QCD lecture 02 - Example 04
Group SU(2) - generators and construction of irreducible representations.
Solution:
SU(2) generators
Group SU(2) - 2 × 2 unitary (U) matrices with unit determinant (S - special). This is also a representation of the group on two dimensional Hilbert space. Generators should be Hermitian matrices 2 × 2 with zero trace. Such general Hermitian matrix can be parametrized with three real numbers a, b, and c
c a − ib . a + ib −c
The SU(2) has therefor three generators which can be chosen as
0 1 0 −i 1 0 σ = , σ = , σ = . 1 1 0 2 i 0 3 0 −1
These are the well known Pauli matrices. It is strightforward to show that these matrices satisfy relation σiσj = δij1 + iεijkσk , with εijk being the Levi-Civita symbol: 0 1 0 1 1 0 0 1 0 −i i 0 σ σ = · = , σ σ = · = = iσ , 1 1 1 0 1 0 0 1 1 2 1 0 i 0 0 −i 3 0 1 1 0 0 −1 σ σ = · = = −iσ , 1 3 1 0 0 −1 1 0 2 0 −i 0 1 −i 0 0 −i 0 −i 1 0 σ σ = · = = −iσ , σ σ = · = , 2 1 i 0 1 0 0 i 3 2 2 i 0 i 0 0 1 0 −i 1 0 0 i σ σ = · = = iσ , 2 3 i 0 0 −1 i 0 1 1 0 0 1 0 1 σ σ = · = = iσ , 3 1 0 −1 1 0 −1 0 2 1 0 0 −i 0 −i 1 0 1 0 1 0 σ σ = · = = −iσ , σ σ = · = . 3 2 0 −1 i 0 −i 0 1 3 3 0 −1 0 −1 0 1 In general, any complex 2×2 matrix can be expressed as a (complex) linear combintaion of the Pauli matrices and unit matrix 1. This is very easy to see. In particlar, for the product of two 2 × 2 Hermitian matrices with zero trace one can explicitly find that the combination is a3 a1 − ia2 b3 b1 − ib2 a1 + ia2 −a3 b1 + ib2 −b3 a b + a b + a b + ia b − ia b a b − a b + ia b − ia b = 1 1 2 2 3 3 1 2 2 1 3 1 1 3 2 3 3 2 a1b3 − a3b1 + ia2b3 − ia3b2 a1b1 + a2b2 + a3b3 − ia1b2 + ia2b1 = (a1b1 + a2b2 + a3b3)1 + i(a2b3 − a3b2)σ1 + i(a3b1 − a1b3)σ2 + i(a1b2 − a2b1)σ3 = ~a ·~b 1 + i ~a ×~b · ~σ,
which leads immediately to the above formula for the product of two Pauli matrices. From this follows directly: [σi, σj] = 2iεijkσk . It is also easy to see from the formula that the base is orthogonal: