements of the Cartan subalgebra. The eigenvalues of the Cartan generators act on the basis of the fundamental representation give the weight vectors of the algebra. For SU(3) we get: H1|u> = h1|u> gives: - - - - - - | 1/2 0 0 || 1 | | 1 | | 0 -1/2 0 || 0 | = (1/2)| 0 | | 0 0 0 || 0 | | 0 | - - - - - - H2|u> = h2|u> gives: - - - - - - | 1/2√3 0 0 || 1 | | 1 | | 0 1/2√3 0 || 0 | = (1/2√3)| 0 | | 0 0 -1/√3 || 0 | | 0 | - - - - - - ∴ w1 = (h1,h2) = (1/2,1/2√3) H1|d> = h1|d> gives: - - - - - - | 1/2 0 0 || 0 | | 0 | | 0 -1/2 0 || 1 | = (-1/2)| 1 | | 0 0 0 || 0 | | 0 | - - - - - - H2|d> = h2|d> gives: - - - - - - | 1/2√3 0 0 || 0 | | 0 | | 0 1/2√3 0 || 1 | = (1/2√3)| 1 | | 0 0 -1/√3 || 0 | | 0 | - - - - - - ∴ w2 = (h1,h2) = (-1/2,1/2√3) H1|s> = h1|s> gives: - - - - - - | 1/2 0 0 || 0 | | 0 | | 0 -1/2 0 || 0 | = (0)| 0 | | 0 0 0 || 1 | | 1 | - - - - - - H2|s> = h2|s> gives: - - - - - - | 1/2√3 0 0 || 0 | | 0 | | 0 1/2√3 0 || 0 | = (-1/√3)| 0 | | 0 0 -1/√3 || 1 | | 1 | - - - - - - ∴ w3 = (h1,h2) = (0,-1/√3) Diagramatically, the weight space for SU(3) in the fundamental representation looks like: Q = I3 + Y/2 Normalize to λ8 ∴ Y => λ8/√3 Q = λ3/2 + λ8/2√3 The coordinates are the ISOSPINs, I3 and √3/2 times the HYPERCHARGEs, Y. Thus we can redraw the above diagram in terms of I3 and Y as follows: Therefore, H1 generates the isospin and H2 generates the hypercharge. Finally, using Q = (2I3 + Y)/2 = I3 + Y/2 - - - - | 1/2 0 0 | | 1/2√3 0 0 | = | 0 -1/2 0 | + (1/√3)| 0 1/2√3 0 | | 0 0 0 | | 0 0 -1/√3 | - - - - - - - - | 1/2 0 0 | | 1/6 0 0 | = | 0 -1/2 0 | + | 0 1/6 0 | | 0 0 0 | | 0 0 -1/3 | - - - - - - | 2/3 0 0 | = | 0 -1/3 0 | | 0 0 -1/3| - - This is referred to as the CHARGE MATRIX. There following relationship is also used in the literature: Y = S + B Where B is the Baryon number and S is the Strangeness quantum number = -(nS - nS') where nS' is the number of anti strange quarks. The above representation is referred to as the '3'. Note: For SU(2) there is only 1 Cartan generator. This generates the WEAK ISOSPIN. In order to get the WEAK HYPERCHARGE it is necessary to consider the SU(2)L ⊗ U(1)Y gauge group and the electroweak model developed by GLASHOW-WEINBERG-SALAM. Thus, gauge group = SU(2)L ⊗ U(1)Y ^ ^ | | weak weak isospin, I3 hypercharge, YW The weak isospin part is: - - - - - - | 1/2 0 || 1 | = (1/2)| 1 | | 0 -1/2 || 0 | | 0 | - - - - - - ∴ w1 = (1/2,?) - - - - - - | 1/2 0 || 0 | = (-1/2)| 0 | | 0 -1/2 || 1 | | 1 | - - - - - - ∴ w2 = (-1/2,?) To get the weak hypercharge we use the formula: YW = 2(Q - I3) Therefore, for the u and d quarks we get:   - - - - YW = 2(| 2/3 0 | - | 1/2 0 |)   | 0 -1/3 | | 0 -1/2 |   - - - -   - -   = 2| 1/6 0 |   | 0 1/6 |   - -   - -   = | 1/3 0 |   | 0 1/3 |   - -   - -   = (1/3)| 1 0 |   | 0 1 |   - -   = (1/3)I Therefore, the eigenvalue is:   - - - - YW| 