# Eigenstates of a bosonic field operator

Even though related questions are discussed here and here, I am still confused about the eigenstates of the field operator of a bosonic field $$\hat{\phi}(\vec{x},t=0)|\phi\rangle=\phi(\vec{x})|\phi\rangle$$ Does this mean that all states of the QFT are related to a field configuration in space? The states are said to fulfill the completeness relation $$\int d\phi(\vec{x})\,|\phi\rangle\langle\phi|=1.$$ The measure here means that we integrate over all field configurations? Or does it mean that we integrate over all values a field can take at position $\vec{x}$? It must be the first case since a state can not already be specified by just the eigenvalue with respect to the field operator at one point, right? This would be in agreement with $$\langle\phi_a|\phi_b\rangle=\prod_\vec{x}\delta(\phi_a(\vec{x})-\phi_b(\vec{x})).$$ So shouldn't one rather write $$\prod_\vec{x}\int d\phi(\vec{x})\,|\phi\rangle\langle\phi|=1.$$

How would one take the trace of an operator? $$\text{tr}\,\hat{\mathcal{O}}=\int d\phi(\vec{x}) \langle\phi|\mathcal{O}|\phi\rangle$$ or $$\text{tr}\,\hat{\mathcal{O}}=\prod_\vec{x}\int d\phi(\vec{x}) \langle\phi|\mathcal{O}|\phi\rangle$$ or even something different?

The formulas I used are from Kapusta "Finite Temperature Field Theory Principles and Applications", p. 12+13.

The $x$ in the measure $D\phi(x)$ of the functional integral $\int D\phi(x) F(\phi(x))$ is a dummy variable that gets integrated over implicitly, so, no, you should not integrated it again. In fact, $D\phi(x) \propto \prod_x \Delta \phi(x)$, assuming a discretized $x$, where $\Delta\phi(x)$ represents the difference between two (close) instances of the function $\phi(x)$.
Likewise $\int D\phi(x) \left| \phi \right\rangle \left\langle \phi\right|$ really means $\int D\phi \left| \phi \right\rangle \left\langle \phi\right|$ where $\phi$ denotes a single possibility for the entire function $x\rightarrow \phi(x)$ at all $x$.
Regarding the question of the relation between $\left|\phi\right\rangle$ and $\phi (x)$: you take an arbitrary function $\phi (x)$ and you assign a state $\left|\phi\right\rangle$ to it. Conversely, the identity of the state $\left|\phi\right\rangle$ is defined by the field configuration $\phi (x)$. These are very abstract states, like the position states in ordinary QM: you just assume the association between $x$ and $\left|x\right\rangle$.