A note on entanglement entropy, coherent states and gravity

Abstract

The entanglement entropy of a free quantum field in a coherent state is independent of its stress energy content. We use this result to highlight the fact that while the Einstein equations for first order variations about a locally maximally symmetric vacuum state of geometry and quantum fields seem to follow from Jacobson’s principle of maximal vacuum entanglement entropy, their possible derivation from this principle for the physically relevant case of finite but small variations remains an open issue. We also apply this result to the context of Bianchi’s identification, independent of unknown Planck scale physics, of the first order variation of Bekenstein–Hawking area with that of vacuum entanglement entropy. We argue that under certain technical assumptions this identification seems not to be extendible to the context of finite but small variations to coherent states. Our particular method of estimation of entanglement entropy variation reveals the existence of certain contributions over and above those of References Jacobson (arXiv:1505.04753, 2015), Bianchi (arXiv:1211.0522 [gr-qc], 2012). We discuss the sense in which these contributions may be subleading to those in References Jacobson (arXiv:1505.04753, 2015), Bianchi (arXiv:1211.0522 [gr-qc], 2012).

Appendix A: Coherent states in curved spacetime

Let (M, g) be a globally hyperbolic d- dimensional spacetime with metric g. Denote a Cauchy slice in (M, g) by $\Sigma$ where $\Sigma$ is a $d-1$ dimensional orientable manifold without boundary. Let ${\mathcal {R}}_I,{\mathcal {R}}_{II}$ be closed $d-1$ dimensional submanifolds of $\Sigma$ such that $\Sigma = {\mathcal {R}}_{I} \cup {\mathcal {R}}_{II}$ and ${\mathcal {R}}_{I}\cap {\mathcal {R}}_{II}={\mathcal {S}}$ where ${\mathcal {S}}= \partial {\mathcal {R}}_{I} =\partial {\mathcal {R}}_{II}$ is a $d-2$ dimensional hypersurface. We restrict attention to the case that ${\mathcal {R}}_I$ is a small geodesic ball [2] and the case $M=R^4$ with ${\mathcal {R}}_I$ the right half plane $z\ge 0$ [6].

Let $\Phi$ be a free scalar field propagating on (M, g). We assume that the free field dynamics on M, g exhibits unique evolution from initial data in any causal domain and that smooth data of compact support evolve to smooth solutions as in Lemma 3.1 of [16]. Let the initial data on $\Sigma$ be $(\phi$, $\pi )$ where $\phi$ denotes the scalar field on $\Sigma$ and $\pi$ its conjugate momentum. Let $\Phi$ be quantized with respect to some choice of complex structure J (i.e. mode decomposition) [16] of this data on $\Sigma$ and let $|0\rangle$ be the vacuum state and ${\mathcal {F}}$ the Fock space, with respect to this choice. Likewise, considering ${\mathcal {R}}_{I}$ and ${\mathcal {R}}_{II}$ as Cauchy slices for their domains of dependence $D_I, D_{II}$, let $J_I, J_{II}$ be complex structures for data on ${\mathcal {R}}_I, {\mathcal {R}}_{II}$ with associated Fock quantizations ${\mathcal {F}}_I, {\mathcal {F}}_{II}$.

Since the field operators ${\hat{\phi }}(x), {\hat{\pi }}(x)$ are operator valued distributions on $\Sigma$, given smooth smearing functions e, f of compact support on $\Sigma$ we have that the operator ${\hat{U}}(e,f)$ is unitary where

$$\begin{aligned} {\hat{U}}(e,f):= e^{i\int _{\Sigma } f(x){\hat{\phi }} (x) - e(x) {\hat{\pi }} (x)} \end{aligned}$$

(6.4)

Define the coherent state $|e,f\rangle$ as

$$\begin{aligned} |e,f\rangle := {\hat{U}}(e,f) |0\rangle \end{aligned}$$

(6.5)

Define ${\hat{\rho }}_I(e,f) = \text {Tr}_{{\mathcal {F}}_{II}}(|e,f\rangle \langle e,f|)$ so that ${\hat{\rho }}_I (0,0)= \text {Tr}_{{\mathcal {F}}_{II}}(|0\rangle \langle 0|)$. We argue below that the entanglement entropy $S(e,f)= -\text {Tr}_{{\mathcal {F}}_I}({\hat{\rho }}_I(e,f)\ln {\hat{\rho }}_I(e,f))$ is the same as the vacuum entanglement entropy $S(0,0)= -\text {Tr}_{{\mathcal {F}}_I}({\hat{\rho }}_I(0,0)\ln {\hat{\rho }}_I(0,0))$ modulo issues of UV divergences.

