# Quantum Energy Inequalities, Part I

I am pretty sure that some of our readers must know of Sabine Hossenfelder, she is a really good researcher and communicator of science, make sure to check out her Youtube channel. A couple a weeks ago, she released a really good video with some exciting news: new developments in a certain class of warp drives. A warp drive is basically something that allows an observer to travel at really large speeds–in some instances, faster than light–by specifying a geometry of spacetime in the context of General Relativity (GR). Miguel Alcubierre was the first to specify one of these geometries, which after being plugged into the Einstein Field Equations (EFE), implies that the energy distribution needed for this to happen is negative–we will come back to this in a moment.

In her video Sabine says:

Which I am pretty sure she meant in a classical setting, as we know that negative energies do occur for quantum fields, and that is what I would like to talk about in this post.

Let us begin by reminding the reader that the EFE in the ${(-,+,-)}$ convention found in Misner, Thorne and Wheeler, can be written as

Whereas the left-hand side of this expression is well-known and every of its components is properly explained within the context of GR, the right-hand side is less clear. The first reason is that a stress-energy tensor ${T_{\mu\nu}}$ by itself is never seen outside gravitational physics–one in general considers differences or derivatives of it, e.g. $\nabla_\mu T^{\mu\nu}$. Secondly, as it is clear from the equality, we can provide an arbitrary geometry and it will be a solution associated to a stress energy tensor. However, GR says little about which of these tensors are deemed to be physically reasonable. This is the main motivation of the energy conditions, to provide certain notion of what a stress-energy tensor should look like.

Let’s say that we want all observers along a timelike trajectory to see a non-negative energy density, this is the so-called Weak Energy Condition ${T_{\mu\nu}u^\mu u^\nu \ge 0}$, where ${u}$ is the tangent vector to the trajectory. Alternatively, one could demand for all the observers to see a causal flux of energy-momentum, this corresponds to the Dominant Energy Condition ${T_{\mu\nu}u^\mu v^\nu \ge 0}$ for all future-pointing timelike ${u^\mu}$ and ${v^\nu}$. We also have the Null Energy Condition ${T_{\mu\nu}u^\mu u^\nu \ge 0}$ for null ${u^\mu}$ and the Strong Energy Condition ${T_{\mu\nu}u^\mu u^\nu -1/2g^{\mu\nu}T_{\mu\nu}\ge 0}$ where ${u^\mu}$ is a timelike unit vector. If the EFE hold and the matter obeys any of this conditions, this will lead to constraints in the geometry–thus preventing to choose an arbitrary one. For instance, the Null Energy Condition and the EFE (1) imply that ${R_{\mu\nu}u^\mu u^\nu}$ for null ${u^\mu}$, which places a condition on the Ricci tensor, and ultimately, on the metric.

It is also worth mentioning that these conditions are paramount when formulating the singularity theorems as they lead to a focusing behaviour of a congruence of geodesics which means that they ultimately will reach a focal point in finite proper time.

However, these conditions fail to hold in the usual sense when considering quantum fields. That is, if we consider two Hadamard states ${\omega}$ and ${\omega_0}$ and a timelike curve ${\gamma(\lambda)}$ with unit tangent vector ${u^\mu}$, it is reasonable to define the energy density along ${\gamma}$ as

$\displaystyle \rho_\omega(\lambda)=\omega(:T_{\mu\nu}(\gamma(\lambda)): u^\mu u^\nu), \ \ \ \ \ (2)$

where ${:T_{\mu\nu}:}$ denotes normal ordering w.r.t. ${\omega_0}$, we will me more precise about this shortly. Epstein, Glasser & Jaffe demonstrated that ${\rho_\omega}$ can not obey any energy condition pointwise as it is unbounded from below. This means that the best next thing one could hope for, is the average doing it, as we expect to recover the original conditions in the classical limit. Ford discovered that although the average over time can be negative, it is bounded from below, where the bound approaches zero as the sampling time becomes infinite. After this, Fewster obtained the following result for the local average along a worldline:

$\displaystyle \inf_\omega \int_\gamma \rho_\omega(\lambda)f(\lambda)^2d\lambda\geq -\mathcal{Q}(f), \ \ \ \ \ (3)$

where ${f}$ is a test function and ${\mathcal{Q}}$ denotes the Sobolev norm, that is a combination of norms of the function and its derivatives. This argument was generalised later by him so that the inequalities hold even when not restricted to a wordline, as a matter of fact we find out that to obtain such inequalities, we just need to use Algebraic Quantum Field Theory and the Microlocal Spectrum Condition–the singularities of the two-point function of Hadamard states are of positive-frequency in the first variable and of negative-frequency in the second.

