# Why choose traceless matrices as basis?

• I
Gold Member
While writing down the basis for SU(2), physicists often choose traceless hermitian matrices as such, often the Pauli matrices. Why is this? In particular why traceless, and why hermitian?

fresh_42
Mentor
Skew-Hermitian is a consequence of unitary, traceless of determinant one:
$$U^\dagger U = I \Longrightarrow D(U^\dagger U) = U^\dagger \cdot I + I \cdot U = U^\dagger + U = D(I) = 0\\ \det U = 1 \Longrightarrow D(\det U) = \operatorname{tr}U = D(1) = 0$$
Here's the computation for the determinant in detail:
https://www.physicsforums.com/insights/pantheon-derivatives-part-iv/
and here are some remarks on ##\operatorname{SU}(2)##:
https://www.physicsforums.com/insights/representations-precision-important/
https://www.physicsforums.com/insights/journey-manifold-su2-part-ii/

phoenix95