Quantum inequality for a scalar field with a background potential

# Quantum inequality for a scalar field with a background potential

Eleni-Alexandra Kontou    Ken D. Olum Institute of Cosmology, Department of Physics and Astronomy,
Tufts University, Medford, MA 02155, USA
###### Abstract

Quantum inequalities are bounds on negative time-averages of the energy density of a quantum field. They can be used to rule out exotic spacetimes in general relativity. We study quantum inequalities for a scalar field with a background potential (i.e., a mass that varies with spacetime position) in Minkowski space. We treat the potential as a perturbation and explicitly calculate the first-order correction to a quantum inequality with an arbitrary sampling function, using general results of Fewster and Smith. For an arbitrary potential, we give bounds on the correction in terms of the maximum values of the potential and its first three derivatives. The techniques we develop here will also be applicable to quantum inequalities in general spacetimes with small curvature, which are necessary to rule out exotic phenomena.

###### pacs:
04.20.Gz 03.70.+k

## I Introduction

General Relativity relates spacetime curvature to the stress-energy tensor , but does not provide any constraints on what might be. Thus relativity alone allows us to construct any spacetime, including those with exotic features, such as wormholes and time machines. However in the context of quantum field theory, while negative energies are possible, for example in the Casimir effect, there are various constraints on the stress energy tensor. One example is averaged energy conditions that provide bounds on integrals of along an entire geodesic. Another example of bounding the stress-energy tensor is quantum inequalities that bound the total energy when averaging over a time period.

Quantum inequalities were introduced by Ford Ford (1978) to avoid the possibility of violating the second law of thermodynamics by sending a flux of negative energy into a black hole. The general form of a quantum inequality is

 ∫∞−∞dτw(τ)Tab(x(τ))VaVb>−B, (1)

where is a timelike path parameterized by proper time with tangent vector , and is a sampling function. The quantity is a bound, depending on the function and the quantum field of interest.

Since the original work of Ford, quantum inequalities have been derived for a wide range of different fields and sampling functions. However, these quantum inequalities apply only to free fields in Minkowski space without boundaries. In other cases, there are difference quantum inequalities Ford and Roman (1995), in which in Eq. (1) is replaced by the difference between in some state of interest and in a reference state. The bound may also then depend on the reference state. However, such difference inequalities cannot be used to rule out exotic spacetimes, at least in the case where the exotic matter that supports the spacetime comes from the vacuum state in the presence of the boundaries.

Nevertheless Ref. Fewster et al. (2007) shows that boundaries do not allow violation of the average null energy condition (ANEC), which states that

 ∫∞−∞dλTab(γ(λ))VaVb≥0, (2)

where the integral is taken on a null geodesic , affinely parameterized by with tangent vector . Reference Graham and Olum (2007) proved that ANEC is sufficient to rule out many exotic spacetimes. The proof made use of quantum inequalities for null contractions of the stress tensor averaged over timelike geodesics Fewster and Roman (2003).

None of this work, however, really addresses the possibility of exotic spacetimes. The quantum inequalities on which it depends apply only in flat spacetime, so they cannot be used to rule out spacetimes with exotic curvature. For that, we need limits on the stress-energy tensor in curved spacetimes. One possible approach is to appeal to the principle of equivalence to say that if the averaging timescale of the quantum inequality is small compared to the curvature radius of the spacetime, then flat-space results should apply approximately Ford and Roman (1996). We used such reasoning in Ref. Kontou and Olum (2013) to conjecture that flat-space quantum inequalities apply, even in curved space, with certain corrections, which we hoped were not too large. From this conjecture, we were able to extend the argument of Ref. Fewster et al. (2007) to curved spacetime. But the truth of our conjecture is not known.

As a first step toward proving the conjecture of Ref. Kontou and Olum (2013), we derive in the present work a quantum inequality in a flat spacetime with a background potential, i.e., a field with a mass depending on spacetime position. This is a simpler system that has many of the important features of quantum fields in curved spacetime. For a scalar field in a background potential, the Lagrangian is

 L=12[∂μΦ∂μΦ−V(x)Φ2], (3)

the equation of motion is

 (□+V(x))Φ=0, (4)

and the classical energy density is

 T00=12[(∂tΦ)2+(∇Φ)2+V(x)Φ2]. (5)

We work only in first order in but don’t otherwise assume that it is small. We can express the maximum values of the background potential and its derivatives as

 |V|≤Vmax|V,a|≤V′% max|V,ab|≤V′′max|V,abc|≤V′′′max, (6)

where , , and are positive numbers, finite but not necessarily small.

