It is often said that NRQM is one dimensional QFT. The Haag-Kastler axioms for QFT should apply to NRQM also then.*

So, for the NRQM system $L^2\left(\mathbb{R}^3\right)$ with time evolution given by $$U_t = \exp\left(-itH\right)$$ $$ = \exp\left(-it\frac{p^2}{2m}\right)$$ $$ = \exp\left(it\frac{\Delta}{2m}\right)$$ What is the corresponding net of algebras? I would have thought that the spacetime regions $\mathcal{O}$ would become time intervals $\left[t,t'\right]$ but then I'm not sure what the algebra of observables of this region should be.

Obviously, the locality axiom about space-like separated regions having commuting algebras is trivial here.

*If the starred statement at the top is wrong, can you explain this?


1 Answer 1


When thinking of NRQM as a QFT in one-dimensional spacetime, the Haag-Kaslter axioms become trivial in the sense that all of the local algebras are equal to each other (not just isomorphic, but equal). Working in the Heisenberg picture as always in AQFT, the observables associated with any finite interval of time — such as the position operators $\vec X(t)$ — generate the full algebra of observables, so the "local" algebras associated with all finite intervals of time are equal to each other. This is just the time-slice axiom (aka primitive causality) applied to the case of zero-dimensional space, where a Cauchy surface consists of a single point.

A more interesting way to relate NRQM to QFT is to think of NRQM as the single-particle sector of a strictly non-relativistic QFT. Non-relativistic QFT can be regarded as a net of algebras of observables similarly to relativistic AQFT, except that instead of open regions of spacetime, we consider regions of the form $(S,t)$ where $S$ is an open region of space and $t$ is a single time. Isotony, the time-slice axiom (primitive causality), microcausality, and the spectrum condition all have fairly obvious analogues in the non-relativistic case. This might not be useful for proving theorems, but it's still useful as an intuitive framework, just like the usual AQFT axioms are in the relativistic case.


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