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Heisenberg Uncertainty Principle states (in the form of the Robertson-Schroedinger Formula) that measurement of two non-commuting observables has a limiting precision, even for flawless measurement devices. Okay, no problem with this. But what about a measurement on a single observable, does Quantum Mechanics imposes some fundamental limit?

p.s.: Googling around I found this paper which states "that an intrinsic quantum uncertainty on a single observable is ineludible in a number of physical situations", talking mainly about bipartite systems.

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Quantum mechanics does not really postulate a limit on the possible precision of measurements of a single observable. Indeed, when you prepare a system in an eigenstate of a given observable, you will always observe the corresponding eigenvalue, in principle without any uncertainty. This must of course be qualified for variables like position whose values range over a continuum, for which states with any finite precision can in principle be prepared.

The paper you quote applies only to entangled bipartite states. This is made clear, for example, in the caption of figure 1: "Quantum correlations trigger local quantum uncertainty". In a bipartite system, it can indeed happen that if one requires the state of both parts to be entangled, then some local observables may be forced to always have nonzero uncertainty. However, you can also prepare unentangled states for which this does not apply.

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  • $\begingroup$ so, although there is no principle/relation that limits precision, I can conclude that (depending on the state and on the observable measured) the error bar of a real measurerement contains "quantum uncertainty"? $\endgroup$ – Rodrigo Thomas Jan 19 '14 at 22:28
  • $\begingroup$ Yes, it can be so concluded. $\endgroup$ – Emilio Pisanty Jan 19 '14 at 22:31

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