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 $p_{B_i}$?


1 Answer 1

  1. Every state which is sensibly described as belonging to a Hilbert space is a pure state. Since you haven't introduced the concept of a mixed state, the desire to refer to only some states in your question as pure probably reflects a misunderstanding. What you get by measuring a complete set of commuting observables is a simultaneous eigenstate of those observables. This is not any more pure than a linear combination of those eigenstates which would itself be an eigenstate of a different set of observables.

  2. To state the Born rule, you need to know that states are vectors which can be projected onto one another with an inner product. But once you can state it, the Born rule alone is what implies (or really just declares) that states encode a probability distribution.

  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.

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    $\begingroup$ I think (the first sentence of) your point 3. is false. Even if two observables do not commute, there can be common eigenvectors. $\endgroup$ May 14, 2023 at 11:59
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    $\begingroup$ Good catch. I just revised it. $\endgroup$ May 14, 2023 at 12:14

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