Why choose traceless matrices as basis?

  • #1
phoenix95
Gold Member
69
20

Main Question or Discussion Point

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?
 

Answers and Replies

  • #2
12,633
9,148
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/
 

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