I A measured spin is still a Qbit

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A Qbit exhibiting an up probability of 1 - (10^-75) can still yield an up reading, indicating that an up result does not equate to a probability of 1. The spin detector, composed of atoms, introduces physical uncertainties, meaning that even if a particle is oriented to produce an up reading, it does not guarantee perfect alignment or the cessation of Qbit spin. The discussion highlights the complexities of quantum measurements and the inherent uncertainties involved. The interplay between probability and physical measurement is emphasized, challenging the notion of absolute certainty in quantum states. Overall, the conversation underscores the nuanced understanding required in quantum mechanics.
Gary Venter
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A Qbit with up probability 1 - (10^-75) would most likely still produce a reading of up, and an up reading does not mean that the up probability is 1.0 to billions of decimal places.
A Qbit with up probability 1 - (10^-75) would most likely still produce a reading of up, so an up reading does not mean that the up probability is 1.0 to millions of decimal places.

The spin detector is a physical device made of atoms. None of its surfaces are pure mathematical planes or lines. Just because it has oriented a particle in the x-direction in a way to produce a reading of up does not mean that it is oriented exactly in that direction with zero uncertainty or has stopped having Qbit spin.
 
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Those are words. What they mean put together like this I cannot fathom.
 
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As Monty Python said "It's not just the total number of words. Getting them in the right order is almost as important."
 

Thread 'Angular Momentum Vector and Its Magnitude'

For the quantum state ##|l,m\rangle= |2,0\rangle## the z-component of angular momentum is zero and ##|L^2|=6 \hbar^2##. According to uncertainty it is impossible to determine the values of ##L_x, L_y, L_z## simultaneously. However, we know that ##L_x## and ## L_y##, like ##L_z##, get the values ##(-2,-1,0,1,2) \hbar##. In other words, for the state ##|2,0\rangle## we have ##\vec{L}=(L_x, L_y,0)## with ##L_x## and ## L_y## one of the values ##(-2,-1,0,1,2) \hbar##. But none of these...
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