*4.3. Resummation*

Equation (23) is a convenient starting point for a resummation of the wave function renormalization constant. To second order, the expansion of the latter reads [51]:

$$
\lambda\_A = \frac{\lambda^{(1)}}{L} + \frac{\lambda^{(2)}}{L^2} + \cdots, \quad \lambda^{(1)} = 4\left(\frac{2}{3} - \zeta\right), \quad \lambda^{(2)} = -8\left(\frac{8}{27} + \left(\frac{2}{3} - \zeta\right)\zeta\right). \tag{29}
$$

As can be seen from Equation (23), the NLO term ∼*β*<sup>2</sup> is proportional to the LO renormalization constant of the wave function. This term, together with the LO term in the gap equation, can be considered as terms of order one and zero, respectively, in the expansion in *λA*. Following Nash, one can then resum the complete expansion of *λA* at the level of the gap equation (see Ref. [2]), leading to:

$$1 = \frac{8\beta}{3L} + \frac{\beta}{4L^2} \left(\lambda^{(2)} - 4\lambda^{(1)} \left(\frac{14}{3} + \zeta\right)\right) + \frac{\Delta(a, \zeta)}{L^2},\tag{30}$$

where <sup>Δ</sup>(*<sup>α</sup>*, *ξ*) = <sup>8</sup>*S*˜(*<sup>α</sup>*, *ξ*) − 4*β* (*ξ*<sup>2</sup> + 4*ξ* + 8/3) + 2*β* (2 + *ξ*) Πˆ . *Particles* **2020**, *3*

Interestingly, the LO term in the Equation (30) is now gauge independent. Using the Equation (29), Equation (30) can now be be written as:

$$1 = \frac{8\beta}{3L} + \frac{1}{L^2} \left[ 8\mathbf{S}(\mathbf{a}, \mathbf{\xi}) - \frac{16}{3} \beta \left( \frac{40}{9} + \mathbf{\Upsilon} \right) \right] + O(L^{-3}),\tag{31}$$

which demonstrates a strong suppression of the gauge dependence, since *ξ*-dependent terms exist, but they enter the equation only through the remainder *S*˜, which is very small in numerical terms.

By analogy with the previous subsection, we now compute the NLO correction to the parameter *β*−<sup>1</sup> of the solution of the SD equation. From (31), this gives:

$$\boldsymbol{\beta}^{-1} = \frac{8}{3L} + \frac{1}{L^2} \left[ \frac{8}{\beta} \boldsymbol{S}(\boldsymbol{a}, \boldsymbol{\xi}) - \frac{16}{3} \left( \frac{40}{9} + \boldsymbol{\Omega} \right) \right] + \boldsymbol{O}(L^{-3}) \,. \tag{32}$$

From this equation it again becomes clear that the first term in brackets is of the order of ∼1/*L* (which can be seen by solving Equation (32) iteratively) and, therefore, its contribution is ∼1/*L*<sup>3</sup> and should be ignored in this analysis. This observation was shown to us by Valery Gusynin. So, we have:

$$\boldsymbol{\beta}^{-1} = \frac{8}{3L} - \frac{1}{L^2} \frac{16}{3} \left(\frac{40}{9} + \boldsymbol{\hat{I}}\right) + O(L^{-3}) \, , \tag{33}$$

which is now completely gauge-independent.

Now consider Equation (33) (or, equivalently, Equation (31) with the condition *<sup>S</sup>*˜(*β*, *ξ*) = 0) at the critical point *α* = 1/4 (*β* = 16), preserving all the terms *<sup>O</sup>*(1/*L*<sup>2</sup>). This gives:

$$L\_c^2 - \frac{128}{3}L\_c + \frac{256}{3}\left(\frac{40}{9} + \text{\textquotedblleft}\text{\textquotedblright}\right) = 0\,\text{.}\tag{34}$$

Solving Equation (34), we have two standard solutions:

$$L\_{\mathfrak{c},\pm} = \frac{64}{3} \left( 1 \pm \sqrt{d\_2(\xi)} \right) \,, \ d\_2(\xi) = 1 - \frac{3}{16} \left( \frac{40}{9} + \mathrm{II} \right) = \frac{1}{6} - \frac{3}{16} \,\mathrm{II} \tag{35}$$

and we have for the " +" solution (the " −" one is nonphysical):

$$
\overline{L}\_{\mathbb{C}} = 28.0981, \qquad \overline{N}\_{\mathbb{C}} = 2.85 \,\text{.}\tag{36}
$$

The results of Equation (36) are completely consistent with the recent results of [29].
