# Generalized Standard Model with higher-order derivatives under Rotor
Mechanism and its Quantization

B.T.T.Wong111Corresponding author. The University of Hong Kong, Pokfulam road.
<EMAIL_ADDRESS>
###### Abstract

The Standard Model is the paradigm of particle physics which gives an accurate
theory for fundamental particle interactions. However, the extension of
Standard Model with higher-order derivatives is not a well-studied subject.
This paper is a follow-up work of the previous study of the generalized
Abelian gauge field theory and Yang-Mills theory under rotor mechanism of
order $n$ of higher order derivatives, and we apply it to the Standard Model
of particle physics. Rotor mechanism on scalar field and Dirac field is also
studied. We will study the quantization of the rotored Standard Model using
path integral approach. We also inherit the previous result from the path
integral quantization of generalized Abelian gauge field and apply it to our
non-Abelian case. Then we carry out the generalized BRST quantization and
prove the existence of the Slavnov-Taylor Identities of the rotor model.
Finally, we discuss the possibility of rotor model on taming the infinities
arise from the self-energy correction of the Higgs boson in high spacetime
dimension, thus this provides a partial solution and new insights to the
Hierarchy problem.

Keywords: Standard Model; Quantum field theory ; Rotor mechanism ; High-order
derivative field theory;Hierarchy Problem

##  1 Introduction

The Standard Model (SM) of particle physics is a well-established theory which
describes particle interaction with great precision, with quantum field theory
as an underlying mathematical foundation. Yet, higher-order derivative SM with
a great potential to tame UV divergence is not a well studied subject.
Promising model includes a recent study in Lee-Wick Standard Model which
stabilizes quadratic divergence [1], of which arises from the generalization
of Lee-Wick electrodynamics [2, 3].

The study of higher-order derivative quantum field theory is particular
interesting because it can eliminate ultraviolet (UV) divergences in
scattering amplitudes [4, 5, 6, 7, 8]. There are studies on higher order
derivative scalar field and gauge field theories, and these theories have
contributions in quantum gravity and modified gravity [9, 10, 11, 12, 13, 14,
15, 16]. Higher-order derivative theory also shows its appearance in string
theory [17, 18, 19, 20, 21]. In our context of study, Yang-Mills theory with
higher-order derivative has been studied in references [22, 23, 24].
Quantization of higher-order derivative quantum field theory using path
integral approach has been studied in [25, 26, 27, 28, 29].

In our previous work, we have established the generalized, higher-order
derivative Yang-Mills theory by rotor mechanism [30], which follows upon our
further previous work in the abelian counterpart [31]. The path integral
quantization of generalized abelian gauge field theory under rotor model is
conducted in our previous work [32]. The Yang-Mills action in $D$ dimensional
spacetime is given by [33]

$S_{\mathrm{YM}}=-\frac{1}{2}\int
d^{D}x\mathrm{Tr}\,G_{\mu\nu}G^{\mu\nu}=-\frac{1}{4}\int
d^{D}x\,G^{a}_{\mu\nu}G^{\mu\nu\,a}\,$ (1)

where

$G_{\mu\nu}=\partial_{\mu}T_{\nu}-\partial_{\nu}T_{\mu}-ig[T_{\mu},T_{\nu}]\,,$
(2)

for which $G_{\mu\nu}=G^{a}_{\mu\nu}t^{a}$ and $T_{\mu}=T_{\mu}^{a}t^{a}$ are
matrices with $t^{a}$ the generators of $\mathrm{SU}(N)$ Lie group. Using the
Lie algebra $[t^{a},t^{b}]=if^{abc}t^{c}$, this gives the gauge field strength
as,

$G_{\mu\nu}^{a}=\partial_{\mu}T_{\nu}^{a}-\partial_{\nu}T_{\mu}^{a}+gf^{abc}T_{\mu}^{b}T_{\nu}^{c}\,.$
(3)

We have $D=4n+4$ for a renormalizable theory with unity gauge field dimension
[32]. By carrying our integration by parts on the kinetic term and expanding
the remaining terms, equation(1) explicitly is,

$S_{\mathrm{YM}}=\int d^{D}x\bigg{(}T^{\mu a}\hat{R}_{\mu\nu}T^{\nu
a}-\frac{g}{2}f^{abc}(\partial^{\mu}T^{\nu a}-\partial^{\nu}T^{\mu
a})T_{\mu}^{b}T_{\nu}^{c}-\frac{g^{2}}{4}f^{abc}f^{ade}T_{\mu}^{b}T_{\nu}^{c}T^{\mu
d}T^{\nu e}\bigg{)}\,,$ (4)

where
$\hat{R}_{\mu\nu}=\frac{1}{2}(\Box\eta_{\mu\nu}-\partial_{\mu}\partial_{\nu})$
is defined as the projection tensor. Under the rotor mechanism we introduced
in our previous work in [31], which is the successive action of the projection
tensors on the original gauge field,

$T_{n\,\mu_{n}}=\hat{R}_{\mu_{n}\mu_{n-1}}\hat{R}^{\mu_{n-1}\mu_{n-2}}\cdots\hat{R}_{\mu_{3}\mu_{2}}\hat{R}^{\mu_{2}\mu_{1}}\hat{R}_{\mu_{1}\mu_{0}}T^{\mu_{0}}=\frac{1}{2^{n-1}}P_{\mu_{n}}^{\,\,\,\mu_{n-1}}P_{\mu_{n-1}}^{\,\,\,\mu_{n-2}}\cdots
P_{\mu_{3}}^{\,\,\,\mu_{2}}P_{\mu_{2}}^{\,\,\,\mu_{1}}\hat{R}_{\mu_{1}\mu_{0}}T^{\mu_{0}}\,,$
(5)

for which

$P_{\mu_{j}}^{\,\,\,\mu_{j-1}}=\Box\delta_{\mu_{j}}^{\,\,\,\mu_{j-1}}$ (6)

is defined as the propagator. This is known as the rotor transformation which
generates high-order derivative gauge fields in the action [31, 32]. The
$n$-th order Yang-Mills action after rotor transformation of gauge field under
Lorentz gauge is [32]

$\displaystyle S^{(n)}_{\mathrm{YM}}$ $\displaystyle=-\frac{1}{4}\int
d^{D}x\,G^{a}_{n\,\mu\nu}G^{\mu\nu\,a}_{n}$ (7) $\displaystyle=\int
d^{D}x\bigg{(}T^{\mu a}_{n}\hat{R}_{\mu\nu}T^{\nu
a}_{n}-\frac{g}{2}f^{abc}(\partial^{\mu}T^{\nu a}_{n}-\partial^{\nu}T^{\mu
a}_{n})T_{\mu}^{b}T_{n\,\nu}^{c}-\frac{g^{2}}{4}f^{abc}f^{ade}T_{n\,\mu}^{b}T_{n\,\nu}^{c}T^{\mu
d}_{n}T^{\nu e}_{n}\bigg{)}$ $\displaystyle=\int
d^{D}x\bigg{(}\frac{1}{4^{n}}\Box^{n}T^{\mu a}\hat{R}_{\mu\nu}\Box^{n}T^{\nu
a}-\frac{1}{2\cdot 8^{n}}gf^{abc}(\partial^{\mu}\Box^{n}T^{\nu
a}-\partial^{\nu}\Box^{n}T^{\mu a})\Box^{n}T_{\mu}^{b}\Box^{n}T_{\nu}^{c}$
$\displaystyle\quad\quad\quad\quad\quad-\frac{g^{2}}{4\cdot
16^{n}}f^{abc}f^{ade}\Box^{n}T_{\mu}^{b}\Box^{n}T_{\nu}^{c}\Box^{n}T^{\mu
d}\Box^{n}T^{\nu e}\bigg{)}$

Therefore, under the rotor mechanism, the gauge field transforms as [30],

$T_{\mu}\rightarrow T_{n\,\mu}=\frac{1}{2^{n}}\Box^{n}T_{\mu}\,.$ (8)

And the gauge field strength becomes [30]

$G_{n\mu\nu}^{a}=\frac{1}{2^{n}}\partial_{\mu}\Box^{n}T_{\nu}^{a}-\frac{1}{2^{n}}\partial_{\nu}\Box^{n}T_{\mu}^{a}+\frac{1}{4^{n}}gf^{abc}\Box^{n}T^{b}_{\mu}\Box^{n}T^{c}_{\nu}\,.$
(9)

The $n-$th ordered covariant derivative is given by [30],

$D_{n\,\mu}=\partial_{\mu}-\frac{i}{2^{n}}g\Box^{n}T_{\mu}^{c}t^{c}\,,$ (10)

