Recent content by clacker

so you're saying that if I measure x1 and x2 that put's me into an eigenstate of Q. What about the fact that Q and P commute. If I now measure P haven't I got x1,x2,p1,p2 and now know each particles position and momentum with certainty thus violating the uncertainty principle.

My question is not the measuring of P and Q together, but how do you measure just Q say. If you measure each particles position separately then subtract I don't think you are actually in an eigenstate of Q but rather in 2 separate 1 particle eigenstates.

Question about measuring observables. If have 2 particle system the particle separation Q=x1-x2 and total momentum P=p1+p2 are observables of the system as a whole and are commuting. How do you measure these observables. It would seem the only way to measure the separation is to measure the...

Quantum mechanics says measurement of observable always produces result that is one of eigenvalues of that observable. Subsequent measurement yields same value. For a particle in a box with infinite potential barriers if measure momentum doesn't that put system in eigenstate of momentum insuring...