1 | = (1/3)| 1 |   | 0 | | 0 |   - - - - YW as the generator of U(1)L is just a phase rotation. For the left handed leptons, eL and νL we get:   - - - - YW = 2(| -1 0 | - | -1/2 0 |)   | 0 0 | | 0 1/2 |   - - - -   - -   = 2| -1/2 0 |   | 0 -1/2 |   - -   - -   = | -1 0 |   | 0 -1 |   - -   - - - - YW| 1 | = -1| 1 |   | 0 | | 0 |   - - - - The Complex Conjugate (Anti-Fundamental) Representation ------------------------------------------------------- In the complex conjugate representation: _ U = exp(iεiTa(C)) The generators of the group are given by Ta(C) = -(Ta(F)). Thus, _ U = exp(-iεiTa(F)) or    _    _ ψ -> exp(iεiTa(F))ψ and ψ = exp(-iεiTa(F))ψ For SU(2) the fundamental and conjugate representation turn out to be equivalent and SU(2) is referred to as a real group. This is not true for the SU(3) and the representations are inequivalent. The generators follow the same commutations rules as for the fundamental case: [Ta(C),Tb(C)] = ifabcTc(C) Proof: [T1(C),T2(C)] = if123T3(C) - -     | 0 -1/2 0 | T1(C) = -λ1/2 = | -1/2 0 0 |     | 0 0 0 | - - - -     | 0 -i/2 0 | T3(C) = -λ3/2 = | i/2 0 0 |     | 0 0 0 | - - - - - - - - - - | 0 -1/2 0 || 0 -i/2 0 | | 0 -i/2 0 || 0 -1/2 0 | | -1/2 0 0 || i/2 0 0 | - | i/2 0 0 || -1/2 0 0 | | 0 0 0 || 0 0 0 | | 0 0 0 || 0 0 0 | - - - - - - - - - - | -i/2 0 0 | = | 0 i/2 0 | | 0 0 0 | - - = if123(-λ3)/2 The weight for SU(3) are: _ _ _ H1|u> = h1|u> gives: - - - - - - | -1/2 0 0 || 1 | | 1 | | 0 1/2 0 || 0 | = (-1/2)| 0 | | 0 0 0 || 0 | | 0 | - - - - - - _ _ _ H2|u> = h2|u> gives: - - - - - - | -1/2√3 0 0 || 1 | | 1 | | 0 -1/2√3 0 || 0 | = (-1/2√3)| 0 | | 0 0 1/√3 || 0 | | 0 | - - - - - - ∴ w1 = (h1,h2) = (-1/2,-1/2√3) _ _ _ H1|d> = h1|d> gives: - - - - - - | -1/2 0 0 || 0 | | 0 | | 0 1/2 0 || 1 | = (1/2)| 1 | | 0 0 0 || 0 | | 0 | - - - - - - _ _ _ H2|d> = h2|d> gives: - - - - - - | -1/2√3 0 0 || 0 | | 0 | | 0 -1/2√3 0 || 1 | = (-1/2√3)| 1 | | 0 0 1/√3 || 0 | | 0 | - - - - - - ∴ w2 = (h1,h2) = (1/2,-1/2√3) _ _ _ H1|s> = h1|s> gives: - - - - - - | -1/2 0 0 || 0 | | 0 | | 0 1/2 0 || 0 | = (0)| 0 | | 0 0 0 || 1 | | 1 | - - - - - - _ _ _ H2|s> = h2|s> gives: - - - - - - | -1/2√3 0 0 || 0 | | 0 | | 0 1/2√3 0 || 0 | = (1/√3)| 0 | | 0 0 1/√3 || 1 | | 1 | - - - - - - ∴ w3 = (h1,h2) = (0,1/√3) Diagramatically, the weight space for SU(3) in the anti-fundamental representation looks like: When acting on anti-quarks we need to use: I+ -> -I+* = -I- I- -> -I-* = -I+ etc. etc. This is because the generators in the anti-fundamental case obey _ T(C) = -T(F) For example, - -   | 0 0 0 | I- = | 1 0 0 |   | 0 0 0 | - - - -    | 0 1 0 | -I-T = | 0 0 0 | = -I+    | 0 0 0 | - - _ _ -I+|d> = -|u> The complete list of raising and lowering operators for the anti-quarks is: _ _ I+|u> = -|d> _ _ V+|u> = -|s> _ _ U+|d> = -|s> _ _ I-|d> = -|u> _ _ V-|s> = -|u> _ _ U-|s> = -|d>