## Tomita-Takesaki in Rindler space

Rindler space provides a convenient example to elucidate some basic properties of AQFT — specifically Tomita-Takesaki theory — in what is arguably the case of greatest interest to high-energy theorists. We shall begin by introducing a few fundamental objects in the theory, and then apply them to Rindler space, loosely following the excellent review by Witten [1], as well as the work of Papadodimas [2] and Papadodimas & Raju [3].

We begin with the ${C^*}$-algebra ${\mathcal{A}_\mathcal{U}}$ of observables in some spacetime region ${\mathcal{U}}$ acting on the Hilbert space ${\mathcal{H}}$ of some quantum field theory (accordingly, we will be concerned with Type III algebras throughout this post). In fact, we shall be slightly more restrictive and assume that ${\mathcal{A}}$ is a von Neumann algebra. Recall that a ${C^*}$-algebra is the set of bounded operators which is closed both in the norm topology and under the adjoint operation. The weak closure of a ${C^*}$-algebra is a von Neumann algebra, which further implies ${(\mathcal{A'})'=\mathcal{A}''=\mathcal{A}}$ by von Neumann’s double commutant theorem. Additionally, we take ${\mathcal{A}}$ to be equipped with the cyclic and separating vector ${\Omega}$. Cyclic means that states ${\mathcal{O}|\Omega\rangle}$, ${\mathcal{O}\in\mathcal{A}}$ are dense in ${\mathcal{H}}$; separating means that ${\mathcal{O}|\Omega\rangle=0}$ iff ${\mathcal{O}=0}$. (Note that, glossing over various representational subtleties, the vector ${\Omega}$ therefore serves as the vacuum state ${|\Omega\rangle}$ of ${\mathcal{H}}$). For concreteness, as well as later convenience, we shall take ${\mathcal{U}}$ to be the right Rindler wedge ${R}$, and denote ${\mathcal{A}\equiv\mathcal{A}_R=\mathcal{A}_\mathcal{U}}$. The commutant is then the algebra in the left Rindler wedge ${L}$, ${\mathcal{A}'\equiv\mathcal{A}_L=\mathcal{A}'_\mathcal{U}=\mathcal{A}_{\mathcal{U}'}}$. (The very last equality is sometimes called Haag duality, and applies when ${\mathcal{U}}$ and ${\mathcal{U}'}$ are causal complements).

Given a von Neumann algebra ${\mathcal{A}}$, Tomita-Takesaki theory admits a canonical construction of the commutant ${\mathcal{A}'}$, which we will use to derive some relationships between the left and right Rindler wedges. The starting point is the antilinear map ${S_\Omega:\mathcal{H}\rightarrow\mathcal{H}}$, defined by

$\displaystyle S_\Omega \mathcal{O}|\Omega\rangle=\mathcal{O}^\dagger|\Omega\rangle~, \ \ \ \ \ (1)$

where recall that an antilinear map ${f:V\rightarrow W}$ between complex vector spaces ${V,W}$ satisfies ${f(aX\!+bY)=a^*f(X)+b^*f(Y)\;\;\forall a,b\!\in\!\mathbb{C}}$ and ${X,Y\!\in\!V}$. Note that ${S_\Omega}$ is a state-dependent operator, in that it is defined in terms of its action on the cyclic and separating state ${|\Omega\rangle}$. To minimize clutter, we shall omit the subscript on ${S}$ henceforth, but one should bear in mind that most of the analysis below does not hold for general (excited) states, which fail the cyclic/separating condition.

Case in point: the fact that ${\Omega}$ is separating is necessary for ${S}$ to be well-defined, otherwise there would exist an annihilation operator ${a\in\mathcal{A}}$ such that ${a|\Omega\rangle=0}$ but ${Sa|\Omega\rangle=a^\dagger|\Omega\rangle\neq0}$, which contradicts the definition (1). The fact that ${\Omega}$ is cyclic then ensures that it maps to a dense set of states on ${\mathcal{H}}$. Additionally, it follows from the definition of the action of ${S}$ on an arbitrary state ${|\psi\rangle=\mathcal{O}|\Omega\rangle}$ that, as an operator, we must have ${S^2=1}$. Note that by the square of the operator, we do not mean ${S^\dagger S}$; rather, this implies that ${S}$ is its own inverse, ${S^{-1}=S}$. It turns out that ${S^\dagger=S'}$, the corresponding operator on the commutant ${\mathcal{A}'}$; see [1] for an elementary proof. Lastly, simply taking ${\mathcal{O}}$ to be the identity operator in (1) shows that ${S}$ leaves the vacuum invariant, ${S|\Omega\rangle=|\Omega\rangle}$.