The equation of motion of this generalized Yang-Mills theory is [30],

$D_{n\,\mu}G_{n}^{\mu\nu a}=\partial_{\mu}G^{\mu\nu
a}_{n}+\frac{1}{2^{n}}gf^{abc}\Box^{n}T_{\mu}^{b}G^{\mu\nu c}_{n}=0\,.$ (11)

The Noether’s current is [30]

$J^{\alpha}=\bigg{(}-\frac{1}{4^{n}}\Box^{n}\tilde{G}^{\alpha\beta\,k}-\frac{1}{2\cdot
8^{n}}gf^{kbc}(\Box^{n}T^{\alpha b}\Box^{n}T^{\beta c}-\Box^{n}T^{\alpha
c}\Box^{n}T^{\beta b})\bigg{)}\delta\Box^{n}T^{k}_{\beta}\,,$ (12)

and the associated Noether’s charge is given by [30]

$Q=\int d^{D-1}xj^{0}=\int
d^{D-1}x\bigg{(}-\frac{1}{4^{n}}\Box^{n}\tilde{G}^{0\beta\,k}-\frac{1}{2\cdot
8^{n}}gf^{kbc}(\Box^{n}T^{0b}\Box^{n}T^{\beta
c}-\Box^{n}T^{0c}\Box^{n}T^{\beta b})\bigg{)}\delta\Box^{n}T^{k}_{\beta}\,,$
(13)

where $\tilde{G}^{\alpha\beta\,k}$ is the Maxwellian gauge field strength. It
can be seen that when $n=0$ (no rotation), we get back the original Yang-Mills
theory.

Under the unitary transformation by SU(N) representation, the spinor
transforms as

$\psi^{\prime}(x)=U(x)\psi(x)\,,$ (14)

where $U(x)=e^{i\alpha^{a}(x)t^{a}}$. The covariant derivative and field
operator under rotor model transform, respectively as,

$\begin{cases}D_{n\,\mu}^{\prime}(x)&=U(x)D_{n\,\mu}(x)U^{\dagger}(x)\\\
G_{n\,\mu\nu}^{\prime}(x)&=U(x)G_{n\,\mu\nu}(x)U^{\dagger}(x)\,.\end{cases}$
(15)

It follows that the rotor gauge field transforms as,

$\Box^{n}T_{\mu}^{\prime}=U\Box^{n}T_{\mu}U^{\dagger}+\frac{i\cdot
2^{n}}{g}U\partial_{\mu}U^{\dagger}\,.$ (16)

Infinitesimally, the rotored gauge field transforms as

$\Box^{n}T_{\mu}^{\prime
a}=\Box^{n}T_{\mu}^{a}+\frac{2^{n}}{g}\partial_{\mu}\alpha^{a}+f^{abc}\Box^{n}T_{\mu}^{b}\alpha^{c}\,.$
(17)

Therefore, the infinitesimal change in rotored gauge field is

$\delta\Box^{n}T_{\mu}^{a}=\Box^{n}T_{\mu}^{\prime
a}-\Box^{n}T_{\mu}^{a}=\frac{2^{n}}{g}(D_{n\,\mu}\alpha)^{a}\,.$ (18)

The $n$-th ordered gauge field strength can be defined through the commutator
of the $n$-th ordered covariant derivative,

$G_{n\,\mu\nu}=\frac{i}{g}[D_{n\,\mu},D_{n\,\nu}]\,.$ (19)

And infinitesimally it transforms as

$G_{n\,\mu\nu}^{\prime}=G_{n\,\mu\nu}-f^{abc}\alpha^{b}G_{n\,\mu\nu}^{c}\,.$
(20)

It is noted that when $n=0$, this gives us back all the properties of the
transformation rules of the original Yang-Mills theory.

We will proceed the quantization of the generalized Yang-Mills theory under
rotor mechanism with Feynman path integral approach. The quantum amplitude can
be computed as an integral of all possible field configurations over the
exponential of the action [34, 35, 36]. Similarly to our previous work for the
abelian case in [32], as in the generalized model it involves the
transformation of field by $T^{\mu}\rightarrow\Box^{n}T^{\mu}$, therefore in
the path integral we sum over all possible configurations of $\Box^{n}T^{\mu}$
instead of $T^{\mu}$, i.e. the integration measure changes as

$\int\mathcal{D}T^{\mu}(x)\rightarrow\int\mathcal{D}\Box^{n}T^{\mu}(x)\,.$
(21)

##  2 Generalized spin-0 scalar field theory under rotor mechanism

In this section, we will complete the study of generalized scalar field theory
under rotor mechanism, this will generate scalar fields with higher-order
derivatives. First consider the massless scalar field theory in $D$-dimension

$S=\int d^{D}x\,\frac{1}{2}\partial_{\mu}\phi\partial^{\mu}\phi=\int
d^{D}x\,\phi\Big{(}-\frac{\Box}{2}\Big{)}\phi\,.$ (22)

According to the definition of rotor mechanism in (5), we define the rotor
mechanism as the successive operations of the operator that couples to the
gauge fields. For the gauge field case, the operator is the projection tensor
$\hat{R}_{\mu\nu}=\frac{1}{2}(\Box\eta_{\mu\nu}-\partial_{\mu}\partial_{\nu})$.
For the scalar field case, we have the rotor operator as
$\hat{\tilde{R}}=-\frac{\Box}{2}$. This scalar rotor operator can be recovered
from tracing the the projection tensor, up to some scaling factor,

$\hat{R}=\eta^{\mu\nu}\hat{R}_{\mu\nu}=\hat{R}^{\mu}_{\,\,\,\mu}=\frac{D-1}{2}\Box\,.$
(23)

It follows that

$\hat{\tilde{R}}=-\frac{1}{D-1}\hat{R}\,.$ (24)

The rotor mechanism on scalar field is simply

$\phi\rightarrow\prod_{j=1}^{n}\bigg{(}-\frac{1}{D-1}\bigg{)}\hat{R}^{\mu_{j}}_{\,\,\,\mu_{j}}\phi=\frac{(-1)^{n}}{2^{n}}\Box\cdots\Box\Box\phi=\frac{(-1)^{n}}{2^{n}}\Box^{n}\phi\,.$
(25)

The generalized massless scalar field theory is therefore

$\displaystyle S$ $\displaystyle=\int
d^{D}x\,\frac{1}{2}\partial_{\mu}\bigg{(}\frac{(-1)^{n}}{2^{n}}\Box^{n}\phi\bigg{)}\partial^{\mu}\bigg{(}\frac{(-1)^{n}}{2^{n}}\Box^{n}\phi\bigg{)}=\int
d^{D}x\,\frac{1}{2}\partial_{\mu}\bigg{(}\frac{1}{2^{n}}\Box^{n}\phi\bigg{)}\partial^{\mu}\bigg{(}\frac{1}{2^{n}}\Box^{n}\phi\bigg{)}$
(26) $\displaystyle=\frac{1}{4^{n}}\int
d^{D}x\,\Box^{n}\phi\Big{(}-\frac{\Box}{2}\Big{)}\Box^{n}\phi\,.$

Notice that when $n=0$, this restores back to the original case. Therefore, we
see that both the gauge field and scalar field transforms under the rotor
mechanism by the same form,

$T_{\mu}\rightarrow
T_{n\,\mu}=\frac{1}{2^{n}}\Box^{n}T_{\mu}\quad\quad,\quad\quad\phi\rightarrow\phi_{n}=\frac{1}{2^{n}}\Box^{n}\phi\,.$
(27)

For the massive scalar field theory, the generalized action under rotor
mechanism is

$S=\frac{1}{2\cdot 4^{n}}\int
d^{D}x\,\partial_{\mu}\Box^{n}\phi\partial^{\mu}\Box^{n}\phi-m^{2}\Box^{n}\phi\Box^{n}\phi=\frac{1}{4^{n}}\int
d^{D}x\,\Box^{n}\phi\Big{(}-\frac{\Box+m^{2}}{2}\Big{)}\Box^{n}\phi\,.$ (28)

The Euler-Lagrangian equation is

$\partial_{\mu}\frac{\partial\mathcal{L}}{\partial_{\mu}\Box^{n}\phi}=\frac{\partial\mathcal{L}}{\partial\Box^{n}\phi}\,.$
(29)

This gives the equation of motion as

$\Box^{n+1}\phi+m^{2}\Box^{n}\phi=0\,.$ (30)

Now consider the complex-scalar field theory. First consider the action with
two scalar fields,

$\displaystyle S$ $\displaystyle=\frac{1}{2\cdot 4^{n}}\int
d^{D}x\,\sum_{i=1,2}(\partial_{\mu}\Box^{n}\phi_{i}\partial^{\mu}\Box^{n}\phi_{i}-m^{2}\Box^{n}\phi_{i}\Box^{n}\phi_{i})$
(31) $\displaystyle=\frac{1}{4^{n}}\int
d^{D}x\,\sum_{i=1,2}\Box^{n}\phi_{i}\Big{(}-\frac{\Box+m^{2}}{2}\Big{)}\Box^{n}\phi_{i}\,.$

