The equivalence between Heisenberg and Schroedinger pictures

I will try to make it as simple and intuitive as possible. In the Schrödinger picture, the expectation value of a given operator $\hat{\xi}$ (which itself is frozen in time) is defined as follows (with $\psi(t)$ the wavefunction of our system at time $t$):

$$\langle \hat{\xi} (t) \rangle = \langle \psi (t) \lvert \hat{\xi} \rvert \psi(t) \rangle$$

Which is just the average value of the observable corresponding to $\hat{\xi}$ if a measurement is made at time $t.$ Now exactly because the expectation value creates a direct link between what we predict with our theory in QM with what we observe experimentally, then logically however one goes about defining quantum mechanics, we should obtain the same values for $\langle \hat{\xi} (t) \rangle$ to ensure that we're going to predict the correct experimentally expected values (and hence be able to claim then that the two pictures are equivalent).

To show this equivalence, we first use an important property of the unitary time evolution operator, namely

$$\psi(t_1) = \hat{U}(t_1,t_0) \psi(t_0)$$

i.e. we propagate our wavefunction in time be acting $\hat{U}$ on it. With this, we can now redefine the wavefunction at time $t$ as its value at time $t=0$ upon which we act $\hat{U}(t,0).$ So we rewrite (by a simple substitution) our original expression for $\langle \hat{\xi} (t) \rangle$ as:

$$ \langle \psi (t) \lvert \hat{\xi} \rvert \psi(t) \rangle = \langle \psi (0) \lvert \hat{U}^{\dagger}(t,0) \hat{\xi} \hat{U}(t,0)\rvert \psi(0) \rangle $$ From the above you can already see the freedom of choice, i.e. to decide to act the time operators either on the wavefunctions or on the operator, by choosing the latter we get:

$$ \langle \psi (0) \lvert \left(\hat{U}^{\dagger}(t,0) \hat{\xi} \hat{U}(t,0)\right)\rvert \psi(0) \rangle = \langle \psi (0) \lvert \hat{\xi}(t) \rvert \psi(0) \rangle $$ Hence we have successfully shown that the time dependence can also be implemented in the operators, instead of wavefunctions while obtaining the same expectation values for our chosen observable, so let's call $\psi(0) = \psi_h$ with $h$ for Heisenberg, and similarly $\hat{\xi}(t) = \hat{\xi}_h(t).$ With this notation then you can easily relate the operator in the Schrödinger picture with that of the Heisenberg picture by:

$$\hat{\xi}_h(t)=\hat{U}^{\dagger}(t,0) \hat{\xi}_{\rm Schrödinger} \hat{U}(t,0)$$ Finally, from here you can straightforwardly obtain the expression of Heisenberg's equation of motion (although you didn't ask for it, but we've come all this way, may as well show it...):

Take the time derivative of $\hat{\xi}_h(t)$ (using the last equation derived) and by using the relation $d\hat{U}/dt=-\frac{i}{\hbar}\hat{H}\hat{U}$ (and also that $[\hat{H},\hat{U}]=0$):

$$ \begin{align*} \frac{d\hat{\xi}_h (t)}{dt} &= \frac{d\hat{U}^{\dagger}}{dt} \hat{\xi} \hat{U} + \hat{U}^{\dagger} \hat{\xi}\frac{d\hat{U}}{dt} \\ &= \frac{-1}{i\hbar}(\hat{U}^{\dagger}\hat{H}\hat{\xi}\hat{U}-\hat{U}^{\dagger}\hat{\xi}\hat{H}\hat{U})\\ &=\frac{1}{i\hbar}[\hat{\xi}_h (t),\hat{H}]. \end{align*} $$

On their non-equivalence. Yes, this is largely a folklore result. There are many ways for them to be non-equivalent. A few examples

https://arxiv.org/abs/1404.6775

https://www.sciencedirect.com/science/article/abs/pii/S0375960102015086

https://arxiv.org/abs/0706.3867

More generally, in curved spacetimes, the Heisenberg Picture treats all coordinates on equal ground, while the Schrödinger Picture has, as a precondition, that there be a universal time variable with respect to which states evolve. There may not be or (if there is) it may lead to a fundamental inconsistency. Indeed, the Problem of Time is, itself, is that very inconsistency: a proof by contradiction that the condition is false and that, hence, no Schrödinger Picture exists at all. So, they are inequivalent in that setting.

The formal equivalence of the two pictures also neglects half the foundation of quantum theory itself. There are not one but two von Neumann postulates to consider: the Evolution Postulate (states evolve in accordance with the Schrödinger equation) and the Projection Postulate (a state upon measurement coughs up an eigenvalue and collapses to an eigenstate, in accordance with Born's Rule). It seems everybody keeps forgetting about that other postulate.

The equivalence of the picture only applies to the first postulate. The Heisenberg Picture version of Evolution, of course, being the Heisenberg Equations. There is no equivalence between the two pictures for the second postulate -- because there is no Heisenberg Picture version of the Born Rule at all! If you try to formulate one, you will see revealed an interesting new infrastructure, that is not present in the Schrödinger Picture, but which is required to properly handle multiple applications of the Born Rule in the Heisenberg Picture. Contained in it is a distinguished "now" and a sense of time flowing with respect to it. But the "flow" is not within the Evolution postulate; rather it is stemming from the Projection postulate!

The question of what the Born Rule is and how it is to be handled, interpreted, explained or explained away is the crux of what's called the Measurement Problem. The different answers to this question then produce the different Interpretations of quantum theory (Bohm, Many Minds, Many Worlds, Consistent Histories, Physical Collapse, each of which may be threaded by the analyses provided of Decoherence).

Here, too, there is a gap. The same question asked of the Born Rule is now passed down to each of their putative replacements: what's the Heisenberg Picture version? And is there even one at all? For instance: Many Worlds and Bohm.