Our proof uses a general absolute quantum inequality proven by Fewster and Smith Fewster and Smith (2008), which we discuss in Sec. II. This inequality gives a bound on the renormalized energy density based on the Fourier transform of the point-split energy density operator applied to the Hadamard series. In Sec. III, we discuss this operator, in Sec. IV we compute the Hadamard series, and in Sec. V, we apply the operator. In Sec. VI, we perform the Fourier transform, leading to the final quantum inequality in Sec. VII. We conclude in Sec. VIII with a discussion of future possibilities.

We use metric signature . Indices denote all spacetime coordinates while denote only spatial coordinates.

## Ii Absolute Quantum Energy Inequality

We start by defining the renormalized energy density according to the renormalization procedure of Wald Wald (1994). Let be the two-point function of the scalar field, and define the Hadamard form

 H(x,x′)=14π2[1σ+(x,x′)+∞∑j=0vj(x,x′)σj+(x,x′)ln(σ+(x,x′))+∞∑j=0wj(x,x′)σj(x,x′)], (7)

where

 σ(x,x′)=−ηab(x−x′)a(x−x′)b, (8)

so that when the separation between and is timelike. By , for some function , we mean the distributional limit

 F(σ+)=limϵ→0+F(σϵ), (9)

where

 σϵ(x,x′)=σ(x,x′)+2iϵ(t−t′)+ϵ2, (10)

with and being the time components of the 4-vectors and . In most of the calculation we consider and separated only in time. In that case, we define

 τ=t−t′, (11)

and write to mean . In general the sums in Eq. (7) do not converge, but we will be concerned only with the first few terms. Following Wald Wald (1994), we will choose .

When the scalar field is in a Hadamard state, the singularity structure of is precisely that of , so the renormalized two-point function is smooth. To this we apply a point-split energy density operator, which is analogous to the classical energy density of Eq. (5),

 Tsplit=12[3∑a=0∂a∂a′+V(x)+V(x′)2], (12)

and take the limit where and coincide. In this limit, the location of evaluation of does not matter, but the form above will be convenient later. Thus we define

 (13)

where is a term added “by hand” to prevent the failure of conservation of the stress-energy tensor. Wald Wald (1978) derived this term for curved spacetime. The calculation for flat space with background potential is essentially the same, giving

 Q(x)=112π2w1(x,x). (14)

Unfortunately, there is an ambiguity in the above procedure. In order to take logarithms, we must divide by the square of some length scale . Changing the scale to some other scale decreases by . This results in increasing by . Using the values for and computed below, this becomes . Thus we see that the definition of must include arbitrary multiple of . This ambiguity can also be understood as the possibility of including in the Lagrangian density a term of the form , where is the scalar curvature. Varying the metric to obtain and then going to flat space yields the above term. The situation is very much analogous to the possible addition of terms of the form and in the case of a field in curved spacetime.

Thus we rewrite Eq. (13) to include the ambiguous term,

 (15)

where is some constant. Whenever definition of one is trying to use, one can pick an arbitrary scale and adjust accordingly.

Now, following Ref. Fewster and Smith (2008) we define

 ~H(x,x′)=12[H(x,x′)+H(x′,x)+iE(x,x′)], (16)

where is the advanced-minus-retarded Green’s function, and thus is the antisymmetric part of the two point function. We use the Fourier transform convention

 ^f(k) or [f]∧(k)=∫∞−∞dxf(x)eixk. (17)

We consider the energy density integrated along a geodesic on the axis with a smooth, positive sampling function . The absolute quantum inequality of Ref. Fewster and Smith (2008) for this case is

 ∫∞−∞dτg(t)2⟨Tren00⟩(t,0)≥−B, (18)

where

 B=∫∞0dξπ^F(−ξ,ξ)+∫∞−∞dtg2(t)(Q−CV,ii), (19)

and

 F(t,t′)=g(t)g(t′)Tsplit~H(5)((t,0),(t′,0)), (20)

denotes the Fourier transform in both arguments according to Eq. (17), and the subscript means that we include only terms through in the sums of Eq. (7).

