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Adjoint Representation. SU(3)

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Adjoint Representation. SU(3) QCD lecture 03 - Group SU(3) Generators, weights, roots and shifting operators of group SU(3). Adjoint representation. SU(3) generators Group SU(3) - 3×3 unitary (U) matrices with unit determinant (S - special). This is also a representation of the group on three dimensional Hilbert space. Generators are Hermitian matrices 3 × 3 with zero trace. Such general Hermitian matrix can be parametrized with eight real numbers a; : : : ; h: 0 a c − id e − if 1 @c + id b g − ih A : e + if g + ih −a − b In analogy with SU(2) and with Pauli matrices, we can take for the base of SU(3) Lie algebra the following generators, so called Gell-Mann matrices: 00 1 01 00 −i 01 01 0 01 λ1 = @1 0 0A λ2 = @i 0 0A λ3 = @0 −1 0A 0 0 0 0 0 0 0 0 0 00 0 11 00 0 −i1 λ4 = @0 0 0A λ5 = @0 0 0 A 1 0 0 i 0 0 00 0 01 00 0 0 1 01 0 0 1 λ = 0 0 1 λ = 0 0 −i λ = p1 0 1 0 : 6 @ A 7 @ A 8 3 @ A 0 1 0 0 i 0 0 0 −2 In general any 3 × 3 complex matrix can be expressed as (complex) linear combination of Gell-Mann matrices and unit matrix 1. Matrix 3 × 3 has nine complex numbers (in general for SU(n) it has n2 complex parameters) and there are 8 generator (for SU(n) we have n2 − 1) plus unit matrix, which gives also 9 terms (n2), i.e. again 9 complex parameters. It is strightforward to find which linear combinations of Gell-Mann matrices are needed to produce desired matrix element. For the product of two Gell-Mann matrices we can therefor write: λa · λb = Aab1 + Babcλc;Aab;Babc 2 C: It is strightforward to check that the Gell-Man matrices give orthogonal basis of SU(3) algebra with normalization + T r λa · λb = 2δab: Canonical choice of normalization is then given for color matrices λ 1 T ≡ a ; with Tr T · T + = δ : a 2 a b 2 ab This gives 1 T · T = δ 1 + B T ;B 2 : a b 6 ab abc c abc C It is easy to see that tensor B stays constant under cyclic permutation of its parameters: 3 Tr (T T T ) = B Tr (T T ) = B ; ) B = B = B : a b c abd d c 2 abc abc bca cab Explicitly expressing this tensor by two real tensors d and f: 1 B = (d + if ) d ; f 2 ; abc 2 abc abc abc abc R and evaluating 1 1 1 1 (T T )+ = δ 1 + (d − if ) T + = δ 1 + (d − if ) T ; a b 6 ab 2 abc abc c 6 ab 2 abc abc c which also equals to 1 1 (T T )+ = T +T + = (T T ) = δ 1 + (d + if ) T ; a b b a b a 6 ab 2 bac bac c we see immediately that dabc is totaly symmetric while fabc is totaly anti-symmetric. Eval- uating the commutator between two color matrices leads to 1 1 [T ;T ] = (d + if )T − (d + if )T = if T a b 2 abc abc c 2 bac bac c abc c i.e. fabc are indeed the structure constants. And for the anti-commutator we get 1 1 1 1 fT ;T g = δ 1 + (d + if )T + (d + if )T = δ 1 + d T : a b 3 ab 2 abc abc c 2 bac bac c 3 ab abc c Finally we can write 1 1 T · T = δ 1 + (d + if )T : a b 6 ab 2 abc abc c This relation will be useful later on for the evaluation of traces of color matrices in calcula- tions of QCD matrix elements without explicit knowledge of values of structure constants fabc and symmetric tensor dabc. This derivation works also for general SU(n) fundamental representation. Weights of fundamental SU(3) representation The rank of group SU(3) is two as no more than two color matrices commute together. These are the two diagonalized generators T3 and T8, which form Cartan subalgebra with H1 = T3 and H2 = T8. The weights of the fundamental SU(3) representation are given by proper numbers of those two generators. As we have made them diagonal already, we can chose the base vectors as 011 001 001 e1 = @0A ; e2 = @1A ; e3 = @0A : 0 0 1 We can find the proper numbers by applying Hi to those vectors: 1 1 1 1 T3e1 = e1;T8e1 = p e1 ) ~µ1 = ; p : 2 2 3 2 2 3 These are basically diagonal members of T3 and T8. For e2 and e3, we get immediately 1 1 1 ~µ2 = − ; p ; and ~µ3 = 0; −p : 2 2 3 3 T8 µ 1 µ 2 2 3 1 - 1 1 T 2 2 3 µ - 1 3 3 Figure 1: Base vectors of Hilbert space for SU(3) fundamental triplet. SU(3) roots and shifting operators Remaining six generators, outside of Cartan subalgebra, could be rearanged into six shifting operators E~α satisfying relation [Hi;E~α] = αiE~α; with H1 = T3 and H2 = T8. We could follow the general prescription and find the roots ~α and the shifting operators E~α as proper numbers and proper vectors of Cartan subalgebra operators Hi. For this, we would need explicit knowledge of SU(3) structure constants. We can, however, guess the form of shifting operators, noting that pairs T1 and T2 has form of Pauli matrices. Similarly, for T4 and T5, and T6 and T7. The shifting operators will have form 1 1 1 p (T1 ± iT2) ; p (T4 ± iT5) ; p (T6 ± iT7) : 2 2 2 We can check the these are indeed shifting operators and find the roots by explicit eval- uation of the commutator with the members of Cartan subalgebra. For (T1 ± iT2) and H1 = T3 we know the results as this is exactly the same as in case of SU(2) [T3; (T1 ± iT2)] = ±1 (T1 ± iT2) : For T8 we get 200 1 01 00 −i 01 00 1 01 00 −i 013 1 [T8; (T1 ± iT2)] = p 4@1 0 0A ± i @i 0 0A − @1 0 0A ∓ i @i 0 0A5 = 0: 4 3 0 0 0 0 0 0 0 0 0 0 0 0 We have found two roots (1; 0) (= ~α3); and (−1; 0) = (−~α3) and corresponding shifting operators 1 E±α~ = p (T1 ± iT2) : 3 2 For (T4 + iT5) we get 2 01 0 01 00 0 213 1 1 [T ; (T + iT )] = 0 −1 0 ; 0 0 0 3 4 5 42 @ A 2 @ A5 0 0 0 0 0 0 200 0 21 00 0 013 1 1 = 0 0 0 − 0 0 0 = (T + iT ) ; 4 4@ A @ A5 2 4 5 0 0 0 0 0 0 2 01 0 0 1 00 0 113 1 [T8; (T4 + iT5)] = 4 p @0 1 0 A ; @0 0 0A5 2 3 0 0 −2 0 0 0 200 0 11 00 0 −213 p 1 3 = p 0 0 0 − 0 0 0 = (T + iT ) : 4@ A @ A5 2 4 5 2 3 0 0 0 0 0 0 We have found another root p ! 1 3 ~α = ; 1 2 2 for shifting operator 1 Eα = p (T4 + iT5) : 1 2 We will omit the vector notation for roots in subscript of the shifting operators. We note that the forth root is −~α and corresponding shifting operator is E+ which is indeed p 1 α1 (T4 − iT5) = 2. Finally, let's evaluate the commutators with Cartan subalgebra for (T6 − iT7): 2 01 0 01 00 0 013 1 1 [T ; (T − iT )] = 0 −1 0 ; 0 0 0 3 6 7 42 @ A 2 @ A5 0 0 0 0 2 0 200 0 01 00 0 013 1 1 = 0 0 0 − 0 0 0 = (T − iT ) ; 4 4@ A @ A5 2 6 7 0 0 0 0 −2 0 2 01 0 0 1 00 0 013 1 [T8; (T6 − iT7)] = 4 p @0 1 0 A ; @0 0 0A5 2 3 0 0 −2 0 1 0 200 0 01 00 0 013 p 1 3 = p 0 0 0 − 0 0 0 = − (T − iT ) : 4@ A @ A5 2 6 7 2 3 0 −2 0 0 1 0 And we have remaining two roots p ! p ! 1 3 1 3 ~α = ; − ; and − ~α = − ; 2 2 2 2 2 2 with shifting operators 1 1 Eα = p (T6 − iT5) and E−α = p (T6 + iT5) : 2 2 2 2 T8 α 1 µ 3 µ 2 2 3 1 - 1 1 T 2 2 3 α α 2 1 µ - 1 3 3 Figure 2: Roots of SU(3) group. To summarize, SU(3) has six roots, three positive roots p ! p ! 1 3 1 3 ~α = ; ; ~α = ; − ; ~α = (1; 0) ; 1 2 2 2 2 2 3 plus three negative directions −~α1; −~α2; −~α3. Corresponding six shifting operators are then 1 1 1 E±α = p (T4 ± iT5) ;E±α = p (T6 ∓ iT7) ;E±α = p (T1 ± iT2) : 1 2 2 2 3 2 Knowing the weights of fundamental representation, we could have find out the values of six roots simply by taking the difference between all pairs of weights. For example ~α3 = ~µ1 − ~µ2, etc. Adjoint representation of SU(3) Adjoint representation lives on the Hilbert space with dimension equal to the number of group generators.
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