## Section5.3Commuting 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.

$$N\vert v\rangle = \alpha \vert v\rangle\tag{5.3.3}$$

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 $$\vert v\rangle\text{.}$$