# 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?

• Technically if you have a degenerate ground state then a mixture of these ground states will again be a state such that $[H,\rho]=0$. However, all eigenstates of a operator are vectors (mathematically speaking) so they will be pure states as long as your QFT is described by an operator acting on a vector space of kets, which you identify as pure states. – Bubble Sep 20 '14 at 11:03
• It is not directly relevant to your question, but there should be a holographic point of view too. Following the classical paper Holographic Derivation of Entanglement Entropy from AdS/CFT, of Ryu, Takayanagi, a thermal state in the $CFT$ (entanglement entropy at finite temperature) is compatible with an area law in the AdS (formula $2.9$). – Trimok Sep 20 '14 at 11:29
• @Trimok Yes, I know the holographic approach, and the extensivity appears for high temperatures. Here I am trying to understand the extensivity and non-extensivity without using holography. – dpravos Sep 20 '14 at 16:35
• At least in $2D$, one may see explicitely the mixed between pure entanglement contributi and thermal contribution, see for instance this paper chapter 2.5, formula $(12)$ and discussion. – Trimok Sep 23 '14 at 10:49
• The fact that the pure state under consideration is the ground state of a local Hamiltonian is crucial. A random pure state taken from the Hilbert space according to the Haar measure has extensive entropy (i.e. increasing proportional to volume until, of course, you reach the point where the region considered becomes as large as half the total system, at which point the the entropy scales like the volume of the complement region). This is Page's conjecture, later proved, and is a consequence of the concentration of measure. arxiv.org/abs/gr-qc/9305007 – Jess Riedel Oct 15 '14 at 19:35

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.

• 1) I have no reference, but it's explained below why it is as I say. I should have specified that this was valid only for highly mixed states, though. For thermal states, the density matrix treats as well-mixed states those with $E<T$ and quenches the ones with $E>T$, then we get an extensive behavior for the former and a vacuum-like (hence, area law) behavior for the latter. The extensive behaviour in the thermal states appears, as you say, as a thermal contribution (i.e., $\sim\text{Vol}(A)T^d$) Now, I don't understand very well how this exactly relates to the extensivity of mixed states. – dpravos Sep 21 '14 at 8:43
• Practical comment: Based on my experience, I would suggest you to be cautious about taking statements at face value, in that line of the literature (EE+holography, etc). I got the feeling that people sometimes overstated the generality of their result which might be based on a very specific computation. – Siva Sep 21 '14 at 16:34