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\)
Section 5.3 Commuting Matrices
Commuting operators have the same (non-degenerate) eigenvectors.
Suppose two operators \(M\) and \(N\) commute, \([M,N]\equiv MN-NM=0\text{.}\) 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{5.3.1}\\
N\left(M\vert v\rangle\right) \amp
= M\left(N\vert v\rangle\right)
=\lambda_v\left(N\vert v\rangle\right)\tag{5.3.2}
\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{5.3.3}
\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 \(\vert v\rangle\text{.}\)
