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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?

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    $\begingroup$ Integrating on the entire spacetime shouldn't work since the probability is only one at a given time. The thing usually done is to make it so that the wavefunction is of unit norm for any Cauchy surface, regardless of foliation. $\endgroup$
    – Slereah
    Commented Jun 15, 2020 at 23:34
  • $\begingroup$ I believe that @Slereah just did express it mathematically $\endgroup$
    – WillO
    Commented Jun 16, 2020 at 3:18
  • $\begingroup$ volume integrals can't be invariant, because volumes transform under lorentz transformations. Any definition of a volume is going to have a trailing timelike lorentz index. $\endgroup$ Commented Apr 24, 2022 at 15:33

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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.

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