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I may be somewhat confused here, but as I understand it:
Energy is a scalar quantity and colloquially objects are said to “possess” say, kinetic energy or potential energy, or rest energy proportional to their mass.
If the “kinetic energy of a particle” is the energy it possesses, I would expect it to be localised at the point of the particle at least from classical intuition. Obviously in QM if the particles aren’t localised in space then I couldn’t say the same thing for energy (if it is defined at all).

I guess what I'm wondering is if energy is more of a mathematical quantity that doesn't really physically exist in the system itself or if it is actually localised (to some extent) within the region of a particle.

I would imagine something like this has to be the case because there are vacuum energy effects, and it's said that the vacuum of space has an intrinsic energy density (dark energy), right?

I probably need to know a lot more about QFT and the like to find a more fundamental answer on this, but to anyone more educated on this I would appreciate it if you could help my struggling brain on this.

It would be great if I could get answers for both frameworks, classical mechanics and quantum.

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    $\begingroup$ I've removed a number of comments that were attempting to answer the question and/or responses to them. Please keep in mind that comments should be used for suggesting improvements and requesting clarification on the question, not for answering. $\endgroup$
    – David Z
    Apr 16 '20 at 7:36
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Energy is a property of a system. It depends on how the system's fields and particles are arranged and on what they're doing. At least in classical physics, energy can be localized in the sense that we can talk about how much of the system's energy is "in" a given region of space. That's called energy density. The situation in quantum theory is similar but with an interesting twist. The rest of this answer addresses how energy density is quantified in both cases.

Energy density in classical physics

In classical field theory, the stress-energy tensor $T^{ab}(x)$ (also called the energy-momentum tensor) is the thing you're looking for. The component $T^{00}(x)$ is the energy density, and integrating $T^{00}(x)$ over all of space gives the usual total energy. The "local" conservation law $\sum_a\partial_a T^{ab}(x)=0$ implies that the total energy is conserved. (The notation $\partial_a$ means the derivative with respect to the coordinate $x^a$.)

This generalizes to any classical field theory with a fixed spacetime metric (like the Minkowski metric). The situation is more subtle in general relativity, where the spacetime metric itself is one of the dynamic fields, but that's probably beyond the scope of the question.

When classical particles are involved, $T^{ab}(x)$ includes contributions from the particles. This involves mathematically-awkward things like the Dirac delta function $\delta^3(\vec x-\vec x_0)$ to account for the fact that a particle (in the classical sense) is localized at a point $\vec x_0$, which may in turn be a function of time.

Within any given model (if it satisfies an action principle), an explicit expression for $T^{ab}(x)$ may be derived using Noether's theorem.

Energy density in quantum field theory

Quantum field theory (QFT) has a reputation for being difficult, but it makes at least one thing easier: in QFT, everything is described in terms of quantum fields. The phenomena we call "particles" are just manifestations of those quantum fields. Understanding how that works (or what it even means) isn't easy, as evidenced by the many questions on this site about the particle aspect of photons, but at least it allows a more uniform treatment of the thing this question is asking about: energy density.

In QFT, the components $T^{ab}(x)$ of the stress-energy tensor are operators (observables) that can be used to generate transformations like rotations and translations in spacetime. The operator $T^{00}(x)$ is QFTs version of "energy density," and integrating this operator over all space gives the total energy operator (an observable, also called the Hamiltonian). Just like in classical physics, the total energy (Hamiltonian) is the conserved quantity associated with time-translation symmetry.

However, the energy density operator $T^{00}(x)$ has a surprising property in relativistic QFT. The general principles of relativistic QFT require that the total energy have a finite lower bound (conventionally zero), but the energy density operator $T^{00}$ (after smearing it over a finite region to make it well-defined) can still be arbitrarily negative. This is related to the fact that the vacuum state in relativistic QFT is entangled with respect to location: when we try to ask "how much energy is in this region of space?" we run into the mathematical fact that observables which are localized in that region are necessarily correlated with observables localized outside that region. As a result, "energy density" has some counter-intuitive properties in relativistic QFT, like the fact that it cannot be strictly positive even if the total energy is.

