Explicit formulas for eigenstates of field operators Let's take the free spin 0 quantum field,
$$\hat{\phi}(\boldsymbol{x},t)=\int d^3p\big(\hat{a}_p u_p(\boldsymbol{x},t)+\hat{a}^\dagger_pu_p^\ast(\boldsymbol{x},t)\big)$$
here $$u_p=\frac{1}{\sqrt{2\omega_p(2\pi)^3}}exp(-i(\omega_pt-\boldsymbol{p}\cdot\boldsymbol{x}))$$
because of the equal time commutation relations, $\hat{\phi}(\boldsymbol{x},t)$ commutes with $\hat{\phi}(\boldsymbol{x},t')$, therefore, we can find simultaneous eigenkets $|\phi,t\rangle$ such that $$\hat{\phi}(\boldsymbol{x},t)|\phi,t\rangle=\phi(\boldsymbol{x})|\phi,t\rangle$$
This is used in path integral quantization of fields.
My questions are:

*

*What are $\phi(\boldsymbol{x})$? Are they arbitrary functions of $\mathbb{R}^3$?

*Is there an explicit formula for $\phi(\boldsymbol{x})$ and $|\phi,t\rangle$ in terms of creation and annilation operators?

 A: *

*Yes, $\phi(x)$ is an arbitrary function of $x$ (maybe with some regularity and boundary conditions if we want to be completely precise)


*This second point is a generalization of the formula expressing a position eigenstate in terms of creation/annihilation operators for the $1$d harmonic oscillator (equation (34) of arXiv:1309.0140), with $\hbar = m = \omega  = 1$ :
$$|x\rangle = \frac{e^{-x^2/2}}{\pi^{1/4}} e^{-\frac 12 {\hat a^\dagger}^2+ \sqrt{2}x\hat a^\dagger}|0\rangle$$
To show that this correct, we use the fact that :
\begin{gather}
e^{\frac 12 {\hat a^\dagger}^2}\hat e^{-\frac 12 {\hat a^\dagger}^2}= \hat a - \hat a^\dagger\\
e^{\sqrt{2}x\hat a^\dagger }\hat a e^{-\sqrt 2x\hat a ^\dagger} = \hat a + \sqrt 2x
\end{gather}
to compute :
\begin{align}
\hat x |x \rangle &= \hat x \frac{e^{-x^2/2}}{\pi^{1/4}} e^{-\frac 12 {\hat a^\dagger}^2+ \sqrt{2}x\hat a^\dagger}|0\rangle \\
&=\frac{e^{-x^2/2}}{\pi^{1/4}} e^{-\frac 12 {\hat a^\dagger}^2+ \sqrt{2}x\hat a^\dagger}\Big( e^{+\frac 12 {\hat a^\dagger}^2-\sqrt{2}x\hat a^\dagger}\frac{\hat a + \hat a ^\dagger}{\sqrt{2}}e^{-\frac 12 {\hat a^\dagger}^2+ \sqrt{2}x\hat a^\dagger}\Big)|0\rangle\\
&=\frac{e^{-x^2/2}}{\pi^{1/4}} e^{-\frac 12 {\hat a^\dagger}^2+ \sqrt{2}x\hat a^\dagger}\left( x  +\frac{1}{\sqrt 2}\hat a \right)|0\rangle\\
&= x|x\rangle
\end{align}
Let us apply this to the scalar field. Let $\hat \phi(p,t), \hat \pi(p,t)$ be the (spatial) Fourier transforms of the quantum scalar field and its canonical conjugate variable.
Let $\phi(x)$ be any (nice enough) function and $\phi(p)$ its Fourier transform. Let :
\begin{align}
\hat{\mathcal O}(t) &= \int \frac{\text d^3 p}{(2\pi)^3} \left(-\frac{e^{2i\omega_p t}}2 (\hat a_p^\dagger)^2 + e^{i\omega_p t}\sqrt{2\omega_p}\phi(p)a_p^\dagger\right) \\
|\phi,t\rangle&=\mathcal N e^{\hat{\mathcal O}(t)}|0\rangle
\end{align}
Then, we have $[\hat{\mathcal O}(t),\hat a_p^\dagger] = 0$ and :
$$[\hat{\mathcal O}(t),a_p] = -a_p^\dagger e^{2i\omega_p t}+\sqrt{2\omega_p} e^{i\omega_p t}\phi(p)$$
Therefore, we have:
\begin{align}
\hat \phi(p,t)|\phi,t\rangle &=\frac{1}{\sqrt{2\omega_p}}\left(a_pe^{-i\omega_p t} + a_p^\dagger e^{i\omega_p t}\right) \mathcal N e^{\hat{\mathcal O}(t)}|0\rangle \\
&= \mathcal Ne^{\hat{\mathcal O}(t)}e^{-\hat{\mathcal O}(t)} \frac{1}{\sqrt{2\omega_p}}\left(a_pe^{-i\omega_p t} + a_p^\dagger e^{i\omega_p t}\right) e^{\hat{\mathcal O}(t)}|0\rangle \\
&=\mathcal Ne^{\hat{\mathcal O}(t)}\frac1{\sqrt{2\omega_p}}\left(e^{-i\omega_p t}\left(a_p - e^{2i\omega_p t} a_p^\dagger  + e^{i\omega_p t}\sqrt{2\omega_p}\phi(p)\right) + a_p^\dagger e^{i\omega_p t}\right) |0\rangle \\ 
&= \phi(p)|\phi,t\rangle
\end{align}
Taking the inverse Fourier transform, we obtain :
$$\hat \phi(x,t)|\phi,t\rangle = \phi(x) |\phi,t\rangle$$
