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  Coupling constants in QFT equations of motion

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My question is about the coupling constants that appear in the quantum analogue to the equations of motion in a classical field theory (the Schwinger-Dyson equations). As a concrete example say we have the Lagrangian $$ \mathcal{L} ~=~ \frac{1}{2} \left ( \partial_{\mu} \phi \partial^{\mu} \phi - m^2 \phi^2\right ) - \frac{\lambda}{4!} \phi^4.$$ Then by a short argument (replace $\phi$ by $\phi+\epsilon$ in the path integral) we should get the vacuum expectation value of the equations of motion vanishing $$ \left \langle (\Box + m^2) \phi + \frac{\lambda}{3!} \phi^3 \right \rangle ~=~ 0.$$ We can also include other operators inside the expectation value if we include the proper contact terms. So in the following (so that the equation is not trivial) imagine that the equation of motion is multiplying some operator.

Now my question is for which $m$ and $\lambda$ is this true? The bare coupling constants don't make sense without a corresponding regularization scheme and scale $\Lambda$, which does not appear in this equation in an obvious way. Likewise the renormalized coupling constants are associated with a scale.

It seems to me these equations must be ill defined and there must be some regularization at some point. And I see that the product of fields at the same point in $\langle\phi^3(x)\rangle$ is problematic. I was wondering if someone could clarify how this equation is interpreted and how the renormalization of the coupling constants comes about?


This post imported from StackExchange Physics at 2016-08-31 10:13 (UTC), posted by SE-user octonion

asked Aug 27, 2016 in Theoretical Physics by octonion (145 points) [ revision history ]
retagged Aug 31, 2016

The correct renormalized field equations are discussed for the massive case in an old paper by Zimmermann. The modern setting of QFT replaces field equations by operator product expansions, which are a more general concept that makes the kind of limits involved precise (at the usual level in QFT). Rigoous results are available only in 2 and 3 dimensions.

1 Answer

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Soon after the mathematical properties of renmormalized relativistic quantum fields were investigated it became clear that the field equations derived from a Lagrangian couldn't make sense as operator equations. The principal reason is that the product of fields at the same point is ill-defined, due to the distributional nature of the fields. Some exactly solvable models in 2 spacetime dimensions (where normal ordering is the only kind of renormalization needed) revealed that in place of the usual field equations one had equations involving the corresponding normally ordered products, but in three dimensions it was found that even these could not be correct.

Gradually it was found that the correct form of the field equations could always be written in a limiting form involving products of fields at infinitesimally different arguments. This lead to Wilson's concept of operator product expansions (OPEs), in which such limits of products of fields are systematically explored and utilized. (Wilson's influential original paper from 1964 was faulty and therefore never published, and the first published papers on the OPE were two 1967 papers by Brandt and by Zimmermann; see the review of Wilson's second paper from 1969.)

For massive $\phi^4$ theory, the correct renormalized field equations are described in W. Zimmermann Local field equation for $A^4$-coupling in renormalized perturbation theory, Comm. Math. Phys. 6 (1967), 161-188. Note that the mass and the coupling constant appearing in the final equations have their physical, renormalized values. For massive $\phi^4$ one needs an OPE for $\phi(x)\phi(y)$ which introduces a renormalized version $\phi_2(x)$ of $\phi(x)^2$, and an OPE for $\phi(x)\phi_2(x)$, which turns out to introduce not a new renormalized field but the d'Alembertian of the original field.

The massless case is significantly more complicated since anomalous exponents must be accounted for correctly. The correct formulation is described in Brandt and Ng Wing-Chiu, Quasicanonical quantum field theory, Phys. Rev. D 9 (1974), 373-386; see also the follow-up paper in Phys. Rev. D 19 (1979), 503-510.

After the usefulness of OPEs was generally recognized (and flourishes till today), the interest in field equations in itself waned, and work on it for other field theories seems virtually inexistent. In terms of OPEs, a field equation is just a nontrivial linear relation between the local fields of sufficiently low scaling dimension and their derivatives. This point of view is discussed (for the case of wick-rotated, Euclidean QFT) in a lecture by E. Witten, Composite operators and operator product expansion (Lecture 3, pp. 445ff of Volume 1 of Quantum Fields and Strings: A Course for Mathematicians); see in particular Section 3.7.

In general, one can define the family of local operators of a theory directly as the collection of all operators satisfying causal commutation relations among themselves and with an assumed family of Wightman fields (e.g., those that figure in the Lagrangian). That this gives a valid definition was proved by Borchers, hence the resulting family is called the Borchers class. Under natural conditions, the smeared fields from the Borchers class form a Hilbert space carrying a unitary representation of the Poincare group augmented by the dilation group. Decomposing it into its irreducible constituents defines the fields figuring in the OPE. They are naturally ordered by the scaling dimension, which is determined by the action of the dilation group. The interesting part of the OPE describes how the tensor product of two fields (in practice, of low scaling dimension) decompose into irreducibles.

In this global view, the complete field content is assumed given at the start, and the family of OPEs is essentially the multiplication law, Just as in a Lie algebra one assumes the generators as given and postulates the commutation laws. On the other hand, in a recursive construction, one starts with the most fundamental fields and obtains the others from the OPEs, just as one can start in so(3) with the two ladder operators $J_\pm$, define $[J_+,J_-]=:J_3$ (corresponding in the analogy to a particular OPE), and then require closure under commutation (which corresponds to the remaining OPEs).

