# Double Stochastic Quantization: Non perturbative approach to Stochastic Quantization $\lambda\Phi^{2n}_d$, $n \ge 2$, $d \ge 4$
model Quantum Field Theory

jaykov foukzon

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jaykov foukzon. Double Stochastic Quantization: Non perturbative approach to Stochastic Quantization $\lambda\Phi^{2n}_d$, $n \ge 2$, $d \ge 4$ model Quantum Field Theory. 2025. hal-05112650v3

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# Double Stochastic Quantization:Non perturbative approach to Stochastic Quantization $\lambda\varphi_d^{2n}, n \ge 2, d \ge 4$ model Eucledean Quantum Field Theory

Jaykov Foukzon

Center for Mathematical Sciences, Israel Institution of Technology, Haifa, Israel.

E-mail: jaikovfoukzon@gmail.com

E-mail: jaykovfoukzon@list.ru

https://orcid.org/0000-0002-6837-4004

ID: INSPIRE-00004491

**Abstract:** In this paper, we show how the finite formulation of quantum field theory based on Langevin equations can be generalized to the case of *nonrenormalizable* theories. The 5th-time stochastic-quantization approach to field theory proposed by Parisi and Wu, is put in a path-integral form in [6]. The procedure of taking the limit $\tau \to \infty$ is analyzed and based on new grounds through the introduction of the vacuum-vacuum generating functional. In this paper non perturbative approach related

to Parisi and Wu stochastic-quantization of the $\lambda\varphi_d^{2n}, n \ge 2, d \ge 4$ model quantum field

theory is considered.

Content.

1.Introduction

2. Non perturbative approach to Stochastic Quantization $\lambda\varphi^{2n}$ model Quantum Field Theory.

2.1. The generating functional.

2.2. Transcendental master equation corresponding to two-point Green function $G(x_1, x_2, \lambda)$.

2.3. Double stochastic quantization of the Free Scalar Fields.

2.4. Double stochastic quantization of the $\lambda\phi_d^4$ theory.

2.5. Double stochastic quantization of the $\lambda\phi_4^6$ theory.

3. Weak coupling. Nonperturbative result.

3.1. Weak coupling. The $\lambda\phi_d^4$ theory.

3.2. Weak coupling. The $\lambda\phi_d^4$ theory. Exact nonperturbative solution.

3.3. Weak coupling. The $\lambda\phi_d^{2n}$ theory. Exact nonperturbative solution.

4.1. Double stochastic quantization of Abelian gauge fields.

4.2. Double stochastic quantization of massive scalar QED.

4.3. Stochastic quantization of scalar QED by canonical perturbation approach.
5. Double stochastic quantization QFT with massive fermions.

5.1. Basic notions.

5.2. Double stochastic quantization of the Eukclidean QED with massive photon.

6. Double stochastic quantization of non-Abelian gauge field and Faddeev-Popov ghost effects.

References.

Apendix 1. Two point function of the euclidean free scalar fields. Main properties.

Apendix 2. Correlation functions for the $(\lambda\phi^4)_d$ scalar theory in canonical Parisi and Wu stochastic quantization.

Apendix 3. Generalized Banach fixed-point theorem. Quasy contractive fixed point (QCFP).

Apendix 4.1. Integration over random interval.

Apendix 4.2. Integration over generalized random interval.

Apendix 5. Generalized random variables and double stochastic processes.

Apendix 6. The Dirac equation.

Apendix 7. Double stochastic quantization of massive free Abelian gauge field

# 1. Introduction

Paris and Wu' proposed the following alternative method to get the quantum averages [1]:

(i) Introduce a 5-th time $\tau$, in addition to the usual four space-time $t$, and postulate the

following Langevin equation for the dynamics of the field $\varphi(\tau, x)$ in this extra time $\tau$

$$ \frac{\partial \varphi(\tau, x)}{\partial \tau} = - \frac{\delta S[\varphi]}{\delta \varphi(\tau, x)} + \eta(\tau, x), $$ 
$$ \langle \eta(\tau, x) \rangle_{\eta} = 0, \tag{1.1} $$

$$ \langle \eta(\tau, x) \eta(\tau', x') \rangle_{\eta} = 2 \delta(\tau - \tau')(x - x'), $$

where the angular brackets denote connected average with respect to the random variable $\eta$.

(ii) Evaluate the stochastic average of fields $\phi_{\eta}(\tau, x)$ satisfying Eq. (1.1), that means

$$ \langle \phi_{\eta}(\tau_1, x_1) \phi_{\eta}(\tau_2, x_2) \dots \phi_{\eta}(\tau_m, x_m) \rangle_{\eta}. \tag{1.2} $$

(iii) Put $\tau_1 = \tau_2 = \dots = \tau_m = \tau$ in (1.2) and take the limit

$$ \lim_{\tau \to \infty} \langle \phi_{\eta}(\tau, x_1) \phi_{\eta}(\tau, x_2) \dots \phi_{\eta}(\tau, x_m) \rangle_{\eta} = G(x_1, x_2, \dots, x_m) \tag{1.3} $$

It is possible to prove, at least perturbatively, that [6]
$$G(x_1, x_2, \dots, x_m) = \frac{\int D[\varphi](\varphi(x_1)\varphi(x_2)\dots\varphi(x_m)) \exp\{-S[\varphi]\}}{\int D[\varphi] \exp\{-S[\varphi]\}}. \quad (1.4)$$

To understand this relation see ref.[6].

In this paper in particular we deal with double stochastic relaxation equations of the form:

$$\begin{align}
\frac{\partial \varphi(\tau, x)}{\partial \tau} &= -\frac{\delta S[\varphi]}{\delta \varphi(\tau, x)} + \eta(\tau, x) + \epsilon \tilde{\eta}(\tau, x), \nonumber \\
\langle \eta(\tau, x) \rangle_{\eta} &= 0, \quad \langle \tilde{\eta}(\tau, x) \rangle_{\tilde{\eta}_1} = 0 \tag{1.5} \\
\langle \eta(\tau, x) \eta(\tau', x') \rangle_{\eta} &= 2\delta(\tau - \tau')(x - x'), \nonumber \\
\langle \tilde{\eta}(\tau, x) \tilde{\eta}(\tau', x') \rangle_{\tilde{\eta}} &= 2\delta(\tau - \tau')(x - x'). \nonumber
\end{align}$$

Here $\eta(\tau, x) = \eta(\tau, x; \omega)$ is a space-time white noise on probability space $\Sigma = (\Omega, S, P)$ and $\tilde{\eta}_{1,2}(\tau, x) = \tilde{\eta}(\tau, x; \omega)$ are space-time white noises on probability space $\tilde{\Sigma} = (\tilde{\Omega}, S, P)$.

(ii) Evaluate the stochastic average of fields $\phi_{\eta}(\tau, x)$ satisfying Eq. (1.5), that means

$$\langle \phi_{\eta, \tilde{\eta}}(\tau_1, x_1; \epsilon) \phi_{\eta, \tilde{\eta}}(\tau_2, x_2; \epsilon) \dots \phi_{\eta, \tilde{\eta}}(\tau_m, x_m; \epsilon) \rangle_{\eta, \tilde{\eta}}. \quad (1.6)$$

(iii) Put $\tau_1 = \tau_2 = \dots = \tau_m = \tau$ in (1.2) and take the limit

$$\lim_{\tau \to \infty} \lim_{\epsilon \to 0} \langle \phi_{\eta, \tilde{\eta}}(\tau, x_1; \epsilon) \phi_{\eta, \tilde{\eta}}(\tau, x_2; \epsilon) \dots \phi_{\eta, \tilde{\eta}}(\tau, x_m; \epsilon) \rangle_{\eta, \tilde{\eta}} = G(x_1, x_2, \dots, x_m; \epsilon) \quad (1.7)$$

To understand this relation we have to introduce the notion of probability (density) $P(\varphi_{\eta}, \tau)$, that is, the probability (density) of having the system in the configuration $\varphi_{\eta}(\tau, x)$ at time $\tau$. There exists for $P(\varphi_{\eta}, \tau)$ an equation that describes its evolution in the time $\tau$. It is called the Stochastic Fokker-Planck (SFP) equation and it has been derived in [8]:

$$\begin{align}
\frac{\partial P[\varphi(\tau, x), \tau]}{\partial \tau} &= \int d^4 x \frac{\delta}{\delta \varphi(\tau, x)} \left[ P[\varphi(\tau, x), \tau] \frac{\delta S^* [\varphi]}{\delta \varphi(\tau, x)} \right] + \int d^4 x \frac{\delta^2 S^* [\varphi]}{\delta \varphi^2(\tau, x)}, \tag{1.8} \\
S^*[\varphi] &= S[\varphi] - \int [\varphi(\tau, x) \eta(\tau, x)] d^4 x. \nonumber
\end{align}$$

It is possible to rewrite this equation in a Schrodinger-type form:

$$\begin{align}
\frac{\partial \Psi[\varphi(\tau, x), \tau]}{\partial \tau} &= -2\mathbf{H}\Psi[\varphi(\tau, x), \tau], \tag{1.9} \\
\Psi &= P[\varphi(\tau, x), \tau] \exp[S^*[\varphi(\tau, x)]/2], \nonumber
\end{align}$$

where
$$ \mathbf{H} = -\frac{1}{2} \frac{\delta^2}{\delta\varphi^2} + \frac{1}{8} \left[ \frac{\delta S^\star[\varphi]}{\delta\varphi} \right]^2 - \frac{1}{4} \frac{\delta^2 S^\star[\varphi]}{\delta\varphi^2}. \quad (1.10) $$

It is a positive semi-definite operator $H\Psi_n = E_n\Psi_n$ which a ground state $E_0$ is $\Psi_0[\varphi(\tau,x),\tau] = \exp[S^\star[\varphi(\tau,x)]/2]$. The solution of Eq.(1.9) is

$$ \Psi[\varphi(\tau,x),\tau] = \sum_{n=0}^{\infty} c_n \Psi_n[\varphi(\tau,x),\tau] \exp(-2E_n\tau), \quad (1.11) $$

where $\{c_n\}_{n=0}^\infty$ are normalizing constants. The probability density $P[\varphi(\tau,x),\tau]$ can be written as

$$ P[\varphi(\tau,x),\tau] = \exp[S^*[\varphi(\tau,x)]/2] \sum_{n=0}^{\infty} c_n \Psi_n[\varphi(\tau,x),\tau] \exp(-2E_n\tau). \quad (1.12) $$

In the limit $\tau \to \infty$ the only term that does not disappear in this expression is $\Psi_0[\varphi(\tau,x),\tau]$, so finally we have

$$ \begin{aligned} \lim_{\tau \to \infty} P[\varphi(\tau,x),\tau] &= c_0 \exp[-S^*[\varphi(\tau,x)]/2] \exp[-S^*[\varphi(\tau,x)]/2] = \\ &= c_0 \exp[-S^*[\varphi(\tau,x)]] \end{aligned} \quad (1.13) $$

This is the formal reason why Eq.(1.4) holds.

## 2. Non perturbative approach to Stochastic Quantization
### $\lambda\varphi^{2n}$ model Quantum Field Theory

#### 2.1. The generating functional

In this paper we deal with a system of the double stochastic relaxation equations of the form:
$$
\begin{align*}
\frac{\partial \varphi_{1,\epsilon}(\tau, x)}{\partial \tau} &= -\frac{\delta S[\varphi_1, \varphi_2]}{\delta \varphi_1(\tau, x)} \Big|_{\varphi_1=\varphi_{1,\epsilon}, \varphi_2=\varphi_{2,\epsilon}} + \eta(\tau, x) + \epsilon \tilde{\eta}_1(\tau, x), \\
\varphi_{1,\epsilon}(0, x) &= 0, x \in \mathbb{R}^4, \tau \in \mathbb{R}_+, \\
\frac{\partial \varphi_{2,\epsilon}(\tau, x)}{\partial \tau} &= -\frac{\delta S[\varphi_1, \varphi_2]}{\delta \varphi_2(\tau, x)} \Big|_{\varphi_1=\varphi_{1,\epsilon}, \varphi_2=\varphi_{2,\epsilon}} - \eta(\tau, x) + \epsilon \tilde{\eta}_2(\tau, x), \\
\varphi_{2,\epsilon}(0, x) &= 0, x \in \mathbb{R}^4, \tau \in \mathbb{R}_+, \\
S[\varphi_1, \varphi_2] &= \\
&\quad \int_{\mathbb{R}^4} d^4 x \Biggl[ \frac{1}{2} (\partial_\mu \varphi_1(\tau, x) \partial_\mu \varphi_1(\tau, x) + m^2 \varphi_1(\tau, x)) + P(\varphi_1(\tau, x)) \Biggr] + \\
&\quad \int_{\mathbb{R}^4} d^4 x \Biggl[ \frac{1}{2} (\partial_\mu \varphi_2(\tau, x) \partial_\mu \varphi_2(\tau, x) + m^2 \varphi_2(\tau, x)) + P(\varphi_2(\tau, x)) \Biggr] + \\
&\quad + \int_{\mathbb{R}^4} d^4 x [\gamma \times \varphi_1(\tau, x) \varphi_2(\tau, x)], \quad \gamma > 0.
\end{align*}
\tag{2.1.1}
$$

where $0 < \epsilon \ll 1$, $P(\cdot)$ is a polynomial degree $2k, k \ge 2$

$$
P(\varphi_{1,2}) = \int_{\mathbb{R}^4} d^4 x [\gamma \times \varphi_1(\tau, x) \varphi_2(\tau, x)], \gamma > 0, \gamma \approx 0. \quad (2.1.2)
$$

and where $\eta(\tau, x; \omega), \tilde{\eta}_1(\tau, x; \omega)$,
$\tilde{\eta}_2(\tau, x; \omega)$ are Gaussian random variables such that

$$
\begin{align}
\langle \eta(\tau, x) \rangle_{\eta} &= 0, \tag{2.1.3} \\
\langle \tilde{\eta}_1(\tau, x) \rangle_{\tilde{\eta}_1} &= 0, \langle \tilde{\eta}_2(\tau, x) \rangle_{\tilde{\eta}_2} = 0, \nonumber
\end{align}
$$

and for the two-point correlation function associated with the random noises fields

$$
\begin{align}
\langle \eta(\tau, x) \eta(\tau', x') \rangle_{\eta} &= 2\delta(\tau - \tau')(x - x'), \notag \\
\langle \tilde{\eta}_1(\tau, x) \tilde{\eta}_1(\tau', x') \rangle_{\tilde{\eta}} &= 2\delta(\tau - \tau')(x - x'), \tag{2.1.4} \\
\langle \tilde{\eta}_2(\tau, x) \tilde{\eta}_2(\tau', x') \rangle_{\tilde{\eta}} &= 2\delta(\tau - \tau')(x - x'). \notag
\end{align}
$$

**Remark 2.1.1.** Here $\eta(\tau, x) = \eta(\tau, x; \omega)$ is a space-time white noise on probability space $\Sigma = (\Omega, S, P)$ and $\tilde{\eta}_{1,2}(\tau, x) = \tilde{\eta}_{1,2}(\tau, x; \omega)$ are space-time white noises on probability space $\tilde{\Sigma} = (\tilde{\Omega}, S, P)$. The angular brackets denote connected average with respect to the random variables $\eta, \tilde{\eta}_{1,2}$.

We want to build a generating functional $Z_{\epsilon}[J_1, J_2]$ from which the correlations
$$ \langle \varphi_{1,\eta}(\tau_1, x_1; \omega) \times \dots \times \varphi_{1,\eta}(\tau_l, x_l; \omega) \varphi_{2,\eta}(\tau_1, x_1; \omega) \times \dots \times \varphi_{2,\eta}(\tau_l, x_l; \omega) \rangle_{\eta} $$

$$ \frac{\partial \varphi_{1,\eta}(\tau, x)}{\partial \tau} = - \frac{\delta S[\varphi_1, \varphi_2]}{\delta \varphi_1(\tau, x)} \Big|_{\varphi_1=\varphi_{1,\eta}, \varphi_2=\varphi_{2,\eta}} + \eta(\tau, x), $$

$$ \frac{\partial \varphi_{2,\eta}(\tau, x)}{\partial \tau} = - \frac{\delta S[\varphi_2, \varphi_2]}{\delta \varphi_2(\tau, x)} \Bigg|_{\varphi_1=\varphi_{1,\eta}, \varphi_2=\varphi_{2,\eta}} + \eta(\tau, x) $$

can be derived by the following fashion:

$$ \langle \varphi_{1,\eta}(\tau_1, x_1; \omega) \times \dots \times \varphi_{1,\eta}(\tau_l, x_l; \omega) \varphi_{2,\eta}(\tau_1, x_1; \omega) \times \dots \times \varphi_{2,\eta}(\tau_l, x_l; \omega) \rangle_{\eta} = \\ \lim_{\epsilon \to 0} \langle \varphi_{1,\eta,\epsilon\tilde{\eta}_1}(\tau_1, x_1; \omega, \tilde{\omega}) \times \dots \times \varphi_{1,\eta,\epsilon\tilde{\eta}_1}(\tau_l, x_l; \omega, \tilde{\omega}) \varphi_{2,\eta,\epsilon\tilde{\eta}_2}(\tau_1, x_1; \omega, \tilde{\omega}) \times \dots $$

$$ \dots \times \varphi_{2,\eta,\epsilon\tilde{\eta}_2}(\tau_l, x_r; \omega, \tilde{\omega}) \rangle_{\eta} = \qquad (2.1.5) $$

$$ = \lim_{\epsilon \to 0} \frac{\delta^l Z_{\epsilon}[J_1, J_2; \omega]}{\delta J_1(\tau_1, x_1) \dots \delta J_1(\tau_l, x_l) \delta J_2(\tau_1, x_1) \dots \delta J_2(\tau_r, x_r)} \Bigg|_{J_1=0, J_2=0} $$

By canonical definition [6] one obtains

$$ Z_{\epsilon}[J_1, J_2; \omega] = $$

$$ N \int D[\eta(\tau', x; \omega)] \left( \left\{ \int D[\varphi_1(\tau', x; \omega)] D[\varphi_2(\tau', x; \omega)] D[\tilde{\eta}_1(\tau', x; \omega)] D[\tilde{\eta}_2(\tau', x; \omega)] \right. \right. \\ \left. \left. \delta(\varphi_1(0, x; \omega)) \delta(\varphi_2(0, x; \omega)) \delta(\varphi_1(\tau', x; \omega) - \varphi_{1,\eta,\epsilon\tilde{\eta}_1,\epsilon\tilde{\eta}_2}) \delta(\varphi_2(\tau', x; \omega) - \varphi_{2,\eta,\epsilon\tilde{\eta}_1,\epsilon\tilde{\eta}_2}) \times \right. \right. \\ \left. \left. \exp\left[-\int_0^{\tau} \int_{\mathbb{R}^4} J_1(\tau', x) \varphi_1(\tau', x; \omega) d^4x d\tau' \right] \exp\left[-\frac{1}{4} \int_0^{\tau} \int_{\mathbb{R}^4} \tilde{\eta}_1^2(\tau', x; \omega) d^4x d\tau' \right] \times \right. \right. \\ \left. \left. \times \exp\left[-\int_0^{\tau} \int_{\mathbb{R}^4} J_2(\tau', x) \varphi_2(\tau', x; \omega) d^4x d\tau' \right] \exp\left[-\frac{1}{4} \int_0^{\tau} \int_{\mathbb{R}^4} \tilde{\eta}_2^2(\tau', x; \omega) d^4x d\tau' \right] \right\} \times \right. \\ \left. \exp\left[-\frac{1}{4} \int_0^{\tau} \int_{\mathbb{R}^4} \eta^2(\tau', x; \omega) d^4x d\tau' \right] \right), $$

where $\varphi_{1,2,\eta,\tilde{\eta}} = \varphi_{1,2}(\tau', x; \omega, \tilde{\omega})$ that appears in Eq.(2.1.1)-Eq.(2.1.2) is the solution of the double stochastic Langevin equations (2.1.1), solved with zero initial condition: $\varphi_{1,2,\eta,\tilde{\eta}}(0, x; \omega, \tilde{\omega}) \equiv 0$ and $N$ is a normalizing constant and

$$ D[\varphi_{1,2}; \omega] = \lim_{M \to \infty} \prod_{i=0}^{M} D[\varphi_{1,2,\tau_i}(x; \omega)] \qquad (2.1.7) $$

where $\varphi_{1,2,\tau_i}(x; \omega)$, are the field configurations at the time $\tau_i$, having sliced the interval $0$ to $\tau$ in $M$ infinitesimal parts $\epsilon$ with $\tau_i = i\epsilon$ and $D[\varphi_{1,2,\tau_i}(x; \omega)]$ is Feinman random measure, that is a product of the usual four-dimensional Feinman random measures, see Appendix 4 for rigorous definitions. Formally such random measure defined by
$$D[\varphi_{1,2,\tau_i}(x; \omega)] = \prod_{x \in \mathbb{R}^4} d[\varphi_{1,2,\tau_i}(x; \omega)]. \qquad (2.1.7')$$

We sort of define it in the following way

$$\prod_{x \in \mathbb{R}^4} d[\varphi_{1,2,\tau_i}(x; \omega)] \triangleq \lim_{\delta \to 0} \prod_{x_i} d[\varphi_{1,2,\tau_i}(x_i; \omega)], \qquad (2.1.7'')$$

where the points $x_i$ belong to a lattice of size $\delta$. Integration over random measure $d[\varphi_{1,2,\tau_i}(x_i; \omega)]$ see Appendix 4. One will first consider a finite volume for this lattice, then take the limit of infinite volume.

**Abbreviation 2.1.1.** $D[\varphi_{1,2}; \omega] \triangleq D[\varphi_1; \omega]D[\varphi_2; \omega]$, $D[\tilde{\eta}_{1,2}(\tau', x; \omega)] \triangleq D[\tilde{\eta}_1(\tau', x; \omega)] \times D[\tilde{\eta}_2(\tau', x; \omega)]$, $\delta(\varphi_{1,2} - \varphi_{1,2,\eta,\epsilon\tilde{\eta}_{1,2}}) \triangleq \delta(\varphi_1 - \varphi_{1,\eta,\epsilon\tilde{\eta}_1})\delta(\varphi_2 - \varphi_{2,\eta,\epsilon\tilde{\eta}_2})$, $J_{1,2}(\tau', x)\varphi_{1,2}(\tau', x; \omega) = J_1(\tau', x)\varphi_1(\tau', x; \omega) + J_2(\tau', x)\varphi_2(\tau', x; \omega)$, etc.

The delta function $\delta(\varphi_{1,2} - \varphi_{1,2,\eta,\tilde{\eta}})$ in Eq.(2.1.6) we can write as

$$\delta(\varphi_{1,2} - \varphi_{1,2,\eta,\tilde{\eta}}) = \delta\left[ \frac{\partial\varphi_{1,2,\epsilon}}{\partial\tau} + \left. \frac{\partial\tilde{S}}{\partial\varphi_{1,2}} \right|_{\varphi_{1,2}=\varphi_{1,2,\epsilon}} - \epsilon\tilde{\eta}_{1,2} \right] \left\| \frac{\delta(\epsilon\tilde{\eta})_{1,2}}{\delta\varphi_{1,2,\epsilon}} \right\|, \quad (2.1.8)$$

where $\tilde{S} = S - \int_{\mathbb{R}^4}(\varphi_{1,2} \times \eta)d^4x$ and where $\|\delta\tilde{\eta}_{1,2}/\delta\varphi_{1,2,\epsilon}\|$ is the Jacobian matrix of the transformation $\tilde{\eta}_{1,2} \to \varphi_{1,2,\epsilon}$, that is

$$\left\| \frac{\delta(\epsilon \tilde{\eta}_{1,2})}{\delta \varphi_{1,2,\epsilon}} \right\| = N_{\epsilon} \left\| \frac{\delta(\tilde{\eta}_{1,2})}{\delta \varphi_{1,2,\epsilon}} \right\|,$$

$$\det \left\| \frac{\delta(\tilde{\eta}_{1,2})}{\delta\varphi_{1,2,\epsilon}} \right\| = N_{\epsilon}^{-1} \det \left[ \left[ \partial_{\tau} + \left. \frac{\partial^2 \tilde{S}}{\partial\varphi_{1,2}(\tau)\partial\varphi_{1,2}(\tau')} \right|_{\varphi_{1,2}=\varphi_{1,2,\epsilon}} \right] \delta(\tau - \tau') \right], \quad (2.1.9)$$

$$N_{\epsilon} = \prod_{i=1}^{\infty} x_i, x_i = \epsilon, i = 1, 2, \dots$$

From Eq.(2.1.9) by canonical calculation we get

$$\begin{align}
\det \left\| \frac{\delta(\tilde{\eta}_{1,2})}{\delta \varphi_{1,2,\epsilon}} \right\| &= \nonumber \\
&= N_{\epsilon}^{-1} \exp \left[ \text{tr} \ln \left[ \left[ \partial_{\tau} + \left. \frac{\partial^2 \tilde{S}}{\partial \varphi_{1,2}(\tau) \partial \varphi_{1,2}(\tau')} \right|_{\varphi_{1,2}=\varphi_{1,2,\epsilon}} \right] \delta(\tau - \tau') \right] \right] \nonumber \\
&= N_{\epsilon}^{-1} \exp \left[ \text{tr} \ln \partial_{\tau} \left[ \delta(\tau - \tau') + \partial_{\tau}^{-1} \left( \left. \frac{\partial^2 \tilde{S}}{\partial \varphi_{1,2}(\tau) \partial \varphi_{1,2}(\tau')} \right|_{\varphi_{1,2}=\varphi_{1,2,\epsilon}} \right) \right] \right], \tag{2.1.10}
\end{align}$$

where $\partial_{\tau}^{-1}$ indicate the Geen's function $G(\tau - \tau')$ that satisfies

$$\partial_{\tau} G(\tau - \tau') = \delta(\tau - \tau'). \qquad (2.1.11)$$

The solutions of the Eq.(2.1.11) are: (i) if we choose propagation forward in time
$$ G(\tau - \tau') = \theta(\tau - \tau') \qquad (2.1.12) $$

(ii) if we choose propagation backward in time

$$ G(\tau - \tau') = -\theta(\tau - \tau') \qquad (2.1.13) $$

In case, propagation forward in time, we get

$$ \det \left\| \frac{\delta(\tilde{\eta}_{1,2})}{\delta\varphi_{1,2,\epsilon}} \right\| = N_{\epsilon}^{-1} \exp \left\{ \text{tr} \left[ \ln \partial_{\tau} + \ln \left[ \delta(\tau - \tau') + \theta(\tau - \tau') \frac{\delta^2 \tilde{S}}{\delta\varphi_{1,2}(\tau)\delta\varphi_{1,2}(\tau')} \bigg|_{\varphi_{1,2}=\varphi_{1,2,\epsilon}} \right] \right] \right\} \quad (2.1.14) \\ = N_{\epsilon}^{-1} \exp(\text{tr} \ln \partial_{\tau}) \exp \left[ \text{tr} \ln \left[ \delta(\tau - \tau') + \theta(\tau - \tau') \frac{\delta^2 \tilde{S}}{\delta\varphi_{1,2}(\tau)\delta\varphi_{1,2}(\tau')} \bigg|_{\varphi_{1,2}=\varphi_{1,2,\epsilon}} \right] \right]. $$

The term $\exp(\text{tr} \ln \partial_{\tau})$ can be dropped, as it cancels with the same term in the denominator of (2.1.6), once we normalize $\tilde{Z}_{\epsilon}[J_{1,2}; \omega] = Z_{\epsilon}[J_{1,2}; \omega]/Z_{\epsilon}[0,0; \omega]$.

**Abbreviation 2.1.2.** $Z_{\epsilon}[J_{1,2}; \omega] \triangleq Z_{\epsilon}[J_{1}, J_{2}; \omega]$

So in Eq.(2.1.14) we are left with

$$ \det \left\| \frac{\delta(\tilde{\eta}_{1,2})}{\delta\varphi_{1,2,\epsilon}} \right\| = N_{\epsilon}^{-1} \exp \left[ \text{tr} \ln \left[ \delta(\tau - \tau') + \theta(\tau - \tau') \frac{\delta^2 \tilde{S}}{\delta\varphi_{1,2}(\tau)\delta\varphi_{1,2}(\tau')} \bigg|_{\varphi_{1,2}=\varphi_{1,2,\epsilon}} \right] \right]. \quad (2.1.15) $$

By using the canonical expansion for the $\ln$, we obtain

$$ \det \left\| \frac{\delta(\tilde{\eta}_{1,2})}{\delta\varphi_{1,2,\epsilon}} \right\| = N_{\epsilon}^{-1} \exp \left[ \text{tr} \left[ \theta(\tau - \tau') \frac{\delta^2 \tilde{S}}{\delta\varphi_{1,2}(\tau)\delta\varphi_{1,2}(\tau')} \bigg|_{\varphi_{1,2}=\varphi_{1,2,\epsilon}} + \theta(\tau - \tau')\theta(\tau' - \tau) \frac{\delta^2 \tilde{S}}{\delta\varphi_{1,2}(\tau)\delta\varphi_{1,2}(\tau')} \frac{\delta^2 \tilde{S}}{\delta\varphi_{1,2}(\tau')\delta\varphi_{1,2}(\tau')} \bigg|_{\varphi_{1,2}=\varphi_{1,2,\epsilon}} + \dots \right] \right] = N_{\epsilon}^{-1} \exp \left[ \int d\tau \int_{\mathbb{R}^4} d^4 x \theta(0) \frac{\delta^2 \tilde{S}}{\delta\varphi_{1,2}^2(\tau)} \bigg|_{\varphi_{1,2}=\varphi_{1,2,\epsilon}} + \int d\tau' \int_{\mathbb{R}^4} d^4 x \theta(\tau - \tau') \theta(\tau' - \tau) \frac{\delta^2 \tilde{S}}{\delta\varphi_{1,2}(\tau)\delta\varphi_{1,2}(\tau')} \frac{\delta^2 \tilde{S}}{\delta\varphi_{1,2}(\tau')\delta\varphi_{1,2}(\tau')} \bigg|_{\varphi_{1,2}=\varphi_{1,2,\epsilon}} + \dots \right] \quad (2.1.16) $$
The second term in this expression is zero because $\theta(\tau - \tau')\theta(\tau' - \tau) = 0$ and the

same for all the subsequent terms. The only one left is the first term and choosing $\theta(0) = 1/2$ we get

$$ \det \left\| \frac{\delta(\tilde{\eta}_{1,2})}{\delta\varphi_{1,2,\epsilon}} \right\| = N^{-1}\epsilon \exp \left[ \frac{1}{2} \int d\tau' \frac{\delta^2\tilde{S}}{\delta\varphi_{1,2,\epsilon}^2} \right]. \quad (2.1.17) $$

Inserting Eq.(2.1.17) and Eq.(2.1.8) into Eq.(2.1.6) and performing the $\tilde{\eta}$ integration, we get

$$ \begin{aligned} Z_{\epsilon}[J_1, J_2; \omega] = N \int & D[\varphi_{1,2}(\tau', x; \omega)] D[\eta(\tau', x; \omega)] \delta(\varphi_{1,2}(0, x; \omega)) \times \\ & \exp\left\{-\int_0^{\tau} \int_{\mathbb{R}^4} \left[ \frac{1}{4\epsilon^2} \left[ \frac{\partial \varphi_1(\tau', x; \omega)}{\partial \tau'} + \frac{\delta \tilde{S}}{\delta \varphi_1(\tau', x; \omega)} \right]^2 - \frac{1}{2} \frac{\delta^2 \tilde{S}}{\delta \varphi_1^2(\tau', x; \omega)} \right] d^4 x d\tau' \right\} \\ & \times \exp\left\{-\int_0^{\tau} \int_{\mathbb{R}^4} \left[ \frac{1}{4\epsilon^2} \left[ \frac{\partial \varphi_2(\tau', x; \omega)}{\partial \tau'} + \frac{\delta \tilde{S}}{\delta \varphi_2(\tau', x; \omega)} \right]^2 - \frac{1}{2} \frac{\delta^2 \tilde{S}}{\delta \varphi_2^2(\tau', x; \omega)} \right] d^4 x d\tau' \right\} \\ & \times \exp\left\{-\int_0^{\tau} \int_{\mathbb{R}^4} J_{1,2}(\tau', x) \varphi_{1,2}(\tau', x; \omega) d^4 x d\tau' \right\} \exp\left[-\frac{1}{4} \int_0^{\tau} \int_{\mathbb{R}^4} \eta^2(\tau', x; \omega) d^4 x d\tau' \right]. \end{aligned} \quad (2.1.18) $$

From Eq.(2.1.18) finally we obtain

$$ \begin{aligned} Z_{\epsilon}[J_{1,2}; \omega] = N \int & D[\varphi_{1,2}(\tau', x; \omega)] D[\eta(\tau', x; \omega)] \delta(\varphi_{1,2}(0, x; \omega)) \times \\ & \exp\left\{-\int_0^{\tau} \int_{\mathbb{R}^4} \left[ \frac{1}{4\epsilon^2} \left[ \frac{\partial \varphi_1(\tau', x; \omega)}{\partial \tau'} + \frac{\delta S}{\delta \varphi_1(\tau', x; \omega)} - \eta \right]^2 - \frac{1}{2} \frac{\delta^2 S}{\delta \varphi_1^2} \right] d^4 x d\tau' \right. \\ & \left. - \int_0^{\tau} \int_{\mathbb{R}^4} \left[ \frac{1}{4\epsilon^2} \left[ \frac{\partial \varphi_2(\tau', x; \omega)}{\partial \tau'} + \frac{\delta S}{\delta \varphi_2(\tau', x; \omega)} - \eta \right]^2 - \frac{1}{2} \frac{\delta^2 S}{\delta \varphi_2^2} \right] d^4 x d\tau' - \right. \\ & \left. - \int_0^{\tau} \int_{\mathbb{R}^4} J_{1,2}(\tau', x) \varphi_{1,2}(\tau', x; \omega) d^4 x d\tau' \right\} \times \exp\left[-\frac{1}{4} \int_0^{\tau} \int_{\mathbb{R}^4} \eta^2(\tau', x; \omega) d^4 x d\tau' \right]. \end{aligned} \quad (2.1.19) $$

If we want also to specify that we are interested only in the correlations at the same 5-th time $\tau_1$, we have just to choose $J_{1,2}(x, \tau')$ of the form $J_{1,2}(x, \tau') = \bar{J}(x)\delta(\tau' - \tau_1)$, $\tau_1 < \tau$ and Eq.(2.1.19) then becomes

$$ \begin{aligned} Z_{\epsilon}[J_{1,2}; \omega] = N \int & D[\varphi_{1,2}(\tau', x; \omega)] D[\eta(\tau', x; \omega)] \delta(\varphi_{1,2}(0, x; \omega)) \times \\ & \exp\left\{-\int_0^{\tau} \int_{\mathbb{R}^4} \left[ \frac{1}{4\epsilon^2} \left[ \frac{\partial \varphi_1(\tau', x; \omega)}{\partial \tau'} + \frac{\delta \tilde{S}}{\delta \varphi_1(\tau', x; \omega)} \right]^2 - \frac{1}{2} \frac{\delta^2 \tilde{S}}{\delta \varphi_1^2(\tau', x; \omega)} \right] d^4 x d\tau' \right. \\ & \left. - \int_0^{\tau} \int_{\mathbb{R}^4} \left[ \frac{1}{4\epsilon^2} \left[ \frac{\partial \varphi_2(\tau', x; \omega)}{\partial \tau'} + \frac{\delta \tilde{S}}{\delta \varphi_2(\tau', x; \omega)} \right]^2 - \frac{1}{2} \frac{\delta^2 \tilde{S}}{\delta \varphi_2^2(\tau', x; \omega)} \right] d^4 x d\tau' - \right. \\ & \left. - \int_{\mathbb{R}^4} J_{1,2}(\tau_1, x) \varphi_{1,2}(\tau_1, x; \omega) d^4 x \right\} \times \exp\left[-\frac{1}{4} \int_0^{\tau} \int_{\mathbb{R}^4} \eta^2(\tau', x; \omega) d^4 x d\tau' \right]. \end{aligned} \quad (2.1.20) $$
**Remark 2.1.2.** In all this we have to remember, of course, that once we set $\tau_1 \to \infty$ we

have also to extend the interval of integration from $[0, \tau]$ to $[0, \infty]$.

