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Local number operators in quantum field theory

+ 3 like - 0 dislike
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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.

This post imported from StackExchange Physics at 2014-08-07 15:36 (UCT), posted by SE-user Issam Ibnouhsein
asked Jul 29, 2014 in Theoretical Physics by Issam Ibnouhsein (120 points) [ no revision ]
Is it a bad idea to post a link to these papers? Not everyone knows them offhand. :)

This post imported from StackExchange Physics at 2014-08-07 15:37 (UCT), posted by SE-user New_new_newbie
You are right! I just added the links.

This post imported from StackExchange Physics at 2014-08-07 15:37 (UCT), posted by SE-user Issam Ibnouhsein

2 Answers

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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.

This post imported from StackExchange Physics at 2014-08-07 15:37 (UCT), posted by SE-user Luboš Motl
answered Jul 29, 2014 by Luboš Motl (10,248 points) [ no revision ]
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?

This post imported from StackExchange Physics at 2014-08-07 15:37 (UCT), posted by SE-user Issam Ibnouhsein
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.

This post imported from StackExchange Physics at 2014-08-07 15:37 (UCT), posted by SE-user Luboš Motl
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!

This post imported from StackExchange Physics at 2014-08-07 15:37 (UCT), posted by SE-user Issam Ibnouhsein
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.

This post imported from StackExchange Physics at 2014-08-07 15:37 (UCT), posted by SE-user Luboš Motl
+ 1 like - 0 dislike

Already in an arbitrary Hilbert space, the product of two unbounded operators need not be defined. The reason is that unbounded operators usually are not defined on the whole Hilbert space but only on a domain dense in the Hilbert space. The product $AB$ is another operator only if there is a dense subspace of the domain of $B$ that is mapped by $B$ into the domain of $A$. This is a nontrivial condition that must be checked in each case. In particular, this holds for Fock space, the arena of free quantum fields.

Now quantum fields are not even operators in Fock space, only operator-valued distributions. In particular, the number operators $N(x):=a^*(x)a(x)$ are well-defined only in a distributional sense. Meaningful as densely defined operators (with a common domain) are only the integrals  $N(f):=\int dx f(x) N(x)$ with a Schwarz test function $f$. Thus products of the form $N(f)N(g)$ can be proved to be well-defined on a suitable domain, while the product $N(x)N(y)$ is questionable.

In nonrelativistic QFT, this proves to be harmless as products of number operators appear only in integrated form, and the integral is nice enough to make operators such as the potential energy well-defined. However, in relativistic QFT, the integral is a local integral, which is not well-defined without renormalization.

answered Aug 27, 2014 by Arnold Neumaier (12,385 points) [ no revision ]

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