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by filmor 4195 days 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>
1 comments
throwaway183839 4195 days ago
I don't think that

  log(a + b) = log(a) log(b)
is a rule I'm familiar with!
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filmor 4195 days 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.
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quarterwave 4195 days 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.

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