# If a Hermitian and Unitary operator commute, do they have a simultanous eigenbasis?

I was learning about Hamiltonian with discrete translational symmetry. We showed that there is a simultaneous eigenbasis of the Hamiltonian $$H$$ and the discrete translation symmetry being the bloch states.

This made me wonder if it can be generalized. If there is a hermitian operator $$H$$ and Unitary operator $$U$$ such that $$[H,U]=0$$. Does that mean there is a simultaneous eigenbasis?

Here is what I've tried: $$H$$ is hermitian therefore it has an eigenbasis $$B= \{|i\rangle\}$$ where $$H|i\rangle = \lambda_i |i\rangle$$

Then consider $$\langle i|HU|j\rangle$$ where $$|i\rangle$$ and $$|j\rangle$$ are eigenvectors of $$H$$ with distinct eigenvalues

$$\langle i|HU|j\rangle=\lambda_i \langle i|U|j\rangle$$ but also: $$\langle i|HU|j\rangle = \langle i|UH|j\rangle=\lambda_j \langle i|U|j\rangle$$. Since $$\lambda_i \neq \lambda_j$$, it means that $$\langle i|U|j\rangle=0$$.

Therefore in this basis $$B$$, we have that $$U$$ is block diagonal, and since this is eigenbasis of $$H$$, $$H$$ is diagonal. Since block matrices multiply block-wise, we can create some matrix that diagonalize $$U$$ blockwise and since $$H$$ is diagonal, it does not do anything to $$H$$. Since $$H,U$$ can be both diagonalized at once, it means that they have simultaneous eigenbasis.

Is this proof above correct?

Edit: It turns out to be a true statement: Both Unitary and Hermitian matrices are normal matrices and therefore commuting implies that they are simultaneously diagonalizable

As Cosmas Zachos remarks, every unitary operator over Hilbert space $$U: \mathcal{H} \rightarrow \mathcal{H}$$ can be written in the form $$\exp(-iA)$$ where $$A$$ is Hermitian. Thus, $$[H, U] = 0 \iff [H, A] = 0$$ using the power series expansion of $$U$$. Two diagonalizable matrices that commute are simultaneously diagonalizable. Hence, share a common eigenbasis.
Sorry, this site does not specialize in "check my proof" or even ab initio strict proof questions. You might get more hidebound answers comporting with your standards and expectations; but, informally, in this trade, a hermitian operator K links to a unitary operator U and vice versa, $$K \leftrightarrow e^{iK}=U\\ \log U =iK,$$ so you really are looking at $$[H,K]=0$$ giving you simultaneous eigenbases...