From Eq.(2.1.21) with $\epsilon \ll 1$ for two-point correlation function $\langle\varphi(\tau_1, x_1)\varphi(\tau_2, x_2)\rangle$ defined by

$$ \langle\varphi_1(\tau_1, x_1)\varphi_2(\tau_2, x_2)\rangle = \lim_{\epsilon \to 0} \left[ \frac{\delta^l Z_\epsilon[J_1, J_2; \omega]}{\delta J_1(\tau_1, x_1) \delta J_2(\tau_2, x_2)} \right] \quad (2.1.21') $$

for mutually two-point correlation function $\langle\varphi_1(\tau_1, x_1)\varphi_2(\tau_2, x_2)\rangle_\eta$ we get

$$ \begin{aligned} \langle\varphi_1(\tau_1, x_1)\varphi_2(\tau_2, x_2)\rangle_\eta &\doteq \langle\varphi_1(\tau_1, x_1)\varphi_2(\tau_2, x_2); \epsilon\rangle \triangleq \\ &\doteq N_\epsilon^{-1} \int D[\eta(\tau', x; \omega)] \exp\left[-\frac{1}{4} \int_0^\tau \int_{\mathbb{R}^4} \eta^2(\tau', x; \omega) d^4x d\tau' \right] \times \\ &\left( \int D[\varphi_{1,2}(\tau', x; \omega)] (\varphi_1(\tau_1, x_1; \omega)\varphi_2(\tau_2, x_2; \omega)) \times \right. \\ &\qquad \left. \delta(\varphi_{1,2}(0, x; \omega)) \times \right. \\ &\qquad \times \exp\left\{ -\frac{1}{4\epsilon^2} \left[ \int_0^\tau \int_{\mathbb{R}^4} \left[ \frac{\partial\varphi(\tau', x; \omega)}{\partial\tau'} + \frac{\delta S}{\delta\varphi_1(\tau', x; \omega)} - \eta(\tau', x; \omega) \right]^2 - \frac{1}{2} \frac{\delta^2 S}{\delta\varphi_1^2(\tau', x; \omega)} \right] d^4x d\tau' \right\} \\ &\qquad \times \exp\left\{ -\frac{1}{4\epsilon^2} \int_0^\tau \int_{\mathbb{R}^4} \left[ \frac{\partial\varphi(\tau', x; \omega)}{\partial\tau'} + \frac{\delta S}{\delta\varphi_2(\tau', x; \omega)} - \eta(\tau', x; \omega) \right]^2 - \frac{1}{2} \frac{\delta^2 S}{\delta\varphi_2^2(\tau', x; \omega)} \right] d^4x d\tau' \right\}. \end{aligned} \quad (2.1.22) $$

Performing the $\varphi_{1,2}(\tau', x; \omega)$ integration in Eq.(2.1.22) by using saddle point approximation we get
$$
\begin{align*}
& \langle \varphi_{1,\eta}(\tau_1, x_1) \varphi_{2,\eta}(\tau_2, x_2) \rangle_{\eta} \times \\
& \det \left\| \frac{\delta(\varphi_1)}{\delta\eta} \right\| \det \left\| \frac{\delta(\varphi_2)}{\delta\eta} \right\| \left\| \int D[\eta(\tau', x; \omega)] (\varphi_{1,\eta}(\tau_1, x_1; \omega) \varphi_{2,\eta}(\tau_2, x_2; \omega)) \right. \\
& \qquad \left. \times \exp \left\{ -\int_0^{\tau} \left[ -\frac{1}{2} \frac{\delta^2 S}{\delta\varphi_1^2(\tau', x; \omega)} \bigg|_{\varphi_{1,2}=\varphi_{1,2,\eta}} \right] d^4x d\tau' \right\} \times \right. \\
& \qquad \left. \exp \left\{ -\int_0^{\tau} \left[ -\frac{1}{2} \frac{\delta^2 S}{\delta\varphi_2^2(\tau', x; \omega)} \bigg|_{\varphi_{1,2}=\varphi_{1,2,\eta}} \right] d^4x d\tau' \right\} \right\} \\
& \qquad \times \exp \left[ -\frac{1}{4} \int_0^{\tau} \eta^2(\tau', x; \omega) d^4x d\tau' \right] = \\
& \qquad \left( \int D[\eta(\tau', x; \omega)] (\varphi_{1,\eta}(\tau_1, x_1; \omega) \varphi_{2,\eta}(\tau_2, x_2; \omega)) \exp \left[ -\frac{1}{4} \int_0^{\tau} \eta^2(\tau', x; \omega) d^4x d\tau' \right] \right) \times \\
& \qquad \times \langle \varphi_{1,\eta}(\tau_1, x_1; \omega) \varphi_{2,\eta}(\tau_2, x_2; \omega) \rangle_{\eta}
\end{align*}
\tag{2.1.23}
$$

$\varphi_{1,2,\eta}(\tau_1, x_1; \omega)$ that appears in Eq.(2.1.23) is the solution of the Langevin equation (2.1.26), solved with zero initial condition. From Eq.(2.1.22)-Eq.(2.1.23) we get

$$
\begin{align}
& \langle \varphi_1(\tau_1, x_1; \omega) \varphi_2(\tau_2, x_2; \omega) \rangle_{\eta} \approx \nonumber \\
& N \int D[\varphi_{1,2}(\tau', x; \omega)] \delta(\varphi_{1,2}(0, x; \omega)) \nonumber \\
& \left( \int D[\eta(\tau', x; \omega)] (\varphi_1(\tau_1, x_1; \omega) \varphi_2(\tau_2, x_2; \omega)) \exp \left[ -\frac{1}{4} \int_0^{\tau} \int_{\mathbb{R}^4} \eta^2(\tau', x; \omega) d^4x d\tau' \right] \right) \times \nonumber \\
& \qquad \times \exp \left\{ -\int_0^{\tau} \left[ \frac{1}{4\epsilon^2} \left[ \frac{\partial \varphi_1(\tau', x; \omega)}{\partial \tau'} + \frac{\delta S}{\delta \varphi_1(\tau', x; \omega)} - \eta(\tau', x; \omega) \right]^2 \right] d^4x d\tau' \right\} \nonumber \\
& \qquad \times \exp \left\{ -\int_0^{\tau} \left[ \frac{1}{4\epsilon^2} \left[ \frac{\partial \varphi_2(\tau', x; \omega)}{\partial \tau'} + \frac{\delta S}{\delta \varphi_2(\tau', x; \omega)} - \eta(\tau', x; \omega) \right]^2 \right] d^4x d\tau' \right\}. \tag{2.1.24}
\end{align}
$$

From Eq.(2.1.24) finally we get
$$
\begin{align}
\langle \phi_{1,\eta}(\tau_1, x_1; \omega) \phi_{2,\eta}(\tau_2, x_2; \omega) \rangle_{\eta} &\approx \langle \phi_1(\tau_1, x_1) \phi_2(\tau_2, x_2); \epsilon \rangle \triangleq \nonumber \\
&\triangleq N_{\epsilon}^{-1} \int D[\phi_{1,2}(\tau', x; \omega)] \langle \phi_1(\tau_1, x_1; \omega) \phi_2(\tau_2, x_2; \omega) \rangle_{\eta} \times \nonumber \\
&\quad \times \exp \left\{ -\frac{1}{4\epsilon^2} \int_0^{\tau_1} \int_{\mathbb{R}^4} \left[ \frac{\partial \phi_1(\tau', x; \omega)}{\partial \tau'} + \frac{\delta S}{\delta \phi_1(\tau', x; \omega)} - \eta(\tau', x; \omega) \right]^2 d^4 x d\tau' \right\} \nonumber \\
&\quad \times \exp \left\{ -\frac{1}{4\epsilon^2} \int_0^{\tau_2} \int_{\mathbb{R}^4} \left[ \frac{\partial \phi_2(\tau', x; \omega)}{\partial \tau'} + \frac{\delta S}{\delta \phi_2(\tau', x; \omega)} - \eta(\tau', x; \omega) \right]^2 d^4 x d\tau' \right\} \tag{2.1.25}
\end{align}
$$

$$
\langle \phi_1(\tau_1, x_1; \omega) \phi_2(\tau_2, x_2; \omega) \rangle_{\eta} \triangleq \\
\int D[\eta(\tau', x; \omega)] (\phi_1(\tau_1, x_1; \omega) \phi_2(\tau_2, x_2; \omega)) \exp \left[ -\frac{1}{4} \int_0^{\tau} \int_{\mathbb{R}^4} \eta^2(\tau', x; \omega) d^4 x d\tau' \right]
$$

**Proposition 2.1.1.** It follows from Eq.(2.1.22)-Eq.(2.1.23) that in Eq.(2.1.22) we can interchange integration on variable $\eta(\tau',x;\omega)$ and integration on variables $\phi_{1,2}(\tau',x;\omega)$ in Eq.(1.24)-Eq.(2.1.25)

**Remark 2.1.3.** Note that for any fixed values of parameters $\tau_1, x_1, \tau_2, x_2$ and $\gamma \simeq 0$ we get

$$
\langle [\phi_1(\tau_1, x_1; \omega) - a][\phi_2(\tau_2, x_2; \omega) + a]; \epsilon \rangle_{\eta} \approx \langle \phi_1(\tau_1, x_1; \omega) \phi_2(\tau_2, x_2; \omega); \epsilon \rangle_{\eta} - a^2, \quad (2.1.26)
$$

where by translation invariance

$$
\langle \phi_1(\tau_1, x_1; \omega) \phi_2(\tau_2, x_2; \omega); \epsilon \rangle_{\eta} - a^2 \approx 0 \Rightarrow a \approx a(\tau_1, \tau_2, x_1 - x_2). \quad (2.1.27)
$$

From Eq.(2.1.25) by the replacement

$$
\begin{align}
\phi_1(\tau, x; \omega) - \theta(\tau)a &= v_-(\tau, x; \omega), \nonumber \\
\phi_2(\tau, x; \omega) + \theta(\tau)a &= v_+(\tau, x; \omega), \nonumber \\
\phi_1(\tau, x; \omega) &= v_-(\tau, x; \omega) + \theta(\tau)a, \nonumber \\
\phi_2(\tau, x; \omega) &= v_+(\tau, x; \omega) - \theta(\tau)a, \nonumber \\
\frac{\partial \phi_1(\tau, x; \omega)}{\partial \tau} &= \frac{\partial v_-(\tau, x; \omega)}{\partial \tau} + \delta(\tau)a, \nonumber \\
\frac{\partial \phi_2(\tau, x; \omega)}{\partial \tau} &= \frac{\partial v_+(\tau, x; \omega)}{\partial \tau} - \delta(\tau)a, \tag{2.1.28}
\end{align}
$$

we obtain
$$
\begin{align}
& \Omega(\tau_1, \tau_2, x_1 - x_2, a; \epsilon) \triangleq \nonumber \\
& N_{\epsilon}^{-1} \int D[\varphi_{1,2}(\tau', x; \omega)] \langle [\varphi_1(\tau_1, x_1; \omega) - a][\varphi_2(\tau_2, x_2; \omega) + a] \rangle_{\eta} \times \nonumber \\
& \times \exp \left\{ -\frac{1}{4\epsilon^2} \int_0^{\tau_1} \int_{\mathbb{R}^4} \left[ \frac{\partial \varphi_1(\tau', x; \omega)}{\partial \tau'} + \frac{\delta S}{\delta \varphi_1(\tau', x; \omega)} - \eta(\tau', x; \omega) \right]^2 d^4 x d\tau' \right\} \times \nonumber \\
& \times \exp \left\{ -\frac{1}{4\epsilon^2} \int_0^{\tau_2} \int_{\mathbb{R}^4} \left[ \frac{\partial \varphi_2(\tau', x; \omega)}{\partial \tau'} + \frac{\delta S}{\delta \varphi_2(\tau', x; \omega)} - \eta(\tau', x; \omega) \right]^2 d^4 x d\tau' \right\} = \nonumber \\
&= N_{\epsilon}^{-1} \int D[v_-(\tau', x; \omega)] D[v_+(\tau', x; \omega)] \langle v_-(\tau_1, x_1) v_+(\tau_1, x_1) \rangle_{\eta} \times \nonumber \\
& \times \exp \left\{ -\frac{1}{4\epsilon^2} \int_0^{\tau_1} \int_{\mathbb{R}^4} \left[ \frac{\partial v_-(\tau', x; \omega)}{\partial \tau'} + \delta(\tau)a + \frac{\delta S}{\delta \varphi_1(\tau', x; \omega)} \bigg|_{\varphi_1=v_-(\tau', x; \omega)+a_-} - \eta(\tau', x; \omega) d^4 x d\tau' \right]^2 \right\} \times \nonumber \\
& \times \exp \left\{ -\frac{1}{4\epsilon^2} \int_0^{\tau_2} \int_{\mathbb{R}^4} \left[ \frac{\partial v_+(\tau', x; \omega)}{\partial \tau'} - \delta(\tau)a + \frac{\delta S}{\delta \varphi_2(\tau', x; \omega)} \bigg|_{\varphi_2=v_+(\tau_1, x_1; \omega)-a} - \eta(\tau', x; \omega) d^4 x d\tau' \right] \right\}. \tag{2.1.29}
\end{align}
$$

**Remark 2.1.4.** Note that

$$
\lim_{\epsilon \to 0} \Omega(\tau_1, \tau_2, x_1 - x_2, a; \epsilon) = 0 \Rightarrow \lim_{\epsilon \to 0} \langle \varphi_1(\tau_1, x_1) \varphi_2(\tau_2, x_2); \epsilon \rangle - a^2 = 0. \quad (2.1.30)
$$

**Definition 2.1.1.** Let $v_{\mp}(\tau, x; \omega)$ be the solution of the Langevin equations (2.1.31)

$$
\begin{align}
\frac{\partial v_-(\tau, x; \omega)}{\partial \tau} &= -\delta(\tau)a - \left. \frac{\delta S[\varphi_{1,2}]}{\delta \varphi_1(\tau, x; \omega)} \right|_{\substack{\varphi_1=v_-(\tau', x; \omega)+a- \\ \varphi_2=v_+(\tau_1, x_1; \omega)-a}} + \eta(\tau, x; \omega), \\
\frac{\partial v_+(\tau, x; \omega)}{\partial \tau} &= \delta(\tau)a - \left. \frac{\delta S[\varphi_{1,2}]}{\delta \varphi_2(\tau, x; \omega)} \right|_{\substack{\varphi_1=v_-(\tau', x; \omega)+a- \\ \varphi_2=v_+(\tau_1, x_1; \omega)-a}} + \eta(\tau, x; \omega), \tag{2.1.31} \\
v_{\mp}(0, x; \omega) &= 0,
\end{align}
$$

Linear stochastic differential *master equation* corresponding to the Langevin
equations

(2.1.31) reads

$$
\begin{align}
\frac{\partial v_-(\tau, x, a; \omega)}{\partial \tau} &= -\delta(\tau)a - \mathcal{L} \left\{ \frac{\delta S[\varphi_{1,2}]}{\delta \varphi_1(\tau, x; \omega)} \bigg|_{\varphi_1=v_-(\tau', x; \omega)+a-} \right\} + \eta(\tau, x; \omega), \nonumber \\
\frac{\partial v_+(\tau, x, a; \omega)}{\partial \tau} &= \delta(\tau)a - \mathcal{L} \left\{ \frac{\delta S[\varphi_{1,2}]}{\delta \varphi_2(\tau, x; \omega)} \bigg|_{\varphi_2=v_+(\tau_1, x_1; \omega)-a} \right\} + \eta(\tau, x; \omega), \tag{2.1.32} \\
v_{\mp}(0, x, a; \omega) &= 0, \nonumber
\end{align}
$$

where
$$ \mathcal{L} \left\{ \delta S[\varphi_{1,2}]/\delta \varphi_1(\tau, x; \omega) \Big|_{\substack{\varphi_1 = v_-(\tau', x; \omega) + a_-, \\ \varphi_2 = v_+(\tau_1, x_1; \omega) - a}} \right\} \quad (2.1.33) $$

is a linear part of variational derivative

$$ \frac{\delta S[\varphi_{1,2}]}{\delta \varphi_1(\tau, x; \omega)} \Big|_{\substack{\varphi_1 = v_-(\tau', x; \omega) + a_-, \\ \varphi_2 = v_+(\tau_1, x_1; \omega) - a}} \quad (2.1.34) $$

and where

$$ \mathcal{L} \left\{ \frac{\delta S[\varphi_{1,2}]}{\delta \varphi_2(\tau, x; \omega)} \Big|_{\substack{\varphi_1 = v_-(\tau', x; \omega) + a_-, \\ \varphi_2 = v_+(\tau_1, x_1; \omega) - a}} \right\}. \quad (2.1.35) $$

is a linear part of variational derivative

$$ \frac{\delta S[\varphi_{1,2}]}{\delta \varphi_2(\tau, x; \omega)} \Big|_{\substack{\varphi_1 = v_-(\tau', x; \omega) + a_-, \\ \varphi_2 = v_+(\tau_1, x_1; \omega) - a}} \quad (2.1.36) $$

## 2.2. Transcendental master equation corresponding to two-point Green function $G(x_1, x_2, \lambda)$.

**Definition 2.2.1.** Let $v_+(\tau, x, a; \omega)$ be the solution of the stochastic differential master equations (2.1.32). Transcendental master equation corresponding to the stochastic Langevin equation (2.1.31) reads

$$ \langle v_-(\tau_1, x_1, a; \omega) v_+(\tau_2, x_2, a; \omega) \rangle_{\eta} = 0. \quad (2.2.1) $$

**Theorem 2.2.1.** Let $a_{1,2}(\bar{\tau}, \bar{x})$ be an solution of the equation (2.2.1) at fixed point $(\bar{\tau}_1, \bar{x}_1; \bar{\tau}_2, \bar{x}_2) \in (\mathbb{R}_+ \times \mathbb{R}^4) \times (\mathbb{R}_+ \times \mathbb{R}^4)$ i.e.,

$$ \langle v_-(\bar{\tau}_1, \bar{x}_1, a(\bar{\tau}_1, \bar{x}_1; \bar{\tau}_2, \bar{x}_2); \omega) v_+(\bar{\tau}_2, \bar{x}_2, a(\bar{\tau}_1, \bar{x}_1; \bar{\tau}_2, \bar{x}_2); \omega) \rangle_{\eta} = 0. \quad (2.2.2) $$

Let $\Delta(\bar{\tau}_1, \bar{x}_1; \bar{\tau}_2, \bar{x}_2)$ be a set such that

$$ a(\bar{\tau}_1, \bar{x}_1; \bar{\tau}_2, \bar{x}_2) \in \Delta(\bar{\tau}_1, \bar{x}_1; \bar{\tau}_2, \bar{x}_2) \iff \\ \langle v_-(\bar{\tau}_1, \bar{x}_1, a(\bar{\tau}_1, \bar{x}_1; \bar{\tau}_2, \bar{x}_2); \omega) v_+(\bar{\tau}_2, \bar{x}_2, a(\bar{\tau}_1, \bar{x}_1; \bar{\tau}_2, \bar{x}_2); \omega) \rangle_{\eta} = 0, \quad (2.2.3) $$

and let $\tilde{\Omega}(\bar{\tau}_1, \bar{\tau}_2, \bar{x}_1 - \bar{x}_2, a(\bar{\tau}_1, \bar{x}_1; \bar{\tau}_2, \bar{x}_2))$ be a set such that

$$ \begin{aligned} a(\bar{\tau}_1, \bar{x}_1; \bar{\tau}_2, \bar{x}_2) &\in \tilde{\Omega}(\bar{\tau}_1, \bar{\tau}_2, \bar{x}_1 - \bar{x}_2, a(\bar{\tau}_1, \bar{x}_1; \bar{\tau}_2, \bar{x}_2)) &\iff \\ &\iff \lim_{\epsilon \to 0} \Omega(\bar{\tau}_1, \bar{\tau}_2, \bar{x}_1 - \bar{x}_2, a(\bar{\tau}_1, \bar{x}_1; \bar{\tau}_2, \bar{x}_2); \epsilon) = 0, \end{aligned} \quad (2.2.4) $$

where the quantity $\Omega(\bar{\tau}_1, \bar{\tau}_2, \bar{x}_1 - \bar{x}_2, a(\bar{\tau}_1, \bar{x}_1; \bar{\tau}_2, \bar{x}_2); \epsilon)$ defined by Eq.(2.1.29). Then

$$ \tilde{\Omega}(\bar{\tau}_1, \bar{\tau}_2, \bar{x}_1 - \bar{x}_2, a(\bar{\tau}_1, \bar{x}_1; \bar{\tau}_2, \bar{x}_2)) \subseteq \Delta(\bar{\tau}_1, \bar{x}_1; \bar{\tau}_2, \bar{x}_2). \quad (2.2.5) $$

## 2.3. Double Stochastic Quantization the Free Scalar
# Fields

For a scalar field theory governed by the action in terms of the Euclidean spacetime is

given by

$$S_E = \int d^4 x \left[ \frac{1}{2} (\partial \varphi(x))^2 + \frac{1}{2} (m \varphi(x))^2 \right] \qquad (2.3.1)$$

differential master equations corresponding to the Langevin equations (2.1.31) reads

$$\begin{aligned}
\frac{\partial v_{-}(x, \tau; \omega, \bar{\omega})}{\partial \tau} &= (\partial^2 - m^2)v_{-}(x, \tau; \omega, \bar{\omega}) + \eta(x, \tau; \omega), \\
\frac{\partial v_{+}(x, \tau; \omega, \bar{\omega})}{\partial \tau} &= (\partial^2 - m^2)v_{+}(x, \tau; \omega, \bar{\omega}) + \eta(x, \tau; \omega), \qquad (2.3.2) \\
v_{\mp}(x, 0; \omega, \bar{\omega}) &= 0.
\end{aligned}$$

Fourier transformed stochastic differential equations (2.3.2) in $k$ and $\tau$ given as

$$\begin{aligned}
\frac{\partial}{\partial \tau} \hat{v}_{-}(k, \tau) &= -(k^2 + m^2) \hat{v}_{-}(k, \tau) - (2\pi)^4 a \delta(\tau) \delta^4(k) - (2\pi)^4 m^2 a \delta^4(k) + \hat{\eta}(k, \tau; \omega), \qquad (2.3.3) \\
\frac{\partial}{\partial \tau} \hat{v}_{+}(k, \tau) &= -(k^2 + m^2) \hat{v}_{+}(k, \tau) - (2\pi)^4 a \delta(\tau) \delta^4(k) + (2\pi)^4 m^2 a \delta^4(k) + \hat{\eta}(k, \tau; \omega).
\end{aligned}$$

Let us consider ODE

$$\dot{x}(\tau, \lambda) + \lambda x(\tau, \lambda) = g(\tau, \lambda), x(0) = 0. \qquad (2.3.4)$$

The corresponding solution $x(\tau, \lambda)$ is

$$x(\tau, \lambda) = e^{-\lambda \tau} \int_{0}^{\tau} e^{\lambda \tau_1} g(\tau_1, \lambda) d\tau_1. \qquad (2.3.5)$$

From Eq.(2.3.3)-Eq.(2.3.5) one obtains
$$
\begin{align*}
\hat{\nu}_{-}(k, \tau, a) &= e^{-(k^2+m^2)\tau} \int_0^{\tau} e^{(k^2+m^2)\tau_1} \Biggl[-(2\pi)^4 a\delta(\tau_1)\delta^4(k) - (2\pi)^4 m^2 a\delta^4(k) + \hat{\eta}(k, \tau_1; \omega)\Biggr] d\tau_1 \\
&= -(2\pi)^4 a\delta^4(k) e^{-(k^2+m^2)\tau} - (2\pi)^4 m^2 a\delta^4(k) e^{-(k^2+m^2)\tau} \int_0^{\tau} e^{(k^2+m^2)\tau_1} d\tau_1 + \\
&\quad + e^{-(k^2+m^2)\tau} \int_0^{\tau} e^{(k^2+m^2)\tau_1} \hat{\eta}(k, \tau_1; \omega) d\tau_1 = \\
&= -(2\pi)^4 a\delta^4(k) e^{-(k^2+m^2)\tau} - (2\pi)^4 m^2 a\delta^4(k) e^{-(k^2+m^2)\tau} \int_0^{\tau} e^{(k^2+m^2)\tau_1} d\tau_1 + \\
&\quad + \int_0^{\tau} e^{-(k^2+m^2)(\tau-\tau_1)} \hat{\eta}(k, \tau_1; \omega) d\tau_1 = \tag{2.3.6} \\
&\quad -(2\pi)^4 a\delta^4(k) e^{-(k^2+m^2)\tau} - (2\pi)^4 m^2 a\delta^4(k) e^{-(k^2+m^2)\tau} \Biggl[ \frac{e^{(k^2+m^2)\tau}}{k^2+m^2} - \frac{1}{k^2+m^2} \Biggr] + \\
&\quad + \int_0^{\tau} e^{-(k^2+m^2)(\tau-\tau_1)} \hat{\eta}(k, \tau_1; \omega) d\tau_1 = \\
&= -(2\pi)^4 a\delta^4(k) e^{-(k^2+m^2)\tau} - (2\pi)^4 \frac{m^2 a\delta^4(k)}{k^2+m^2} \Bigl[ 1 - e^{-(k^2+m^2)\tau} \Bigr] + \\
&\quad + \int_0^{\tau} e^{-(k^2+m^2)(\tau-\tau_1)} \hat{\eta}(k, \tau_1; \omega) d\tau_1 = \\
&= -(2\pi)^4 a\delta^4(k) \Biggl[ e^{-(k^2+m^2)\tau} + \frac{m^2}{k^2+m^2} \Bigl[ 1 - e^{-(k^2+m^2)\tau} \Bigr] \Biggr] + \\
&\quad + \int_0^{\tau} e^{-(k^2+m^2)(\tau-\tau_1)} \hat{\eta}(k, \tau_1; \omega) d\tau_1.
\end{align*}
$$

and
$$
\begin{align*}
\hat{v}_+(k, \tau, a) &= e^{-(k^2+m^2)\tau} \int_0^\tau e^{(k^2+m^2)\tau_1} \Biggl[ -(2\pi)^4 a\delta(\tau_1)\delta^4(k) + (2\pi)^4 m^2 a\delta^4(k) + \hat{\eta}(k, \tau_1; \omega) \Biggr] d\tau_1 \\
&= -(2\pi)^4 a\delta^4(k) e^{-(k^2+m^2)\tau} - (2\pi)^4 m^2 a\delta^4(k) e^{-(k^2+m^2)\tau} \int_0^\tau e^{(k^2+m^2)\tau_1} d\tau_1 + \\
&\quad + e^{-(k^2+m^2)\tau} \int_0^\tau e^{(k^2+m^2)\tau_1} \hat{\eta}(k, \tau_1; \omega) d\tau_1 = \\
&= -(2\pi)^4 a\delta^4(k) e^{-(k^2+m^2)\tau} + (2\pi)^4 m^2 a\delta^4(k) e^{-(k^2+m^2)\tau} \int_0^\tau e^{(k^2+m^2)\tau_1} d\tau_1 + \\
&\quad + \int_0^\tau e^{-(k^2+m^2)(\tau-\tau_1)} \hat{\eta}(k, \tau_1; \omega) d\tau_1 = \\
&\quad -(2\pi)^4 a\delta^4(k) e^{-(k^2+m^2)\tau} - (2\pi)^4 a\delta^4(k) e^{-(k^2+m^2)\tau} + \\
&\quad (2\pi)^4 m^2 a\delta^4(k) e^{-(k^2+m^2)\tau} \left[ \frac{e^{(k^2+m^2)\tau}}{k^2+m^2} - \frac{1}{k^2+m^2} \right] + \\
&\quad + \int_0^\tau e^{-(k^2+m^2)(\tau-\tau_1)} \hat{\eta}(k, \tau_1; \omega) d\tau_1 = \\
&= -(2\pi)^4 a\delta^4(k) e^{-(k^2+m^2)\tau} - (2\pi)^4 \frac{m^2 a\delta^4(k)}{k^2+m^2} \left[ 1 - e^{-(k^2+m^2)\tau} \right] + \\
&\quad + \int_0^\tau e^{-(k^2+m^2)(\tau-\tau_1)} \hat{\eta}(k, \tau_1; \omega) d\tau_1 = \\
&= -(2\pi)^4 a\delta^4(k) \left[ e^{-(k^2+m^2)\tau} + \frac{m^2}{k^2+m^2} \left[ 1 - e^{-(k^2+m^2)\tau} \right] \right] + \\
&\quad + \int_0^\tau e^{-(k^2+m^2)(\tau-\tau_1)} \hat{\eta}(k, \tau_1; \omega) d\tau_1.
\end{align*}
\tag{2.3.7}
$$

From Eq.(2.3.6)-Eq.(2.3.7) one obtains
$$
\begin{equation}
\begin{split}
\hat{v}_-(k_1, \tau, a) \hat{v}_+(k_2, \tau', a') = & \\
& \left\{ -(2\pi)^4 a \delta^4(k_1) \left[ e^{-(k_1^2+m^2)\tau} + \frac{m^2}{k_1^2+m^2} \left[ 1 - e^{-(k_1^2+m^2)\tau} \right] \right] + \right. \\
& \qquad \left. + \int_0^\tau e^{-(k_1^2+m^2)(\tau-\tau_1)} \hat{\eta}(k_1, \tau_1; \omega) d\tau_1 \right\} \times \\
& \left\{ (2\pi)^4 a' \delta^4(k_2) \left[ e^{-(k_2^2+m^2)\tau'} + \frac{m^2}{k_2^2+m^2} \left[ 1 - e^{-(k_2^2+m^2)\tau'} \right] \right] + \right. \\
& \qquad \left. + \int_0^\tau' e^{-(k_2^2+m^2)(\tau'-\tau_1)} \hat{\eta}(k_2, \tau_1; \omega) d\tau_1 \right\}
\end{split}
\tag{2.3.8}
\end{equation}
$$

From Eq.(3.2.11) one obtains

$$
\begin{equation}
\begin{split}
\langle \hat{v}_-(k_1, \tau, a) \hat{v}_+(k_2, \tau', a') \rangle_{\eta} = & \\
& \left\{ -(2\pi)^4 a \delta^4(k_1) \left[ e^{-(k_1^2+m^2)\tau} + \frac{m^2}{k_1^2+m^2} \left[ 1 - e^{-(k_1^2+m^2)\tau} \right] \right] \right\} \times \\
& \left\{ (2\pi)^4 a' \delta^4(k_2) \left[ e^{-(k_2^2+m^2)\tau'} + \frac{m^2}{k_2^2+m^2} \left[ 1 - e^{-(k_2^2+m^2)\tau'} \right] \right] + \right. \\
& \left. \left\langle \left( \int_0^\tau e^{-(k_1^2+m^2)(\tau-\tau_1)} \hat{\eta}(k_1, \tau_1; \omega) d\tau_1 \right) \left( \int_0^{\tau'} e^{-(k_2^2+m^2)(\tau'-\tau_2)} \hat{\eta}(k_2, \tau_2; \omega) d\tau_2 \right) \right\rangle_{\eta} \right. \\
& \left. - (2\pi)^8 a a' \delta^4(k_1) \delta^4(k_2) \left[ e^{-(k_1^2+m^2)\tau} + \frac{m^2}{k_1^2+m^2} \left[ 1 - e^{-(k_1^2+m^2)\tau} \right] \right] \times \right. \\
& \left. \left[ e^{-(k_2^2+m^2)\tau'} + \frac{m^2}{k_2^2+m^2} \left[ 1 - e^{-(k_2^2+m^2)\tau'} \right] \right] + \right. \\
& \left. + \int_0^{\tau'} e^{-(k_2^2+m^2)(\tau'-\tau_2)} \int_0^{\tau'} e^{-(k_1^2+m^2)(\tau-\tau_1)} \langle \hat{\eta}(k_1, \tau_1; \omega) \hat{\eta}(k_2, \tau_2; \omega) \rangle_{\eta} d\tau_1 d\tau_2 \right]
\end{split}
\tag{2.3.9}
\end{equation}
$$

We set now $\tau' = \tau, a' = a$. Note that
$$ -(2\pi)^8 a a' \delta^4(k_1) \delta^4(k_2) \left[ e^{-(k_1^2+m^2)\tau} + \frac{m^2}{k_1^2+m^2} [1-e^{-(k_1^2+m^2)\tau}] \right] \times \\ \left[ e^{-(k_2^2+m^2)\tau'} + \frac{m^2}{k_2^2+m^2} [1-e^{-(k_1^2+m^2)\tau'}] \right] \Big|_{\tau'=\tau} = \\ -(2\pi)^8 a^2 \delta^4(k_1) \delta^4(k_2) \left[ e^{-(k_1^2+m^2)\tau} + \frac{m^2}{k_1^2+m^2} [1-e^{-(k_1^2+m^2)\tau}] \right] \times \\ \left[ e^{-(k_1^2+m^2)\tau} + \frac{m^2}{k_2^2+m^2} [1-e^{-(k_1^2+m^2)\tau}] \right] \qquad (2.3.10) $$

and

$$ \begin{aligned} & \int_0^{\tau'} e^{-(k_2^2+m^2)(\tau'-\tau_2)} \int_0^{\tau} e^{-(k_1^2+m^2)(\tau-\tau_1)} \langle \hat{\eta}(k_1, \tau_1; \omega) \hat{\eta}(k_2, \tau_2; \omega) \rangle_{\eta} d\tau_1 d\tau_2 \Big|_{\tau'=\tau} = \\ & \qquad \delta(k_1+k_2) \int_0^{\tau} e^{-(k_2^2+m^2)(\tau-\tau_2)} \int_0^{\tau} e^{-(k_1^2+m^2)(\tau-\tau_1)} \delta(\tau_1-\tau_2) d\tau_1 d\tau_2 = \\ & \qquad 2(2\pi)^4 \delta(k_1+k_2) \int_0^{\tau} e^{-(k_1^2+k_2^2+2m^2)(\tau-\tau_1)} d\tau_1 = \end{aligned} \qquad (2.3.11) $$

$$ 2(2\pi)^4 \delta(k_1 + k_2) e^{-(k_1^2+k_2^2+2m^2)\tau} \int_0^{\tau} e^{(k_1^2+k_2^2+2m^2)\tau_1} d\tau_1 = $$

$$ \begin{aligned} & 2(2\pi)^4 \delta(k_1 + k_2) e^{-(k_1^2+k_2^2+2m^2)\tau} \left[ \frac{1}{(k_1^2 + k_2^2 + 2m^2)} e^{(k_1^2+k_2^2+m^2)\tau} - \frac{1}{(k_1^2 + k_2^2 + 2m^2)} \right] = \\ & = \frac{2(2\pi)^4 \delta(k_1 + k_2)}{k_1^2 + k_2^2 + 2m^2} - 2(2\pi)^4 \delta(k_1 + k_2) \frac{e^{-(k_1^2+k_2^2+2m^2)\tau}}{(k_1^2 + k_2^2 + 2m^2)} \end{aligned} $$

From Eq.(2.3.9)-Eq.(2.3.11) we get

$$ \lim_{\tau \to \infty} \langle \hat{v}(k_1, \tau, a) \hat{v}(k_2, \tau, a) \rangle_{\eta} = (2\pi)^8 a^2 \delta^4(k_1) \delta^4(k_2) \frac{m^4}{(k_1^2 + m^2)(k_2^2 + m^2)} - \frac{2(2\pi)^4 \delta(k_1 + k_2)}{k_1^2 + k_2^2 + 2m^2}. \qquad (2.3.12) $$

Therefore
$$
\begin{align}
\lim_{\tau \to \infty} \langle \hat{v}(x_1, \tau, a) \hat{v}(x_2, \tau, a) \rangle_{\eta} &= \nonumber \\
& \quad \times a^2 (2\pi)^{-8} \times \int d^4 k_1 e^{ik_1 x_1} \int d^4 k_2 e^{ik_2 x_2} \delta^4(k_1) \delta^4(k_2) \frac{m^4}{(k_1^2 + m^2) \times (k_2^2 + m^2)} \nonumber \\
& \quad - (2\pi)^{-8} \times \int d^4 k_1 e^{ik_1 x_1} \int d^4 k_2 e^{ik_2 x_2} \frac{2(2\pi)^4 \delta(k_1 + k_2)}{k_1^2 + k_2^2 + 2m^2} = \tag{2.3.13} \\
&= \left[ \frac{a^2 m^4}{m^2 \times m^2} + (2\pi)^{-4} \int d^4 k \frac{e^{ik(x_1-x_2)}}{k^2 + m^2} \right] = a^2 - (2\pi)^{-4} \int d^4 k \frac{e^{ik(x_1-x_2)}}{k^2 + m^2} \nonumber
\end{align}
$$

Two point function $G(x_1,x_2)$ of euclidean QFT corresponding to the action (2.3.1) is

$$
G(x_1, x_2) = \lim_{\tau \to \infty} \langle \varphi(x_1, \tau; \omega) \varphi(x_2, \tau; \omega) \rangle_{\eta} \quad (2.3.14)
$$

Master equation corresponding to two-point function $G(x_1, x_2)$ reads

$$
\lim_{\tau \to \infty} \langle \varphi(x_1, \tau; \omega) \varphi(x_2, \tau; \omega) \rangle - a^2 = \lim_{\tau \to \infty} \langle \hat{v}(k_1, \tau, a) \hat{v}(k_2, \tau, a) \rangle_{\eta} = 0. \quad (2.3.15)
$$

From (2.3.15) we get

$$
\lim_{\tau \to \infty} \langle \varphi(x_1, \tau; \omega) \varphi(x_2, \tau; \omega) \rangle_{\eta} = a^2. \quad (2.3.16)
$$

From Eq.(2.3.13)-Eq.(2.3.16) we get

$$
\lim_{\tau \to \infty} \langle \hat{v}(k_1, \tau, a) \hat{v}(k_2, \tau, a') \rangle_{\eta} = 0 \iff a^2 - (2\pi)^{-4} \int d^4 k \frac{e^{ik(x_1-x_2)}}{k^2 + m^2} = 0 \quad (2.3.17)
$$

and therefore

$$
a^2 = (2\pi)^{-4} \int d^4 k \frac{e^{ik(x_1-x_2)}}{k^2 + m^2}. \qquad (2.3.18)
$$

From Eq.(2.3.16) and Eq.(2.3.21) finally we get desired result

$$
\begin{align}
G_F(x_1, x_2) &= \lim_{\tau \to \infty} \langle \varphi(x_1, \tau; \omega) \varphi(x_2, \tau; \omega) \rangle_{\eta} = \nonumber \\
&\quad (2\pi)^{-4} \int d^4 k \frac{e^{ik(x_1-x_2)}}{k^2 + m^2} = (2\pi)^{-2} \left( \frac{m}{|x|} \right) K_1(m|x_1 - x_2|), \tag{2.3.19}
\end{align}
$$

where $K_1$ is the modified Bessel functions of the second kind, integer order 1, and
where we used formula 6.566.2 of [12].
<figure><img src="image_2.png" /><figcaption>Figure 2.3.1. Plot of the modified Bessel functions of the second kind, integer order 1.</figcaption></figure>

## 2.4. Double stochastic quantization of the $\lambda\phi_d^4$ theory.

In this section we consider a neutral scalar field with a $\frac{\lambda}{4!}\phi_d^4$, $d \ge 4$, self-interaction, defined in a $d$-dimensional Minkowski spacetime. The vacuum persistence functional is

the generating functional of all vacuum expectation value of time-ordered products of the

theory. The Euclidean field theory can be obtained by analytic continuation to imaginary

time supported by the positive energy condition for the relativistic field theory. In the Euclidean field theory, we have the Euclidean counterpart for the vacuum persistence

functional, that is, the generating functional of complete Schwinger functions. Actually,

the $(\lambda\phi^4)_d$ Euclidean theory is defined by these Euclidean Green's functions. The Euclidean generating functional $Z[h]$ is formally defined by the following functional integral:

$$Z[h] = \int [d\varphi] \exp(-S_0 - S_I + \int d^dx h(x)\varphi(x)), \qquad (2.4.1)$$

where the action that usually describes a free scalar field is

$$S_0[\varphi] = \int d^dx \left( \frac{1}{2}(\partial\varphi)^2 + \frac{1}{2}m_0^2\varphi^2(x) \right), \qquad (2.4.2)$$

and the interacting part, defined by the non-Gaussian contribution, is

$$S_I[\varphi] = \int d^dx \frac{\lambda}{4!} \varphi^4(x). \qquad (2.4.3)$$

In Eq. (2.4.1), $[d\varphi]$ is a translational invariant measure, formally given by

$[d\varphi] = \prod_{x \in \mathbb{R}^d} d\varphi(x)$. The terms $\lambda$ and $m_0^2$ are respectively the bare coupling constant
and the squared mass of the model. Finally, $h(x)$ is a smooth function that we introduce

to generate the Schwinger functions of the theory by functional derivatives. In the weak-coupling perturbative expansion, which is the conventional procedure, we perform a formal perturbative expansion with respect to the non-Gaussian terms of the action. As a consequence of this formal expansion, all the n-point unrenormalized

Schwinger functions are expressed in a powers series of the bare coupling constant $\lambda$.

The aim of this section is to discuss the double stochastic quantization of a free scalar

field. It can be shown that it is equivalent to the usual path integral quantization. The starting point of the stochastic quantization to obtain the Euclidean field theory is a Markovian Langevin equation. Assume an Euclidean d-dimensional manifold, where we are choosing periodic boundary conditions for a scalar field and also a random noise. In other words, they are defined in a d-torus $\Omega = T^d$. To implement the stochastic quantization we supplement the scalar field $\varphi(x)$ and the random noises $\eta(x)$ and $\hat{\eta}(\tau, x)$ with an extra coordinate $\tau$, the Markov parameter, such that $\varphi(x) \to \varphi(\tau, x)$ and $\eta(x) \to \eta(\tau, x)$.

