## 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.

$$N\vert v\rangle = \alpha \vert v\rangle\tag{6.3.3}$$

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}