The issue is very simple, but it can be surprisingly hard to find a clear discussion, because standard quantum field theory books choose to avoid discussing position operators at all, for reasons that will be clear below. Mathematical physicists do still discuss it today, but they like to use lots of jargon, which obscures the simplest and most physically important cases. I will give a short answer using notation familiar from undergraduate quantum mechanics.
The setup
Let $|\mathbf{p} \rangle$ be unit-normalized momentum eigenstates for a relativistic particle. In standard relativistic quantum field theory textbooks, one begins with field operators
$$\hat{\psi}(\mathbf{x}) = \int_{\mathbf{p}} \frac{1}{\sqrt{2 E_p}} \, (a_{\mathbf{p}} e^{i \mathbf{p} \cdot \mathbf{x}} + b_{\mathbf{p}}^\dagger e^{-i \mathbf{p} \cdot \mathbf{x}})$$
and defines relativistically normalized states by their action,
$$|\mathbf{x} \rangle_{\text{rel}} = \hat{\psi}^\dagger (\mathbf{x}) |0 \rangle = \int_{\mathbf{p}} \frac{e^{- i \mathbf{p} \cdot \mathbf{x}}}{\sqrt{2 E_p}} \, |\mathbf{p} \rangle.$$
Alternatively, one may define the so-called Newton-Wigner states, which match the definition in nonrelativistic quantum mechanics,
$$|\mathbf{x} \rangle_{\text{nw}} = \int_{\mathbf{p}} e^{- i \mathbf{p} \cdot \mathbf{x}}\, |\mathbf{p} \rangle.$$
Using each set of states, one can define a position operator,
$$\hat{\mathbf{x}}_{\text{rel}} |\mathbf{x} \rangle_{\text{rel}} = \mathbf{x} |\mathbf{x} \rangle_{\text{rel}}, \quad \hat{\mathbf{x}}_{\text{nw}} |\mathbf{x} \rangle_{\text{nw}} = \mathbf{x} |\mathbf{x} \rangle_{\text{nw}}$$
as well as position-space wavefunctions
$$\psi(\mathbf{x})_{\text{rel}} = {}_{\text{rel}}\langle \mathbf{x} | \psi \rangle, \quad \psi(\mathbf{x})_{\text{nw}} = {}_{\text{nw}} \langle \mathbf{x} | \psi \rangle. $$
So it's easy to define two separate quantities, each of which one might naively call "the amplitude for a particle to be at position $\mathbf{x}$".
The problems
So what's the problem? Why don't these simple expressions show up in introductory textbooks?
- The relativistically normalized position states are not orthogonal, as
$${}_{\text{rel}} \langle \mathbf{y} | \mathbf{x} \rangle_{\text{rel}} = \int_{\mathbf{p}} \frac{e^{i \mathbf{p} \cdot (\mathbf{y} - \mathbf{x})}}{2 E_p} \sim e^{-m |\mathbf{x} - \mathbf{y}|}$$
for $|\mathbf{x} - \mathbf{y}| \gg 1/m$. So you can't say a particle is definitely localized at $\mathbf{x}$ if it's in the state $|\mathbf{x} \rangle_{\text{rel}}$, because such a particle also has some amplitude to be in the state $|\mathbf{y}\rangle_{\text{rel}}$.
- The Newton-Wigner position states don't have this problem,
$${}_{\text{nw}} \langle \mathbf{y} | \mathbf{x} \rangle_{\text{nw}} = \delta(\mathbf{x} - \mathbf{y}).$$
But they are not Lorentz covariant: if you boost a particle that's in a Newton-Wigner position state, then it won't remain in such a state. That also seems to contradict what it means to be perfectly localized.
- The worst issue is that both types of position-space wavefunction obey the equation of motion
$$i \dot{\psi}(\mathbf{x}) = \sqrt{- \nabla^2 + m^2} \, \psi(\mathbf{x})$$
which is simply the Klein-Gordan equation restricted to positive frequencies. This is fatal, as solutions of this equation generically spread superluminally! For instance, if the support of $\psi(\mathbf{x})$ is restricted to a finite region, then after an infinitesimal time, it will spread out infinitely far. So you cannot say that $|\psi(\mathbf{x})|^2$ is the probability density to detect a particle at $\mathbf{x}$. If that were actually true, then one could signal faster than light, violating causality.
Because of this final issue, it is very difficult to see how relativistic quantum mechanics can be causal, at the level of particle operators. Causality is only straightforward if you write all interactions in terms of local products of Lorentz covariant fields -- which is the approach taken by all modern textbooks. The tradeoff is that this language doesn't let you speak of the precise positions of particles.