To learn more about energy density in QFT, the papers Quantum Energy Inequalities and Stability Conditions in Quantum Field Theory and Lectures on quantum energy inequalities are a place to start. For learning more about entanglement properties of the vacuum, the review Notes on Some Entanglement Properties of Quantum Field Theory is excellent. These papers assume some background in quantum field theory, but even without that background, skimming them might still provide some inspiration.

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    $\begingroup$ This is extremely helpful, thank you. I only have a cursory background in scalar quantum fields at this point (klein-gordon) and some elementary examples with QED that i don’t fully follow. $\endgroup$ Apr 15 '20 at 16:22
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    $\begingroup$ @Thatpotatoisaspy Since you mentioned scalar fields, here's the energy density ($T^{00}$) for a scalar field $\phi$: $$T^{00}(x)=\frac{\dot\phi^2(x) + \big(\nabla\phi(x)\big)^2+m^2\phi^2(x)}{2}+V\big(\phi(x)\big)$$ where $\dot\phi$ is its time-derivative, $\nabla\phi$ is its gradient, $m$ is a mass parameter, and $V$ accounts for interaction terms. This has the usual "kinetic + potential" pattern, but localized at a point $x$. The total energy is $$H=\int d^3\ T^{00}(x).$$ This is the same in both classical and quantum field theory, except that in QFT, $\phi(x)$ is an operator (observable). $\endgroup$ Apr 15 '20 at 18:26
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    $\begingroup$ @Thatpotatoisaspy ... and in the QFT case, we need to be careful about the ordering of non-commuting operators in order to enforce the requirement that $H$ have a finite lower bound (conventionally zero). $\endgroup$ Apr 15 '20 at 18:42
  • $\begingroup$ Isn't it true that local, not just global, energy is conserved? That is, the change in energy contained within a local volume equals the energy that's left the enclosing local boundary. $\endgroup$ Apr 21 '20 at 6:59
  • $\begingroup$ @Physikslover Yes, you're right. The stress-energy tensor satisfies $\partial_a T^{ab}=0$, and this equation captures exactly what you said. The conservation of total energy and total momentum can be derived from this "local" conservation law. In the quantum case, we need to be careful how we define products of operators at a single point ($T^{ab}$ involves such products), but that's another story. $\endgroup$ Apr 21 '20 at 17:03
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I would argue that the answer to the OP's question is largely a matter of perspective, but I'm going to respectfully disagree with Chiral Anomaly's answer. That answer is neglecting a critical detail, which is that there is not always a unique natural choice of stress-energy tensor in either classical or quantum field theory. For example, in classical EM, it's sometimes more convenient to use the canonical stress tensor and sometimes the Belinfante tensor. They differ by a total divergence, so their total energies agree when integrated over any time slice. Only the total integrated energy/Hamiltonian is unambiguous.

We can make this esoteric-seeming point concrete by considering electrostatics; the straightforward guess for the electrostatic energy density is just $\frac{1}{2} \rho V$, but in practice it's often easier to add a total divergence $\nabla \cdot (V{\bf E})$ which changes the energy density to $\frac{1}{2} E^2$. Griffiths E&M discusses this exact question in section 2.4.4, where he says that the question of where the energy is stored "is simply an unanswerable question ... it is unnecessary to worry about where the energy is located ... the difference is purely a matter of bookkeeping."

So I would say that in some special situations, one particular choice of energy density is so overwhelmingly natural that we can reasonably say that it does describe the energy's "location", but in general we cannot say that the energy is anywhere in particular; only its integral is physically meaningful. But what counts as "overwhelmingly natural" is somewhat a matter of opinion: some people might categorically rule out the canonical stress-tensor because it's gauge-dependent, but others might not.

