What happens when we measure spin of a fermion?

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Measuring the spin of a fermion in the direction of a unit vector results in either +0.5 or -0.5, indicating that the fermion adopts an eigenstate aligned with that direction. This aligns with the principles outlined in Bell's paper on the Einstein-Podolsky-Rosen paradox. The discussion confirms that the measurement does not yield a value of 0.5 cos θ, as initially questioned. Participants express appreciation for clarifying the concept, particularly in relation to understanding Bell's work. The focus remains on the implications of spin measurement in quantum mechanics.
andrewkirk
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I am trying to understand the measurement of spin, in order to understand Bell's paper on the Einstein-Podolsky-Rosen paradox.

When we measure the spin of a fermion in the direction of unit vector a, will the result be:

1. a value of either +0.5 or -0.5, and upon measurement, the fermion adopts an eigenstate of the spin operator in which the spin is aligned with a

or

2. a value of 0.5 cos θ, where θ is the angle between a and the direction of spin the particle has adopted as a result of the measurement, and θ may be anywhere in [0, π)


or is it something else?

Thank you for any help.
 
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1. a value of either +0.5 or -0.5, and upon measurement, the fermion adopts an eigenstate of the spin operator in which the spin is aligned with a
This one is correct.
 
Thank you Bill.
I meant to say thanks earlier, but I got a bit carried away reading the Bell paper when I found I could actually understand it.
 
Time reversal invariant Hamiltonians must satisfy ##[H,\Theta]=0## where ##\Theta## is time reversal operator. However, in some texts (for example see Many-body Quantum Theory in Condensed Matter Physics an introduction, HENRIK BRUUS and KARSTEN FLENSBERG, Corrected version: 14 January 2016, section 7.1.4) the time reversal invariant condition is introduced as ##H=H^*##. How these two conditions are identical?

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