## Iii General considerations

### iii.1 Smooth, symmetrical contributions

Let , and . Let

 A(τ)=∫∞−∞d¯tF(¯t+τ2,¯t−τ2). (21)

Then .

Suppose contains some term that is symmetrical in and . Let be the corresponding term in according to Eq. (21). Then will be even in , so will be even also. If , then , and we can perform the integral of this term separately, giving an inverse Fourier transform,

 ∫∞0dξπ^f(−ξ,ξ)=∫∞−∞dξ2π^a(ξ)=a(0). (22)

In particular, if

 limt→t′f(t,t′)=f(t), (23)

then

 ∫∞0dξπ^f(−ξ,ξ)=∫∞−∞dtg(t)2f(t), (24)

and if there is no contribution.

Terms arising from appear symmetrically in . At orders they have at least 4 powers of , so they vanish in the coincidence limit even when differentiated twice by the operators of . Thus such terms make no contribution to Eq. (19).

### iii.2 Simplification of Tsplit

We would like to write the operator in terms of separate derivatives on the centerpoint and the difference between the points. First we separate the derivatives in into time and space,

 3∑a=0∂a∂a′=∂t∂t′+∇x⋅∇x′. (25)

We can expand the spatial derivative with respect to111When a derivative is with respect to or , we mean to keep the other of these fixed, while when the derivative is with respect to or , we mean to keep the other of these fixed. ,

 ∇2¯x=∇2x+2∇x⋅∇x′+∇2x′, (26)

and Eqs. (12,25,26) give

 Tsplit = 12[∂t∂t′+12(∇2¯x−∇2x−∇2x′)+12(V(x)+V(x′))]= (27) = 14[∇2¯x+□x−∂2t+□x′−∂2t′+2∂t∂t′+V(x)+V(x′)],

where and denote the D’Alembertian operator with respect to and . Then using

 ∂2τ=14[∂2t−2∂t∂t′+∂2t′], (28)

we can write

 Tsplit~H=14[(□x+V(x))~H+(□x′+V(x′))~H+∇2¯x~H]−∂2τ~H. (29)

Consider the first term. The function obeys the equation of motion in , and so does . Thus

 (□x+V(x))~H=12(□x+V)H(x′,x). (30)

The only asymmetrical part of comes from the , so

 H(x′,x)=H(x,x′)+14π2∑j(wj(x′,x)−wj(x,x′))σj(x,x′). (31)

Terms involving both and are second order in , so we can ignore them, giving

 (□x+V(x))~H=14π2□x∑j(wj(x′,x)−wj(x,x′))σj(x,x′). (32)

Similarly,

 (□x′+V(x))~H=14π2□x′∑j(wj(x,x′)−wj(x′,x))σj(x,x′). (33)

Adding together Eqs. (32,33), we get something which is symmetric in and and vanishes in the coincidence limit. Thus according to the analysis of Sec. III.1, it makes no contribution and for our purposes we can take

 Tsplit~H=[14∇2¯x−∂2τ]~H. (34)

## Iv Computation of ~H

Examining Eq. (34) we see that is sufficient to compute for purely temporal separation as a function of , , and , the common spatial position of the points. The function is a series of terms with decreasing degree of singularity at coincidence: , , , etc. For the first term in Eq. (34), terms in that have any positive powers of will not contribute by the analysis of Sec. III.1. For the second term we need to keep terms in up to order , because the derivatives will reduce the order by 2.

The symmetrical combination , will lead to something whose Fourier transform does not decline rapidly for positive , so that if this alone were put into Eq. (19) the integral over would not converge. But each term in will combine with a term coming from to give something whose Fourier transform does decline rapidly.

We will work order by order in and write , , to denote the term in involving (with or without ), and to to denote the sum of all terms up through . We will split up into terms labeled that are proportional to , define a “remainder term”

 Rj=E−j∑k=−1Ek, (35)

and let

 ~Hj(x,x′) = 12[Hj(x,x′)+Hj(x′,x)+iEj(x,x′))] (36a) ~H(j)(x,x′) = 12[H(j)(x,x′)+H(j)(x′,x)+iE(x,x′))]. (36b)

### iv.1 General computation of E

We will need the Green’s functions for the background potential, including only first order in , so we write

 G=G(0)+G(1)+⋯. (37)

The equation of motion is

 (□+V(x))G(x,x′)=δ(4)(x−x′). (38)

