On both the CMV and the IMV the members of E are (most likely) literally false since the quantum state of the universe is (most likely) not an eigenstate
of pointer positions, etc.
57], "it is quite legitimate to write down an eigenstate
of energy at a well-defined time".
12) So on the orthodox interpretation, the set of truths about a system's intrinsic state at a time is exhausted by listing its state-independent properties (mass, charge, spin-type), the eigenvalues of the observables of which it is in an eigenstate
, and the expectation values for all observables of which it is not in an eigenstate
But the two-particle system is in an eigenstate
for the particles' having opposite values for spin when measured along this axis.
We are asked to imagine a being able to carry out a measurement of an observable for a system in an eigenstate
of the corresponding operator simply by means of a single particle in the being's brain whose position becomes correlated with the state of the system.
With [alpha] = 0 the left hand side is just the Hamiltonian with the plane wave its eigenstate
The expression that an observable 'has a particular value' for a particular state is permissible in quantum mechanics in the special case when a measurement of the observable is certain to lead to the particular value, so that the state is an eigenstate
of the observable .
But this is not the case, in virtue of the fact that a superposition and a eigenstate
are states on equal footing.
1) and (2)) because, as Cartwright argues, he never believes in the reduction of wave packet but only the transition from one eigenstate
This just means that all but a finite number of the systems in this infinite array are in the spin-up eigenstate
of spin in the z-direction.
In order to actually realize such a which-path experiment we need to calculate the joint probability associated with a recording of the photon in the state |x} (or |k>) in coincidence with a measurement of the detector in the eigenstate
|[lambda]} corresponding to one of its observable.
1]>) as this would not result in an eigenstate
of the strongly spin-dependent Hamiltonian, entering the energy eigenvalue equation