Since ${S}$ is invertible, it admits a unique polar decomposition,

$\displaystyle S=J\Delta^{1/2}~, \ \ \ \ \ (2)$

where ${J}$ is an antiunitary operator (meaning ${\langle JA,JB\rangle=\langle A,B\rangle^\dagger\;\;\forall A,B\in\mathcal{H}}$) called the modular conjugation, and ${\Delta}$ is a self-adjoint operator called the modular operator. The decomposition (2), combined with the fact that ${\Delta^\dagger=\Delta}$, implies that

$\displaystyle \Delta= S^\dagger S~, \ \ \ \ \ (3)$

which is sometimes taken as its definition. Additionally, since ${\Delta}$ is a positive Hermitian operator, we may define the modular Hamiltonian ${K\equiv-\log(S^\dagger S)}$ such that

$\displaystyle \Delta=e^{-K}~. \ \ \ \ \ (4)$

(N.b., some authors — e.g., Witten [1] — define ${K}$ with an additional factor of ${2\pi}$ such that ${\Delta=e^{-2\pi K}}$. Our notation here is consistent with [2,3]). Since both ${S}$ and ${S^\dagger}$ leave the vacuum state invariant, this property extends to the modular operator as well, ${\Delta|\Omega\rangle=|\Omega\rangle}$. It it important to note that ${\Delta}$ does not have support purely within the right Rindler wedge ${R}$, i.e., on the algebra ${\mathcal{A}}$, but acts on the full Hilbert space defined over both the left and right wedges. This is immediate from (4), and the fact that the domain of ${S^\dagger\!=\!S'}$ is ${\mathcal{A}'}$. We shall return to this point below. (By extension, ${K|\Omega\rangle=0}$, but this does not contradict the fact that ${\Omega}$ is separating since ${K}$ is not localized to any subregion).

For the last of the preliminaries, we note the following useful identities:

$\displaystyle J\Delta^{1/2}=\Delta^{-1/2}J\quad\mathrm{and}\quad J^2=1\,\iff\,J^{-1}=J~. \ \ \ \ \ (5)$

These essentially follow from the fact that ${S^2\!=\!1}$ in conjunction with the polar decomposition. Combining (5) with the aforementioned fact that ${S'=S^\dagger}$ leads to ${J'=J}$ and ${\Delta'=\Delta^{-1}}$. Finally, expressing ${J}$ in terms of ${S}$ and ${\Delta}$ via (2) enables one to show that it likewise leaves the vacuum invariant; i.e., we have

$\displaystyle S|\Omega\rangle=J|\Omega\rangle=\Delta|\Omega\rangle=|\Omega\rangle~. \ \ \ \ \ (6)$

Now, the fundamental result of Tomita-Takesaki theory is comprised of the following two facts: first, the modular operator ${\Delta}$ defines a 1-parameter family of modular automorphisms

$\displaystyle \Delta^{it}\mathcal{A}\Delta^{-it}=\mathcal{A}~,\;\;\forall t\in\mathbb{R}~. \ \ \ \ \ (7)$

In other words, the algebra ${\mathcal{A}}$ is invariant under modular flow. Second, the modular conjugation induces an isomorphism between the algebra and its commutant ${\mathcal{A}'}$,

$\displaystyle J\mathcal{A}J=\mathcal{A}'~, \ \ \ \ \ (8)$

i.e., ${\forall\mathcal{O}\in\mathcal{A}}$, ${J}$ defines an element ${\mathcal{O}'\equiv J\mathcal{O}J}$ such that ${[\mathcal{O},\mathcal{O}']=0}$. The isomorphism (8) is rather remarkable: it allows one to map operators between the left and right Rindler wedges, and features crucially in the works of Papadodimas and Raju on reconstructing black hole interiors [2,3].

