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Center for Particle Astrophysics

arXiv:hep-th/9303048v2 20 Mar 1993March 8, 1993CfPA–93–02LBL–33754hep-th/9303048Entropy and Area∗Mark Srednicki†Center for Particle AstrophysicsUniversity of CaliforniaBerkeley, CA 94720andTheoretical Physics GroupLawrence Berkeley Laboratory1 Cyclotron RoadBerkeley, CA 94720ABSTRACT:The ground state density matrix for a massless free field istraced over the degrees of freedom residing inside an imaginary sphere; the resultingentropy is shown to be proportional to the area (and not the volume) of the sphere.Possible connections with the physics of black holes are discussed.∗This work was supported in part by the Director, Office of Energy Research, Office of High Energyand Nuclear Physics, Division of High Energy Physics of the U.S. Department of Energy under ContractDE-AC03-76SF00098, and in part by the National Science Foundation under Grant Nos. AST91-20005 andPHY91-16964.† E-mail: mark@tpau.physics.ucsb.edu.

On leave from Department of Physics, University of California,Santa Barbara, CA 93106.

This document was prepared as an account of work sponsored by the United States Government. Neitherthe United States Government nor any agency thereof, nor The Regents of the University of California, norany of their employees, makes any warranty, express or implied, or assumes any legal liability or responsibilityfor the accuracy, completeness, or usefulness of any information, apparatus, product, or process disclosed, orrepresents that its use would not infringe privately owned rights.

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A free, massless, scalar, quantum field (which could just as well represent, say, theacoustic modes of a crystal, or any other three-dimensional system with dispersion rela-tion ω = c|⃗k| ) is in its nondegenerate ground state, |0⟩. We form the ground state densitymatrix, ρ0 = |0⟩⟨0|, and trace over the field degrees of freedom located inside an imaginarysphere of radius R. The resulting density matrix, ρout, depends only on the degrees of free-dom outside the sphere.

We now compute the associated entropy, S = −Tr ρout log ρout.How does S depend on R ?Entropy is usually an extensive quantity, so we might expect that S ∼R3. However,this is not likely to be correct, as can be seen from the following argument.

Consider tracingover the outside degrees of freedom instead, to produce a density matrix ρin which dependsonly on the inside degrees of freedom. If we now compute S′ = −Tr ρin log ρin, we wouldexpect that S′ scales like the volume outside the sphere.

However, it is straightforward toshow that ρin and ρout have the same eigenvalues (with extra zeroes for the larger, if theyhave different rank), so that in fact S′ = S [1]. This indicates that S should depend onlyon properties which are shared by the two regions (inside and outside the sphere).

Theone feature they have in common is their shared boundary, so it is reasonable to expectthat S depends only on the area of this boundary, A = 4πR2. S is dimensionless, so to geta nontrivial dependence of S on A requires another dimensionful parameter.

We have twoat hand: the ultraviolet cutoffM and the infrared cutoffµ, both of which are necessaryto give a precise definition of the theory. (For a crystal, M would be the inverse atomicspacing, and µ the inverse linear size, in units with ¯h = c = 1.) Physics in the interiorregion should be independent of µ, which indicates that perhaps S will be as well.

Wetherefore expect that S is some function of M2A.In fact, as will be shown below, S = κM2A, where κ is a numerical constant whichdepends only on the precise definition of M that we adopt.This result bears a striking similarity to the formula for the intrinsic entropy of a blackhole, SBH = 14M2PlA, where MPl is the Planck mass and A is the surface area of the horizonof the black hole [2]. The links in the chain of reasoning establishing this formula areremarkably diverse, involving, in turn, classical geometry, thermodynamic analogies, andquantum field theory in curved space.

The result is thus rather mysterious. In particular,we would like to know whether or not SBH has anything to do with the number of quantumstates accessible to the black hole.As a black hole evaporates and shrinks, it produces Hawking radiation whose entropy,SHR, can be computed by standard methods of statistical mechanics.

