Ward–Schwinger–Dyson equations in $\phi ^3_6$ quantum field theory

Abstract

We develop a system of equations for the propagators and three-point functions of the $\phi ^3$ quantum field theory in six dimensions. Inspired from a refinement by Ward on the Schwinger–Dyson equations, the main characteristics of this system are to be formulated purely in terms of renormalized quantities and to give solutions satisfying renormalization group equations. These properties were difficult to get together, due to the overlapping divergences in the propagator. The renormalization group equations are an integral part of any efficient resolution scheme of this system and will be instrumental in the study of the resurgent properties of the solutions. It is our belief that this method can be generalized to the case of gauge fields, shedding some light on their quantum properties.

Appendices

Parametric representations

It is well known that it is possible to rewrite a Feynman integral as a projective integral. The first step is to use the so called “Schwinger trick”:

$$\begin{aligned} x^{-\alpha }\Gamma (\alpha ) = \int _\mathbb {R_+} dt \, t^{\alpha -1} e^{-tx} \end{aligned}$$

(93)

on each propagator.

Then we need some definitions. A graph is a collection of vertices and edges, a tree (T) is a graph with no loops, and a forest is a disjoint union of trees. A spanning tree (forest) is a tree (forest) which contains all the vertices of the graph. Let $\mathcal {T}$ be the set of spanning trees and $\mathcal {F}_k$ the set of spanning forests with exactly $k$ components. Let I be the set of internal edges $e_i$, V the set of vertices. We associate to each edge a mass parameter $m_i$ and a power of the propagator $\alpha _i$. Finally, let $p_T$ be the sum of the external momenta entering the subgraph $T$. In the case of a graph with only one connected component, $h_1(\Gamma ) = |I| - |V| + 1$ is the number of loops.

It is then possible to associate to a given graph $\Gamma$ the two Symanzik polynomials $\psi _\Gamma$ and $\phi _\Gamma$ with the following definitions:

$$\begin{aligned}&\psi _{\Gamma }(\{t_i\}) := \sum _{T\in \mathcal {T}}\prod _{e_i\notin T}t_i\\&\phi _{\Gamma }\left( \{t_i\},\{p_i \cdot p_j\},\{m_i\}\right) := \sum _{(T_1,T_2)\in \mathcal {F}^2} p_{T_1}^2 \biggl (\prod _{e_i\notin (T_1,T_2)}t_i\biggr ) + \psi _{\Gamma }(\{t_i\})\sum _{i=1}^{|I|}t_i{m^2_i}. \end{aligned}$$

The Feynman integral for the graph $\Gamma$ can then be written as

$$\begin{aligned} \mathscr {I}\left( \{p_i \cdot p_j\},\{\alpha _i\}, \{m_i\}, D\right) = \int _{\mathbb {R}_+^{|I|}} \prod _{i\in I} \Bigl (\frac{dt_i t_i^{\alpha _i-1}}{\Gamma (\alpha _i)}\Bigr ) \frac{e^{-\phi /\psi }}{\psi ^{D/2}} \end{aligned}$$

where we have omitted the variables on which $\psi$ and $\phi$ depend for the sake of readability. From their definition, we can see that the Symanzik polynomials $\psi$ and $\phi$ are homogeneous in the $t_i$ variables with respective degrees $h_1(\Gamma )$ and $h_1(\Gamma ) +1$. If we insert the equality

$$\begin{aligned} 1 = \int _0^{+\infty } d\lambda \, \delta (\lambda - H(t)) \end{aligned}$$

(94)

for any nonzero hyperplane equation $H(t)= H^j t_j$, we obtain:

$$\begin{aligned} \mathscr {I}\left( \{p_i\},\{\alpha _i\}, \{m_i\}, D\right) = \prod _i^{|I|} \int _\mathbb {R_+} \frac{dt_i t_i^{\alpha _i-1}}{\Gamma (\alpha _i)} \frac{\delta (1-H(t))}{\psi ^{D/2}} \int _0^{+\infty } d\lambda \, \lambda ^{\sum \alpha _i - D/2 \, h_1(\Gamma ) -1 } e^{-\lambda \phi /\psi } \end{aligned}$$

with a small abuse of notation due to the scaling of $t_i$. In terms of the superficial degree of divergence $\omega$,

$$\begin{aligned} \omega := h_1 \frac{D}{2} - \sum _i^{|I|} \alpha _i , \end{aligned}$$

the integral on $\lambda$ gives $\Gamma (-\omega ) (\phi /\psi )^\omega$. The presence of the hyperplane in the delta function induces a split on the domain and on the volume form which brings us to

$$\begin{aligned} \frac{\Gamma (-\omega )}{\prod _i \Gamma (\alpha _i)} \int _{\mathbb {RP}^{|I|-1}_+} \Omega _H \, \, H^{|I|} \frac{1}{\psi ^{D/2}} \Bigg (\frac{\phi }{\psi }\Bigg )^\omega \prod _i t_i^{\alpha _i-1} \end{aligned}$$

written as an integral on the projective space $\mathbb {RP}^{|I|-1}_+$ with the volume form $\Omega _H$ given by

$$\begin{aligned} \Omega _H:= \sum _i^{|I|} (-1)^{i-1}\frac{t_i}{H} \quad d\left( \frac{t_1}{H}\right) \wedge \cdots \widehat{ d\left( \frac{t_i}{H}\right) } \cdots \wedge d\left( \frac{t_{|I|}}{H}\right) . \end{aligned}$$

