How do you describe a quantum field theory state?

Question

Intuitively, quantizing a field implies that there exists a probability for each state of the field denoted as $$\Omega[\psi]$$ So the vector $|\Omega\rangle$ must be a superposition of all possible field states weight with an amount relating to probability $$|\Omega\rangle=\Omega[\psi_0]|\psi_0\rangle + \Omega[\psi_0+\delta\psi]|\psi_0+\delta\psi\rangle+\Omega[\psi_0-\delta\psi]|\psi_0-\delta\psi\rangle+...$$ $$|\Omega\rangle=\Omega[\psi_0]|\psi_0\rangle + \sum_{n=1}^{\infty} \left[ \Omega[\psi_0+n\delta\psi]|\psi_0+n\delta\psi\rangle + \Omega[\psi_0-n\delta\psi]|\psi_0-n\delta\psi\rangle\right]$$ Or written with integral notation $$|\Omega\rangle=\int d\phi \, \Omega[\psi_0 + \phi] |\psi_0 + \phi\rangle$$ Is this the correct formula for the field state and is there any other way of writing the field state in terms of any other basis like energy, particles, and other properties.

Answer 1

Particles should be identified with states in the Hilbert space that transform in unitary irreducible representations of the Poincaré group. Single- and multi-particle states are eigenstates of the momentum operator $\hat P^\mu$, with $\hat P^\mu \vert X \rangle = p^\mu \vert X \rangle$ for a set of real numbers $p^\mu$ with $p^0 \gt 0$ and $p^2 \ge 0$, which transform in the 4-vector representation of the Lorentz group. Single-particle states $\vert X \rangle$ transform under unitary irreducible representations of the Lorentz group as well, as $\vert X \rangle \to exp (i \theta_{\mu \nu} S^{\mu \nu}) \vert X \rangle$ where $\theta_{\mu \nu}$ are the boost and rotation angles and $S^{\mu \nu}$ are the generators of the Lorentz group in the representation of that particle. The vacuum $\Omega$ is assumed to be Lorentz invariant and to have zero momentum: $\hat P^\mu \vert \Omega \rangle = 0$.

The completeness relation is
$1 = \sum_X \int d \Pi_X \vert X \rangle \langle X \vert$
where the sum is over all of the single- and multi-particle states $\vert X \rangle$ and
$d \Pi_X = \prod_{j \in X} \frac{d^3 p_j}{(2 \pi)^3} \frac{1}{2 E_j}$

Quantum fields are operators and are defined as integrals over creation and annihilation operators, respectively $a_p^\dagger$ and $a_p$, for each momentum
$\phi (x) = \int \frac{d^3 p}{(2 \pi)^3} \frac{1}{\sqrt{2 \omega_p}} [a_p (t) e^{-i p x} + a_p^\dagger (t) e^{i p x}]$
where
$x = (\vec x, t)$
$signature = (+, -, -, -)$
$a_p^\dagger \vert \Omega \rangle = \frac{1}{\sqrt{2 \omega_p}} \vert p \rangle$
and $\vert p \rangle$ is a state with a single particle of momentum $p$.

You have $\phi (x) \vert \Omega \rangle = \vert x \rangle$, that is $\phi (x)$ creates a particle at position $x$.

Note: In QFT the symbol $\Omega$ is reserved for the vacuum state in an interacting theory.