One finds, after the1

black hole has shrunk to negligible size, that SHR is a number of order one (depending onthe masses and spins of the elementary particles) times the original black hole entropy [3].This calculation of SHR is done by counting quantum states, and the fact that SBH ≃SHRlends support to the idea that SBH should also be related to a counting of quantum states.It is then tempting think of the horizon as a kind of membrane [4], with approximatelyone degree of freedom per Planck area. However, in classical general relativity, the horizondoes not appear to be a special place to a nearby observer, so it is hard to see why itshould behave as an object with local dynamics.

The new result quoted above indicatesthat S ∼A is a much more general formula than has heretofore been realized. It showsthat the amount of missing information represented by SBH is about right, in the sensethat we would get the same answer in the vacuum of flat space if we did not permitourselves access to the interior of a sphere with surface area A, and set the ultravioletcutoffto be of order MPl (perfectly reasonable for comparison with a quantum theory thatincludes gravity).

Furthermore, getting S ∼A clearly does not require the boundary ofthe inaccessible region to be dynamical, since in our case it is entirely imaginary.To establish that S = κM2A for the problem at hand, let us begin with the simplestpossible version of it: two coupled harmonic oscillators, with hamiltonianH = 12p21 + p22 + k0(x21 + x22) + k1(x1 −x2)2. (1)The normalized ground state wave function isψ0(x1, x2) = π−1/2(ω+ω−)1/4 exp−(ω+x2+ + ω−x2−)/2,(2)where x± = (x1 ± x2)/√2, ω+ = k1/20, and ω−= (k0 + 2k1)1/2.

We now form the groundstate density matrix, and trace over the first (“inside”) oscillator, resulting in a densitymatrix for the second (“outside”) oscillator alone:ρout(x2, x′2) =Z +∞−∞dx1 ψ0(x1, x2)ψ∗0(x1, x′2)= π−1/2(γ −β)1/2 exp−γ(x22 + x′22 )/2 + βx2x′2,(3)where β = 14(ω+ −ω−)2/(ω+ + ω−) and γ −β = 2ω+ω−/(ω+ + ω−). We would like tofind the eigenvalues pn of ρout(x, x′):Z +∞−∞dx′ ρout(x, x′)fn(x′) = pnfn(x) ,(4)2

because in terms of them the entropy is simply S = −Pn pn log pn. The solution to Eq.

(4)is found most easily by guessing, and ispn = (1 −ξ)ξn ,fn(x) = Hn(α1/2x) exp(−αx2/2) ,(5)where Hn is a Hermite polynomial, α = (γ2 −β2)1/2 = (ω+ω−)1/2, ξ = β/(γ + α), and nruns from zero to infinity. Eq.

(5) imples that ρout is equivalent to a thermal density matrixfor a single harmonic oscillator specified by frequency α and temperature T = α/ log(1/ξ).The entropy isS(ξ) = −log(1 −ξ) −ξ1 −ξ log ξ ,(6)where ξ is ultimately a function only of the ratio k1/k0.We can easily expand this analysis to a system of N coupled harmonic oscillators withhamiltonianH = 12NXi=1p2i + 12NXi,j=1xiKijxj ,(7)where K is a real symmetric matrix with positive eigenvalues. The normalized groundstate wave function isψ0(x1, .

. ., xN) = π−N/4(det Ω)1/4 exp−x·Ω·x/2,(8)where Ωis the square root of K: if K = UTKDU, where KD is diagonal and U is orthogonal,then Ω= UTK1/2DU.

We now trace over the first n (“inside”) oscillators to getρout(xn+1, . .

., xN; x′n+1, . .

., x′N) =ZnYi=1dxi ψ0(x1, . .

. , xn, xn+1, .

. ., xN)× ψ∗0(x1, .

. ., xn, x′n+1, .

. .

, x′N) . (9)To carry out these integrals explicitly, we writeΩ= ABBTC,(10)where A is n × n and C is (N −n) × (N −n).

We findρout(x, x′) ∼exp−(x·γ·x + x′·γ·x′)/2 + x·β·x′,(11)where x now has N −n components, β = 12BTA−1B, and γ = C −β. In general β andγ will not commute, which implies that Eq.

