By David Atkinson
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Sample text
9). 2 Actually, the generators of a Lie group can even be defined without making any reference to a specific representation. One makes use of the fact that a Lie group is also a manifold, parametrized by the coordinates θ a , and defines the generators as a basis of the tangent space at the origin. One then proves that their commutator (defined as a Lie bracket) is again a tangent vector, and therefore it must be a linear combination of the basis vector. In this approach no specific representation is ever mentioned, so it becomes obvious that the structure constants are independent of the representation.
Observe that, because of the factor i in eq. 52), a representation of SU (2) × SU (2) with J± hermitian induces a representation of SO(3, 1) with J hermitian but K antihermitian. e. it is obtained taking the time variable t purely imaginary. The universal covering group of SO(3, 1) is SL(2, C). −,i +,i ,J ,J −,j −,j =i ijk J +,k ijk −,k = 0. 55), including spinor representations. Since we know the representations of SU (2), and here we have two commuting SU (2) factors, we find that: • The representations of the Lorentz algebra can be labeled by two half-integers: (j− , j+ ).
114) (using eq. 100) and the antisymmetry of µνρσ ), and then Wµ W µ commutes also with P ν . Since Wµ W µ is Lorentz-invariant, we can compute it in the frame that we prefer. If m = 0, it is convenient to choose the rest frame of the particle; in this frame W µ = (−m/2) µνρ0 Jνρ = (m/2) 0µνρ Jνρ , so W 0 = 0 while m ijk jk m 0ijk jk Wi = J = J = mJ i . 115) 2 2 Therefore on a one-particle state with mass m and spin j we have −Wµ W µ = m2 j(j + 1) , (m = 0) . 116) If instead m = 0 the rest frame does not exist, but we can choose a frame where P µ = (ω, 0, 0, ω); in this frame a straightforward computation gives W 0 = W 3 = ωJ 3 , W 1 = ω(J 1 − K 2 ) and W 2 = ω(J 2 + K 1 ).