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What does it mean for 2 quantum states to be "observationally indistinguishable"?

If I may venture a guess: Does that mean that the set of possible measured values are the same though the probability of measuring each eigenstate does not necessarily have to be equal?

References will also be appreciated. Thanks.

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In what context did this notion arise ? Are you sure it is "observationally indistinguishable" states, bud not sets of states ? – Frédéric Grosshans Jun 4 '12 at 21:25
@FrédéricGrosshans: I was reading something online and came across this... unfortunately I don't seem to be able to relocate it... what would the definition be if it were applied to "sets of states"? and is it defined for simply "2 states"? Actually, I think by "states" the article meant a superposition of eigenstates, if that is what you mean... – Jose Jun 4 '12 at 22:06
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The expression "Observationally indistinguishable" is sometimes used (although rarely) in the context of "Gauge symmetry" and its breakdown, Please see for example: Ward Struyve article. This concept is related to the question here: "Gauge symmetry is not a symmetry" and used to advocate the use of gauge invariant quantities such as holonomies in the description of gauge theories.

According to this principle orbits of the gauge potentials under gauge transformations must be considered as "Observationally indistinguishable", and only gauge invariant quantities can be considered as "Observationally distinguishable".

For example, the standard heuristic explanation of the Higgs mechanism involves gauge variant quantities. Now, if gauge symmetry is not a symmetry, then what is the meaning of gauge symmetry breakdown. An explanation in Struyve's article is given through an example of the Higgs mechanism within the framework of classical field theory which involves only gauge invariant quantities.

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I doubt that this phrase has a rigorous and commonly agreed-upon definition. Nonetheless, I will venture a guess as to what it ought to mean. Given an algebra of observables $\mathcal{A}$, then two states $\rho$ and $\sigma$ are observationally indistinguishable with respect to $\mathcal{A}$ if and only if for all $A \in \mathcal{A}$, $$\mathrm{Tr}(A\rho) = \mathrm{Tr}(A\sigma) .$$ I can't imagine it could mean anything else.

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