Loop vertex expansion for higher-order interactions

Abstract

This note provides an extension of the constructive loop vertex expansion to stable interactions of arbitrarily high order, opening the way to many applications. We treat in detail the example of the $(\bar{\phi } \phi )^p$ field theory in zero dimension. We find that the important feature to extend the loop vertex expansion is not to use an intermediate field representation, but rather to force integration of exactly one particular field per vertex of the initial action.

Appendix: explicit formulas for p small

1.1 The $(\bar{\phi } \phi )^2$ theory

In this case, $g= (-\lambda )^{\frac{1}{2}} = i \sqrt{\lambda }$, $\lambda = - g^2$ and $z = g^2 \bar{\phi } \phi = - \lambda \bar{\phi } \phi$. Equation (2.14) takes the form

$$\begin{aligned} zT_2^2(z) -T_2(z) +1 =0 \end{aligned}$$

(6.1)

with solution the ordinary Catalan function

$$\begin{aligned} T_2(z) = \frac{1- \sqrt{1 -4z}}{2z}. \end{aligned}$$

(6.2)

We find

$$\begin{aligned} F_2(z) = z T_2' +T_2 = (1- 4z)^{-1/2} = \left( 1- 4g^2 \bar{\phi } \phi \right) ^{-1/2} = \left( 1 + 4 \lambda \bar{\phi } \phi \right) ^{-1/2},\nonumber \\ \end{aligned}$$

(6.3)

and the LVR action is

$$\begin{aligned} S_2 = - \frac{1}{2}\log (1 -4z) = - \frac{1}{2}\log \left( 1- 4g^2 \bar{\phi } \phi \right) = - \frac{1}{2}\log \left( 1+4 \lambda \bar{\phi } \phi \right) . \end{aligned}$$

(6.4)

Its first-order derivative is

$$\begin{aligned} \frac{\partial S_2}{\partial g}= \frac{4g \bar{\phi } \phi }{1- 4g^2 \bar{\phi } \phi } \end{aligned}$$

(6.5)

so that the loop vertex representation of the partition function is

$$\begin{aligned} Z_2(\lambda )= & {} \int \mathrm{d}\mu \left( \phi , \bar{\phi }\right) e^{ - \frac{1}{2}\log \left( 1-4 \lambda \bar{\phi } \phi \right) } , \end{aligned}$$

(6.6)

$$\begin{aligned} G^c_{2,1} (\lambda )= & {} 1 - \frac{\lambda }{Z_2(\lambda )}\int \mathrm{d}\mu \left( \phi ,\bar{\phi }\right) \frac{4 \bar{\phi } \phi }{1+ 4\lambda \bar{\phi } \phi } e^{ - \frac{1}{2}\log \left( 1-4 \lambda \bar{\phi } \phi \right) } \end{aligned}$$

(6.7)

$$\begin{aligned}= & {} 1 - 4\lambda + \cdots . \end{aligned}$$

(6.8)

We recover the familiar logarithmic form of the action and resolvent of the intermediate field theory. However, our LVR representation is not the intermediate field representation. Indeed, in the LVR representation the argument of the log is quadratic in complex fields similar to the initial fields although it would be linear in the single real field $\sigma$ of the intermediate field representation. We should rather think to the fields of the LVR as what remains of the initial fields after having forced integration of one particular marked $\bar{\phi }$ field per vertex.

1.2 The $(\bar{\phi } \phi )^3$ theory

In this case, $g= (-\lambda )^{\frac{1}{3}} = e^{i\pi /3 } \lambda ^{1/3}$, $\lambda = - g^3$ and $z = g^3 (\bar{\phi } \phi )^2 = - \lambda (\bar{\phi } \phi )^2$. Equation (2.14) is now

$$\begin{aligned} zT_3^3(z) -T_3(z) +1 =0 . \end{aligned}$$

(6.9)

which is soluble by radicals. Introducing

$$\begin{aligned} u:= - \frac{27 z}{4} = -\frac{27 }{4}g^3 \left( \bar{\phi } \phi \right) ^2 = \frac{27 }{4}\lambda \left( \bar{\phi } \phi \right) ^2, \end{aligned}$$

(6.10)

Cardano’s solution is

$$\begin{aligned} T_3(z) =\frac{\Delta _+(u) - \Delta _{-}(u)}{\sqrt{-3z}} = 1 +z + 3z^2 + \cdots , \end{aligned}$$

(6.11)

where

$$\begin{aligned} \Delta _\pm (u) := \biggl (\sqrt{1 +u} \pm \sqrt{u} \biggr )^{1/3} = 1 \pm \frac{1}{3} \sqrt{u} + \frac{u}{18} \mp \frac{4 u^{3/2}}{81} - \frac{35u^2}{1944} + \cdots .\nonumber \\ \end{aligned}$$

(6.12)

Defining $h (u) := \frac{1}{ \sqrt{1 +u}}$, we can compute the derivatives

$$\begin{aligned} \Delta '_\pm =\frac{\mathrm{d}}{\mathrm{d}u}\Delta _\pm (u)= & {} \frac{1}{6} \biggl ((1 + u)^{-1/2} \pm u^{-1/2} \biggr ) \biggl (\sqrt{1 + u} \pm \sqrt{u} \biggr )^{-2/3}\nonumber \\= & {} \pm \frac{1}{6\sqrt{u(1+u)}} \Delta _\pm (u) = \pm \frac{h }{6\sqrt{u}} \Delta _\pm (u) . \end{aligned}$$

(6.13)

Hence

$$\begin{aligned} zT_3'(z) =\frac{27 \sqrt{-z}}{4 \sqrt{3}}\left[ \Delta '_+(u) - \Delta '_{-}(u)\right] - \frac{1}{2\sqrt{-3z}}\left[ \Delta _+ (u) - \Delta _{-}(u)\right] . \end{aligned}$$