(11) is not equivalent to a thermal densitymatrix for a system of oscillators.3

We need not keep track of the normalization of ρout, since we know that its eigenvaluesmust sum to one. To find them, we note that the appropriate generalization of Eq.

(4)implies that (det G) ρout(Gx, Gx′) has the same eigenvalues as ρout(x, x′), where G is anynonsingular matrix. Let γ = V TγDV , where γD is diagonal and V is orthogonal; then letx = V Tγ−1/2Dy.

(The eigenvalues of γ are guaranteed to be positive, so this transformationis well defined.) We then haveρout(y, y′) ∼exp−(y·y + y′·y′)/2 + y·β′·y′,(12)where β′ = γ−1/2DV βV Tγ−1/2D.

If we now set y = Wz, where W is orthogonal and W Tβ′Wis diagonal, we getρout(z, z′) ∼NYi=n+1exp−(z2i + z′2i )/2 + β′iziz′i,(13)where β′i is an eigenvalue of β′. Each term in this product is identical to the ρout of Eq.

(3),with γ →1 and β →β′i. Therefore, the entropy associated with the ρout of Eq.

(13) is justS = Pi S(ξi), where S(ξ) is given by Eq. (6), and ξi = β′i1 + (1 −β′2i )1/2.We now wish to apply this general result to a quantum field with hamiltonianH = 12Zd3xπ2(⃗x ) + |∇ϕ(⃗x )|2.

(14)To regulate this theory, we first introduce the partial wave componentsϕlm(x) = xZdΩZlm(θ, φ)ϕ(⃗x ) ,πlm(x) = xZdΩZlm(θ, φ)π(⃗x) ,(15)where x = |⃗x | and the Zlm are real spherical harmonics: Zl0 = Yl0, Zlm =√2 Re Ylm form > 0, and Zlm =√2 Im Ylm for m < 0; the Zlm are orthonormal and complete. Theoperators defined in Eq.

(15) are hermitian, and obey the canonical commutation relationsϕlm(x), πl′m′(x′)= iδll′δmm′δ(x −x′) . (16)In terms of them, we can write H = Plm Hlm, whereHlm = 12Z ∞0dx(π2lm(x) + x2 ∂∂xϕlm(x)x2+ l(l + 1)x2ϕ2lm(x)).

(17)4

So far we have made no approximations or regularizations.Now, as an ultraviolet regulator, we replace the continuous radial coordinate x bya lattice of discrete points with spacing a; the ultraviolet cutoffM is thus a−1. As aninfrared regulator, we put the system in a spherical box of radius L = (N + 1)a, where Nis a large integer, and demand that ϕlm(x) vanish for x ≥L; the infrared cutoffµ is thusL−1.

All together, this yieldsHlm = 12aNXj=1"π2lm,j + (j + 12)2ϕlm,jj−ϕlm,j+1j + 12+ l(l + 1)j2ϕ2lm,j#,(18)where ϕlm,N+1 = 0; ϕlm,j and πlm,j are dimensionless, hermitian, and obey the canonicalcommutation relationsϕlm,j, πl′m′,j′= iδll′δmm′δjj′ . (19)Thus, Hlm has the general form of Eq.

(7), and for a fixed value of N we can compute(numerically) the entropy Slm(n, N) produced by tracing the ground state of Hlm over thefirst n sites. The ground state of H is a direct product of the ground states of each Hlm,and so the total entropy is found by summing over l and m: S(n, N) = Plm Slm(n, N).As can be seen from Eq.

(18), Hlm is actually independent of m, and therefore so isSlm(n, N) = Sl(n, N). Summing over m just yields a factor of 2l + 1, and so we haveS(n, N) = Pl(2l + 1)Sl(n, N).From Eq.

(18) we also see that the l-dependent termdominates if l ≫N, and in this case we can compute Sl(n, N) perturbatively. The resultis that, for l ≫N, Sl(n, N) is independent of N, and is given bySl(n, N) = ξl(n)−log ξl(n) + 1,(20)whereξl(n) = n(n + 1)(2n + 1)264 l2(l + 1)2+ O(l−6) .

