Can quantum vacuum carry entropy? So, we know that the state of quantum vacuum does carry energy, as it was measured in the Casimir effect. This energy comes from particles almost instantaneous creation and annihilation. Even if they only exist for a short time, those particle are, for instance, in two possible states, e.g. if $e^-$ and $e^+$ are created, then they can be either in 
$$ |e^-, \uparrow; e^+, \downarrow \rangle $$ 
or 
$$ |e^-, \downarrow; e^+, \uparrow \rangle $$ 
Or in general a superposition of the two. Then, for a very brief moment, the system has a finite entropy(something like $\ln(2) $). 
What you guys think about this?
[UPDATE] It is probably right to state that empty space carry no entropy, since it is a pure quantum state, $S = \text{Tr} (\rho \ln \rho) $. However, to add a bit to the question, i did some research, and  found this paper on
Casimir entropy, where they calculate the entropy of the Casimir effect. Their calculation proves that the force between the plates is of entropic nature, simply because the particles created inside the plates are constrained on the wavelength. 
In this paper, they calculate the entropy of a black hole, and it turns out that most of this entropy comes from 'vacuum'. 
 A: Correct me if I'm wrong, but your line of thinking goes like this... Since quantum fields do not commute in general one can have finite variances for, e.g., particle number. Since the vacuum states defines a probability distribution we can find the corresponding entropy. However, here we are dealing with quantum physics. The entropy is in general $S =-\mathrm{tr}(\rho\ln\rho)$, where $\rho$ is your state. If you have a pure state, $\rho=|0\rangle \langle 0 |$, it is easy to see that the entropy is 0. The vacuum is a pure state and hence its entropy is zero.
What about virtual particles and quantum fluctuations? 
You're essentially imagining that the fluctuations are classical. If these fluctuations were classical you would have a mixed state density matrix whose entropy would not vanish. This would be the case if you performed a measurement then the density matrix would, through decoherence, become a diagonal mixed state in the eigenbasis of the operator being measured. Then the entropy will depend on what you want to measure. 