(6.14)

(2.16) gives

$$\begin{aligned} F_{3}= & {} \sum _n {{3n}\atopwithdelims (){n}} z^n = 2z T_3' +T_3 = h \frac{ \Delta _+ + \Delta _{-} }{2} , \end{aligned}$$

(6.15)

$$\begin{aligned} S_{3}= & {} \log F_3 = -\frac{1}{2} \log (1+u) + \log \frac{ \Delta _+ + \Delta _{-} }{2} \end{aligned}$$

(6.16)

The u derivatives of $S_3$ give access to its g derivatives since $u= -\frac{27 }{4}g^3 (\bar{\phi } \phi )^2$, hence

$$\begin{aligned} \frac{\partial u}{\partial g} = -\frac{81 }{4}g^2 \left( \bar{\phi } \phi \right) ^2 . \end{aligned}$$

(6.17)

For instance,

$$\begin{aligned} \frac{\partial S_3}{\partial g}= & {} \frac{81 g^2 \left( \bar{\phi } \phi \right) ^2}{4} \biggl (\frac{1}{2(1 + u)} -\frac{ \Delta '_+ + \Delta '_-}{ \Delta _+ + \Delta _-}\biggr ) \nonumber \\= & {} \frac{81 g^2 \left( \bar{\phi } \phi \right) ^2}{8} \biggl (h^2 - \frac{h }{3\sqrt{u}} \frac{ \Delta _+ - \Delta _-}{ \Delta _+ + \Delta _-}\biggr ). \end{aligned}$$

(6.18)

We remark that the quotient $\frac{ \Delta _+ - \Delta _-}{ \Delta _+ + \Delta _-}= \frac{A-B}{A+B}$ for $A = \Delta _+$, $B= \Delta _-$ simplifies, using that $(A+B)(A^2 - AB + B^2) = A^3 + B^3$ and $(A-B)(A^2 - AB + B^2) = A^3 - B^3 -2AB(A-B)$. Remarking that in our case $AB = \Delta _+ \Delta _-= 1$, we find

$$\begin{aligned} \frac{ \Delta _+ - \Delta _-}{ \Delta _+ + \Delta _-}= & {} h \left[ \sqrt{u} - ( \Delta _+ - \Delta _-)\right] \end{aligned}$$

(6.19)

$$\begin{aligned} \frac{\partial S_3}{\partial g}= & {} \frac{81 g^2 \left( \bar{\phi } \phi \right) ^2}{8} \biggl (h^2 - \frac{h^2 }{3\sqrt{u}} \left[ \sqrt{u} - ( \Delta _+ - \Delta _-)\right] \biggr )\nonumber \\= & {} \frac{81 g^2 \left( \bar{\phi } \phi \right) ^2}{8} \biggl [\frac{2}{3} h^2 + \frac{h^2 }{3\sqrt{u}} ( \Delta _+ - \Delta _-) \biggr ]\nonumber \\= & {} \frac{27 g^2 \left( \bar{\phi } \phi \right) ^2}{4 (1+u)} \biggl [ 1 + \frac{ \Delta _+ - \Delta _- }{2 \sqrt{u}} \biggr ] \end{aligned}$$

(6.20)

from which we find

$$\begin{aligned} G^c_{3,1} (\lambda )= & {} 1 + \frac{g}{Z_3(\lambda )}\int \mathrm{d}\mu \left( \phi ,\bar{\phi }\right) \frac{\partial S_3 }{\partial g} e^{S_3}\nonumber \\= & {} 1 - \frac{27 \lambda }{4Z_3(\lambda )}\int \mathrm{d}\mu \left( \phi ,\bar{\phi }\right) \frac{\left( \bar{\phi } \phi \right) ^2}{1+u} \biggl [ 1 + \frac{ \Delta _+ - \Delta _- }{2 \sqrt{u}} \biggr ] e^{S_3}\nonumber \\= & {} 1 - 18 \lambda +\cdots . \end{aligned}$$

(6.21)

1.3 The $(\bar{\phi } \phi )^4$ theory

In this case, $g= (-\lambda )^{\frac{1}{4}} = e^{i\pi /4 } \lambda ^{1/4}$, $\lambda = - g^4$ and $z = g^4 (\bar{\phi } \phi )^3 = - \lambda (\bar{\phi } \phi )^3$. Equation (2.14) is now

$$\begin{aligned} zT_4^4(z) -T_4(z) +1 =0 . \end{aligned}$$

(6.22)

which is still soluble by radicals. Denoting

$$\begin{aligned} v= \frac{z^{1/3}}{ 2^{1/3} } \left[ \left( 1+\sqrt{ 1 - \frac{2^8}{3^3} z } \right) ^{1/3} + \left( 1 -\sqrt{1 - \frac{2^8}{3^3} z} \right) ^{1/3} \right] , \end{aligned}$$

(6.23)

then

$$\begin{aligned} T_4(z)= \frac{ (1+4v)^{1/4} - \bigl [2-(1+4v)^{1/2} \bigr ]^{1/2} }{2 (vz)^{1/4}}. \end{aligned}$$

(6.24)

We can then compute

$$\begin{aligned} F_{4}= & {} \frac{T_4}{4 - 3T_4} = 3z T_4' +T_4 \nonumber \\= & {} \frac{ (1+4v)^{1/4} - \bigl [2-(1+4v)^{1/2} \bigr ]^{1/2} }{8 (vz)^{1/4} - 3 \bigl [ (1+4v)^{1/4} - \bigl [2-(1+4v)^{1/2} \bigr ]^{1/2}\bigr ] } \end{aligned}$$

(6.25)

from which $S_4 = \log F_4$ can be derived explicitly.