I have just started studying the basics of QFT from Matthew Schwartz's book and it says that once we've second quantized a field, that the field operator at point x creates a particle at point x.
Now I'm pretty confused by this because aren't particles described by wavefunctions, so how can a particle have a definite position? Or do I misunderstand something?
I would appreciate any explanation.
Are particles localized in quantum field theory?
That depends on how well-localized you want them to be.
In a strictly nonrelativistic QFT, a field operator $\varphi(x)$ may indeed create a particle that is strictly localized at $x$, with the same caveats in nonrelativistic single-particle quantum mechanics: those states don't really belong to the Hilbert space because they're not normalizable. We can fix that by using the smeared field operator $\int d^3x'\ f(x')\varphi(x')$ instead, where the support of the smearing function $f(x')$ is limited to an arbitrarily small neighborhood of $x$.
However, Schwartz's book is presumably talking about relativistic QFT. In that case, there is a deeper obstacle: particles cannot arbitrarily well-localized like they can in nonrelativistic QM. They can still be approximately localized, but not arbitrarily. The requirement for wavefunctions to be normalizable is not the issue.
In QFT, "location" is defined by the field operators. Observables are constructed from the field operators, and that defines the relationship between observables and locations. Consider the simplest QFT: a free scalar field. Try to construct an observable $\Omega$ with two properties:
$\Omega$ should be entirely contained within a finite region of space at time $t=0$, which means it should be constructed from field operators that are localized in that region at time $t=0$.
$\Omega$ should reliably count particles. In particular, it shouldn't count anything in the vacuum state, so it should annihilate the vacuum state: $\Omega|0\rangle=0$.
You can't do it. An observable $\Omega$ with these two properties does not exist — not for the free scalar field, and not in any other relativistic QFT. Not even in approximately-relativistic QFTs like lattice QED, which are completely unambiguous mathematically. The general reason is reviewed by Witten in Notes on Some Entanglement Properties of Quantum Field Theory, but the free-scalar-field exercise is still worth doing so that you own the insight.
The fact that such an observable cannot exist means that the concept of a strictly-localized particle just doesn't make sense in relativistic QFT. The concept of an approximately localized particle still makes sense, and that's what you get when you apply the field operator $\phi(x)$ to the vacuum state: a particle that is approximately localized at $x$. If the particle has mass $m$, then its spatial profile has tails that fall off exponentially like $\sim \exp(-mD)$ where $D$ is the distance from $x$, with implied factors of $\hbar$ and $c$ to make the exponent dimensionless. If the particle is massless, then the tails fall off like a power of $D$ instead. A symptom of these tails can be seen in the form of the two-point vacuum correlation function when the two points are spacelike-separated. I said more about this subject in another answer.
The Newton-Wigner position operator is a famous attempt to construct an operator that resembles a particle-position observable in some ways, but the Newton-Wigner position operator is nonlocal, so it doesn't really address the issue. It only obscures the issue.
By the way, field operators don't always create individual particles, localized or not. Free fields are the excepetion, not the rule. In a theory with interacting fields, applying a field operator to the vacuum state might give a state that has a single-particle term as part of a more complicated superposition (the usual formulation of scattering theory exploits this fact), but even that isn't always true. This is reviewed in more detail by Accidental Fourier Transform in another answer.