Now define a n-rotored complex scalar field,

$\Box^{n}\Phi=\frac{1}{\sqrt{2}}(\Box^{n}\phi_{1}+i\Box^{n}\phi_{2})\quad\text{and
}\quad\Box^{n}\Phi^{\dagger}=\frac{1}{\sqrt{2}}(\Box^{n}\phi_{1}-i\Box^{n}\phi_{2})\,.$
(32)

Then the action in 31 can be written as

$S=\frac{1}{4^{n}}\int
d^{D}x\,\partial_{\mu}\Box^{n}\Phi^{\dagger}\partial^{\mu}\Box^{n}\Phi-m^{2}\Box^{n}\Phi^{\dagger}\Box^{n}\Phi=\frac{1}{4^{n}}\int
d^{D}x\,\Box^{n}\Phi^{\dagger}(-\Box^{n}-m^{2})\Box^{n}\Phi\,.$ (33)

The two Euler-Lagrangian equations are

$\partial_{\mu}\frac{\partial\mathcal{L}}{\partial_{\mu}\Box^{n}\Phi}=\frac{\partial\mathcal{L}}{\partial\Box^{n}\Phi}\quad\text{and}\quad\partial_{\mu}\frac{\partial\mathcal{L}}{\partial_{\mu}\Box^{n}\Phi^{\dagger}}=\frac{\partial\mathcal{L}}{\partial\Box^{n}\Phi^{\dagger}}\,.$
(34)

This gives the two equations of motion as follows:

$\Box^{n+1}\Phi^{\dagger}+m^{2}\Box^{n}\Phi^{\dagger}=0\quad\text{and}\quad\Box^{n+1}\Phi+m^{2}\Box^{n}\Phi=0\,.$
(35)

Similar to the argument in our previous paper [31], the scalar field action
under rotor mechanism is renormalizable in $D=4n+4$ dimension with unity
scalar field dimension.

##  3 Generalized spin-1/2 Dirac field theory theory under rotor mechanism

Next, we will investigate how the Dirac field transforms under rotor
mechanism. First we consider the massless Dirac action in $D$-dimensional
spacetime,

$S=\int d^{D}x\,\bar{\psi}i\gamma^{\mu}\partial_{\mu}\psi=\int
d^{D}x\,\bar{\psi}i\not{\partial}\psi\equiv\int
d^{D}x\,\bar{\psi}_{a}i\gamma^{\mu}_{ab}\partial_{\mu}\psi_{b}\,.$ (36)

So in analogy to the case of gauge field and scalar field, we have the rotor
operator as the matrix $\hat{R}_{ab}=i\gamma^{\mu}_{ab}\partial_{\mu}$. Hence
under rotor transformation, the spinor field transforms as

$\psi_{a_{0}}\rightarrow(i^{n})\gamma^{\mu_{n}}_{a_{n}a_{n-1}}\gamma^{\mu_{n-1}}_{a_{n-1}a_{n-2}}\cdots\gamma^{\mu_{1}}_{a_{1}a_{0}}\partial_{\mu_{n}}\partial_{\mu_{n-1}}\cdots\partial_{\mu_{1}}\psi_{a_{0}}\,.$
(37)

Now let’s see how the adjoint spinor transforms. Consider ,

$\gamma^{0}_{a_{0}b_{0}}\psi_{b_{0}}\rightarrow(i^{n})\gamma^{\mu_{n}}_{a_{n}a_{n-1}}\gamma^{\mu_{n-1}}_{a_{n-1}a_{n-2}}\cdots\gamma^{\mu_{1}}_{a_{1}a_{0}}\partial_{\mu_{n}}\partial_{\mu_{n-1}}\cdots\partial_{\mu_{1}}\gamma^{0}_{a_{0}b_{0}}\psi_{b_{0}}\,.$
(38)

Then by taking the Hermitian conjugate

$(\gamma^{0}_{a_{0}b_{0}}\psi_{b_{0}})^{\dagger}\rightarrow(-i)^{n}(\partial_{\mu_{n}}\partial_{\mu_{n-1}}\cdots\partial_{\mu_{1}}\psi_{b_{0}}^{\dagger}\gamma^{0\dagger}_{b_{0}a_{0}})\gamma^{\mu_{1}\dagger}_{a_{0}a_{1}}\cdots\gamma^{\mu_{n-1}\dagger}_{a_{n-2}a_{n-1}}\gamma^{\mu\dagger}_{a_{n-1}a_{n}}\,.$
(39)

As
$(\gamma^{0}_{a_{0}b_{0}}\psi_{b_{0}})^{\dagger}=\psi_{b_{0}}^{\dagger}\gamma^{0\dagger}_{b_{0}a_{0}}$
and because $(\gamma^{0})^{\dagger}=\gamma^{0}$, therefore
$\bar{\psi}_{a_{0}}=\psi_{b_{0}}^{\dagger}\gamma^{0}_{b_{0}a_{0}}$. Hence the
adjoint spinor transforms as follow:

$\bar{\psi}_{a_{0}}\rightarrow(-i)^{n}(\partial_{\mu_{n}}\partial_{\mu_{n-1}}\cdots\partial_{\mu_{1}}\bar{\psi}_{a_{0}})\gamma^{\mu_{1}\dagger}_{a_{0}a_{1}}\cdots\gamma^{\mu_{n-1}\dagger}_{a_{n-2}a_{n-1}}\gamma^{\mu\dagger}_{a_{n-1}a_{n}}\,.$
(40)

Therefore, the generalized Dirac field theory under rotor mechanism is

$S=\int
d^{D}x(\partial_{\nu_{n}}\partial_{\nu_{n-1}}\cdots\partial_{\nu_{1}}\bar{\psi})\gamma^{\nu_{1}\dagger}\gamma^{\nu_{2}\dagger}\cdots\gamma^{\nu_{n}\dagger}\,i\not{\partial}\,\gamma^{\mu_{n}}\gamma^{\mu_{n-1}}\cdots\gamma^{\mu_{1}}\partial_{\mu_{n-1}}\partial_{\mu_{n-2}}\cdots\partial_{\mu_{1}}\psi\,.$
(41)

Now, let’s define short-hand tensor notation by

$\gamma^{\alpha\beta\gamma\cdots}\equiv\gamma^{\alpha}\gamma^{\beta}\gamma^{\gamma}\cdots\quad\text{and}\quad\nabla_{\alpha\beta\gamma\cdots}\equiv\partial_{\alpha}\partial_{\beta}\partial_{\gamma}\cdots\,.$
(42)

Then the action in 41 can be formally written as

$S=\int
d^{D}x(\nabla_{\nu_{n}\cdots\nu_{1}}\bar{\psi})\,\gamma^{\dagger\nu_{1}\cdots\nu_{n}}i\not{\partial}\,\gamma^{\mu_{n}\cdots\mu_{1}}\nabla_{\mu_{n}\cdots\mu_{1}}\psi\,.$
(43)

The Dirac action with the mass term is given by

$S=\int d^{D}x\,\bar{\psi}i\gamma^{\mu}\partial_{\mu}\psi-m\bar{\psi}\psi\,.$
(44)

Under rotor mechanism, the whole action transforms as

$S=\int
d^{D}x(\nabla_{\nu_{n}\cdots\nu_{1}}\bar{\psi})\,\gamma^{\dagger\nu_{1}\cdots\nu_{n}}i\not{\partial}\,\gamma^{\mu_{n}\cdots\mu_{1}}\nabla_{\mu_{n}\cdots\mu_{1}}\psi-m(\nabla_{\nu_{n}\cdots\nu_{1}}\bar{\psi})\,\gamma^{\dagger\nu_{1}\cdots\nu_{n}}\,\gamma^{\mu_{n}\cdots\mu_{1}}\nabla_{\mu_{n}\cdots\mu_{1}}\psi\,.$
(45)

The Euler-Lagrangian equation is given by

$\partial_{\mu}\frac{\partial\mathcal{L}}{\partial_{\mu}(\nabla_{\nu_{n}\cdots\nu_{1}}\bar{\psi})\,\gamma^{\dagger\nu_{1}\cdots\nu_{n}}}=\frac{\partial\mathcal{L}}{\partial(\nabla_{\nu_{n}\cdots\nu_{1}}\bar{\psi})\,\gamma^{\dagger\nu_{1}\cdots\nu_{n}}}\,.$
(46)

This gives the equation of motion as

$i\not{\partial}\,\gamma^{\mu_{n}\cdots\mu_{1}}\nabla_{\mu_{n}\cdots\mu_{1}}\psi-m\gamma^{\mu_{n}\cdots\mu_{1}}\nabla_{\mu_{n}\cdots\mu_{1}}\psi=0\,.$
(47)