To see why the the energy density fails to be positive we will give a synthesis of the first and second sections of Fewster’s Lectures on Quantum Energy Inequalities, we do encourage the reader to stop reading this post and dive into the real thing. Although if the reader is looking for a streamlined mobile-friendly version, then let us begin. Recall that the energy density in this case is a monomial of ${\Phi}$ and its derivatives, bearing this in mind we turn our attention to the so-called two-point function ${\omega_0(\Phi(x)\Phi(y))}$ where ${\omega_0}$ is a positive linear functional that preserves the unit, which we will consider to be a Hadamard state. This object fails to be well-defined which one could attribute to the fact that ${\Phi(x)}$ has contributions of modes of arbitrarily high frequency in Fourier space, which indicates that we can’t think of it as an object defined sharply at ${x}$–but ultimately is because fields are distributions, which means that products can’t be taken for granted. This is solved by the following prescription: define ${\mathpunct{:} A(x,y)\mathpunct{:}=A(x,y)-\omega_0(A(x,y))\mathbf{1}}$, take the limit ${y \rightarrow x}$ and smear it against a test function ${f}$–in particular, for ${A(x,y)={\Phi}(x) {\Phi}(y)}$ this amounts to put the annihilation operator–which will be introduced in a moment–to the right. The quantity ${\mathpunct{:}\Phi^2\mathpunct{:}(x)}$ is known as the Wick square, consequently its smeared version is ${\mathpunct{:}\Phi^2\mathpunct{:}(f)=\int d^4x~\mathpunct{:}\Phi^2\mathpunct{:}(x)f(x)}$ which happens to be not really a square, because as we will see shortly, we have traded positive-definiteness for finiteness. To be more explicit, consider the free real scalar quantum field on four dimensional Minkowski spacetime in its usual mode expansion for ${k_\mu=(\omega_k,\vec{k})}$ with ${\omega_k^2=||\vec{k}||^2+m^2}$, that is

$\displaystyle \Phi(x)=\int\frac{d^3\vec{k}}{(2\pi)^{3}\sqrt{2\omega_k}}\big(e^{-ik\cdot x}a(\vec{k})+e^{ik\cdot x}a(\vec{k})^\star\big). \ \ \ \ \ (4)$

Where ${a(\vec{k})^{*}}$ and ${a(\vec{k})}$ are the creation and annihilation operators that obey the following commutation relations: ${[a(\vec{k}),a(\vec{\ell})]=0}$ and ${[a(\vec{k}),a(\vec{\ell})^*]=(2\pi)^3\delta^3(\vec{k}-\vec{\ell})\mathbf{1}}$. The name of the latter comes from the fact that there is a unit vector ${\Omega}$ named the vacuum such that for any ${\vec{k}}$ we have ${a(\vec{k})\Omega=0}$. The action of the smeared Wick square on the vacuum state is given by

$\displaystyle \mathpunct{:}\Phi^2\mathpunct{:}(f)\Omega=\frac{1}{(2\pi)^6}\int \frac{d^3\vec{k}}{\sqrt{2\omega_k}}\frac{d^3\vec{\ell}}{\sqrt{2\omega_\ell}}\hat{f}(-\vec{k}-\vec{\ell}) a(\vec{k})^*a(\vec{\ell})^*\Omega, \ \ \ \ \ (5)$

where the hat denotes the Fourier transform. Clearly, we have a vanishing vacuum expectation value for this quantity, i.e. ${\langle\Omega|\mathpunct{:}\Phi^2\mathpunct{:}(f)\Omega\rangle}$ where ${\langle | \rangle}$ denotes the standard ${L^2}$ inner product. However, the norm of the vector ${\mathpunct{:}\Phi^2\mathpunct{:}(f)\Omega}$ is in general not zero, as it can be seen from

$\displaystyle ||\mathpunct{:}\Phi^2\mathpunct{:}(f)\Omega||^2=\int d\mu \hat{f}(-\vec{k}'-\vec{\ell}')^*\hat{f}(-\vec{k}-\vec{\ell})\langle a(\vec{k}')^*a(\vec{\ell}')^*\Omega|a(\vec{k})^*a(\vec{\ell})^*\Omega\rangle, \ \ \ \ \ (6)$

where ${(2\pi)^{12}d\mu=\frac{d^3\vec{k}'}{\sqrt{2\omega_{k'}}}\frac{d^3\vec{\ell}'}{\sqrt{2\omega_{\ell'}}}\frac{d^3\vec{k}}{\sqrt{2\omega_k}}\frac{d^3\vec{\ell}}{\sqrt{2\omega_\ell}}}$. We manipulate the inner product to obtain two products of pairs of deltas, which after integration yields

