B Difference between opposite states

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Let's say there are two 1/2 spin particles, one in state

1/√2 |up> + 1/√2 |down>

and the other in the state

- 1/√2 |up> - 1/√2 |down>

Both particles then have an equal chance of being measured to be in either the up or down states. Is there any physical difference between these or are they indistinguishable?
 
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These are the same state. Only the relative phase matters.
 
You can take out a global phase factor of -1 or ( e^(i*pi) ) from the second one and return the first one you mentioned. (By which I simply mean you can take out a factor of -1). This phase factor does not effect something called the expectation values of the Hermitian operator, which means, that the possible outcomes of either system should be the same. Or in this case, these two vector spaces are the same.
 
A pure state is represented not by a vector but by a ray in Hilbert space or, equivalently, a projection operator ##\hat{P}_{\psi}=|\psi \rangle \langle \psi |## with a normalized vector ##|\psi \rangle##. The most general state is given by an arbitrary statistical operator, i.e., a positive semidefinite self-adjoint operator with trace 1.
 
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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