###  3.1 Generalized Quantum electrodynamics under rotor mechanism

Now we proceed to develop the theory of higher-order derivative quantum
electrodynamics (QED) by rotor mechanism. From [31], the general Maxwell
action under rotor mechanism is

$S=-\frac{1}{4^{n+1}}\int d^{D}x\,\Box^{n}G_{\mu\nu}\Box^{n}G^{\mu\nu}\,.$
(48)

From the Dirac action under rotor mechanism, we expect the interactive term
can be achieved by replacing the ordinary partial derivative into covariant
derivative $D_{n}$

$S=\int
d^{D}x(\nabla_{\nu_{n}\cdots\nu_{1}}\bar{\psi})\,\gamma^{\dagger\nu_{1}\cdots\nu_{n}}i\not{D}_{n}\,\gamma^{\mu_{n}\cdots\mu_{1}}\nabla_{\mu_{n}\cdots\mu_{1}}\psi\,,$
(49)

where

$D_{n\,\alpha}=\partial_{\alpha}+\frac{ie}{2^{n}}\Box^{n}T_{\alpha}$ (50)

under Lorentz gauge. Therefore, the full QED action with interaction under
rotor mechanism is given by

$\displaystyle S_{\mathrm{QED}}$ $\displaystyle=\int
d^{D}x\,\bigg{(}-\frac{1}{4^{n+1}}\Box^{n}G_{\mu\nu}\Box^{n}G^{\mu\nu}+(\nabla_{\nu_{n}\cdots\nu_{1}}\bar{\psi})\,\gamma^{\dagger\nu_{1}\cdots\nu_{n}}(i\gamma^{\alpha}\partial_{\alpha}-m)\,\gamma^{\mu_{n}\cdots\mu_{1}}\nabla_{\mu_{n}\cdots\mu_{1}}\psi$
(51)
$\displaystyle\quad\quad\quad\quad\quad-\frac{e}{2^{n}}(\nabla_{\nu_{n}\cdots\nu_{1}}\bar{\psi})\,\gamma^{\dagger\nu_{1}\cdots\nu_{n}}\gamma^{\alpha}\Box^{n}T_{\alpha}\gamma^{\mu_{n}\cdots\mu_{1}}\nabla_{\mu_{n}\cdots\mu_{1}}\psi\bigg{)}\,.$

##  4 Path integral quantization under rotor mechanism

###  4.1 Generalized Yang-Mills Theory

In this section, we will study the quantization of general Yang-Mills theory
by path integral approach in detail. From now on, we take the transformed
$\Box^{n}T^{\mu}$ field as the field variable. Simply speaking, the physics is
changed by $T^{\mu}\rightarrow\Box^{n}T^{\mu}$. The quantum amplitude of the
$\Box^{n}T^{\mu}$ field in the renormalizable $4n+4$ dimension is

$\langle\Box^{n}T^{\mu}_{f}(t_{f},\boldsymbol{\mathrm{x}})|e^{-i\hat{H}(t_{f}-t_{i})}|\Box^{n}T^{\mu}_{i}(t_{i},\boldsymbol{\mathrm{x}})\rangle=\int\mathcal{D}\Box^{n}T^{\mu}(x)\,\exp\Big{(}iS_{\mathrm{YM}}^{(n)}[\Box^{n}T^{\mu}]\Big{)}\,,$
(52)

where $|\Box^{n}T^{\mu}_{i}(t_{i},\boldsymbol{\mathrm{x}})\rangle$ is the
field state at initial time $t_{i}$ and
$|\Box^{n}T^{\mu}_{i}(t_{f},\boldsymbol{\mathrm{x}})\rangle$ is the field
state at final time $t_{f}$, and $\hat{H}$ is the Hamiltonian. The sourced
generating functional is a functional of 4-(covariant) transformed vector
current $\Box^{n}J_{\mu}(x)$,

$Z[\Box^{n}J_{\mu}(x)]=\int\mathcal{D}\Box^{n}T^{\mu}(x)\,\exp\bigg{(}iS_{\mathrm{YM}}^{(n)}[\Box^{n}T^{\mu}]\,+\,2i\int
d^{4n+4}x\,\mathrm{Tr}\Big{(}\Box^{n}J_{\mu}(x)\Box^{n}T^{\mu}(x)\Big{)}\bigg{)}\,.$
(53)

The normalized generating functional for gauge field is

$\mathcal{Z}[\Box^{n}J_{\mu}]=\frac{Z[\Box^{n}J_{\mu}]}{Z[0]}=\frac{\int\mathcal{D}\Box^{n}T^{\mu}(x)\,\exp\bigg{(}iS_{\mathrm{YM}}^{(n)}[\Box^{n}T^{\mu}]\,+\,2i\int
d^{4n+4}x\mathrm{Tr}\Big{(}\Box^{n}J_{\mu}(x)\Box^{n}T^{\mu}(x)\Big{)}\bigg{)}}{\int\mathcal{D}\Box^{n}T^{\mu}(x)\,\exp\Big{(}iS_{\mathrm{YM}}^{(n)}[\Box^{n}T^{\mu}]\Big{)}}\,.$
(54)

The free $n$-point correlation function is given by

$\langle
0|\mathrm{T}\Box^{n}\hat{T}^{\nu_{1}}(x_{1})\Box^{n}\hat{T}^{\nu_{2}}(x_{2})\cdots\Box^{n}\hat{T}^{\nu_{n}}(x_{n})|0\rangle=\frac{1}{i^{n}}\frac{\delta^{n}}{\delta\Box^{n}J_{\nu_{1}}(x_{1})\delta\Box^{n}J_{\nu_{2}}(x_{2})\cdots\delta\Box^{n}J_{\nu_{n}}(x_{n})}\mathcal{Z}[\Box^{n}J_{\mu}]\bigg{|}_{\Box^{n}J_{\mu}=0}\,,$
(55)

where noting that each gauge field has different Lorentz indices and
$\mathrm{T}$ means the time ordering operator. Then the path integral
representation is

$\displaystyle\quad\langle
0|\mathrm{T}\Box^{n}\hat{T}^{\nu_{1}}(x_{1})\cdots\Box^{n}\hat{T}^{\nu_{n}}(x_{n})|0\rangle$
(56)
$\displaystyle=\frac{\int\mathcal{D}\Box^{n}T^{\mu}(x)\,\Box^{n}T^{\nu_{1}}(x_{1})\cdots\Box^{n}T^{\nu_{n}}(x_{n})\exp\Big{(}iS_{\mathrm{YM}}^{(n)}[\Box^{n}T^{\mu}]\Big{)}}{\int\mathcal{D}\Box^{n}T^{\mu}(x)\,\exp\Big{(}iS_{\mathrm{YM}}^{(n)}[\Box^{n}T^{\mu}]\Big{)}}\,.$

Since the projection tensor in the action $S_{\mathrm{YM}}^{(n)}$ is not
invertible, we need to perform gauge fixing. We will use the Fadeev-Popov
method. Recall the gauge field transforms as (16)

$\Box^{n}T_{\mu}^{\prime}=\Box^{n}T_{\mu}^{U}=U\Box^{n}T_{\mu}U^{\dagger}+\frac{i\cdot
2^{n}}{g}U\partial_{\mu}U^{\dagger}\,.$ (57)

Now we need to choose a gauge fixing functional,

$G(\Box^{n}T^{\mu})=\partial_{\mu}\Box^{n}T^{\mu}(x)-w(x)\,,$ (58)

where $w(x)$ is some arbitrary matrix ($w(x)=w^{a}(x)t^{a}$ ) . Now we use the
identity,

$1=\int\mathcal{D}U\delta[G(\Box^{n}T^{\mu U})]\det\bigg{(}\frac{\delta
G[\Box^{n}T^{\mu U}(x)]}{\delta U(y)}\bigg{)}$ (59)

Next we insert this into (53),

$\displaystyle Z[\Box^{n}J_{\mu}(x)]$
$\displaystyle=\int\mathcal{D}U\mathcal{D}\Box^{n}T^{\mu}(x)\,\delta[G(\Box^{n}T^{U\mu})]\det\bigg{(}\frac{\delta
G[\Box^{n}T^{U\mu}(x)]}{\delta U(y)}\bigg{)}$ (60)
$\displaystyle\quad\quad\quad\quad\quad\quad\times\exp\bigg{(}iS_{\mathrm{YM}}^{(n)}[\Box^{n}T^{\mu}]\,+\,2i\int
d^{4n+4}x\,\mathrm{Tr}\Big{(}\Box^{n}J_{\mu}(x)\Box^{n}T^{\mu}(x)\Big{)}\bigg{)}\,.$