Using and keeping only first-order terms we have

 □G(1)(x,x′)=−V(x)G(0)(x,x′), (39)

so

 G(1)(x,x′)=−∫d4x′′G(0)(x,x′′)V(x′′)G(0)(x′′,x′). (40)

For we have for the retarded Green’s function,

 G(0)R(x′′,x′)=12πδ((t′′−t′)2−|x′′−x′|2)=14πδ(t′′−t′−|x′′−x′|)|x′′−x′|. (41)

So we can write

 G(1)R(x,x′)=−18π2∫d3x′′∫dt′′δ((t−t′′)2−|x−x′′|2)δ(t′′−t′−|x′′−x′|)|x′′−x′|V(t′′,x′′). (42)

Integrating over the second delta function we find . Again considering purely temporal separation and defining and , we find

 G(1)R(t,t′)=−18π2∫dΩ∫dz′′z′′2δ(τ2−2τz′′)z′′V(t′+z′′,x′+z′′^Ω), (43)

where denotes integration over solid angle, and varies over all unit vectors. We can integrate over to get and

 G(1)R(t,t′)=−132π2∫dΩV(¯t,x′+τ2^Ω). (44)

If we define a 4-vector we can write

 G(1)R(t,t′)=−132π2∫dΩV(¯x+τ2Ω). (45)

The advanced Green’s functions are the same with and reversed. Since is the advanced minus the retarded function, we have

 E(1)(t,t′)=132π2∫dΩV(¯x+|τ|2Ω)sgnτ. (46)

### iv.2 Terms of order τ−2

We now compute the various , , and , starting with terms that go as or . These terms are exactly what one would have for flat space without potential. Equation (7) gives

 H−1(x,x′)=14π2σ+(x,x′)=−14π2(τ2−−z2), (47)

where

 z=x−x′ (48)

and

 z=|z|. (49)

Similarly, the advanced minus retarded Green’s function to this order is

 E−1(x,x′)=GA(x,x′)−GR(x,x′)=δ(τ−z)−δ(τ+z)4πz, (50)

so

 ~H−1(t,t′)=limz→018π2[−1τ2+−z2−1τ2−−z2+iπδ(τ+z)−δ(τ−z)z], (51)

where

 F(τ+)=limϵ→0F(τ+iϵ). (52)

Taking the limit in and gives the formula

 −1τ2+−z2+1τ2−−z2=−iπδ(τ+z)−δ(τ−z)z (53)

so

 ~H−1(t,t′)=−14π2τ2−=H−1(t,t′) (54)

as discussed in Ref Fewster and Smith (2008).

### iv.3 Terms with no powers of τ

The Hadamard coefficients are given by the Hadamard recursion relations, which are the solutions to , giving

 V(x′′)+2ηabv0,aσ,b+4v0+v0□σ=0 (55a) (□+V(x′′))vj+2(j+1)ηabvj+1,aσ,b−4j(j+1)vj+1+(j+1)vj+1□σ=0. (55b)

In Eqs. (55), , and their derivatives are functions of and , and all derivatives act on .

To find the zeroth order of the Hadamard series we need only . For flat space, and . Putting these in Eq. (55a) we have

 (x′′−x′)av0,a+v0=V(x′′)4, (56)

Now let to integrate along the geodesic going from to . We observe that

 dv0(x′′,x′)dλ=(x−x′)av0,a(x′′,x′). (57)

So Eq. (56) gives

 λdv0(x′′,x′)dλ+v0(x′′,x′)=V(x′′)4, (58)

or

 d(λv0(x′′,x′))dλ=V(x′′)4, (59)

from which we immediately find

 v0(x,x′)=∫10dλV(x′+λ(x−x′))4. (60)

Now we consider purely temporal separation so the background potential is evaluated at . We expand in a Taylor series in around with fixed,

 V(t′+λτ)=V(¯t)+τ(λ−12)V,t(¯t)+τ22(λ−12)2V,tt(¯t)+⋯. (61)

We are calculating the zeroth order so we keep only the first term of Eq. (61), and Eq. (60) gives

 v0(t,t′)=14V(¯t)+O(τ2) (62)

and thus

 H0(x,x′)=116π2V(¯x)ln(−τ2−), (63)

and

 H0(x,x′)+H0(x′,x)=14π2V(¯x)ln|τ|. (64)

