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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

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2 Answers 2

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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.

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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...

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