Is the ground state of a QFT always a pure state? And excited states are mixed? I am studying entanglement entropy.
It's fullfilled for any local quantum system that the entanglement entropy of a region $A$ in a highly mixed state is extensvie,
$$ S_A \sim \frac{\text{Vol}(A)}{\epsilon^d} $$
where $\epsilon$ is the length between sites (or the UV cutoff for a regularized QFT). This is because 
$$S_A=\log[\text{dim}(\mathcal{H}_A)]$$
where $\mathcal{H}_A$ is the Hilbert space of the degrees of freedom in $A$, and the dimension scales as the number of degrees of freedom per site by the number of the sites, $\sim \exp(\text{Vol}(A)/\epsilon^d)$.
On the other hand, for a pure state the entanglement entropy is not extensive, since $S_A=S_B$, where $B$ is the complement of $A$. In fact it is proven that it follows an 'area law'.
An usual intuituve argument for this area law is that to compute the entanglement entropy we count the pairs entangled in both sides of the boundary of $A$. Since the theory is local, the most entangled pairs will be those who lay at $\sim\epsilon$ of the boundary, while the further sites won't count for the entropy. So this way we have that the EE must scale as $\text{Vol}(\partial A)$.
My problem is that I don't know if I am understanding why are we counting all the states un $A$ in one case and only the states near to the boundary in the other.
The reason I found is that in the case of the pure state we are talking about the vaccum, the system is in the ground state of energy and a site can only see what is near around it, since the interactions are local.
But if we excite the system (for example, if we consider a thermal state $\rho = e^{-\beta H}/\text{tr }e^{-\beta H}$, which is a mixed state), the system has enough energy to go beyond the local behavior and then we need to consider all the sites for the EE.
Am I right with that?
 A: 
It's fullfilled for any local quantum system that the entanglement entropy of a region A in a mixed state is extensvie,

Could you give a reference, please?
The entanglement entropy measures not just the entanglement but also the classical entropy. It is a measure of entanglement, only if you start with a pure state and then trace over some subsystem. (It's been a while since I read the literature on this, but)IIRC, when people talk about mixed (thermal) states, there is a classical/thermal entropy of your mixed state that scales extensively -- while the entanglement (described using some measure other than the naive entanglement entropy) might scale differently. IMHO, I would expect that to scale with an area law, just like you said. The entanglement entropy is a sum of those two, and it looks like the "leading" behaviour is extensive.

On the other hand, for a pure state the entanglement entropy is not extensive, since SA=SB, where B is the complement of A. In fact it is proven that it follows an 'area law'.

I don't see how one can say this for a generic pure state. This is believed to hold (strong evidence adding up) only for ground states of local field theories.
Unless you have a homogeneous (unbroken translational symmetry) and isotropic+boost-invariant (unbroken Lorentz symmetry), there are a lot of qualifiers before you talk about scaling of entanglement entropy. EG: where your region sits, it's shape, the frame you're sitting in and then its size. The vacuum of a relativistic field theory respects all these symmetries and hence, is convenient to consider.
For generic pure states (eg: localized lump as an excitation on top of the vacuum) the area scaling will of course be affected when the boundary of your "expanding region" crosses the lump. So you cannot talk about a general monotonic "scaling behaviour" for such states, even if they're pure.
Thus, to my understanding, it is difficult to make any definitive statements about the zoo of excited (albeit pure) states. Once can always consider one particular state and compute entanglement for that case, but I don't see how that will generalize to statements about other states.