But since the action is gauge invariant,
$S_{\mathrm{YM}}^{(n)}[\Box^{n}T^{U\mu}]=S_{\mathrm{YM}}^{(n)}[\Box^{n}T^{\mu}]$,
also the integration measure remains unchanged, $\mathcal{D}\Box^{n}T^{\mu
U}=\mathcal{D}\Box^{n}T^{\mu}$, therefore we can write

$\displaystyle Z[\Box^{n}J_{\mu}(x)]$
$\displaystyle=\int\mathcal{D}U\mathcal{D}\Box^{n}T^{U\mu}(x)\,\delta[G(\Box^{n}T^{U\mu})]\det\bigg{(}\frac{\delta
G[\Box^{n}T^{U\mu}(x)]}{\delta U(y)}\bigg{)}$ (61)
$\displaystyle\quad\quad\quad\quad\quad\quad\times\exp\bigg{(}iS_{\mathrm{YM}}^{(n)}[\Box^{n}T^{U\mu}]\,+\,2i\int
d^{4n+4}x\,\mathrm{Tr}\Big{(}\Box^{n}J_{\mu}(x)\Box^{n}T^{\mu}(x)\Big{)}\bigg{)}\,.$

It is remarked that the source term is not gauge invariant unless
$D^{\mu}\Box^{n}T_{\mu}=0$. Now we can relabel the transformed, rotored gauge
field variable to $\Box^{n}T^{U\mu}=\Box^{n}T^{\prime\mu}$,

$\displaystyle Z[\Box^{n}J_{\mu}(x)]$
$\displaystyle=\int\mathcal{D}U\mathcal{D}\Box^{n}T^{\prime\mu}(x)\,\delta[G(\Box^{n}T^{\prime\mu})]\det\bigg{(}\frac{\delta
G[\Box^{n}T^{\prime\mu}(x)]}{\delta U(y)}\bigg{)}$ (62)
$\displaystyle\quad\quad\quad\quad\quad\quad\times\exp\bigg{(}iS_{\mathrm{YM}}^{(n)}[\Box^{n}T^{\prime\mu}]\,+\,2i\int
d^{4n+4}x\,\mathrm{Tr}\Big{(}\Box^{n}J_{\mu}(x)\Box^{n}T^{\mu}(x)\Big{)}\bigg{)}\,.$

Now we choose some normalization $N(\xi)$ which is dependent on the gauge
fixing parameter $\xi$ such that

$1=N(\xi)\int\mathcal{D}w(x)\exp\bigg{(}-i\int
d^{4n+4}x\,\mathrm{Tr}\Big{(}\frac{w^{2}(x)}{2\xi}\Big{)}\bigg{)}\,.$ (63)

Then we relate the rotored field variable $\Box^{n}T^{\prime\mu}(x)$ to
$\Box^{n}T^{\mu}(x)$. And inserting equation(63) to (62), now we obtain,

$\displaystyle Z[\Box^{n}J_{\mu}(x)]$
$\displaystyle=\Big{(}N(\xi)\int\mathcal{D}U\Big{)}\int\mathcal{D}w(x)\mathcal{D}\Box^{n}T^{\mu}(x)\,\delta[G(\Box^{n}T^{\mu})]\det\bigg{(}\frac{\delta
G[\Box^{n}T^{\mu}(x)]}{\delta U(y)}\bigg{)}$ (64)
$\displaystyle\times\exp\bigg{(}iS_{\mathrm{YM}}^{(n)}[\Box^{n}T^{\mu}]\,+\,2i\int
d^{4n+4}x\,\mathrm{Tr}\Big{(}\Box^{n}J_{\mu}(x)\Box^{n}T^{\mu}(x)\Big{)}-i\int
d^{4n+4}x\,\mathrm{Tr}\Big{(}\frac{w^{2}(x)}{\xi}\Big{)}\bigg{)}\,,$

where $N(\xi)\int\mathcal{D}U$ is some constant and can be integrated out. Now
employing the gauge condition in 58, in which the Dirac delta function picks
out the term of $\partial_{\mu}\Box^{n}T^{\mu}(x)$, we obtain,

$\displaystyle Z[\Box^{n}J_{\mu}(x)]$
$\displaystyle\propto\int\mathcal{D}\Box^{n}T^{\mu}(x)\,\det\bigg{(}\frac{\delta
G[\Box^{n}T^{\mu}(x)]}{\delta U(y)}\bigg{)}$ (65)
$\displaystyle\times\exp\bigg{(}iS_{\mathrm{YM}}^{(n)}[\Box^{n}T^{\mu}]\,+\,2i\int
d^{4n+4}x\,\mathrm{Tr}\Big{(}\Box^{n}J_{\mu}(x)\Box^{n}T^{\mu}(x)\Big{)}$
$\displaystyle\quad\quad\quad\quad-\frac{i}{\xi}\int
d^{4n+4}x\,\mathrm{Tr}\Big{(}\partial_{\mu}\Box^{n}T^{\mu}(x)\partial^{\mu}\Box^{n}T^{\mu}(x)\Big{)}\bigg{)}\,,$

where the last term in equation (65) is the gauge fixing term. We define the
gauge-fixed action as

$S_{\xi,\mathrm{YM}}^{(n)}=\int
d^{4n+4}x\bigg{(}-\frac{1}{2}\mathrm{Tr}\,G_{n\,\mu\nu}G_{n}^{\mu\nu}-\frac{1}{\xi}\mathrm{Tr}\,(\partial_{\mu}\Box^{n}T^{\mu}(x)\partial^{\mu}\Box^{n}T^{\mu}(x))\bigg{)}\,.$
(66)

Now we proceed to calculate the terms in the determinant. First notice that

$\delta\Box^{n}T^{U}_{\mu}=-\frac{1}{g}\big{(}\partial_{\mu}\alpha+ig\big{[}\frac{1}{2^{n}}\Box^{n}T^{U}_{\mu},\alpha\big{]}\big{)}=-\frac{1}{g}D_{n\,\mu}\alpha\,.$
(67)

For $G[\Box^{n}T_{\mu}(x)]=g\partial^{\mu}\Box^{n}T_{\mu}(x)$, then

$\delta G[\Box^{n}T_{\mu}(x)]=-\partial^{\mu}D_{n\,\mu}\alpha\,.$ (68)

Since $i\alpha=\delta U\cdot U^{-1}$, we have

$\frac{\delta G[\Box^{n}T^{\mu}(x)]}{\delta
U(y)}=i\delta^{4n+4}(x-y)\partial^{\mu}D_{n\,\mu}\cdot U(y).$ (69)

Since the determinant of $U(y)$ is unity, it follows that

$\det\bigg{(}\frac{\delta G[\Box^{n}T^{\mu}(x)]}{\delta
U(y)}\bigg{)}=\det\bigg{(}i\delta^{4n+4}(x-y)\partial^{\mu}D_{n\,\mu}\bigg{)}\,.$
(70)

Now we can implement the Grassmann action as follow

$\det(i\partial^{\mu}D_{n\,\mu})=\int\mathcal{D}\Box^{n}c^{*}\mathcal{D}\Box^{n}c\exp\bigg{(}-2i\int
d^{4n+4}x\,\mathrm{Tr}(\Box^{n}c^{*}\partial^{\mu}D_{n\,\mu}\Box^{n}c)\bigg{)}\,,$
(71)

where $\Box^{n}c^{*},\Box^{n}c$ are rotored transformed Grassmann fields,
known as the Faddeev-Popov ghost fields. Therefore, the full sourced partition
functional now gives

$\displaystyle Z[\Box^{n}J_{\mu}(x)]$
$\displaystyle\propto\int\mathcal{D}\Box^{n}T^{\mu}(x)\mathcal{D}\Box^{n}c^{*}(x)\mathcal{D}\Box^{n}c(x)$
(72)
$\displaystyle\times\exp\bigg{(}iS_{\mathrm{YM}}^{(n)}[\Box^{n}T^{\mu}]\,+\,2i\int
d^{4n+4}x\,\mathrm{Tr}\Big{(}\Box^{n}J_{\mu}(x)\Box^{n}T^{\mu}(x)\Big{)}$
$\displaystyle\quad\quad\quad\quad-\frac{i}{\xi}\int
d^{4n+4}x\,\mathrm{Tr}\Big{(}\partial_{\mu}\Box^{n}T^{\mu}(x)\partial^{\mu}\Box^{n}T^{\mu}(x)\Big{)}\bigg{)}$
$\displaystyle\quad\quad\quad\quad-2i\int
d^{4n+4}x\,\mathrm{Tr}\big{(}\Box^{n}c^{*}(x)\partial^{\mu}D_{n\,\mu}\Box^{n}c(x)\big{)}\bigg{)}\,.$

###  4.2 Generalized Scalar field theory

The quantization of generalized scalar field theory follows similarly to that
of the abelian gauge field case in our previous paper [30], where we will not
repeat the steps here. Generally speaking, the quantum amplitude follows the
form of

