(Note: Those who are already familiar with the notion of OPE's might prefer to skip directly to the Question section below)

Setup: Consider, for simplicity, the quantum theory of the Euclidean K-G field. That is, the "quantization" of the the classical field which obeys the equations of motion,

\begin{align} (\partial^2-m^2)\phi=0, \end{align} where $\partial^2\equiv \sum^4_{i=1} \partial^2/\partial (x^i)^2$ with $x^i$ denoting elements of 4-dimensional Cartesian coordinates.

Following the "canonical" quantization procedure for the free field a la Peskin and Schroeder, I find the 2-point (Schwinger?) function

\begin{align} \langle 0| \phi(x_1)\phi(x_2)|0\rangle=\Delta_E(x_1-x_2), \tag{1} \label{2point} \end{align} where $\Delta_E$ denotes the Euclidean propagator/Green's function,

\begin{align} \Delta_E (x_1-x_2)= \int \frac{d^4p}{(2\pi)^4}\frac{e^{i p\cdot (x_1-x_2)}}{p^2+m^2}, \end{align} where $p,x_1,x_2$ are 4-vectors and "$\cdot$" denotes the usual Euclidean inner product.

Now, the 2-point function \eqref{2point} clearly diverges in the coincidence limit $x_1\to x_2$, so it is customary to define the quadratic field operator via a "normal ordering" procedure. This procedure is described in Peskin and Schroeder as "moving all the annihilation operators, $a_\vec{p}$, to the right of the creation operators, $a^\dagger_\vec{p}$" by hand, but this is essentially equivalent to subtracting off the vacuum contribution as,

\begin{equation} :\phi(x_1)\phi(x_2):\equiv \phi(x_1)\phi(x_2)-\Delta_E(x_1-x_2), \tag{2}\label{normal} \end{equation} so that the vacuum expectation of the normal-ordered operators vanish: $\langle 0|:\phi(x_1)\phi(x_2):0\rangle=0$, by definition. Hence, one now has a well-defined quadratic operator,

\begin{align} :\phi^2(x_2):\equiv \lim_{x_1\to x_2} \left\{\phi(x_1)\phi(x_2)-\Delta_E (x_1-x_2)\right\}, \end{align}

in the sense that $\langle0|:\phi^2(x):|0\rangle$ is finite--and, in fact, zero for this choice of normal ordering. (Of course, one could get a different vacuum expectation value via a different choice of normal ordering)

Definition of OPE coefficients: By an "operator product expansion" (OPE), I have in mind a short-distance asymptotic relation (see Scholarpedia article) which holds in any arbitrary (well-behaved) state,

\begin{align} \langle \phi(x_1)\phi(x_2)\rangle \sim \sum_{A} C^{\phi\phi}_{A}(x_1,x_2) \langle \mathcal{O}^A(x_2) \rangle, \end{align}

where $C^{\phi\phi}_A$ are the OPE coefficients of this expansion around $x_1\to x_2$ and the summation runs over all monomials of the fields and its derivatives.

For the free theory, it would appear such an expansion can be obtained via Taylor expanding the terms inside the normal ordering sign of \eqref{normal} as

\begin{align} \phi(x_1)\phi(x_2)= \sum_{\alpha} \frac{(x_1-x_2)^\alpha}{\alpha!} :\partial_\alpha \phi(x_2)\phi(x_2):+\Delta_E (x_1-x_2), \label{expansion} \tag{3} \end{align} where $\alpha$ is a multi-index.

Clearly, the preceding expansion holds trivially in the vacuum state. Because the particle states of the free theory can be expressed in terms of the vacuum, I believe such an expansion will hold for an arbitrary particle state.

Reading off the nontrivial coefficients of the expansion \eqref{expansion}, I find

\begin{align} C^{\phi\phi}_{I}(x_1,x_2)&=\Delta_E (x_1-x_2)\\ C^{\phi\phi}_{\partial_\alpha\phi\phi}(x_1,x_2)&=\frac{(x_1-x_2)^\alpha}{\alpha!}. \tag{4} \\ \end{align}

I expect to obtain more complicated OPE's in the free theory simply from applications of Wick's theorem.

Question: In the "canonical" approach, I seem to obtain the OPE coefficients for the free theory via normal ordering and Wick's theorem. However, it's not clear to me how one defines/obtains the OPE coefficients working instead strictly in the path integral approach or even how to get started in this direction.

