Is it just a mnemonic to call $\phi (x)|0\rangle$ a particle at position $x$? We often take $\phi (x) |0\rangle$ to mean preparing a particle at position $x$. We also take $\langle 0|\phi(x) \phi(y)|0\rangle$ to mean the probability of creating a particle at $y$ and observing it at $x$.
I believe these are just mnemonics to get intuitive meanings of terms in the propagator? I don't think one ever talks about real particles at a position using $\phi (x)|0\rangle$. It only shows up as a mathematical term in the S-matrix calculation.
I do get that $\phi (x)|0\rangle$ is a superposition of all momentum eigenkets. But still the coefficients in the superposition are pretty weird.

*

*The coefficents in the superposition have a $e^{-i\omega _kt}$ or $e^{i\omega_k t}$ factor, and a $\frac{1}{\sqrt {2\omega _k}}$ factor. The time dependent exponential factor can make the position representation spread out in space.


*Its not clear to me what it means to create a particle at a time point. The wavefunction looks like $\int \frac{1}{\sqrt {2\omega_k}} e^{-ipx} e^{i \omega t}dp$. This wavefunction seems to imply that a particle was created at $(x,0)$ and was evolved with negative energy $-\omega_k$ till time $(x,t)$. Aren't negative energy particles not allowed?


*The norm of $\phi (x) |0\rangle$ isn't the delta function.


*If we wanted to talk about a real particle at a position $x$, we could just manually write the state $\int e^{-ipx} |p\rangle dp$. There's no reason to involve $\phi (x)$


*The mnemonic only seems to work for scalar field operators $\phi (x)|0\rangle$. How would one interpret something like $A^{\mu} (x) |0\rangle$? $A^{\mu} (x) |0\rangle$ is also a superposition of all momentum states, but it returns four different wavefunctions because of the superscript $\mu$.
For these reasons, I believe it's just a mnemonic. If it is supposed to be taken more literally, then how do we address the above points ? Please help.
 A: $\phi(x)|0\rangle$ is not the state of a particle (I stress that $\phi(x)|0\rangle$ is a one-particle state since I am referring to a free field) with position $x$ (when the temporal component of $x$ is zero in particular).
The situation is different from the momentum representation. Indeed, $a_p^\dagger|0\rangle$ is a momentum-defined one-particle state.
The position representation of the particles of QFT is quite a delicate issue. It is still unclear and actually there are a number of no-go theorems about its existence either in terms of projection valued measures (PVMs) or positive-operator valued measures (POVMs).
An apparent standard definition of the position representation is the famous Newton-Wigner one. It is however plagued by a number of issues concerning locality.
A modern treatise on the issues about the position representation and the various no-go theorems, in relativistic QM can be found here, here, and more recently here.
Probably the most powerful no-go result is the following one.
I should premit some facts. The position representation is defined by a set of operators $P_{E}$ labeled
by sets $E \subset \mathbb{R}^3$ in the 3D rest space of a Minkowski reference space. These operators may be orthogonal projectors as in the case of a spectral measure (a PVM):
$$\vec{X} = \int_{\mathbb{R}^3} \vec{\lambda} dP(\vec{\lambda})$$
is the triple of position operators.
For instance the Newton-Wigner position operator has this form.
A weaker formulation is the one where the $P_E$ simply define a POVM (Positive-operator valued measure) and one is dealing with the modern formulation of observables.
In both cases $\langle \psi| P_E\psi \rangle$ is the probability to find the particle in $E$ at time $t=0$.
In the case of a POVM, every $P_E$ is simply a positive operator bounded by $I$ instead of an orthogonal projector as in the spectral decomposition where one can also take advantage of some quantum logic formulation.
One of the no-go theorems proved in 2 has the following form.
THEOREM(Halvorson Clifton theorem) There is no POVM (or PVM) $P_E$, labeled by sets in the 3D space of a Minkowski reference frame such that:
1) It is covariant under the action of spacetime translations:
$$U^{-1}_{t,a} P_E U_{t,a} = P^{(t)}_{E-a}$$
where $P^{(t)}$ denotes the analogous operator defined at time $t$ (the Heisenberg evolution of the initial one);
2) the generator $H$ of the time translations $U(-t,0)= e^{itH}$ is positive;
3) the operators $P_E$ satisfy locality if $(t,E)$ and $(t',E')$ are spacelike separated then
$$[P^{(t)}_E, P^{(t')}_{E'}]=0$$
The last requirement can be refined or weakened and it corresponds to the requirement that we cannot transmit superluminal information with the outcomes of these detectors (a version of the no-signaling requirement)
For instance the Newton-Wigner operator (obtained by integrating its PVM)  violates (3) and thus cannot be considered a physical observable.
The theorem above (I stated it in a quite rough way actually, for a precise statement see the reference I posted) is a refinement of a number of previous results due to Hegerfeldt, Borchers, Malament, Castrigiano, Busch, in particular.
All that should answer items 3th and 4th.
Regarding the 1st and 2nd points, barring $1/\sqrt{2\omega_k}$ the only exponential appearing in $\phi(x)|0\rangle$ is the one of $a^\dagger_k$. The phase $e^{-it\omega_k}$ (not  $e^{+it\omega_k}$ as it seems you wrote in (2) ) is the correct one since the Hamiltonian is just the factor $\omega_k$ in the momentum basis. $\phi(t,\vec{x})|0\rangle$ is just the Schroedinger evolution up to time t of $\phi(0,\vec{x})|0\rangle$.
Your last point (5) is quite difficult due to the presence of gauge degrees of freedom. In the Hilbert space of the photon there are only two-degrees of freedom whereas the associated quantum field has apparently 4 degrees of freedom.
The formally added 2 degrees of freedom have their reason in making explicit the Lorentz covariance of the theory.
The one-particle state of the photon is obtained by $A_\mu(x)|0\rangle$ after removing the gauge redundancies and it can be done with several procedures (e.g. the Gupta-Bleuler one).
However, the definition of the position operator for photons is even more difficult than for massive particles.
