2nd Quantization of KG Field

The space of solutions of Klein Gordon-equation Klein Gordon-equation is spanned
by eigenbasis consisting of plane wave solutions
$Amathrm{e}^{-i(vec{p}cdotvec{x} pm E_pt)}$ which correspond to
positive (negative) energy eigenvalues $pm E_p, E_p >0$. Therefore a general solution
is represented as integral (as generalization of linear combinations) over all
plane wave solutions weighted by certain coefficients $a(vec{p}), b(vec{p})$.
After change of variables $p to -p$ in second summand the classical solution becomes

$$phi(vec{x},t) = int c cdot d^3pleft[a(vec{p})mathrm{e}^{-i(vec{p}cdotvec{x}-E_pt)} + b(vec{p})mathrm{e}^{+i(vec{p}cdotvec{x}-E_pt))}right] $$

with $c$ an appropriate normalization constant and $a(vec{p})$ and $b(vec{p})$
regarded as coefficients or "weights". Passing to field quantization $hat{phi}(vec{x},t)$ of
$phi(vec{x},t)$ the coefficients $a(vec{p}),b(vec{p})$ become
operators $hat{a}(vec{p}),hat{b}(vec{p})$ and in this question I tried to
elaborate a hint given by my lecturer why $a(vec{p})$ becomes
creation operator and $b(vec{p})$ the anihilation and not vice versa.

We consider now the promoted operator

$$hat{phi}(vec{x},t) = int c cdot d^3pleft[hat{a}(vec{p})mathrm{e}^{-i(vec{p}cdotvec{x}-E_pt)} + hat{b}(vec{p})mathrm{e}^{+i(vec{p}cdotvec{x}-E_pt))}right] $$

and the question was if creation op belongs to negative exponential
or positive one. And searching for an argument I came to following conjecture how the
problem can be attacked. By constuction $hat{phi}(vec{x},t) $ acts on Fock space
of momenta . Let now $|0rangle$ be the vacuum state and the question is which state is

$$hat{phi}(vec{x},t) |0rangle$$

if we don’t know a priori which of the two $hat{a}(vec{p}),hat{b}(vec{p})$ is creation
and which is anihilation. My conjecure and motivation is that I’m searching for
a reason why if $|xrangle$ is a state
corresponding to an eigenvector of position operator $X$, then up to a constant
$hat{phi}(vec{x},t) |0rangle = |xrangle$

unfortunatelly I haven’t found a reason why this should be true. As far as we know that it holds, we can
exploit the relation $langle p |vec{x} rangle =mathrm{e}^{-i(vec{p}cdotvec{x})}$
between eigen vectors of momentum and place operators in order to deduce that
$hat{a}(vec{p}) |0rangle = |prangle$ and not $hat{a}(vec{p}) |0rangle = |0rangle$
by evaluation of

$$langle p | hat{phi}(vec{x},t) |0rangle$$

Then we win since then consequence is that $hat{a}(vec{p})$
is indeed creation operator. But the unsolved problem is still why
$hat{phi}(vec{x},t) |0rangle = |xrangle$? Maybe we may assume $hat{phi}(vec{x},t) vert _{t=0}$ and ask if $hat{phi}(vec{x},0) |0rangle = |xrangle$ is true?