Absolutely! (whoops- accidental pun)
When you can make a list of possible distinct (orthogonal) quantum states i=1,2,3.... that some system might be in, consistent with your knowledge of it, with probabilities P1
,...., then the entropy sigma = sum(Pi
)). For the simple case where you can make a complete list of N equally likely states (each with P=1/N) that just becomes ln(N).
Some caveats: If you pick some basis states in which particular phases of superpositions are more likely than others, a more general form is
-Trace (rho ln(rho)) where rho is the density matrix.
Often entropy is given in traditional units where you take this fundamental dimensionless sigma and multiply by Boltzmann's constant k to get S=k*sigma.
Now why is this quantum entropy 'absolute'? We've given a complete exact specification for it in terms of whatever is known about some system. Classical entropy, in contrast, always had to deal with the problem of how to divide up the continuum of possible classical states into some discrete collection. There's an arbitrary constant involved in how finely you divide up that continuum. So classical entropy differences could, in some cases, be calculated, but never classical absolute entropies.
(published on 10/03/06)