Define

$$\begin{aligned} {\hat{U}}_I(e_I,f_I)= & {} e^{i\int _{{\mathcal {R}}_I} f_{I} (x)\phi (x) + e_{I}(x) \pi (x)}, \end{aligned}$$

(6.6)

$$\begin{aligned} e_I= e \;\;\;\mathrm{on} \;\;\;{\mathcal {R}}_I&e_I=0 \;\;\;\mathrm{on}\;\;\;{\mathcal {R}}_{II}. \end{aligned}$$

(6.7)

$$\begin{aligned} f_I= f \;\;\;\mathrm{on} \;\;\;{\mathcal {R}}_I&f_I=0 \;\;\;\mathrm{on}\;\;\;{\mathcal {R}}_{II}. \end{aligned}$$

(6.8)

and likewise for $I\leftrightarrow II$.

Let e, f be supported away from ${\mathcal {S}}$. Then we have that ${\hat{U}}_I(e_I,f_I)$ and ${\hat{U}}_{II}(e_{II},f_{II})$ are unitary operators on ${\mathcal {F}}, F_I$ and ${\mathcal {F}}, F_{II}$ respectively and that

$$\begin{aligned} {\hat{U}}(e,f)= {\hat{U}}_{II}(e_{II},f_{II}){\hat{U}}_{I}(e_I,f_I)= {\hat{U}}_{I}(e_{I},f_{I}){\hat{U}}_{II}(e_{II},f_{II}) \end{aligned}$$

(6.9)

We have that

$$\begin{aligned} {\hat{\rho }}_I(e,f)= & {} \text {Tr}_{{\mathcal {F}}_{II}} ({\hat{U}}_{II}(e_{II},f_{II}){\hat{U}}_{I}(e_I,f_I)|0\rangle \langle 0| {\hat{U}}_{I}(e_I,f_I)^{\dagger }{\hat{U}}_{II}(e_{II},f_{II})^{\dagger })\nonumber \\ \end{aligned}$$

(6.10)

$$\begin{aligned}= & {} \text {Tr}_{{\mathcal {F}}_{II}}({\hat{U}}_{I}(e_I,f_I)|0\rangle \langle 0| {\hat{U}}_{I}(e_I,f_I)^{\dagger }) = {\hat{U}}_{I}(e_I,f_I)\text {Tr}_{{\mathcal {F}}_{II}}(|0\rangle \langle 0|){\hat{U}}_{I}(e_I,f_I)^{\dagger }\nonumber \\ \end{aligned}$$

(6.11)

where we have used the invariance of the Trace under unitary tranformations, and the fact that $U_{I}(e,f)$ acts as the identity operator on ${\mathcal {F}}_{II}$ in the last line. The equality of S(e, f) and S(0, 0) then follows, once again from the invariance of the Trace under unitary transformations. Our demonstration suffers from exactly the same UV divergences as in the case of flat spacetime discussed in Sect. 2. That discussion clearly applies here and, while we do not provide an explicit regularization, indicates the any UV regulation softens the sharp boundary ${\mathcal {S}}$ and strongly suggests that the above argument can be made well defined.

Next, let the vacuum state be Hadamard. It is straightforward to check, for example using Sects. 3, 4 of [16], that the two point function in the coherent state $|e,f\rangle$ evaluates to

$$\begin{aligned} \langle e,f| {\hat{\Phi }}(X){\hat{\Phi }}(X^{\prime })|e,f\rangle = \langle 0| {\hat{\Phi }}(X){\hat{\Phi }}(X^{\prime })|0\rangle + H(X)H(X^{\prime }) \end{aligned}$$

(6.12)

where H is scalar field solution obtained from the initial data $(e=\phi (x), f=\pi (x)), \; x\in \Sigma$ and $H(X), H(X^{\prime })$ are the values of H at the spacetime points $X, X^{\prime } \in M$. From our assumptions on free field evolution it follows that H is smooth. Equation (6.12) then implies that $|e,f\rangle$ is also Hadamard for any choice of smooth and compactly supported (e, f). Since we are interested in stress energy expectation value relative to the vacuum and since the stress energy operator is quadratic in the field, we adopt a normal ordering prescription with respect to our choice of annihilation and creation operators on ${\mathcal {F}}$. We are interested in coherent states which are first order departures from the vacuum. For some small positive parameter $\delta$, consider the coherent state $|\delta e, \delta f\rangle$ so that

$$\begin{aligned} |\delta e, \delta f\rangle = e^{i\delta \int _{\Sigma } f(x){\hat{\phi }} (x) - e(x) {\hat{\pi }} (x)}|0\rangle \end{aligned}$$

(6.13)

Expanding the exponential out and noting that the field operators are linear in the annihilation- creation modes, we obtain an expansion of the coherent state in terms of n- particle states with the norm of the m-particle component of order $\delta ^m$. In particular the one particle component is of order $\delta$ so that this coherent state constitutes a first order variation from the vacuum. Since the (normal ordered) stress energy tensor is quadratic in the field operators, the order $\delta$ contribution to its expectation value which is from the off diagonal matrix element between its vacuum component and its 1 particle component, vanishes, so that the leading order contribution is at second order¹⁰

More generally, since the stress energy operator is quadratic in the modes, the only way for any state to be both a first order variation of the vacuum and exhibit a non-trivial normal ordered stress energy expectation value at first order is if its two particle component is of order $\delta$.