The \(\mathfrak{s u}(2)\) algebra consists of three operators in the basis of its Lie algebra which are not mutually commutative. We could also consider a Lie algebra for which the basis is a single Hermitian operator, \(\hat{A}\), but that necessarily commutes with itself, \([\hat{A}, \hat{A}]=0\), so describes an Abelian Lie group. We can also imagine Lie algebras with more than three basis elements, but what about a Lie algebra with two elements, call them \(\hat{A}\) and \(\hat{B}\) ?

(a) In general, we can write the commutation relation of the Lie algebra spanned by \(\hat{A}\) and \(\hat{B}\) as

\[\begin{equation*}[\hat{A}, \hat{B}]=i \alpha \hat{A}+i \beta \hat{B} \tag{8.152}\end{equation*}\]

where \(\alpha, \beta\) are the real-valued structure constants of the algebra. Further, if \(\hat{A}\) and \(\hat{B}\) are orthogonal, this is expressed through the trace of the product of \(\hat{A}\) and \(\hat{B}\) :

\[\begin{equation*}\operatorname{tr}[\hat{A} \hat{B}]=0 \tag{8.153}\end{equation*}\]

Show that this orthogonality constraint requires that \(\alpha=\beta=0\), or that \(\hat{A}\) and \(\hat{B}\) necessarily commute. Thus, the smallest dimension of a non-Abelian Lie algebra is 3 .

(b) What if \(\hat{A}\) and \(\hat{B}\) are not initially orthogonal, \(\operatorname{tr}[\hat{A} \hat{B}] eq 0\) ? Show that you can always define a new operator \(\hat{C}\) in the Lie algebra which is orthogonal to \(\hat{A}\); that is, an element of the Lie algebra spanned by \(\hat{A}\) and \(\hat{B}\) for which
\[\begin{equation*}\operatorname{tr}[\hat{A} \hat{C}]=0 \tag{8.154}\end{equation*}\]