Monotonicity of Relative Entropy In Quantum
Field Theory
Edward Witten

Strings 2018

The goal of this lecture is to explain two things:(1) The Reeh-Schlieder theorem (1961), which is the basic result
showing that entanglement is unavoidable in quantum field theory.
(2) A very specific version of monotonicity of relative entropy inquantum field theory: relative entropy is monotonic underdecreasing the size of a region.
The second statement was used by A. Wall in proving a generalized
second law of thermodynamics in the presence of a black hole.
If one is familiar with the proof of this second statement, it is not
too difficult to understand the general proof of strong subadditivityof quantum entropy (which has had many additional applications)and its close cousin, monotonicity of entropy in a general quantum
channel, but we will not have time for this. (This and many related
matters are described in my article “Notes On Some Entanglement
Properties Of Quantum Field Theory,” on the arXiv.)

We begin with the Reeh-Schlieder theorem, which must haveseemed paradoxical when it was first discovered in 1961. We
consider a quantum field theory in Minkowski spacetime M, with a
Hilbert space H that contains a vacuum state Ω. There is analgebra of local operators, whose action can produce “all” states(or at least all states in a superselection sector) from the vacuum.
For simplicity in the notation, we will assume that this operator
algebra is generated by a hermitian scalar field φ(x). So statesφ(x1 )φ(x2 ) · · · φ(xn )|Ωiwith arbitrary n and points x1 , . . . , xn ∈ M, are dense in H.

The Reeh-Schlieder theorem says that actually, we get a dense setof states (in the vacuum sector of H) if we restrict the points
x1 , · · · , xn to any possibly very small open set U ⊂ M:

If this is false, there is a state χ in the vacuum sector such thathχ|φ(x1 )φ(x2 ) · · · φ(xn )|Ωi = 0
whenever x1 , · · · , xn ∈ U.

We will show that any such χ actually satisfieshχ|φ(x1 )φ(x2 ) · · · φ(xn )|Ωi = 0
for all xi ∈ M. Since states created by the φ’s are dense (in thevacuum sector) this implies that χ = 0.

Let us definef (x1 , x2 , · · · xn ) = hχ|φ(x1 )φ(x2 ) · · · φ(xn )|Ωi.
We are given that this function vanishes if the xi are in U and wewant to prove that it vanishes for all xi .

As a first step, pick a future-pointing timelike vector t and consider
shifting xn by a real multiple of t:
xn → xn + ut.
Letg (u) = hχ|φ(x1 )φ(x2 ) · · · φ(xn−1 )φ(xn + ut)|Ωiwith xi ∈ U. We have g (u) = 0 for sufficiently small real u
because then xn + ut is still in u. Also with H the Hamiltonian fortranslation in the t direction,g (u) = hχ|φ(x1 )φ(x2 ) · · · exp(iHu)φ(xn ) exp(−iHu)|Ωi.
Since HΩ = 0 this isg (u) = hχ|φ(x1 )φ(x2 ) · · · exp(iHu)φ(xn )|Ωiand since H is nonnegative, g (u) is holomorphic in the upper half
u-plane.

Such a function is zero. If we knew g (u) to be holomorphic on thereal axis and vanishing on a segment I of the real axis, we would
say that g (u) has a convergent Taylor series expansion around apoint p ∈ I and this expansion would have to be identically zero to
make g (u) vanish on the axis. To begin with we only know thatg (u) is holomorphic above the real axis, not on it, but we can getaround this using the Cauchy integral formula:

u

a)
u

u

b)
γ

γ

u

So now we know that (keeping x1 , · · · , xn−1 ∈ U)hχ|φ(x1 ) · · · φ(xn−1 )φ(xn0 )|Ωivanishes if xn0 = xn + ut where t is a timelike vector and u is any
real number. Now we do this again, picking another timelike vector
t0 and replacing xn0 by xn00 = xn0 + u 0 t0 with u 0 real. Repeating theargument, we learn thathχ|φ(x1 ) · · · φ(xn−1 )φ(xn00 )|Ωi = 0
for any such xn00 . But since any point in M can be reached from U
by zigzagging backwards and forwards in various timelike
directions, we learn thathχ|φ(x1 ) · · · φ(xn−1 )φ(xn )|Ωi = 0
for x1 , · · · , xn−1 ∈ U with no restriction on xn .

