## Section6.3Commuting 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}