Defining spacetime as a Lorentzian manifold, how to define the property $\iiint_{E^3}\Psi^*\Psi d^3{\bf x}=1$? If we define spacetime as a Lorentzian manifold $(E^4,g)$, how can we define the property that
$$\iiint_{E^3}\Psi^*\Psi\ d^3\mathbf{x}=1$$
for a wavefunction $\Psi$? Because there's no global sense of time.
With my approach, I started with a Lorentzian manifold $(M,\,g)$, the frame bundle $LM\xrightarrow{\pi_{LM}} M$, and the associated bundle $E\equiv (LM\times\mathbb{C})/GL(\mathrm{dim}\,M,\mathbb{R})\xrightarrow{\pi_E}M$. Then, a wavefunction is a section $\psi:M\to E$ of this $\mathbb{C}$-vector bundle $E\xrightarrow{\pi_E}M$. This is so $\psi(p)=[\mathbf{e},z]=\{(\mathbf{e}\lhd g,g^{-1}\rhd z)\in L_pM\times \mathbb{C}\ |\ g\in GL(\mathrm{dim}\,M,\mathbb{R})\}$ is independent of frame.
We can also look at a wavefunction as a function $\Psi:LM\to\mathbb{C}$ defined by $\psi(p)=[\mathbf{e},\Psi(\mathbf{e})]$. We can additionally pull $\Psi$ back using a local section $\sigma:U\to LU$ to view it as a local complex-valued function $\sigma^*\Psi:U\to\mathbb{C}$. Whichever global method we take, how do we parameterize $\Psi$ over time so that we can normalize? I tried using the volume form with respect to the metric but this integrates over spacetime instead of just space.
$$\int_M \Psi^*\Psi\ vol_g\neq 1$$
How do we define this property of a wavefunction? Do we need quantum field theory for this task?
 A: If you quantize the manifestly covariant Polyakov action for the point particle, written in its 1st-order form as
$$
S[x^{\mu}(\tau), p_{\mu}(\tau), \lambda(\tau)] = \int d\tau \left( p_{\mu} \dot{x}^{\mu} - \lambda \left( g^{\mu \nu} p_{\mu} p_{\nu} - m^2 \right) \right),
$$
you will end up with wavefunctions which are functions over the spacetime like you wanted, with $x$ and $p$ acting by
$$
\hat{x}^{\mu} \Psi(x) = x^{\mu} \Psi(x), \quad
\hat{p}_{\mu} \Psi(x) = - i \hbar \partial_{\mu} \Psi(x).
$$
However, these functions aren't elements of the physical Hilbert space.
You have to implement the constraint that is fixed by the Lagrange multiplier $\lambda$ as a quantum operator:
$$
\left( \hbar^2 \Box + m^2 \right) \Psi(x) = 0.
$$
This is of course just the Klein-Gordon equation.
Naively, to recover the physical Hilbert space you should look at the rigged Hilbert space of functions over space-time, and select the kernel of the constraint operator to be your physical Hilbert space. However, this operation is only well defined when $0$ is in the discrete spectrum of the contraint operator, and it is easy to see that $0$ is in the continuous spectrum of the Klein-Gordon operator. You therefore need additional input solve this theory.
The solution comes from noticing that the solutions of the Klein-Gordon equation are completely determined by $\Psi(t_0, \vec{x})$ and $\dot{\Psi}(t_0, \vec{x})$ for some $t_0$. Therefore your wavefunctions are two-component functions over the spatial slice parametrized by $\vec{x}$. The two components correspond to particles and anti-particles.
However, it isn't possible to define a positive definite inner product on the space of solutions.
That is not the case for Dirac's equation, where a positive definite inner product can be defined.
In either case, second quantization makes the problem of defining an inner product on the space of solutions to a relativistic field equation obsolete. The modern treatment is to start with the field Lagrangian instead of a particle Lagrangian and derive a positive Hilbert space of the QFT.