Therefore, the fields and the random noises $\eta(\tau, x)$ and $\hat{\eta}(\tau, x)$ are defined in a domain:

$T^d \times R^{(+)}$. Let us consider that this dynamical system is out of equilibrium, being described by the following equation of evolution:

$$ \frac{\partial}{\partial \tau} \varphi(\tau, x) = - \left. \frac{\delta S_0[\varphi]}{\delta \varphi(x)} \right|_{\varphi(x)=\varphi(\tau, x)} + \eta(\tau, x) + \epsilon \hat{\eta}(\tau, x), \quad (2.4.4) $$

where $\tau$ is a Markov parameter, $\eta(\tau, x)$ is a random noise field and $S_0$ is the usual free

action defined in Eq.(2.4.2). For a free scalar field, the double stochastic Langevin equation reads

$$ \frac{\partial}{\partial \tau} \varphi(\tau, x) = -(-\Delta + m_0^2) \varphi(\tau, x) + \eta(\tau, x) + \epsilon \hat{\eta}(\tau, x), \quad (2.4.5) $$

where $\Delta$ is the d-dimensional Laplace operator. The Eq.(2.4.5) describes a Ornstein-Uhlenbeck process and we are assuming the Einstein relations, that is:

$$ \begin{aligned} \langle \eta(\tau, x) \rangle_{\eta} &= 0, \\ \langle \hat{\eta}(\tau, x) \rangle_{\hat{\eta}} &= 0, \end{aligned} \quad (2.4.6) $$

and for the two-point correlation function associated with the random noise fields

$$ \begin{aligned} \langle \eta(\tau, x) \eta(\tau', x') \rangle_{\eta} &= 2\delta(\tau - \tau')(x - x'), \\ \langle \hat{\eta}(\tau, x) \hat{\eta}(\tau', x') \rangle_{\hat{\eta}} &= 2\delta(\tau - \tau')(x - x'), \end{aligned} \quad (2.4.7) $$
where $\langle \dots \rangle_\eta$ means stochastic averages. In a generic way, the stochastic average for any functional of $\varphi$ given by $F[\varphi]$ is defined by

$$ \langle F[\varphi] \rangle_{\eta, \hat{\eta}} = \frac{\int D[\eta] D[\hat{\eta}] F[\varphi] \exp\left[-\frac{1}{4} \int d^d x \int d\tau \eta^2(\tau, x, \omega)\right] \exp\left[-\frac{1}{4} \int d^d x \int d\tau \tilde{\eta}^2(\tau, x, \tilde{\omega})\right]}{\left(\int D[\eta] \exp\left[-\frac{1}{4} \int d^d x \int d\tau \eta^2(\tau, x, \omega)\right]\right) \left(\int D[\hat{\eta}] \exp\left[-\frac{1}{4} \int d^d x \int d\tau \tilde{\eta}^2(\tau, x, \tilde{\omega})\right]\right)}. \quad (2.4.8) $$

Let us define the retarded Green function for the diffusion problem that we call $G(\tau - \tau', x - x')$. The retarded Green function satisfies $G(\tau - \tau', x - x') = 0$ if $\tau - \tau' < 0$ and also

$$ \left[ \frac{\partial}{\partial \tau} + (-\Delta_x + m_0^2) \right] G(\tau - \tau', x - x') = \delta^d(x - x') \delta^d(\tau - \tau'). \quad (2.4.9) $$

Using the retarded Green function and the initial condition $\varphi(\tau, x)|_{\tau=0} = 0$, the solution for Eq.(3.5.5) reads

$$ \varphi(\tau, x) = \int_0^\tau d\tau' \int_\Omega d^d x' G(\tau - \tau', x - x') [\eta(\tau', x') + \epsilon \hat{\eta}(\tau', x)]. \quad (2.4.10) $$

In the following we are interested in calculating the quantity $\langle \varphi(\tau, x)\varphi(\tau', x') \rangle_{\eta, \hat{\eta}}$. Using Eq.(2.4.6), Eq.(2.4.7) and Eq.(2.4.10), we have

$$ \langle \varphi(\tau_1, x_1) \varphi(\tau_2, x_2) \rangle_{\eta, \hat{\eta}} = 2 \int_0^{\min(\tau_1, \tau_2)} d\tau' \int_{\Omega} d^d x' G(\tau_1 - \tau', x_1 - x') G(\tau_2 - \tau', x_2 - x'), \quad (2.4.11) $$

where $\min(\tau_1, \tau_2)$ means the minimum of $\tau_1$ and $\tau_2$. Using a Fourier representation, the two-point correlation function $\langle \varphi(\tau, x)\varphi(\tau', x') \rangle_\eta = D(\tau, x; \tau', x')$ is given by

$$ D(\tau, x; \tau', x') = \frac{1}{(2\pi)^d} \int d^d p \frac{e^{-ip(x-x')}}{(p^2 + m_0^2)} e^{-(p^2+m_0^2)(\tau-\tau')}. \quad (2.4.12) $$

It is not difficult to show that Eq.(2.4.12) can be written as:

$$ D(\tau, x; \tau', x') = \frac{1}{(2\pi)^d} \sum_{n=0}^{\infty} \frac{(-1)^n}{\Gamma(1-n)n!} (\tau - \tau')^n \left(\frac{m_0}{r}\right)^{\frac{d}{2}+n-1} K_{\frac{d}{2}+n-1}(m_0 r). \quad (2.4.13) $$

where $r = |x - x'|$ and $K_v$ is the modified Bessel function of order $v$. We can use the Fourier analysis to show that when the Markov parameters $\tau$ and $\tau'$ go to infinity we recover the standard Euclidean free field theory. Therefore let us define the Fourier transforms for the field and the noises given by $\varphi(\tau, k)$ and $\eta(\tau, k)$. We have respectively

$$ \varphi(\tau, k) = \frac{1}{(2\pi)^{\frac{d}{2}}} \int d^d x e^{-ikx} \varphi(\tau, x), \quad (2.4.14) $$

and
$$ 
\begin{aligned}
\hat{\eta}(\tau, k) &= \frac{1}{(2\pi)^{d/2}} \int e^{-ikx} \eta(\tau, x), \\
\hat{\tilde{\eta}}(\tau, k) &= \frac{1}{(2\pi)^{d/2}} \int e^{-ikx} \hat{\eta}(\tau, x)
\end{aligned}
\qquad (2.4.15) $$

Substituting Eq.(2.4.14) in Eq.(2.4.2), the free action for the scalar field in the $(d+1)$-dimensional space writing in terms of the Fourier coefficients reads

$$ S_0[\varphi(k)]|_{\varphi(k)=\varphi(\tau,k)} = \frac{1}{2} \int d^d k \varphi(\tau,k) (k^2 + m_0^2) \varphi(\tau,k). \qquad (2.4.16) $$

Substituting Eq.(2.4.14) and Eq.(2.4.15) in Eq.(2.4.5) we have that each Fourier coefficient satisfies a Langevin equation given by

$$ \frac{\partial}{\partial \tau} \varphi(\tau, k) = -(k^2 + m_0^2)\varphi(\tau, k) + \eta(\tau, k) + \epsilon \hat{\eta}(\tau, k). \qquad (2.4.17) $$

The solution for this equation reads

$$ \varphi(\tau, k) = \exp(-(k^2 + m_0^2)\tau)\varphi(0, k) + \int_0^\tau d\tau' \exp(-(k^2 + m_0^2)(\tau - \tau'))[\eta(\tau', k) + \epsilon\hat{\eta}(\tau', k)]. \qquad (2.4.18) $$

Using the Einstein relation, we get that the Fourier coefficients for the random noise satisfies

$$ \begin{aligned} \langle \eta(\tau, k) \rangle_{\eta} &= 0, \\ \langle \tilde{\eta}(\tau, k) \rangle_{\eta} &= 0 \end{aligned} \qquad (2.4.19) $$

and

$$ \begin{aligned} \langle \eta(\tau, k) \eta(\tau', k') \rangle_{\eta} &= 2(2\pi)^d \delta^d(\tau - \tau')(k + k'), \\ \langle \hat{\eta}(\tau, k) \hat{\eta}(\tau', k') \rangle_{\eta} &= 2(2\pi)^d \delta^d(\tau - \tau')(k + k') \end{aligned} \qquad (2.4.20) $$

Before investigate the interacting field theory, let us calculate the Fourier representation for the two-point correlation function, i.e., $\langle \varphi(\tau, k) \varphi(\tau', k') \rangle_{\eta}$. Using Eq.(2.4.18), we obtain three contributions to the scalar two-point correlation function. The first one is given by

$$ \exp(-(k^2 + m_0^2)\tau + (k'^2 + m_0^2)\tau')\varphi(0, k)\varphi(0, k'), \qquad (2.4.21) $$

and decay to zero at long time. Let us assume that $\varphi(\tau, k)|_{\tau=0} = 0$. There are also two crossed terms, each first order in the noise Fourier component given by

$$ 2 \varphi(0, k) \exp(-(k^2 + m_0^2)\tau) \int_0^{\tau'} ds \exp(-(k'^2 + m_0^2)(\tau' - s)) [\eta(s, k') + \epsilon \hat{\eta}(\sigma, k')] \qquad (2.4.22) $$

Since we are assuming the Einstein relations, i.e., $\langle \eta(\tau, x) \rangle_{\eta} = 0, \langle \tilde{\eta}(\tau, x) \rangle_{\eta} = 0$ on averaging on noise, these cross terms vanish. The final term is second-order in the noise Fourier component. Again, the solution subject to the initial condition $\varphi(\tau, k)|_{\tau=0} = 0$ can be used to give
$$ \left\{ \int_{0}^{\tau} ds \exp(-(k^2 + m_0^2)(\tau - s)) \left[ \eta(s, k) + \epsilon \hat{\eta}(s, k) \right] \right\} \times \left\{ \int_{0}^{\tau'} d\sigma \exp(-(k'^2 + m_0^2)(\tau' - \sigma)) \left[ \eta(\sigma, k') + \epsilon \hat{\eta}(\sigma, k') \right] \right\}. \quad (2.4.23) $$

Again averaging on noises and using the Einstein relation given by Eq.(2.4.20) we have that this term becomes

$$ 2\delta^d(k + k'_0) \int_0^{\min(\tau, \tau')} ds \exp(-(k^2 + m_0)(\tau + \tau' - 2s)). \quad (2.4.24) $$

Assuming that $\tau = \tau'$ and using $\langle \varphi(\tau, k)\varphi(\tau', k') \rangle_{\eta}|_{\tau=\tau'} \equiv D(k, k'; \tau, \tau')$ we have

$$ D(k; \tau, \tau) = (2\pi)^d \delta^d(k + k') \frac{1}{(k^2 + m_0^2)} (1 - \exp(-2\tau(k^2 + m_0^2))). \quad (2.4.25) $$

In the following, we are redefining the two-point correlation function as $D(k; \tau, \tau) \rightarrow (2\pi)^d D(k; \tau, \tau)$. In the limit when $\tau \to \infty$ we recover the standard two-point function of the Euclidean free field theory. Before going to the next section, we would

like to mention the existence of more general Markovian Langevin equations. We can

introduce a kernel defined in the d-torus. The kerneled Langevin equation reads:

$$ \frac{\partial}{\partial \tau} \varphi(\tau, x) = - \int d^d y K(x, y) \frac{\delta S_0}{\delta \varphi(y)} |_{\varphi(y)=\varphi(\tau, y)} + \eta(\tau, x) + \epsilon \hat{\eta}(\tau, x). \quad (2.4.26) $$

The second moment of the noise fields will be modified to:

$$ \begin{aligned} \langle \eta(\tau, x) \eta(\tau', x') \rangle_{\eta} &= 2\delta(\tau - \tau') K(x, x'), \\ \langle \tilde{\eta}(\tau, x) \tilde{\eta}(\tau', x') \rangle_{\eta} &= 2\delta(\tau - \tau') K(x, x'). \end{aligned} \quad (2.4.27) $$

Choosing an appropriate kernel, it can be shown that all the above conclusions remain unchanged.

The double stochastic Langevin equation reads

$$ \frac{\partial}{\partial \tau} \varphi_{\epsilon}(\tau, x) = (\Delta - m_0^2) \varphi_{\epsilon}(\tau, x) - \frac{\lambda}{3!} \varphi_{\epsilon}^3(\tau, x) + \eta(\tau, x) + \epsilon \tilde{\eta}(\tau, x). \quad (2.4.28) $$

By the replacements

$$ \begin{aligned} \varphi_{\epsilon}(x, \tau; \omega, \bar{\omega}) &= v_{\epsilon-}(x, \tau; \omega, \bar{\omega}) + \theta(\tau)a, \\ \varphi_{\epsilon}(x, \tau; \omega, \bar{\omega}) &= v_{\epsilon+}(x, \tau; \omega, \bar{\omega}) - \theta(\tau)a, \\ \theta(\tau) &= \begin{cases} 0 & \text{if } \tau \le 1 \\ 1 & \text{if } \tau > 1 \end{cases} \end{aligned} \quad (2.4.29) $$

we obtain from Eqs.(2.4.28)
$$
\begin{align*}
\frac{\partial[v_{\epsilon-}(x, \tau; \omega, \omega) + \theta(\tau)a]}{\partial\tau} &= \frac{\partial v_{\epsilon-}(x, \tau; \omega, \omega)}{\partial\tau} + a\delta(\tau) = \\
&= (\partial^2 - m^2)[v_{\epsilon-}(x, \tau; \omega, \omega) + \theta(\tau)a] - \frac{\lambda}{3!}[v_{\epsilon-}(x, \tau; \omega, \omega) + a]^3 + \\
&\quad + \eta(x, \tau; \omega) + \epsilon\tilde{\eta}(x, \tau; \omega) = \\
&\quad (\partial^2 - m^2)[v_{\epsilon-}(x, \tau; \omega, \omega) + \theta(\tau)a] - \frac{\lambda}{3!}(v_{\epsilon-}^3 + 3av_{\epsilon-}^2 + 3a^2v_{\epsilon-} + a^3) + \\
&\quad + \eta(x, \tau; \omega) + \epsilon\tilde{\eta}(x, \tau; \omega) \tag{2.4.30}
\end{align*}
$$

and

$$
\begin{align*}
\frac{\partial[v_{\epsilon+}(x, \tau; \omega, \omega) - \theta(\tau)a]}{\partial\tau} &= \frac{\partial v_{\epsilon+}(x, \tau; \omega, \omega)}{\partial\tau} - a\delta(\tau) = \\
&\quad (\partial^2 - m^2)[v_{\epsilon+}(x, \tau; \omega, \omega) - \theta(\tau)a] - \frac{\lambda}{3!}(v_{\epsilon+}^3 - 3av_{\epsilon+}^2 + 3a^2v_{\epsilon+} - a^3) + \\
&\quad + \eta(x, \tau; \omega) + \epsilon\tilde{\eta}(x, \tau; \omega)
\end{align*}
$$

Differential master equations corresponding to double stochastic Langevin equations (2.4.30) reads

$$
\begin{align}
\frac{\partial v_{-}(x, \tau; \omega)}{\partial \tau} &= -a\delta(\tau) + [\partial^2 - (m^2 + 0.5\lambda a^2)]v_{-}(x, \tau; \omega) - \nonumber \\
&\quad - \frac{\lambda}{6}a^3 - m^2a + \eta(x, \tau; \omega) = \nonumber \\
&\quad -a\delta(\tau) + [\partial^2 - m_1^2]v_{-}(x, \tau; \omega) - \left(\frac{\lambda}{6}a^3 + m^2a\right) + \eta(x, \tau; \omega) \nonumber \\
&\text{and} \nonumber \\
\frac{\partial v_{+}(x, \tau; \omega)}{\partial \tau} &= a\delta(\tau) + [\partial^2 - (m^2 + 0.5\lambda a^2)]v_{+}(x, \tau; \omega) + \nonumber \\
&\quad + \frac{\lambda}{6}a^3 + m^2a + \eta(x, \tau; \omega) = \nonumber \\
&\quad a\delta(\tau) + [\partial^2 - m_1^2]v_{+}(x, \tau; \omega) + \left(\frac{\lambda}{6}a^3 + m^2a\right) + \eta(x, \tau; \omega) \nonumber \\
&\qquad m_1^2 = m^2 + 0.5\lambda a^2. \tag{2.4.31}
\end{align}
$$

Consider the Fourier transformed stochastic differential equation (2.4.31) in $k$ and $\tau$ given as
$$
\begin{align}
\frac{\partial}{\partial \tau} \hat{v}_{-}(k, \tau) &= -(k^2 + m_1^2) \hat{v}_{-}(k, \tau) - (2\pi)^4 a \delta(\tau) \delta^4(k) - (2\pi)^4 \left(m^2 a + \frac{\lambda}{6} a^3\right) \delta^4(k) + \hat{\eta}(k, \tau; \omega) \nonumber \\
&\quad \text{and} \tag{2.4.32} \\
\frac{\partial}{\partial \tau} \hat{v}_{+}(k, \tau) &= -(k^2 + m_1^2) \hat{v}_{+}(k, \tau) + (2\pi)^4 a \delta(\tau) \delta^4(k) + (2\pi)^4 \left(m^2 a + \frac{\lambda}{6} a^3\right) \delta^4(k) + \hat{\eta}(k, \tau; \omega) \nonumber
\end{align}
$$

Let us consider ODE

$$
\dot{x}(\tau, \lambda) + \lambda x(\tau, \lambda) = g(\tau, \lambda), x(0) = 0. \quad (2.4.33)
$$

The corresponding solution $x(t, \lambda)$ reads

$$
x(\tau, \lambda) = e^{-\lambda \tau} \int_{0}^{\tau} e^{\lambda \tau_1} g(\tau_1, \lambda) d\tau_1. \quad (2.4.34)
$$

From Eq.(2.4.32)-Eq.(2.4.34) one obtains
$$
\begin{align*}
\hat{v}_-(k, \tau, a) &= e^{-(k^2+m_1^2)\tau} \times \\
&\quad \int_0^\tau e^{(k^2+m_1^2)\tau_1} \Biggl[ -(2\pi)^4 a\delta(\tau_1)\delta^4(k) - (2\pi)^4 \left(m^2a + \frac{\lambda}{6}a^3\right)\delta^4(k) + \hat{\eta}(k, \tau_1; \omega) \Biggr] d\tau_1 = \\
&\quad -(2\pi)^4 a\delta^4(k) e^{-(k^2+m_1^2)\tau} + (2\pi)^4 \left(m^2a + \frac{\lambda}{6}a^3\right)\delta^4(k) e^{-(k^2+m_1^2)\tau} \int_0^\tau e^{(k^2+m_1^2)\tau_1} d\tau_1 + \\
&\quad + e^{-(k^2+m_1^2)\tau} \int_0^\tau e^{(k^2+m_1^2)\tau_1} \hat{\eta}(k, \tau_1; \omega) d\tau_1 = \\
&\quad -(2\pi)^4 a\delta^4(k) e^{-(k^2+m_1^2)\tau} - (2\pi)^4 a \left(m^2 + \frac{\lambda}{6}a^2\right)\delta^4(k) e^{-(k^2+m_1^2)\tau} \int_0^\tau e^{(k^2+m_1^2)\tau_1} d\tau_1 + \\
&\quad + \int_0^\tau e^{-(k^2+m_1^2)(\tau-\tau_1)} \hat{\eta}(k, \tau_1; \omega) d\tau_1 = \\
&\quad -(2\pi)^4 a\delta^4(k) e^{-(k^2+m_1^2)\tau} - (2\pi)^4 a \left(m^2 + \frac{\lambda}{6}a^2\right)\delta^4(k) e^{-(k^2+m_1^2)\tau} \left[ \frac{e^{(k^2+m_1^2)\tau}}{k^2+m_1^2} - \frac{1}{k^2+m_1^2} \right] + \\
&\quad + \int_0^\tau e^{-(k^2+m_1^2)(\tau-\tau_1)} \hat{\eta}(k, \tau_1; \omega) d\tau_1 = \\
&\quad -(2\pi)^4 a\delta^4(k) e^{-(k^2+m_1^2)\tau} - (2\pi)^4 \frac{\left(m^2 + \frac{\lambda}{6}a^2\right)\delta^4(k)}{k^2+m_1^2} \left[ 1 - e^{-(k^2+m_1^2)\tau} \right] + \\
&\quad + \int_0^\tau e^{-(k^2+m_1^2)(\tau-\tau_1)} \hat{\eta}(k, \tau_1; \omega) d\tau_1 = \\
&\quad -(2\pi)^4 a\delta^4(k) \left[ e^{-(k^2+m_1^2)\tau} + \frac{\left(m^2 + \frac{\lambda}{6}a^2\right)}{k^2+m_1^2} \left[ 1 - e^{-(k^2+m_1^2)\tau} \right] \right] + \\
&\quad + \int_0^\tau e^{-(k^2+m_1^2)(\tau-\tau_1)} \hat{\eta}(k, \tau_1; \omega) d\tau_1,
\end{align*}
\tag{2.4.35}
$$

and
$$
\begin{align*}
& \hat{v}_+(k, \tau, a) = e^{-(k^2+m_1^2)\tau} \times \\
& \quad \int_0^\tau e^{(k^2+m_1^2)\tau_1} \left[ (2\pi)^4 a \delta(\tau_1) \delta^4(k) - (2\pi)^4 \left(-m^2a - \frac{\lambda}{6}a^3\right) \delta^4(k) \pm \hat{\eta}(k, \tau_1; \omega) \right] d\tau_1 = \\
& \quad (2\pi)^4 a \delta^4(k) e^{-(k^2+m_1^2)\tau} - (2\pi)^4 \left(-m^2a - \frac{\lambda}{6}a^3\right) \delta^4(k) e^{-(k^2+m_1^2)\tau} \int_0^\tau e^{(k^2+m_1^2)\tau_1} d\tau_1 + \\
& \quad + e^{-(k^2+m_1^2)\tau} \int_0^\tau e^{(k^2+m_1^2)\tau_1} \hat{\eta}(k, \tau_1; \omega) d\tau_1 = \\
& \quad + (2\pi)^4 a \delta^4(k) e^{-(k^2+m_1^2)\tau} + (2\pi)^4 a \left(m^2 + \frac{\lambda}{6}a^3\right) \delta^4(k) e^{-(k^2+m_1^2)\tau} \int_0^\tau e^{(k^2+m_1^2)\tau_1} d\tau_1 + \\
& \quad + \int_0^\tau e^{-(k^2+m_1^2)(\tau-\tau_1)} \hat{\eta}(k, \tau_1; \omega) d\tau_1 = \\
& \quad + (2\pi)^4 a \delta^4(k) e^{-(k^2+m_1^2)\tau} + \left[ (2\pi)^4 a \left(m^2 + \frac{\lambda}{6}a^3\right) \delta^4(k) e^{-(k^2+m_1^2)\tau} - \frac{1}{k^2+m_1^2} \right] + \\
& \quad + \int_0^\tau e^{-(k^2+m_1^2)(\tau-\tau_1)} \hat{\eta}(k, \tau_1; \omega) d\tau_1 = \\
& \quad + (2\pi)^4 a \delta^4(k) e^{-(k^2+m_1^2)\tau} + (2\pi)^4 \frac{\left(m^2 + \frac{\lambda}{6}a^3\right) \delta^4(k)}{k^2+m_1^2} \left[ 1 - e^{-(k^2+m_1^2)\tau} \right] + \\
& \quad + \int_0^\tau e^{-(k^2+m_1^2)(\tau-\tau_1)} \hat{\eta}(k, \tau_1; \omega) d\tau_1 = \\
& \quad + (2\pi)^4 a \delta^4(k) \left[ e^{-(k^2+m_1^2)\tau} + \frac{\left(m^2 + \frac{\lambda}{6}a^3\right)}{k^2+m_1^2} \left[ 1 - e^{-(k^2+m_1^2)\tau} \right] \right] + \\
& \quad + \int_0^\tau e^{-(k^2+m_1^2)(\tau-\tau_1)} \hat{\eta}(k, \tau_1; \omega) d\tau_1.
\end{align*}
\tag{2.4.35'}
$$

From Eq.(2.4.35)-Eq.(2.4.35') one obtains
$$
\begin{equation}
\begin{split}
\hat{v}_-(k_1, \tau, a) \hat{v}_+(k_2, \tau', a') = & \\
& \left\{ -(2\pi)^4 a \delta^4(k_1) \left[ e^{-(k_1^2+m_1^2)\tau} + \frac{m^2+\frac{\lambda}{6}a^2}{k_1^2+m_1^2} [1-e^{-(k_1^2+m_1^2)\tau}] \right] + \right. \\
& \qquad \left. + \int_0^\tau e^{-(k_1^2+m_1^2)(\tau-\tau_1)} \hat{\eta}(k_1, \tau_1; \omega) d\tau_1 \right\} \times \\
& \left\{ (2\pi)^4 a' \delta^4(k_2) \left[ e^{-(k_2^2+m_2^2)\tau'} + \frac{m^2+\frac{\lambda}{6}a'^2}{k_2^2+m_2^2} [1-e^{-(k_2^2+m_2^2)\tau'}] \right] + \right. \\
& \qquad \left. + \int_0^\tau e^{-(k_2^2+m_2^2)(\tau'-\tau_1)} \hat{\eta}(k_2, \tau_1; \omega) d\tau_1 \right\}
\end{split}
\tag{2.4.36}
\end{equation}
$$

From Eq.(2.4.36) one obtains

$$
\begin{equation}
\begin{split}
\langle \hat{v}_-(k_1, \tau, a) \hat{v}_+(k_2, \tau', a') \rangle_\eta = & \\
& \left\{ -(2\pi)^4 a \delta^4(k_1) \left[ e^{-(k_1^2+m^2)\tau} + \frac{m^2+\frac{\lambda}{6}a^2}{k_1^2+m_1^2} [1-e^{-(k_1^2+m_1^2)\tau}] \right] \right\} \times \\
& \left\{ (2\pi)^4 a' \delta^4(k_2) \left[ e^{-(k_2^2+m^2)\tau'} + \frac{m^2+\frac{\lambda}{6}a'^2}{k_2^2+m_2^2} [1-e^{-(k_2^2+m_2^2)\tau'}] \right] \right\} \\
& - \left\langle \left( \int_0^\tau e^{-(k_1^2+m_1^2)(\tau-\tau_1)} \hat{\eta}(k_1, \tau_1; \omega) d\tau_1 \right) \left( \int_0^{\tau'} e^{-(k_2^2+m_2^2)(\tau'-\tau_2)} \hat{\eta}(k_2, \tau_2; \omega) d\tau_2 \right) \right\rangle_\eta \\
& - \left\{ -(2\pi)^8 a a' \delta^4(k_1) \delta^4(k_2) \left[ e^{-(k_1^2+m_1^2)\tau} + \frac{m^2+\frac{\lambda}{6}a^2}{k_1^2+m_1^2} [1-e^{-(k_1^2+m_1^2)\tau}] \right] \right. \\
& \qquad \left. + \frac{m^2+\frac{\lambda}{6}a'^2}{k_2^2+m_2^2} [1-e^{-(k_2^2+m_2^2)\tau'}] \right. \\
& \qquad \left. - \int_0^{\tau'} e^{-(k_2^2+m_2^2)(\tau'-\tau_2)} \int_0^\tau e^{-(k_1^2+m_1^2)(\tau-\tau_1)} \langle \hat{\eta}(k_1, \tau_1; \omega) \hat{\eta}(k_2, \tau_2; \omega) \rangle_\eta d\tau_1 d\tau_2 \right\}
\end{split}
\tag{2.4.37}
\end{equation}
$$

Note that
$$
-(2\pi)^8 aa' \delta^4(k_1) \delta^4(k_2) \left[ e^{-(k_1^2+m^2)\tau} + \frac{m^2 + \frac{\lambda}{6}a^2}{k_1^2 + m_1^2} [1 - e^{-(k_1^2+m_1^2)\tau}] \right] \times \\
\left[ e^{-(k_2^2+m^2)\tau'} + \frac{m^2 + \frac{\lambda}{6}a'^2}{k_2^2 + m_1^2} [1 - e^{-(k_1^2+m_1^2)\tau'}] \right] \Bigg|_{\tau'=\tau, a'=a} = \\
-(2\pi)^8 aa' \delta^4(k_1) \delta^4(k_2) \left[ e^{-(k_1^2+m_1^2)\tau} + \frac{m^2 + \frac{\lambda}{6}a^2}{k_1^2 + m_1^2} [1 - e^{-(k_1^2+m_1^2)\tau}] \right] \times \\
\times \left[ e^{-(k_1^2+m^2)\tau} + \frac{m^2 + \frac{\lambda}{6}a'^2}{k_2^2 + m_1^2} [1 - e^{-(k_1^2+m_1^2)\tau}] \right]
\qquad (2.4.38)
$$

and note that

$$
\begin{align*}
& \int_0^{\tau'} e^{-(k_2^2+m^2)(\tau'-\tau_2)} \int_0^{\tau} e^{-(k_1^2+m^2)(\tau-\tau_1)} \langle \hat{\eta}(k_1, \tau_1; \omega) \hat{\eta}(k_2, \tau_2; \omega) \rangle_{\eta} d\tau_1 d\tau_2 \Bigg|_{\tau'=\tau, a'=a} = \\
& \qquad \delta(k_1+k_2) \int_0^{\tau} e^{-(k_2^2+m_1^2)(\tau-\tau_2)} \int_0^{\tau} e^{-(k_1^2+m_1^2)(\tau-\tau_1)} \delta(\tau_1-\tau_2) d\tau_1 d\tau_2 = \\
& \qquad 2(2\pi)^4 \delta(k_1+k_2) \int_0^{\tau} e^{-(k_1^2+k_2^2+2m_1^2)(\tau-\tau_1)} d\tau_1 = \tag{2.4.39}
\end{align*}
$$

$$
\begin{align*}
& 2(2\pi)^4 \delta(k_1 + k_2) e^{-(k_1^2+k_2^2+2m_1^2)\tau} \left[ \frac{1}{(k_1^2 + k_2^2 + 2m_1^2)} e^{(k_1^2+k_2^2+m_1^2)\tau} - \frac{1}{(k_1^2 + k_2^2 + 2m_1^2)} \right] = \\
&= \frac{2(2\pi)^4 \delta(k_1 + k_2)}{k_1^2 + k_2^2 + 2m_1^2} - 2(2\pi)^4 \delta(k_1 + k_2) \frac{e^{-(k_1^2+k_2^2+2m_1^2)\tau}}{(k_1^2 + k_2^2 + 2m_1^2)}
\end{align*}
$$

From Eq.(2.4.38)-Eq.(2.4.39) we get

$$
\begin{equation}
\begin{split}
\lim_{\tau \to \infty} \langle \hat{v}_-(k_1, \tau, a) \hat{v}_+(k_2, \tau, a) \rangle_{\eta} ={}& (2\pi)^8 a^2 \delta^4(k_1) \delta^4(k_2) \frac{\left(m^2 + \frac{\lambda}{6}a^2\right)^2}{(k_1^2 + m_1^2)(k_2^2 + m_1^2)} \\
& - \frac{2(2\pi)^4 \delta(k_1 + k_2)}{k_1^2 + k_2^2 + 2m_1^2} \\
& m_1^2 = m^2 + 0.5\lambda a^2
\end{split}
\tag{2.4.40}
\end{equation}
$$

Therefore
$$
\lim_{\tau \to \infty} \langle \hat{v}_-(x_1, \tau, a) \hat{v}_+(x_2, \tau, a) \rangle_{\eta} = \\
-a^2 (2\pi)^{-8} \times \int d^4 k_1 e^{ik_1 x_1} \int d^4 k_2 e^{ik_2 x_2} \delta^4(k_1) \delta^4(k_2) \frac{\left(m^2 + \frac{\lambda}{6} a^2\right)^2}{(k_1^2 + m_1^2) \times (k_2^2 + m_2^2)} +
$$

$$
(2\pi)^{-8} \times \int d^4 k_1 e^{ik_1 x_1} \int d^4 k_2 e^{ik_2 x_2} \frac{2(2\pi)^4 \delta(k_1 + k_2)}{k_1^2 + k_2^2 + 2m_1^2} = \\
= \left[ \frac{a^2 \left(m^2 + \frac{\lambda}{6} a^2\right)^2}{m_1^2 \times m_1^2} - (2\pi)^{-4} \int d^4 k \frac{e^{ik(x_1-x_2)}}{k^2 + m_1^2} \right] = \qquad (2.4.41)
$$

$$
\frac{a^2 \left(m^2 + \frac{\lambda}{6} a^2\right)^2}{m_1^2 \times m_1^2} - (2\pi)^{-4} \int d^4 k \frac{e^{ik(x_1-x_2)}}{k^2 + m_1^2}
$$

$m_1^2 = m^2 + 0.5\lambda a^2$

Transcendental master equation corresponding to two-point Eucledian Green
function

$G_{(4)}(x_1, x_2, m, \lambda)$ in Euclidean space $E_4$ with $\dim E_4 = 4$ reads

$$
\frac{[a(x_1 - x_2)^2] (m^2 + \frac{\lambda}{6} a(x_1 - x_2)^2)^2}{(m^2 + 0.5\lambda a(x_1 - x_2)^2)^2} - (2\pi)^{-4} \int d^4 k \frac{\theta_{\delta}(|x_1 - x_2|) e^{ik(x_1 - x_2)}}{k^2 + m^2 + 0.5\lambda a(x_1 - x_2)^2} = 0. \quad (2.4.41)
$$

Transcendental master equation corresponding to two-point Eucledian Green
function

$G_{(D)}(x_1, x_2, m, \lambda)$ in Euclidean space $E_D$ with $\dim E_D = D$ reads

$$
\frac{[a(x_1 - x_2)^2] (m^2 + \frac{\lambda}{6} a(x_1 - x_2)^2)^2}{(m^2 + 0.5\lambda a(x_1 - x_2)^2)^2} - (2\pi)^{-D} w \lim_{\eta \to 0^+} \int d^D k \frac{e^{-\eta k + ik(x_1 - x_2)}}{k^2 + m^2 + 0.5\lambda a(x_1 - x_2)^2} = 0. \quad (2.4.42)
$$
$$
\eta > 0,
$$

where weak limit taken in $\mathcal{L}'(\mathbb{R}_x^D)$, see Appendix 1.

Note that
$$ 
\begin{aligned}
& \int d^4 k \frac{e^{ik(x_1-x_2)}}{[k^2 + m^2 + 0.5\lambda a(x_1 - x_2)^2]^2} = \\
& (4\pi)^{-2} \left( \frac{m_1}{|x_1 - x_2|} \right) K_0(m_1|x_1 - x_2|) \\
& m_1 = \sqrt{m^2 + 0.5\lambda a(x_1 - x_2)^2},
\end{aligned} 
\quad (2.4.43) 
$$

$$ K_0(m_1|x_1 - x_2|) \propto -\ln\left(\frac{m_1|x_1 - x_2|}{2}\right) - \gamma \quad \text{as} \quad m_1|x_1 - x_2| \to 0 $$

see Appendix 1. In order to derive Eq.(2.4.43) we applied Eq.(2.4.44) (see formula 13.6 (2) from [13]).

$$ 
\begin{aligned}
\int_0^\infty \frac{y^{\nu+1} J_\nu(by) dy}{(y^2 + p^2)^{\mu+1}} &= \frac{b^\mu p^{\nu-\mu}}{2^\mu \Gamma(\mu+1)} K_{\nu-\mu}(bp) \\
&\quad -1 < \text{Re}\nu < 2\text{Re}\mu + 1.5.
\end{aligned} 
\quad (2.4.44) 
$$

## 2.5. Double stochastic quantization of the $\lambda\phi_4^6$ theory.

In this section we consider a neutral scalar field with a $\frac{\lambda}{6!}\phi_d^6$, $d \ge 4$, self-interaction, defined in a 4-dimensional Minkowski spacetime. It well known that all these theories is

nonrenormalizable [14]. The vacuum persistence functional is the generating functional

of all vacuum expectation value of time-ordered products of the theory. Thus we deal

now with simple *nonrenormalizable* theory with the Lagrangian

$$ \mathcal{L} = S_0[\varphi] + S_I[\varphi], \quad (2.5.1) $$

where the action that usually describes a free scalar field is

$$ S_0[\varphi] = \int d^4 x \left( \frac{1}{2} (\partial_\mu \varphi \partial^\mu \varphi) + \frac{1}{2} m_0^2 \varphi^2(x) \right), \quad (2.5.2) $$

and the interacting part, defined by the non-Gaussian contribution, is

$$ S_I[\varphi] = \int d^4 x \frac{\lambda}{6!} \varphi^6(x). \quad (2.5.3) $$

The double stochastic Langevin equation reads

$$ \frac{\partial}{\partial \tau} \varphi_{\epsilon}(\tau, x) = (\Delta - m_0^2) \varphi_{\epsilon}(\tau, x) - \frac{\lambda}{5!} \varphi_{\epsilon}^5(\tau, x) + \eta(\tau, x) + \epsilon \bar{\eta}(\tau, x). \quad (2.5.4) $$

By the replacements
$$
\begin{align}
\varphi_{\epsilon}(x, \tau; \omega, \bar{\omega}) &= v_{\epsilon-}(x, \tau; \omega, \bar{\omega}) + \theta(\tau)a, \nonumber \\
\varphi_{\epsilon}(x, \tau; \omega, \bar{\omega}) &= v_{\epsilon+}(x, \tau; \omega, \bar{\omega}) - \theta(\tau)a, \nonumber \\
\theta(\tau) &= \begin{cases} 0 & \text{if } \tau \le 1 \\ 1 & \text{if } \tau > 1 \end{cases} \tag{2.5.5}
\end{align}
$$

we obtain from Eq.(2.5.4)

$$
\begin{align}
\frac{\partial[v_{\epsilon-}(x, \tau; \omega, \bar{\omega}) + \theta(\tau)a]}{\partial \tau} &= \frac{\partial v_{\epsilon-}(x, \tau; \omega, \bar{\omega})}{\partial \tau} + a\delta(\tau) = \nonumber \\
&= (\partial^2 - m^2)[v_{\epsilon-}(x, \tau; \omega, \bar{\omega}) + \theta(\tau)a] - \frac{\lambda}{5!}[v_{\epsilon-}(x, \tau; \omega, \bar{\omega}) + a]^5 + \nonumber \\
&\quad + \eta(x, \tau; \omega) + \epsilon\bar{\eta}(x, \tau; \omega) = \nonumber \\
&= (\partial^2 - m^2)[v_{\epsilon-}(x, \tau; \omega, \bar{\omega}) + \theta(\tau)a] - \nonumber \\
&\quad - \frac{\lambda}{5!}(a^5 + 5a^4v_{\epsilon-} + 10a^3v_{\epsilon-}^2 + 10a^2v_{\epsilon-}^3 + 5av_{\epsilon-}^4 + v_{\epsilon-}^5) + \nonumber \\
&\quad + \eta(x, \tau; \omega) + \epsilon\bar{\eta}(x, \tau; \omega) \tag{2.5.6}
\end{align}
$$

and

$$
\begin{align*}
\frac{\partial[v_{\epsilon+}(x, \tau; \omega, \bar{\omega}) - \theta(\tau)a]}{\partial \tau} &= \frac{\partial v_{\epsilon+}(x, \tau; \omega, \bar{\omega})}{\partial \tau} - a\delta(\tau) = \\
&\quad (\partial^2 - m^2)[v_{\epsilon+}(x, \tau; \omega, \bar{\omega}) - \theta(\tau)a] - \\
&\quad - \frac{\lambda}{5!}(-a^5 + 5a^4v_{\epsilon+} - 10a^3v_{\epsilon+}^2 + 10a^2v_{\epsilon+}^3 - 5av_{\epsilon+}^4 + v_{\epsilon+}^5) + \\
&\quad + \eta(x, \tau; \omega) + \epsilon\bar{\eta}(x, \tau; \omega)
\end{align*}
$$

Differential master equations corresponding to double stochastic Langevin
equations (2.5.6) reads

$$
\begin{align}
\frac{\partial v_{-}(x, \tau; \omega)}{\partial \tau} &= -a\delta(\tau) + \left[ \partial^2 - \left(m^2 + \frac{\lambda}{4!}a^4\right) \right] v_{-}(x, \tau; \omega) - \nonumber \\
&\quad -\frac{\lambda}{5!}a^5 - m^2a + \eta(x, \tau; \omega) = \nonumber \\
&-a\delta(\tau) + \left[ \partial^2 - m_1^2 \right] v_{-}(x, \tau; \omega) - \left(\frac{\lambda}{5!}a^5 + m^2a\right) + \eta(x, \tau; \omega) \nonumber \\
&\text{and} \nonumber \\
\frac{\partial v_{+}(x, \tau; \omega)}{\partial \tau} &= a\delta(\tau) + \left[ \partial^2 - \left(m^2 + \frac{\lambda}{5!}a^5\right) \right] v_{+}(x, \tau; \omega) + \nonumber \\
&\quad +\frac{\lambda}{5!}a^5 + m^2a + \eta(x, \tau; \omega) = \nonumber \\
&+a\delta(\tau) + \left[ \partial^2 - m_1^2 \right] v_{+}(x, \tau; \omega) + \left(\frac{\lambda}{5!}a^5 + m^2a\right) + \eta(x, \tau; \omega) \nonumber \\
&m_1^2 = m^2 + \frac{\lambda}{4!}a^4. \tag{2.5.7}
\end{align}
$$