$\langle\Box^{n}\phi_{f}(t_{f},\boldsymbol{\mathrm{x}})|e^{-i\hat{H}(t_{f}-t_{i})}|\Box^{n}\phi_{i}(t_{i},\boldsymbol{\mathrm{x}})\rangle=\int\mathcal{D}\Box^{n}\phi(x)\,\exp\Big{(}\frac{i}{4^{n}}\int
d^{4n+4}x\,\Box^{n}\phi\Big{(}-\frac{\Box+m^{2}}{2}\Big{)}\Box^{n}\phi\Big{)}\,,$
(73)

The sourced partition functional is then given by

$Z[\Box^{n}J(x)]=\int\mathcal{D}\Box^{n}\phi(x)\,\exp\Big{(}\frac{i}{4^{n}}\int
d^{4n+4}x\,\Box^{n}\phi\Big{(}-\frac{\Box+m^{2}}{2}\Big{)}\Box^{n}\phi+i\int
d^{4n+4}x\,\Box^{n}J\Box^{n}\phi\Big{)}\,,$ (74)

The normalized generating functional for the scalar field is

$\mathcal{Z}[\Box^{n}J]=\frac{Z[\Box^{n}J]}{Z[0]}=\frac{\int\mathcal{D}\Box^{n}\phi(x)\,\exp\Big{(}\frac{i}{4^{n}}\int
d^{4n+4}x\,\Box^{n}\phi\Big{(}-\frac{\Box+m^{2}}{2}\Big{)}\Box^{n}\phi+i\int
d^{4n+4}x\,\Box^{n}J\Box^{n}\phi\Big{)}}{\int\mathcal{D}\Box^{n}\phi(x)\,\exp\Big{(}\frac{i}{4^{n}}\int
d^{4n+4}x\,\Box^{n}\phi\Big{(}-\frac{\Box+m^{2}}{2}\Big{)}\Box^{n}\phi\Big{)}}\,.$
(75)

The free $n$-point correlation function is given by

$\langle
0|\mathrm{T}\Box^{n}\phi(x_{1})\Box^{n}\phi(x_{2})\cdots\Box^{n}\phi(x_{n})|0\rangle=\frac{1}{i^{n}}\frac{\delta^{n}}{\delta\Box^{n}J(x_{1})\delta\Box^{n}J(x_{2})\cdots\delta\Box^{n}J(x_{n})}\mathcal{Z}[\Box^{n}J]\bigg{|}_{\Box^{n}J=0}\,,$
(76)

which is evaluated to be

$\displaystyle\quad\langle
0|\mathrm{T}\Box^{n}\phi(x_{1})\Box^{n}\phi(x_{2})\cdots\Box^{n}\phi(x_{n})|0\rangle$
(77)
$\displaystyle=\frac{\int\mathcal{D}\Box^{n}(x)\Box^{n}\phi(x_{1})\Box^{n}\phi(x_{2})\cdots\Box^{n}\phi(x_{n})\exp\Big{(}\frac{i}{4^{n}}\int
d^{4n+4}x\,\Box^{n}\phi\Big{(}-\frac{\Box+m^{2}}{2}\Big{)}\Box^{n}\phi\Big{)}}{\int\mathcal{D}\Box^{n}\phi(x)\exp\Big{(}\frac{i}{4^{n}}\int
d^{4n+4}x\,\Box^{n}\phi\Big{(}-\frac{\Box+m^{2}}{2}\Big{)}\Box^{n}\phi\Big{)}}\,.$

Using the same technique in [32] similar to the case of gauge field, one finds
the Feynman propagator as

$\langle
0|\mathrm{T}\Box^{n}\phi(x)\Box^{n}\phi(y)|0\rangle=\int\frac{d^{4n+4}p}{(2\pi)^{4n+4}}\frac{i\cdot
4^{n}}{p^{4n}(p^{2}-m^{2})}e^{-ip\cdot(x-y)}\,.$ (78)

Or in the most generalized case, for the case of inhomogeneous order, we have

$\langle
0|\mathrm{T}\Box^{n}\phi(x)\Box^{k}\phi(y)|0\rangle=\int\frac{d^{2n+2k+4}p}{(2\pi)^{2n+2k+4}}\frac{i\cdot
2^{n+k}}{p^{2n+2k}(p^{2}-m^{2})}e^{-ip\cdot(x-y)}\,.$ (79)

From 78, when $n=0$ which is the unrotored case this restores us back to the
original two point correlation and the Feynman propagator of the scalar
fields,

$\langle
0|\mathrm{T}\phi(x)\phi(y)|0\rangle=\int\frac{d^{4}p}{(2\pi)^{4}}\frac{i}{(p^{2}-m^{2})}e^{-ip\cdot(x-y)}\,.$
(80)

###  4.3 Generalized Dirac field and Electrodynamics

For the quantization of Dirac field under rotor mechanism, first we consider
the quantum amplitude as follow:

$\displaystyle\quad\langle\gamma^{\mu_{n}\cdots\mu_{1}}\nabla_{\mu_{n}\cdots\mu_{1}}\psi(t_{f},\boldsymbol{\mathrm{x}})|e^{-i\hat{H}(t_{f}-t_{i})}|\nabla_{\nu_{n}\cdots\nu_{1}}\bar{\psi}(t_{i},\boldsymbol{\mathrm{x}})\,\gamma^{\dagger\nu_{1}\cdots\nu_{n}}\rangle$
(81)
$\displaystyle=\int\mathcal{D}[(\nabla_{\nu_{n}\cdots\nu_{1}}\bar{\psi})\,\gamma^{\dagger\nu_{1}\cdots\nu_{n}}]\mathcal{D}[\gamma^{\mu_{n}\cdots\mu_{1}}\nabla_{\mu_{n}\cdots\mu_{1}}\psi]$
$\displaystyle\quad\quad\quad\quad\times\exp\bigg{(}i\int
d^{D}x\,(\nabla_{\nu_{n}\cdots\nu_{1}}\bar{\psi})\,\gamma^{\dagger\nu_{1}\cdots\nu_{n}}(i\not{\partial}-m)\,\gamma^{\mu_{n}\cdots\mu_{1}}\nabla_{\mu_{n}\cdots\mu_{1}}\psi\bigg{)}\,.$

The sourced partition functional is then given by

$\displaystyle\quad
Z[\nabla_{\beta_{n}\cdots\beta_{1}}\bar{J}(x)\,\gamma^{\dagger\beta_{1}\cdots\beta_{n}}\,,\,\gamma^{\alpha_{n}\cdots\alpha_{1}}\nabla_{\alpha_{n}\cdots\alpha_{1}}J(x)]$
(82)
$\displaystyle=\int\mathcal{D}[(\nabla_{\nu_{n}\cdots\nu_{1}}\bar{\psi})\,\gamma^{\dagger\nu_{1}\cdots\nu_{n}}]\mathcal{D}[\gamma^{\mu_{n}\cdots\mu_{1}}\nabla_{\mu_{n}\cdots\mu_{1}}\psi]$
$\displaystyle\quad\quad\times\exp\bigg{(}i\int
d^{D}x\,(\nabla_{\nu_{n}\cdots\nu_{1}}\bar{\psi})\,\gamma^{\dagger\nu_{1}\cdots\nu_{n}}(i\not{\partial}-m)\,\gamma^{\mu_{n}\cdots\mu_{1}}\nabla_{\mu_{n}\cdots\mu_{1}}\psi$
$\displaystyle\quad\quad\quad+i\int
d^{D}x(\nabla_{\nu_{n}\cdots\nu_{1}}\bar{\psi})\,\gamma^{\dagger\nu_{1}\cdots\nu_{n}}\gamma^{\alpha_{n}\cdots\alpha_{1}}\nabla_{\alpha_{n}\cdots\alpha_{1}}J(x)$
$\displaystyle\quad\quad\quad+i\int
d^{D}x\nabla_{\beta_{n}\cdots\beta_{1}}\bar{J}(x)\,\gamma^{\dagger\beta_{1}\cdots\beta_{n}}\gamma^{\mu_{n}\cdots\mu_{1}}\nabla_{\mu_{n}\cdots\mu_{1}}\psi\bigg{)}\,.$

And it is remarked that

$\begin{aligned} &\quad
iS[(\nabla_{\nu_{n}\cdots\nu_{1}}\bar{\psi})\,\gamma^{\dagger\nu_{1}\cdots\nu_{n}},\gamma^{\mu_{n}\cdots\mu_{1}}\nabla_{\mu_{n}\cdots\mu_{1}}\psi]\\\
&=\iint
d^{D}xd^{D}y(\nabla_{\nu_{n}\cdots\nu_{1}}\bar{\psi}(x))\,\gamma^{\dagger\nu_{1}\cdots\nu_{n}}i\delta^{D}(x-y)(i\not{\partial}-m)\gamma^{\mu_{n}\cdots\mu_{1}}\nabla_{\mu_{n}\cdots\mu_{1}}\psi(y)\end{aligned}\,.$
(83)