The next step is to remove the restriction on xn−1 . We pick t as
before and now consider a common shift of xn−1 and xn in the tdirection0(xn−1 , xn ) → (xn−1, xn0 ) = (xn−1 + ut, xn + ut)

Now we look ath(u) = hχ|φ(x1 )φ(x2 ) · · · φ(xn−1 + ut)φ(xn + ut)|Ωi
It vanishes for small real u, and it can be writtenh(u) = hχ|φ(x1 )φ(x2 ) · · · exp(iuH)φ(xn−1 )φ(xn )|Ωi,
which implies that h(u) is holomorphic in the upper half plane.
Hence h(u) is identically 0.

Repeating the process by shifting xn−1 and xn in some othertimelike direction, we learn thathχ|φ(x1 ) · · · φ(xn−1 )φ(xn )|Ωivanishes for x1 , · · · , xn−2 ∈ U with no restriction on xn−1 , xn . The
next step is to remove the restriction on xn−2 . We do this inexactly the same way, by considering what happens when we shift
the last three coordinates by a common timelike vector. And so on.

So we end up proving the Reeh-Schlieder theorem: an “arbitrary”
state (more exactly, a dense set of states in the vacuum sector of
Hilbert space) can be created from the vacuum by acting with aproduct of local operators in a small open set U ⊂ M.

Now I want to discuss the interpretation of this theorem. The firstquestion that I want to dispose of is whether it contradicts
causality. It certainly sounds unintuitive at first sight. Consider astate of the universe that on some initial time slice looks like thevacuum near U, but contains the planet Jupiter at a distant region
spacelike V separated from U. Let J be a “Jupiter” operator
whose expectation value in a state that contains the planet Jupiterin region V is close to 1, while its expectation value is close to 0otherwise. The Reeh-Schlieder theorem says that there is anoperator X in region V such that the state X Ω contains the planet
Jupiter in region V. So
hΩ|J|Ωi ∼= 0,

hX Ω|J|X Ωi ∼= 1.

Is this a contradiction? Since X is supported in U and J in thespacelike separated region V , X † and J commute. So
1∼= hX Ω|J|X Ωi = hΩ|X † JX |Ωi = hΩ|JX † X |Ωi.

If X were unitary there would be a contradiction between thestatements0∼= hΩ|J|Ωiand1∼= hΩ|JX † X |Ωi,
because if X is unitary, then X † X = 1. But the Reeh-Schlieder
theorem does not tell us that we can pick X to be unitary; it justtells us that there is some X in region U that will create the planet
Jupiter in a distant region V.

In comparing the above formulas, all we have found is that in thevacuum, the operators J and X † X have a nonzero correlation
function in the vacuum at spacelike separation. There is nocontradiction there; spacelike correlations in quantum field theory
are ubiquitous, even in free field theory.

The intuitive interpretation of the Reeh-Schlieder theorem involves
entanglement between the degrees of freedom inside an open set U
and those at spacelike separation from U. To explain the intuitive
picture, let us imagine that the Hilbert space H of our QFT has afactorization
H = HU ⊗ H U 0
where HU describes the degrees of freedom in region U and HU 0
describes all of the degrees of freedom outside of U. Then any
state in H, such as the vacuum state Ω, would have a
decompositionX√Ω=pi ψUi ⊗ ψUi 0
iwhere we can assume the states ψUi and also ψUi 0 to beorthonormal and we assume the pi are all positive (otherwise wedrop some terms from the sum).

In general when we writeΩ=X√pi ψUi ⊗ ψUi 0i

the ψUi and ψUi 0 do not form a basis of HU or of HU 0 , because
there are not enough of them. However, something like theReeh-Schlieder theorem will be true for any state Ω such that theψUi and the ψUi 0 do form bases of their respective spaces. Using thefact that the ψUi 0 are a basis of HU 0 , we would be able to expand
any state Ψ ∈ H asXΨ=λiU ⊗ ψUi 0 ,λi ∈ HU .
i

Then because the ψUi are a basis and the pi are nonzero, we candefine a linear operator X acting on HU by√
X ( p i ψUi ) = λiand we see that we have found an operator X acting only on
degrees of freedom in U such thatX Ω = Ψ.