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OP's title question "Is Energy Localized in space?" translates into physics speak as "Does the system have a stress-energy-momentum (SEM) tensor at each spacetime point?", at least if we're discussing a local relativistic theory.

The answer is often yes, e.g. for the standard model, but there is 1 notable exception: Gravitational energy of a curved spacetime!

One can assign a gravitational energy to a (region of) spacetime, but there is usually not a satisfactory/consistent definition of a gravitational energy-density for a local point of spacetime. (See however, the Landau–Lifshitz SEM pseudotensor.) This is Einstein's equivalence principle at work: At a single point, one may choose Riemann normal coordinates so that the gravitational energy-density vanishes at that point!

Here is an oversimplified non-serious picture from daily life to help visualizing it: Think of a carpet ("spacetime") that is slightly too big for a room, so that it makes a bump ("curvature/energy") somewhere. The bump is not confined to any particular point. By stepping on it you can make the bump go elsewhere, or even split into several bumps. In this somewhat imperfect analogy, a bump movement represents an unphysical/gauge coordinate transformation, i.e. the energy is not localized to a point. However, the total gravitational energy of the curved carpet still makes sense.

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    $\begingroup$ Yes I quite agree. The fact that gravitation allows energy to be moved around via g. waves, yet there is no purely local way to define gravitational energy, is, I think, quite a deep puzzle, hinting at non-local (or at least non-localisable) features of physics even before we invoke quantum theory. $\endgroup$ Apr 16 '20 at 22:52
  • $\begingroup$ Far above my pay grade, but: Isn't the (e.g. gravitational) energy in a region of spacetime the integral over that region of the (e.g. gravitational) energy density at each point? Wouldn't that require that the density at each "point" is defined? $\endgroup$ Apr 17 '20 at 10:31
  • $\begingroup$ I updated the answer. $\endgroup$
    – Qmechanic
    Apr 17 '20 at 10:48
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Since you mentioned $E=mc^2$ we're talking relativistically, and in that case, energy is not a scalar. It is the time-like component of 4-momentum:

$$ p^{\mu} = (E/c, \vec p) = mu^{\mu}$$

where the 4-velocity is:

$$ u^{\mu} = \gamma(c, \vec v)$$

In the rest frame:

$$ p^{\mu} \rightarrow (mc, 0,0,0) $$ $$ u^{\mu} \rightarrow (c, 0,0,0) $$

If you jump over to a quantum description, remember:

$$ p^{\mu} = \hbar k^{\mu} $$

where the 4-wavevector is:

$$ k^{\mu} = (\omega/c, \vec k)$$

In the particle's rest frame:

$$ k^{\mu}\rightarrow (\omega_{\rm cutoff}/c, 0,0,0) $$

there is a cut-off frequency

$$ \omega_{\rm cutoff} = \frac{mc^2}{\hbar}$$

below which excitations cannot exist (this is analogous to microwave propagation in a waveguide, where the wavelength goes to infinity at finite frequency).

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    $\begingroup$ What about Potential Energy? $\endgroup$ Apr 15 '20 at 15:36
  • $\begingroup$ @JamesArathoon there's $j^{\mu}A_{\mu} = qu^{\mu}(\phi, \vec A)$ $\endgroup$
    – JEB
    Apr 15 '20 at 23:35
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On the mechanics level I would also say that one should think of energy as something abstract and not localized in space. On the level of field theory and therefore also QFT one talks about the energy density of a field which is again a scalar field in space. So on a fundamental level I guess one can say that it is localized.

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  • $\begingroup$ Thank you for your answer. That’s interesting, i wonder if there’s any sort of philosophy work on this, i’ve seen a lot on the nature of space and time. I just need a big book on the philosophy of physics. Would be the best birthday present for me $\endgroup$ Apr 15 '20 at 13:44
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    $\begingroup$ You are welcome! $\endgroup$
    – Paul G.
    Apr 15 '20 at 13:56

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