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Physics
Quantum Physics
The clarity of Heisenberg uncertainty
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[QUOTE="vanhees71, post: 6846688, member: 260864"] That's NOT what the Heisenberg uncertainty states. It's a misinterpretation, unfortunately found in most popular-science publications and even in some textbooks. The quantum state operationally describes a preparation procedure. E.g., you use some trap to localize an electron in some region of space. The meaning of the quantum state is to provide the probablities for the outcome of measurements on the particle. You can measure any observable of the particle as accurately as you wish. There are no fundamental restrictions to the accuracy from quantum mechanics. E.g., you can measure the momentum or the position of the electron as precisely as you wish. The Heisenberg-Robertson uncertainty principle then states that for any quantum state (preparation of the system) the standard deviations of two observables ##A## and ##B## represented by the corresponding self-adjoint operators ##\hat{A}## and ##\hat{B}## obey the uncertainty relation, $$\Delta A \Delta B \geq \frac{1}{2} |\langle [\hat{A},\hat{B}] \rangle|.$$ For components of the position and momentum in the same ##x##-direction this implies $$\Delta x \Delta p_x \geq \frac{\hbar}{2}.$$ As I said you can measure ##x## as precisely as you wish or ##p##. To test the uncertainty relation you have to prepare many particles in the same state (e.g., putting it in a trap with a certain spatial extent) and measure very accurately its position and do the statistical analysis to get the standard deviation ##\Delta x##. To ensure that you measure the true uncertainty due to quantum fluctuations you must ensure that the resolution of your position measurement is much smaller than this quantum-mechanical standard deviation. Then you can again prepare many particles in the same state and do the same procedure with the momentum measurement. Another question is the influence of the measurements on the system, i.e., the disturbance of the system due to measurements. This is a much more complicated issue than the Heisenberg-Robertson uncertainty relation, which can be derived after a few weeks in the QM1 introductory lecture. For more on the measurement-disturbance relations, see P. Busch et al, Quantum Measurement, Springer (2016) [URL]https://dx.doi.org/10.1007/978-3-319-43389-9[/URL] Also the issue of how to measure incompatible observables at the same time on a single particle. For this you have to extent the description of measurements from the standard idealized von Neumann (projective) measurements to socalled positive-operator-valued-measure measurements, which is also a quite complicated issue. For more on this, see A. Peres, Quantum Theory: Concepts and Methods, Kluwer Academic Publishers, New York, Boston, Dordrecht, London, Moscow (2002). It's due to the nature of the quantum system itself, as detailed above. [/QUOTE]
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Quantum Physics
The clarity of Heisenberg uncertainty
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