Interpretation of creation of a localized particle in space in QFT

Question

The mode expansion of a field in QFT is $$\phi(\vec x)= \int \frac{d^3p}{(2\pi)^3}\frac{1}{\sqrt{w_p}}\left(a_p^\dagger e^{-i\vec p\cdot\vec x}+a_p e^{i\vec p\cdot\vec x}\right)\quad .$$

From what I have understood, we have taken the analogy from SHO in QM to construct this and here, every mode expansion represents a particle of definite momenta in momentum space. Talking about the field $\phi$, it doesn't refer to a single particle localized in space sort of thing. $\vec x$ here is a parameter here.

My problem arises from the interpretation of the propagator given in P&S (pg.24)

In Eq.(2.41), the RHS implies that $\phi(\vec x)|0\rangle$ is the superposition of definite momenta particles in momentum space. The next line pulling out the NR interpretation is throwing me off.

Firstly, In Quantum Mechanics, we said that there is nothing such as a free particle with definite momenta. Rather, we proceeded with the wave packet description. We did not interpret $|p\rangle$ in the Fourier expansion of $|x\rangle$ as particles in momentum space. So, the one-to-one mapping between $\phi(x)|0\rangle$ in QFT with $|x\rangle$ in QM -- just because it looks similar -- is not sitting well with me.

Moreover, the position eigenstates in QM were localized and we expanded our states in this basis. The $|x\rangle$s were not physical states -- they were not normalized. In QFT, our system is not a single particle to begin with. I'm very confused about how to make sense of creation of a particle at some localized point in space in QFT which in turn is making the description of the propagator being $\langle0|\phi(x)\phi(y)|0\rangle$ difficult to understand because i don't see how $\phi(y)|0\rangle$ is creating a particle at point y in space.

A very similar question has been asked before but the answer to it was along the lines that it looks like the equation in QM. This answer doesn't satisfy me.

Answer 1

There is no such thing as creating a particle at a point in space in QFT. The reason is that if you try to create a particle localised in a region smaller than the Compton wavelength leads to the creation of particle-antiparticle pairs that vitiate the localization, see Section 6.5 of "The Conceptual Framework of Quantum Field Theory" by Anthony Duncan. Field operators are defined at points because of locality of interactions between fields, not because particles are defined at a single point in space.

QFT allows for a notion of effective localization that is good enough for any practical purpose, e.g. - saying a detector is in a particular location so it is measuring a quantum field in that location. A state $|\psi\rangle$ is localised in some region $R$ if the difference between the expectation values of field observables in $|\psi\rangle$ and in the vacuum state is negligible outside $R$. For a discussion of this see Section 5 of

https://arxiv.org/abs/quant-ph/0112148

Answer 2

This passage in P&S is a part where the authors are very carefully stepping around some sensitive issues in QFT. What they are trying to say (I think) without actually saying it is that $$\phi(\mathbf{x})|0\rangle = |\mathbf{x}\rangle . $$ They way they "proof" it, is by showing that $$\langle 0|\phi(\mathbf{x})|\mathbf{p}\rangle = \exp(i\mathbf{x}\cdot\mathbf{p}) \equiv \langle \mathbf{x}|\mathbf{p}\rangle . $$ The single-particle nature of their analysis is revealed by the ladder operators and the fact that each only creates or annihilates a single particle at a time.

The important difference here compared to non-relativistic equivalent is that $|\mathbf{x}\rangle$ is not a state. It cannot be normalized. But that is also true for $|\mathbf{p}\rangle$. However, both $|\mathbf{x}\rangle$ and $|\mathbf{p}\rangle$ are coordinate basis (respectively in configuration space and Fourier space) in terms of which single-particle states can be expressed. In other words, one can define a general single-particle state as $$ |\psi\rangle = \int|\mathbf{p}\rangle \tilde{\psi}(\mathbf{p}) dp \equiv \int|\mathbf{x}\rangle \psi(\mathbf{x}) dx , $$ where $dp$ and $dx$ represent the appropriate integration measures for the respective domains, and $\tilde{\psi}(\mathbf{p})$ and $\psi(\mathbf{x})$ are the single-particle wave functions in the respective domains, related by Fourier transforms.

Of course P&S is not concerned about single-particle states per se. Their discussion is to provide a justification for identifying the field operator $\phi(\mathbf{x})$ as an operator that is associated with a specific point $\mathbf{x}$. Therefore it correctly represents the dynamics that relates different points in space as required for the propagator.