We can expand around ,

 V(¯x+τ2Ω)=V(¯x)+V(1)(¯x+τ2Ω), (65)

where is the remainder of the Taylor series

 V(1)(¯x+τ2Ω)=V(¯x+τ2Ω)−V(¯x)=∫τ/20drV,i(¯x+rΩ)Ωi. (66)

Then from Eq. (46),

 E0(x,x′) = 18πV(¯x)sgnτ (67a) R0(x,x′) = 132π2∫dΩV(1)(¯x+|τ|2Ω)sgnτ. (67b)

Using

 2ln|τ|+πisgnτ=ln(−τ2−), (68)

we combine Eqs. (64,67a) to find

 ~H0(t,t′)=116π2V(¯x)ln(−τ2−). (69)

Combining all terms through order gives

 ~H(0)(t,t′)=~H−1(t,t′)+~H0(t,t′)+12iR0(t,t′). (70)

### iv.4 Terms of order τ2

Now we compute the terms of order in and . First we need at this order, so we use Eq. (61) in Eq. (60), to get

 v0(x,x′)=14V(¯x)+τ2196V,tt(¯x)+⋯. (71)

Next we need to know , but since is multiplied by in , we need only the -independent term . From Eq. (55b),

 (□+V(x))v0(x,x′)+2ηabv1,a(x,x′)σ,b(x,x′)+v1(x,x′)□xσ(x,x′)=0. (72)

We neglect the term because it is second order in . At , , so

 v1(x,x)=18limx′→x□xv0(x,x′). (73)

Using Eq. (60) we find

 □xv0(x,x′)=14∫10dλ□xV(x′+λ(x−x′))=14∫10dλλ2(□V)(x′+λ(x−x′)), (74)

and Eq. (73) gives

 v1(x,x)=196□V(¯x). (75)

We also need to know , but again only at coincidence. Reference Wald (1978) gives

 w1(x,x)=−32v1(x,x)=−164□V(x). (76)

Combining the second term of Eq. (71) with Eqs. (75,76) gives

 H1(t,t′)=τ2128π2[13V,ii(¯x)ln(−τ2−)+12□V(¯x)]. (77)

Then is given by symmetry, so

 H1(x,x′)+H1(x′,x)=τ264π2[23V,ii(¯x)ln|τ|+12□V(¯x)]. (78)

The calculation of is similar to that of , but now we have to include more terms in the Taylor expansion of around . So we expand

 V(¯x+τ2Ω)=V(¯x)+12V,i(¯x)Ωiτ+18V,ij(¯x)ΩiΩjτ2+V(3)(¯x+τ2Ω), (79)

where the remainder of the Taylor series is

 V(3)(¯x+τ2Ω)=12∫τ/20drV,ijk(¯x+rΩ)(τ2−r)2ΩiΩjΩkdr. (80)

Since and , Eq. (46) gives

 E1(x,x′) = 1192πV,ii(¯x)τ2sgnτ, (81a) R1(x,x′) = 132π2∫dΩV(3)(¯x+|τ|2Ω)sgnτ. (81b)

Again using Eq. (68), we combine Eqs. (78,81a) to get

 ~H1(x,x′)=τ2128π2[13ln(−τ2−)V,ii+12□V(¯x)]. (82)

Combining all terms through order 1 gives

 ~H(1)(t,t′)=~H−1(t,t′)+~H0(t,t′)+~H1(t,t′)+12iR1(t,t′). (83)

## V The Tsplit~H

Using Eqs. (20,34), we need to compute

 ∫∞0dξπ^F(−ξ,ξ′), (84)

where

 F(t,t′)=g(t)g(t′)[14∇2¯x~H(0)(t,t′)−∂2τ~H(1)(t,t′)]. (85)

Using Eqs. (54,67b,69,70,81b,82,83) we can write this

 F(t,t′)=g(t)g(t′)6∑i=1fi(t,t′), (86)

with

 f1 = 32π2τ4− (87a) f2 = 18π2τ2−V(¯x) (87b) f3 = 196π2V,ii(¯x)ln(−τ2−) (87c) f4 = −1128π2[V,tt(¯x)+V,ii(¯x)] (87d) f5 = 1256π2∫dΩ∇2¯x[V(1)(¯x+|τ|2Ω)]isgnτ (87e) f6