(21)Eqs. (20) and (21) demonstrate that the sum over l will converge, and also provide a usefulcheck on the numerical results.Let us define R = (n + 12)a, a radius midway between the outermost point which wastraced over, and the innermost point which was not.

The computed values of S(n, N) areshown for N = 60 and 1 ≤n ≤30 as a function of R2 in Fig. 1.

As can be seen, the pointsare beautifully fit by a straight line:S = 0.30 M2R2 ,(22)5

where M = a−1. Furthermore, S(n, N) turns out to be independent of N (and hence thevalue of the infrared cutoff).

Specifically, for fixed n, with n ≤12N, the values of S(n, N)turn out to be identical (in the worst case, to within 0.5%) for N = 20, 40, and 60. Therestriction to n ≤12N is necessary, since the linear behavior in Fig.

1 cannot continue allthe way to n = N: at this point we will have traced over all the degrees of freedom, andmust find S = 0. S must therefore start falling as R begins to approach the wall of thebox at radius L = (N + 1)a.Of course, similar calculations can be done for one- and two-dimensional systems aswell.

For d = 2, our introductory arguments would lead us to expect that S = κMR,since the relevant “area” is the circumference of the dividing circle of radius R. This isconfirmed by the numerical results, which will be presented in detail elsewhere [5]. Ford = 1, our arguments must break down: they would lead to the conclusion that S isindependent of R, and this is clearly impossible.

In fact, the numerical results indicatethat S = κ1 log(MR)+κ2 log(µR) in one dimension; for the first time, we see a dependenceon the infrared cutoffµ [6]. For d ≥4, regularization by a radial lattice turns out to beinsufficient; the sum over partial waves does not converge.

Regularization by a full, d-dimensional lattice would certainly produce a finite S, but this procedure would greatlyincrease the computational complexity.To summarize, a straightforward counting of quantum states in a simple, well-definedcontext has produced an entropy proportional to the surface area of the inaccessible region,inaccessible in the sense that we ignore the information contained there.Eq. (22) isstrikingly similar to the formula for the entropy of a black hole, SBH = 14M2PlA, and so mayprovide some clues as to its deeper meaning.I am grateful to Orlando Alvarez, Steve Giddings, David Gross, and Andy Stromingerfor helpful discussions.Note added: After this paper was completed, I learned of related work by Bombelliet al [7].

Also motivated by the black hole analogy, these authors find an equivalent resultfor the entropy of a coupled system of oscillators. They also argue that, for a quantumfield, the entropy should be proportional to the area of the boundary; the argument theygive is different from those presented here, and is valid only if the field has a mass m whichis large enough to make the Compton wavelength 1/m much less than R. I thank ErikMatinez for bringing this paper to my attention.6

REFERENCES[1] Let |0⟩= Pia ψia|i⟩in|a⟩out, so that (ρin)ij = (ψψ†)ij and (ρout)ab = (ψTψ∗)ab. Nowit is clear that Tr ρkin = Tr ρkout for any positive integer k. This can only be true if ρinand ρout have the same eigenvalues, up to extra zeroes.

[2] J. D. Bekenstein, Phys. Rev.

D 7, 2333 (1973); Phys. Rev.

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Phys. 43, 199 (1975); G. W. Gibbons and S. W. Hawking,Phys.

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S. W. Hawking and W. Israel (Cambridge, 1979); R. Kallosh,T. Ort´ın, and A. Peet, Stanford Univ.

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[4] G. ’t Hooft, Nucl. Phys.

B355, 138 (1990), and references therein. [5] M. Srednicki, in preparation.

[6] This formula bears no apparent relation to that for the entropy of a one-dimensionalblack hole; see G. W. Gibbons and M. J. Perry, Int. J. Mod.

Phys. D1, 335 (1992);C. Nappi and A. Pasquinucci, Mod.

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200400600800 2R050100150200250SFig. 1.

The entropy S resulting from tracing the ground state of a massless scalar field overthe degrees of freedom inside a sphere of radius R. The points shown correspond toregularization by a radial lattice with N = 60 sites; the line is the best linear fit. R ismeasured in lattice units, and is defined to be n + 12, where n is the number of tracedsites.8

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