I read that if one takes a quantum harmonic oscillator system, not externally driven, and performs a position measurement (measurement in position basis) that reduces the oscillator to an eigenstate of position (or, I suppose, to very "close" to that, so that it is localized between $x + \epsilon$ and $x - \epsilon$ for arbitrarily small $\epsilon$), the system is no longer an oscillator.

Does anybody know why that would be?

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    $\begingroup$ Why would it not be an oscillator? The measurement simply puts it into a mix of excited states, but that doesn't change the Hamiltonian. $\endgroup$ – CuriousOne Mar 10 '16 at 6:35
  • $\begingroup$ See p 70 Arno Bohm (not David Bohm) "Quantum Mechanics" It wasn't clear to me, either, so I am asking books.google.com/… $\endgroup$ – David Mar 10 '16 at 6:42
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    $\begingroup$ may be the following is happening.....The quantum Zeno effect (also known as the Turing paradox) is a situation in which an unstable particle, if observed continuously, will never decay. One can "freeze" the evolution of the system by measuring it frequently enough in its known initial state.....Ref.>espace.library.uq.edu.au/view/UQ:157843/n03chapter2.pdf< $\endgroup$ – drvrm Mar 10 '16 at 7:09
  • $\begingroup$ @ drvrm: Could be. Maybe also the expectation value for its energy approaches infinity, as the localization interval continues to shrink, or the complete uncertainty in momentum is incompatible with energy eigenstates. Not really sure, but was surprised to read it. Never saw that anywhere else. $\endgroup$ – David Mar 10 '16 at 7:15
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    $\begingroup$ After reading the passage I would suggest to repurpose the book as a door stop... that doesn't make any sense, whatsoever. $\endgroup$ – CuriousOne Mar 10 '16 at 7:51

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