In $(d+1)$-dimensional quantum systems described by conformal field theories at criticality, I have been under the impression that the entanglement entropy is described by a non-universal area law plus universal logarithmic corrections.

This area+log is true in $(1+1)$-dimensional systems, which has $$S(l) = \frac{c}{3}\ln \left(\frac{L}{\pi} \sin{\frac{\pi l}{L} }\right) +C$$ as seen in the transverse field ising model in another question, though please see another question of mine about this when $c$ is large compared to the local degrees of freedom. Note that the area law in $1d$ is just a constant law.

Fradkin and Moore in "Hearing the Shape of a Quantum Drum" showed that in $(2+1)$-dimensional systems, the area+log statement is still true, though the log correction can sometimes vanish depending on the geometry of the region:

$$S(R) = 2f_s(l/a) + \alpha c \log(l/a) + O(1)$$

where $l$ is the length of the boundary of the region $R$, $a$ is the ultraviolet cutoff, $f_s$ is the non-universal leading area-law term, and there is the requisite universal $c \log(l/a)$ term. In said paper, they speculate that the $O(1)$ term is not universal. $\alpha$ here depends on the shape of the region $R$ and can be $0$.

However, I was recently perusing a paper that noted that for a a (d-1 + 1)-dimensional theory, the entanglement entropy of a spherically shaped region with radius $r$ is

$$S(r) = \mu_{d-2}r^{d-2} + \mu_{d-4}r^{d-4} + ... + \begin{cases} (-1)^{d/2-1}4A\log(r/a), & \text{if $d$ is even} \\ (-1)^{(d-1)/2}F, & \text{if $d$ is odd} \end{cases} $$ where this last term is the universal term. Note that the above predicts a logarithmic term in the case $(1+1)$-dimensions with $d=2$ but not in $(2+1)$ dimensions with $d=3$.

This seems to partially contradict Fradkin and Moore's result above. While it's likely that the lack of a log term in $(2+1)$-dimensions for a sphere comes from that particular geometry suffering $\alpha=0$ in Fradkin and Moore's result, it seems that the $O(1)$ piece is indeed universal.

Is it that in $(2+1)$-dimensions both the logarithmic and constant terms are universal? Or is the constant term only universal in some geometries of the region? How about for other dimensions?


1 Answer 1


Indeed, it has to do with the geometry of the entangling region. In (2+1)d, when the boundary of the region is smooth (e.g. a disk region in (2+1)d), the subleading correction to the area law is a universal constant and is related to the free energy of the theory on $S^3$. If the region contains sharp corners (e.g. rectangle), the subleading term is logarithmic in $l$. In that case, the constant piece is not universal as it can be absorbed into $\ln (l/a)$, by redefining $a$. But the coefficient of the log is a universal function of the opening angles of the corners.

The higher-dimensional result you quote applies to sphere boundary, which is all smooth. When there are singularities on the boundary additional logarithmic corrections occur.

  • $\begingroup$ This is perfect! I hadn't realized that when there's a nonzero logarithmic term, the constant piece is no longer universal because of the $\ln(a)$ part of the logarithmic term. It's clear in hindsight but I was missing this completely. Please let me know if you have any insight on my question on $1+1$-dimensional systems with large central charge $c$. $\endgroup$
    – user196574
    Dec 7, 2021 at 22:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.