On the equivalence of the Schrödinger and Heisenberg (and all other) pictures The Schrödinger and Heisenberg (and, indeed, infinitely many other pictures) are often referred to as equivalent descriptions of quantum dynamics in a given system. I'm wondering two things in particular:

*

*What exactly do we mean by equivalent?


*What is the proof of their equivalence?
Elaborating briefly now, I think that the answer to (1) is simply that they reproduce the same predictions (i.e. probabilities) for all observable experiments. That then begets question number (2); how does one prove that this is indeed the case? In particular, it's usually shown (quite simply) in textbooks that expectation values are preserved under different shifts to different pictures. However, that is not sufficient to say that they are equivalent. Is there a most general statement/proof of this commonly made (and admittedly intuitively expected) claim?
 A: Consider two operators $A$, $B$ such that $B=UAU^\dagger   $
$|e_k\rangle$ and $U$ is unitary. If the eigentstates of $A$ are
$$A|e_k\rangle=\mu_k|e_k\rangle$$
the states $|e'_k\rangle=U|e_k\rangle$ are the eigenstates of $B$ (and vice versa):
$$B|e_k\rangle=UAU^\dagger(U|e_k\rangle)=U\mu_k|e_k\rangle=\mu_k|e'_k\rangle$$
Let $O(t)$ be an observable in the Heisenberg picture and $S(t,t_0)$ the time evolution such that:
$$O(t)=S^{-1}(t,t_0)O(t_0)S(t,t_0)$$
Now let $|n(t)\rangle$ be an eigenstate of $O(t)$, in the Heisenberg picture to get the probability to measure at time $t$ the eigenvalue associated (at any time) with the eigenstate  $|n\rangle$ you calculate:
$$\langle n(t)|\psi(t_0)\rangle$$
Referring to the proof above $S^{-1}=U$ so:
$$ |n(t)\rangle= S(t,t_0)^{-1} |n(t_0)\rangle$$
and:
$$\langle n(t)|\psi(t_0)\rangle =\langle n(t_0)|S(t,t_0)|\psi(t_0)\rangle$$
as in the Schroedinger picture.
A: (1) Indeed, you say it correctly: both representations ("pictures") are equivalent because there is a one-to-one correspondence between their predictions. More precisely, there is an isomorphism whereby any solution in the Schrödinger picture admits one solution and only one solution in the Heisenberg picture.
(2) Regarding the proof, it is based on the uniqueness of the solution of the linear differential equations as in the case of the Schrödinger equation. If $|\psi(t,\boldsymbol{x})\rangle$ is a solution of the Schrödinger equation with given initial conditions, then it follows that $\hat{U}(t)|\psi(0,\boldsymbol{x})\rangle$ is also a solution of the Schrödinger equation with the same initial conditions (where $\hat{U}(t)$ is the unitary group associated with the Hamiltonian):
$$\frac{\text{d}\hat{U}}{\text{d}t} = -\frac{i}{\hbar}\hat{H}\hat{U}, \qquad \qquad \hat{U}(0) = \mathbf{1}$$
