What happens if the Compton wavelength of a particle exceeds the inverse square root of the Ricci tensor (call it curvature) precisely?

Can somebody please use some formal QFT in curved spacetime to show what happens at least to a scalar field in this case and what the implications would be?

  • $\begingroup$ About the equivalence of our examples, well, I guess at least classically, geometry=matter+energy, therefore dynamic geometry= dynamic (matter+energy), but I don't know if such equality holds if matter behaves quantum mechanically and space remains classical. Perhaps one is forced to go beyond QFT in curved spacetime to account for all possibilities( any theory that quantizes spacetime as well as matter). But at first glance, I can agree with your proposed fabrication of the problem. @ChiralAnomaly $\endgroup$ Sep 24 at 14:25
  • $\begingroup$ @ChiralAnomaly One observation of mine is that, if one inverses the process you mentioned in time, one can see particle creation but this time from an unknown(the one I'm asking for) "initial state" which somehow reminds me of a near big-bang situation where spacetime curvature is so high or a situation in which a particle is moving further from a black-hole. $\endgroup$ Sep 24 at 22:42

The precise details depend on the spacetime metric, the specific QFT model, and the initial state, none of which was specified in the question. This answer focuses on the key ideas, at least for the case of a free scalar field, with enough detail that it could be turned into precise (but usually very difficult) calculations after those other inputs are specified.

Equation of motion for the field

In a generic curved spacetime, the equation of motion for a free quantum scalar field $\phi$ has the form $$ \newcommand{\pl}{\partial} A^{ab}(x)\pl_a \pl_b\phi(x) +B^a(x)\pl_a\phi(x) +C(x)\phi(x) = 0, \tag{1} $$ where the coefficients $A,B,C$ depend on the metric. More explicitly, it has the form $$ g^{ab}(x)\nabla_a\nabla_b\phi(x)+C'(x)\phi(x)=0, \tag{2} $$ where $g^{ab}$ are the components of the inverse metric, $\nabla$ is the metric-compatible covariant derivative, and $C'$ may depend on the Ricci scalar. I won't try to be more explicit about the form of $C'$ here, because I won't be explicitly solving the equation of motion anyway. Remember that we're talking about quantum field theory (QFT), so the field $\phi(x)$ is an operator on a Hilbert space, not a real-valued function. I'm working in the heisenberg picture, where all time-dependence is carried by the field instead of by the state.

Defining particles in generic spacetime

Particles are defined relative to the vacuum state. The vacuum state is usually defined as the state that minimizes the total energy, which in turn is defined by the Hamiltonian $H$ — the operator that generates translations in time. But this all depends on which coordinate system we use to define "time"! A generic spacetime does not have any specially-distinguished time coordinate, which means it doesn't have any specially-distinguished definition of "particle." A state with $N$ particles as defined using one coordinate system generally won't have $N$ particles as defined using a different coordinate system. "Particle" is not a coordinate-independent concept.

Even after we choose a time-coordinate, another complication remains: the corresponding Hamiltonian will generally be time-dependent, so the state that minimizes the expectation value of $H$ at one time may be different than the state that minimizes the expectation value of $H$ at another time. Therefore, the number of particles can change over time, even though the equation of motion is linear in the field $\phi$.

Defining particles in static spacetime

The complications highlighted above could be avoided if we could choose a coordinate system in which the coefficients in equation (1) are independent of the "time" coordinate, because then $H$ is time-independent. In a generic spacetime, no such coordinate system exists, but now let's consider a case where such a coordinate system does exist. Suppose the metric is $$ dt^2-h_{ab}(x) dx^a\,dx^b \tag{3} $$ where the coefficients $h$ are independent of the time-coordinate $t$. They may still be functions of the spatial coordinates $x^a$, so space can still be curved. With this metric and this choice of "time," $H$ is time-independent, and therefore the state that qualifies as the "vacuum state" at one time does so at all times. In this special setting, we can define "particle" without any complications. Write the field operator as $$ \phi(x)=\phi_+(x)+\phi_-(x), \tag{4} $$ where $\phi_\pm(x)$ are the positive- and negative-frequency parts of $\phi(x)$, respectively, with respect to the chosen time coordinate $t$. More explicitly, $$ \phi_+(x)=\sum_f f^*(x)a(f) \hspace{2cm} \phi_-(x)=\sum_f f(x)a^\dagger(f), \tag{5} $$ where the sum is over a "complete" set of negative-frequency complex-valued solutions $f(x)$ of the equation of motion (1), and $a(f)$ is an operator — one for each $f$ in the "complete" set. When we write $a(f)$, we're using the function $f$ as in index. (This streamlines the notation.) I won't bother defining "complete" here, except to mention that in flat spacetime, we usually (but not always) take the functions $f$ to be plane waves, and then "complete" means a complete set of wavenumbers.

