# Why is $\vert \phi \vert ^2$ infinite in QFT?

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I've read here¹ that for a scalar field $\phi$, the square $\vert \phi \vert ^2$ is infinite (which gives an infinite contribution to mass), more precisely:

the square of the field – a quantity which diverges in QFT as necessary consequence of the commutation rules of the theory and unitarity.

It seems that this is related to the fact that $\phi$ is a distribution and the square would be the correlation function.

if we compute in quantum field theory a correlation function like $\langle A(x)A(y)\rangle$ and let $x\to y$, we find a divergent quantity.

Is there a simple way to see why is this true, with little knowloedge of QFT?

This post imported from StackExchange Physics at 2014-03-31 16:03 (UCT), posted by SE-user jinawee
retagged Mar 31, 2014
Minor comment to the post (v1): Please consider to mention explicitly author, title, etc. of link, so it is possible to reconstruct link in case of link rot.

This post imported from StackExchange Physics at 2014-03-31 16:03 (UCT), posted by SE-user Qmechanic

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Let $\Phi(x)$ be a free scalar field. If $\Phi^2(x)$ were a well-defined operator then the commutator $[\Phi(x),\Phi(y)]$ would vanish in the limit $y\to x$, while in fact it diverges. The latter follows by considering the Fourier transform; see, e.g., the beginning of the QFT book by Peskin/Schroeder.

answered Mar 31, 2014 by (15,448 points)
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It does not really say that. All the fields (not just scalars) are finite at any given point in space. The correct statement (in the context of the mentioned paper) is that for a scalar field, the correction to the squared mass term involves the average over all space of another field $\langle|A(x)|^2\rangle$ where <> indicates averaging over all space. Now the average over all space of any quantity might be divergent (infinity) and if this is the case then we have to make sure we understand what that infinite correction in the mass means. This is explained partly in the paper as well as in any QFT book that has a chapter on renormalization.

This post imported from StackExchange Physics at 2014-03-31 16:03 (UCT), posted by SE-user Heterotic
answered Feb 23, 2014 by (525 points)
Thanks. And why does he mention the commutation relations?

This post imported from StackExchange Physics at 2014-03-31 16:03 (UCT), posted by SE-user jinawee
@jinawee : This is not easy to explain without going into the dirty details. The big picture is that a QFT is a quantum theory so all the observables are turned into operators. The commutation relations between these operators are a fundamental ingredient of the theory and will of course affect many quantities in the theory. As a general comment, I would say you can only learn that much about renormalization from a single paper. I would recommend having a look in some QFT books for more details.

This post imported from StackExchange Physics at 2014-03-31 16:03 (UCT), posted by SE-user Heterotic

Free relativistic quantum fields in 4 dimensions are always operator-valued distributions, without definite operator values at particular points. Would they have definite values then the (anti)commutators would so, too, but these commutators are c-number distributions that diverge at equal time, by simple symmetry arguments.

This is completely wrong.

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