Now perform fourier transform,

$\psi(x)=\int\frac{d^{D}p}{(2\pi)^{D}}\tilde{\psi}(p)e^{-ip\dot{x}}\quad\text{and}\quad\bar{\psi}(x)=\int\frac{d^{D}p}{(2\pi)^{D}}\bar{\tilde{\psi}}(p)e^{+ip\dot{x}}\,.$
(84)

Equation 83 yields

$\displaystyle\quad
iS[(\nabla_{\nu_{n}\cdots\nu_{1}}\bar{\psi})\,\gamma^{\dagger\nu_{1}\cdots\nu_{n}},\gamma^{\mu_{n}\cdots\mu_{1}}\nabla_{\mu_{n}\cdots\mu_{1}}\psi]$
(85) $\displaystyle=\iint
d^{D}xd^{D}y\iint\frac{d^{D}p}{(2\pi)^{D}}\frac{d^{D}q}{(2\pi)^{D}}(i^{n})(-i)^{n}p_{\nu_{n}}\cdots
p_{\nu_{1}}\bar{\tilde{\psi}}(p)\gamma^{\dagger\nu_{1}\cdots\nu_{n}}i\delta^{D}(x-y)(\not{q}-m)\gamma^{\mu_{n}\cdots\mu_{1}}$
$\displaystyle\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\times
q_{\mu_{n}}\cdots q_{\mu_{1}}\tilde{\psi}(q)e^{-ip\cdot x+iq\cdot y}$
$\displaystyle=\int
d^{D}x\iint\frac{d^{D}p}{(2\pi)^{D}}\frac{d^{D}q}{(2\pi)^{D}}p_{\nu_{n}}\cdots
p_{\nu_{1}}\bar{\tilde{\psi}}(p)\gamma^{\nu_{1}\dagger}\cdots\gamma^{\nu_{n}\dagger}i(\not{q}-m)\gamma^{\mu_{n}}\cdots\gamma^{\mu_{1}}q_{\mu_{n}}\cdots
q_{\mu_{1}}\tilde{\psi}(q)e^{-i(p-q)\cdot x}$
$\displaystyle=\iint\frac{d^{D}p}{(2\pi)^{D}}\frac{d^{D}q}{(2\pi)^{D}}(\not{p}^{\dagger})^{n}\bar{\tilde{\psi}}(p)i(2\pi)^{D}\delta^{D}(p-q)(\not{q}-m)\not{q}^{n}\tilde{\psi}(q)$
$\displaystyle=\int\frac{d^{D}p}{(2\pi)^{D}}\bar{\tilde{\psi}}(p)(\not{p}^{\dagger})^{n}i(\not{p}-m)(\not{p})^{n}\tilde{\psi}(p)\,.$

Therefore, we have the matrix element as

$\tilde{M}(p,q)=-i(2\pi)^{D}\delta^{D}(p-q)(\not{p}^{\dagger})^{n}(\not{q}-m)(\not{q})^{n}\,.$
(86)

Therefore, the momentum green’s function for Dirac field under rotor mechanism
is

$\tilde{M}^{-1}(p,q)=\frac{i}{(\not{p}^{\dagger})^{n}(\not{q}-m)(\not{q})^{n}}(2\pi)^{D}\delta^{D}(p-q)\,.$
(87)

Hence, the Feynman propagator in momentum space is

$S_{F}^{(n)}(p)=\frac{i}{(\not{p}^{\dagger})^{n}(\not{p}-m)(\not{p})^{n}}\,.$
(88)

Now, let’s evaluate the sourceless partition functional, which is

$\displaystyle\quad
Z[0,0]=\int\mathcal{D}[(\nabla_{\nu_{n}\cdots\nu_{1}}\bar{\psi})\,\gamma^{\dagger\nu_{1}\cdots\nu_{n}}]\mathcal{D}[\gamma^{\mu_{n}\cdots\mu_{1}}\nabla_{\mu_{n}\cdots\mu_{1}}\psi]$
(89) $\displaystyle\quad\quad\quad\quad\quad\quad\quad\times\exp\bigg{(}i\int
d^{D}x\,(\nabla_{\nu_{n}\cdots\nu_{1}}\bar{\psi})\,\gamma^{\dagger\nu_{1}\cdots\nu_{n}}(i\not{\partial}-m)\,\gamma^{\mu_{n}\cdots\mu_{1}}\nabla_{\mu_{n}\cdots\mu_{1}}\psi\bigg{)}\,,$

which is

$\displaystyle\quad
Z[0,0]=\int\mathcal{D}[(\nabla_{\nu_{n}\cdots\nu_{1}}\bar{\psi})\,\gamma^{\dagger\nu_{1}\cdots\nu_{n}}]\mathcal{D}[\gamma^{\mu_{n}\cdots\mu_{1}}\nabla_{\mu_{n}\cdots\mu_{1}}\psi]$
(90) $\displaystyle\quad\quad\times\exp\bigg{(}i\iint
d^{D}xd^{D}y\,(\nabla_{\nu_{n}\cdots\nu_{1}}\bar{\psi}(x))\,\gamma^{\dagger\nu_{1}\cdots\nu_{n}}\delta^{D}(x-y)(i\not{\partial}-m)\,\gamma^{\mu_{n}\cdots\mu_{1}}\nabla_{\mu_{n}\cdots\mu_{1}}\psi(y)\bigg{)}\,,$

They by using the identity from Gaussian integral,

$\int\bigg{(}\prod_{j=1}^{n}d\theta_{j}^{*}d\theta_{j}\bigg{)}\exp(-\theta^{*T}M\theta)=\det(M)\,,$
(91)

we obtain

$Z[0,0]=\det[\delta^{D}(x-y)(i\not{\partial}-m)]\,.$ (92)

Now, using the usual fact from Gaussian integral with sourced term,

$\int\bigg{(}\prod_{j=1}^{n}d\theta_{j}^{*}d\theta_{j}\bigg{)}\exp(-\theta^{*T}M\theta+\eta^{*T}\theta+\theta^{*T}\eta)=\det(M)\exp(\eta^{*T}M^{-1}\eta)\,,$
(93)

and with equations (82) and (83), the sourced partition functional for the
Dirac field is evaluated to be

$\displaystyle\quad
Z[\nabla_{\beta_{n}\cdots\beta_{1}}\bar{J}(x)\,\gamma^{\dagger\beta_{1}\cdots\beta_{n}}\,,\,\gamma^{\alpha_{n}\cdots\alpha_{1}}\nabla_{\alpha_{n}\cdots\alpha_{1}}J(x)]$
(94) $\displaystyle=\det[\delta^{D}(x-y)(i\not{\partial}-m)]\exp\bigg{(}\iint
d^{D}xd^{D}y\nabla_{\beta_{n}\cdots\beta_{1}}\bar{J}(x)\,\gamma^{\dagger\beta_{1}\cdots\beta_{n}}M^{-1}(x,y)\gamma^{\alpha_{n}\cdots\alpha_{1}}\nabla_{\alpha_{n}\cdots\alpha_{1}}J(y)\bigg{)}$
$\displaystyle=Z[0,0]\exp\bigg{(}\iint
d^{D}xd^{D}y\nabla_{\beta_{n}\cdots\beta_{1}}\bar{J}(x)\,\gamma^{\dagger\beta_{1}\cdots\beta_{n}}M^{-1}(x,y)\gamma^{\alpha_{n}\cdots\alpha_{1}}\nabla_{\alpha_{n}\cdots\alpha_{1}}J(y)\bigg{)}\,.$

The normalized generating functional for the Dirac field is

$\displaystyle\mathcal{Z}[\nabla_{\beta_{n}\cdots\beta_{1}}\bar{J}(x)\,\gamma^{\dagger\beta_{1}\cdots\beta_{n}}\,,\,\gamma^{\alpha_{n}\cdots\alpha_{1}}\nabla_{\alpha_{n}\cdots\alpha_{1}}J(x)]=\frac{Z[\nabla_{\beta_{n}\cdots\beta_{1}}\bar{J}(x)\,\gamma^{\dagger\beta_{1}\cdots\beta_{n}}\,,\,\gamma^{\alpha_{n}\cdots\alpha_{1}}\nabla_{\alpha_{n}\cdots\alpha_{1}}J(x)]}{Z[0,0]}\,.$
(95)