To demonstrate how this works, let ${\mathcal{O}\!\in\!\mathcal{A}}$ be a unitary operator with support in ${R}$. This creates a state ${|\psi\rangle=\mathcal{O}|\Omega\rangle}$ which is indistinguishable from the vacuum state for all observers (i.e., all operators ${\mathcal{O}'\in\mathcal{A}'}$) in ${L}$:

$\displaystyle \langle\psi|\mathcal{O}'|\psi\rangle =\langle\Omega|\mathcal{O}^\dagger\mathcal{O}'\mathcal{O}|\Omega\rangle =\langle\Omega|\mathcal{O}'|\Omega\rangle~, \ \ \ \ \ (9)$

where in the last step, we used the fact that ${[\mathcal{O},\mathcal{O}']=0}$ (note that this wouldn’t have worked if ${\mathcal{O}}$ were non-unitary). Now consider the state

$\displaystyle |\phi\rangle=\Delta^{1/2}\mathcal{O}|\Omega\rangle~. \ \ \ \ \ (10)$

Perhaps surprisingly, this state is indistinguishable from the vacuum for an observer in ${R}$, i.e., it represents an excitation localized entirely outside ${R}$! (One has to be careful in saying that it’s localized in ${L}$, since under suitable time evolution it may evolve through the left Rindler horizon). This is straightforward to show, using the properties of the modular conjugation above:

$\displaystyle |\phi\rangle=J^2\Delta^{1/2}\mathcal{O}|\Omega\rangle =JS\mathcal{O}|\Omega\rangle =J\mathcal{O}^\dagger|\Omega\rangle =J\mathcal{O}^\dagger J|\Omega\rangle =\mathcal{O}'|\Omega\rangle~,\ \ \ \ \ (11)$

where ${\mathcal{O}'\equiv J\mathcal{O}^\dagger J}$ is an operator in the commutant. Indistinguishably of ${|\phi\rangle}$ from the vacuum for observers in ${R}$ then follows by the same logic as (9). However, while it is true that the state ${|\phi\rangle}$ is (initially) localized entirely in ${L}$, the same is not true for the operator ${\Delta^{1/2}\mathcal{O}}$! This is due to the fact, mentioned above, that ${\Delta}$ has support in both ${L}$ and ${R}$, and hence the operator ${\Delta^{1/2}\mathcal{O}\notin\mathcal{A},\mathcal{A}'}$. This example highlights the subtle yet crucial distinction between localized states vs. localized operators. Another way to emphasize this is that, as operators,

$\displaystyle \mathcal{O}'\neq\Delta^{1/2}\mathcal{O} \ \ \ \ \ (12)$

(since these exist in different algebras), even though as states,

$\displaystyle \mathcal{O}'|\Omega\rangle=\Delta^{1/2}\mathcal{O}|\Omega\rangle~. \ \ \ \ \ (13)$

as shown in (11). [I am grateful to Kyriakos Papadodimas for clarifying this distinction to me].

Note the similarity to the Reeh-Schlieder theorem at play here: it appears that we’ve created an excitation in ${L}$ by acting with a unitary operator in the spacelike separated region ${R}$; but this isn’t quite true, since we relied on the modular operator ${\Delta}$ to do so, which lives in both ${\mathcal{A}}$ and ${\mathcal{A}'}$. To accomplish this feat purely with operators whose support lies entirely in ${\mathcal{A}}$ would require these operators to be non-unitary, as explained in the eponymous post. This suggests a deep relationship between locality and unitarity embedded within this framework.

Now, let’s make the above formalism a bit more concrete. In the case of Rindler space specifically, it turns out that the modular Hamiltonian is proportional to the generator of Lorentz boosts. Witten [1] offers an intuitive path integral explanation of this fact. The derivation is non-rigorous, since it assumes a factorized Hilbert space ${\mathcal{H}=\mathcal{H}_L\!\otimes\!\mathcal{H}_R}$, but in the course of doing so it provides some insight into where and why this factorization fails, and highlights which operators do and do not exist—the reduced density matrix for either wedge being a prime example of the latter.

