# Topology and Quantum mechanics

I have a very simple question. Can we know about the topology of the underlying space-time manifolds from Quantum mechanics calculations? If the Space-time is not simply connected, how can one measure the effects of this from calculating transition amplitudes in the path integral formalism?

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Great question! I wonder if you wanted to generalize it slightly by saying "if the spacetime is not simply connected", since this also covers the interesting case where the spacetime is in one piece but has holes in it? – twistor59 Dec 31 '12 at 7:50
I think there have to be two cases here: (1) non-simply-connected spacetimes without closed timelike curves, and (2) those with. Transitioning from one to the other, you get a closed light-like curve, which according to Thorne causes bad things to happen. – Peter Shor Dec 31 '12 at 15:59
– Qmechanic Dec 31 '12 at 16:35
There is a great question IMO on this site, I can't seem to find it, where the OP asks if one can deduce the space-time structure by computing commutators a bunch of times by making measurements in a lab. Since the commutators vanish for space-like separation, one could deduce the light-cone. – kηives Dec 31 '12 at 20:43

Let us for simplicity consider non-relativistic quantum mechanics of a particle on a spatial, $3$-dimensional, connected (possibly curved, possibly non-simply connected) manifold $M$ in the path integral formalism. Then the Feynman propagator

$$\tag{1} K({\bf q}_f, t_f; {\bf q}_i, t_i)~\sim ~ \int_{{\bf q}(t_i)={\bf q}_i}^{{\bf q}(t_f)={\bf q}_f} \!{\cal D}{\bf q}~ e^{\frac{i}{\hbar}S[q]}$$

in principle carries detailed information about the full manifold $M$, since we should sum over all histories in the path integral (1). For instance, if the manifold $M$ is not simply connected, then the overlap (1) will depend on path-histories that are not homotopic.

How the information about the manifold $M$ can be extracted from only knowing the Feynman propagator $K({\bf q}_f, t_f; {\bf q}_i, t_i)$ is an example of an inverse scattering problem.$^1$

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$^1$ Note that these hearing-the-shape-of-a-drum-type problems may not always have unique solutions for the manifold $M$.

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