$\displaystyle =\frac{1}{(2\pi)^6}\int \frac{d^3\vec{k}}{2\omega_k}\frac{d^3\vec{\ell}}{2\omega_\ell}|\hat{f}(-\vec{\ell}-\vec{k})|^2, \ \ \ \ \ (7)$
which is zero if and only ${f\equiv 0}$. Given that the expectation value of the operator ${\mathpunct{:}\Phi^2\mathpunct{:}(f)}$ vanishes but its norm is different from zero, we must expect negative spectrum.

Next, we will derive a inequality for ${\mathpunct{:}\Phi^2\mathpunct{:}}$, although this is not the full-fledged quantum energy inequality, it encapsulates the essence of the argument. Henceforth, consider that ${\omega_0}$ is no longer the Minkowski vacuum but any reference Hadamard state, we do this because the argument has little change and the generality of our result will be wider. The local average of ${\omega(\mathpunct{:}\Phi^2\mathpunct{:})}$ along a causal curve ${\gamma}$ is given by

$\displaystyle \omega(\mathpunct{:}\Phi^2\mathpunct{:}(f^2))=\int d\lambda~ \omega\left[\Phi(\gamma(\lambda))\Phi(\gamma(\lambda))-\omega_0(\Phi(\gamma(\lambda))\Phi(\gamma(\lambda)))\mathbf{1}\right]f(\gamma(\lambda))^2, \ \ \ \ \ (8)$

we follow the point-splitting recipe by inserting a delta and then expressing it in terms of its Fourier transform to get

$\displaystyle =\int\frac{dk}{2\pi}\int d\lambda~d\lambda'~\omega\left[\Phi(\gamma(\lambda))\Phi(\gamma(\lambda'))-\omega_0(\Phi(\gamma(\lambda))\Phi(\gamma(\lambda')))\mathbf{1}\right]f(\gamma(\lambda))f(\gamma(\lambda'))e^{-ik(\lambda-\lambda')}. \ \ \ \ \ (9)$

Introducing the distribution ${G(\lambda,\lambda')=\omega\left[\Phi(\gamma(\lambda))\Phi(\gamma(\lambda')\right],~}$ and recalling that ${\omega}$ is linear and unital, we arrive to

$\displaystyle =\int\frac{dk}{2\pi}\int d\lambda~d\lambda'~\left[G(\lambda,\lambda')-G_0(\lambda,\lambda')\right]f(\gamma(\lambda))f(\gamma(\lambda'))e^{-ik(\lambda-\lambda')}, \ \ \ \ \ (10)$

since ${\omega}$ and ${\omega_0}$ are Hadamard states, their difference must be a smooth and symmetric function, this allows us to write

$\displaystyle =\int^\infty_0\frac{dk}{\pi}\int d\lambda~d\lambda'~\left[G(\lambda,\lambda')-G_0(\lambda,\lambda')\right]f(\gamma(\lambda))f(\gamma(\lambda'))e^{-ik(\lambda-\lambda')}, \ \ \ \ \ (11)$

as both ${G}$ and ${G_0}$ are positive, this leads to our final result

$\displaystyle \omega(\mathpunct{:}\Phi^2\mathpunct{:}(f^2))\ge-\int^\infty_0\frac{dk}{\pi}\int d\lambda~d\lambda'~G_0(\lambda,\lambda')f(\gamma(\lambda))f(\gamma(\lambda'))e^{-ik(\lambda-\lambda')}. \ \ \ \ \ (12)$

It is also possible to define ${\omega(\mathpunct{:}\Phi^2\mathpunct{:}(x))=(\omega_2-H)(x,x)}$ where ${H}$ is the Hadamard parametrix, which is specified by the local geometry of the manifold. This makes the RHS independent from the reference state, as long as the integrals we wrote exist, but that will be the matter of a future post where we will cover that and explore an important consequence of these inequalities: they impose bounds on the duration of the negative energy densities.

A final observation: although we mentioned that these results can be generalised outside of a timelike curve, this can not be done for null curves or averaging over spacelike manifolds. However, it certainly can be done for Lorentzian submanifolds.

If you have any questions or feel a little negative (but bounded), feel free to drop me a Tweet!