First, consider the Euclidean path integral that prepares the vacuum state ${\Omega}$. This is illustrated in figure (a) of the image below, which we’ve taken from [1]. The shaded region represents an integral over the ${\tau<0}$ half-plane which produces the vacuum state ${|\Omega\rangle}$ on the boundary. Similarly, integrating over the upper half-plane produces the bra ${\langle\Omega|}$. Stitching these together by integrating over all boundary values at ${\tau=0}$ yields the density matrix ${\rho=|\Omega\rangle\langle\Omega|}$. As we shall shortly see, this is related to the modular operator ${\Delta}$.

Now pretend that we could factorize the Hilbert space as ${\mathcal{H}=\mathcal{H}_L\!\otimes\!\mathcal{H}_R}$, where the fields ${\phi_L,\phi_R}$ at ${\tau=0}$ are respectively supported in ${x<0}$, ${x>0}$. Strictly speaking, this defines ${L}$ and ${R}$ as open subregions of ${\mathbb{R}^2}$, but taking the weak closure to construct the associated von Neumann algebras entails including the bifurcation point ${x=0}$, as well as the lightlike Rindler horizons themselves. In some sense, this is the source of the problem, since we need to regulate these boundary regions when performing the path integral over either subspace. Specifically, consider the reduced density matrix for the right Rindler wedge,

$\displaystyle \rho_R(\phi_R^+,\phi_R^-)=\int_{\phi_L^+=\phi_L^-}\!\mathcal{D}\phi_L|\Omega(\phi_L^-,\phi_R^-)\rangle\langle\Omega(\phi_L^+,\phi_R^+)|~, \ \ \ \ \ (14)$

which is illustrated in figure (b). The boundary condition ${\phi_L^+\!=\!\phi_L^-}$ indicates that we identify the fields in the upper and lower half-plane when computing the partial trace, while the cut along ${x\!>\!0}$ indicates that ${\phi_R^{\pm}}$ are kept free. The regularization issue mentioned above arises from the fact that local operators at the boundary point ${x\!=\!0}$ are not well-defined, but must be treated as smeared distributions over some finite region. In typical calculations of entanglement entropy, one excises some small region around the endpoint of the cut, which acts as a UV cutoff that regulates the divergent result. [I am grateful to Jamie Sully for discussions on this issue]. But we can’t strictly do this while preserving the (von Neumann) algebraic structure above, which is one reason this procedure doesn’t rigorously hold.

Now consider figure (c), which represents the Euclidean wedge with opening angle ${\theta}$. Leaving both boundary conditions unfixed implies that this computes some operator, which acts as a rotation ${R_\theta}$ in the Euclidean ${\tau\!-\!x}$ plane:

$\displaystyle R_\theta\binom{\tau}{x}=\begin{pmatrix} \cos\theta & \sin\theta \\ -\sin\theta & \cos\theta \end{pmatrix}\binom{\tau}{x} \;\implies\; R_\theta\binom{t}{x}=\begin{pmatrix} \cosh i\theta & -\sinh i\theta \\ -\sinh i\theta & \cosh i\theta \end{pmatrix}\binom{t}{x}~, \ \ \ \ \ (15)$

where in the second expression we’ve gone to Lorentzian time ${t=-i\tau}$. But the latter is precisely a Lorentz boost of the real ${t\!-\!x}$ plane with boost parameter ${-i\theta}$, and we know that the generator of such a boost can be expressed as

$\displaystyle M=\int_{t=0}\!\mathrm{d} x\,x\,T_{00}~. \ \ \ \ \ (16)$

Quite independent of the above analysis, ${M}$ generates a Lorentz boost by a real parameter ${\eta}$, i.e., the operator ${\exp(-i\eta M)}$. Now if the factorization of the Hilbert space were legitimate, we could split ${M=M_R-M_L}$ (with domains of integration restricted to the associated wedges), such that the path integral over the wedge in figure (c) yields the operator ${\exp(-\theta M_R)}$, where ${\eta=-i\theta}$. Such an operator propagates degrees of freedom from the upper boundary of the wedge at ${x\!>\!0}$, ${t\!=\!0}$ to the lower boundary. By setting ${\theta=2\pi}$, one sees that we recover the path integral in figure (b), and therefore

