quantum field theory - Why should the modes of the linearized metric perturbation be "wavefunctions" of gravitons (in the Randall-Sundrum model)?

In "An Alternative to Compactification" by Randall and Sundrum, they discuss the localization of "graviton modes" around the Planck brane in the Randall-Sundrum model where we have a compact fifth interval-like dimension (an $S^1/\mathbb{Z}_2$ orbifold, to be precise) at whose two ends two branes (the Planck and the TeV brane) sit:

Taking a linearized metric perturbation

$$g_{MN} = \eta_{MN} + h_{MN}$$ and expanding $h_{\mu\nu} = \sum_n h^{(n)}_{\mu\nu}(x)\psi^{(n)}(y)$ where $x^\mu$ are the 4D coordinates and $y$ is the fifth compact dimension, a Schrödinger-like equation is derived for the modes $\psi^{(n)}$ (using the form from Gabella's notes): $$\left(-\partial_y^2 + \frac{15}{4}\frac{k^2}{(k\lvert y \rvert + 1)^2} - \frac{3k(\delta(y) - \delta(y - L)}{k\lvert y \rvert + 1}\right)\psi^{(n)}(y) = m_n^2 \psi^{(n)}(y)$$ This is solved for the $\psi^{(n)}$ and it is claimed that the $\psi^{(n)}$ are "wavefunctions", showing by inspection of the solution that gravitons are localized around the Planck brane, indicating the weakness of gravity at the TeV brane.

The question is: Why would the $\psi^{(n)}$ be the "wavefunctions" of gravitons?

Either we are in a classical theory, and we have shown localization of classical metric perturbations around the Planck brane, or we are in a quantum field theory and the $\psi$ should be operator-valued rather than "wavefunctions", and should only obey the classical equation of motion as an operator equation. In the latter case, it is not obvious to me that localization of the solution of the classical equation of motion should imply localization of the quantum states belonging to the field.

Answer

This is a slight abuse of terminology, related to talking about 'second quantization.' The word 'wave function' in this case really refers to the 'one particle wave function,' which happens to correspond to the solutions of the (linear) classical equations of motion. It does not refer to the 'wave functional,' ie the Schrodigner representation of the full quantum field theory which is of course a rather complicated object.

Let's work with a scalar field $\phi(x)$ living on some fixed geometry described by $g_{\mu\nu}$ (it's not hard to generalize this discussion to include spin). We will discuss the free theory, which can be thought of as the first step to setting up a perturbative treatment of an interacting theory.

For a free scalar obeying the operator equation

$$\begin{equation} \square \phi = \frac{1}{\sqrt{-g}}\partial_\mu\left(g^{\mu\nu}\partial_\nu \phi\right) = 0, \end{equation}$$ you can expand in mode functions $$\begin{equation} \phi(x) = \sum_{n} a_n u_n(x) + a_n^\dagger u_n^{\star}(x), \end{equation}$$ where $u_n(x)$ are the mode functions associated with the wave operator, appropriately normalized. As an example, for Minkowski spacetime, we can replace the sum over $n$ with an integral $\sum_n \rightarrow \int d^3 p / (2\pi)^3$ and the appropriately normalized mode functions are $u_n(x)\rightarrow u_p(x) = e^{i p\cdot x}/\sqrt{2 E_p}$.

You can then define a position eigenstate (at least formally) by

$$\begin{equation} |x\rangle = \phi(x) |0\rangle = \sum_n u_n^\star a_n^\dagger |0\rangle = \sum_n u_n^\star |n\rangle. \end{equation}$$

In an interacting relativistic quantum field theory the notion of a one particle position space eigenstate isn't obviously well defined because you can't localize particles arbitrarily precisely without having a probability to create particle/anti-particle pairs. But, at the level of the free theory it is ok and if we work perturbatively and squint we can imagine that the notion of position eigenstate will be approximately correct so long as we keep in mind that this is an idealization and we don't try to compute physical quantities that rely on a particle being localized to within its compton wavelength (or you could be more rigorous and define everything in terms of wavepackets).

Carrying on ignoring that subtlety, we can define the wave function of the one particle state $|n\rangle$ by its projection onto the position eigenstate

$$\begin{equation} \psi_n(x) \equiv \langle x | n \rangle = u_n(x), \end{equation}$$ where $|n\rangle = a_n^\dagger |0 \rangle$ (note that $n$ here is not an occupation number, it is a label for the modes). Intuitively a particle is a quanta of the mode $n$, roughly corresponding to a small vibration in a given classical mode of the field which gives some sense of where you might be likely to find that particle.

Thus, there is a direct relationship between the one particle position eigenstate (the 'wavefunction'), and the modes of the classical equations of motion.

What you typically want in extra dimensional models like Randall-Sundrum is for the one particle eigenstates to be localized in the extra dimensions. Let's say that spacetime is $4+1$ dimensional, and our universe lives on a brane with 3 spatial dimensions living in this larger space. The idea is that if an observer on the brane creates a graviton, that graviton should 'look like' a graviton appropriate for 3+1 dimensions which in turn will guarantee that scattering amplitudes computed by an observer on the brane will 'look like' scattering amplitudes for an observer in a 3+1 dimensional spacetime. (The exact definition of 'looks like' depends on your setup and on the experimental constraints on extra dimensions). If this condition is not satisfied, the model would be ruled out, for example by tests of the inverse square law. We can satisfy the condition if the graviton wavefunction (=the classical mode associated with the graviton wave operator = the 1 graviton state projected into the position basis) is sufficiently localized on the brane. In that case, acting on the (4+1) vacuum with a one particle graviton creation operator creates a state that 'looks like' the state you would get in 3+1 dimensions by acting with a graviton creation operator.