4
$\begingroup$

I know that in Quantum Field Theory in curved spacetime particle creation is possible. My question is whether the opposite "particle vanishing" is also possible? I don't like to use the word particle annihilation because that term is used to describe in particular partice-antiparticle annihilation.

$\endgroup$
1
  • 2
    $\begingroup$ Yes, both of those are possible. If we consider it in a covariant quantum gravity way, this is roughly equivalent to pairs of particle-antiparticle being produced or annihilating from graviton vertices. $\endgroup$
    – Slereah
    Aug 13, 2019 at 11:26

1 Answer 1

2
$\begingroup$

Short answer: Yes, the opposite of particle creation (particle "vanishing") is also possible.

Particles and observables

The question demonstrates that the OP already knows this, but I'll start with this anyway to set the stage: The presence or absence of "particles" is not an intrinsic property of the state itself. It also depends on which observables we're using to probe the state.

Which observables represent particle detectors? The answer depends on which time-coordinate we use (details below). Since each observer has its own proper time, different observers will tend to identify different observables as particle-detectors, so they will detect different numbers of particles in the same state — because they're using different observables.

In everyday life, we tend to think of "particle" as being defined independently of any observers/observables, just like we tend to think of "time" as being the same for all observers. Everyday experience is limited, though, and after we broaden our experience, we realize that time is not the same for all observers, and neither is the number of particles.

The observer-dependence of time is visible in classical physics (special/general relativity), but the observer-dependence of particle-number is visible only in quantum physics. Quantum physics is all about observables (operators representing things that could be measured), and only questions expressed in terms of observables are physically meaningful. To ask physically-meaningful questions about particles, we need to express them in terms of the observables that we would measure in order to detect/count particles.

An example and an answer

To illustrate this, consider a famous example: Hawking radiation. The derivation of Hawking radiation goes something like this:

    1. Consider a spacetime corresponding to a collapsing star, and consider a quantum field in this prescribed spacetime. (The star is treated as a classical object in this model, so it doesn't contribute to the number of "particles" in this model.)
    1. Consider a quantum field in the Heisenberg picture, where observables are time-dependent and states are not.
    1. Consider a collection of observers drifting in space in the distant past (long before the black hole has formed), and consider the state of the quantum field that doesn't have any particles according to the observables that those observers would naturally use as particle-detectors. Call this state $|0\rangle$.
    1. Using this same state $|0\rangle$, consider hovering observers (not falling into the black hole) in the distant future. Consider the observables that those observers would naturally use as particle-detectors. With respect to those observables, the same state $|0\rangle$ has a steady stream of particles emanating from the vicinity of the black hole. This is Hawking radiation.

That's an example of what is usually described as particle "creation." That language is reasonable, because a single observer that was drifting in space in the distant past may end up hovering somwhere (near or far) outside the black hole in the distant future.

Now, to answer the question, consider the time-reverse of the same situation. In the time-reversed situation, the spacetime metric corresponds to a white hole that decays into a star! That will never happen in the real world, for essentially the same reason that balloons never un-pop in the real world: it is statistically absurdly unlikely. Technically, though, it is allowed by the same laws of physics that allowed the original situation, and it provides an example of particle "vanishing."

More generally, given a state that has no particles according to one collection of particle-detection observables will have a non-zero particle content according to most other particle-detection observables. Different time-coordinates are associated with different particle-counting observables.

Given an arbitrary state, can we always find a particle-detection observable that would count zero particles in that state? I'm not sure if this is always possible (or how to make the question well-posed), but even if it is possible, there is a caveat...

A caveat

The idea of a particle-detection observable associated with a specific localized observer is only approximately valid in quantum field theory. As explained in my answer to another recent question, in relativistic quantum field theory, the Reeh-Schlieder theorem implies that an observable that annihilates the vacuum state (or any other state that is "analytic" with respect to energy) is necessarily not localized in any bounded region of spacetime. Conversely, localized particle-detectors are necessarily noisy: they can't reliably register zero particles in any physically-reasonable state. Only non-local particle-detectors can do that.

In practice, this isn't really noticeable for particle-detectors of ordinary dimensions, so it doesn't really affect the spirit of this answer. I just had to mention it to clear my conscience.

Particle detectors and the relativity of time

The preceding answer was just words. Here, I'll briefly explain why different observers will tend to interpret different observables as particle detectors.

Each observer, regarded as a worldline in spacetime, has its own proper time. We can consider a tube-like neighborhood of the observer's worldline and construct a coordinate system in which the time coordinate coincides with the proper time along the given worldline. Working in the Heisenberg picture (where observables are time-dependent and states are not), we can separate any time-dependent observable $A(t)$ into its positive- and negative-frequency components $A_\pm(t)$, where $t$ is the time-coordinate in the coordinate system that was just described. An observable of the form $A_+(t)A_-(t)$ is an example of a natural "particle detector" for the given observer. That's because the operator $A_-(t)$ reduces the energy of any state on which it acts (by construction — see my Math SE question about this construction). The definition of "energy" is tied to the notion of time, so a state that has the lowest possible energy with respect to a given observer's proper time would naturally be interpreted by that observer as a state with no particles. The operator $A_-(t)$ annihilates that state, so the self-adjoint operator $A_+(t)A_-(t)$ is a natural candidate for a particle-detection observable.

[By the way, notice that even if the original observables $A(t)$ are localized in a strictly bounded neighborhood of the given worldline, the operators $A_\pm(t)$ are not. The definition of positive/negative frequency implicitly refers to the infinite past/future, and because of how things propagate, this makes the operators $A_\pm(t)$ non-local in space as well as in time. If we replace these non-local particle-detectors with local observables, then the local versions are necessarily noisy. This is the caveat that was highlighted above.]

Here's the point: since proper time is different for different observers (different worldlines), the set of observables that qualify as "particle detectors" will also be different for different observers. The same state may have non-zero particle-number according to one observer even if it has zero particle-number according to another observer, and conversely.

$\endgroup$
4
  • 1
    $\begingroup$ Thank you for the detailed answer. I am not sure if I have understood everything. But let me briefly write what I understood: I understood that different observers use different observables and therefore detect different number of particles. But my question was whether particles can vanish with respect to one observer: So does the answer to that question lie in the fact, that the observables change with time even for the same observer and therefore the number of particles can change as well? Can you give realistic example for particle vanishing as opposed to a white hole? $\endgroup$
    – eeqesri
    Aug 13, 2019 at 22:47
  • $\begingroup$ @user139383 Here's an example: Let $|0\rangle$ be the state with no particles according to inertial observers in Minkowski (flat) spacetime. A uniformly-accelerating observer will detect a non-zero density of particles in the same state. If an observer is uniformly accelerating at first and then turns off the rocket engine to become an inertial observer, the density of particles seen by that observer will drop from a non-zero value to zero. This could be considered an example of particle-vanishing (particle density changing from non-zero to zero according to a single observer). $\endgroup$ Aug 14, 2019 at 0:34
  • $\begingroup$ @user139383 For a more careful analysis, see Unruh and Wald (1984), What happens when an accelerating observer detects a Rindler particle, Physical Review D 29: 1047 $\endgroup$ Aug 14, 2019 at 0:38
  • $\begingroup$ makes sense. thanks a lot $\endgroup$
    – eeqesri
    Aug 14, 2019 at 8:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.