# State space of a quantum system

I think about the "state space" like this:

• A pure state of a quantum system can be created by measuring a sufficient amount of commuting observables of the system. So one pure state corresponds to a set of commuting observables.

• Those pure states are vectors in the state space of a system.

• The state space is a Hilbert space, thus all linear combinations (including infinite sums) of pure states are also possible states of the quantum system.

• It follows from the first three points (due to each possible state being a superposition of pure states) and "Born rule" that each state provides a probability distribution for each possible observable of the system.

My question is: Is this a right way to think about the "state space" or does this miss out anything?

A follow up question is: Can i think about non-commutating observables in the following way:

Say $$A_i$$ and $$B_i$$ are the possible measurement values for two observables. These observables are non-commutating means:

When we know the probability distribution of a certain state $$|\psi\rangle$$ of the first observable $$p_{A_i}$$ we can deduct from this the probability distribution of the second observable $$p_{B_i}$$?

3. Non-commutation of $$A$$ and $$B$$ means there will be at least one eigenvalue $$a \in \text{Spec}(A)$$ such that a state giving a 100% probability to $$a$$ cannot also give 100% probability to a single $$b \in \text{Spec}(B)$$ or vice versa. If we know $$\left | \psi \right >$$, then the probabilities of $$a$$ and $$b$$ (not both 1) can be computed as $$| \left < a | \psi \right > |^2$$ and $$| \left < b | \psi \right > |^2$$. If we just know $$| \left < a | \psi \right > |^2$$ then there's a single component of the state vector we know (up to a phase) and this is not enough to deduce what any other component of it is.