Consider the Fourier transformed stochastic differential equation (2.5.7) in *k* given
as

$$
\begin{align}
\frac{\partial}{\partial \tau} \hat{v}_{-}(k, \tau) = & \nonumber \\
& - (k^2 + m_1^2) \hat{v}_{-}(k, \tau) - (2\pi)^4 a \delta(\tau) \delta^4(k) - (2\pi)^4 \left(m^2 a + \frac{\lambda}{5!} a^5\right) \delta^4(k) + \hat{\eta}(k, \tau; \omega) \nonumber \\
& \text{and} \tag{2.4.32} \\
\frac{\partial}{\partial \tau} \hat{v}_{+}(k, \tau) = & \nonumber \\
& - (k^2 + m_1^2) \hat{v}_{-}(k, \tau) + (2\pi)^4 a \delta(\tau) \delta^4(k) + (2\pi)^4 \left(m^2 a + \frac{\lambda}{5!} a^5\right) \delta^4(k) + \hat{\eta}(k, \tau; \omega) \nonumber
\end{align}
$$

Let us consider ODE

$$
\dot{x}(\tau, \lambda) + \lambda x(\tau, \lambda) = g(\tau, \lambda), x(0) = 0. \quad (2.5.8)
$$

The corresponding solution $x(t, \lambda)$ reads

$$
x(\tau, \lambda) = e^{-\lambda \tau} \int_{0}^{\tau} e^{\lambda \tau_1} g(\tau_1, \lambda) d\tau_1. \quad (2.5.9)
$$

From Eq.(2.4.32)-Eq.(2.4.34) one obtains
$$
\begin{align*}
\hat{v}_-(k, \tau, a) &= e^{-(k^2+m_1^2)\tau} \times \\
&\quad \int_0^\tau e^{(k^2+m_1^2)\tau_1} \left[ -(2\pi)^4 a\delta(\tau_1)\delta^4(k) - (2\pi)^4 \left(m^2a + \frac{\lambda}{5!}a^5\right)\delta^4(k) + \hat{\eta}(k, \tau_1; \omega) \right] d\tau_1 = \\
&\quad -(2\pi)^4 a\delta^4(k) e^{-(k^2+m_1^2)\tau} + (2\pi)^4 \left(m^2a + \frac{\lambda}{5!}a^5\right) \delta^4(k) e^{-(k^2+m_1^2)\tau} \int_0^\tau e^{(k^2+m_1^2)\tau_1} d\tau_1 + \\
&\quad + e^{-(k^2+m_1^2)\tau} \int_0^\tau e^{(k^2+m_1^2)\tau_1} \hat{\eta}(k, \tau_1; \omega) d\tau_1 = \\
&\quad -(2\pi)^4 a\delta^4(k) e^{-(k^2+m_1^2)\tau} - (2\pi)^4 a \left(m^2 + \frac{\lambda}{5!}a^4\right) \delta^4(k) e^{-(k^2+m_1^2)\tau} \int_0^\tau e^{(k^2+m_1^2)\tau_1} d\tau_1 + \\
&\quad + \int_0^\tau e^{-(k^2+m_1^2)(\tau-\tau_1)} \hat{\eta}(k, \tau_1; \omega) d\tau_1 = \\
&\quad -(2\pi)^4 a\delta^4(k) e^{-(k^2+m_1^2)\tau} - (2\pi)^4 a \left(m^2 + \frac{\lambda}{5!}a^4\right) \delta^4(k) e^{-(k^2+m_1^2)\tau} \left[ \frac{e^{(k^2+m^2)\tau}}{k^2+m_1^2} - \frac{1}{k^2+m_1^2} \right] + \\
&\quad + \int_0^\tau e^{-(k^2+m_1^2)(\tau-\tau_1)} \hat{\eta}(k, \tau_1; \omega) d\tau_1 = \\
&\quad -(2\pi)^4 a\delta^4(k) e^{-(k^2+m_1^2)\tau} - (2\pi)^4 \frac{a \left(m^2 + \frac{\lambda}{5!}a^4\right) \delta^4(k)}{k^2+m_1^2} \left[ 1 - e^{-(k^2+m_1^2)\tau} \right] + \\
&\quad + \int_0^\tau e^{-(k^2+m_1^2)(\tau-\tau_1)} \hat{\eta}(k, \tau_1; \omega) d\tau_1 = \\
&\quad -(2\pi)^4 a\delta^4(k) \left[ e^{-(k^2+m_1^2)\tau} + \frac{\left(m^2 + \frac{\lambda}{5!}a^4\right)}{k^2+m_1^2} \left[ 1 - e^{-(k^2+m_1^2)\tau} \right] \right] + \\
&\quad + \int_0^\tau e^{-(k^2+m_1^2)(\tau-\tau_1)} \hat{\eta}(k, \tau_1; \omega) d\tau_1,
\end{align*}
$$
$$
m_1^2 = m^2 + \frac{\lambda}{4!}a^4,
$$

and
$$
\begin{align*}
& \hat{V}_+(k, \tau, a) = e^{-(k^2+m_1^2)\tau} \times \\
& \quad \times \int_0^\tau e^{(k^2+m_1^2)\tau_1} \left[ (2\pi)^4 a \delta(\tau_1) \delta^4(k) - (2\pi)^4 \left(-m^2a - \frac{\lambda}{5!}a^5\right) \delta^4(k) + \hat{\eta}(k, \tau_1; \omega) \right] d\tau_1 = \\
& \quad (2\pi)^4 a \delta^4(k) e^{-(k^2+m_1^2)\tau} - (2\pi)^4 \left(-m^2a - \frac{\lambda}{5!}a^5\right) \delta^4(k) e^{-(k^2+m_1^2)\tau} \int_0^\tau e^{(k^2+m_1^2)\tau_1} d\tau_1 + \\
& \quad + e^{-(k^2+m_1^2)\tau} \int_0^\tau e^{(k^2+m_1^2)\tau_1} \hat{\eta}(k, \tau_1; \omega) d\tau_1 = \\
& \quad + (2\pi)^4 a \delta^4(k) e^{-(k^2+m_1^2)\tau} + (2\pi)^4 a \left(m^2 + \frac{\lambda}{5!}a^4\right) \delta^4(k) e^{-(k^2+m_1^2)\tau} \int_0^\tau e^{(k^2+m_1^2)\tau_1} d\tau_1 + \\
& \quad + \int_0^\tau e^{-(k^2+m_1^2)(\tau-\tau_1)} \hat{\eta}(k, \tau_1; \omega) d\tau_1 = \\
& \quad + (2\pi)^4 a \delta^4(k) e^{-(k^2+m_1^2)\tau} + \left[ (2\pi)^4 a \left(m^2 + \frac{\lambda}{5!}a^4\right) \delta^4(k) e^{-(k^2+m_1^2)\tau} \left[ \frac{e^{(k^2+m_1^2)\tau}}{k^2+m_1^2} - \frac{1}{k^2+m_1^2} \right] + \right. \\
& \quad \left. + \int_0^\tau e^{-(k^2+m_1^2)(\tau-\tau_1)} \hat{\eta}(k, \tau_1; \omega) d\tau_1 = \right. \\
& \quad + (2\pi)^4 a \delta^4(k) e^{-(k^2+m_1^2)\tau} + \left( (2\pi)^4 \frac{a \left(m^2 + \frac{\lambda}{5!}a^4\right) \delta^4(k)}{k^2+m_1^2} \left[ 1 - e^{-(k^2+m_1^2)\tau} \right] + \right. \\
& \quad \left. + \int_0^\tau e^{-(k^2+m_1^2)(\tau-\tau_1)} \hat{\eta}(k, \tau_1; \omega) d\tau_1 = \right. \\
& \quad + (2\pi)^4 a \delta^4(k) \left[ e^{-(k^2+m_1^2)\tau} + \frac{\left(m^2 + \frac{\lambda}{5!}a^4\right)}{k^2+m_1^2} \left[ 1 - e^{-(k^2+m_1^2)\tau} \right] + \right. \\
& \quad \left. + \int_0^\tau e^{-(k^2+m_1^2)(\tau-\tau_1)} \hat{\eta}(k, \tau_1; \omega) d\tau_1. \right] \\
& m_1^2 = m^2 + \frac{\lambda}{4!}a^4
\end{align*}
$$

From Eq.(2.5.10)-Eq.(2.5.11) one obtains
$$
\begin{equation}
\begin{split}
\hat{v}_-(k_1, \tau, a) \hat{v}_+(k_2, \tau', a') = & \\
& \left\{ -(2\pi)^4 a \delta^4(k_1) \left[ e^{-(k_1^2+m_1^2)\tau} + \frac{m^2 + \frac{\lambda}{5!}a^4}{k_1^2+m_1^2} [1-e^{-(k_1^2+m_1^2)\tau}] \right] + \right. \\
& \qquad \left. + \int_0^\tau e^{-(k_1^2+m_1^2)(\tau-\tau_1)} \hat{\eta}(k_1, \tau_1; \omega) d\tau_1 \right\} \times \\
& \left\{ (2\pi)^4 a' \delta^4(k_2) \left[ e^{-(k_2^2+m_1^2)\tau'} + \frac{m^2 + \frac{\lambda}{5!}a^4}{k_2^2+m_1^2} [1-e^{-(k_2^2+m_1^2)\tau'}] \right] + \right. \\
& \qquad \left. + \int_0^\tau e^{-(k_2^2+m_1^2)(\tau'-\tau_1)} \hat{\eta}(k_2, \tau_1; \omega) d\tau_1 \right\}
\end{split}
\tag{2.5.12}
\end{equation}
$$

From Eq.(2.5.12) one obtains

$$
\begin{equation}
\begin{split}
\langle \hat{v}_-(k_1, \tau, a) \hat{v}_+(k_2, \tau', a') \rangle_\eta = & \\
& \left\{ -(2\pi)^4 a \delta^4(k_1) \left[ e^{-(k_1^2+m_1^2)\tau} + \frac{m^2 + \frac{\lambda}{5!}a^4}{k_1^2+m_1^2} [1-e^{-(k_1^2+m_1^2)\tau}] \right] \right\} \times \\
& \left\{ (2\pi)^4 a' \delta^4(k_2) \left[ e^{-(k_2^2+m_1^2)\tau'} + \frac{m^2 + \frac{\lambda}{5!}a^4}{k_2^2+m_1^2} [1-e^{-(k_2^2+m_1^2)\tau'}] \right] \right\} - \\
& - \left\langle \left( \int_0^\tau e^{-(k_1^2+m_1^2)(\tau-\tau_1)} \hat{\eta}(k_1, \tau_1; \omega) d\tau_1 \right) \left( \int_0^{\tau'} e^{-(k_2^2+m_1^2)(\tau'-\tau_2)} \hat{\eta}(k_2, \tau_2; \omega) d\tau_2 \right) \right\rangle_\eta \\
& - (2\pi)^8 a a' \delta^4(k_1) \delta^4(k_2) \left[ e^{-(k_1^2+m_1^2)\tau} + \frac{m^2 + \frac{\lambda}{5!}a^4}{k_1^2+m_1^2} [1-e^{-(k_1^2+m_1^2)\tau}] \right] \times \\
& \left[ e^{-(k_2^2+m_1^2)\tau'} + \frac{m^2 + \frac{\lambda}{5!}a^4}{k_2^2+m_1^2} [1-e^{-(k_2^2+m_1^2)\tau'}] \right] - \\
& - \int_0^{\tau'} e^{-(k_2^2+m_1^2)(\tau'-\tau_2)} \int_0^\tau e^{-(k_1^2+m_1^2)(\tau-\tau_1)} \langle \hat{\eta}(k_1, \tau_1; \omega) \hat{\eta}(k_2, \tau_2; \omega) \rangle_\eta d\tau_1 d\tau_2
\end{split}
\tag{2.5.13}
\end{equation}
$$

Note that
$$ -(2\pi)^8 aa' \delta^4(k_1)\delta^4(k_2) \left[ e^{-(k_1^2+m^2)\tau} + \frac{m^2 + \frac{\lambda}{5!}a^4}{k_1^2+m_1^2} [1-e^{-(k_1^2+m_1^2)\tau}] \right] \times \\ \left[ e^{-(k_2^2+m^2)\tau'} + \frac{m^2 + \frac{\lambda}{5!}a'^4}{k_2^2+m_1^2} [1-e^{-(k_1^2+m_1^2)\tau'}] \right] \Bigg|_{\tau'=\tau, a'=a} = \\ -(2\pi)^8 aa' \delta^4(k_1)\delta^4(k_2) \left[ e^{-(k_1^2+m_1^2)\tau} + \frac{m^2 + \frac{\lambda}{5!}a^4}{k_1^2+m_1^2} [1-e^{-(k_1^2+m_1^2)\tau}] \right] \times \\ \times \left[ e^{-(k_1^2+m^2)\tau} + \frac{m^2 + \frac{\lambda}{6}a'^2}{k_2^2+m_1^2} [1-e^{-(k_1^2+m_1^2)\tau}] \right] \\ m_1^2 = m^2 + \frac{\lambda}{4!}a^4 \qquad (2.5.14) $$

and note that

$$ \begin{aligned} & \int_0^{\tau'} e^{-(k_2^2+m_1^2)(\tau'-\tau_2)} \int_0^{\tau} e^{-(k_1^2+m_1^2)(\tau-\tau_1)} \langle \hat{\eta}(k_1, \tau_1; \omega) \hat{\eta}(k_2, \tau_2; \omega) \rangle_{\eta} d\tau_1 d\tau_2 \Bigg|_{\tau'=\tau, a'=a} = \\ & \qquad \delta(k_1+k_2) \int_0^{\tau} e^{-(k_2^2+m_1^2)(\tau-\tau_2)} \int_0^{\tau} e^{-(k_1^2+m_1^2)(\tau-\tau_1)} \delta(\tau_1-\tau_2) d\tau_1 d\tau_2 = \\ & \qquad 2(2\pi)^4 \delta(k_1+k_2) \int_0^{\tau} e^{-(k_1^2+k_2^2+2m_1^2)(\tau-\tau_1)} d\tau_1 = \\ & \qquad 2(2\pi)^4 \delta(k_1+k_2) e^{-(k_1^2+k_2^2+2m_1^2)\tau} \int_0^{\tau} e^{(k_1^2+k_2^2+2m_1^2)\tau_1} d\tau_1 = \\ & \qquad 2(2\pi)^4 \delta(k_1+k_2) e^{-(k_1^2+k_2^2+2m_1^2)\tau} \left[ \frac{1}{(k_1^2+k_2^2+2m_1^2)} e^{(k_1^2+k_2^2+m_1^2)\tau} - \frac{1}{(k_1^2+k_2^2+2m_1^2)} \right] = \\ & \qquad = \frac{2(2\pi)^4 \delta(k_1+k_2)}{k_1^2+k_2^2+2m_1^2} - 2(2\pi)^4 \delta(k_1+k_2) \frac{e^{-(k_1^2+k_2^2+2m_1^2)\tau}}{(k_1^2+k_2^2+2m_1^2)} \\ & \qquad m_1^2 = m^2 + \frac{\lambda}{4!}a^4 \end{aligned} \qquad (2.5.15) $$

Transcendental master equation corresponding to two-point Euclidian Green function

$G_{(4)}(x_1, x_2, m, \lambda)$ in Euclidean space $E_4$ with $\dim E_4 = 4$ reads
$$ \frac{\left[a(x_1 - x_2)^2\right] \left(m^2 + \frac{\lambda}{5!} a(x_1 - x_2)^4\right)^2}{\left(m^2 + \frac{\lambda}{4!} a(x_1 - x_2)^4\right)^2} - (2\pi)^{-4} \int d^4 k \frac{e^{ik(x_1 - x_2)}}{k^2 + m^2 + \frac{\lambda}{4!} a(x_1 - x_2)^4} = 0. \qquad (2.5.16) $$

Transcendental master equation corresponding to two-point Euclidian Green function

$G_D(x_1, x_2, m, \lambda)$ in Euclidean space $E_D$ with $\dim E_D = D$ reads

$$ \frac{\left[a(x_1 - x_2)^2\right] \left(m^2 + \frac{\lambda}{5!} a(x_1 - x_2)^4\right)^2}{\left(m^2 + \frac{\lambda}{4!} a(x_1 - x_2)^4\right)^2} - (2\pi)^{-D} w \lim_{\eta \to 0_+} \int d^D k \frac{e^{-\eta k + ik(x_1 - x_2)}}{k^2 + m^2 + \frac{\lambda}{4!} a(x_1 - x_2)^4} = 0. \qquad (2.5.17) $$
$$ \eta > 0, $$

where a weak limit taken in $\mathcal{L}'(\mathbb{R}_x^D)$, see Appendix 1.

Note that

$$ \int d^4 k \frac{e^{ik(x_1 - x_2)}}{\left[k^2 + m^2 + \frac{\lambda}{4!} a(x_1 - x_2)^4\right]^2} = (4\pi)^{-2} \left(\frac{m_1}{|x_1 - x_2|}\right) K_0(m_1|x_1 - x_2|) \qquad (2.5.18) $$
$$ m_1 = \sqrt{m^2 + \frac{\lambda}{4!} a(x_1 - x_2)^4}, $$
$$ K_0(m_1|x_1 - x_2|) \propto -\ln\left(\frac{m_1|x_1 - x_2|}{2}\right) - \gamma \quad \text{as } m_1|x_1 - x_2| \to 0 $$

see Appendix 1. In order to derive Eq.(2.5.18) we applied Eq.(2.5.19) (see formula 13.6 (2) from [13]).

$$ \int_0^\infty \frac{\nu^{\nu+1} J_\nu(by) dy}{(\nu^2 + p^2)^{\mu+1}} = \frac{b^\mu p^{\nu-\mu}}{2^\mu \Gamma(\mu+1)} K_{\nu-\mu}(bp) \qquad (2.5.19) $$
$$ -1 < \text{Re} \nu < 2 \text{Re} \mu + 1.5. $$

### 3. Weak coupling. Nonperturbative result.
## 3.1.Weak coupling.The $\lambda\phi_d^4$ theory

We assume now that

$$ \varepsilon = \frac{\lambda \theta_{\delta}(|x|) a(x_1 - x_2)^2}{m^2} \ll 1, \qquad (3.1.1) $$

where $\theta_{\delta}(|x|) = \theta_{\delta}(|x_1 - x_2|) = \theta(|x_1 - x_2| - \delta), x = |x_1 - x_2|$.

From Eq.(2.4.41) and Eq.(3.1.1) we get

$$ \theta_{\delta}(|x|)a^2(x) \frac{\left(1 + \frac{\lambda a^2(x)}{6m^2}\right)^2}{\left(1 + \frac{0.5\lambda a^2(x)}{m^2}\right)^2} - (2\pi)^{-4} \int \frac{d^4 k e^{ikx} \theta_{\delta}(|x|)}{(m^2 + k^2) \left[1 + \frac{0.5\lambda a^2(x)}{m^2 + k^2}\right]} \quad (3.1.2) $$

and

$$ \begin{aligned} \theta_{\delta}(|x|)a^2(x) = (2\pi)^{-4} \int \frac{d^4 k e^{ikx} \theta_{\delta}(|x|)}{(m^2 + k^2)} \left[ 1 - \frac{0.5\lambda a^2(x)}{m^2 + k^2} + \left( \frac{0.5\lambda a^2(x)}{m^2 + k^2} \right)^2 + \dots \right] = & \\ (2\pi)^{-4} \int \frac{d^4 k e^{ikx} \theta_{\delta}(|x|)}{(m^2 + k^2)} - 0.5\lambda(2\pi)^{-4} a^2(x) \int \frac{d^4 k e^{ikx} \theta_{\delta}(|x|)}{(m^2 + k^2)^2} + & \\ + 0.25\lambda^2(2\pi)^{-4} a^4(x) \int \frac{d^4 k e^{ikx} \theta_{\delta}(|x|)}{(m^2 + k^2)^3} + \dots & \end{aligned} \qquad (3.1.3) $$

Therefore under condition (3.1.41) we get

$$ a^2(x) \left[ 1 + 0.5\lambda(2\pi)^{-4} \int \frac{d^4 k e^{ikx} \theta_{\delta}(|x|)}{(m^2 + k^2)^2} \right] = (2\pi)^{-4} \int \frac{\theta_{\delta}(|x|) d^4 k e^{ikx}}{(m^2 + k^2)} + O(\lambda^2), \qquad (3.1.4) $$

where constant in symbol $O(\lambda^2)$ depend on $m^2$ and $\delta$.

**Remark 3.1.1.** Note that for a given values of the parameters $m^2$ and $\delta$ we can choose value of the parameter $\lambda$ such that the inequality (3.1.1) is satisfied.

Thus finally for two-point for $\phi_d^4$ theory in the Euclidean QFT we obtain non perturbative result
$$
\begin{align*}
\theta_{\delta}(|x_1 - x_2|)G(x_1 - x_2) &= \theta_{\delta}(|x_1 - x_2|)a^2(x_1 - x_2) = \\
&\quad (2\pi)^{-4} \int \frac{d^4 k e^{ik(x_1-x_2)} \theta_{\delta}(|x_1 - x_2|)}{(m^2 + k^2)} \times \\
&\quad \left[ 1 + 0.5\lambda(2\pi)^{-4} \int \frac{d^4 k e^{ik(x_1-x_2)} \theta_{\delta}(|x_1 - x_2|)}{(m^2 + k^2)^2} \right]^{-1} = \tag{3.1.5} \\
&= (2\pi)^{-4} \int \frac{d^4 k e^{ik(x_1-x_2)} \theta_{\delta}(|x_1 - x_2|)}{(m^2 + k^2)} - \\
&\quad -0.5\lambda(2\pi)^{-8} \int \frac{d^4 k e^{ik(x_1-x_2)} \theta_{\delta}(|x_1 - x_2|)}{(m^2 + k^2)^2} \times \\
&\quad \left( \int \frac{d^4 k e^{ik(x_1-x_2)} \theta_{\delta}(|x_1 - x_2|)}{(m^2 + k^2)} \right) + o(\varepsilon).
\end{align*}
$$

**Remark 3.1.2.** To first order in $\lambda$, and in coordinate space, the two point function $G_{(4)}(x_1 - x_2; \delta) = \theta_{\delta}(|x_1 - x_2|)G_{(4)}(x_1 - x_2)$ bounded on region $\mathbb{R}^4 \setminus [-\delta, \delta]^4$ in Euclidean space with dim = 4 is

$$
\begin{align}
G_{(4)}(x_1 - x_2; \delta) &= \int \frac{d^4 k e^{ik(x_1-x_2)} \theta_{\delta}(|x_1 - x_2|)}{(2\pi)^4 (m^2 + k^2)} - \nonumber \\
&\quad -\frac{\lambda}{2} \int \frac{d^4 k e^{ik(x_1-x_2)} \theta_{\delta}(|x_1 - x_2|)}{(2\pi)^4 (m^2 + k^2)^2} \times \int \frac{d^4 k e^{ik(x_1-x_2)} \theta_{\delta}(|x_1 - x_2|)}{(2\pi)^4 (m^2 + k^2)} = \tag{3.1.6} \\
&\quad \times \frac{d^4 k e^{ik(x_1-x_2)} \theta_{\delta}(|x_1 - x_2|)}{(2\pi)^4 (m^2 + k^2)} \nonumber \\
&\quad \left\{ 1 - \frac{\lambda}{2} \frac{G_{(4)F}(x_1 - x_2; \delta)}{(m^2 + k^2)} \right\}, \nonumber
\end{align}
$$

where $G_{(4)F}(x_1 - x_2; \delta) = \theta_{\delta}(|x_1 - x_2|)G_{(4)F}(x_1 - x_2)$. For $\lambda \ll 1$ we get

$$
\begin{align}
\left\{ 1 - \frac{\lambda}{2} \frac{G_{(4)F}(x_1 - x_2; \delta)}{(m^2 + k^2)} \right\} &\approx \left\{ 1 + \frac{0.5\lambda G_{(4)F}(x_1 - x_2; \delta)}{(m^2 + k^2)} \right\}^{-1} \tag{3.1.7} \\
&= \frac{m^2 + k^2}{m^2 + 0.5\lambda G_{(4)F}(x_1 - x_2; \delta) + k^2}. \nonumber
\end{align}
$$

From Eq.(3.1.6) and Eq.(3.1.7) we get

$$
G_{(4)}(x_1 - x_2; \delta) = \frac{1}{(2\pi)^4} \int \frac{d^4 k e^{ik(x_1-x_2)} \theta_{\delta}(|x_1 - x_2|)}{m^2 + 0.5\lambda G_{(4)F}(x_1 - x_2; \delta) + k^2}. \quad (3.1.8)
$$

From Eq.(3.1.8) finally we get

$$
G_{(4)}(x_1 - x_2; \delta) \approx \frac{1}{(2\pi)^4} \int \frac{d^4 k e^{ik(x_1-x_2)} \theta_{\delta}(|x_1 - x_2|)}{m^2 + 0.5\lambda G_{(4)F}(\delta; \delta) + k^2}. \quad (3.1.9)
$$

This expression leads us to define $\mu_{\text{ren}}^2$ by
$$ \mu_{\text{ren}}^2 = m^2 + 0.5\lambda G_{(4)F}(\delta; \delta) = m^2 + \delta m^2. \qquad (3.1.10) $$

Note that in contrast with canonical perturbative calculation (see Appendix 2,eq.2.28)
$\delta m^2 = 0.5\lambda G_{(4)F}(\delta; \delta)$ is finite.

**Remark 3.1.3.** To first order in $\lambda$, and in coordinate space, the two point function

$$ G_{(D)}(x_1 - x_2; \delta) \uparrow \mathbb{R}^D \setminus [-\delta, \delta]^D = \theta_{\delta}(|x_1 - x_2|) G_{(4)}(x_1 - x_2) \text{ bounded on region } \mathbb{R}^D \setminus [-\delta, \delta]^D $$

in Euclidean space $E_D$ with $\dim E_D = D$ is

$$ G_{(D)}(x_1 - x_2; \delta) = \int \frac{d^D k e^{ik(x_1-x_2)} \theta_{\delta}(|x_1 - x_2|)}{(2\pi)^D (m^2 + k^2)} - \frac{\lambda}{2} \int \frac{d^D k e^{ik(x_1-x_2)} \theta_{\delta}(|x_1 - x_2|)}{(2\pi)^D (m^2 + k^2)^2} \times \int \frac{d^D k e^{ik(x_1-x_2)} \theta_{\delta}(|x_1 - x_2|)}{(2\pi)^D (m^2 + k^2)} = \frac{\lambda}{2} \int \frac{d^D k e^{ik(x_1-x_2)} \theta_{\delta}(|x_1 - x_2|)}{(2\pi)^D (m^2 + k^2)} \left\{ 1 - \frac{\lambda}{2} \frac{G_{(D)F}(x_1 - x_2; \delta)}{(m^2 + k^2)} \right\}, \qquad (3.1.11) $$

where $G_{(D)F}(x_1 - x_2; \delta) = \theta_{\delta}(|x_1 - x_2|)G_{(D)F}(x_1 - x_2)$. For $\lambda \ll 1$ we get

$$ \left\{ 1 - \frac{\lambda}{2} \frac{G_{(D)F}(x_1 - x_2; \delta)}{(m^2 + k^2)} \right\} \approx \left\{ 1 + \frac{0.5\lambda G_{(D)F}(x_1 - x_2; \delta)}{(m^2 + k^2)} \right\}^{-1} = \frac{m^2 + k^2}{m^2 + 0.5\lambda G_{(D)F}(x_1 - x_2; \delta) + k^2}. \qquad (3.1.12) $$

From Eq.(3.1.11) and Eq.(3.1.12) we get

$$ G_{(D)}(x_1 - x_2; \delta) = \frac{1}{(2\pi)^D} \int \frac{d^D k e^{ik(x_1-x_2)} \theta_{\delta}(|x_1 - x_2|)}{m^2 + 0.5\lambda G_{(D)F}(x_1 - x_2; \delta) + k^2}. \qquad (3.1.13) $$

From Eq.(3.1.13) finally we get

$$ G_{(D)}(x_1 - x_2; \delta) \approx \frac{1}{(2\pi)^D} \int \frac{d^D k e^{ik(x_1-x_2)} \theta_{\delta}(|x_1 - x_2|)}{m^2 + 0.5\lambda G_{(D)F}(\delta; \delta) + k^2}. \qquad (3.1.14) $$

This expression leads us to define $\mu_{\text{ren}}^2$ by

$$ \mu_{\text{ren}}^2 = m^2 + 0.5\lambda G_{(D)F}(\delta; \delta) = m^2 + \delta m^2. \qquad (3.1.15) $$

Note that in contrast with canonical perturbative calculation (see Appendix 2,eq.2.28)
$\delta m^2 = 0.5\lambda G_{(D)F}(\delta; \delta)$ is finite.

## 3.2. Weak coupling. The $\lambda\phi_d^4$ theory. Exact nonperturbative solution.

Trancendental master equation corresponding to $\lambda\phi_d^4$ theory [see Eq.(3.1.2)] reads
$$ \theta_{\delta}(|x|)a^2(x) \frac{\left(1 + \frac{\lambda a^2(x)}{6m^2}\right)^2}{\left(1 + \frac{0.5\lambda a^2(x)}{m^2}\right)^2} - (2\pi)^{-4} \int \frac{d^4 k e^{ikx} \theta_{\delta}(|x|)}{(m^2 + k^2) \left[1 + \frac{0.5\lambda a^2(x)}{m^2 + k^2}\right]} \quad (3.2.1) $$

Assuming now for simplicity that $\lambda/m^2 \ll 1$, then from Eq.(3.2.1) we obtain

$$ \theta_{\delta}(|x|)a^2(x) \approx (2\pi)^{-4} \int \frac{d^4 k e^{ikx} \theta_{\delta}(|x|)}{(m^2 + k^2) \left[1 + \frac{0.5\lambda a^2(x)}{m^2 + k^2}\right]} \quad (3.2.2) $$

Note that

$$ \frac{1}{1+z} = 1 - z + z^2 - z^3 + z^4 - \dots + \sum_{n=1}^{\infty} (-1)^n z^n \quad (3.2.3) $$
$$ z > 0, z < 1. $$

From Eq.(3.2.3) by setting

$$ z := \frac{0.5\lambda a^2(x)}{m^2 + k^2} \quad (3.2.4) $$

we obtain

$$ \left(1 + \frac{0.5\lambda a^2(x)}{m^2 + k^2}\right)^{-1} = 1 - \frac{0.5\lambda a^2(x)}{m^2 + k^2} + \left(\frac{0.5\lambda a^2(x)}{m^2 + k^2}\right)^2 - \dots = \\ \qquad 1 + \sum_{n=1}^{\infty} (-1)^n \frac{0.5\lambda a^2(x)}{m^2 + k^2}. \quad (3.2.5) $$

From Eq.(3.2.2) using Eq.(3.2.5) we obtain

$$ \begin{aligned} \theta_{\delta}(|x|)a^2(x) = (2\pi)^{-4} \int & \frac{d^4 k e^{ikx} \theta_{\delta}(|x|)}{(m^2 + k^2)} \left[ 1 - \frac{0.5\lambda a^2(x)}{m^2 + k^2} + \left( \frac{0.5\lambda a^2(x)}{m^2 + k^2} \right)^2 + \dots \right] = \\ & (2\pi)^{-4} \int \frac{d^4 k e^{ikx} \theta_{\delta}(|x|)}{(m^2 + k^2)} - 0.5\lambda(2\pi)^{-4} a^2(x) \int \frac{d^4 k e^{ikx} \theta_{\delta}(|x|)}{(m^2 + k^2)^2} + \\ & - 0.25\lambda^2(2\pi)^{-4} a^4(x) \int \frac{d^4 k e^{ikx} \theta_{\delta}(|x|)}{(m^2 + k^2)^3} + \dots = \\ = (2\pi)^{-4} & \int \frac{d^4 k e^{ikx} \theta_{\delta}(|x|)}{(m^2 + k^2)} + (2\pi)^{-4} \left\{ \sum_{n=1}^{\infty} (-1)^n (\lambda/2)^n a^{2n}(x) \int \frac{d^4 k e^{ikx} \theta_{\delta}(|x|)}{(m^2 + k^2)^{n+1}} \right\}. \end{aligned} \quad (3.2.6) $$

Let $\mathbb{R}^1$ be metric space equipped with distance $\rho[x_1, y_1] = |x_1 - y_1|$, $x_1, y_1 \in \mathbb{R}$.
Define now a map $\mathcal{F}(x, a(x)) : \mathbb{R}^1 \to \mathbb{R}^1$ by
$$
\begin{align}
\mathcal{F}(x, a^2(x)) = \nonumber \\
& (2\pi)^{-4} \int \frac{d^4 k e^{ikx} \theta_{\delta}(|x|)}{(m^2 + k^2)} + (2\pi)^{-4} \left\{ \sum_{n=1}^{\infty} (-1)^n (\lambda/2)^n a^{2n}(x) \int \frac{d^4 k e^{ikx} \theta_{\delta}(|x|)}{(m^2 + k^2)^{n+1}} \right\} = \nonumber \\
& (2\pi)^{-4} \int \frac{d^4 k e^{ikx} \theta_{\delta}(|x|)}{(m^2 + k^2)} - (2\pi)^{-4} \lambda/2 \int \frac{d^4 k e^{ikx} \theta_{\delta}(|x|)}{(m^2 + k^2)^{n+2}} + \nonumber \\
& + (2\pi)^{-4} \left\{ \sum_{n=2}^{\infty} (-1)^n (\lambda/2)^n a^{2n}(x) \int \frac{d^4 k e^{ikx} \theta_{\delta}(|x|)}{(m^2 + k^2)^{n+1}} \right\} \tag{3.2.7}
\end{align}
$$

We rewrite now Eq.(3.2.7) of the form

$$
\begin{align}
\mathcal{F}(x, b(x)) = & \nonumber \\
& g_0(x) + (2\pi)^{-4} \left\{ \sum_{n=1}^{\infty} (-1)^n (\lambda/2)^n b^n(x) \int \frac{d^4 k e^{ikx} \theta_{\delta}(|x|)}{(m^2 + k^2)^{n+1}} \right\} = \nonumber \\
& + g_0(x) - \lambda g_1(x) b(x) + \nonumber \\
& + (2\pi)^{-4} \left\{ \sum_{n=2}^{\infty} (-1)^n (\lambda/2)^n b^n(x) \int \frac{d^4 k e^{ikx} \theta_{\delta}(|x|)}{(m^2 + k^2)^{n+1}} \right\}, \tag{3.2.8}
\end{align}
$$

where we let for a shortness

$$
\begin{align}
b(x) &= a^2(x), \notag \\
g_0(x) &= (2\pi)^{-4} \int \frac{d^4 k e^{ikx} \theta_{\delta}(|x|)}{(m^2 + k^2)}, \tag{3.2.9} \\
g_1(x) &= (2\pi)^{-4} \int \frac{d^4 k e^{ikx} \theta_{\delta}(|x|)}{2(m^2 + k^2)^2}. \notag
\end{align}
$$

In Eq.(3.2.8) - Eq.(3.2.9) we set for simplisity but without loss of generality $m^2 = 1$.
We define now infinite sequence by

$$
\begin{align}
q_n(x) &= \mathcal{F}(x, q_{n-1}(x)), n = 1, 2, \dots \tag{3.2.11} \\
q_0(x) &= g_0(x). \notag
\end{align}
$$

From Eq.(3.2.8) and Eq.(3.2.9) we obtain

$$
\begin{align}
q_1(x) = \mathcal{F}(x, g_0(x)) = \nonumber \\
& g_0(x) + (2\pi)^{-4} \left\{ \sum_{n=1}^{\infty} (-1)^n (\lambda/2)^n g_0^n(x) \int \frac{d^4 k e^{ikx} \theta_{\delta}(|x|)}{(m^2 + k^2)^{n+1}} \right\} = \nonumber \\
& \qquad + g_0(x) - \lambda g_1(x) g_0(x) + \nonumber \\
& \qquad + (2\pi)^{-4} \left\{ \sum_{n=2}^{\infty} (-1)^n (\lambda/2)^n g_0^n(x) \int \frac{d^4 k e^{ikx} \theta_{\delta}(|x|)}{(m^2 + k^2)^{n+1}} \right\}. \tag{3.2.12}
\end{align}
$$
From Eq.(3.2.12) we obtain

$$
\begin{align}
\rho[q_1(x), g_0(x)] = \rho[\mathcal{F}(x, g_0(x)), g_0(x)] &= |q_1(x) - q_0(x)| \nonumber \\
&= (2\pi)^{-4} \left| \sum_{n=1}^{\infty} (-1)^n (\lambda/2)^n g_0^n(x) \int \frac{d^4 k e^{ikx} \theta_\delta(|x|)}{(m^2 + k^2)^{n+1}} \right| \nonumber \\
&\le \left| -\lambda g_1(x) g_0(x) + (2\pi)^{-4} \left\{ \sum_{n=2}^{\infty} (-1)^n (\lambda/2)^n g_0^n(x) \int \frac{d^4 k e^{ikx} \theta_\delta(|x|)}{(m^2 + k^2)^{n+1}} \right\} \right| \nonumber \\
&= \lambda g_1(x) g_0(x) + (2\pi)^{-4} \left| \sum_{n=2}^{\infty} (-1)^n (\lambda/2)^n g_0^n(x) \int \frac{d^4 k e^{ikx} \theta_\delta(|x|)}{(m^2 + k^2)^{n+1}} \right| \le \tag{3.2.13} \\
&= \lambda g_1(x) g_0(x) + (2\pi)^{-4} \left( \sum_{n=2}^{\infty} (\lambda/2)^n g_0^n(x) \int \frac{d^4 k e^{ikx} \theta_\delta(|x|)}{(m^2 + k^2)^{n+1}} \right) \le \nonumber \\
&= \lambda g_1(x) g_0(x) + (2\pi)^{-4} \left( \sum_{n=2}^{\infty} (\lambda/2)^n g_0^n(x) \theta_\delta(|x|) \int \frac{d^4 k}{(m^2 + k^2)^{n+1}} \right). \nonumber
\end{align}
$$

From Eq.(3.2.8) we obtain

$$
\begin{align}
q_2(x) &= \mathcal{F}(x, q_1(x)) = \nonumber \\
g_0(x) + (2\pi)^{-4} &\left\{ \sum_{n=1}^{\infty} (-1)^n (\lambda/2)^n q_1^n(x) \int \frac{d^4 k e^{ikx} \theta_{\delta}(|x|)}{(m^2 + k^2)^{n+1}} \right\} = \tag{3.2.14} \\
&= g_0(x) - \lambda g_1(x) q_1(x) + \nonumber \\
&\quad + (2\pi)^{-4} \left\{ \sum_{n=2}^{\infty} (-1)^n (\lambda/2)^n q_1^n(x) \int \frac{d^4 k e^{ikx} \theta_{\delta}(|x|)}{(m^2 + k^2)^{n+1}} \right\}. \nonumber
\end{align}
$$

