Consider three mutually commuting observables from Mermin's square, A_1=σ_x⊗I, A_2=I⊗σ_x, A_3=σ_x⊗σ_x, where σ_x is the Pauli operator and I is the identity operator. We can see that R=(A_1)(A_2)(A_3)=I.
According to the Non-Contextual Realist model v(R)=v(A_1)v(A_2)v(A_3) and is valid for both simultaneous and subsequent measurements. Here v(X) is the outcome of individual measurement of X. As R=I, v(R)=v(I)=+1 only. This implies, v(A_1)=1/(v(A_2)v(A_3)). In any measurement the outcome is random, it can be either +1 or -1.
But in case we measure all the three observables simultaneously, the outcomes of at most two observables can be random because the outcome of other observable is constrained by the previous equation and is therefore definite.
According to this analysis, in the case of subsequent measurements, once we measure two observables the outcome of third one can be known even without measuring it.
But does it really happen?
And please let me know if there are any mistakes in my interpretation.
Yes the third outcome is known definitely if the first two outcomes are known because all the three states are entangled. A measurement of the state causes a collapse of the of wavefunction, and since the three particles describe a "system" they collapse into an eigenstate.
Try reading Mermin's paper: "does the moon exist if no one looks at it" for a good insight into a similar problem but for two states. After that you can try GHZ's paper: "Bell's theorem without inequalities" for the three state system you mentioned.
Hi Jaskaran,
I think you have misunderstood my question. I am considering three observables and not three particles. These observables could correspond only to a single particle which has a Hilbert space of dimension four. Please have a look at it again.
Still do read Mermin's paper and then GHZ's paper. It addresses your question very nicely. There are many things that cannot be written here, and many more which 'we' might not be able to convey to you. And you are also doing the same mistake Neumann made: assuming that observables will commute and will have simultaneous eigenvalues.
However the answer to your question is that such a scenario does happen in quantum mechanics.
The question , according to Kumar, is correct. Two spin measurements relating two different particles always are commutative. This is a difefernt situation respect to the case to consider three observables of the same quantum object. The question is easily solved if you look at a non standard formulation of quantum mechanics , as example as in a Clifford algebraic formulation as in my papers and realizing an appropriate basis with three basic elements. It is not so difficult to answer in this case.
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