# Square-Integrable Functions in Curved Spacetime

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1. Dec 23, 2013

### stevendaryl

Staff Emeritus
In non-relativistic quantum mechanics, an important set of functions are the normalized square-integrable ones. Those are functions on $\mathcal{R}^3$ such that

$\int |\Psi(x,y,z)|^2 dx dy dz = 1$

I'm just curious as to whether there is some analogous concept for curved spacetime. One complication in curved spacetime is that an integral over "all space" requires a choice of a way to divide spacetime into spatial slices. Is the above condition on $\Psi$ independent of how one slices up spacetime?

2. Dec 23, 2013

### ChrisVer

I also think you can use the invariant volume element in curved spacetimes. Since it's invariant it must have some certain form, which is generally
$dV= \sqrt{-g} d^{D}x$
with $g=detg$
I think it comes out of the Jacobian

3. Dec 23, 2013

### WannabeNewton

What you have written down isn't even a Lorentz scalar in Minkowski space-time. Different time-like congruences of inertial observers determine different foliations of space-time into space-like hypersurfaces; two different foliations (amounting to two different families of inertial observers) will be related through Lorentz boosts. Geometrically, the foliations correspond to planes of simultaneity relative to a given family of inertial observers and Lorentz boosts will tilt the planes of simultaneity by angles related to the rapidity when going from one family of inertial observers to another.

Quantities that are invariant under such Lorentz boosts are of course Lorentz scalars. For example if we have a charged fluid with charge 4-current density $j^{\mu}$ then $\partial^{\mu}j_{\mu} = 0$ implies that $Q = \int_{\Sigma} \rho d^{3}x$ is a Lorentz invariant (here $\Sigma$ is a single plane of simultaneity relative to a given family of inertial observers).

In relativistic QM, the Lorentz invariant total probability* is given by $\int \rho d^{3}x = \int i[(\partial_{0}\varphi)\varphi^{\dagger} - (\partial_{0}\varphi^{\dagger})\varphi]d^{3}x = 1$. This value of unity for the total probability is preserved in the same manner the total charge $Q$ is due to the conservation of the probability 4-current density $j^{\mu} = i[(\partial^{\mu}\varphi )\varphi^{\dagger} - (\partial^{\mu}\varphi^{\dagger} )\varphi]$.

As you noted, what you have written down for the total probability is only valid in non-relativistic QM which uses Galilean relativity as a meta-theory of space-time, not special relativity. As such you can't even use that if you want a Lorentz invariant total probability let alone one that is valid for all space-time foliations, not just those determined by families of inertial observers in Minkowski space-time.

*of course once we go to QFT, the concepts of probability 4-current density and total probability are replaced, for obvious reasons, by those of charge 4-current density (operators) and total charge (operators) as per $j^{\mu} \rightarrow qj^{\mu}$.

4. Dec 23, 2013

### bcrowell

Staff Emeritus