Attempt to define a momentum space wavefunction for a superposition of of 1-particle states Consider an arbitrary state in the Fock space constructed by superposing 1-particle states: $$|\psi\rangle=\mathbb{1}|\psi\rangle=\int\frac{d^3\textbf{p}}{(2\pi)^32E_p}|p\rangle\langle p|\psi\rangle$$ where I used the completeness relation for 1-particle states. Let me call $\langle p|\psi\rangle\equiv\psi(p)$. Therefore, $$|\psi\rangle=\int\frac{d^3\textbf{p}}{(2\pi)^32E_p}\psi(p)|p\rangle$$ I'm interested in calculating the object $\langle 0|\phi(x)|\psi\rangle$. Since, $\langle 0|\phi(x)|p\rangle=e^{ip\cdot x}$, I get, $$\langle 0|\phi(x)|\psi\rangle=\int\frac{d^3\textbf{p}}{(2\pi)^32E_p}\psi(p)\langle 0|\phi(x)|p\rangle=\int\frac{d^3\textbf{p}}{(2\pi)^32E_p}\psi(p)e^{ip\cdot x}$$
$\bullet$ How to invert this expression to express $\psi(p)$ (in terms of an integral over the LHS expression)? 
$\bullet$ Can $\psi(p)$ be regarded as the momentum space wavefunction corresponding to the state $|\psi\rangle$?
 A: Let
$$
\langle 0|\phi(x)|\psi\rangle=\int d^3\textbf{p}\ f(p)\ e^{ip\cdot x}\tag{1}
$$
where
$$
f(p)\equiv \frac{1}{(2\pi)^32E_p}\psi(p)
$$
Multiply both sides of $(1)$ with $\mathrm e^{i\vec q\cdot\vec x}$, and integrate over $\mathrm d\vec x$:
$$
\int\mathrm d\vec x\ \mathrm e^{i\vec q\cdot \vec x}\langle 0|\phi(x)|\psi\rangle=\int d^3\textbf{p}\ f(p)\ \mathrm e^{iE_pt}\int\mathrm d\vec x\ e^{i(\vec q-\vec p)\vec x}
$$
The integral of the exponential generates a delta function, and so
$$
\int\mathrm dx\ \mathrm e^{-i\vec q\cdot\vec x}\langle 0|\phi(x)|\psi\rangle=(2\pi)^3\int d^3\textbf{p}\ f(p)\mathrm e^{iE_pt}\delta(\vec q-\vec p)
$$
so that
$$
\int\mathrm dx\ \mathrm e^{-i\vec q\cdot\vec x}\langle 0|\phi(x)|\psi\rangle=(2\pi)^3 f(q)\mathrm e^{iE_qt}
$$
This means that
$$
\psi(p)=2E_p\int\mathrm dx\ \mathrm e^{ip x}\langle 0|\phi(x)|\psi\rangle
$$
Your second question is meaningless unless you define what you mean by "momentum space wavefunction". You can for example define it by the expression above, in which case the answer is trivially yes: that expression is the momentum space wavefunction, by definition.