From Eq.(3.2.12) and Eq.(3.2.14) we obtain
$$\rho[q_2(x), q_1(x)] = \rho[\mathcal{F}(x, q_1(x)), \mathcal{F}(x, q_0(x))] =$$

$$\rho[\mathcal{F}(x, q_1(x)), q_1(x)] = |\mathcal{F}(x, q_1(x)) - q_1(x)| =$$

$$\left| \left( g_0(x) - \lambda g_1(x) q_1(x) + (2\pi)^{-4} \sum_{n=2}^{\infty} (-1)^n (\lambda/2)^n q_1^n(x) \int \frac{d^4 k e^{ikx} \theta_\delta(|x|)}{(m^2 + k^2)^{n+1}} \right) - \left( g_0(x) - \lambda g_1(x) q_0(x) + (2\pi)^{-4} \sum_{n=2}^{\infty} (-1)^n (\lambda/2)^n q_0^n(x) \int \frac{d^4 k e^{ikx} \theta_\delta(|x|)}{(m^2 + k^2)^{n+1}} \right) \right| =$$

$$|\lambda g_1(x)(q_0(x) - q_1(x)) +$$

$$+(2\pi)^{-4} \left\{ \sum_{n=2}^{\infty} (-1)^n (\lambda/2)^n [q_1^n(x) - q_0^n(x)] \int \frac{d^4 k e^{ikx} \theta_\delta(|x|)}{(m^2 + k^2)^{n+1}} \right\} =$$

$$|\lambda g_1(x)(q_0(x) - q_1(x)) -$$

$$-(2\pi)^{-4} \lambda (q_0(x) - q_1(x)) \times$$

$$\left\{ \sum_{n=2}^{\infty} (-1)^n (\lambda/2)^{n-1} [q_0^{n-1}(x) + q_0^{n-2}(x)q_1(x) + \dots + q_0(x)q_1^{n-2}(x) + q_1^{n-1}(x)] \right\} \times \left. \left| \sum_{n=2}^{\infty} (-1)^n (\lambda/2)^{n-1} [q_0^{n-1}(x) + q_0^{n-2}(x)q_1(x) + \dots + q_0(x)q_1^{n-2}(x) + q_1^{n-1}(x)] \right| \right\} \times$$

$$\times \left| \int \frac{d^4 k e^{ikx} \theta_\delta(|x|)}{(m^2 + k^2)^{n+1}} \right| =$$

$$\sqrt{\lambda} |(q_0(x) - q_1(x))| \left| \left[ \sqrt{\lambda} g_1(x) - (2\pi)^{-4} \sqrt{\lambda} \times \sum_{n=2}^{\infty} (-1)^n (\lambda/2)^{n-1} [q_0^{n-1}(x) + q_0^{n-2}(x)q_1(x) + \dots + q_0(x)q_1^{n-2}(x) + q_1^{n-1}(x)] \right] \right| \le$$

$$\left\{ \sum_{n=2}^{\infty} (-1)^n (\lambda/2)^{n-1} [q_0^{n-1}(x) + q_0^{n-2}(x)q_1(x) + \dots + q_0(x)q_1^{n-2}(x) + q_1^{n-1}(x)] \right\} \times \left. \left| \int \frac{d^4 k e^{ikx} \theta_\delta(|x|)}{(m^2 + k^2)^{n+1}} \right| \right\} \le$$

$$\times \left| \int \frac{d^4 k e^{ikx} \theta_\delta(|x|)}{(m^2 + k^2)^{n+1}} \right| \right\} \le$$

$$\sqrt{\lambda} |(q_0(x) - q_1(x))| \left| \left( \sqrt{\lambda} g_1(x) - (2\pi)^{-4} \sqrt{\lambda} \right) \left| \sum_{n=2}^{\infty} (-1)^n (\lambda/2)^{n-1} (n-1) q_0^{n-1}(x) \right| \right| \le$$

$$\sqrt{\lambda} |(q_0(x) - q_1(x))| O(\lambda), O(\lambda) < 1.$$

In the last lines we applied formulas

$$a^n - b^n = (a-b)(a^{n-1} + a^{n-2}b + a^{n-3}b^2 + \dots + ab^{n-2} + b^{n-1}),$$

$$\int_0^\infty \frac{dr^3}{(m^2 + r^2)^{n+1}} = \left( \frac{-1}{2(n-1)(m^2 + r^2)^{n-1}} + \frac{m^2}{2n(m^2 + r^2)^n} \right) \Big|_0^\infty = \qquad (3.2.16)$$

$$\frac{1}{2(n-1)m^2(n-1)} - \frac{1}{2nm^2(n-1)} = \frac{1}{2m^2(n-1)} \left( \frac{1}{n-1} - \frac{1}{n} \right),$$

and Euler–Maclaurin sum formula
$$ \sum_{n=a}^{\infty} f(n) = \int_{a}^{\infty} f(x)dx + 0.5f(a) - \sum_{i=2}^{k} \frac{b_i}{i!} f^{(i-1)}(a) - \int_{a}^{\infty} \frac{B_k(\{1-t\})}{k!} dt \quad (3.2.17) $$

where $b_i$ is Bernoulli's numbers and $\{x\}$ is the fractional part of $x$. Note that

$$ |q_1(x)| = |\mathcal{F}(x, g_0(x))| \le g_0(x) + \lambda g_1(x)g_0(x) + \frac{(2\pi)^{-4}(\lambda/2)^2 g_0^2(x)}{1 - (\lambda/2)g_0(x)}. \quad (3.2.18) $$

From Eq.(3.2.8) we obtain

$$ \begin{aligned} |\mathcal{F}(x, b(x))| = & \left| g_0(x) + (2\pi)^{-4} \left\{ \sum_{n=1}^{\infty} (-1)^n (\lambda/2)^n b^n(x) \int \frac{d^4 k e^{ikx} \theta_\delta(|x|)}{(m^2 + k^2)^{n+1}} \right\} \right| \le \\ & g_0(x) + \lambda g_1(x) b(x) + (2\pi)^{-4} \left| \sum_{n=2}^{\infty} (-1)^n (\lambda/2)^n b^n(x) \int \frac{d^4 k e^{ikx} \theta_\delta(|x|)}{(m^2 + k^2)^{n+1}} \right| \le \\ & g_0(x) + \lambda g_1(x) b(x) + (2\pi)^{-4} \sum_{n=2}^{\infty} (\lambda/2)^n b^n(x) \int \frac{\theta_\delta(|x|) d^4 k}{(m^2 + k^2)^{n+1}} \le \quad (3.2.19) \\ & g_0(x) + \lambda g_1(x) b(x) + (2\pi)^{-4} \sum_{n=2}^{\infty} (\lambda/2)^n b^n(x) = \\ & g_0(x) + \lambda g_1(x) b(x) + (2\pi)^{-4} (\lambda/2)^2 b^2(x) \sum_{n=0}^{\infty} (\lambda/2)^n b^n(x) = \\ & = g_0(x) + \lambda g_1(x) b(x) + \frac{(2\pi)^{-4} (\lambda/2)^2 b^2(x)}{1 - (\lambda/2) b(x)}. \end{aligned} $$

From Eq.(3.2.8) and Eq.(3.2.14) we obtain
$$
\rho[q_3(x), q_2(x)] = \rho[\mathcal{F}(x, q_2(x)), \mathcal{F}(x, q_1(x))] =
$$

$$
|\mathcal{F}(x, q_2(x)) - \mathcal{F}(x, q_1(x))| =
$$

$$
\left| \left( g_0(x) - \lambda g_1(x) q_2(x) + (2\pi)^{-4} \left\{ \sum_{n=2}^{\infty} (-1)^n (\lambda/2)^n q_2^n(x) \int \frac{d^4 k e^{ikx} \theta_\delta(|x|)}{(m^2 + k^2)^{n+1}} \right\} \right) - \right. \\
\left. \left( g_0(x) - \lambda g_1(x) q_1(x) + (2\pi)^{-4} \left\{ \sum_{n=2}^{\infty} (-1)^n (\lambda/2)^n q_1^n(x) \int \frac{d^4 k e^{ikx} \theta_\delta(|x|)}{(m^2 + k^2)^{n+1}} \right\} \right) \right| =
$$

$$
|\lambda g_1(x)(q_2(x) - q_1(x)) +
$$

$$
+ (2\pi)^{-4} \left\{ \sum_{n=2}^{\infty} (-1)^n (\lambda/2)^n [q_2^n(x) - q_1^n(x)] \int \frac{d^4 k e^{ikx} \theta_\delta(|x|)}{(m^2 + k^2)^{n+1}} \right\} =
$$

$$
|\lambda g_1(x)(q_2(x) - q_1(x)) +
$$

$$
-(2\pi)^{-4}\lambda(q_2(x) - q_1(x)) \times
$$

$$
\left\{ \sum_{n=2}^{\infty} (-1)^n (\lambda/2)^{n-1} [q_1^{n-1}(x) + q_1^{n-2}(x)q_2(x) + \dots + q_1(x)q_2^{n-2}(x) + q_2^{n-1}(x)] \right\} \times \tag{3.2.20}
$$

$$
\times \left| \int \frac{d^4 k e^{ikx} \theta_{\delta}(|x|)}{(m^2 + k^2)^{n+1}} \right\rangle =
$$

$$
\sqrt{\lambda} |(q_2(x) - q_1(x))| \left| \left[ \sqrt{\lambda} g_1(x) - (2\pi)^{-4} \sqrt{\lambda} \times \right. \right.
$$

$$
\left\{ \sum_{n=2}^{\infty} (-1)^n (\lambda/2)^{n-1} [q_1^{n-1}(x) + q_1^{n-2}(x)q_2(x) + \dots + q_1(x)q_2^{n-2}(x) + q_2^{n-1}(x)] \right\} \times
$$

$$
\times \left| \int \frac{d^4 k e^{ikx} \theta_{\delta}(|x|)}{(m^2 + k^2)^{n+1}} \right\rangle \right| \le
$$

$$
\sqrt{\lambda} |(q_1(x) - q_2(x))| (\sqrt{\lambda} g_1(x) - (2\pi)^{-4} \sqrt{\lambda}) \left| \sum_{n=2}^{\infty} (-1)^n (\lambda/2)^{n-1} (n-1) q_1^{n-1}(x) \right| \le
$$

$$
\sqrt{\lambda} |(q_1(x) - q_2(x))| O(\lambda), O(\lambda) < 1.
$$

Processing inductively we get
$$
\rho[q_{n+1}(x), q_n(x)] = \rho[\mathcal{F}(x, q_{n+1}(x)), \mathcal{F}(x, q_n(x))] =
$$

$$
|\mathcal{F}(x, q_{n+1}(x)) - \mathcal{F}(x, q_n(x))| =
$$

$$
\left| \left( g_0(x) - \lambda g_1(x) q_{n+1}(x) + (2\pi)^{-4} \left\{ \sum_{n=2}^{\infty} (-1)^n (\lambda/2)^n q_{n+1}^n(x) \int \frac{d^4 k e^{ikx} \theta_\delta(|x|)}{(m^2 + k^2)^{n+1}} \right\} \right) - \right. \\
\left. \left( g_0(x) - \lambda g_1(x) q_n(x) + (2\pi)^{-4} \left\{ \sum_{n=2}^{\infty} (-1)^n (\lambda/2)^n q_n^n(x) \int \frac{d^4 k e^{ikx} \theta_\delta(|x|)}{(m^2 + k^2)^{n+1}} \right\} \right) \right| =
$$

$$
|\lambda g_1(x)(q_{n+1}(x) - q_n(x)) +
$$

$$
+ (2\pi)^{-4} \left\{ \sum_{n=2}^{\infty} (-1)^n (\lambda/2)^n [q_{n+1}^n(x) - q_n^n(x)] \int \frac{d^4 k e^{ikx} \theta_\delta(|x|)}{(m^2 + k^2)^{n+1}} \right\} ] =
$$

$$
|\lambda g_1(x)(q_{n+1}(x) - q_n(x)) +
$$

$$
-(2\pi)^{-4}\lambda(q_{n+1}(x) - q_n(x)) \times
$$

$$
\left\{ \sum_{n=2}^{\infty} (-1)^n (\lambda/2)^{n-1} [q_n^{n-1}(x) + q_n^{n-2}(x)q_{n+1}(x) + \dots + q_n(x)q_{n+1}^{n-2}(x) + q_{n+1}^{n-1}(x)] \right\} \times \quad (3.2.21)
$$

$$
\times \left| \int \frac{d^4 k e^{ikx} \theta_{\delta}(|x|)}{(m^2 + k^2)^{n+1}} \right| =
$$

$$
\sqrt{\lambda} |(q_{n+1}(x) - q_n(x))| \left[ \sqrt{\lambda} g_1(x) - (2\pi)^{-4} \sqrt{\lambda} \times \right.
$$

$$
\left\{ \sum_{n=2}^{\infty} (-1)^n (\lambda/2)^{n-1} [q_n^{n-1}(x) + q_n^{n-2}(x)q_{n+1}(x) + \dots + q_n(x)q_{n+1}^{n-2}(x) + q_{n+1}^{n-1}(x)] \right\} \times
$$

$$
\times \left| \int \frac{d^4 k e^{ikx} \theta_{\delta}(|x|)}{(m^2 + k^2)^{n+1}} \right| \le
$$

$$
\sqrt{\lambda} |(q_{n+1}(x) - q_n(x))| (\sqrt{\lambda} g_1(x) - (2\pi)^{-4} \sqrt{\lambda}) \left| \sum_{n=2}^{\infty} (-1)^n (\lambda/2)^{n-1} (n-1) q_n^{n-1}(x) \right| \le
$$

$$
\sqrt{\lambda} |(q_{n+1}(x) - q_n(x))| O(\lambda), O(\lambda) < 1.
$$

**Theorem 3.2.1.** *Infinite sequence defined by*

$$
q_n(x) = \mathcal{F}(x, q_{n-1}(x)), n = 1, 2, \dots \tag{3.2.22}
$$

$q_0(x) = g_0(x)$

for any $x, |x| > \delta$ has a limit: $\lim_{n\to\infty} q_n(x) = \bar{b}(x)$ such that $\mathcal{F}(x, \bar{b}(x)) = \bar{b}(x)$, where
$$
\begin{equation}
\begin{aligned}
\mathcal{F}(x, b(x)) = & g_0(x) + (2\pi)^{-4} \left\{ \sum_{n=1}^{\infty} (-1)^n (\lambda/2)^n b^n(x) \int \frac{d^4 k e^{ikx} \theta_{\delta}(|x|)}{(m^2 + k^2)^{n+1}} \right\} = \\
& g_0(x) - \lambda g_1(x) b(x) + \\
& \qquad + (2\pi)^{-4} \left\{ \sum_{n=2}^{\infty} (-1)^n (\lambda/2)^n b^n(x) \int \frac{d^4 k e^{ikx} \theta_{\delta}(|x|)}{(m^2 + k^2)^{n+1}} \right\},
\end{aligned}
\tag{3.2.23}
\end{equation}
$$

and where

$$
\begin{align}
b(x) &= a^2(x), \nonumber \\
g_0(x) &= (2\pi)^{-4} \int \frac{d^4 k e^{ikx} \theta_{\delta}(|x|)}{(m^2 + k^2)}, \nonumber \\
g_1(x) &= (2\pi)^{-4} \int \frac{d^4 k e^{ikx} \theta_{\delta}(|x|)}{2(m^2 + k^2)^2}, \tag{3.2.24} \\
|x| &> \delta. \nonumber
\end{align}
$$

**Proof.** Immediate from (3.2.21) by Generalized Banach fixed-point theorem, see Appendix3 Theorem 3.2..

3.3.Weak coupling.The $\lambda\phi_d^{2n}, n > 2$ theory.Exact nonperturbative solution.

Transcendental master equation (2.5.16) corresponding to two-point Eucledian
Green

function $G_{(4)}(x_1, x_2, m, \lambda)$ in Euclidean space $E_4$ with $\dim E_4 = 4$ reads

$$
\begin{equation}
\begin{split}
& \frac{\left[a(x_1 - x_2)^2\right] \left(m^2 + \frac{\lambda}{5!} a(x_1 - x_2)^4\right)^2}{\left(m^2 + \frac{\lambda}{4!} a(x_1 - x_2)^4\right)^2} - \\
& \quad - (2\pi)^{-4} \int d^4 k \frac{e^{ik(x_1-x_2)}}{k^2 + m^2 + \frac{\lambda}{4!} a(x_1 - x_2)^4} = 0.
\end{split}
\tag{3.3.1}
\end{equation}
$$

Assuming now for simplisy that $\lambda/m^2 \ll 1$, then from Eq.(3.3.1) we obtain

$$
\theta_{\delta}(|x|)a^2(x) \approx (2\pi)^{-4} \int \frac{d^4 k e^{ikx} \theta_{\delta}(|x|)}{(m^2 + k^2) \left[ 1 + \frac{\lambda a^4(x)}{4!(m^2 + k^2)} \right]} . \quad (3.3.2)
$$

Note that
$$ \frac{1}{1+z} = 1 - z + z^2 - z^3 + z^4 - \dots + \sum_{n=1}^{\infty} (-1)^n z^n \quad (3.3.3) \\ z > 0, z < 1. $$

From Eq.(3.3.3) by setting

$$ z := \frac{\lambda a^4(x)}{4!(m^2 + k^2)} \quad (3.3.4) $$

we obtain

$$ \left(1 + \frac{\lambda a^4(x)}{4!(m^2 + k^2)}\right)^{-1} = 1 - \frac{\lambda a^4(x)}{4!(m^2 + k^2)} + \left(\frac{\lambda a^4(x)}{4!(m^2 + k^2)}\right)^2 - \dots = \\ = 1 + \sum_{n=1}^{\infty} (-1)^n \frac{\lambda a^4(x)}{4!(m^2 + k^2)}. \quad (3.3.5) $$

From Eq.(3.3.2) using Eq.(3.3.5) we obtain

$$ \begin{aligned} \theta_{\delta}(|x|)a^2(x) = & (2\pi)^{-4} \int \frac{d^4 k e^{ikx} \theta_{\delta}(|x|)}{(m^2 + k^2)} \left[ 1 - \frac{\lambda a^4(x)}{4!(m^2 + k^2)} + \left( \frac{\lambda a^4(x)}{4!(m^2 + k^2)} \right)^2 + \dots \right] = \\ & (2\pi)^{-4} \int \frac{d^4 k e^{ikx} \theta_{\delta}(|x|)}{4!(m^2 + k^2)} - \lambda (2\pi)^{-4} a^4(x) \int \frac{d^4 k e^{ikx} \theta_{\delta}(|x|)}{4!(m^2 + k^2)^2} + \\ & - \lambda^2 (2\pi)^{-4} a^8(x) \int \frac{d^4 k e^{ikx} \theta_{\delta}(|x|)}{(4!)^2 (m^2 + k^2)^3} + \dots = \\ = & (2\pi)^{-4} \int \frac{d^4 k e^{ikx} \theta_{\delta}(|x|)}{(m^2 + k^2)} + (2\pi)^{-4} \left\{ \sum_{n=1}^{\infty} (-1)^n (\lambda/4!)^n a^{4n}(x) \int \frac{d^4 k e^{ikx} \theta_{\delta}(|x|)}{(m^2 + k^2)^{n+1}} \right\}. \end{aligned} \quad (3.3.6) $$

Let $\mathbb{R}^1$ be metric space equipped with distance $\rho[x_1, y_1] = |x_1 - y_1|$, $x_1, y_1 \in \mathbb{R}$.
Define now a map $\mathcal{F}(x, a(x)) : \mathbb{R}^1 \to \mathbb{R}^1$ by

$$ \begin{aligned} \mathcal{F}(x, a^2(x)) = & (2\pi)^{-4} \int \frac{d^4 k e^{ikx} \theta_{\delta}(|x|)}{(m^2 + k^2)} + (2\pi)^{-4} \left\{ \sum_{n=1}^{\infty} (-1)^n (\lambda/4!)^n a^{4n}(x) \int \frac{d^4 k e^{ikx} \theta_{\delta}(|x|)}{(m^2 + k^2)^{n+1}} \right\} = \\ & (2\pi)^{-4} \int \frac{d^4 k e^{ikx} \theta_{\delta}(|x|)}{(m^2 + k^2)} - (2\pi)^{-4} (\lambda/4!) \int \frac{d^4 k e^{ikx} \theta_{\delta}(|x|)}{(m^2 + k^2)^{n+2}} + \\ & + (2\pi)^{-4} \left\{ \sum_{n=2}^{\infty} (-1)^n (\lambda/4!)^n a^{4n}(x) \int \frac{d^4 k e^{ikx} \theta_{\delta}(|x|)}{(m^2 + k^2)^{n+1}} \right\} \end{aligned} \quad (3.3.7) $$

We rewrite now Eq.(3.3.7) of the form
$$
\begin{equation}
\begin{aligned}
\mathcal{F}(x, b(x)) = & g_0(x) + (2\pi)^{-4} \left\{ \sum_{n=1}^{\infty} (-1)^n (\lambda/4!)^n b^{4n}(x) \int \frac{d^4 k e^{ikx} \theta_{\delta}(|x|)}{(m^2 + k^2)^{n+1}} \right\} \\
& + (2\pi)^{-4} \left\{ \sum_{n=2}^{\infty} (-1)^n (\lambda/2)^n b^{2n}(x) \int \frac{d^4 k e^{ikx} \theta_{\delta}(|x|)}{(m^2 + k^2)^{n+1}} \right\},
\end{aligned}
\tag{3.2.8}
\end{equation}
$$

where we let for a shortness

$$
\begin{align}
b(x) &= a^2(x), \nonumber \\
g_0(x) &= (2\pi)^{-4} \int \frac{d^4 k e^{ikx} \theta_{\delta}(|x|)}{(m^2 + k^2)}, \tag{3.3.9} \\
g_1(x) &= (2\pi)^{-4} \int \frac{d^4 k e^{ikx} \theta_{\delta}(|x|)}{2(m^2 + k^2)^2}. \nonumber
\end{align}
$$

In Eq.(3.3.8) - Eq.(3.3.9) we set for simplisity but without loss of generality $m^2 = 1$.
We define now infinite sequence by

$$
\begin{align}
q_n(x) &= \mathcal{F}(x, q_{n-1}(x)), n = 1, 2, \dots \tag{3.3.11} \\
q_0(x) &= g_0(x). \notag
\end{align}
$$

From Eq.(3.3.8) and Eq.(3.3.9) we obtain

$$
\begin{equation}
\begin{aligned}
q_1(x) = \mathcal{F}(x, g_0(x)) = & \\
g_0(x) + (2\pi)^{-4} \left\{ \sum_{n=1}^{\infty} (-1)^n (\lambda/4!)^n g_0^{2n}(x) \int \frac{d^4 k e^{ikx} \theta_{\delta}(|x|)}{(m^2 + k^2)^{n+1}} \right\} = & \\
& + (2\pi)^{-4} \left\{ \sum_{n=2}^{\infty} (-1)^n (\lambda/4!)^n g_0^{2n}(x) \int \frac{d^4 k e^{ikx} \theta_{\delta}(|x|)}{(m^2 + k^2)^{n+1}} \right\}.
\end{aligned}
\tag{3.3.12}
\end{equation}
$$

From Eq.(3.3.12) we obtain
$$ \rho[q_1(x), g_0(x)] = \rho[\mathcal{F}(x, g_0(x)), g_0(x)] = |q_1(x) - q_0(x)| = |q_1(x) - g_0(x)| = $$

$$ (2\pi)^{-4} \left| \sum_{n=1}^{\infty} (-1)^n (\lambda/4!)^n g_0^{2n}(x) \int \frac{d^4 k e^{ikx} \theta_\delta(|x|)}{(m^2 + k^2)^{n+1}} \right| = $$

$$ \left| -(\lambda/4!)g_1(x)g_0^2(x) + (2\pi)^{-4} \left\{ \sum_{n=2}^{\infty} (-1)^n (\lambda/4!)^n g_0^{2n}(x) \int \frac{d^4 k e^{ikx} \theta_\delta(|x|)}{(m^2 + k^2)^{n+1}} \right\} \right| \le \quad (3.3.13) $$

$$ (\lambda/4!)g_1(x)g_0^2(x) + (2\pi)^{-4} \left( \sum_{n=2}^{\infty} (\lambda/4!)^n g_0^{2n}(x) \int \frac{d^4 k e^{ikx} \theta_\delta(|x|)}{(m^2 + k^2)^{n+1}} \right) \le $$

$$ (\lambda/4!)g_1(x)g_0^2(x) + (2\pi)^{-4} \left( \sum_{n=2}^{\infty} (\lambda/4!)^n g_0^{2n}(x) \theta_\delta(|x|) \int \frac{d^4 k}{(m^2 + k^2)^{n+1}} \right). $$

From Eq.(3.3.8) we obtain

$$ q_2(x) = \mathcal{F}(x, q_1(x)) = $$

$$ g_0(x) + (2\pi)^{-4} \left\{ \sum_{n=1}^{\infty} (-1)^n (\lambda/2)^n q_1^{2n}(x) \int \frac{d^4 k e^{ikx} \theta_\delta(|x|)}{(m^2 + k^2)^{n+1}} \right\} = \quad (3.3.14) \\ = g_0(x) - \lambda g_1(x) q_1^2(x) + $$

$$ +(2\pi)^{-4} \left\{ \sum_{n=2}^{\infty} (-1)^n (\lambda/2)^n q_1^{2n}(x) \int \frac{d^4 k e^{ikx} \theta_\delta(|x|)}{(m^2 + k^2)^{n+1}} \right\}. $$

From Eq.(3.3.12) and Eq.(3.3.14) we obtain
$$
\begin{align*}
\rho[q_2(x), q_1(x)] &= \rho[\mathcal{F}(x, q_1(x)), \mathcal{F}(x, q_0(x))] = \\
&= \rho[\mathcal{F}(x, q_1(x)), q_1(x)] = |\mathcal{F}(x, q_1(x)) - q_1(x)| = \\
&\quad \left| \left( g_0(x) - \lambda g_1(x) q_1^2(x) + (2\pi)^{-4} \sum_{n=2}^{\infty} (-1)^n (\lambda/2)^n q_1^{2n}(x) \int \frac{d^4 k e^{ikx} \theta_\delta(|x|)}{(m^2 + k^2)^{n+1}} \right) - \right. \\
&\quad \left. - \left( g_0(x) - \lambda g_1(x) q_0^2(x) + (2\pi)^{-4} \sum_{n=2}^{\infty} (-1)^n (\lambda/2)^n q_0^{2n}(x) \int \frac{d^4 k e^{ikx} \theta_\delta(|x|)}{(m^2 + k^2)^{n+1}} \right) \right| = \\
&\quad \left| \lambda g_1(x) (q_0^2(x) - q_1^2(x)) + \right. \\
&\quad \left. + (2\pi)^{-4} \left\{ \sum_{n=2}^{\infty} (-1)^n (\lambda/2)^n [q_1^{2n}(x) - q_0^{2n}(x)] \int \frac{d^4 k e^{ikx} \theta_\delta(|x|)}{(m^2 + k^2)^{n+1}} \right\} \right| = \\
&\quad \left| \lambda g_1(x) (q_0^2(x) - q_1^2(x)) - \right. \\
&\quad \left. - (2\pi)^{-4} \lambda (q_0(x) - q_1(x)) \times \right. \\
&\quad \left. \left\{ \sum_{n=2}^{\infty} (-1)^n (\lambda/2)^{n-1} [q_0^{2n-1}(x) + q_0^{2n-2}(x) q_1(x) + \dots + q_0(x) q_1^{2n-2}(x) + q_1^{2n-1}(x)] \times \right. \right. \\
&\quad \left. \left. \times \int \frac{d^4 k e^{ikx} \theta_\delta(|x|)}{(m^2 + k^2)^{n+1}} \right\} \right| = \\
&\quad \left| \sqrt{\lambda} |(q_0(x) - q_1(x))|(q_0(x) + q_1(x))| \left[ \sqrt{\lambda} g_1(x) - (2\pi)^{-4} \sqrt{\lambda} \times \right. \right. \\
&\quad \left. \left. \left\{ \sum_{n=2}^{\infty} (-1)^n (\lambda/2)^{n-1} [q_0^{2n-1}(x) + q_0^{2n-2}(x) q_1(x) + \dots + q_0(x) q_1^{2n-2}(x) + q_1^{2n-1}(x)] \times \right. \right. \right. \\
&\quad \left. \left. \left. \times \int \frac{d^4 k e^{ikx} \theta_\delta(|x|)}{(m^2 + k^2)^{n+1}} \right\} \right] \right| \le \sqrt{\lambda} |(q_1(x) - q_0(x))| \times c_1, \right. \\
&\quad \left. c_1 \le 1. \right.
\end{align*}
$$

In the last lines we applied formula

$$
a^{2n} - b^{2n} = (a-b)(a^{2n-1} + a^{2n-2}b + a^{2n-3}b^2 + \dots + ab^{2n-2} + b^{2n-1}) \quad (3.3.16)
$$

and Euler–Maclaurin sum formula

$$
\sum_{n=a}^{\infty} f(n) = \int_{a}^{\infty} f(x) dx + 0.5 f(a) - \sum_{i=2}^{k} \frac{b_i}{i!} f^{(i-1)}(a) - \int_{a}^{\infty} \frac{B_k(\{1-t\})}{k!} dt \quad (3.3.17)
$$

where $b_i$ is Bernoulli's numbers and $\{x\}$ is the fractional part of $x$. Note that

$$
|q_1(x)| = |\mathcal{F}(x, g_0(x))| \le g_0(x) + \lambda g_1(x) g_0^2(x) + \frac{(2\pi)^{-4} (\lambda/2)^2 g_0^2(x)}{1 - (\lambda/2) g_0^2(x)}. \quad (3.3.18)
$$

From Eq.(3.3.8) we obtain
$$
\begin{align*}
|\mathcal{F}(x, b(x))| &= \left| g_0(x) + (2\pi)^{-4} \left\{ \sum_{n=1}^{\infty} (-1)^n (\lambda/2)^n b^{2n}(x) \int \frac{d^4 k e^{ikx} \theta_{\delta}(|x|)}{(m^2 + k^2)^{n+1}} \right\} \right| \le \\
&\le g_0(x) + \lambda g_1(x) b^2(x) + (2\pi)^{-4} \left| \sum_{n=2}^{\infty} (-1)^n (\lambda/2)^n b^{2n}(x) \int \frac{d^4 k e^{ikx} \theta_{\delta}(|x|)}{(m^2 + k^2)^{n+1}} \right| \le \\
&\le g_0(x) + \lambda g_1(x) b^2(x) + (2\pi)^{-4} \sum_{n=2}^{\infty} (\lambda/2)^n b^{2n}(x) \int \frac{\theta_{\delta}(|x|) d^4 k}{(m^2 + k^2)^{n+1}} \le \tag{3.3.19} \\
&= g_0(x) + \lambda g_1(x) b^2(x) + (2\pi)^{-4} \sum_{n=2}^{\infty} (\lambda/2)^n b^{2n}(x) = \\
&= g_0(x) + \lambda g_1(x) b(x) + (2\pi)^{-4} (\lambda/2)^2 b^2(x) \sum_{n=0}^{\infty} (\lambda/2)^n b^{2n}(x) = \\
&= g_0(x) + \lambda g_1(x) b^2(x) + \frac{(2\pi)^{-4} (\lambda/2)^2 b^2(x)}{1 - (\lambda/2) b^2(x)}
\end{align*}
$$

Processing inductively we get
$$
\rho[q_{n+1}(x), q_n(x)] = \rho[\mathcal{F}(x, q_{n+1}(x)), \mathcal{F}(x, q_n(x))] =
$$

$$
|\mathcal{F}(x, q_{n+1}(x)) - \mathcal{F}(x, q_n(x))| =
$$

$$
\left| \left( g_0(x) - \lambda g_1(x) q_{n+1}^2(x) + (2\pi)^{-4} \left\{ \sum_{n=2}^{\infty} (-1)^n (\lambda/2)^n q_{n+1}^{2n}(x) \int \frac{d^4 k e^{ikx} \theta_{\delta}(|x|)}{(m^2 + k^2)^{n+1}} \right\} \right) - \right. \\
\left. - \left( g_0(x) - \lambda g_1(x) q_n^2(x) + (2\pi)^{-4} \left\{ \sum_{n=2}^{\infty} (-1)^n (\lambda/2)^n q_n^{2n}(x) \int \frac{d^4 k e^{ikx} \theta_{\delta}(|x|)}{(m^2 + k^2)^{n+1}} \right\} \right) \right|
$$

$$
|\lambda g_1(x)(q_{n+1}(x) - q_n(x))(q_{n+1}(x) + q_n(x)) + \\
\qquad + (2\pi)^{-4} \left\{ \sum_{n=2}^{\infty} (-1)^n (\lambda/2)^n [q_{n+1}^{2n}(x) - q_n^{2n}(x)] \int \frac{d^4 k e^{ikx} \theta_{\delta}(|x|)}{(m^2 + k^2)^{n+1}} \right\}| =
$$

$$
|\lambda g_1(x)(q_{n+1}(x) - q_n(x))(q_{n+1}(x) + q_n(x)) -
$$

$$
-(2\pi)^{-4}\lambda(q_{n+1}(x) - q_n(x)) \times
$$

$$
\left\{ \sum_{n=2}^{\infty} (-1)^n (\lambda/2)^{n-1} [q_n^{2n-1}(x) + q_n^{2n-2}(x)q_{n+1}(x) + \dots + q_n(x)q_{n+1}^{2n-2}(x) + q_{n+1}^{2n-1}(x)] \times \right. \\
\left. \qquad \times \left| \int \frac{d^4 k e^{ikx} \theta_{\delta}(|x|)}{(m^2 + k^2)^{n+1}} \right| \right\} = \tag{3.3.21}
$$

$$
\times \left| \int \frac{d^4 k e^{ikx} \theta_{\delta}(|x|)}{(m^2 + k^2)^{n+1}} \right| =
$$

$$
\sqrt{\lambda} |(q_{n+1}(x) - q_n(x))|(q_{n+1}(x) + q_n(x))|\left| \left[ \sqrt{\lambda} g_1(x) - (2\pi)^{-4} \sqrt{\lambda} \times \right.\right.
$$

$$
\left\{ \sum_{n=2}^{\infty} (-1)^n (\lambda/2)^{n-1} [q_n^{2n-1}(x) + q_n^{2n-2}(x)q_{n+1}(x) + \dots + q_n(x)q_{n+1}^{2n-2}(x) + q_{n+1}^{2n-1}(x)] \times \right. \\
\left. \qquad \left. \times \left| \int \frac{d^4 k e^{ikx} \theta_{\delta}(|x|)}{(m^2 + k^2)^{n+1}} \right| \right] \right| \le
$$

$$
\sqrt{\lambda} |(q_{n+1}(x) - q_n(x))| \times |(q_{n+1}(x) + q_n(x))| \times
$$

$$
\left| \left( \sqrt{\lambda} g_1(x) - (2\pi)^{-4} \sqrt{\lambda} \right) \left| \sum_{n=2}^{\infty} (-1)^n (\lambda/2)^{n-1} (n-1) q_n^{2n-1}(x) \right| \le
$$

$$
\sqrt{\lambda} |(q_{n+1}(x) - q_n(x))|c_n,
$$

$c_n \le 1.$

**Theorem 3.3.1.** *Infinite sequence defined by*

$$
\begin{align}
q_n(x) &= \mathcal{F}(x, q_{n-1}(x)), n = 1, 2, \dots \tag{3.3.22} \\
q_0(x) &= g_0(x) \nonumber
\end{align}
$$

for any $x, |x| > \delta$ has a limit: $\lim_{n\to\infty} q_n(x) = \bar{b}(x)$ such that $\mathcal{F}(x, \bar{b}(x)) = \bar{b}(x)$, where
$$
\begin{align}
A_{\mu}^{T} &= \left( \delta_{\mu\nu} - \frac{k_{\mu}k_{\nu}}{k^2} \right) A_{\nu} = O_{\mu\nu}^{T} A_{\nu} \nonumber \\
A_{\mu}^{L} &= \frac{k_{\mu}k_{\nu}}{k^2} A_{\nu} = O_{\mu\nu}^{L} A_{\nu} \tag{4.1.3} \\
\eta_{\mu}^{T} &= \left( \delta_{\mu\nu} - \frac{k_{\mu}k_{\nu}}{k^2} \right) \eta_{\nu} = O_{\mu\nu}^{T} \eta_{\nu} \nonumber \\
\eta_{\mu}^{L} &= \frac{k_{\mu}k_{\nu}}{k^2} \eta_{\nu} = O_{\mu\nu}^{L} \eta_{\nu} \nonumber
\end{align}
$$

This splits the Parisi-Wu equation as

$$
\begin{align}
\frac{\partial A^T(k,t)}{\partial t} &= -k^2 A^T(k,t) + [\eta^T(k,t) + \epsilon \tilde{\eta}^T(k,t)] \tag{4.1.4} \\
\frac{\partial A^L(k,t)}{\partial t} &= \eta^L(k,t) + \epsilon \tilde{\eta}^L(k,t). \notag
\end{align}
$$

This equation has solutions

$$
\begin{align}
A^T(k,t) &= \exp(-k^2 t) A_0^T + \exp(-k^2 t) \int_0^t \exp(k^2 \tau) [\eta^T(k, \tau) + \epsilon \tilde{\eta}^T(k, \tau)] d\tau \tag{4.1.5} \\
A^L(k,t) &= A_0^L + \int_0^t \exp(k^2 (t-\tau)) [\eta^L(k, \tau) + \epsilon \tilde{\eta}^L(k, \tau)] d\tau. \notag
\end{align}
$$

From this, we observe that the initial distribution in the transverse modes get
dissipated

at infinite time due to the presence of the damping term, while in the case of
longitudinal mode, there is no damping term and hence the initial distribution
persists

even in equilibrium, i.e. there exists no stationary distribution in the transverse
modes.

This is directly the consequence of the gauge invariance. To see how this affects
calculations, let us look at the propagator for the gauge fields.

$$
\begin{align}
D_{\mu\nu}(k,t|k',t') &= \langle A_{\mu}(k,t)^{L} A_{\nu}(k',t')^{L} \rangle = \nonumber \\
&= \delta^{4}(k+k') \frac{1}{k^2} O_{\mu\nu}^{T} (\exp(k^2(t-t'^2(t+t')))) + \tag{4.1.6} \\
&\quad + 2t_{<} \delta^{4}(k+k') \frac{k_{\mu}k_{\nu}}{k^2} + A_{\mu}(k,0)^{L} A_{\nu}(k',0)^{L} \nonumber
\end{align}
$$

The translational invariance of the propagator requires that the propagator be
proportional to δ⁴(k + k'), which is broken by the last term in the above-derived
propagator. One way to circumvent the problem would be to select the initial
configuration of the longitudinal mode to be A_μ(k, 0) = 0. But such a distribution is
exceptional and therefore we consider a more general distribution that is symmetric
around k = 0 as
$$A_{\mu}^{L} = \frac{k_{\mu}}{k^2} \phi(k) \qquad (4.1.7)$$

and for the propagator, we take another average over the distributions $\phi(k)$. Knowing

$$\langle \phi(k) \phi(k') \rangle_{\phi} = -\alpha \delta^4(k + k') \qquad (4.1.8)$$

where $\alpha$ is the width of $\phi(k)$, we get the limit of the propagator as

$$\lim_{t \to \infty} D_{\mu\nu}(k, t|k', t') = \delta^4(k+k') \left\{ \frac{1}{k^2} \left( \delta_{\mu\nu} - (1-\alpha) \frac{k_{\mu}k_{\nu}}{k^2} \right) + 2t \frac{k_{\mu}k_{\nu}}{k^2} \right\} \qquad (4.1.9)$$

which is nothing but the usual gauge fixed propagator, with a diverging term proportional to $t$. Therefore one can see that the stochastic quantization prescription allows one to quantize gauge fields and get the propagator of the theory without having

to resort to gauge fixing explicitly, but as seen from the above calculations, the gauge

freedom is reflected as a choice in the initial distribution for the longitudinal mode of the

gauge fields, the width of the distribution $\alpha$ playing the role of the gauge fixing parameter. Another subtlety worth mentioning is that for gauge-invariant quantities, the

term proportional to $t$ in the equilibrium limit goes to zero. But the gauge-variant quantities diverge in the limit $t \to \infty$. This is mainly due to the absence of drift force in the transverse part of Langevin's equation, meaning that the stationary distribution for

the transverse part doesn't exist, and a particle undergoing Brownian motion in the configuration space will drift forever in this direction. Speaking in terms of forces, the action provides a force that constrains a particle to a gauge orbit in the configuration space, but the particle is free to move along the gauge orbit indefinitely. It is this free motion along the gauge orbits that causes the gauge-variant quantities to diverge in the

equilibrium limit. Therefore, for gauge-invariant quantities, the perturbative stochastic

quantization prescription gives the same results as the regular perturbative quantization

without having to resort to explicit gauge fixing procedures, but in the case of gauge-invariant observables, the theory again leads to divergences.

