Let $\mathcal{H}$ be a Hilbert space. Consider an arbitrary and non-diagonal unitary operator $O: \mathcal{H} \to \mathcal{H}$ that acts on an initial quantum state $|\psi_0\rangle \in \mathcal{H}$ producing a new quantum state
$$
|\psi\rangle = O|\psi_0\rangle.
$$
Now, assume that either $O$ is easily diagonalizable or that someone can efficiently diagonalize it for us. Let $O'$ be the diagonilization of $O$.
Now, I want to understand what happens when I measure the expectation value of this observable in either basis of $O$. I.e., I want to understand under what conditions
$$
\langle O \rangle = \langle \psi_0|O|\psi_0\rangle = \langle \psi|O|\psi\rangle =\langle O' \rangle .
$$
Furthermore I am interested in understanding if, given a second unitary operator $H$, equality of the expectation values $\langle O \rangle = \langle O' \rangle $ means that also
$$
\langle HO \rangle = \langle HO' \rangle.
$$
Best Answer
Diagonalizing an operator means simply changing the basis you are using in the Hilbert space. Essentially, the idea is that instead of writing, for example, the states as $$|\psi\rangle = a|+\rangle + b|-\rangle,$$ you'd write $$|\psi\rangle = \alpha|0\rangle + \beta|1\rangle,$$ so that the expression of the action of $\mathcal{O}$ would be simpler. Namely, you could have, for example, $$\mathcal{O}|\psi\rangle = \alpha\mathcal{O}|0\rangle + \beta\mathcal{O}|1\rangle = \beta|1\rangle,$$ where I assumed $\mathcal{O}|0\rangle = 0$ and $\mathcal{O}|1\rangle = |1\rangle$ just for the sake of an example.
In short, the operator doesn't change. Its matrix elements (the numbers $\langle n | \mathcal{O} | m \rangle$, where $\lbrace| n \rangle\rbrace$ is the chosen basis) do change, but $\mathcal{O}$ itself is an abstract operator which does not depend on the basis.