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This seems to be a key reference (article titled "Complementarity and uncertainty"): http://www.nature.com/nature/journal/v375/n6530/pdf/375367b0.pdf http://ww

This seems to be a key reference (article titled "Complementarity and uncertainty"): http://www.nature.com/nature/journal/v375/n6530/pdf/375367b0.pdf http://www.nature.com/nature/journal/v375/n6530/pdf/375367b0... In particular, they have a description of why they find the traditional connection between uncertainty and wave-particle duality to be unsatisfactory. Unfortunately the technical details are beyond me, but hopefully someone else can sort them out? Edit: Okay, some more digging suggests the following story: (i) We have wave-particle duality => Heisenberg uncertainty, but not the other way around (ii) With entropic uncertainty [1] (stronger than Heisenberg uncertainty), the authors show that we can obtain wave-particle duality Again, can somebody more knowledgeable can clarify? :) [1] http://en.wikipedia.org/wiki/Entropic_uncertainty http://en.wikipedia.org/wiki/Entropic_uncertainty

quarterwave 12y ago
@obastani : very good comment. Aside: If |Psi> is a state with wavefunction <x|Psi>, then is there a linear operator A such that <x|A|Psi> = log(<x|Psi>)? Note: Logarithm of a complex function.
- filmor 12y ago
  Well, the logarithm is all but linear. If there was such an A you'd have <x|A(|Psi> + |Chi>) = log(<x|(|Psi> + |Chi>)) = log(<x|Psi> + <x|Chi>) = log(<x|Psi>) log(<x|Chi>) /= <x|A|Psi> + <x|A|Chi>
  - throwaway183839 12y ago
    I don't think that log(a + b) = log(a) log(b) is a rule I'm familiar with!
    - filmor 12y ago
      You're right, I messed that one up. But as log(a + b) /= log(a) + log(b) in general, the result is still correct.
      - quarterwave 12y ago
        Yes, agree that log() wouldn't result from a linear operator. The idea behind my question: does the Shannon entropic integral correspond to the L2 length of some projected state? Leading to a prescription to prepare a state with minimum uncertainty product of canonically conjugate physical quantities. Afaik, in classical statistical physics the log() shows up when the N! in a binomial probability distribution limits to large N via the Stirling approximation. It would be interesting to find a different route for log() to enter the picture from a quantum standpoint. All admittedly vague and hand waving speculation.

quarterwave 12y ago

filmor 12y ago
Well, the logarithm is all but linear. If there was such an A you'd have <x|A(|Psi> + |Chi>) = log(<x|(|Psi> + |Chi>)) = log(<x|Psi> + <x|Chi>) = log(<x|Psi>) log(<x|Chi>) /= <x|A|Psi> + <x|A|Chi>
- throwaway183839 12y ago
  I don't think that log(a + b) = log(a) log(b) is a rule I'm familiar with!
  - filmor 12y ago
    You're right, I messed that one up. But as log(a + b) /= log(a) + log(b) in general, the result is still correct.
    - quarterwave 12y ago
      Yes, agree that log() wouldn't result from a linear operator. The idea behind my question: does the Shannon entropic integral correspond to the L2 length of some projected state? Leading to a prescription to prepare a state with minimum uncertainty product of canonically conjugate physical quantities. Afaik, in classical statistical physics the log() shows up when the N! in a binomial probability distribution limits to large N via the Stirling approximation. It would be interesting to find a different route for log() to enter the picture from a quantum standpoint. All admittedly vague and hand waving speculation.