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.
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    $\begingroup$ This is found in any elementary book on linear algebra... $\endgroup$ – ZeroTheHero Jul 18 '20 at 3:21
  • $\begingroup$ 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. $\endgroup$ – mithusengupta123 Jul 18 '20 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 $\mathbf{v}$.


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