  • $\begingroup$ To get you started, look at section 2.9 of Polchinski volume 1. $\endgroup$
    – user159249
    Sep 26, 2017 at 0:11

1 Answer 1


The OPE is a statement about the asymptotic expansion when $x_1\longrightarrow x_2$ inside correlations. For this you need to introduce other fields just sitting there called spectator fields. It does not matter which formalism you use for calculating these correlations: canonical or path integral. For more mathematical precision about the OPE see my article https://arxiv.org/abs/1604.05259 If that's too much math look-up ref [114] I cited by Witten.

Edit with a bit more detail:

The correlations for the free field are given by the Isserlis-Wick's theorem:

$$ \langle \phi(x_1)\cdots\phi(x_n)\rangle=\sum_{\mathcal{W}}\prod_{\{a,b\}\in\mathcal{W}} \langle\phi(x_a)\phi(x_b)\rangle $$ where $\mathcal{W}$ runs over all set partitions of $\{1,2,\ldots,n\}$ which only contain blocks of size two.

One the other hand $$ \langle :\phi(x_1)\phi(x_2):\ \phi(x_3)\cdots\phi(x_n)\rangle= \sum_{\mathcal{W}\ {\rm nonsingular}}\prod_{\{a,b\}\in\mathcal{W}} \langle\phi(x_a)\phi(x_b)\rangle $$ where by "nonsingular" I mean all pair partitions $\mathcal{W}$ that do not contain the pair $\{1,2\}$. For these contributions there is no divergence when you take the limit $x_1\rightarrow x_2$ with $x_2,\ldots,x_n$ distinct and fixed. The full expansion is indeed obtained by writing the Taylor expansion (in multiindex notation) $$ \phi(x_1)=\sum_{\alpha\in\mathbb{N}_{0}^d}\frac{(x_1-x_2)^{\alpha}}{\alpha!}\ \partial^{\alpha}\phi(x_2) $$ "inside" these expressions.

The full OPE is obtained from $$ \langle \phi(x_1)\cdots\phi(x_n)\rangle= \langle \phi(x_1)\phi(x_2)\rangle\ \langle\phi(x_3)\cdots\phi(x_n)\rangle+ \langle :\phi(x_1)\phi(x_2):\ \phi(x_3)\cdots\phi(x_n)\rangle $$ namely $$ \langle \phi(x_1)\cdots\phi(x_n)\rangle=\sum_{A}\mathcal{C}_{A}^{\phi\phi}(x_1,x_2) \langle \mathcal{O}_{A}(x_2)\phi(x_3)\cdots\phi(x_n)\rangle $$ with $$ \mathcal{C}_{1}^{\phi\phi}(x_1,x_2)=\langle \phi(x_1)\phi(x_2)\rangle $$ and $$ \mathcal{C}_{:(\partial^{\alpha}\phi)\phi:}^{\phi\phi}(x_1,x_2)=\frac{(x_1-x_2)^{\alpha}}{\alpha!} $$

  • $\begingroup$ I'm confused: Are the OPE coefficients I gave in (4) above incorrect for the free field? I would think what I've done above is analogous--modulo neglecting the distributional nature of the field--to some of the calculations you gave in section 1.1 of that paper. $\endgroup$
    – user143410
    Sep 26, 2017 at 12:24
  • $\begingroup$ In the spirit of my original question, I've seen papers like this Holland and Hollands one (arxiv.org/abs/1401.3144) where the OPE coefficients are seemingly compactly defined in terms of "regularized amputated Green's functions (AG's) with insertions" (eq. 2.62 of that paper). Does something akin to their definition hold for the free theory? They are considering perturbative phi^4, so presumably an analogous definition for the free theory shouldn't require as much technology, but I'm not sure how to extract the free theory amidst all of their machinery. $\endgroup$
    – user143410
    Sep 26, 2017 at 12:24
  • $\begingroup$ Yes the coefficients you gave in (4) are correct. I will try to give more details in my answer in a bit but the OPE simply follows from the explicit formulas for correlations given by Wick's theorem. These formulas can be derived in the operator or path integral formalism. That's why I said it does not matter which one you use. $\endgroup$ Sep 26, 2017 at 14:12
  • 1
    $\begingroup$ @user143410: I tried to put a bit more details. The reference by Holland and Hollands you mentioned is the wrong one for your purposes. This one on the other hand arxiv.org/abs/0906.5313 has an extensive discussion of the OPE for free fields. Note that there is an ambiguity in OPE due to the field equation. In the massless free case, the field satisfies the Laplace equation. This motivates the use of traceless tensor monomials which are also needed in the description of irreducible representations of $SO(d)$. $\endgroup$ Sep 26, 2017 at 15:44

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