TITLE: What makes a one particle state?
QUESTION [5 upvotes]: I'm trying to understand free particle states in quantum field theory but I'm having trouble with one thing: what exactly defines a one particle state?
For example, we can define a 'plane wave' as a momentum creation operator acting on the vacuum: $a^{\dagger}_p|0\rangle$, and we can write a "two particle 'plane wave'" state as two different creation operators: $a^{\dagger}_p a^{\dagger}_q |0 \rangle$. We could continue to develop three, four etc. states using the same method.
But why is a localized particle state not a state containing infinitely many particles?
It is meant to represent a single, localized particle. But following the same logic as above it should be uncountably infinite particles i.e. 
$$\int_{-\infty}^\infty \! \! dp \, \, e^{-px^2} a^{\dagger}_p \rvert 0 \rangle$$
or something of the like. Why is this considered a one particle state and not a state of a collection of plane wave particles?

REPLY [3 votes]: It depends, roughly speaking, on what is the operator that you interpret as the number operator of the theory. It is customary to choose $N$ as the operator such that: a vector $\phi$ of the Fock space is in the $n$-particle sector if $N\phi=n\phi$, i.e. if it is an eigenvector of the number operator $N$ with eigenvalue $n\in\mathbb{N}$.
A possible choice is $N=\int a^\dagger a$. Observe that the number operator is not just $a^\dagger(\cdot)a(\cdot)$, but the integration of this over the variable. I am not specifying the variable because you can think of $a^{\#}(\cdot)$ as the creation/annihilation operators in the position or momentum representation, it does not matter (the two representations of $N$ are unitarily equivalent, and related by the Fourier transform).
This choice is related to the mathematical theory of Fock spaces, and depends, roughly speaking, from the fact that $a^\#(p)$ (or $a^\#(x)$) are not operators, but make sense as an operator only when they are integrated with a square integrable function, e.g.
$$a^{\dagger}(f)=\int_{-\infty}^\infty f(p) a^\dagger(p)dp$$
with $f(p)\in L^2$ is an operator on the Fock space ($a^\#(p)$ are often called operator valued distributions). You superposition for example is not a vector of the Fock space, because $e^{-px^2}$ is not square integrable w.r.t. $p$. 
You see that the choice above implies: $N\lvert 0\rangle=0$, and formally (it is easy to see using the canonical commutation relations):
$$N\int_{-\infty}^\infty dp e^{-px^2}a_p^\dagger\lvert 0\rangle=\int_{-\infty}^\infty dp e^{-px^2}a_p^\dagger\lvert 0\rangle$$
So your "vector" (as I said it does not belong to the Fock space) has only one particle (and the vacuum zero). It is also easy to see (again using the commutation relations) that for any $n\in\mathbb{N}$, $f\in L^2(\mathbb{R}^{n})$ the vector:
$$\phi_n=\int f(p_1,\dotsc,p_n)a^\dagger(p_1)\dotsm a^\dagger(p_n)dp_1\dotsm dp_n\lvert 0\rangle$$
belongs to the $n$-particle subspace, i.e. $N\phi_n=n\phi_n$.
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