Re:Wave-particle duality is the uncertainty principle in dis...
by filmor on Saturday December 20, 2014 @03:33AM
(#8776160)
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>
Re:Wave-particle duality is the uncertainty principle in dis...
by throwaway183839 on Saturday December 20, 2014 @08:15AM
(#8776557)
I don't think that
log(a + b) = log(a) log(b)
is a rule I'm familiar with!
Re:Wave-particle duality is the uncertainty principle in dis...
by filmor on Saturday December 20, 2014 @08:36AM
(#8776599)
You're right, I messed that one up. But as
log(a + b) /= log(a) + log(b)
in general, the result is still correct.
Re:Wave-particle duality is the uncertainty principle in dis...
by quarterwave on Saturday December 20, 2014 @11:47AM
(#8777082)
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.
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.