# 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.

• 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