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Section 6.3 Commuting Matrices

Suppose two operators \(M\) and \(N\) commute, \([M,N]=0\text{.}\) Then if \(M\) has an eigenvector \(\vert v\rangle\) with non-degenerate eigenvalue \(\lambda_v\text{,}\) we will show that \(\vert v\rangle\) is also an eigenvector of \(N\text{.}\)

\begin{align} M\vert v\rangle \amp = \lambda_v\vert v\rangle\tag{6.3.1}\\ NM\vert v\rangle \amp = MN\vert v\rangle=\lambda_vN\vert v\rangle\tag{6.3.2} \end{align}

The last equality shows that \(N\vert v\rangle\) is also an eigenvector of \(M\) with the same non-degenerate eigenvalue \(\lambda_v\text{.}\) But if this is true, then \(N\vert v\rangle\) must be proportional to \(\vert v\rangle\text{,}\) i.e.

\begin{equation} N\vert v\rangle = \alpha \vert v\rangle\tag{6.3.3} \end{equation}

which is just the statement that \(\vert v\rangle\) is also an eigenvector of \(N\) with eigenvalue \(\alpha\text{.}\)

Note that if \(\lambda_v\) were a degenerate eigenvalue, then we would not have been able to assume that \(N\vert v\rangle\) is proportional to \(\vert v\rangle\text{.}\)\include{unitary}