The advantage of the global view is that it always works, while the recursive view requires a Lagrangian as a starting point. In addition, it requires lots of hocus pocus to turn the Lagrangian formalism into a well-defined renormalized perturbation theory - and even then it works only on the perturbative level. But perturbation theory works in the textbook fashion only in the absence of bound states and other infrared obstacles. Under these restrictions, there is (perturbatively, as Witten shows) a 1-1 correspondence between normally ordered products of the free field and interacting local field operators.

The latter can be obtained by composite operator renormalization. This process uses only a small part of the information in the OPEs. For example, to renormalize the composite operator $:\phi(x)^2:$, one needs to find a corresponding local field in the OPE for $\phi(x)\phi(y)$. Now, perturbatively, this OPE contains just one new field with a singular prefactor - which is declared to be the renormalization of $:\phi(x)^2:$ (defined only up to certain renormalization constants, as explained by Witten).

Note that many CFTs have no Lagrangian, so that composite operator renormalization makes no sense! (Even the notion of compositeness becomes problematic.) The global OPE view is also preferable in curved space; see work by Hollands, e.g., http://arxiv.org/abs/1105.3375 who starts with the OPE rather with a Lagrangian.

Also note that all papers and results mentioned (except for the properties of the Borchers class, whcih is a rigorous result) work at the relaxed level of rigor conventionally used in QFT. Rigorous results on field equations are available only in 2 and 3 space-time dimensions (and even there only sparingly).

answered Sep 4, 2016 by Arnold Neumaier (15,787 points) [ revision history ]
edited Sep 8, 2016 by Arnold Neumaier

At first sight the only troubling term $\phi^3(x)$ can be defined by composite operator renormalization. Is there an obvious reason you need the full power of OPE here? 

@JiaYiyang: An OPE is an expansion of some product of operators. ''The OPE'' is the collection of all OPE's. Of course the latter is far more than is needed for the field equations; it is coextensive with all renormalized compositie operators and their products.

For massive $\phi^4$ one needs an OPE for $\phi(x)\phi(y)$ which introduces a renormalized version $\phi_2(x)$ of $\phi(x)^2$, and an OPE for $\phi$ times $\phi_2$ which turns out to introduce not a new renormalized field but the d'Alembertian of the original field. Whether one discusses the details in terms of the OPE language or in terms of composite operator renormalization is a matter of taste as it leads to precisely the same results.

Thank you very much for pointing out these papers and explaining their significance.

For massive $\phi^{4}$ one needs an OPE for $\phi(x)\phi(y)$ which introduces a renormalized version $\phi_{2}(x)$  of  $\phi(x)^2$

But shouldn't the logic go the other way? One needs to define composite operators before one can do OPE, since composite operators may appear in the result of an OPE.

@JiaYiyang: The logic cannot go the other way since the only way to define the composite operators is through some OPE.

Note that one can define the family of local operators of a theory directly as the collection of all operators satisfying causal commutation relations among themselves and with the Lagrangian fields. That this gives a valid definition was proved by Borchers, hence the resulting family is called the Borchers class.

Under natural conditions, the smeared fields from the Borchers class form a Hilbert space carrying a unitary representation of the Poincare group augmented by the dilation group. Decomposing it into its irreducible constituents defines the fields figuring in the OPE. They are naturally ordered by the dimension of the action of the dilation group. The interesting part of the OPE is how the products of the fields of low dimension decompose into irreducibles.

Suppose I have an OPE $A(x)B(y)=\frac{1}{|x-y|^\Delta} C(x)+\cdots$ where $A,B,C$ are all composite, how can this OPE make any real sense before composite operator renormalization?

@JiaYiyang: How are $A(x),B(x),C(x)$ defined in the first place?

Given a Wightman field theory one has naturally a Borchers class of local operators, gives names to the generators in irreducible representations of low dimensions, then one knows (on the conceptual level) what $A,B,C$ are. Then one forms the tensor products and decomposes them to find the OPE. Nowhere the concept of a composite operator appears.

If one does not assume the Borchers class beforehand but only a single Wightman field, one takes $A$ and $B$ as components of already known fields, finds the singular part of the decomposition of the tensor product, and may find in this way additional fields $C$ (in addition to already known fields and their derivatives). This leads in the free case to the normally ordered products, in the interacting case to renormalized composite operators. Thus the latter appear automatically in a recursive way. However, the latter holds only if there is at most one new singular field on the right hand side of the corresponding OPE; otherwise (e.g., for massless $\phi^4$) the traditional process of renormalizing composite operators breaks down. Thus starting with the OPE is the more general approach.

 How are A(x),B(x),C(x) defined in the first place?\

By composite operator renormalization, that in fact was my main point.

I'm quite unfamiliar with the language you used in your second paragraph. But in general my confusion is that, for me the term OPE is a nontrivial relation between objects A,B and C, a QFT version of Taylor/Laurent expansion. But for the last sentence to be meaningful one must make sure A, B, C are first well defined, through renormalization. But it seems somehow you are saying OPE itself can be used as a defining relation, that simply doesn't look like the "OPE"  in my dictionary, hence the confusion.

@JiaYiyang: Well, once one studies a bit of conformal field theory (where the OPE of a chiral CFT figures under the name of vertex operator algebra) one gets a more global view of the OPE along the lines I mentioned.

Indeed, one can use the OPEs to fully define a QFT from scratch, without any Lagrangian input, though so far, this has been proved only in perturbation theory and in certain situations (by Hollands). This is analogous to defining Lie algebras by generators and commutation rules. I added substantial explanations in my answer.

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