This integral is independent on the particular choice of the hyperplane thanks to its homogeneity, since for a different choice $H'$, $\Omega _{H'}=(H/H')^{|I|}\Omega _H$.

In the case of the one-loop massless diagram ($m_i=0 \quad \forall i\in I$)

the polynomials are very simple:

$$\begin{aligned} \psi =&t_1 + t_2\\ \phi =&t_1 t_2 p^2 . \end{aligned}$$

By the choice of the hyperplane $H(t)=t_2$ we get

$$\begin{aligned} \mathscr {I}(p^2,\alpha _1, \alpha _2, D)&:= (p^2)^{\omega } \frac{\Gamma ( -\omega )}{\Gamma (\alpha _1)\Gamma (\alpha _2)}\int dt_1 dt_2 \, \delta (1-t_2) \frac{t_1^{D/2 -\alpha _1-1}t_2^{D/2 -\alpha _2-1}}{(t_1 +t_2)^{D-\alpha _1-\alpha _2}}\\&=(p^2)^{ \omega } \frac{\Gamma ( -\omega )}{\Gamma (\alpha _1)\Gamma (\alpha _2)} B(D/2-\alpha _1, D/2-\alpha _2)\\&=(p^2)^{D/2-\alpha _1-\alpha _2}\frac{\Gamma (D/2-\alpha _2)\Gamma (D/2-\alpha _1)\Gamma (-\omega )}{\Gamma (\alpha _1)\Gamma (\alpha _2)\Gamma (D/2 +\omega )} \end{aligned}$$

Since we are studying a model in 6 dimensions, in the main text we will use this value.

Let us also report here the explicit calculation mentioned in Sect. 3. Let us consider the integral

$$\begin{aligned} \int _{\mathbb {R}^6} dq \, \left( \frac{1}{(p+q)^2} \right) ^{1-x} 2q^\nu \left( \frac{1}{q^2} \right) ^{2-y}. \end{aligned}$$

(95)

We can perform the Schwinger trick  (93) and have

$$\begin{aligned} \frac{2}{\Gamma (1-x)\Gamma (2-y)} \int _{\mathbb {R}_+} dt_1\int _{\mathbb {R}_+} dt_2 \, t_1^{-x}t_2^{1-y}\int _{\mathbb {R}^6} dq\, q^\nu e^{-t_1(p+q)^2-t_2q^2}. \end{aligned}$$

(96)

We can complete the square in the exponent

$$\begin{aligned} t_1(p+q)^2 +t_2q^2 = t_{12} \left( q + \frac{t_1}{t_{12}}p\right) ^2 + \frac{t_1t_2}{t_{12}} p^2 \end{aligned}$$

(97)

and then perform a change of variable $u= q+\frac{t_1}{t_{12}}p$ to get

$$\begin{aligned} \int _{\mathbb {R}^6} dq\, q^\nu e^{-t_1(p+q)^2-t_2q^2} = \int _{\mathbb {R}^6} du\, \left( u^\nu -\frac{t_1}{t_{12}}p^\nu \right) e^{-t_{12}u^2} e^{-\frac{t_1t_2}{t_{12}}p^2}. \end{aligned}$$

(98)

In all these formula, we use the abbreviation $t_{12} := t_1 + t_2$. The term proportional to $u^\nu$ vanishes since it is the integral of an odd function, and the second term is a gaussian integral which gives

$$\begin{aligned} \int _{\mathbb {R}^6} dq\, q^\nu e^{-t_1(p+q)^2-t_2q^2} = -\pi ^3p^\nu \frac{t_1}{t_{12}^4}e^{-\frac{t_1t_2}{t_{12}}p^2}. \end{aligned}$$

(99)

Going back to Eq. (95)

$$\begin{aligned}&\int _{\mathbb {R}^6} dq \, \left( \frac{1}{(p+q)^2} \right) ^{1-x} 2q^\nu \left( \frac{1}{q^2} \right) ^{2-y}\nonumber \\&\quad = \frac{-2\pi ^3p^\nu }{\Gamma (1-x)\Gamma (2-y)} \int _{\mathbb {R}_+} dt_1\int _{\mathbb {R}_+} dt_2 \, \frac{t_1^{1-x}t_2^{1-y} }{t_{12}^4}e^{-\frac{t_1t_2}{t_{12}}p^2} \end{aligned}$$