A stateΩ=X√pi ψUi ⊗ ψUi 0

iwhere the pi are all positive and the ψUi , ψUi 0 are bases might becalled a “fully” entangled state. (I don’t think this is standard
terminology.) We call a state “maximally” entangled if the pi areall equal (this is not possible for Hilbert spaces of infinite
dimension, as in quantum field theory). The Reeh-Schlieder
theorem means intuitively that the vacuum state Ω of a quantum
field theory is fully entangled in this sense, between the inside andoutside of an arbitrary open set U.

However, the decomposition
H = HU ⊗ H U 0that we started with is certainly not literally valid in quantum field
theory. If it were, then in H there would be an unentangled purestate ψ ⊗ χ, ψ ∈ HU , χ ∈ HU 0 .This contradicts the fact that inquantum field theory there is a universal ultraviolet divergence inthe entanglement entropy: the entanglement entropy of thevacuum between degrees of freedom in U and those outside of U is
ultraviolet divergent, and the leading ultraviolet divergence isuniversal, that is it is the same for any state. The leading
divergence is universal because any state looks like the vacuum atshort distances.

Now let us discuss an important corollary of the Reeh-Schlieder
theorem. Let U and V be spacelike separated open sets in
Minkowski spacetime:

Let b be an operator supported in V. Suppose thatbΩ = 0.
Then if a is supported in U, we haveb(aΩ) = abΩ = 0,
where I use the fact that [a, b] = 0 since U and V are spacelike
separated. But the states aΩ are dense in H (according toReeh-Schlieder) so b identically vanishes.

Thus if b 6= 0 is supported in a spacelike open set V that is smallenough that it is spacelike separated from another open set U, thenbΩ 6= 0.
The roles of U and V are symmetrical, so also for a 6= 0 supported
in U,aΩ 6= 0.

Let AU be the algebra of operators in region U. We have proved
two facts about the algebra AU acting on the vacuum sector H:(1) States aΩ, a ∈ AU , are dense in H. This is described by sayingthat Ω is a “cyclic” vector for the algebra AU .
(2) For any nonzero a ∈ AU , aΩ 6= 0. This is described by sayingthat Ω is a “separating” vector of AU .
In short, the Reeh-Schlieder theorem and its corollary say that thevacuum is a cyclic separating vector for AU .

Consider a quantum system with a Hilbert space H = H1 ⊗ H2
and let A be the algebra of operators on H1 . A little thoughtshows that a general vector Ψ expanded in the usual way
X√Ψ=
pi ψ1i ⊗ ψ2ii

is cyclic for A if the ψ2i are a basis of H2 , and it is separating for
A if the ψ1i are a basis for H1 . So that is the meaning of the cyclic
separating property if the Hilbert space is a tensor product.

What about quantum field theory? A mathematical machinerythat can be useful for analyzing entanglement when the Hilbert
space does not factorize is called Tomita-Takesaki theory. It
applies whenever one has an algebra A acting on a Hilbert space
H with a cyclic separating vector.

The starting point in Tomita-Takesaki theory is that, given an
algebra A with cyclic separating vector Ψ, we define an antilinear
operatorSΨ : H → H
bySΨ aΨ = a† Ψ.
The definition makes sense because of the separating property (ifwe could have aΨ = 0 with a† Ψ 6= 0, we would get acontradiction) and it does define SΨ on a dense set of states in H,
because of the cyclic property (states aΨ are dense in H). A
couple of obvious facts are that2SΨ=1

(which in particular says that SΨ is invertible) andSΨ |Ψi = |Ψi.

The modular operator is a linear, self-adjoint operator defined by†∆Ψ = SΨSΨ .

(The definition of the adjoint of an antilinear operator is
hα|S|βi = hβ|S † |αi.) ∆Ψ is positive-definite because SΨ isinvertible.

We will also need the relative modular operator. Let the state Ψbe cyclic separating, and let Φ be any other state. The relative
modular operator SΨ|Φ is an antilinear operator defined bySΨ|Φ a|Ψi = a† |Φi.
Again the well-definedness of the definition depends on the cyclic
separating nature of Ψ, but no property of Φ is needed. In definingSΨ|Φ , we assume that Ψ and Φ are unit vectorshΨ|Ψi = hΦ|Φi = 1.
The relative modular operator is defined by†∆Ψ|Φ = SΨ|ΦSΨ|Φ .

