quantum mechanics - How to interpret this construction of the states in QFT?

Non-Relativistic Quantum Mechanics

To make this question clear it might be useful to contrast with non-relativistic quantum mechanics.

In any quantum theory, the states of a system are unit rays in a Hilbert space and hence can be described by unit vectors $\Psi$.

In the non-relativistic quantum mechanics of a particle, the appropriate Hilbert space is $\mathscr{H}=L^2(\mathbb{R},dx)$. The position operator $X$ is simply $X\Psi(x)=x\Psi(x)$. This allows to interpret $|\Psi(x)|^2$ as a probability density for position.

In any quantum theory, the time evolution is given by a unitary $U(t,t_0)$ which is generated by the Hamiltonian.

Once some state $\Psi_0(x)$ is given, one evolves it to $\Psi(t,x)=U(t,t_0)\Psi_0(x)$. So our conclusions are:

1. 
   

   A state at one fixed time instant is a map $\Psi_0 : \mathbb{R}\to \mathbb{C}$;

   
   


2. 
   
   

   The time-evolution of the system is a path of states $\Psi : \mathbb{R}\times \mathbb{R}\to \mathbb{C}$ so that for each $t$ fixed we have one such state as in (1) which is $\Psi(t,\cdot)$.

   
   

Free QFT in Minkowski Spacetime

Here we consider a free scalar QFT in Minkowski spacetime. This already is able to show the essence of the doubt.

One way to construct the space of states is outlined in Wald's GR book. The question is about this specific construct because it can be generalized to other spacetimes easier than the one based on representations of the Poincare group.

We define the one-particle Hilbert space, $\mathscr{H}$, to be the vector space composed of positive frequency solutions of the Klein-Gordon equation whose Klein-Gordon norm is finite, with inner product on $\mathscr{H}$ defined by

$$(\alpha,\beta)_{KG}=i\int_{\Sigma}(\bar{\alpha}\nabla_a\beta -\beta \nabla_a \bar{\alpha})n^a d\Sigma$$

The Hilbert space of all possible states of the Klein-Gordon scalar field is taken to be the symmetric Fock space, $\mathscr{F}_S(\mathscr{H})$, constructed from $\mathscr{H}$.

So focus on the one-particle Hilbert space $\mathscr{H}$ so constructed. Its elements are the states of one particle. But now these are maps $\Psi : M\to \mathbb{C}$ where $M$ is Minkowski spacetime. In coordinates such a map is $\Psi(t,\mathbf{x})$.

But wait a minute. Now this is confusing. Compare to non-relativistic quantum mechanics. There the state has no time-dependence. The curve of states representing the time-evolution which has time-dependence.

Here the state itself has time-dependence.

So the evolution would be something like $\Psi(t')(t,\mathbf{x})$ with "two time parameters"? This is extremely weird.

Or these states already carry some idea of time evolution into them?

It is also hard to interpret such a state $\Psi(t,\mathbf{x})$. There is no position operator in QFT so it is not possible to say that $|\Psi(t,\mathbf{x})|^2$ is the probability density for the particle to be at $\mathbf{x}$ at time $t$ as in the non-relativistic case.

So how do we interpret this construction of states? How such a $\Psi(t,\mathbf{x})$ is to be interpreted and what justifies the interpretation?