6 ms·
@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>)? Not
by 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 agoWell, 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 agoI don't think that log(a + b) = log(a) log(b) is a rule I'm familiar with!
- filmor 12y agoYou'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 agoYes, 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.