We set now

$$\begin{aligned} A_{\mu}(x, t) - a_{\mu}\theta(t) &= \underline{A}_{\mu}(x, t, a_{\mu}), \\ A_{\mu}(x, t) &= \underline{A}_{\mu}(x, t, a_{\mu}) + a_{\mu}\theta(t). \end{aligned} \qquad (4.1.10)$$
Substitution Eq.(4.1.10)-Eq.(4.3.13) into Eq.(4.1.1) gives

$$ \frac{\partial(\underline{A}_{\mu}^{T}(x, t, a_{\mu}) + a_{\mu}\theta(t))}{\partial t} = (\delta_{\mu\nu}\partial^{2} - \partial_{\mu}\partial_{\nu})\underline{A}_{\mu}^{T}(x, t, a_{\mu}) + \eta_{\mu}^{T}(x, t; \omega, \omega') + \epsilon\underline{\eta}_{\mu}^{T}(x, t; \omega, \omega') \quad (4.1.11) $$

and thus

$$ \frac{\partial \underline{A}_{\mu}^{T}(x, t, a_{\mu})}{\partial t} = -a_{\mu}\delta(t) + (\delta_{\mu\nu}\partial^{2} - \partial_{\mu}\partial_{\nu})\underline{A}_{\mu}^{T}(x, t, a_{\mu}) + \eta_{\mu}^{T}(x, t; \omega, \omega') + \epsilon\underline{\eta}_{\mu}^{T}(x, t; \omega, \omega') \quad (4.1.12) $$

Performing a Fourier transformation to momentum space, we get

$$ \frac{\partial \underline{A}_{\mu}^{T}(k, t, a_{\mu})}{\partial t} = -(2\pi)^{4}a_{\mu}\delta^{4}(k)\delta(t) - k^{2}T_{\mu\nu}\underline{A}_{\nu}^{T}(k, t, a_{\mu}) + \eta_{\mu}^{T}(k, t; \omega, \omega') + \epsilon\underline{\eta}_{\mu}^{T}(k, t; \omega, \omega'), \quad (4.1.13) $$ $$ T_{\mu\nu} = \delta_{\mu\nu} - \frac{k_{\mu}k_{\nu}}{k^{2}}, $$

where we introduced the transverse projection operator $T_{\mu\nu}$. In order to solve Eq. (4.1.13) one studies, the Green function $G_{\mu\nu}$ which satisfies Eq. (4.1.14)

$$ (\delta_{\mu\nu} \frac{\partial}{\partial t} + T_{\mu\nu} k^2) G_{\mu\nu}(k,t) = \delta_{\mu\nu} \delta(t). \quad (4.1.14) $$

This equation is immediately solved, and solution reads

$$ G_{\mu\nu}(k,t) = \theta(t)(\exp(-tk^2 T))_{\mu\nu} = \theta(t)T_{\mu\nu}\exp(-tk^2). \quad (4.1.15) $$

The general solution for $A^T(k, t)$ follows:

$$ A_{\mu}^{T}(k, t, a_{\mu}) = \exp(-k^2 t)A_{0}^{T} - a_{\mu}\delta^{4}(k)\exp(-k^2 t)\int_{0}^{t}\delta(\tau)\exp(k^2\tau)d\tau + \exp(-k^2 t)\int_{0}^{t}\exp(k^2\tau)[\eta_{\mu}^{T}(k, \tau, \omega, \omega') + \epsilon\underline{\eta}_{\mu}^{T}(k, \tau, \omega, \omega)]d\tau \quad (4.1.16) $$ $$ \eta_{\mu}^{T}(k, \tau, \omega, \omega') = \left(\delta_{\mu\nu} - \frac{k_{\mu}k_{\nu}}{k^2}\right)\eta_{\nu}(k, \tau, \omega, \omega'). $$

## 4.2. Double stochastic quantization of massive scalar QED

We now consider scalar QED with the Euclidean action given by
$$S = \int d^4 x \left[ (D_\mu \varphi)^* (D_\mu \varphi) - \frac{1}{2} m^2 \varphi^{*2} - \frac{1}{2} m^2 \varphi^2 + \frac{1}{4} F_{\mu\nu} F_{\mu\nu} \right], \quad (4.2.1)$$

where $F_{\mu\nu} = \partial_\mu A_\nu - \partial_\nu A_\mu$ is the electromagnetic field strength, $D_\mu\varphi = (\partial_\mu - ieA_\mu)\varphi$ is the covariant derivative of the field $\varphi$ and $D_\mu\varphi^* = (\partial_\mu - ieA_\mu)\varphi^*$ is the covariant derivative of the field $\varphi^*$. Associated double stochastic Langevin equations reads

$$ \begin{aligned} \frac{\partial}{\partial t} A_{\mu} &= \partial_{\nu} F_{\mu\nu} - ie[\varphi^* \partial_{\mu} \varphi - (\partial_{\mu} \varphi^*) \varphi - 2ieA_{\mu} \varphi^* \varphi] + \eta_{\mu}(x,t; \omega, \omega') + \epsilon \tilde{\eta}_{\mu}(x,t; \omega, \omega') = \\ &= (\delta_{\mu\nu} \partial^2 - \partial_{\mu} \partial_{\nu}) A_{\nu}(x,t) - ie[\varphi^* \partial_{\mu} \varphi - (\partial_{\mu} \varphi^*) \varphi - 2ieA_{\mu} \varphi^* \varphi] + \\ & \qquad + \eta_{\mu}(x,t; \omega, \omega') + \epsilon \tilde{\eta}_{\mu}(x,t; \omega, \omega') \end{aligned} \quad (4.2.2) $$

$$ \begin{aligned} \frac{\partial}{\partial t} \varphi &= D_{\mu} D^{\mu} \varphi - m^2 \varphi + \eta(x, \tau; \omega) + \epsilon \tilde{\eta}(x, t; \omega, \omega') = \\ &= \partial^2 \varphi - ieA_{\mu} \partial_{\mu} \varphi - ie \partial_{\mu} (A_{\mu} \varphi) - e^2 A_{\mu} A_{\mu} \varphi - m^2 \varphi + \eta(x, t; \omega, \omega') + \epsilon \tilde{\eta}(x, t; \omega, \omega'), \end{aligned} \quad (4.2.3) $$

$$ \begin{aligned} \frac{\partial}{\partial t} \varphi^* &= (D_{\mu} D^{\mu} \varphi)^* - m^2 \varphi^* + \eta^*(x,t;\omega,\omega') + \epsilon \tilde{\eta}^*(x,t;\omega,\omega') = \\ &= \partial^2 \varphi^* + ieA_{\mu} \partial_{\mu} \varphi^* + ie \partial_{\mu} (A_{\mu} \varphi^*) - e^2 A_{\mu} A^{\mu} \varphi^* - m^2 \varphi^* + \eta^*(x,t;\omega,\omega') + \\ & \qquad + \epsilon \tilde{\eta}^*(x,t;\omega,\omega') \end{aligned} \quad (4.2.4) $$

with the noise correlations

$$ \begin{aligned} \langle \eta_{\mu}(x, t; \omega, \omega') \rangle &= \langle \eta(x, t; \omega, \omega') \rangle = \langle \eta^*(x, t; \omega, \omega') \rangle = 0, \\ \langle \tilde{\eta}_{\mu}(x, t; \omega, \omega') \rangle &= \langle \tilde{\eta}(x, t; \omega, \omega') \rangle = \langle \tilde{\eta}^*(x, t; \omega, \omega') \rangle = 0, \end{aligned} \quad (4.2.5) $$

$$ \begin{aligned} \langle \eta_{\mu}(x,t) \eta_{\mu}(x',t') \rangle &= 2 \delta_{\mu\nu} (x-x') \delta(t-t'), \\ \langle \tilde{\eta}_{\mu}(x,t) \tilde{\eta}_{\mu}(x',t') \rangle &= 2 \delta_{\mu\nu} (x-x') \delta(t-t'), \end{aligned} \quad (4.2.6) $$

$$ \begin{aligned} \langle \eta(x,t) \eta^*(x,t) \rangle &= 2\delta(x-x')\delta(t-t'), \\ \langle \tilde{\eta}(x,t) \tilde{\eta}^*(x,t) \rangle &= 2\delta(x-x')\delta(t-t'). \end{aligned} \quad (4.2.7) $$

We set now

$$ A_{\mu}(x,t) - a_{\mu}\theta(t) = \underline{A}_{\mu}(x,t, a_{\mu}), \quad (4.2.8) $$

$$ \varphi(x,t) - a\theta(t) = \underline{\varphi}(x,t, a), \quad (4.2.9) $$

$$ \varphi^*(x,t) - a^*\theta(t) = \underline{\varphi^*}(x,t, a^*). \quad (4.2.10) $$
Thus

$$A_{\mu}(x,t) = \underline{A}_{\mu}(x,t,a_{\mu}) + a_{\mu}\theta(t), \qquad (4.2.11)$$

$$\varphi(x,t) = \underline{\varphi}(x,t,a) + a\theta(t), \qquad (4.2.12)$$

$$\varphi^*(x,t) = \underline{\varphi}^*(x,t,a^*) + a^*\theta(t). \qquad (4.2.13)$$

Substitution Eq.(4.2.11)-Eq.(4.2.13) into Eq.(4.2.2) gives

$$\begin{aligned} \frac{\partial}{\partial t}[\underline{A}_{\mu}(x,t,a_{\mu}) + a_{\mu}\theta(t)] = \frac{\partial}{\partial t}\underline{A}_{\mu}(x,t,a_{\mu}) + a_{\mu}\delta(t) = & \\ (\delta_{\mu\nu}\partial^2 - \partial_{\mu}\partial_{\nu})[\underline{A}_{\nu}(x,t) + a_{\mu}] - ie[&\underline{(\varphi^* + a^*)}\partial_{\mu}\underline{\varphi} + \\ &\underline{(\partial_{\mu}\underline{\varphi}^* + a^*)}(\underline{\varphi}(x,t,a) + a) - & \\ -2ie(\underline{A}_{\mu}(x,t,a_{\mu}) + a_{\mu})(\underline{\varphi}^*(x,t,a^*) + a^*)(\underline{\varphi}(x,t,a) + a)] + & \\ +\eta_{\mu}(x,t;\omega) + \epsilon\underline{\eta}_{\mu}(x,t;\omega) & \end{aligned} \qquad (4.2.14)$$

From Eq.(4.2.14) we get

$$\begin{aligned} \frac{\partial}{\partial t}\underline{A}_{\mu}(x,t,a_{\mu}) + a_{\mu}\delta(t) = & \\ (\delta_{\mu\nu}\partial^2 - \partial_{\mu}\partial_{\nu})\underline{A}_{\nu}(x,t) - ie[&\underline{(\varphi^* + a^*)}\partial_{\mu}\underline{\varphi} - \\ &\underline{(\partial_{\mu}\underline{\varphi}^*)}(\underline{\varphi}(x,t,a) + a) - & \\ -2ie(\underline{A}_{\mu}(x,t,a_{\mu}) + a_{\mu})(\underline{\varphi}^*(x,t,a^*) + a^*)(\underline{\varphi}(x,t,a) + a)] + & \\ +\eta_{\mu}(x,t;\omega) + \epsilon\underline{\eta}_{\mu}(x,t;\omega) = & \\ (\delta_{\mu\nu}\partial^2 - \partial_{\mu}\partial_{\nu})\underline{A}_{\nu}(x,t) - ie[&\underline{(\varphi^*\partial_{\mu}\underline{\varphi} + a^*\partial_{\mu}\underline{\varphi})} - \\ &\underline{(\partial_{\mu}\underline{\varphi}^*)}\underline{\varphi}(x,t,a) + a\partial_{\mu}\underline{\varphi}^*(x,t,a^*) - & \\ -2ie(\underline{A}_{\mu}(x,t,a_{\mu})\underline{\varphi}^*(x,t,a^*) + a^*\underline{A}_{\mu}(x,t,a_{\mu}) + a_{\mu}\underline{\varphi}^*(x,t,a^*) + aa_{\mu}) \times & \\ &\underline{(\varphi(x,t,a) + a)]} + \eta_{\mu}(x,t;\omega) + \epsilon\underline{\eta}_{\mu}(x,t;\omega). & \end{aligned} \qquad (4.2.15)$$

From Eq.(4.2.15) finally we get

$$\begin{aligned} \frac{\partial}{\partial t}\underline{A}_{\mu}(x,t,a_{\mu}) = & \\ -a_{\mu}\delta(t) + (\delta_{\mu\nu}\partial^2 - \partial_{\mu}\partial_{\nu})\underline{A}_{\nu}(x,t,a_{\mu}) - ie[&\underline{(\varphi^*(x,t,a^*)\partial_{\mu}\underline{\varphi}(x,t,a) + a^*\partial_{\mu}\underline{\varphi}(x,t,a))} - \\ &\underline{-(\partial_{\mu}\underline{\varphi}^*(x,t,a^*))}\underline{\varphi}(x,t,a) + a\partial_{\mu}\underline{\varphi}^*(x,t,a^*) - & \\ -2ie(\underline{A}_{\mu}(x,t,a_{\mu})\underline{\varphi}^*(x,t,a^*)\underline{\varphi}(x,t,a) + a^*\underline{A}_{\mu}(x,t,a_{\mu})\underline{\varphi}(x,t,a) + a_{\mu}\underline{\varphi}^*(x,t,a^*)\underline{\varphi}(x,t,a) + & \\ +aa_{\mu}\underline{\varphi}(x,t,a) + a\underline{A}_{\mu}(x,t,a_{\mu})\underline{\varphi}^*(x,t,a^*) + aa^*\underline{A}_{\mu}(x,t,a_{\mu}) + aa_{\mu}\underline{\varphi}^*(x,t,a^*) + a^2a_{\mu})] + & \\ +\eta_{\mu}(x,t;\omega) + \epsilon\underline{\eta}_{\mu}(x,t;\omega). & \end{aligned} \qquad (4.2.16)$$

From Eq.(4.2.16) we obtain differential master equation corresponding to
$\underline{A}_{\mu}(x, t, a_{\mu}) :$

$$
\begin{align}
\frac{\partial}{\partial t} \underline{A}_{\mu}(x,t,a_{\mu}) = & \nonumber \\
& -a_{\mu} \delta(t) + (\delta_{\mu\nu} \partial^2 - \partial_{\mu} \partial_{\nu}) \underline{A}_{\nu}(x,t,a_{\mu}) - ie \left[ (a^* \partial_{\mu} \underline{\varphi}(x,t,a) + a \partial_{\mu} \underline{\varphi}^*(x,t,a^*)) - \right. \tag{4.2.17} \\
& \left. -2ie(\underline{a}_{a\mu} \underline{\varphi}(x,t,a) + \underline{a}a^* \underline{A}_{\mu}(x,t,a_{\mu}) + \underline{a}a_{\mu} \underline{\varphi}^*(x,t,a^*) + \underline{a}^2 a_{\mu}) \right] + \eta_{\mu}(x,t; \omega) \nonumber
\end{align}
$$

Substitution Eq.(4.2.11)-Eq.(4.2.13) into Eq.(4.2.3) gives

$$
\begin{align*}
\frac{\partial}{\partial t} (\underline{\varphi}(x,t,a) + a\theta(t)) &= \\
&= \partial^2 (\underline{\varphi}(x,t,a) + a) - ie(\underline{A}_\mu(x,t,a_\mu) + a_\mu)\partial_\mu (\underline{\varphi}(x,t,a) + a) - \\
&\quad -ie\partial_\mu ((\underline{A}_\mu(x,t,a_\mu) + a_\mu)(\underline{\varphi}(x,t,a) + a)) - e^2(\underline{A}_\mu(x,t,a_\mu) + a_\mu)^2 (\underline{\varphi}(x,t,a) + a) \\
&\quad -m^2 (\underline{\varphi}(x,t,a) + a) + \eta(x,t;\omega) + \epsilon\underline{\eta}(x,t;\omega) = \\
&\partial^2 \underline{\varphi}(x,t,a) - m^2 \underline{\varphi}(x,t,a) - am^2 - ie(\underline{A}_\mu(x,t,a_\mu)\partial_\mu \underline{\varphi}(x,t,a) + a_\mu\partial_\mu \underline{\varphi}(x,t,a)) - \\
&\quad -ie\partial_\mu (\underline{A}_\mu(x,t,a_\mu)\underline{\varphi}(x,t,a) + a\underline{A}_\mu(x,t,a_\mu) + a_\mu\underline{\varphi}(x,t,a) + aa_\mu) - \\
&\quad -e^2(\underline{A}_\mu(x,t,a_\mu)\underline{A}_\mu(x,t,a_\mu) + 2a_\mu\underline{A}_\mu(x,t,a_\mu) + a_\mu a_\mu)(\underline{\varphi}(x,t,a) + a) + \\
&\quad +\eta(x,t;\omega) + \epsilon\underline{\eta}(x,t;\omega) = \\
&\partial^2 \underline{\varphi}(x,t,a) - m^2 \underline{\varphi}(x,t,a) - am^2 - ie(\underline{A}_\mu(x,t,a_\mu)\partial_\mu \underline{\varphi}(x,t,a) + a_\mu\partial_\mu \underline{\varphi}(x,t,a)) - \\
&\quad -ie\partial_\mu (\underline{A}_\mu(x,t,a_\mu)\underline{\varphi}(x,t,a)) - ie\partial_\mu (\underline{A}_\mu(x,t,a_\mu) - ie\partial_\mu \underline{\varphi}(x,t,a) - \\
&\quad -e^2(\underline{A}_\mu(x,t,a_\mu)\underline{A}_\mu(x,t,a_\mu)\underline{\varphi}(x,t,a) + 2a_\mu\underline{A}_\mu(x,t,a_\mu)\underline{\varphi}(x,t,a) + a_\mu a_\mu\underline{\varphi}(x,t,a)) - \\
&\quad -e^2(a\underline{A}_\mu(x,t,a_\mu)\underline{A}_\mu(x,t,a_\mu) + 2aa_\mu\underline{A}_\mu(x,t,a_\mu) + aa_\mu a_\mu) + \eta(x,t;\omega) + \epsilon\underline{\eta}(x,t;\omega).
\end{align*}
$$

From Eq.(4.3.18) we obtain differential master equation corresponding to $\underline{\varphi}(x,t,a)$

$$
\begin{align}
\frac{\partial}{\partial t} \underline{\varphi}(x,t,a) = & -a\delta(t) + \partial^2 \underline{\varphi}(x,t,a) - m^2 \underline{\varphi}(x,t,a) - am^2 - \nonumber \\
& -ie(a_\mu \partial_\mu \underline{\varphi}(x,t,a)) - ie\alpha \partial_\mu \underline{A}_\mu(x,t,a_\mu) - ie\alpha_\mu \partial_\mu \underline{\varphi}(x,t,a) - \tag{4.3.19} \\
& -e^2(a_\mu a_\mu \underline{\varphi}(x,t,a) + 2aa_\mu \underline{A}_\mu(x,t,a_\mu) + aa_\mu a_\mu) + \eta(x,t;\omega) + \epsilon\underline{\eta}(x,t;\omega). \nonumber
\end{align}
$$

Similarly we obtain differential master equation corresponding to $\underline{\varphi}^*(x,t,a)$

$$
\begin{align}
\frac{\partial}{\partial t} \underline{\varphi}^*(x,t,a^*) = & -a^* \delta(t) + \partial^2 \underline{\varphi}^*(x,t,a^*) - m^2 \underline{\varphi}^*(x,t,a^*) - a^* m^2 - \nonumber \\
& + ie(a_\mu \partial_\mu \underline{\varphi}^*(x,t,a^*)) + ie a^* \partial_\mu \underline{A}_\mu(x,t,a_\mu) - ie a_\mu \partial_\mu \underline{\varphi}^*(x,t,a^*) - \tag{4.2.20} \\
& -e^2 (a_\mu a_\mu \underline{\varphi}^*(x,t,a^*) + 2a^* a_\mu \underline{A}_\mu(x,t,a_\mu) + a^* a_\mu a_\mu) + \eta^*(x,t;\omega) + \epsilon \underline{\eta}^*(x,t;\omega). \nonumber
\end{align}
$$

Note that

$$
\underline{A}_{\mu}(k, t, a_{\mu}) = \int d^{4}k \exp(ikx) \underline{A}_{\mu}(x, t, a_{\mu}), \qquad (4.2.21)
$$
$$ \varphi(k, t, a) = \int d^4 k \exp(-ikx) \varphi(x, t, a), \qquad (4.2.22) $$

$$ \varphi^*(k, t, a^*) = \int d^4 k \exp(ikx) \varphi^*(x, t, a). \qquad (4.2.23) $$

Performing in differential master equations (4.2.22), (4.2.23) a Fourier transformation (4.2.22)-(4.2.23) to momentum space, we get

$$ \begin{aligned} \frac{\partial}{\partial t} \underline{\mathcal{A}}_{\mu}(k, t, a_{\mu}) = & -a_{\mu} \delta(t) \delta^{4}(k) - (\delta_{\mu\nu} k^{2} + k_{\mu} k_{\nu}) \underline{\mathcal{A}}_{\nu}(k, t, a_{\mu}) - i e \left[ (a^{*} k_{\mu} \underline{\varphi}(k, t, a) + a k_{\mu} \underline{\varphi}^{*}(k, t, a^{*})) \right. \\ & \left. - 2 i e (a a_{\mu} \underline{\varphi}(x, t, a) + a a^{*} \underline{\mathcal{A}}_{\mu}(x, t, a_{\mu}) + a a_{\mu} \underline{\varphi}^{*}(x, t, a^{*}) + a^{2} a_{\mu} \delta^{4}(k)) \right] + \\ & + \eta_{\mu}(x, t; \omega, \omega'). \end{aligned} \qquad (4.2.24) $$

$$ \begin{aligned} \frac{\partial}{\partial t} \varphi(k, t, a) = & -a\delta(t)\delta^4(k) - (k^2 + m^2)\varphi(k, t, a) - am^2\delta^4(k) - \\ & -ie(a_\mu k_\mu \underline{\varphi}(k, t, a)) + ieak_\mu \underline{\mathcal{A}}_\mu(x, t, a_\mu) - iea_\mu k_\mu \underline{\varphi}(x, t, a) - \\ & -e^2(a_\mu a_\mu \underline{\varphi}(k, t, a) + 2aa_\mu \underline{\mathcal{A}}_\mu(k, t, a_\mu) + aa_\mu a_\mu \delta^4(k)) + \eta(k, t; \omega, \omega'). \end{aligned} \qquad (4.2.25) $$

$$ \begin{aligned} \frac{\partial}{\partial t} \varphi^*(k, t, a^*) = & -a^* \delta(t) \delta^4(k) - (k^2 + m^2) \underline{\varphi}^*(k, t, a^*) - a^* m^2 \delta^4(k) - \\ & + i e (a_\mu k_\mu \underline{\varphi}^*(k, t, a^*)) + i e a^* k_\mu \underline{\mathcal{A}}_\mu(k, t, a_\mu) - i e a_\mu k_\mu \underline{\varphi}^*(k, t, a^*) - \\ & - e^2 (a_\mu a_\mu \underline{\varphi}^*(k, t, a^*) + 2 a^* a_\mu \underline{\mathcal{A}}_\mu(k, t, a_\mu) + a^* a_\mu a_\mu \delta^4(k)) + \eta^*(k, t; \omega, \omega'). \end{aligned} \qquad (4.2.26) $$

From (4.2.25), (4.2.26) we get

$$ \begin{aligned} \underline{\mathcal{A}}_{\mu}(k, t, a_{\mu}) = & \exp(-k^2 t) \int_0^{\tau} d\tau \exp(k^2 t) \{-a_{\mu} \delta(t) \delta^4(k) \\ & -i e \left[ (a^* k_{\mu} \underline{\varphi}(k, t, a) + a k_{\mu} \underline{\varphi}^*(k, t, a^*)) \right. \\ & \left. - 2i e (a a_{\mu} \underline{\varphi}(x, t, a) + a a^* \underline{\mathcal{A}}_{\mu}(x, t, a_{\mu}) + a a_{\mu} \underline{\varphi}^*(x, t, a^*) + a^2 a_{\mu} \delta^4(k)) \right] \} + \\ & + \exp(-k^2 t) \int_0^{\tau} \exp(k^2 \tau) \left( \delta_{\mu\nu} - \frac{k_{\mu} k_{\nu}}{k^2} \right) \eta_{\nu}(k, \tau; \omega, \omega') d\tau. \end{aligned} \qquad (4.2.27) $$
$$
\begin{equation}
\begin{aligned}
\varphi(k,t,a) = {}& -a\delta^4(k) \exp[-(k^2+m^2)t] \int_0^t \delta(\tau) \exp[(k^2+m^2)\tau] d\tau + \\
& \qquad + \exp[-(k^2+m^2)t] \int_0^t d\tau \exp[(k^2+m^2)\tau] \{-am^2\delta^4(k) - \\
& \qquad -ie(a_\mu k_\mu \varphi(k,t,a)) + ieak_\mu A_\mu(x,t,a_\mu) - iea_\mu k_\mu \varphi(x,t,a) - \\
& \qquad -e^2(a_\mu a_\mu \varphi(k,t,a) + 2aa_\mu A_\mu(k,t,a_\mu) + aa_\mu a_\mu \delta^4(k)) \} + \\
& \qquad + \exp[-(k^2+m^2)t] \int_0^t d\tau \exp[(k^2+m^2)\tau] \eta(k,\tau; \omega, \omega')
\end{aligned}
\tag{4.2.28}
\end{equation}
$$

$$
\begin{equation}
\begin{aligned}
\varphi^*(k,t,a) = -(2\pi)^4 a\delta^4(k) \exp[-(k^2+m^2)t] &\int_0^t \delta(\tau) \exp[(k^2+m^2)\tau] d\tau + \\
& + \exp[-(k^2+m^2)t] \int_0^t d\tau \exp[(k^2+m^2)\tau] \{-(-2\pi)^4 a m^2 \delta^4(k) - \\
& - ie(a_\mu k_\mu \varphi^*(k,t,a)) + ieak_\mu A_\mu(x,t,a_\mu) - iea_\mu k_\mu \varphi^*(x,t,a) - \\
& - e^2(a_\mu a_\mu \varphi^*(k,t,a) + 2aa_\mu A_\mu(k,t,a_\mu) + aa_\mu a_\mu (2\pi)^4 \delta^4(k)) \} + \\
& + \exp[-(k^2+m^2)t] \int_0^t d\tau \exp[(k^2+m^2)\tau] \eta^*(k,\tau; \omega, \omega')
\end{aligned}
\tag{4.2.29}
\end{equation}
$$

Treating the coupling constant *e* as a small parameter we can solve the equations Eq.(4.2.27)-Eq.(4.2.29) by iteration. We rewrite now Eq.(4.2.27)-Eq.(4.2.29) of the following form

$$
\begin{align}
\underline{A}_{\mu}(k,t,a_{\mu},e) &= \Xi_{\mu} \left( \left\{ \underline{A}_{\nu}(k,t,a_{\mu},e) \right\}_{\nu=1}^{4}, \underline{\varphi}(k,t,a,e), \underline{\varphi}^{*}(k,t,a^{*},e) \right) \tag{4.2.30} \\
\Xi_{\mu} \left( \left\{ \underline{A}_{\nu}(k,t,a_{\mu},e) \right\}_{\nu=1}^{4}, \underline{\varphi}(k,t,a,e), \underline{\varphi}^{*}(k,t,a^{*},e) \right) = \nonumber
\end{align}
$$

$$
\underline{\varphi}(k,t,a,e) = \Phi\left(\left\{\underline{A}_v(k,t,a_\mu,e)\right\}_{v=1}^4, \underline{\varphi}(k,t,a,e), \underline{\varphi}^*(k,t,a^*,e)\right) \quad (4.2.31)
$$

$$
\underline{\varphi}^*(k,t,a^*,e) = \Phi^*(\{\underline{A}_v(k,t,a_\mu,e)\}_{v=1}^4, \underline{\varphi}(k,t,a,e), \underline{\varphi}^*(k,t,a^*,e)) \quad (4.2.32)
$$

We define now countable sequences
$$ \{\underline{\mathcal{A}}_{\mu}^{(n)}(k,t,a_{\mu},e)\}_{n=0}^{\infty}, \{\underline{\varphi}^{(n)}(k,t,a_{\mu},e)\}_{n=0}^{\infty}, \{\underline{\varphi}^{*(n)}(k,t,a_{\mu},e)\}_{n=0}^{\infty}, \quad (4.2.33) $$

by requsions

$$ \underline{\mathcal{A}}_{\mu}^{(n)}(k,t,a_{\mu},e) = \Xi_{\mu} (\{\underline{\mathcal{A}}_{\nu}^{(n-1)}(k,t,a_{\mu},e)\}_{\nu=1}^{4}, \underline{\varphi}^{(n-1)}(k,t,a,e), \underline{\varphi}^{*(n-1)}(k,t,a^{*},e)), \\ \underline{\mathcal{A}}_{\mu}^{(0)}(k,t,a_{\mu},e) = \underline{\mathcal{A}}_{\mu}^{(0)}(k,t,a_{\mu},e=0) = \underline{\mathcal{A}}_{\mu}^{(0)}(k,t); $$

$$ \underline{\varphi}^{(n)}(k,t,a,e) = \Phi(\{\underline{\mathcal{A}}_{\nu}^{(n-1)}(k,t,a_{\mu},e)\}_{\nu=1}^{4}, \underline{\varphi}^{(n-1)}(k,t,a,e), \underline{\varphi}^{*(n-1)}(k,t,a^{*},e)), \\ \underline{\varphi}^{(0)}(k,t,a,e) = \underline{\varphi}^{(0)}(k,t,a,e=0) = \underline{\varphi}^{(0)}(k,t); \quad (4.2.34) $$

$$ \underline{\varphi}^{*(n)}(k,t,a^{*},e) = \Phi^{*}(\{\underline{\mathcal{A}}_{\nu}^{(n-1)}(k,t,a_{\mu},e)\}_{\nu=1}^{4}, \underline{\varphi}^{(n-1)}(k,t,a,e), \underline{\varphi}^{*(n-1)}(k,t,a^{*},e)), \\ \underline{\varphi}^{*(0)}(k,t,a^{*},e) = \underline{\varphi}^{*(0)}(k,t,a^{*},e=0) = \underline{\varphi}^{*(0)}(k,t). $$

In order to obtain $\underline{\mathcal{A}}_{\mu}^{(0)}(k,t), \underline{\varphi}^{(0)}(k,t), \underline{\varphi}^{*(0)}(k,t)$ we set $e = 0$ in Eq.(4.2.27)-Eq.(4.2.29)

$$ \frac{\partial}{\partial t} \underline{\mathcal{A}}_{\mu}(k,t) = -k^2 T_{\mu\nu}(k) \underline{\mathcal{A}}_{\mu} + \eta_{\mu}(x,t;\omega,\omega'), \quad (4.2.35) $$

$$ \begin{aligned} \frac{\partial}{\partial t} \underline{\varphi}(k,t,a) &= -(2\pi)^4 a \delta(t) \delta^4(k) - p^2 \underline{\varphi}(k,t,a) - m^2 \underline{\varphi}(k,t,a) \\ &\quad -(2\pi)^4 a m^2 \delta^4(k) + \eta(k,t;\omega,\omega') = \\ &\quad -a \delta(t) \delta^4(k) - (p^2 + m^2) \underline{\varphi}(k,t,a) - (2\pi)^4 a m^2 \delta^4(k) + \eta(k,t;\omega,\omega'), \end{aligned} \quad (4.2.36) $$

$$ \begin{aligned} \frac{\partial}{\partial t} \underline{\varphi}^*(k,t,a^*) = &-(2\pi)^4 a^* \delta(t) \delta^4(k) - p^2 \underline{\varphi}(k,t,a^*) - m^2 \underline{\varphi}^*(k,t,a^*) \\ &-(2\pi)^4 a m^2 \delta^4(k) + \eta(k,t;\omega,\omega^*) = \\ &-a^* \delta(t) \delta^4(k) - (p^2 + m^2) \underline{\varphi}(k,t,a^*) - (2\pi)^4 a m^2 \delta^4(k) + \eta(k,t;\omega,\omega^*), \end{aligned} \quad (4.2.37) $$

This equations is immediately solved, and solutions reads

$$ \underline{\mathcal{A}}_{\mu}(k,t,0) = \int_{0}^{t} d\tau [T_{\mu\nu} \exp[-k^2(t-\tau)]] \eta_{\nu}(k,\tau) \quad (4.2.38) $$
$$
\begin{align}
\varphi(k,t,a) &= -(2\pi)^4 a \delta^4(k) e^{-(k^2+m^2)t} \int_0^t d\tau e^{[-(k^2+m^2)\tau]} \delta(\tau) - \nonumber \\
&\quad -(2\pi)^4 a m^2 \delta^4(k) e^{-(k^2+m^2)t} \int_0^t d\tau e^{[-(k^2+m^2)\tau]} - \nonumber \\
&\quad -ieae^{-(k^2+m^2)t} \int_0^t d\tau e^{[-(k^2+m^2)\tau]} \{k_\mu \underline{A}_\mu(k, \tau, 0) + \eta(k, \tau; \omega)\} = \nonumber \\
&\quad -(2\pi)^4 a \delta^4(k) e^{-(k^2+m^2)t} - (2\pi)^4 a m^2 \delta^4(k) e^{-(k^2+m^2)t} \int_0^t d\tau e^{[-(k^2+m^2)\tau]} - \nonumber \\
&\quad -ieae^{-(k^2+m^2)t} \int_0^t d\tau e^{[-(k^2+m^2)\tau]} \{k_\mu \underline{A}_\mu(k, \tau, 0) + \eta(k, \tau; \omega)\} = \tag{4.2.1} \\
&\quad -(2\pi)^4 a \delta^4(k) e^{-(k^2+m^2)t} - (2\pi)^4 a m^2 \delta^4(k) e^{-(k^2+m^2)t} \int_0^t d\tau e^{[-(k^2+m^2)\tau]} - \nonumber \\
&\quad -ieae^{-(k^2+m^2)t} \int_0^t d\tau e^{[-(k^2+m^2)\tau]} \{k_\mu \underline{A}_\mu(k, \tau, 0)\} \nonumber \\
&\quad -ieae^{-(k^2+m^2)t} \int_0^t d\tau e^{[-(k^2+m^2)\tau]} \eta(k, \tau; \omega) = \nonumber
\end{align}
$$

4.3. Stochastic quantization of scalar QED by canonical perturbation approach.

We now consider scalar QED with the Euclidean action given by

$$
S = \int d^4 x \left[ (D_\mu \varphi)^* (D_\mu \varphi) - \frac{1}{2} m^2 \varphi^{*2} - \frac{1}{2} m^2 \varphi^2 + \frac{1}{4} F_{\mu\nu} F_{\mu\nu} \right], \quad (4.3.1)
$$

where $F_{\mu\nu} = \partial_{\mu}A_{\nu} - \partial_{\nu}A_{\mu}$ is the electromagnetic field strength,
$D_{\mu}\varphi = (\partial_{\mu} - ieA_{\mu})\varphi$ is the covariant derivative of the field $\varphi$. Associated Langevin
equations reads
$$
\begin{align}
\frac{\partial}{\partial t} A_{\mu} &= \notag \\
&\partial_{\nu} F_{\mu\nu} - ie[\varphi^* \partial_{\mu} \varphi - (\partial_{\mu} \varphi^*) \varphi - 2ieA_{\mu} \varphi^* \varphi] + \eta_{\mu}(x, t; \omega) = \tag{4.3.2} \\
&\qquad (\delta_{\mu\nu} \partial^2 - \partial_{\mu} \partial_{\nu}) A_{\nu}(x, t) - ie[\varphi^* \partial_{\mu} \varphi - (\partial_{\mu} \varphi^*) \varphi - 2ieA_{\mu} \varphi^* \varphi] \notag \\
&\qquad \qquad + \eta_{\mu}(x, t; \omega) \notag
\end{align}
$$

$$
\begin{align}
\frac{\partial}{\partial t} \varphi &= D_{\mu} D^{\mu} \varphi - m^2 \varphi + \eta(x, \tau; \omega) = \tag{4.3.3} \\
&= \partial^2 \varphi - ieA_{\mu} \partial_{\mu} \varphi - ie \partial_{\mu} (A_{\mu} \varphi) - e^2 A_{\mu} A_{\mu} \varphi - m^2 \varphi + \eta(x, t; \omega). \nonumber
\end{align}
$$

$$
\begin{align}
\frac{\partial}{\partial t} \varphi^* &= (D_{\mu} D^{\mu} \varphi)^* - m^2 \varphi^* + \eta^*(x, t; \omega) = \tag{4.3.4} \\
&= \partial^2 \varphi^* + ie A_{\mu} \partial_{\mu} \varphi^* + ie \partial_{\mu} (A_{\mu} \varphi^*) - e^2 A_{\mu} A^{\mu} \varphi^* - m^2 \varphi^* + \eta^*(x, t; \omega) \nonumber
\end{align}
$$

with the noise correlations

$$
\begin{align}
\langle \eta_{\mu}(x,t) \rangle &= 0, \notag \\
\langle \eta_{\mu}(x,t) \eta_{\nu}(x,t') \rangle &= 2 \delta_{\mu\nu} \delta(x-x') \delta(t-t'), \tag{4.3.5} \\
\langle \eta(x,t) \rangle &= \langle \eta^*(x,t) \rangle = 0, \notag \\
\langle \eta(x,t) \eta^*(x',t') \rangle &= 2 \delta(x-x') \delta(t-t'). \notag
\end{align}
$$

Note that

$$
A_{\mu}(k, t) = (2\pi)^{-4} \int d^4 k \exp(ikx) A_{\mu}(x, t). \quad (4.3.6)
$$

Going to Fourier space, with the additional conventions

$$
\varphi(k, t, a) = (2\pi)^{-4} \int d^4 k \exp(-ikx) \varphi(x, t, a), \quad (4.3.7)
$$

$$
\varphi^*(k, t, a^*) = (2\pi)^{-4} \int d^4 k \exp(ikx) \varphi^*(x, t), \quad (4.3.8)
$$

from Eq.(4.3.2)-Eq.(4.3.8) we obtain the following set of integral equations:

$$
\begin{equation}
\begin{split}
A_{\mu}(k,t) = \int_{0}^{t} d\tau [T_{\mu\nu} \exp[-(k^2 + m^2)(t-\tau)] + L_{\mu\nu}] & \{\eta_{\nu}(k,\tau) - \\
& -e \int \frac{d^4 p d^4 q}{(2\pi)^4} \varphi^*(p,\tau) \varphi(q,\tau) (q_{\mu} + p_{\mu}) - \\
& -2e^2 \int \frac{d^4 p d^4 q d^4 r}{(2\pi)^8} \delta^4(k+p-q+r) A_{\nu}(p,\tau) \varphi^*(q,\tau) \varphi(r,\tau)
\end{split}
\tag{4.3.9}
\end{equation}
$$
$$
\begin{equation}
\begin{split}
\varphi(k,t) = \int_0^t d\tau \exp[-(k^2+m^2)(t-\tau)] \times \\
& \qquad \times \left\{ \left[\eta(k,t)\right] - e \int \frac{d^4p d^4q}{(2\pi)^4} \delta^4(k-p-q) A_\mu(p,t) \right. \\
& \qquad \left. \qquad \times \varphi(q,\tau)(k_\mu+q_\mu) - e^2 \int \frac{d^4p d^4q d^4r}{(2\pi)^8} \delta^4(k-p-q-r) A_\mu(p,t) A_\mu(q,t) \varphi(r,\tau) \right\},
\end{split}
\tag{4.3.10}
\end{equation}
$$

$$
\begin{equation}
\begin{split}
\varphi^*(k,t) = \int_0^t d\tau \exp[-(k^2+m^2)(t-\tau)] \times \\
& \quad \times \left\{ \left[\eta^*(k,t)\right] - e \int \frac{d^4p d^4q}{(2\pi)^4} \delta^4(-k-p+q) A_\mu(p,t) \right. \\
& \qquad \left. \qquad \times \varphi^*(q,\tau)(k_\mu+q_\mu) - e^2 \int \frac{d^4p d^4q d^4r}{(2\pi)^8} \delta^4(k-p-q+r) A_\mu(p,t) A_\mu(q,t) \varphi^*(r,\tau) \right\},
\end{split}
\tag{4.3.11}
\end{equation}
$$

We note that with conventions as in Eq.(4.3.7)-Eq.(4.3.8), the two-point correlations
of
the scalar fields read

$$
\langle \eta(k,t) \eta^*(k',t') \rangle = 2\delta(k-k')\delta(t-t'). \quad (4.3.12)
$$

For the assignment of momenta in diagrams, let us explicitly draw the vertices, and
indicate the flow of momenta by arrows as in figs. 4.3.1 to 4.3.6.

