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Why isn't the renormalization of composite operators determined by the renormalization of elementary fields?

+ 7 like - 0 dislike
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In typical expositions of renormalization of composite operators, one needs such renormalization to tame the divergences of some matrix elements such as (say in an interacting scalar theory)

\[\langle \alpha|\phi^2(x)|\beta\rangle\]

One renormalizes $\phi^2$ by introducing the renormalized composite operator

\[(\phi^2)_R=Z_{(\phi^2)}\phi^2.\]

My question is, doesn't the above relation imply $Z_{(\phi^2)}=Z^2_{\phi}$, where $Z_{\phi}$ is the field strength renormalization of $\phi$? That is, once the elementary field is renormalized, how come one can still have the freedom to renormalize the composite operator?

Useful scholarpedia introduction on the subject(still scratching my head reading it): http://www.scholarpedia.org/article/Local_operator#Definition_of_normal_products_and_operator-mixing

asked Mar 8, 2015 in Theoretical Physics by Jia Yiyang (2,465 points) [ revision history ]
edited Apr 26, 2015 by Jia Yiyang

In the book ''The theory of quark and gluon interactions'' by F.J. Yndurain (4th edition), there is on p.76 beginning with (3.4.2) an explicit example for the 1-loop renormalization of a composite operator in QCD. 

@ArnoldNeumaier, thanks for the new reference!

1 Answer

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One has no extra freedom in renormalizing the composite fields (i.e., no new parameters in the renormalization prescription).

However, the normalization factors are not multiplicative since the operator product expansion is singular, and the limits do not commute as it would be needed to show multiplicativity. Therefore the renormalization of $\phi^2$ is quite different from the square of the renormalization of $\phi$ - which is not even defined, since all local fields are distributions only that cannot be multiplied. (In conformal field theories, not even the scaling dimensions add, as one would expect from a naive argument.)

answered Mar 8, 2015 by Arnold Neumaier (12,355 points) [ revision history ]
edited Mar 9, 2015 by Arnold Neumaier
Most voted comments show all comments

 I must admit there must be something very elementary that I don't understand: In the renormalized perturbation theory, when we write down the renormalized Lagrangian, the $\phi^2$ terms simply become $Z^2\phi^2$, where $Z$ is the field strength renormalization factor. However in composite operator renormalization this does not look so. What's the difference in the two cases?

@JiaYiyang: This is because a composite field is not just a product, but a renormalized version of the deformed normally ordered product.

Even in the free case, the product $\phi^2$ is ill-defined, and only $:\phi^2:$ makes sense. In the interacting case, everything is deformed, the vacuum, the field, and the recipe for normal ordering, by infinitely many terms to be constructed perturbatively.

If you use your expression, hence $:(Z\phi)^2:$ in the renormalized Lagrangian, you renormalized $\phi$ but not $\phi^2$, since the normal ordering is still that of the free procedure.

In a Lagrangian setting, what determines the renormalization is the success in getting a meaningful asymptotic perturbation theory. So one just follows the practitioners in the field, who had worked out what works, and does the same to verify that it indeed works.

I know that this is somewhat unsatisfactory, but at present nobody has better rules. A somewhat better understanding of field products is given by the operator product expansion. The cleanest procedure is perhaps to take this as the starting point, as Hollands and Olbermann do; then only physical constants arise.

@JiaYiyang: 

 Weinberg used a very arbitrary (to my eyes) condition in equation (18.1.11)(Vol2 page 117), which basically requires ⟨α|ϕ^2|β⟩ to be normalized to 1 at zero momentum, what does this mean?

Since $Z_{\phi^2}$ is infinite in the limit of infinite cutoff, any finite rescaling of it has the same property. hence the scale of the renormalized field is determined only after fixing some expectation of it. This can be arbitrary, as different ways or choosing the scale just select different multiples of the same field and call this multiple the renormalized field.

The thing is you don't have such arbitrariness in the renormalization of field strength, mass and other coupling constants. Due to LSZ, S-matrix becomes a very solid anchoring point for us to write down the renormalization conditions, and seemingly different renormalization schemes either result from different regularization schemes, or different schemes of relating correlation function to S-matrix, so the freedom of such renormalizations is very limited. What would be the "anchoring point" for the renormalization of local products of operators?

 I gradually realize it is really a question about the meaning of local products, maybe I'll start a new thread if necessary.

Stefan Hollands has several papers on perturbative OPE construction (for scalar fields), which ''proves'' the existence of all local products (i.e., demonstrates their existence at the level of rigor of theoretical physics).

Most recent comments show all comments

@JiaYiyang: As the whole procedure and how to talk about it is ill-defined, it is difficult for me to identify in full detail how to make it well-defined. My way of processing the literature on ill-defined objects is simply to take for granted the properties and tricks that proved to work, and trust an authour unless another one contradicts. So on that level I can say little more than Weinberg, apart from providing my own ill-defined intuition.

Independent of that I am looking for a coherent way to organize things in a more consistent way. In the case of operator products, the rigorous interpretation (though nowhere rigorously completed) is along the lines of Hollands: For free fields, the well-defined variations of the products are the normally ordered products with respect to a vacuum state (or more generally a coherent state - these typically provide inequivalent representations of the CCR). Thus $\phi^2$ should be understood as $:\phi^2:$, and $\phi^4\sim(\phi^2)^2$ as $:\phi^4:=:(:\phi^2:)^2:$. In the case with interactions, everything is deformed according to finite rules that were derived by Hollands in perturbation theory. This deformation defines the renormalized interaction, and $\phi$ and $\phi^2$ are defomed differently. I believe that there lies the future, and some version of this might become fully rigorous one day.

How these two pictures (the traditional one and the near-rigorous one) relate to each other is again given only by some ill-defined intuition that I tried to convey. Don't attach too much meaning to it.

@ArnoldNeumaier, thanks, but I'm hardly convinced, let me think about it.

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