# 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?

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 $$$$S(A) \propto N_A \propto \left(\frac{L}{\epsilon} \right)^{D-1}$$$$ which is called a volume-law growth with $$D-1$$ the number of spatial dimensions.
However, physical states usually retain some notion of locality so we can expect a short-range entanglement and an entropy dominated by the entanglement across the entangling surface.
$$$$S(A) \propto \left(\frac{L}{\epsilon} \right)^{D-2}$$$$ which is a area-law growth and is given by the amount of bonds cut by the entangling surface.
Clearly, for most $$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

• This is the kind of answer I was looking for. Thanks. I have a little doubt: can you explain a little more the first equation? I read through the paper you linked but didn't understand that approximation. Commented Feb 14, 2021 at 23:01
• Consider a hypercube with side $L$, then its total volume is $L^{D-1}$. Now you can imagine that since the spacing between lattice sites is $\epsilon$ each lattice site takes up a volume of a hypercube with side $\epsilon /2$. Hence the total amount of lattice sites within $A$ is just the volume divided by the volume one lattice site takes up, which is $\propto \epsilon^{D-1}$ Commented Feb 15, 2021 at 9:17
• I am sorry, the hypercube around a lattice site also has $\epsilon$ as its length. Commented Feb 15, 2021 at 11:19
• Thanks, now I get it. Commented Feb 15, 2021 at 13:34

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.

I think the best way of making sense of this divergent entanglement entropy is to work with von Neumann algebras. Typical QFTs are type III von Neumann algebras, due to which the definition of a trace is no longer valid. As Guliano discussed in their answer, the area-law growth best depicts why lattice settings are used. Since the entanglement entropy scales like $$$$S(A)\sim \frac{1}{\epsilon ^{D-2}}\;,$$$$ the continuum limit $$\epsilon \to 0$$ becomes the usual setting for QFTs. When $$\epsilon \neq 0$$, one gets a type I algebra for the QFT, and entanglement entropy is meaningful. In the vanishing $$\epsilon$$ limit, the type III nature comes back, and one no longer has good" entanglement entropy. A good instance of this situation is the Ryu-Takayanagi formula for holographic entanglement entropy in AdS/CFT, where one has a cut-off $$\epsilon$$, due to which the formula becomes $$S\sim \text{Area of RT surface}$$. In the vanishing cut-off scenario, one would simply have divergent entanglement entropy. There are more technicalities to this but this suffices for this answer.