How should this be possible? We should start with the simple case first, where we have position as an observable, e.g., for a massive particle as an electron. In contradistinction to position time is never an observable, but a parameter in QT. The reason is simply that otherwise ##\hat{t}## would be the conjugate observable to ##\hat{H}##, the Hamiltonian (usually representing energy), and then both ##\hat{t}## and ##\hat{H}## would have both ##\mathbb{R}## as their spectra. That's, because from the commutator relations between two observables (from here on I use the sloppy language and identify observables with their representing self-adjoint operators) ##[\hat{A},\hat{B}]=\mathrm{i} \hbar \hat{1}##, you can proove that then both ##\hat{A}## and ##\hat{B}## have the entire real numbers as spectra.
For time and energy that's in clear contradiction to observations for at least two reasons:
(a) our world is pretty stable, i.e., the the energy is bounded from below. If this were wrong, any system could get to ever lower energy states, and nothing would be stable. That was Pauli's argument against time being an observable in his famous Handbuch article about wave mechanics.
(b) one of the key successes of early QM was the explanation for the spectral lines of many atoms beyond hydrogen, i.e., it solved the very puzzle, which was among the strongest motivations for physicists to look for a "new quantum mechanics", going beyond the Bohr-Sommerfeld quantization prescriptions, which worked for hydrogen (and the harmonic oscillator or rotator), but for nearly nothing else.
So space and time are distinct on a very fundamental level. Time is an parameter, and the Hamiltonian describes the dynamics of the system. According to the formalism of quantum theory, what's observable in principle are probabilities for the outcome of measurements on ensembles of equally prepared quantum systems: For a complete set of compatible observables ##\{\hat{A}_1,\ldots,\hat{A}_n \}## there are 1D eigenspaces, i.e., the possible outcomes when measuring these compatible observables are ##(a_1,\ldots,a_n)## with the ##a_j## in the spectra of ##\hat{A}_j##. Then let ##|t,a_1,\ldots,a_n \rangle## be a arbitrary set of common normalized eigenvectors. These are all determined up to a phase factor. The same holds for the state vector ##|\psi(t) \rangle##, which describes the initial preparation of the system.
The time evolution of the observables ##\hat{A}_j(t)## and ##|\psi(t) \rangle## is pretty arbitrary, i.e., there are two unitary operators ##\hat{A}(t)## and ##\hat{C}(t)## describing the time evolution,
$$\hat{A}_j(t)=\hat{A}(t) \hat{A}_j(0) \hat{A}^{\dagger}(t), \quad |\psi(t) \rangle=\hat{C}(t) |\psi(0) \rangle.$$
Note that for the eigenvectors the time evolution is
$$|a_1,\ldots,a_n;t \rangle=\hat{A}(t) |a_1,\ldots,a_n; 0 \rangle,$$
so that for all ##t##
$$\hat{A}_j(t) |a_1,\ldots,a_n;t \rangle = a_1 |a_1,\ldots,a_n; t \rangle.$$
The unitary operators ##\hat{A}(t)## and ##\hat{C}(t)## fulfill the equations of motion
$$\dot{\hat{A}}(t)=\frac{\mathrm{i}}{\hbar} \hat{X}(t) \hat{A}(t), \quad \dot{\hat{C}}=-\frac{\mathrm{i}}{\hbar} \hat{Y}(t) \hat{C}(t),$$
and the only thing necessary for the self-adjoint operators ##\hat{X}(t)## and ##\hat{Y}(t)## is that
$$\hat{X}(t) + \hat{Y}(t)=\hat{H}(t).$$
The Schrödinger picture is defined by ##\hat{X}(t)=0## and ##\hat{Y}(t)=\hat{H}##, the Heisenberg picture in the opposite way ##\hat{X}(t)=\hat{H}## and ##\hat{Y}(t)=0##.
Then there are many "intermediate pictures", which split ##\hat{H}## in any arbitrary, convenient way. E.g., to do "perturbation theory" you choose ##\hat{X}=\hat{H}_0##, where you can solve the equation of motion for ##\hat{A}## exactly and then ##\hat{Y}=\hat{H}_1## to do a perturbative expansion in powers of ##\hat{H}_1##. For scattering theory it's often convenient to choose ##\hat{H}_0## as the Hamiltonian of free particles (or fields) and ##\hat{H}_1## as the part that describes interactions.
No matter, how you choose the picture of time evolution, the observable (probabilistic) facts about the system come out the same (at least as long as you solve everything exactly), i.e., the probabilities,
$$P(t,a_1,\ldots,a_n)=|\langle a_1,\ldots,a_n;t|\psi(t) \rangle|^2,$$
are always the same no matter how you split ##\hat{H}## in ##\hat{X}(t)## and ##\hat{Y}(t)##. So are all other derived quantities like expectation values,
$$\langle O(t) \rangle=\langle \psi(t)|\hat{O}(t) \psi(t).$$
If you take as a complete set of observables for a single (spin-0) particle the three components of the position vector you get the wave function,
$$\psi(t,\vec{x})=\langle x(t)|\psi(t) \rangle,$$
which is also invariant under a change of the picture of time evolution, modulo an overall phase factor, which never occurs in any calculation of observable quantities like probabilities etc.
The reason is that the transformation from one picture to another is a unitary transformation ##\hat{B}_{12}## for the states and the observables, i.e., if you have two splits,
$$\hat{H}=\hat{X}_1(t) +\hat{Y}_1(t)=\hat{X}_2(t) +\hat{Y}_2(t),$$
then the states and observables transform from one to the other picture with
$$|\psi_2(t)=\hat{B}_{12}(t) |\psi_1(t) \rangle, \quad \hat{O}_{2}(t)=\hat{B}_{12}(t) \hat{O}_{1}(t) \hat{B}_{12}^{\dagger}(t),$$
where
$$\hat{B}_{12}(t)=\hat{C}_2(t) \hat{C}_1^{\dagger}(t) =\hat{A}_2(t) \hat{A}_1^{\dagger}.$$
The latter equality follows from the fact that
$$\hat{U}_1(t)=\hat{A}_1^{\dagger}(t) \hat{C}_1(t)$$
and
$$\hat{U}_2(t)=\hat{A}_2^{\dagger}(t) \hat{C}_2(t)$$
both fulfill the same equation of motion,
$$\partial_t \hat{U}=-\frac{\mathrm{i}}{\hbar} \hat{H} \hat{U},$$
and the same initial condition ##\hat{U}(0)=\hat{1}##, such that they are equal.
