Section 5.3 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 \(N\text{.}\)
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.
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}