# For two commuting operators $A$ and $B$ and in absence of any degeneracy, is every eigenstate of $A$ is also an eigenstate of $B$ and vice-versa?

Two commuting operators $$\hat{A}$$ and $$\hat{B}$$ always share a complete set of common eigenfunctions. However, in the presence of degeneracy, every eigenstate of $$\hat{A}$$ need not an eigenstate of $$\hat{B}$$ and vice-versa. For instance, in case of $${\rm 1D}$$ free particle motion, the Hamiltonian $$\hat{H}=\frac{p_x^2}{2m}$$ commutes with momentum $$\hat{p}_x$$ but the energy eigenfunctions $$\phi_1(x)=\sin(p_xx/\hbar)$$ and $$\phi_2(x)=\cos(p_xx/\hbar)$$ are not eigenfunctions of $$p_x$$. Other examples can also be given. I read this post which refers to the case I mention above but another question needs to be answered.

• When there is no degeneracy, is it true that every eigenstate of $$\hat{A}$$ must also be an eigenstate of $$\hat{B}$$ and vice-versa? If false, I would prefer a counterexample.
• This is found in any elementary book on linear algebra... Jul 18, 2020 at 3:21
• Do you mean a counterexample? Please note that I know that two commuting operators share a complete set of common eigenfunctions. What I want to know is whether every eigenstate of A is also an eigenstate of B, in the absence of any degeneracy. Jul 18, 2020 at 3:24

Yes it is true. Let $$A$$ be a non-degenerate operator which commutes with another operator $$B$$ $$A\mathbf{v}=\lambda_v\mathbf{v}\tag{1}$$ where $$\mathbf{v}$$ is an eigenvector with corresponding eigenvalue $$\lambda_v$$
$$AB=BA\Rightarrow AB\mathbf{v}=BA\mathbf{v}$$ $$\Rightarrow A(B\mathbf{v})=\lambda_v(B\mathbf{v})$$ This implies $$B\mathbf{v}$$ is an eigenvector (more generally, eigenvector times a scalar) of A with eigenvalue $$\lambda_v$$. Since $$\mathbf{v}$$ is only eigenvector corresponding to $$\lambda_v$$, $$B\mathbf{v}=\mathbf{v}$$ or more generally $$B\mathbf{v}=\alpha\mathbf{v}$$ where $$\alpha$$ is a scalar, which is now an eigenvalue of $$B$$ with eigenvector $$\mathbf{v}$$.