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

  • $\begingroup$ 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? $\endgroup$ Jan 18, 2022 at 17:02
  • $\begingroup$ 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. $\endgroup$
    – Marion
    Jan 19, 2022 at 8:14
  • $\begingroup$ But you understand this change of basis is not included in the expectation value you wrote, a mere matrix element, right? $\endgroup$ Jan 19, 2022 at 14:36
  • $\begingroup$ Sure. Totally understood. $\endgroup$
    – Marion
    Jan 20, 2022 at 10:12

1 Answer 1


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.

  • $\begingroup$ Thanks. Essentially what I am asking is as to wether there are any ceveats to this. You are saying something very fundamental, the expectation value of the observable does not depend on the basis you measure it. I am wondering, are there any potential exceptions? For example, some people discuss a relaxation of unitarity sometimes. $\endgroup$
    – Marion
    Jan 19, 2022 at 8:16
  • $\begingroup$ @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. $\endgroup$ Jan 19, 2022 at 12:25
  • $\begingroup$ Thanks a lot. This was helpful. $\endgroup$
    – Marion
    Jan 20, 2022 at 10:13

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