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