# Local number operators in quantum field theory

Redhead claims in his paper "More ado about nothing" (http://link.springer.com/article/10.1007%2FBF02054660) that number operators associated with different space points (at fixed time) fail to commute, and hence are not physically meaningful.

However, Halvorson, in his paper "Reeh-Schlieder defeats Newton-Wigner" (http://arxiv.org/abs/quant-ph/0007060), section 3.1, claims that operators $N(x)=a^\dagger(x)a(x)$ are not even mathematically well-defined. However I can't understand in what sense his argument using phase invariance proves that such operators are not well defined: we are simply taking the product of two unbounded operators. This product might indeed not have a clear physical sense (more precisely no "nice" localisation properties), but this was more or less Redhead's claim.

So basically I'm trying to understand if $N(x)$ is not associated to any local algebra and hence is not physically meaningful or really mathematical ill-defined, and if so what would be a clear argument to prove it.

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Is it a bad idea to post a link to these papers? Not everyone knows them offhand. :) –  The Dark Side Jul 29 '14 at 17:52
You are right! I just added the links. –  Issam Ibnouhsein Jul 29 '14 at 18:30

Such operators are ill-defined in an interacting theory because whatever counterterms we try to subtract, their expectation value in any finite-energy state will diverge.

The closest operators that are well-defined are densities of charge – number operators with signs labeling antiparticles – because the divergent contributions naturally cancel for them.

In free quantum field theories, you may define the number operator and write it as an integral but the integrand won't really be commuting with itself at other points so the attribution of the particles into different points will be misleading.

In the non-relativistic limit of quantum field theory, all these problems go away under some extra assumptions.

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If I understand you well, since in Halvorson's paper we are concerned with a free Bose field, Redhead is closer to be correct than Halvorson: we can define number operators $N(x)=a(x)a(x)^\dagger$, they just fail to commute for different points $x$ hence provide no clear mapping between points of space and the number of particles at each point. What do you think he is trying to say with his argument on phase invariance then? –  Issam Ibnouhsein Jul 29 '14 at 18:33
I think he's effectively saying that the number density operator is defined in the phase space, with the cell-of-phase-space resolution. So if you compute the number of particles in a region, you always have to make assumptions about its momentum i.e. allowed relative phases, too. –  Luboš Motl Jul 29 '14 at 19:21
I don't fully understand your last comment, but I guess it would be too long to develop all the details. Anyway, thanks a lot for your help! –  Issam Ibnouhsein Jul 29 '14 at 20:52
Hi, I actually don't understand it fully, either - I think that in the case of non-relativistic quantum field theory, the statement is really wrong. One may define the number density operator in that limit. The number of particles in a region is nothing else than the "second quantization" (promotion of kets and bras to annihilation and creation operators) of the bilinear expression in bra-kets that would calculate the probability in the 1-particle Hilbert space. There's nothing wrong about it. The nonlocalities, problems, and divergences only start with relativity and loop corrections. –  Luboš Motl Jul 30 '14 at 6:43