In QFT we have an action of the restricted Lorentz group which is implemented via a unitary transformation. In other words, if $\Lambda\in SO(1,3)^\uparrow$, then the corresponding unitary operator is $U(\Lambda)\colon \mathcal{H}\to \mathcal{H}$ sends $|\psi\rangle\to U(\Lambda)|\psi\rangle$.
Suppose now that I have an operator $\hat{O}\colon \mathcal{H}\to \mathcal{H}$. If I perform a Lorentz transformation, should this operator also change? Namely, should I also have $\hat{O} \to U(\Lambda)\hat{O}U(\Lambda)^\dagger$? I ask this because I have the following conceptual problems.
Perspective I: Some books say that, when I have a matrix element $\langle \phi|\hat{\mathcal{O}}|\psi\rangle$ and I perform a Lorentz transformation, I simply set $|\phi\rangle \to U(\Lambda)|\phi\rangle$ and $|\psi\rangle \to U(\Lambda)|\psi\rangle$, having thus
$$ \langle \phi|\hat{\mathcal{O}}|\psi\rangle\to \langle \phi|U(\Lambda)^\dagger \hat{\mathcal{O}}U(\Lambda)|\psi\rangle$$
and they say that we can also view this as a transformation of the operator
$$ \hat{\mathcal{O}}\to U(\Lambda)^\dagger \hat{\mathcal{O}}U(\Lambda)$$
Basically this is the same thing we do in QM when we switch from the Schrödinger to the Heisenberg picture. In the former we let states evolve in time and operators stay fixed, whereas on the second we fix states and let operators evolve. Here I just discovered I have a conceptual issue. Why do we not evolve both operators and states in time? This is related to my second perspective.
Perspective II: If I have an operator $\hat{\mathcal{O}}$ acting on a state $|\psi\rangle$, Peskin says that, since $\hat{\mathcal{O}}|\psi\rangle \to U(\Lambda)\hat{\mathcal{O}}|\psi\rangle$, then we can view the operator $\hat{\mathcal{O}}$ transforming as $\hat{\mathcal{O}}\to U(\Lambda)\hat{\mathcal{O}}U(\Lambda)^\dagger$. Is this because
$$ \hat{\mathcal{O}}|\psi\rangle \to (U(\Lambda)\hat{\mathcal{O}}U(\Lambda)^\dagger)(U(\Lambda)|\psi\rangle)= U(\Lambda)\hat{\mathcal{O}}|\psi\rangle$$
or does the state remain invariant under this transformation? Should it not remain invariant, then the matrix element would be invariant, namely
$$ \langle \phi|\hat{\mathcal{O}}|\psi\rangle\to \langle \phi|U(\Lambda)^\dagger (U(\Lambda)\hat{\mathcal{O}}U(\Lambda)^\dagger)U(\Lambda)|\psi\rangle = \langle \phi|\hat{\mathcal{O}}|\psi\rangle$$
Comments: under a Lorentz transformation, which of the two perspectives is correct? How do operators transform, as $U(\Lambda)\hat{\mathcal{O}}U(\Lambda)^\dagger$ or as $U(\Lambda)^\dagger\hat{\mathcal{O}}U(\Lambda)$? Do both states and operators transform, or only one of them? If I had
$$ \langle \phi| \hat{P}_\mu |\psi\rangle$$
since it has a Lorentz index, I would expect it not to be invariant under Lorentz transformations, thus discarding Perspective II. This would leave me with Perspective I. But, if I decompose the operator as:
$$\hat{\mathcal{O}} = \sum_{I}\mathcal{O}_i |\delta_i\rangle\langle \delta_i| $$
then since $|\delta_i\rangle \to U(\Lambda)|\delta_i\rangle$, then
$$ \hat{\mathcal{O}} = \sum_{I}\mathcal{O}_i |\delta_i\rangle\langle \delta_i| \to\sum_{I}\mathcal{O}_i U(\Lambda)|\delta_i\rangle\langle \delta_i|U(\Lambda)^\dagger = U(\Lambda)\hat{\mathcal{O}}U(\Lambda)^\dagger$$
namely the operator would transform (which contradicts Perspective I) and it would transform as depicted under Perspective II, which would make me reconsider its validity.
Conclusion: As you can see, I am a bit lost. Could some please help me clarify these concepts? I am aware of the question Confusion on symmetry and basis transformation, but this one talks explicitly about a change of basis. It is clear that under a change of basis $|\psi\rangle$ should not change, because it is a vector, What I think they derive there is the fact that components transform contravariantly (as one already knows), so I do not see this question of help.