Let us now calculating the one-loop correction to the scalar propagator

$$
\langle \varphi(k,t) \varphi^*(k',t') \rangle.
$$

To do so we list first all contributing diagrams in fig. 4.4.1(a)-(g). For the sake of
demonstration, we choose this gauge variant quantity on purpose so that we are
facing

divergences linear in $t$ when performing the equilibrium limit. We would like to split
the

calculations into two parts, and first obtain the well-converging expression with just
transverse gauge field contributions. Subsequently, we will extract the linearly
diverging

contribution and at the end of this calculation take the limits $\lim_{t \to \infty} \lim_{x \to y} \varphi(x, t) \varphi(y, t)$ upon which the diverging terms disappear explicitly, as generally expected from gauge

invariant quantities. Thus we have already acquired all the knowledge to obtain the
large *t* contribution of the diagrams (a) to (e) immediately, avoiding any lengthy
calculation. Starting by considering the gauge field as purely transverse, the large *t*
contribution is given by multiplying together vertex factors and projection operators
from
gauge field lines (at the moment just transverse projection operators); dividing by crossed-line contributions (i.e. the momentum² of the corresponding line); multiplying

with the fictitious time integration factors (which are identical to the ones discussed in

the scalar-field case, as the transverse gauge field has the same exponential fictitious time dependence).

<figure><img src="image_3.png" /><figcaption>Pic.4.3.1. Vertices in stochastic perturbation theory of scalar QED.</figcaption></figure>

<figure><img src="image_4.png" /><figcaption>Pic.4.3.2. Vertices in stochastic perturbation theory of scalar QED.</figcaption></figure>

<figure><img src="image_5.png" /><figcaption>Pic.4.3.3. Vertices in stochastic perturbation theory of scalar QED.</figcaption></figure>
<figure><img src="image_6.png" /><figcaption>Pic.4.3.4. Vertices in stochastic perturbation theory of scalar QED.</figcaption></figure>

<figure><img src="image_7.png" /><figcaption>Pic.4.3.5. Vertices in stochastic perturbation theory of scalar QED.</figcaption></figure>

<figure><img src="image_8.png" /><figcaption>Pic.4.3.6. Vertices in stochastic perturbation theory of scalar QED.</figcaption></figure>
<figure><img src="image_9.png" /></figure>

Stochastic diagrams contributing to the one-loop correction to the $\phi^*\phi$ propagator in scalar QED.

Diagram (a) has two different time orderings. The diagram being completely symmetric,

it suffices to take twice the contribution of one of these, say when $\tau_1 < \tau_2$.

Apart from the overall $\int \int d^4p d^4q \delta(k-p-q) \delta(k-k')$ we then get

$$ (a) = -2e^2 \frac{(k_\mu + q_\mu) T_{\mu\nu} (k_\nu + q_\nu)}{p^2 q^2} \frac{1}{p^2 + q^2 + k^2} \frac{1}{2k^2}, $$

$$ (a) = (b) = -e^2 \frac{(k_\mu + q_\mu) T_{\mu\nu} (k_\nu + q_\nu)}{p^2 q^2} \frac{1}{p^2 + q^2 + k^2} \frac{1}{2k^2}, \quad (4.3.13) $$

$$ (d) = (e) = -e^2 \frac{(k_\mu + q_\mu) T_{\mu\nu} (k_\nu + q_\nu)}{p^2 q^2} \frac{1}{p^2 + q^2 + k^2} \frac{1}{2k^2} $$

and (not unexpectedly, in fact) the sum gives the conventional result in Landau gauge

$$ -2e^2 \int \int d^4p d^4q \delta(k-p-q) \delta(k-k') \frac{(k_\mu + q_\mu) T_{\mu\nu} (k_\nu + q_\nu)}{k^4 p^2 q^2}. \quad (4.3.14) $$

The remaining two diagrams can easily be evaluated in a similar way (note that the loop momentum does not contribute in the fictitious time integrations, as its line is connected to just identical fictitious times),

$$ -2e^2 \int d^4p \frac{\mathrm{Tr}(T_{\mu\nu})}{k^4 p^2} \quad (4.3.15) $$

Generalizing the results of this little exercise, one concludes that the equivalence proof

for scalar fields carries over directly to scalar QED with transverse gauge bosons, having identical fictitious time dependence. It therefore follows that the sum of stochastic diagrams with transverse gauge field contributions sum up to the corresponding Feynman diagram in Landau gauge.

In scalar QED a general vertex involves always at least one scalar field line which
carries exponential time dependence. Given some fixed time ordering, fictitious time integrations have the general structure

$$ \int_{0}^{\tau_{i+1}} d\tau_i (\tau_i^n \exp(\sum p^2 \tau_i^2)) = (4.3.16) $$

# 5. Double stochastic quantization QFT with massive fermions.

## 5.1. Basic notions.

Remin that originally developed for bosonic models, it is well known many years that stochastic quantization of fermions received a physically consistent treatment, see ref. [16],[17]. The starting point is a generalization of the original Langevin equation by

means of the introduction of a kernel $k_{ij}(x,y)$

$$ \frac{\partial \varphi_i(\tau, x)}{\partial \tau} = - \int d^D y k_{ij}(x, y) \frac{\delta S[\varphi]}{\delta \varphi_i(\tau, y)} + \eta_i(\tau, x), \quad (5.1.1) $$

where $S[\varphi] \triangleq S[\{\varphi_i\}_{i=1}^N]$ is the Euclidean action, and $\{\eta_i(\tau, x)\}_{i=1}^N$ a distributional noise

field, with correlations

$$ \begin{gathered} \langle \eta_i(\tau, x) \rangle = 0, \quad (5.1.2) \\ \langle \eta_i(\tau, x) \eta_j(\tau', x') \rangle = \delta_{ij} k_{ij}(x, x') \delta(\tau - \tau'). \end{gathered} $$

In the case of free fermions with the classical Euclidean action $S[\bar{\psi}(x)\psi(x)]$, is given by

$$ S[\bar{\psi}(x)\psi(x)] = -i \int d^4 x \bar{\psi}(x) (\delta_x + im) \psi(x), \quad (5.1.3) $$

the kernel is given by [1]-[2]:

$$ \begin{gathered} k_{ij}(x,y) = \delta_{ij}(\delta_x + m)\delta^4(x-y) \quad (5.1.4) \\ \delta = \gamma^\mu \partial_\mu, \mu = 0,1,2,3 \end{gathered} $$

and the Langevin equation reads

$$ \frac{\partial \psi(\tau, x)}{\partial \tau} = (\partial^2 + m^2) \psi(\tau, x) + g. \quad (5.1.5) $$

The Green function in momentum space reads [1]:

$$ G_{\alpha\beta}(k, \tau - \tau') = \delta_{\alpha\beta} e^{-(k^2+m^2)(\tau-\tau')} \hat{g}(k, \tau - \tau'). \quad (5.1.6) $$

where
$$ 
\begin{aligned}
\langle \hat{\mathcal{G}}(\mathbf{k}, \tau) \hat{\mathcal{G}}(\mathbf{k}', \tau') \rangle &= (-\mathbf{k} + m) \delta^4(\mathbf{k} + \mathbf{k}') \delta(\tau - \tau'), \\
\langle \hat{\eta}_{\mu}(\mathbf{k}, \tau) \hat{\eta}_{\nu}(\mathbf{k}', \tau') \rangle &= {}^4\delta_{\mu\nu} \delta(\mathbf{k} + \mathbf{k}') \delta(\tau - \tau').
\end{aligned} 
\qquad (5.1.7) 
$$

Notice that 2-point function $\langle \psi(x, \tau) \bar{\psi}(x', \tau') \rangle$ corresponding to Euclidean action (5.1.3) reads

$$ 
\begin{aligned}
\langle \psi(x, \tau) \bar{\psi}(x', \tau') \rangle = & \int_0^\infty dt_1 \int_0^\infty dt_2 \int_{\mathbb{R}^4} d^4 x_1 \int_{\mathbb{R}^4} d^4 x_2 G(x-x_1, \tau-t_1) G(x'-x_2, \tau'-t_2) \times \\
& \times \langle \mathcal{G}(x_1, t_1) \mathcal{G}(x_2, t_2) \rangle.
\end{aligned} 
\qquad (5.1.8) 
$$

Using the stochastic expectation values (4.1.7) and the momentum-space representation (4.1.6) of the Green function, this immediately leads to the momentum-space propagator

$$ 
\begin{aligned}
\langle \psi(\mathbf{k}, \tau) \bar{\psi}(\mathbf{k}', \tau') \rangle = & -(2\pi)^4 \delta^4(\mathbf{k} - \mathbf{k}') \left( \frac{\mathbf{k} - \mathbf{m}}{\mathbf{k}^2 + \mathbf{m}^2} \right) \times \\
& \times \exp\{-(\mathbf{k}^2 + \mathbf{m}^2)(\tau + \tau')\} \times [\exp\{2(\mathbf{k}^2 + \mathbf{m}^2)\tilde{\tau}\} - 1],
\end{aligned} 
\qquad (5.1.9) 
$$

where $\tilde{\tau} = \min(\tau_1, \tau_2)$. For practical purposes the 'time ordered' propagator $\tau < \tau'$ is useful:

$$ 
\begin{aligned}
\Delta(\mathbf{k}; \tau, \tau') = & -\left(\frac{\mathbf{k} - \mathbf{m}}{\mathbf{k}^2 + \mathbf{m}^2}\right) \times \\
& \times \exp\{-(\mathbf{k}^2 + \mathbf{m}^2)(\tau - \tau')\} \times [1 - \exp\{2(\mathbf{k}^2 + \mathbf{m}^2)\tau'\}].
\end{aligned} 
\qquad (5.1.10) 
$$

Note that for equal times this simply reduces to the standard Euclidean fermion propagator:

$$ 
\lim_{\tau=\tau' \to \infty} \Delta(\mathbf{k}; \tau, \tau') = \frac{\mathbf{k} - \mathbf{m}}{\mathbf{k}^2 + \mathbf{m}^2}. 
\qquad (5.1.11) 
$$

The four-dimensional action for Euclidean QCD is given by

$$ S[A, \psi, \bar{\psi}] = \int d^4 x \left[ \frac{1}{4} (F_{\mu\nu}^a)^2 - \bar{\psi}(x)(\mathbf{D}_x + im)\psi(x) \right] \qquad (5.1.12) $$

where we should use modified covariant kernels

$$ k_{\alpha\beta}(x, y) = (\mathbf{D}_x + im)_{\alpha\beta} \delta^4(x - y) \qquad (5.1.13) $$

Thus, the corresponding Langevin equations hold,

$$ 
\begin{aligned}
\frac{\partial}{\partial \tau} \psi_{\alpha}(x, \tau) &= -[(\mathbf{D} - im)(\mathbf{D} + im)]_{\alpha\beta} \psi_{\beta}(x, \tau) + \mathcal{G}_{\alpha}(x, \tau), \\
\frac{\partial}{\partial \tau} \bar{\psi}_{\alpha}(x, \tau) &= -\bar{\psi}_{\beta}(x, \tau)[(\mathbf{D}' - im)^{\top}(\mathbf{D}' - im)^{\top}]_{\alpha\beta} + \bar{\mathcal{G}}_{\alpha}(x, \tau),
\end{aligned} 
\qquad (5.1.14) 
$$
where

$$ \mathbf{D}_{\mu} = \partial_{\mu} - ieA_{\mu}, \qquad (5.1.15) $$

$$ \mathbf{D}'_{\mu} = -\partial_{\mu} - ieA_{\mu}, \qquad (5.1.16) $$

and for the gauge fields

$$ A_{\mu} = -\partial_{\nu}F_{\mu\nu} + e\bar{\psi}\gamma_{\mu}\psi \qquad (5.1.16) $$

The Langevin equations for the general non-Abelian case reads

$$ \frac{\partial}{\partial \tau} A_{\lambda}^{a}(x, \tau) = D_{\nu} F_{\nu\lambda}^{a}(x, \tau) + J_{\lambda}^{a}(x, \tau) + \eta_{\lambda}^{a}(x, \tau) \qquad (5.1.17) \\ J_{\lambda}^{a}(x, \tau) = \frac{1}{2} e \bar{\psi}_{\alpha}(x, \tau) \gamma^{\lambda} A_{\lambda}^{a}(x, \tau) \psi_{\alpha}(x, \tau) $$

This corresponds to the Euclidean action

$$ S[A, \psi, \bar{\psi}] = \int d^4 x \left[ \frac{1}{4} (F_{\mu\nu}^a)^2 - \bar{\psi}(x) (i\mathcal{D} - m)\psi(x) \right] \qquad (5.1.18) $$

with the standard notation

$$ F_{\lambda\nu}^{a}(x, \tau) = \partial_{\lambda}A_{\nu}^{a}(x, \tau) - \partial_{\nu}A_{\lambda}^{a}(x, \tau) - f^{abc}A_{\lambda}^{b}(x, \tau)A_{\nu}^{c}(x, \tau), \qquad (5.1.19) \\ \mathcal{D} = \partial - ieA. $$

The stochastic expectation values for the $\eta_{\lambda}^{a}(x, \tau)$ fields are given by

$$ \langle \eta_{\lambda}^{a}(x, \tau) \eta_{\nu}^{b}(x', \tau') \rangle_{\eta} = 2\delta^{ab}\delta_{\lambda\nu}\delta^{c}(x-x')\delta(\tau-\tau'). \qquad (5.1.20) $$

## 5.2. Double stochastic quantization of the Euklidean QED with massive photon.

The Euklidean action for QED with massive photon reads

$$ S = \int d^4 x \left[ \frac{1}{4} F^{\mu\nu} F_{\mu\nu} + \frac{1}{2} m_{\text{reg}}^2 A^{\mu} A_{\mu} - i \bar{\psi} (\partial - i e A + i m) \psi \right] \qquad (5.2.1) $$

where $F_{\mu\nu} = \partial_{\mu}A_{\nu} - \partial_{\nu}A_{\mu}$. The Langevin equations reads

$$ \frac{\partial \psi}{\partial t} = \int d^4 y K(x, y) [i(\mathcal{D} + im)\psi + \eta_1](y, t) + \eta_2(x, t), \\ \frac{\partial \bar{\psi}}{\partial t} = \int d^4 y \bar{K}(x, y)^{\dagger} [i(\mathcal{D}' + im)\bar{\psi} + \bar{\eta}_2](y, t) + \bar{\eta}_1(x, t), \qquad (5.2.2) $$

$$ \frac{\partial A_{\mu}}{\partial t} = \partial_{\nu} F_{\nu\mu} - m_{\text{reg}}^2 A_{\mu} + e \bar{\psi} \gamma_{\mu} \psi + \eta_{\mu}, $$

$$ \mathbf{D}_{\mu} = \partial_{\mu} - ieA_{\mu}, \quad \mathbf{D}'_{\mu} = -\partial_{\mu} - ieA_{\mu}, $$

where $\top$ means matrix transposition and $\eta_1 = \text{diag}\{\eta_{1,0}, \eta_{1,1}, \eta_{1,2}, \eta_{1,3}\}$, $\eta_2 = \text{diag}\{\eta_{2,0}, \eta_{2,1}, \eta_{2,2}, \eta_{2,3}\}$ are diagonal matrix and where $\bar{\eta}_1, \bar{\eta}_2$ its Dirac adjoint is defined as $\bar{\eta}_1 = \eta_1^\dagger \gamma_0, \bar{\eta}_2 = \eta_2^\dagger \gamma_0$.
$$
\begin{align}
\langle \eta_1(x,t) \rangle &= \langle \bar{\eta}_2(x,t) \rangle = 0 \nonumber \\
\langle \eta_i(x,t) \bar{\eta}_k(x',t') \rangle &= 2\delta_{ik}\delta^4(x-x')\delta(t-t') \tag{5.2.3} \\
i &= 1,2; k = 1,2 \nonumber
\end{align}
$$

From Eq.(5.2.2) one obtains

$$
\begin{align}
\frac{\partial \psi(x,t)}{\partial t} &= -(\mathbf{D} - im)(\mathbf{D} + im)\psi(x,t) + i(\mathbf{D} - im)\eta_1(x,t) + \eta_2(x,t) \nonumber \\
\frac{\partial \bar{\psi}(x,t)}{\partial t} &= -(\mathbf{D}' - im)^{\top}(\mathbf{D}' + im)^{\top}\bar{\psi}(x,t) + i(\mathbf{D}' - im)^{\top}\bar{\eta}_2(x,t) + \bar{\eta}_1(x,t) \tag{5.2.4} \\
\frac{\partial A_{\mu}(x,t)}{\partial t} &= \partial_{\nu}F_{\nu\mu} - m_{reg}^2 A_{\mu}(x,t) + e\bar{\psi}(x,t)\gamma_{\mu}\psi(x,t) + \eta_{\mu}(x,t) \nonumber
\end{align}
$$

The Fourier transform is defined as

$$
\begin{align}
\mathcal{F}[f(\mathbf{x})](\mathbf{k}) &= \int f(\mathbf{x}) e^{-i\mathbf{k}\cdot\mathbf{x}} d^4\mathbf{x} = g(\mathbf{k}), \tag{5.2.4'} \\
\mathcal{F}^{-1}[g(\mathbf{k})](\mathbf{x}) &= (2\pi)^{-4} \int g(\mathbf{k}) e^{-i\mathbf{k}\cdot\mathbf{x}} d^4\mathbf{k}. \nonumber
\end{align}
$$

After standard computations, one finds the following set of integral equations,
corresponding to (5.2.4) in momentum space, as:

$$
\begin{align}
\psi(\mathbf{k}, t) &= \int_0^t d\tau e^{-(\mathbf{k}^2+m^2)(t-\tau)} \left\{ (\mathbf{k}+m)\eta_1(\mathbf{k}, \tau) + \eta_2(\mathbf{k}, \tau) + \right. \nonumber \\
&\qquad \left. + \frac{e}{(2\pi)^4} \int d^4\mathbf{p} \int d^4\mathbf{q} \delta(\mathbf{k}-\mathbf{p}-\mathbf{q}) \times \right. \nonumber \\
&\qquad \left. \times [(\not\mathbf{p}\mathbf{A}(\mathbf{p}, \tau) - 2\mathbf{q}_\nu \mathbf{A}_\nu(\mathbf{p}, \tau))\psi(\mathbf{q}, \tau) + \mathbf{A}(\mathbf{p}, \tau)\eta_1(\mathbf{q}, \tau)] - \right. \nonumber \\
&\qquad \left. - \frac{e^2}{(2\pi)^8} \int d^4\mathbf{p} \int d^4\mathbf{q} \int d^4\mathbf{r} \delta(\mathbf{k}-\mathbf{p}-\mathbf{q}-\mathbf{r}) \mathbf{A}_\nu(\mathbf{p}, \tau) \mathbf{A}_\nu(\mathbf{q}, \tau) \psi(\mathbf{r}, \tau) \right\} = \nonumber \\
&= \int_0^t d\tau e^{-(\mathbf{k}^2+m^2)(t-\tau)} \left\{ (\mathbf{k}+m)\eta_1(\mathbf{k}, \tau) + \eta_2(\mathbf{k}, \tau) + \right. \nonumber \\
&\qquad \left. + \frac{e}{(2\pi)^4} \int d^4\mathbf{p} \times [(\not\mathbf{p}\mathbf{A}(\mathbf{p}, \tau) - 2(\mathbf{k}_\nu - \mathbf{p}_\nu)\mathbf{A}_\nu(\mathbf{p}, \tau))\psi(\mathbf{k}-\mathbf{p}, \tau) + \mathbf{A}(\mathbf{p}, \tau)\eta_1(\mathbf{k}-\mathbf{p}, \tau)] \right. \nonumber \\
&\qquad \left. - \frac{e^2}{(2\pi)^8} \int d^4\mathbf{p} \int d^4\mathbf{q} \mathbf{A}_\nu(\mathbf{p}, \tau) \mathbf{A}_\nu(\mathbf{q}, \tau) \psi(\mathbf{k}-\mathbf{p}-\mathbf{q}, \tau) \right\}. \tag{5.2.5}
\end{align}
$$
$$
\begin{equation}
\begin{aligned}
\psi(\mathbf{k}, t) = & \int_0^t d\tau e^{-(\mathbf{k}^2+m^2)(t-\tau)} \{ (\mathbf{k}+m)\eta_1(\mathbf{k}, \tau) + \eta_2(\mathbf{k}, \tau) + \\
& \qquad \qquad \qquad \qquad \qquad + \frac{e}{(2\pi)^4} \int d^4\mathbf{p} \times \\
& \qquad \qquad \qquad \qquad \qquad \times [(\mathbf{p}\mathbf{A}(\mathbf{p}, \tau) - 2(\mathbf{k}_v - \mathbf{p}_v)\mathbf{A}_v(\mathbf{p}, \tau))\psi(\mathbf{k}-\mathbf{p}, \tau) + \mathbf{A}(\mathbf{p}, \tau)\eta_1(\mathbf{k}-\mathbf{p}, \tau)] \\
& \qquad \qquad \qquad \qquad \qquad - \frac{e^2}{(2\pi)^8} \int d^4\mathbf{p} \int d^4\mathbf{q} \mathbf{A}_v(\mathbf{p}, \tau)\mathbf{A}_v(\mathbf{q}, \tau)\psi(\mathbf{k}-\mathbf{p}-\mathbf{q}, \tau) \}
\end{aligned}
\tag{5.2.5}
\end{equation}
$$

$$
\begin{equation}
\begin{aligned}
\bar{\psi}(\mathbf{k}, t) = & \int_0^t d\tau e^{-(\mathbf{k}^2+m^2)(t-\tau)} \{ (\mathbf{k}+m)\eta_2(\mathbf{k}, \tau) + \bar{\eta}_1(\mathbf{k}, \tau) + \\
& \qquad + \frac{e}{(2\pi)^4} \int d^4\mathbf{p} \int d^4\mathbf{q} \delta(\mathbf{k}-\mathbf{p}-\mathbf{q}) \times \\
& \qquad \times [\bar{\psi}(\mathbf{q}, \tau)(-\mathbf{A}(\mathbf{p}, \tau)\mathbf{p} - 2q_v A_v(\mathbf{p}, \tau)) + \bar{\eta}_2(\mathbf{k}, \tau)\mathbf{A}(\mathbf{p}, \tau)] - \\
& \qquad - \frac{e^2}{(2\pi)^8} \int d^4\mathbf{p} \int d^4\mathbf{q} \delta(-\mathbf{k}-\mathbf{p}-\mathbf{q}+\mathbf{r}) A_v(\mathbf{p}, \tau)A_v(\mathbf{q}, \tau)\bar{\psi}(\mathbf{r}, \tau) = \\
& \qquad \int_0^t d\tau e^{-(\mathbf{k}^2+m^2)(t-\tau)} \{ (\mathbf{k}+m)\eta_2(\mathbf{k}, \tau) + \eta_1(\mathbf{k}, \tau) + \\
& \qquad + \frac{e}{(2\pi)^4} \int d^4\mathbf{p} [\bar{\psi}(\mathbf{k}-\mathbf{p}, \tau)(-\mathbf{A}(\mathbf{p}, \tau)\mathbf{p} - 2(\mathbf{k}_v - \mathbf{p}_v)\mathbf{A}_v(\mathbf{p}, \tau)) + \bar{\eta}_2(\mathbf{k}, \tau)\mathbf{A}(\mathbf{p}, \tau)] - \\
& \qquad - \frac{e^2}{(2\pi)^8} \int d^4\mathbf{p} \int d^4\mathbf{q} A_v(\mathbf{p}, \tau)A_v(\mathbf{q}, \tau)\bar{\psi}(\mathbf{k}+\mathbf{p}+\mathbf{q}, \tau) \}
\end{aligned}
\tag{5.2.6}
\end{equation}
$$

$$
\begin{equation}
\begin{aligned}
\bar{\psi}(\mathbf{k}, t) = & \int_0^t d\tau e^{-(\mathbf{k}^2+m^2)(t-\tau)} \{ (\mathbf{k}+m)\eta_2(\mathbf{k}, \tau) + \bar{\eta}_1(\mathbf{k}, \tau) + \\
& \qquad + \frac{e}{(2\pi)^4} \int d^4\mathbf{p} [\bar{\psi}(\mathbf{k}-\mathbf{p}, \tau)(-\mathbf{A}(\mathbf{p}, \tau)\mathbf{p} - 2(\mathbf{k}_v - \mathbf{p}_v)\mathbf{A}_v(\mathbf{p}, \tau)) + \bar{\eta}_2(\mathbf{k}, \tau)\mathbf{A}(\mathbf{p}, \tau)] - \\
& \qquad - \frac{e^2}{(2\pi)^8} \int d^4\mathbf{p} \int d^4\mathbf{q} A_v(\mathbf{p}, \tau)A_v(\mathbf{q}, \tau)\bar{\psi}(\mathbf{k}+\mathbf{p}+\mathbf{q}, \tau) \}
\end{aligned}
\tag{5.2.6'}
\end{equation}
$$
$$
\begin{align}
A_{\mu}(\mathbf{k}, t) &= \int_{0}^{t} d\tau \left[ e^{-(\mathbf{k}^2+m_{\text{reg}}^2)(t-\tau)} T_{\mu\nu} + L_{\mu\nu} \right] \{ \eta_{\nu}(\mathbf{k}, \tau) + \nonumber \\
&\qquad + \frac{e}{(2\pi)^4} \int d^4\mathbf{p} \int d^4\mathbf{q} \delta(\mathbf{k}+\mathbf{p}-\mathbf{q}) \bar{\psi}(\mathbf{p}, t) \gamma_{\mu} \psi(\mathbf{q}, t) \} = \tag{5.2.7} \\
&\qquad \int_{0}^{t} d\tau \left[ e^{-(\mathbf{k}^2+m_{\text{reg}}^2)(t-\tau)} T_{\mu\nu}^{\text{reg}} + L_{\mu\nu}^{\text{reg}} \right] \left\{ \eta_{\nu}(\mathbf{k}, \tau) + \frac{e}{(2\pi)^4} \int d^4\mathbf{p} \bar{\psi}(\mathbf{p}, t) \gamma_{\mu} \psi(\mathbf{k}+\mathbf{p}, \tau) \right\}, \nonumber \\
T_{\mu\nu}^{\text{reg}} &= \delta_{\mu\nu} - \frac{\mathbf{k}_{\mu}\mathbf{k}_{\nu}}{\mathbf{k}^2+m_{\text{reg}}^2}, \quad L_{\mu\nu}^{\text{reg}} = \frac{\mathbf{k}_{\mu}\mathbf{k}_{\nu}}{\mathbf{k}^2+m_{\text{reg}}^2}. \nonumber
\end{align}
$$

$$
\begin{align}
A_{\mu}(\mathbf{k}, t) &= \int_{0}^{t} d\tau \left[ e^{-(\mathbf{k}^2+m_{\text{reg}}^2)(t-\tau)} T_{\mu\nu}^{\text{reg}} + L_{\mu\nu}^{\text{reg}} \right] \left\{ \eta_{\nu}(\mathbf{k}, \tau) + \frac{e}{(2\pi)^4} \int d^4\mathbf{p} \bar{\psi}(\mathbf{p}, t) \gamma_{\mu} \psi(\mathbf{k}+\mathbf{p}, \tau) \right\}, \tag{5.2.7'} \\
T_{\mu\nu}^{\text{reg}} &= \delta_{\mu\nu} - \frac{\mathbf{k}_{\mu}\mathbf{k}_{\nu}}{\mathbf{k}^2+m_{\text{reg}}^2}, \quad L_{\mu\nu}^{\text{reg}} = \frac{\mathbf{k}_{\mu}\mathbf{k}_{\nu}}{\mathbf{k}^2+m_{\text{reg}}^2}. \nonumber
\end{align}
$$

We derive Fourier transformed differential master equations directly from
the Eq.(5.2.5')-Eq.(5.2.7') By the replacement
$$
\begin{align*}
\psi(x,t) - \theta(t)\psi^* &= \psi_-(x,t), \bar{\psi}(x,t) - \theta(t)\bar{\psi}^* = \bar{\psi}_-(x,t), \\
\bar{\psi}(x,t) &= \bar{\psi}_+(x,t) + \theta(t)\psi^*, \bar{\psi}(x,t) = \bar{\psi}_+(x,t) - \theta(t)\bar{\psi}^*, \\
\psi(\mathbf{p},t) &= \mathcal{F}[\psi(x,t)](\mathbf{p},t) = \mathcal{F}[\psi_-(x,t)](\mathbf{p},t) + \theta(t)(2\pi)^4\delta(\mathbf{p})\psi^* = \\
&= \psi_-(\mathbf{p},t) + \theta(t)(2\pi)^4\delta(\mathbf{p})\psi^*, \\
\bar{\psi}(\mathbf{p},t) &= \mathcal{F}[\bar{\psi}(x,t)](\mathbf{p},t) = \mathcal{F}[\bar{\psi}_+(x,t)](\mathbf{p},t) - \theta(t)(2\pi)^4\delta(\mathbf{p})\bar{\psi}^* = \\
&= \bar{\psi}_+(\mathbf{p},t) - \theta(t)(2\pi)^4\delta(\mathbf{p})\bar{\psi}^*, \\
\frac{\partial\psi(x,t)}{\partial t} &= \frac{\partial\psi_-(x,t)}{\partial t} + \delta(t)\psi^*, \\
\frac{\partial\psi(\mathbf{p},t)}{\partial t} &= \frac{\partial\psi_-(\mathbf{p},t)}{\partial t} + \delta(t)(2\pi)^4\delta(\mathbf{p})\psi^*, \\
\frac{\partial\bar{\psi}(x,t)}{\partial t} &= \frac{\partial\bar{\psi}_-(x,t)}{\partial t} + \delta(t)\bar{\psi}^*, \\
\frac{\partial\bar{\psi}(\mathbf{p},t)}{\partial t} &= \frac{\partial\bar{\psi}_+(\mathbf{p},t)}{\partial t} - \delta(t)(2\pi)^4\delta(\mathbf{p})\bar{\psi}^*, \tag{5.2.8} \\
A_\mu(x,t) - \theta(t)a_\mu &= A_{\mu-}(x,t), \\
A_\mu(x,t) &= A_{\mu-}(x,t) + \theta(t)a_\mu, \\
A_\mu(\mathbf{p},t) &= \mathcal{F}[A_{\mu-}(x,t)](\mathbf{p},t) = \mathcal{F}[A_{\mu-}(x,t)](\mathbf{p},t) + \theta(t)(2\pi)^4\delta(\mathbf{p})a_\mu = \\
&= A_{\mu-}(\mathbf{p},t) + \theta(t)(2\pi)^4a_\mu, \\
A_\mu(x,t) + \theta(t)a_\mu &= A_{\mu+}(x,t), \\
A_\mu(x,t) &= A_{\mu+}(x,t) - \theta(t)a_\mu, \\
\frac{\partial A_\mu(x,t)}{\partial t} &= \frac{\partial A_{\mu-}(x,t)}{\partial t} + \delta(t)a_\mu, \\
\frac{\partial A_\mu(\mathbf{p},t)}{\partial t} &= \frac{\partial A_{\mu-}(\mathbf{p},t)}{\partial t} + \delta(t)(2\pi)^4\delta(\mathbf{p})a_\mu, \\
\frac{\partial A_\mu(x,t)}{\partial t} &= \frac{\partial A_{\mu+}(x,t)}{\partial t} - \delta(t)a_\mu, \\
\frac{\partial A_\mu(\mathbf{p},t)}{\partial t} &= \frac{\partial A_{\mu+}(\mathbf{p},t)}{\partial t} - \delta(t)(2\pi)^4\delta(\mathbf{p})a_\mu.
\end{align*}
$$

from Eq.(5.2.4) one obtains
$$
\begin{align}
\frac{\partial \psi_-(x,t)}{\partial t} &= -\delta(t)\psi^* - (\underline{\mathbf{D}} - im)(\underline{\mathbf{D}} + im)[\psi_-(x,t) + \psi^*] + i(\underline{\mathbf{D}} - im)\eta_1(x,t) + \eta_2(x,t) \nonumber \\
\frac{\partial \bar{\psi}_-(x,t)}{\partial t} &= -\delta(t)\bar{\psi}^* - (\underline{\mathbf{D}}' - im)^\top(\underline{\mathbf{D}}' + im)^\top[\bar{\psi}_-(x,t) + \bar{\psi}^*] + i(\underline{\mathbf{D}}' - im)^\top\bar{\eta}_2(x,t) + \bar{\eta}_1(x,t) \tag{5.2.9} \\
\frac{\partial A_{\mu-}(x,t)}{\partial t} &= -\delta(t)a_\mu - \partial_v F_{v\mu-} - m_{reg}^2[A_{\mu-}(x,t) + a_\mu] + e[\bar{\psi}_-(x,t) + \bar{\psi}^*]\gamma_\mu[\psi_-(x,t) + \bar{\psi}^*] + \eta_\mu(x,t) \nonumber \\
\underline{\mathbf{D}}_\mu &= \partial_\mu - ie(A_{\mu-} + a_\mu), \quad \underline{\mathbf{D}}'_\mu = -\partial_\mu - ie(A_{\mu-} + a_\mu), \nonumber
\end{align}
$$

From Eq.(5.2.9) and Eq.(5.2.5') one obtains

$$
\begin{align*}
\psi_-(\mathbf{k}, t) &= -(2\pi)^4 \delta(\mathbf{k}) \psi^* \int_0^t d\tau \delta(\tau) e^{-(\mathbf{k}^2+m^2)(t-\tau)} + \\
&\quad + \int_0^t d\tau e^{-(\mathbf{k}^2+m^2)(t-\tau)} \{(\mathbf{k}+m)\eta_1(\mathbf{k}, \tau) + \eta_2(\mathbf{k}, \tau) + \\
&\quad + \frac{e}{(2\pi)^4} \int d^4\mathbf{p} \left[ (\mathbf{p}\mathbf{A}_-(\mathbf{p}, \tau) + (2\pi)^4 \delta(\mathbf{p})\mathbf{a}_\mu) - 2(\mathbf{k}_\nu - \mathbf{p}_\nu)(\mathbf{A}_{\nu-}(\mathbf{p}, \tau) + (2\pi)^4 \delta(\mathbf{p})\mathbf{a}_\nu) \right] \times \\
&\quad \times [\psi_-(\mathbf{k}-\mathbf{p}, \tau) + (2\pi)^4 \delta(\mathbf{k}-\mathbf{p})\psi^*] + (\mathbf{A}_-(\mathbf{p}, \tau) + (2\pi)^4 \delta(\mathbf{p})\mathbf{a}_\mu)\eta_1(\mathbf{k}-\mathbf{p}, \tau)] - \\
&\quad - \frac{e^2}{(2\pi)^8} \int d^4\mathbf{p} \int d^4\mathbf{q} \times \\
&\quad (\mathbf{A}_{\nu-}(\mathbf{p}, \tau) + (2\pi)^4 \delta(\mathbf{p})\mathbf{a}_\mu)(\mathbf{A}_{\nu-}(\mathbf{q}, \tau) + (2\pi)^4 \delta(\mathbf{p})\mathbf{a}_\mu)[\psi_-(\mathbf{k}-\mathbf{p}-\mathbf{q}, \tau) + (2\pi)^4 \delta(\mathbf{k}-\mathbf{p}-\mathbf{q})\psi^*] \} \\
&= -(2\pi)^4 \delta(\mathbf{k}) e^{-(\mathbf{k}^2+m^2)t} \psi^* + \int_0^t d\tau e^{-(\mathbf{k}^2+m^2)(t-\tau)} \{(\mathbf{k}+m)\eta_1(\mathbf{k}, \tau) + \eta_2(\mathbf{k}, \tau) + \\
&\quad + \frac{e}{(2\pi)^4} \int d^4\mathbf{p} \times [(\mathbf{p}\mathbf{A}_-(\mathbf{p}, \tau) + (2\pi)^4 \delta(\mathbf{p})\mathbf{p})\mathbf{a}_\mu - 2(\mathbf{k}_\nu - \mathbf{p}_\nu)\mathbf{A}_{\nu-}(\mathbf{p}, \tau) - 2(2\pi)^4(\mathbf{k}_\nu - \mathbf{p}_\nu)\delta(\mathbf{p})\mathbf{a}_\nu] \times \\
&\quad \times [\psi_-(\mathbf{k}-\mathbf{p}, \tau) + (2\pi)^4 \delta(\mathbf{k}-\mathbf{p})\psi^*] + (\mathbf{A}_-(\mathbf{p}, \tau) + (2\pi)^4 \delta(\mathbf{p})\mathbf{a}_\mu)\eta_1(\mathbf{k}-\mathbf{p}, \tau)] - \\
&\quad - \frac{e^2}{(2\pi)^8} \int d^4\mathbf{p} \int d^4\mathbf{q} (\mathbf{A}_{\nu-}(\mathbf{p}, \tau) + (2\pi)^4 \delta(\mathbf{p})\mathbf{a}_\nu)(\mathbf{A}_{\nu-}(\mathbf{q}, \tau) + (2\pi)^4 \delta(\mathbf{q})\mathbf{a}_\nu) \times \\
&\quad \times (\psi_-(\mathbf{k}-\mathbf{p}-\mathbf{q}, \tau) + (2\pi)^4 \delta(\mathbf{k}-\mathbf{p}-\mathbf{q})\psi^*) \}
\end{align*}
$$