It is still self-adjoint and positive semi-definite, but it is notnecessarily invertible. If Φ = Ψ then the definitions reduce to theprevious ones
SΨ|Ψ = SΨ , ∆Ψ|Ψ = ∆Ψ .

Now we are ready to define relative entropy in quantum field
theory. We fix an open set U (small enough so that the vacuum iscyclic separating), and consider the algebra AU . Let Ψ be any
cyclic separating vector for AU , and Φ any other vector. The
relative entropy between the states Ψ and Φ, for measurements inregion U (as defined by Araki in the 1970’s) isSΨ|Φ (U) = −hΨ| log ∆Ψ|Φ |Ψi.
It is not immediately obvious that this has anything to do withrelative entropy as defined for a general quantum system (and as
used for example in Casini’s proof of the Bekenstein bound), but it
is not hard to show that the definitions coincide. (The main stepin doing this is to understand the operators SΨ|Φ and ∆Ψ|Φ when
the Hilbert space does factorize. See for example section 4 of mynotes.)

First let us discuss positivity properties of relative entropy, definedbySΨ|Φ (U) = −hΨ| log ∆Ψ|Φ |Ψi.
First of all, if Φ = Ψ then we had∆Ψ Ψ = Ψso(log ∆Ψ )Ψ = 0
and hence the relative entropy between Ψ and itself is 0:SΨ|Ψ (U) = 0.
But more than that, suppose that Φ = a0 Ψ, where a0 is unitary and[a0 , AU ] = 0, so that measurements in region U cannot distinguishΦ from Ψ. A short computation shows that in this case∆Ψ|Φ = ∆Ψ so again
SΨ|a0 Ψ (U) = 0.

Now consider a completely general state Φ. The inequality− log λ ≥ 1 − λ for a positive real number λ implies an operator
inequality − log ∆ ≥ 1 − ∆, implying
SΨ|Φ (U) = −hΨ| log ∆Ψ|Φ |Ψi ≥ hΨ|(1 − ∆Ψ|Φ )|Ψi†
= hΨ|Ψi − hΨ|SΨ|Φ
SΨ|Φ |Ψi = hΨ|Ψi − hΦ|Φi = 0.

So in generalSΨ|Φ (U) ≥ 0.

Now we consider a smaller open set Ue ⊂ U. Now we have twodifferent algebras AUe ⊂ AU and two different operators SΨ|Φ;Ue andSΨ|Φ;U and associated modular operators ∆Ψ|Φ;Ue and ∆Ψ|Φ;U . The
relative entropy beween Ψ and Φ for measurements in U isSΨ|Φ (U) = −hΨ| log ∆Ψ|Φ;U |Ψi.
The corresponding relative entropy for measurements in Ue ise = −hΨ| log ∆SΨ|Φ (U)|Ψi.
Ψ|Φ;UeWe want to prove that relative entropy is monotonic underincreasing the region considered:e ≤ SΨ|Φ (U).
SΨ|Φ (U)This is an important statement for applications; I alreadymentioned its use by A. Wall in proving a generalized second law.

The states Ψ and Φ will be held fixed in this discussion, so tollighten the notation we will omit subscripts and denote theoperators just as SU , SUe and likewise ∆U and ∆Ue . The main pointof the proof is to show that as an operator∆Ue ≥ ∆U .
As I will explain in a moment, this implies
log ∆Ue ≥ log ∆U .

(∗)

The inequality we want−hΨ| log ∆Ue |Ψi ≤ −hΨ| log ∆U |Ψi
is just a matrix element of inequality (∗) in the state Ψ.

To show that if P and Q are positive self-adjoint operators and
P≥Q

(∗)

then alsolog P ≥ log Q
let
R(t) = tP + (1 − t)Qso (by virtue of (∗)), R is an increasing function of t, in the sensethat Ṙ(t) ≥ 0. We have

Z ∞ 
1
1
log R(t) =
ds
−
.
s
s +R
0
So

Z ∞
d
1
1
log R(t) =
Ṙ
.
ds
dt
s
+
R
s
+
R
0
The integrand is positive since it is BAB with A, B positive(A = Ṙ, B = 1/(s + R)), so the integral is positive and thusddt log R ≥ 0. Hence R(1) ≥ R(0) or
log P ≥ log Q.

