Let's say I'm given a two state system that consists of base states $|1 \rangle$ and $|2 \rangle$, those being eigenstates of an hermitean operator $\hat{O}$ that commutes with the hamiltonian, and as well being eigenstates of the hamiltonian, with eigenvalues $o_1$ and $o_2$, or $h_1$ and $h_2$, respectively.
In the Schrödinger picture, the state is evoluted according to the Schrödinger equation. I begin with a state $|\Psi_0 \rangle$ that is evoluted to a state $|\Psi(t)\rangle$.
The time-independent operator $\hat{O}$ defines $|1\rangle$ and $|2\rangle$ as its eigenstates, and then I can write $|\Psi(t) \rangle = c_1(t) |1 \rangle + c_2 |2(t) \rangle$. So in the Schrödinger picture, we know the the relative phases of $|1\rangle$ and $|2\rangle$, and because $c_i(t) = e^{\hbar h_i t}$
In the Heisenberg picture, with our complete set of observables (that commutes with the hamiltonian), $|\Psi_0\rangle$ won't change during time evolution. $\hat{O}$ won't change as well, it commutes with the hamiltonian, and thus its eigenstates $|1(t)\rangle = |1\rangle$ and $|2(t)\rangle = |2\rangle$ won't change either.In this picture, and using only our "complete set of observables" to deduce the time evolution, we can't deduce the time evolution of the relative phases of $|\Psi_0 \rangle = c_1 |1\rangle + c_2 |2\rangle$.
Does that mean our complete set of observables doesn't contain everything that is knowable about the system? Or am I making a mistake somewhere?
Best Answer
I'm not sure what the OP means by "complete set of observables". But what I can say is that (1) in the Heisenberg equations of motion you can in principle* determine the expectation value of any operator ($\sigma_x$, $\sigma_y$, $\sigma_z$, $\sigma_x^2$, $\sigma_x \sigma_y$, etc.) at any moment in time. Any repeated measurement will be related to some combination of such expectation values. (2) I'm pretty sure that knowledge of all of these expectations values is equivalent to knowledge of the wavefunction. That the wavefunction allows you to predict the expectation value of such operators is clear. That the wavefunction can be inferred from the expectation values is more challenging but I believe it is so. I believe it's related to a quantum generalization of Bochner's Theorem. See also this question of mine: Quantum Probability, what makes quantum characteristic functions quantum?
To answer the OPs question: Does knowing information about some set of observables that commutes with the Hamiltonian give us as much information about the state of the system as the wavefunction? No. You also need information about some observables that don't commute with the Hamiltonian for that.
*it may be computationally difficult