What are the momentum states in the second quantization of a free field?
I am reading through <PERSON>, and have become quite confused by his discussion of second quantization of a free field (Section 2.3, page 20-21).
To be clear, I think I understand second quantization in general, for instance as described by Wikipedia. Here, a basis of states is agreed upon beforehand, then creation and annihilation operators are defined as those operators which raise the occupation number of a given basis state by 1, in a symmetrized/antisymmetrized way (which implies the well-known commutation relations for creation/annihilation operators).
In <PERSON>, however, it seems the creation operators $a_p^{\dagger}$ and the associated basis states $|p\rangle$ are defined simultaneously. More precisely, inspired by the form of the solution in classical field theory, <PERSON> defines the free field Hamiltonian $$H_0 = \int \frac{d^3p}{(2\pi)^3}\omega_p(a_p^{\dagger}a_p + \frac{1}{2}V),$$ in terms of creation and annihilation operators $a_p^{\dagger}$ and $a_p$ for each $\vec{p}$. 669
But as far as I can tell, he never actually specifies what the state created by $a_p^{\dagger}$, $|p\rangle = \sqrt{2\omega_p}a_p^{\dagger}|0\rangle$, is. He calls these states "momentum" states (again, I assume, inspired by the classical field theory case), but this is far from a definition: at this point in the book, no momentum operator has been defined on Fock space, so I don't know what it means for a quantum state to have momentum $\vec{p}$ (of course I know what this means in single-particle quantum mechanics, but is it really well-defined to define a state in Fock space in terms of a state in the Hilbert space of single-particle quantum mechanics?).
So my question is: What is the defining property that <PERSON> has in mind when he declares a single-particle state $|p\rangle$ to have a momentum $\vec{p}$?
(One potentially related piece of the puzzle is that on the following page, <PERSON> defines the $\hat{\phi_0}(\vec{x})$ operator, then notices that $\langle \vec{p} | \hat{\phi_0}(\vec{x}) | 0 \rangle = e^{-i\vec{p}\cdot \vec{x}}$. <PERSON> interprets this finding to mean that the operator $\hat{\phi_0}(\vec{x})$ must create a particle at position $\vec{x}$, since from single-particle quantum mechanics we know that $\langle \vec{x} | \vec{p} \rangle = e^{-i\vec{p}\cdot \vec{x}}.$ So it seems like <PERSON> is certainly implicitly assuming that $| \vec{p} \rangle$ should satisfy certain properties that we know from single-particle quantum mechanics. Still, single-particle quantum mechanics is not typically relativistic, so I am left not totally certain of what $| \vec{p} \rangle$ is meant to mean here.) 669