The commutator $[a(f),a^\dagger(g)]$ is proportional to the identity operator, with a proportionality factor that depends on the functions $f$ and $g$ and on the metric. The operators $\phi_+(x)$ and $\phi_-(x)$ act as energy-lowering and -raising operators, respectively, so the vacuum state $|0\rangle$ satisfies $$ \phi_+(x)|0\rangle = 0 \tag{6} $$ for all $x$. Equivalently, $$ a(f)|0\rangle = 0 \tag{7} $$ for all $f$. Any state of the form $$ \sum_f c(f)a^\dagger(f)|0\rangle \tag{8} $$ is a single-particle state, with complex coefficients $c(f)$ indexed by $f$. In the flat spacetime case, if we use plane waves for the $f$s, then this is the same as using the wavenumber as the index.

Behavior of particles

First consider the special situation described by equations (3)-(8), so that the number of particles is well-defined and constant, like it is for a free field in flat spacetime in Minkowski coordinates. The particle can propagate and disperse, much like it does in flat spacetime, but now the details of propagation/dispersion are affected by the metric, because the functions $f$ depend on the metric. (Remember: the functions $f$ are solutions of the equation of motion (1).) As an example, if the particle starts in a region of space that is approximately flat compared to the size of the particle's wavepacket and then encounters a region of intense curvature, the particle can scatter — much like a particle can scatter from an external potential in nonrelativistic quantum mechanics. All of this behavior is built into the functions $f$, because we're using the heisenberg picture.

More generally, if the metric does not have the form (3), we run into the complications that I described earlier: the number of particles in a given state depends on which coordinate system we use (the "particle" concept is not coordinate-independent), and even in a given coordinate system, the number of particles can change with time. To see what this time-dependence looks like mathematically, recall that equation (4) refers to positive- and negative-frequency solutions of (1). That's ambiguous when the coefficients in (1) are themselves time-dependent. We can try to avoid that ambiguity by subdividing time into short intervals so that the coefficients are approximately time-independent within each interval, and then we can define approximate notions of positive- and negative-frequency, but this is only an approximation. It can hold asymptotically in some cases, like in asymptotically flat spacetimes, but generically the ambiguity is not avoidable. A state which we deemed to have $N$ particles at one time, or with respect to one observer, may have a different number (typically an ill-defined number) of particles at other times, or with respect to other observers.

For references and some details about a few special cases where this has been studied carefully, see Su's 2017 thesis Quantum effects in non-inertial frames and curved spacetimes (link to pdf).

  • $\begingroup$ Thanks. But still, I might pose a stupid question that seems strange. What if, by any means, I keep the particle localized, perhaps by means of continuous observation( suppose I have a large tube, that on one side the spacetime is flat and on the other side the curvature is so high that its inverse square root is smaller than the inverse of the particle's mass and the particle moves through the tube on a straight line( the geodesic)). Then given the initial and final states, what should I expect to observe as the final state? Suppose the spacetime is static. $\endgroup$ Sep 27 at 9:29
  • $\begingroup$ @BastamTajik If the curvature scale is much shorter than the particle's wavelength, then the particle's wavepacket isn't localized well enough to follow any single geodesic. The particle-follows-a-geodesic concept only makes sense when the wavepacket is localized much more tightly than the scale of the curvature. Even if the particle is artificially confined to a narrow tube, it can still scatter backward: think of transmission/reflection coefficients in the case of an ordinary external potential. (A tube can't keep a particle confined near a single geodesic. Only a moving 3d box can do that.) $\endgroup$ Sep 28 at 13:59

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