(100)

As before, we can insert the equality (94) and choose the hyperplane $H(t) = \lambda \,t_{12}$ to get

$$\begin{aligned} \int _{\mathbb {R}_+} dt_1\int _{\mathbb {R}_+} dt_2 \, \frac{t_1^{1-x}t_2^{1-y} }{t_{12}^4}e^{-\frac{t_1t_2}{t_{12}}p^2} =\left( p^2\right) ^{x+y} \Gamma (-x-y)\int _0^1 dt_1\, t_1^{1+y} (1-t_1)^{1+x} \end{aligned}$$

(101)

which finally brings us to

$$\begin{aligned} \int _{\mathbb {R}^6} dq \, \left( \frac{1}{(p+q)^2} \right) ^{1-x} 2q^\nu \left( \frac{1}{q^2} \right) ^{2-y}= -2p^\nu \pi ^3\left( p^2\right) ^{x+y} \frac{\Gamma (-x-y)\Gamma (2+x)\Gamma (2+y)}{\Gamma (1-x) \Gamma (2-y)\Gamma (4+x+y)} \end{aligned}$$

(102)

Massless loop trick

In this article, we are studying a massless model, so the propagators that appear in the whole text are powers of $p^2$. With the Feynman rules for momentum space, a simple loop is the convolution of two propagators. In the massless case though, the Fourier transform back in position space will also be a power of $x^2$ by homogeneity reasons and the convolution can be evaluated as a multiplication in position space of the Fourier transforms. Apart from some $\pi$ factors independent on $\alpha$, we have that

$$\begin{aligned} \mathcal {F}\left[ \left( \frac{1}{p^2}\right) ^a\right] (x) \propto \frac{\Gamma (D/2-a)}{\Gamma (a)} \left( \frac{1}{x^2}\right) ^{D/2-a} \end{aligned}$$

(103)

so that in the case of a loop

(104)

(105)

(106)

where we have used $\mathcal {F}(\alpha *\beta )= \mathcal {F}(\alpha ) \cdot \mathcal {F}(\beta )$. Repetitively using these transformations allows to evaluate the diagram appearing in Sect. 5.1.

In this section, we also have used the following construction: starting from the diagram

(107)

we can add another line of index c

(108)

Now, if we want to consider this graph as a vacuum one, the dependence on the exterior momentum $p$ should disappear. We set therefore $c=D-a-b$, which is also the condition that $\omega =0$ for the completed graph, making it logarithmically divergent. The preceding formula becomes

$$\begin{aligned} \frac{\Gamma (D/2-a)\Gamma (D/2-b)\Gamma (a+b-D/2)}{\Gamma (a)\Gamma (b)\Gamma (D -a-b)} \quad \frac{\Gamma (0)}{\Gamma (D/2)} \end{aligned}$$

(109)

The first factor is the same we would have found if we had transformed the a, b loop. There is however a $\Gamma (0)/\Gamma (D/2)$ additional factor, which is infinite, reflecting the divergence of the whole diagram. This factor can, however, be interpreted as the integral

$$\begin{aligned} \int _{\mathbb R^D} \frac{d^D u}{(u^2)^{D/2}} \end{aligned}$$

the simplest scale invariant integral in $D$ dimensions. This same integral can appear either in $x$-space, since the product of the three propagators gives the power $D/2$ of $x^2$ or in $p$-space, where the combination of any two of the propagators and the last one combine to give $(p^2)^{D/2}$. The scale invariance can be broken by fixing the momentum in any of the propagators while giving the same number, giving another approach to the completion invariance of the residues of propagator graphs.

In this work, we have been interested in the pole structure when one of the propagator of a completed graph becomes scale invariant. Since the whole structure is scale invariant, the complementary of this line becomes also scale invariant, giving two infinite factors. We therefore understand that there should be a pole in the evaluation of the diagram, but it is not so clear how to evaluate the residue of this pole. In a previous work [5], a procedure was devised from the parametric representation, which has the advantage of generalizing to poles associated to a propagator with a power larger than $D/2$ by any positive integer, but another approach is possible in this simple case. We consider the diagram in $x$-space: The scale invariance is broken by fixing the distance between the two vertices while the almost scale invariant link contributes only by its normalization, $\Gamma (\varepsilon )/\Gamma (D/2-\varepsilon )$. In the limit of vanishing $\varepsilon$, $\Gamma (\varepsilon )$ gives a pole of residue 1 while all other terms have a smooth limit. Differentiating with respect to $L$ compensate the pole, so that we end up with $1/\Gamma (D/2)$ times the residue of the remaining scale invariant diagram. From its scale invariance, it comes that the choice of any two fixed vertices will give the same value for the residue, allowing for a simpler evaluation through suitable choices of these vertices.

The same result on the residue of the pole has been previously obtained in the appendix of [40], by a slightly different derivation. We thank Andrei Kataev for pointing out this reference.