$\displaystyle \rho_R=\exp(-2\pi M_R)~. \ \ \ \ \ (17)$

Of course, the same analysis holds for the left wedge, with ${M_L}$ generating ${\rho_L}$. We then recover the modular operator ${\Delta}$ by appealing to the tensor factor structure, which implies ${[M_L,M_R]=0}$, and hence

$\displaystyle \Delta=\rho_R\otimes\rho_L^{-1}=\exp(-2\pi M_R)\exp(2\pi M_L)=\exp(-2\pi M)=e^{-K}~, \ \ \ \ \ (18)$

where the promised proportionality between the modular Hamiltonian and the generator of Lorentz boosts is ${K\!=\!2\pi M}$ (note that many authors absorb the factor of ${2\pi}$ into the definition of ${K}$).

The formulation in terms of the wedge operator provides a cute illustration of how ${\Delta^{1/2}}$ maps states in ${R}$ to states in ${L}$ and vice-versa, cf. (13). The interpretation is that ${{\Delta^\alpha\!=\!\exp(-2\pi\alpha M)}}$ removes a wedge of angle ${2\pi\alpha}$ from the ${t\!=\!0}$ slice in ${R}$ as depicted in figure (c), and adds an equal wedge in ${L}$ (in the positive half-plane; i.e., it rotates the ${t\!=\!0}$ slice). Since we started with vacuum in ${L}$, one can think of this as rotating excitations in ${R}$ by an angle ${\alpha}$; setting ${\alpha=1/2}$ thus moves an operator insertion from ${R}$ to ${L}$, which is precisely the action of ${\Delta^{1/2}}$ in equation (13) above.

As mentioned above, this path integral picture is not strictly valid since the Hilbert space of Type III factors does not admit a tensor product structure. One way in which this manifests is that ${K\!\in\!\mathcal{H}}$ cannot be split into ${K_{L,R}=2\pi M_{L,R}}$ in such a way as to yield operators whose support lies solely in ${L,R}$. This implies that the reduced density matrices ${\rho_{L,R}}$ do not really exist! Given how much of our understanding of entanglement in quantum field theory is based on the reduced density matrix — and more recently, at least in certain holographic applications, the modular Hamiltonian ${K_{L,R}}$ itself — this is in principle a rather serious problem. It remains to be seen how much of our ideas about entanglement must accordingly be revised, or whether this is analagous to Haag’s theorem about the non-existence of the interaction picture: theoretically damning, but practically irrelevant.

To see how one might get around this issue, consider two arbitrary, well-defined states ${{|\psi_1\rangle,\,|\psi_2\rangle}}$. The non-existence of the operator ${K_R}$ manifests in a universal UV divergence in ${K_R|\psi_1\rangle}$ at ${x\!=\!0}$, as explained above. Nonetheless, the matrix elements ${\langle\psi_2|K_R|\psi_1\rangle}$ are still well-defined. An intuitive argument for this is that in the energy eigenbasis, correlation functions between well-defined (read: finite) states simply don’t see the problematic UV modes. [I am grateful to Ben Freivogel for this explanation]. Of course, one could certainly inquire as to the ontological status of matrix elements of an operator which does not technically exist. But, at least at an epistemic level, this offers a potential back-door to entanglement entropy. That is, taking the trace of the reduced density matrix ${\rho_{L,R}}$ as is typically done is not a rigorous method for computing entanglement, since as discussed above ${\rho_{L,R}}$ is not a valid operator in the theory. However, insofar as entanglement is essentially a measure of correlation functions, one could imagine performing the sum over all possible correlators instead. While the resulting sum may still be divergent, the individual terms are well-defined, and this may lend some support to the practical validity of many existing results despite the false foundation.

References

1. E. Witten, “Notes on Some Entanglement Properties of Quantum Field Theory,”, arXiv:1803.04993
2. K. Papadodimas, “A class of non-equilibrium states and the black hole interior,” arXiv:1708.06328
3. K. Papadodimas and S. Raju, “State-Dependent Bulk-Boundary Maps and Black Hole Complementarity,” arXiv:1310.6335
This entry was posted in Physics. Bookmark the permalink.