Note that
$$
\begin{align}
& (\mathbf{p}\mathbf{A}_{-}(\mathbf{p}, \tau) + (2\pi)^4 \delta(\mathbf{p})\mathbf{p}\mathbf{a}_{\mu} - 2(\mathbf{k}_{\nu} - \mathbf{p}_{\nu})\mathbf{A}_{\nu-}(\mathbf{p}, \tau) - 2(2\pi)^4(\mathbf{k}_{\nu} - \mathbf{p}_{\nu})\delta(\mathbf{p})a_{\nu}) \nonumber \\
& \times [\psi_{-}(\mathbf{k}-\mathbf{p}, \tau) + (2\pi)^4 \delta(\mathbf{k}-\mathbf{p})\psi^*] + (\mathbf{A}_{-}(\mathbf{p}, \tau) + (2\pi)^4 \delta(\mathbf{p})\mathbf{a}_{\mu})\eta_1(\mathbf{k}-\mathbf{p}, \tau) = \nonumber \\
& = (\mathbf{p}\mathbf{A}_{-}(\mathbf{p}, \tau) + (2\pi)^4 \delta(\mathbf{p})\mathbf{p}\mathbf{a}_{\mu})[\psi_{-}(\mathbf{k}-\mathbf{p}, \tau) + (2\pi)^4 \delta(\mathbf{k}-\mathbf{p})\psi^*] - \nonumber \\
& -(2(\mathbf{k}_{\nu} - \mathbf{p}_{\nu})\mathbf{A}_{\nu-}(\mathbf{p}, \tau) + 2(2\pi)^4(\mathbf{k}_{\nu} - \mathbf{p}_{\nu})\delta(\mathbf{p})a_{\nu})[\psi_{-}(\mathbf{k}-\mathbf{p}, \tau) + (2\pi)^4 \delta(\mathbf{k}-\mathbf{p})\psi^*] = \nonumber \\
& \underline{\mathbf{p}\mathbf{A}_{-}(\mathbf{p}, \tau)\psi_{-}(\mathbf{k}-\mathbf{p}, \tau)} + \nonumber \\
& +(2\pi)^4 \delta(\mathbf{p})\mathbf{p}\mathbf{a}_{\mu}\psi_{-}(\mathbf{k}-\mathbf{p}, \tau) + (2\pi)^4 \delta(\mathbf{k}-\mathbf{p})\mathbf{p}\mathbf{A}_{-}(\mathbf{p}, \tau)\psi^* - \tag{5.2.11} \\
& -2(\mathbf{k}_{\nu} - \mathbf{p}_{\nu})\mathbf{A}_{\nu-}(\mathbf{p}, \tau)\psi_{-}(\mathbf{k}-\mathbf{p}, \tau) - \nonumber \\
& -2(2\pi)^4(\mathbf{k}_{\nu} - \mathbf{p}_{\nu})\delta(\mathbf{p})a_{\nu}\psi_{-}(\mathbf{k}-\mathbf{p}, \tau) - 2(2\pi)^4(\mathbf{k}_{\nu} - \mathbf{p}_{\nu})\mathbf{A}_{\nu-}(\mathbf{p}, \tau)\delta(\mathbf{k}-\mathbf{p})\psi^* - \nonumber \\
& -2(2\pi)^4(\mathbf{k}_{\nu} - \mathbf{p}_{\nu})\delta(\mathbf{p})a_{\nu}\delta(\mathbf{k}-\mathbf{p})\psi^* + (2\pi)^4 \delta(\mathbf{p})\mathbf{p}\mathbf{a}_{\mu}\delta(\mathbf{k}-\mathbf{p})\psi^* + \nonumber \\
& +\mathbf{A}_{-}(\mathbf{p}, \tau)\eta_1(\mathbf{k}-\mathbf{p}, \tau) + (2\pi)^4 \delta(\mathbf{p})\mathbf{a}_{\mu}\eta_1(\mathbf{k}-\mathbf{p}, \tau) \nonumber
\end{align}
$$

$$
\begin{align}
\Delta_1(\mathbf{k}, \mathbf{p}, \tau, e) &= \underline{\mathbf{p}\mathbf{A}_{-}(\mathbf{p}, \tau)\psi_{-}(\mathbf{k}-\mathbf{p}, \tau)} - 2(\mathbf{k}_{\nu}-\mathbf{p}_{\nu})\mathbf{A}_{\nu-}(\mathbf{p}, \tau)\psi_{-}(\mathbf{k}-\mathbf{p}, \tau) \nonumber \\
\Delta_2(\mathbf{k}, \mathbf{p}, \tau, e) &= (2\pi)^4 \delta(\mathbf{p})\underline{\mathbf{p}\mathbf{a}_{\mu}\psi_{-}(\mathbf{k}-\mathbf{p}, \tau)} + (2\pi)^4 \delta(\mathbf{k}-\mathbf{p})\underline{\mathbf{p}\mathbf{A}_{-}(\mathbf{p}, \tau)\psi^*} - \tag{5.2.12} \\
&\quad -2(2\pi)^4(\mathbf{k}_{\nu}-\mathbf{p}_{\nu})\delta(\mathbf{p})a_{\nu}\delta(\mathbf{k}-\mathbf{p})\psi^* + (2\pi)^4 \delta(\mathbf{p})\underline{\mathbf{p}\mathbf{a}_{\mu}\delta(\mathbf{k}-\mathbf{p})\psi^*} + \nonumber \\
&\quad +\mathbf{A}_{-}(\mathbf{p}, \tau)\eta_1(\mathbf{k}-\mathbf{p}, \tau) + (2\pi)^4 \delta(\mathbf{p})\mathbf{a}_{\mu}\eta_1(\mathbf{k}-\mathbf{p}, \tau) \nonumber
\end{align}
$$

Note that

$$
\begin{align}
& (\mathbf{A}_v(\mathbf{p}, \tau) + (2\pi)^4 \delta(\mathbf{p}) a_v) (\mathbf{A}_v(\mathbf{q}, \tau) + (2\pi)^4 \delta(\mathbf{q}) a_v) \times \nonumber \\
& \qquad \times (\psi_{-}(\mathbf{k}-\mathbf{p}-\mathbf{q}, \tau) + (2\pi)^4 \delta(\mathbf{k}-\mathbf{p}-\mathbf{q}) \psi^*) = \nonumber \\
&= [\mathbf{A}_v(\mathbf{p}, \tau) \mathbf{A}_v(\mathbf{q}, \tau) + (2\pi)^4 \delta(\mathbf{p}) a_v \mathbf{A}_v(\mathbf{q}, \tau) + (2\pi)^4 \delta(\mathbf{q}) a_v \mathbf{A}_v(\mathbf{p}, \tau) + (2\pi)^8 \delta(\mathbf{p}) \delta(\mathbf{q}) a_v^2] \times \nonumber \\
& \qquad \times [\psi_{-}(\mathbf{k}-\mathbf{p}-\mathbf{q}, \tau) + (2\pi)^4 \delta(\mathbf{k}-\mathbf{p}-\mathbf{q}) \psi^*] = \nonumber \\
&\underline{\mathbf{A}_v(\mathbf{p}, \tau) \mathbf{A}_v(\mathbf{q}, \tau) \psi_{-}(\mathbf{k}-\mathbf{p}-\mathbf{q}, \tau)} + (2\pi)^4 \delta(\mathbf{p}) a_v \underline{\mathbf{A}_v(\mathbf{q}, \tau) \psi_{-}(\mathbf{k}-\mathbf{p}-\mathbf{q}, \tau)} + \nonumber \\
&\quad +(2\pi)^4 \delta(\mathbf{q}) a_v \underline{\mathbf{A}_v(\mathbf{p}, \tau) \psi_{-}(\mathbf{k}-\mathbf{p}-\mathbf{q}, \tau)} + \nonumber \\
&\quad \# + (2\pi)^8 \delta(\mathbf{p}) \delta(\mathbf{q}) a_v^2 \psi_{-}(\mathbf{k}-\mathbf{p}-\mathbf{q}, \tau) + \nonumber \\
&+ [ (2\pi)^4 \delta(\mathbf{p}) a_v \mathbf{A}_v(\mathbf{q}, \tau) + (2\pi)^4 \delta(\mathbf{q}) a_v \mathbf{A}_v(\mathbf{p}, \tau) + (2\pi)^8 \delta(\mathbf{p}) \delta(\mathbf{q}) a_v^2 ] (2\pi)^4 \delta(\mathbf{k}-\mathbf{p}-\mathbf{q}) \psi^* = \nonumber \\
&= (2\pi)^8 \delta(\mathbf{p}) \delta(\mathbf{q}) a_v^2 \psi_{-}(\mathbf{k}-\mathbf{p}-\mathbf{q}, \tau) + \nonumber \\
&\quad +(2\pi)^8 \delta(\mathbf{p}) \delta(\mathbf{k}-\mathbf{p}-\mathbf{q}) \psi^* a_v \mathbf{A}_v(\mathbf{q}, \tau) + (2\pi)^8 \delta(\mathbf{q}) \delta(\mathbf{k}-\mathbf{p}-\mathbf{q}) \psi^* a_v \mathbf{A}_v(\mathbf{p}, \tau) + \nonumber \\
&\quad +(2\pi)^{12} \delta(\mathbf{p}) \delta(\mathbf{q}) \delta(\mathbf{k}-\mathbf{p}-\mathbf{q}) \psi^* a_v^2 \nonumber
\end{align}
$$
$$
\begin{equation}
\begin{aligned}
\Delta_3(\mathbf{k}, \mathbf{p}, \tau, e) &= A_\nu(\mathbf{p}, \tau)A_\nu(\mathbf{q}, \tau)\psi_-(\mathbf{k}-\mathbf{p}-\mathbf{q}, \tau) + \\
&\quad +(2\pi)^4\delta(\mathbf{p})a_\nu A_\nu(\mathbf{q}, \tau)\psi_-(\mathbf{k}-\mathbf{p}-\mathbf{q}, \tau) + (2\pi)^4\delta(\mathbf{q})a_\nu A_\nu(\mathbf{p}, \tau)\psi_-(\mathbf{k}-\mathbf{p}-\mathbf{q}, \tau) + \\
&\quad + (2\pi)^8\delta(\mathbf{p})\delta(\mathbf{k}-\mathbf{p}-\mathbf{q})\psi^*a_\nu A_\nu(\mathbf{q}, \tau) + (2\pi)^8\delta(\mathbf{q})\delta(\mathbf{k}-\mathbf{p}-\mathbf{q})\psi^*a_\nu A_\nu(\mathbf{p}, \tau) + \\
&\quad +(2\pi)^{12}\delta(\mathbf{p})\delta(\mathbf{q})\delta(\mathbf{k}-\mathbf{p}-\mathbf{q})\psi^*a_\nu^2.
\end{aligned}
\tag{5.2.14}
\end{equation}
$$

We rewrite Eq.(2.5.10) of the short form

$$
\begin{equation}
\begin{aligned}
\psi_{-}(\mathbf{k}, t) = & - (2\pi)^4 \delta(\mathbf{k}) \psi^* e^{-(\mathbf{k}^2+m^2)t} + \int_0^t d\tau e^{-(\mathbf{k}^2+m^2)(t-\tau)} \{ (\mathbf{k}+m)\eta_1(\mathbf{k}, \tau) + \eta_2(\mathbf{k}, \tau) + \\
& \qquad + \frac{e}{(2\pi)^4} \int d^4\mathbf{p} [\Delta_1(\mathbf{k}, \mathbf{p}, \tau, e) + \Delta_2(\mathbf{k}, \mathbf{p}, \tau, e)] - \\
& \qquad - \frac{e^2}{(2\pi)^8} \int d^4\mathbf{p} \int d^4\mathbf{q} [\Delta_3(\mathbf{k}, \mathbf{p}, \tau, e) + \Delta_4(\mathbf{k}, \mathbf{p}, \tau, e)] \}
\end{aligned}
\tag{5.2.15}
\end{equation}
$$

From Eq.(5.2.6') by the Eq.(5.2.8) we obtain

$$
\begin{equation}
\begin{aligned}
\bar{\psi}_{-}(\mathbf{k}, t) = & \int_{0}^{t} d\tau e^{-(\mathbf{k}^2+m^2)(t-\tau)} \{ (\mathbf{k}+m)\bar{\eta}_2(\mathbf{k}, \tau) + \bar{\eta}_1(\mathbf{k}, \tau) + \\
& \qquad + \frac{e}{(2\pi)^4} \int d^4\mathbf{p} \left[ (\bar{\psi}(\mathbf{k}-\mathbf{p}, \tau) + (2\pi)^4\delta(\mathbf{k}-\mathbf{p})) \times \right. \\
& \qquad \left. \left( (-\mathbf{A}(\mathbf{p}, \tau)\mathbf{p}) - (2\pi)^4\delta(\mathbf{p})\mathbf{p} \right) - 2(\mathbf{k}_v - \mathbf{p}_v)\mathbf{A}_v(\mathbf{p}, \tau) - 2(2\pi)^4(\mathbf{k}_v - \mathbf{p}_v)\delta(\mathbf{p}) \right) \right. \\
& \qquad \left. + \bar{\eta}_2(\mathbf{k}+\mathbf{p}, \tau) (\mathbf{A}(\mathbf{p}, \tau) + (2\pi)^4\delta(\mathbf{p})) \right] - \\
& \qquad - \frac{e^2}{(2\pi)^8} \int d^4\mathbf{p} \int d^4\mathbf{q} \left( A_v(\mathbf{p}, \tau) + (2\pi)^4\delta(\mathbf{p}) \right) \left( A_v(\mathbf{q}, \tau) + (2\pi)^4\delta(\mathbf{q}) \right) \times \\
& \qquad \times (\bar{\psi}(\mathbf{k}+\mathbf{p}+\mathbf{q}, \tau) + (2\pi)^4\delta(\mathbf{k}+\mathbf{p}+\mathbf{q})) \}.
\end{aligned}
\tag{5.2.16}
\end{equation}
$$

Note that
$$
\begin{align}
&(-\mathbf{A}_{-}(\mathbf{p}, \tau)\mathbf{p} + (2\pi)^4 \delta(\mathbf{p})\mathbf{p}\dot{a}_{\mu} - 2(\mathbf{k}_{\nu} - \mathbf{p}_{\nu})\mathbf{A}_{\nu-}(\mathbf{p}, \tau) - 2(2\pi)^4(\mathbf{k}_{\nu} - \mathbf{p}_{\nu})\delta(\mathbf{p})a_{\nu}) \nonumber \\
&\quad \times[\psi_{-}(\mathbf{k} - \mathbf{p}, \tau) + (2\pi)^4 \delta(\mathbf{k} - \mathbf{p})\psi^{*}] + (\mathbf{A}_{-}(\mathbf{p}, \tau) + (2\pi)^4 \delta(\mathbf{p})\dot{a}_{\mu})\eta_1(\mathbf{k} - \mathbf{p}, \tau) = \nonumber \\
&= (\mathbf{p}\mathbf{A}_{-}(\mathbf{p}, \tau) + (2\pi)^4 \delta(\mathbf{p})\mathbf{p}\dot{a}_{\mu})[\psi_{-}(\mathbf{k} - \mathbf{p}, \tau) + (2\pi)^4 \delta(\mathbf{k} - \mathbf{p})\psi^{*}] - \nonumber \\
&- (2(\mathbf{k}_{\nu} - \mathbf{p}_{\nu})\mathbf{A}_{\nu-}(\mathbf{p}, \tau) + 2(2\pi)^4(\mathbf{k}_{\nu} - \mathbf{p}_{\nu})\delta(\mathbf{p})a_{\nu})[\psi_{-}(\mathbf{k} - \mathbf{p}, \tau) + (2\pi)^4 \delta(\mathbf{k} - \mathbf{p})\psi^{*}] = \nonumber \\
&\quad \underline{\mathbf{p}\mathbf{A}_{-}(\mathbf{p}, \tau)\psi_{-}(\mathbf{k} - \mathbf{p}, \tau)} + \nonumber \\
&\quad +(2\pi)^4 \delta(\mathbf{p})\mathbf{p}\dot{a}_{\mu}\psi_{-}(\mathbf{k} - \mathbf{p}, \tau) + (2\pi)^4 \delta(\mathbf{k} - \mathbf{p})\mathbf{p}\mathbf{A}_{-}(\mathbf{p}, \tau)\psi^{*} - \tag{5.2.11} \\
&\quad -2(\mathbf{k}_{\nu} - \mathbf{p}_{\nu})\mathbf{A}_{\nu-}(\mathbf{p}, \tau)\psi_{-}(\mathbf{k} - \mathbf{p}, \tau) - \nonumber \\
&\quad -2(2\pi)^4(\mathbf{k}_{\nu} - \mathbf{p}_{\nu})\delta(\mathbf{p})a_{\nu}\psi_{-}(\mathbf{k} - \mathbf{p}, \tau) - 2(2\pi)^4(\mathbf{k}_{\nu} - \mathbf{p}_{\nu})\mathbf{A}_{\nu-}(\mathbf{p}, \tau)\delta(\mathbf{k} - \mathbf{p})\psi^{*} - \nonumber \\
&\quad -2(2\pi)^4(\mathbf{k}_{\nu} - \mathbf{p}_{\nu})\delta(\mathbf{p})a_{\nu}\delta(\mathbf{k} - \mathbf{p})\psi^{*} + (2\pi)^4 \delta(\mathbf{p})\mathbf{p}\dot{a}_{\mu}\delta(\mathbf{k} - \mathbf{p})\psi^{*} + \nonumber \\
&\quad +\mathbf{A}_{-}(\mathbf{p}, \tau)\eta_1(\mathbf{k} - \mathbf{p}, \tau) + (2\pi)^4 \delta(\mathbf{p})\dot{a}_{\mu}\eta_1(\mathbf{k} - \mathbf{p}, \tau) \nonumber
\end{align}
$$

Note that

$$
\begin{align}
& (\mathbf{A}_v(\mathbf{p}, \tau) + (2\pi)^4 \delta(\mathbf{p})a_v) (\mathbf{A}_v(\mathbf{q}, \tau) + (2\pi)^4 \delta(\mathbf{q})a_v) \times \nonumber \\
& \quad \times (\psi_{-}(\mathbf{k} - \mathbf{p} - \mathbf{q}, \tau) + (2\pi)^4 \delta(\mathbf{k} - \mathbf{p} - \mathbf{q})\psi^*) = \nonumber \\
&= [\mathbf{A}_v(\mathbf{p}, \tau)\mathbf{A}_v(\mathbf{q}, \tau) + (2\pi)^4 \delta(\mathbf{p})a_v\mathbf{A}_v(\mathbf{q}, \tau) + (2\pi)^4 \delta(\mathbf{q})a_v\mathbf{A}_v(\mathbf{p}, \tau) + (2\pi)^8 \delta(\mathbf{p})\delta(\mathbf{q})a_v^2] \times \nonumber \\
& \quad \times [\psi_{-}(\mathbf{k} - \mathbf{p} - \mathbf{q}, \tau) + (2\pi)^4 \delta(\mathbf{k} - \mathbf{p} - \mathbf{q})\psi^*] = \nonumber \\
&\quad \underline{\mathbf{A}_v(\mathbf{p}, \tau)\mathbf{A}_v(\mathbf{q}, \tau)\psi_{-}(\mathbf{k} - \mathbf{p} - \mathbf{q}, \tau)} + (2\pi)^4 \delta(\mathbf{p})a_v\underline{\mathbf{A}_v(\mathbf{q}, \tau)\psi_{-}(\mathbf{k} - \mathbf{p} - \mathbf{q}, \tau)} + \nonumber \\
&\quad +(2\pi)^4 \delta(\mathbf{q})a_v\underline{\mathbf{A}_v(\mathbf{p}, \tau)\psi_{-}(\mathbf{k} - \mathbf{p} - \mathbf{q}, \tau)} + \tag{5.2.13} \\
& \quad \# + (2\pi)^8 \delta(\mathbf{p})\delta(\mathbf{q})a_v^2\psi_{-}(\mathbf{k} - \mathbf{p} - \mathbf{q}, \tau) + \nonumber \\
&+[(2\pi)^4 \delta(\mathbf{p})a_v\mathbf{A}_v(\mathbf{q}, \tau) + (2\pi)^4 \delta(\mathbf{q})a_v\mathbf{A}_v(\mathbf{p}, \tau) + (2\pi)^8 \delta(\mathbf{p})\delta(\mathbf{q})a_v^2](2\pi)^4 \delta(\mathbf{k} - \mathbf{p} - \mathbf{q})\psi^* = \nonumber \\
&= (2\pi)^8 \delta(\mathbf{p})\delta(\mathbf{q})a_v^2\psi_{-}(\mathbf{k} - \mathbf{p} - \mathbf{q}, \tau) + \nonumber \\
&+ (2\pi)^8 \delta(\mathbf{p})\delta(\mathbf{k} - \mathbf{p} - \mathbf{q})\psi^*a_v\mathbf{A}_v(\mathbf{q}, \tau) + (2\pi)^8 \delta(\mathbf{q})\delta(\mathbf{k} - \mathbf{p} - \mathbf{q})\psi^*a_v\mathbf{A}_v(\mathbf{p}, \tau) + \nonumber \\
&+ (2\pi)^{12} \delta(\mathbf{p})\delta(\mathbf{q})\delta(\mathbf{k} - \mathbf{p} - \mathbf{q})\psi^*a_v^2 \nonumber
\end{align}
$$

We rewrite Eq.(2.5.16) of the short form
$$
\begin{equation}
\begin{aligned}
\bar{\psi}_{-}(\mathbf{k}, t) = & - (2\pi)^4 \delta(\mathbf{k}) \psi^* e^{-(\mathbf{k}^2+m^2)t} + \int_0^t d\tau e^{-(\mathbf{k}^2+m^2)(t-\tau)} \{ (\mathbf{k}+m)\bar{\eta}_1(\mathbf{k}, \tau) + \bar{\eta}_2(\mathbf{k}, \tau) + \\
& \qquad + \frac{e}{(2\pi)^4} \int d^4\mathbf{p} [\Delta_1(\mathbf{k}, \mathbf{p}, \tau, e) + \Delta_2(\mathbf{k}, \mathbf{p}, \tau, e)] - \\
& \qquad - \frac{e^2}{(2\pi)^8} \int d^4\mathbf{p} \int d^4\mathbf{q} [\Delta_3(\mathbf{k}, \mathbf{p}, \tau, e) + \Delta_4(\mathbf{k}, \mathbf{p}, \tau, e)] \}
\end{aligned}
\tag{5.2.17}
\end{equation}
$$

where

$$
\begin{align}
\Delta_2(\mathbf{k}, \mathbf{p}, \tau, e) &= \tag{5.2.18} \\
\Delta_4(\mathbf{k}, \mathbf{p}, \tau, e) &= \nonumber
\end{align}
$$

Therefore Fourier transformed differential master equations corresponding to
Eq.(5.2.10) and Eq.(5.2.12) reads

$$
\begin{equation}
\begin{aligned}
\psi_{-}(\mathbf{k}, t) = &-(2\pi)^4 \delta(\mathbf{k}) \psi^* e^{-(\mathbf{k}^2+m^2)t} + \\
&+ \int_0^t d\tau e^{-(\mathbf{k}^2+m^2)(t-\tau)} \{ (\mathbf{k}+m)\eta_1(\mathbf{k}, \tau) + \eta_2(\mathbf{k}, \tau) + \\
& \qquad + \frac{e}{(2\pi)^4} \int d^4\mathbf{p} \Delta_2(\mathbf{k}, \mathbf{p}, \tau, e) - \frac{e^2}{(2\pi)^8} \int d^4\mathbf{p} \int d^4\mathbf{q} \Delta_4(\mathbf{k}, \mathbf{p}, \tau, e) \}
\end{aligned}
\tag{5.2.19}
\end{equation}
$$

and

$$
\begin{equation}
\begin{aligned}
\bar{\psi}_{-}(\mathbf{k}, t) ={}& (2\pi)^4 \delta(\mathbf{k}) \bar{\psi}^* e^{-(\mathbf{k}^2+m^2)t} + \\
& + \int_0^t d\tau e^{-(\mathbf{k}^2+m^2)(t-\tau)} \{ (\mathbf{k}+m)\bar{\eta}_1(\mathbf{k}, \tau) + \bar{\eta}_2(\mathbf{k}, \tau) + \\
& \qquad + \frac{e}{(2\pi)^4} \int d^4\mathbf{p} \Delta_2(\mathbf{k}, \mathbf{p}, \tau, e) - \frac{e^2}{(2\pi)^8} \int d^4\mathbf{p} \int d^4\mathbf{q} \Delta_4(\mathbf{k}, \mathbf{p}, \tau, e) \}
\end{aligned}
\tag{5.2.20}
\end{equation}
$$

correspondingly.

By the replacement (5.2.8) from (5.2.7') one obtains
$$
\begin{align}
A_{\mu-}(\mathbf{k}, t) = & - (2\pi)^4 a_{\mu} \delta(\mathbf{k}) \times \int_0^t \delta(\tau) e^{-(\mathbf{k}^2+m_{\text{reg}}^2)(t-\tau)} d\tau - m_{\text{reg}}^2 a_{\mu} \delta(\mathbf{k}) \int_0^t e^{-(\mathbf{k}^2+m_{\text{reg}}^2)(t-\tau)} d\tau + \nonumber \\
& + \int_0^t \left[ e^{-(\mathbf{k}^2+m_{\text{reg}}^2)(t-\tau)} T_{\mu\nu} + L_{\mu\nu} \right] \left\{ \eta_v(\mathbf{k}, \tau) + \frac{e}{(2\pi)^4} \int d^4\mathbf{p} \left[ \bar{\psi}_{-}(\mathbf{p}, \tau) + (2\pi)^4 \delta(\mathbf{p}) \bar{\psi}^* \right] \times \right. \nonumber \\
& \qquad \left. \times \gamma_{\mu} \left[ \psi_{-}(\mathbf{k}+\mathbf{p}, \tau) + (2\pi)^4 \delta(\mathbf{k}+\mathbf{p}) \psi^* \right] \right\} d\tau = \nonumber \\
& - (2\pi)^4 a_{\mu} e^{-(\mathbf{k}^2+m^2)t} \delta(\mathbf{k}) \int_0^t \delta(\tau) e^{-(\mathbf{k}^2+m^2)(t-\tau)} d\tau - \nonumber \\
& - m_{\text{reg}}^2 a_{\mu} \delta(\mathbf{k}) \int_0^t e^{-(\mathbf{k}^2+m_{\text{reg}}^2)(t-\tau)} d\tau + \nonumber \\
& + \int_0^t d\tau \left[ e^{-(\mathbf{k}^2+m_{\text{reg}}^2)(t-\tau)} T_{\mu\nu}^{\text{reg}} + L_{\mu\nu}^{\text{reg}} \right] \left\{ \eta_v(\mathbf{k}, \tau) + \right. \nonumber \\
& \qquad \left. + \frac{e}{(2\pi)^4} \int d^4\mathbf{p} \left[ \bar{\psi}_{-}(\mathbf{p}, \tau) \gamma_{\mu} \psi_{-}(\mathbf{k}+\mathbf{p}, \tau) + (2\pi)^4 \delta(\mathbf{p}) \bar{\psi}^* \psi_{-}(\mathbf{k}+\mathbf{p}, \tau) + \right. \right. \nonumber \\
& \qquad \left. \left. + (2\pi)^4 \bar{\psi}_{-}(\mathbf{p}, \tau) \delta(\mathbf{k}+\mathbf{p}) \psi^* + (2\pi)^8 \delta(\mathbf{p}) \delta(\mathbf{k}+\mathbf{p}) \bar{\psi}^* \psi^* \right] \right\} \nonumber \\
& T_{\mu\nu}^{\text{reg}} = \delta_{\mu\nu} - \frac{\mathbf{k}_{\mu} \mathbf{k}_{\nu}}{\mathbf{k}^2 + m_{\text{reg}}^2}, \quad L_{\mu\nu}^{\text{reg}} = \frac{\mathbf{k}_{\mu} \mathbf{k}_{\nu}}{\mathbf{k}^2 + m_{\text{reg}}^2}. \tag{5.2.21}
\end{align}
$$

Therefore Fourier transformed linear differential master equation corresponding to Eq.(5.2.10) and Eq.(5.2.12) reads

$$
\begin{equation}
\begin{split}
A_{\mu-}(\mathbf{k}, t) = -(2\pi)^4 & a_{\mu} e^{-(\mathbf{k}^2+m^2)t} \delta(\mathbf{k}) \int_0^t \delta(\tau) e^{-(\mathbf{k}^2+m^2)(t-\tau)} d\tau - \\
& -m_{\text{reg}}^2 a_{\mu} \delta(\mathbf{k}) \int_0^t e^{-(\mathbf{k}^2+m_{\text{reg}}^2)(t-\tau)} d\tau + \int_0^t d\tau \Biggl[ e^{-(\mathbf{k}^2+m_{\text{reg}}^2)(t-\tau)} T_{\mu\nu} + L_{\mu\nu} \Biggr] \Biggl\{ \eta_v(\mathbf{k}, \tau) + \\
& + \frac{e}{(2\pi)^4} \int d^4\mathbf{p} \Biggl[ (2\pi)^4 \delta(\mathbf{p}) \bar{\psi}^* \psi_{-}(\mathbf{k}+\mathbf{p}, \tau) + \\
& \phantom{+ \frac{e}{(2\pi)^4} \int d^4\mathbf{p} \Biggl[} (2\pi)^4 \bar{\psi}_{-}(\mathbf{p}, \tau) \delta(\mathbf{k}+\mathbf{p}) \psi^* + (2\pi)^8 \delta(\mathbf{p}) \delta(\mathbf{k}+\mathbf{p}) \bar{\psi}^* \psi^* \Biggr] \Biggr\}
\end{split}
\tag{5.2.22}
\end{equation}
$$

By the replacement (5.2.8) from (5.2.7') one obtains
$$
\begin{align}
A_{\mu+}(\mathbf{k}, t) = & \nonumber \\
& (2\pi)^4 a_{\mu} \delta(\mathbf{k}) \times \int_0^t \delta(\tau) e^{-(\mathbf{k}^2+m_{\text{reg}}^2)(t-\tau)} d\tau + m_{\text{reg}}^2 a_{\mu} \delta(\mathbf{k}) \int_0^t e^{-(\mathbf{k}^2+m_{\text{reg}}^2)(t-\tau)} d\tau + \nonumber \\
& + \int_0^t \left[ e^{-(\mathbf{k}^2+m_{\text{reg}}^2)(t-\tau)} T_{\mu\nu} + L_{\mu\nu} \right] \left\{ \eta_v(\mathbf{k}, \tau) + \frac{e}{(2\pi)^4} \int d^4\mathbf{p} \left[ \bar{\psi}_{-}(\mathbf{p}, \tau) + (2\pi)^4 \delta(\mathbf{p}) \bar{\psi}^* \right] \times \right. \nonumber \\
& \qquad \left. \times \gamma_{\mu} \left[ \psi_{-}(\mathbf{k}+\mathbf{p}, \tau) + (2\pi)^4 \delta(\mathbf{k}+\mathbf{p}) \psi^* \right] \right\} d\tau = \nonumber \\
& (2\pi)^4 a_{\mu} e^{-(\mathbf{k}^2+m^2)t} \delta(\mathbf{k}) \int_0^t \delta(\tau) e^{-(\mathbf{k}^2+m^2)(t-\tau)} d\tau + \nonumber \\
& + m_{\text{reg}}^2 a_{\mu} \delta(\mathbf{k}) \int_0^t e^{-(\mathbf{k}^2+m_{\text{reg}}^2)(t-\tau)} d\tau + \nonumber \\
& + \int_0^t d\tau \left[ e^{-(\mathbf{k}^2+m_{\text{reg}}^2)(t-\tau)} T_{\mu\nu}^{\text{reg}} + L_{\mu\nu}^{\text{reg}} \right] \left\{ \eta_v(\mathbf{k}, \tau) + \right. \nonumber \\
& \qquad \left. + \frac{e}{(2\pi)^4} \int d^4\mathbf{p} \left[ \bar{\psi}_{-}(\mathbf{p}, \tau) \gamma_{\mu} \psi_{-}(\mathbf{k}+\mathbf{p}, \tau) + (2\pi)^4 \delta(\mathbf{p}) \bar{\psi}^* \psi_{-}(\mathbf{k}+\mathbf{p}, \tau) + \right. \right. \nonumber \\
& \qquad \left. \left. + (2\pi)^4 \bar{\psi}_{-}(\mathbf{p}, \tau) \delta(\mathbf{k}+\mathbf{p}) \psi^* + (2\pi)^8 \delta(\mathbf{p}) \delta(\mathbf{k}+\mathbf{p}) \bar{\psi}^* \psi^* \right] \right\} \nonumber \\
& T_{\mu\nu}^{\text{reg}} = \delta_{\mu\nu} - \frac{\mathbf{k}_{\mu} \mathbf{k}_{\nu}}{\mathbf{k}^2 + m_{\text{reg}}^2}, \quad L_{\mu\nu}^{\text{reg}} = \frac{\mathbf{k}_{\mu} \mathbf{k}_{\nu}}{\mathbf{k}^2 + m_{\text{reg}}^2}. \tag{5.2.16}
\end{align}
$$

We define now countable sequences $\{\psi_{-}^{(n)}(\mathbf{k}, t, e)\}_{n \in \mathbb{N}}$, $\{\bar{\psi}_{-}^{(n)}(\mathbf{k}, t, e)\}_{n \in \mathbb{N}}$, $\{A_{\mu-}^{(n)}(\mathbf{k}, t, e)\}_{n \in \mathbb{N}}$, $\{A_{\mu+}^{(n)}(\mathbf{k}, t, e)\}_{n \in \mathbb{N}}$ by

$$
\begin{align}
\psi_{-}^{(n)}(\mathbf{k}, t, e) ={}& \Psi\left[\psi_{-}^{(n-1)}(\mathbf{k}, t, e), \bar{\psi}_{-}^{(n-1)}(\mathbf{k}, t, e), A_{\mu-}^{(n-1)}(\mathbf{k}, t, e)\right] = \nonumber \\
& - (2\pi)^4 \delta(\mathbf{k}) \psi^* e^{-(\mathbf{k}^2+m^2)t} + \nonumber \\
& + \int_0^t d\tau e^{-(\mathbf{k}^2+m^2)(t-\tau)} \left\{ (\mathbf{k}+\mathbf{m})\eta_1(\mathbf{k}, \tau) + \eta_2(\mathbf{k}, \tau) + \right. \nonumber \\
& \qquad \left. + \frac{e^2}{(2\pi)^4} \int d^4\mathbf{p} \Delta_2^{(n-1)}(\mathbf{k}, \mathbf{p}, \tau, e) - \frac{e^2}{(2\pi)^8} \int d^4\mathbf{p} \int d^4\mathbf{q} \Delta_4^{(n-1)}(\mathbf{k}, \mathbf{p}, \tau, e) \right\} \tag{5.2.17}
\end{align}
$$
$$
\begin{equation}
\begin{split}
\bar{\psi}_{-}^{(n)}(\mathbf{k}, t, e) ={}& \Psi[\psi_{-}^{(n-1)}(\mathbf{k}, t, e), \bar{\psi}_{-}^{(n-1)}(\mathbf{k}, t, e), A_{\mu-}^{(n-1)}(\mathbf{k}, t, e)] = \\
& -(2\pi)^4 \delta(\mathbf{k}) \psi^{*} e^{-(\mathbf{k}^2+m^2)t} + \\
& + \int_0^t d\tau e^{-(\mathbf{k}^2+m^2)(t-\tau)} \{(\mathbf{k}+m)\bar{\eta}_1(\mathbf{k}, \tau) + \bar{\eta}_2(\mathbf{k}, \tau) + \\
& + \frac{e}{(2\pi)^4} \int d^4\mathbf{p} \Delta_2^{(n-1)}(\mathbf{k}, \mathbf{p}, \tau, e) - \frac{e^2}{(2\pi)^8} \int d^4\mathbf{p} \int d^4\mathbf{q} \Delta_4^{(n-1)}(\mathbf{k}, \mathbf{p}, \tau, e)\}
\end{split}
\tag{5.2.17'}
\end{equation}
$$

where

$$
\begin{align}
\Delta_2^{(n-1)}(\mathbf{k}, \mathbf{p}, \tau, e) ={}& \nonumber \\
& (2\pi)^4 \delta(\mathbf{p}) \mathbf{p} \dot{a}_\mu \psi_-^{(n-1)}(\mathbf{k}-\mathbf{p}, \tau) + (2\pi)^4 \delta(\mathbf{k}-\mathbf{p}) \mathbf{p} \mathbf{A}_-^{(n-1)}(\mathbf{p}, \tau) \psi^* - \tag{5.2.12} \\
& -2(2\pi)^4 (\mathbf{k}_v - \mathbf{p}_v) \delta(\mathbf{p}) a_v \delta(\mathbf{k}-\mathbf{p}) \psi^* + (2\pi)^4 \delta(\mathbf{p}) \mathbf{p} \dot{a}_\mu \delta(\mathbf{k}-\mathbf{p}) \psi^* + \nonumber \\
& + \mathbf{A}_-^{(n-1)}(\mathbf{p}, \tau) \eta_1(\mathbf{k}-\mathbf{p}, \tau) + (2\pi)^4 \delta(\mathbf{p}) \dot{a}_\mu \eta_1(\mathbf{k}-\mathbf{p}, \tau) \nonumber
\end{align}
$$

$$
\begin{equation}
\begin{split}
\Delta_4^{(n-1)}(\mathbf{k}, \mathbf{p}, \tau, e) ={}& \\
& (2\pi)^8 \delta(\mathbf{p}) \delta(\mathbf{q}) a_v^2 \psi_-^{(n-1)}(\mathbf{k}-\mathbf{p}-\mathbf{q}, \tau) + \\
& (2\pi)^8 \delta(\mathbf{p}) \delta(\mathbf{k}-\mathbf{p}-\mathbf{q}) \psi^* a_v \mathbf{A}_{v-}^{(n-1)}(\mathbf{q}, \tau) + \\
& (2\pi)^8 \delta(\mathbf{q}) \delta(\mathbf{k}-\mathbf{p}-\mathbf{q}) \psi^* a_v \mathbf{A}_{v-}^{(n-1)}(\mathbf{p}, \tau) + \\
& +(2\pi)^{12} \delta(\mathbf{p}) \delta(\mathbf{q}) \delta(\mathbf{k}-\mathbf{p}-\mathbf{q}) \psi^* a_v^2.
\end{split}
\tag{5.2.14}
\end{equation}
$$

$$
\begin{align*}
\Delta_2^{(n-1)}(\mathbf{k}, \mathbf{p}, \tau, e) &= \\
\Delta_4^{(n-1)}(\mathbf{k}, \mathbf{p}, \tau, e) &=
\end{align*}
\qquad (5.2.14)
$$

$$
\begin{equation}
\begin{split}
\psi_{-}^{(1)}(\mathbf{k}, t, e) = \psi_{-}(\mathbf{k}, t, e = 0) ={}& \\
& -(2\pi)^4 \delta(\mathbf{k}) \psi^* \int_0^t d\tau \delta(\tau) e^{-(\mathbf{k}^2+m^2)(t-\tau)} - m^2 \psi^* \int_0^t d\tau e^{-(\mathbf{k}^2+m^2)(t-\tau)} + \\
& + \int_0^t d\tau e^{-(\mathbf{k}^2+m^2)(t-\tau)} \{(\mathbf{k}+m)\eta_1(\mathbf{k}, \tau) + \eta_2(\mathbf{k}, \tau)\} =
\end{split}
\tag{5.2.18}
\end{equation}
$$

$$
\bar{\psi}_{-}^{(n)}(\mathbf{k}, t, e) = \bar{\Psi}[\psi_{-}^{(n-1)}(\mathbf{k}, t, e), \bar{\psi}_{+}^{(n-1)}(\mathbf{k}, t, e), A_{\mu-}^{(n-1)}(\mathbf{k}, t, e)] = \quad (5.2.19)
$$