While explaining 'quantum state' to a beginner, is it scientifically accurate to say that "just like '$v$' represents velocity and '$p$' represents the momentum of an object, $|ψ\rangle$ represents the quantum state, which kind of encompasses everything there is to know (spin, orbital number, momentum, etc.) about the quantum object"?

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1This question doesn't ask what you are asking, But the answers might help you. Does the collapse of the wave function happen immediately everywhere? – mmesser314 Dec 25 '21 at 10:02
1 Answers
Pretty much. A slightly more accurate description is to compare it with the classical notion of a state. Namely, the pair $(x,p)$ formed by the positions and momenta of all particles in the system. Notice that with this data you can compute whatever you want in Classical Mechanics (energies are given in terms of these, so are angular momenta, so are velocities, etc). Furthermore, these are the initial conditions we use to solve the equations of motion.
As you said, $\vert \psi \rangle$ plays a similar role: if you have a state, you can compute the expectation value of any observable you want. Given $\vert\psi\rangle$ you can obtain $\langle\hat{p}\rangle$, for example, despite not being able to obtain an exact value for the momentum due to the Uncertainty Principle. Nevertheless, the state also allows you to compute the uncertainty through $\Delta p^2 = \left\langle\hat{p}^2 - \langle\hat{p}\rangle^2\right\rangle$. Hence, knowing the state means knowing everything about the system (up to quantum uncertainty).

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