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Sorry if this is a silly question (engineer here), but I was wondering if the math in particle physics assumes that unitarity applies even between measurements. In other words, I take it that the evolution of quantum states is governed by an operator that ensures the probability of all possible events adds up to 1 at all times. What I'm wondering -- is there an operator for which this does not apply at all times but still gives measurement results with probability between 0 and 1?

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You are asking if you can violate unitarity in intermediate steps? THe answer is yes, using ghosts. –  Ron Maimon Sep 2 '12 at 22:16
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Yes, you can make a unitary asymptotic S-matrix (so asymptotic measurements) when the intermediate states do not evolve in a unitary way. This is what ghost fields do--- the intermediate states in ghost-descriptions include negative probability objects, but when you make asymptotic measurements you don't see the ghosts, you only see the positive probability objects.

In cases where you have a ghost description, there are often no-ghost formulations, like light-cone or axial gauges. In these formulations, the Hamiltonian is well defined, so that you can ask about measurements on the intermediate states and get well defined answers. These formulations have a reduced symmetry compared to the ghost formulation, but they are manifestly unitary.

In ghost formulations, you assume that every measurement is made on asymptotic states which have no ghosts. Even if it is not true that every measurement is of an S-matrix quantity, the existence of the unitary formulation guarantees that anything you build out of asymptotic states will only end up measuring a quantity which has a reasonable positive probability interpretation.

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But these intermediates involving ghosts are only virtual, without an associated state vector. Any state vector must always evolve unitarily, as this is dictated by Schroedinger's equation, which remains valid in QFT. –  Arnold Neumaier Sep 3 '12 at 9:01
    
@ArnoldNeumaier: The ghosts are "virtual" for S-matrix calculations, the fields themselves have wavefunctions, and the ghost fields intermediate states evolve nonunitarily. –  Ron Maimon Sep 3 '12 at 17:46
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The physical wave functions are functionals of all fields, annihilating the BRST operator $Q$. These evolve unitarily according to Schroedingers equation. The Hamiltonian is the generator of the time translations of the corresponding unitary representation of the Poincare group defined on $Ker~Q$. There is no room for a separate non-unitary dynamics of some states - unless they are unphysical states without any relevance. –  Arnold Neumaier Sep 3 '12 at 17:55
    
@ArnoldNeumaier: You can consider the full state space to include unphysical states which are not annihilated by the BRST operator, and consider the BRST condition as an asymptotic condition on incoming states. The nonunitary dynamics is on states which are not physical, I agree, but they are fine intermediate states. The question was about theories where intermediate states are not unitary, but asymptotic states are. –  Ron Maimon Sep 4 '12 at 3:41
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yes, but the full state space is physically irrelevant, as only states annihilated by $Q$ can be prepared or observed. Therefore they are also the only states that exist at finite time, i.e., between measurements. - The others are just mathematical constructs. They may appear in perturbative calculations, but not in the dynamics. –  Arnold Neumaier Sep 4 '12 at 7:09
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