So monotonicity of relative entropy under increasing the region
considered will follow from an inequality∆Ue ≥ ∆U .
If we try to understand this inequality, we may get confused at
first. We have
∆Ue = S †e SUe ,U∆U = SU† SU ,Here the two S’s were defined, naively, by the same formula
SUe aΨ = a† Φ,SU aΨ = a† Ψwith the sole difference that a is in AUe in one case and in AU in
the other. The algebra AU is bigger, so SU is defined on morestates. But states aΨ with a ∈ AUe are already dense in Hilbert
space so actually SUe and SU coincide on a dense set of states.

If one is careless, one might assume that two operators that agree
on a dense subspace of Hilbert space actually coincide. This is nottrue, however, for unbounded operators such as SUe and SU . We
have to remember that an unbounded operator is never defined on
all states in Hilbert space, only (at most) on a dense subspace.
The proper statement is that SU is an extension of SUe , meaningthat SU is defined whenever SUe is defined and, on states on which
they are both defined, they coincide. In our problem, SU is aproper extension, because there are states aΨ, a ∈ AU , that are
not of the form aΨ, a ∈ AUe . Anyway, the fact that SU is anextension of SUe implies, as a general Hilbert space statement, thatSUe † SUe ≥ SU † SU ,which is what we need for monotonicity of relative entropy.

The intuitive idea of the inequality
SUe † SUe ≥ SU † SU
is that the fact that SUe is defined on fewer states than SU is
defined on corresponds to a constraint that has been placed on thestates in the case of SUe , and this constraint raises the energy. I will
give an analogy that aims to make this obvious. Instead of SU , wewill consider the exterior derivative d mapping zero-forms(functions) on a manifold M to 1-forms. But we will assume that
M has a boundary N, and we will consider two different versions of
the operator d.

The first will be the operator d acting on differentiable functions
that are constrained to vanish on the boundary of X :


∂f ∂f
∂f
b
d : f (x1 , . . . , xn ) →,, · · · , n , f |∂X = 0.
∂x1 ∂x2∂x
We also consider the same operator d without the constraint thatf vanishes on the boundary. Differentiable functions that vanish on
the boundary are dense in Hilbert space, so bd and d are eachdefined on a dense subspace of Hilbert space; moreover, obviously,
d is an extension of bd since it is defined whenever b
d is defined andthey agree when they are both defined.

Associated to b
d is the Dirichlet Laplacian
b =b
∆
d† b
d
and associated in the same way to d is the Neumann Laplacian∆ = d† d.
b and ∆ are nonnegative operators that coincide on a dense
Here ∆b is more positive than ∆ because of theset of states, but ∆constraint that the wavefunction should vanish on the boundary.

Indeed, ∆ is associated to the energy functionZ√1hf |∆|f i =dn x g |df |2
2 M
b we should add a boundary term to the energy. In
but to get ∆fact, we can consider a family of operators ∆t , 0 ≤ t ≤ ∞associated to the energy function
Z
Z
√
1
2
n √
hf |∆t |f i =d x g |df | + t
dn−1 x g |f |2 .
2 M
NClearly the operator ∆t is an increasing function of t. For t = 0,∆t is the Neumann Laplacian, and for t → ∞, ∆t goes over to thebDirichlet Laplacian ∆.

b ≥ ∆, which is analogous to our desired ∆ e ≥ ∆U . A
So ∆U
constraint on the state always raises the energy.

Just to make sure the analogy is clear, ∆Ue is the operator
associated to the energy function
hSUe Λ|SUe Λifor a state Λ that should be in the domain of SUe . ∆U is similarly
associated tohSU Λ|SU Λifor a state Λ that should be in the larger domain of SU . The
second energy function is the same as the first except that it isdefined on a larger space of states; we can get the second from thefirst by a constraint that removes some states. Such a constraint
can only raise the energy so ∆Ue ≥ ∆U . (The precise proof uses the
projection onto the graph of an operator. See for example section
3.6 of my notes.)