# Operator diagonalization in Quantum Mechanics

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 $$\langle HO \rangle = \langle HO' \rangle.$$

• Your unitary operator is $O=\exp (iM)$ for M hermitian, hence $M=U N U^{-1}$ for N diagonal. So $O=U O' U^{-1}$. You may now compare respective matrix elements. What do you conclude? Jan 18, 2022 at 17:02
• Well, as discussed also by the answer below, I conclude that diagonalization is equivalent to transforming the original state as $|\psi \rangle \to U^{-1}|\psi \rangle$. So, as he says it is a change of basis. Naively this is equivalent to a (non-observableas always) gauge transformation. My questions essentially boils down to as there exist any ceveats to this. Jan 19, 2022 at 8:14
• But you understand this change of basis is not included in the expectation value you wrote, a mere matrix element, right? Jan 19, 2022 at 14:36
• Sure. Totally understood. Jan 20, 2022 at 10:12

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.
• @Marion There are no exceptions, the expectation value is basis-independent. A way of noticing this is because the action of an operator on a state is something defined abstractly, without any mention to a choice of basis (you wrote $|\psi\rangle = O |\psi_0\rangle$, for example, without ever choosing a basis), and so is the inner product. A change of basis never changes the operator, only it's matrix elements, so the expectation values all continue to be the very same ones. Jan 19, 2022 at 12:25