B Can the probabilities of state vectors |r⟩ and |i⟩ be determined from |ψ⟩?

hongseok
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∣r⟩,∣l⟩,∣i⟩, and ∣o⟩ can all be expressed as expressions for ∣u⟩ and ∣d⟩. So, given the state vector ∣ψ⟩ = α∣u⟩ + β∣d⟩, is it possible to know not only the probability of ∣u⟩ but also the probability of ∣r⟩ and ∣i⟩? ∣ψ⟩ can be expressed as an expression for ∣r⟩, ∣l⟩ or ∣i⟩, ∣o⟩.
∣r⟩,∣l⟩,∣i⟩, and ∣o⟩ can all be expressed as expressions for ∣u⟩ and ∣d⟩. So, given the state vector ∣ψ⟩ = α∣u⟩ + β∣d⟩, is it possible to know not only the probability of ∣u⟩ but also the probability of ∣r⟩ and ∣i⟩? ∣ψ⟩ can be expressed as an expression for ∣r⟩, ∣l⟩ or ∣i⟩, ∣o⟩.
 
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hongseok said:
So, given the state vector ∣ψ⟩ = α∣u⟩ + β∣d⟩, is it possible to know not only the probability of ∣u⟩ but also the probability of ∣r⟩ and ∣i⟩? ∣ψ⟩ can be expressed as an expression for ∣r⟩, ∣l⟩ or ∣i⟩, ∣o⟩.
You haven't said what any if any observable corresponds to any of these vectors, so all we can give you is the general answer:
If we can write $$|\psi\rangle=\alpha_1|A_1\rangle+\alpha_2|A_2\rangle=\beta_1|B_1\rangle+\beta_2|B_2\rangle$$ where ##|A_1\rangle## and ##|A_2\rangle## are orthogonal eigenvectors of the Hermitian operator ##\hat{A}## and likewise for ##|B_1\rangle## and ##|B_2\rangle## and the observable ##\hat{B}## then ...

Yes, we know the probability of getting any of these four possible results if we were to perform the measurement. The catch is that if ##\hat{A}## and ##\hat{B}## do not commute we only get to measure one of them.
 
hongseok said:
TL;DR Summary: ∣r⟩,∣l⟩,∣i⟩, and ∣o⟩ can all be expressed as expressions for ∣u⟩ and ∣d⟩. So, given the state vector ∣ψ⟩ = α∣u⟩ + β∣d⟩, is it possible to know not only the probability of ∣u⟩ but also the probability of ∣r⟩ and ∣i⟩? ∣ψ⟩ can be expressed as an expression for ∣r⟩, ∣l⟩ or ∣i⟩, ∣o⟩.

∣r⟩,∣l⟩,∣i⟩, and ∣o⟩ can all be expressed as expressions for ∣u⟩ and ∣d⟩. So, given the state vector ∣ψ⟩ = α∣u⟩ + β∣d⟩, is it possible to know not only the probability of ∣u⟩ but also the probability of ∣r⟩ and ∣i⟩? ∣ψ⟩ can be expressed as an expression for ∣r⟩, ∣l⟩ or ∣i⟩, ∣o⟩.
The probability that the state is |r> in observation is
|<r|\psi>|^2=|\alpha<r|u>+\beta<r|d>|^2
in condition that all these bras and kets are normalized.
 
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Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. Towards the end of the first lecture for the Qiskit Global Summer School 2025, Foundations of Quantum Mechanics, Olivia Lanes (Global Lead, Content and Education IBM) stated... Source: https://www.physicsforums.com/insights/quantum-entanglement-is-a-kinematic-fact-not-a-dynamical-effect/ by @RUTA
If we release an electron around a positively charged sphere, the initial state of electron is a linear combination of Hydrogen-like states. According to quantum mechanics, evolution of time would not change this initial state because the potential is time independent. However, classically we expect the electron to collide with the sphere. So, it seems that the quantum and classics predict different behaviours!

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