Why is entanglement entropy in QFT infinite? I have been reading some slides of Ed Witten about the Reeh-Schlieder theorem and entanglement in QFT (pdf). In it says that entanglement entropy in quantum field theory have a universal ultraviolet divergence. That is, the entanglement entropy of the vacuum between degrees of freedom in a spacetime region $\mathcal{U}$ and those outside of $\mathcal{U}$ (its causal complement) is ultraviolet divergent, and the leading ultraviolet divergence is universal, that is it is the same for any state.
I don't know if this is a known property of QFT. Can somebody explain why is entanglement entropy in QFT ultraviolet divergent? Does anyone known another reference where this issue is explained or proved?
Thanks in advance.
 A: Consider a lattice with spacing $\epsilon$ and label its sites with $x$, the full Hilbert space of our system is then the tensor product of the Hilbert space at each lattice site: $\bigotimes_x \mathcal{H}_x$
We can now divide our lattice in a region $A$ and its complement $A^c$ where $\partial A $ is the boundary dividing the two surfaces, also called the entangling surface. The Hilbert space associated to region $A$ is the same tensor product of the individual Hilbert spaces but now with the restriction $x \in A$
This procedure leads to the factorisation $\mathcal{H} = \mathcal{H}_A \otimes \mathcal{H}_{A^c}$
If our subregion $A$ has a size of $L \gg \epsilon$ and consists of $N_A$ lattice sites then the entropy can be as large as $N_A \ln(\mathcal{H}_i)$ which is approximately the log of the dimension of $\mathcal{H}_A$.
In this case, when a random state is picked we get
\begin{equation}
S(A) \propto N_A \propto \left(\frac{L}{\epsilon} \right)^{D-1}
\end{equation}
which is called a volume-law growth with $D-1$ the number of spatial dimensions.
However, physical states usually retain some notion of locality so we can expect a short-range entanglement and an entropy dominated by the entanglement across the entangling surface.
\begin{equation}
S(A) \propto \left(\frac{L}{\epsilon} \right)^{D-2}
\end{equation}
which is a area-law growth and is given by the amount of bonds cut by the entangling surface.
Clearly, for most $D$ these diverge in the continuum limit $\epsilon \to 0$.
Which is not that surprising as in a QFT the Hilbert space becomes infinite-dimensional and the amount of fields across the entangling surface grows with its area.
A good introduction is https://arxiv.org/pdf/1907.08126.pdf
A: It is often explained along with the Unruh effect. If you write the Minkowski vacuum state $|0_M\rangle$ in terms of Rindler modes, which only have support on the right $(z > 0)$ and left $(z < 0)$ halves of the $t = 0$ time slice, then
$$
|0_M\rangle = \prod_j (1 - e^{-2 \pi \omega_j})^{1/2} \sum_{n_j = 0}^\infty e^{- \pi n_j \omega_j }|n_j \rangle_{\rm right} \otimes |n_j\rangle_{\rm left}.
$$
A derivation of this formula can be found in many texts on the Unruh effect. Here, $j$ labels all the differnt particle modes. $\omega_j$ is the dimensionless "Rindler energy" of the boost vector field. (Often times, it is instead writte as $\omega_j/a$, where now $\omega_j$ corresponds to the dimensionful energy that would be measured by an observer accelerating with acceleration $a$. ) Here $|n_j \rangle_{\rm right}$ is the state with $n$ quanta in the $j^{\rm th}$ mode on the right half.
One can then compute the density matrix for the right half by tracing out the left half.
$$
\rho_{\rm right} = \mathrm{tr}_{\rm left}(|0_m \rangle \langle 0_M |) = \prod_j (1 - e^{-2 \pi \omega_j}) \sum_{n_j = 0}^\infty e^{- 2\pi n_j \omega_j}|n_j \rangle_{\rm right} \otimes \langle n_j|_{\rm right}.
$$
This takes the form of a thermal density matrix. Because the $\omega_j$ range over all positive real numbers, if one tries to compute the entanglement entropy of this density matrix, they'll find that it diverges due to the unboundedly large $\omega_j$ one is summing over.
This should help guide your reading into the issue.