The two point correlation function is given by

$\displaystyle\quad\langle
0|\mathrm{T}[\gamma^{\mu_{n}\cdots\mu_{1}}\nabla_{\mu_{n}\cdots\mu_{1}}\psi(x_{1})][(\nabla_{\nu_{n}\cdots\nu_{1}}\bar{\psi}(x_{2}))\,\gamma^{\dagger\nu_{1}\cdots\nu_{n}}]|0\rangle$
(96)
$\displaystyle=\frac{1}{i^{2}}\frac{\delta^{2}\mathcal{Z}[\nabla_{\beta_{n}\cdots\beta_{1}}\bar{J}(x)\,\gamma^{\dagger\beta_{1}\cdots\beta_{n}}\,,\,\gamma^{\alpha_{n}\cdots\alpha_{1}}\nabla_{\alpha_{n}\cdots\alpha_{1}}J(x)]}{\delta[\gamma^{\mu_{n}\cdots\mu_{1}}\nabla_{\mu_{n}\cdots\mu_{1}}J(x_{1})]\delta[(\nabla_{\nu_{n}\cdots\nu_{1}}\bar{J}(x_{2}))\,\gamma^{\dagger\nu_{1}\cdots\nu_{n}}]}\bigg{|}_{\gamma^{\mu_{n}\cdots\mu_{1}}\nabla_{\mu_{n}\cdots\mu_{1}}J(x_{1})=(\nabla_{\nu_{n}\cdots\nu_{1}}\bar{J}(x_{2}))\,\gamma^{\dagger\nu_{1}\cdots\nu_{n}}=0}\,.$

The variational principle follows that

$\frac{\delta\gamma^{\alpha_{n}\cdots\alpha_{1}}\nabla_{\alpha_{n}\cdots\alpha_{1}}J(x)}{\delta\gamma^{\mu_{n}\cdots\mu_{1}}\nabla_{\mu_{n}\cdots\mu_{1}}J(y)}=\delta^{D}(x-y)$
(97)

and

$\frac{\delta\nabla_{\beta_{n}\cdots\beta_{1}}\bar{\psi}(x)\,\gamma^{\dagger\beta_{1}\cdots\beta_{n}}}{\delta\nabla_{\nu_{n}\cdots\nu_{1}}\bar{\psi}(y)\,\gamma^{\dagger\nu_{1}\cdots\nu_{n}}}=\delta^{D}(x-y)\,.$
(98)

This would give

$\displaystyle\quad\langle
0|\mathrm{T}[\gamma^{\mu_{n}\cdots\mu_{1}}\nabla_{\mu_{n}\cdots\mu_{1}}\psi(x_{1})][(\nabla_{\nu_{n}\cdots\nu_{1}}\bar{\psi}(x_{2}))\,\gamma^{\dagger\nu_{1}\cdots\nu_{n}}]|0\rangle=M^{-1}(x,y)$
(99)
$\displaystyle=\int\frac{d^{D}p}{(2\pi)^{D}}\frac{i}{(\not{p}^{\dagger})^{n}(\not{p}-m)(\not{p})^{n}}e^{-ip\cdot(x-y)}\,.$

For the inhomogeneous case, we will have,

$\langle
0|\mathrm{T}[\gamma^{\mu_{n}\cdots\mu_{1}}\nabla_{\mu_{n}\cdots\mu_{1}}\psi(x_{1})][(\nabla_{\nu_{k}\cdots\nu_{1}}\bar{\psi}(x_{2}))\,\gamma^{\dagger\nu_{1}\cdots\nu_{k}}]|0\rangle=\int\frac{d^{D}p}{(2\pi)^{D}}\frac{i}{(\not{p}^{\dagger})^{n}(\not{p}-m)(\not{p})^{k}}e^{-ip\cdot(x-y)}\,.$
(100)

From equation (99), when $n=0$, this restores us back the fermion two-point
correlation function and Feynman propagator

$\langle
0|\mathrm{T}\psi(x)\bar{\psi}(y)|0\rangle=\int\frac{d^{D}p}{(2\pi)^{D}}\frac{i}{\not{p}-m}e^{-ip\cdot(x-y)}\,.$
(101)

To quantize the generalized electrodynamics under rotor mechanism, we first
consider the sourceless functional as follow:

$Z[0,0,0]=\int\mathcal{D}[(\nabla_{\nu_{n}\cdots\nu_{1}}\bar{\psi})\,\gamma^{\dagger\nu_{1}\cdots\nu_{n}}]\mathcal{D}[\gamma^{\mu_{n}\cdots\mu_{1}}\nabla_{\mu_{n}\cdots\mu_{1}}\psi]\mathcal{D}\Box^{n}T^{\mu}\exp\big{(}iS_{\mathrm{QED}}\big{)}\,,$
(102)

where $S_{\mathrm{QED}}$ is the quantum electrodynamic action in 51. The
normalized generating functional for QED is

$\displaystyle\quad\mathcal{Z}_{\mathrm{QED}}[\nabla_{\beta_{n}\cdots\beta_{1}}\bar{J}(x)\,\gamma^{\dagger\beta_{1}\cdots\beta_{n}}\,,\,\gamma^{\alpha_{n}\cdots\alpha_{1}}\nabla_{\alpha_{n}\cdots\alpha_{1}}J(x),\Box^{n}J^{\mu}]$
(103) $\displaystyle=\frac{\exp{(i\int
d^{D}z\mathcal{L}_{\mathrm{int}}[\mathrm{source}])}Z[\nabla_{\beta_{n}\cdots\beta_{1}}\bar{J}(x)\,\gamma^{\dagger\beta_{1}\cdots\beta_{n}}\,,\,\gamma^{\alpha_{n}\cdots\alpha_{1}}\nabla_{\alpha_{n}\cdots\alpha_{1}}J(x),\Box^{n}J^{\mu}]}{\\{\exp{(i\int
d^{D}z\mathcal{L}_{\mathrm{int}}[\mathrm{source}])}Z[\nabla_{\beta_{n}\cdots\beta_{1}}\bar{J}(x)\,\gamma^{\dagger\beta_{1}\cdots\beta_{n}}\,,\,\gamma^{\alpha_{n}\cdots\alpha_{1}}\nabla_{\alpha_{n}\cdots\alpha_{1}}J(x),\Box^{n}J^{\mu}]\\}|_{\mathrm{sourced\,\,terms=0}}}\,,$

where

$\begin{aligned} &\quad\exp{\bigg{(}i\int
d^{4}z\,.\mathcal{L}_{\mathrm{int}}[\mathrm{source}]\bigg{)}}\\\
&=\exp\bigg{(}i\int
d^{D}z\frac{1}{i\delta\nabla_{\beta_{n}\cdots\beta_{1}}\bar{J}(z)\,\gamma^{\dagger\beta_{1}\cdots\beta_{n}}}\frac{1}{i\delta\gamma^{\alpha_{n}\cdots\alpha_{1}}\nabla_{\alpha_{n}\cdots\alpha_{1}}J(y)}\frac{1}{i\delta\Box^{n}J^{\mu}(z)}\bigg{)}\end{aligned}\\\
\,.$ (104)

Then the full physical two-point correlation function is

$\langle\Omega|\mathrm{T}\psi(x)\bar{\psi}(y)|\Omega\rangle=\frac{1}{i^{2}}\frac{\delta^{2}\mathcal{Z}_{\mathrm{QED}}}{\delta\nabla_{\beta_{n}\cdots\beta_{1}}\bar{J}(x)\,\gamma^{\dagger\beta_{1}\cdots\beta_{n}}\delta\gamma^{\alpha_{n}\cdots\alpha_{1}}\nabla_{\alpha_{n}\cdots\alpha_{1}}J(y)}\bigg{|}_{\mathrm{source\,\,terms=0}}$
(105)

with $|\Omega\rangle$ the physical vacuum.

###  4.4 A summary of Feynman propagators under rotor mechanism

Using the path integral quantization technique approach, we can obtain the
Feynman propagators under rotor mechanism ( n-rotors of $\Box^{n}$ operators)
as follow:

#### Scalar spin-0 boson:

$\Delta_{F}^{(n)}(x-y)=\int\frac{d^{D}p}{(2\pi)^{D}}\frac{i\cdot
4^{n}}{p^{4n}(p^{2}-m^{2})}e^{-ip\cdot(x-y)}\,.$ (106)

#### Massless spin-1 gauge boson:

$D_{\mu\nu}^{(n)}(x-y)=\int\frac{d^{D}p}{(2\pi)^{D}}\frac{-i\cdot
4^{n}g_{\mu\nu}}{p^{2+4n}}e^{-ip\cdot(x-y)}\,.$ (107)

#### Massless spin-1 gauge boson in Lorentz gauge:

$D_{\mu\nu}^{(n)}(x-y)=\int\frac{d^{D}p}{(2\pi)^{D}}\frac{-i\cdot
4^{n}}{p^{2+4n}}\bigg{(}g_{\mu\nu}-(1-\xi)\frac{p_{\mu}p_{\nu}}{p^{2}}\bigg{)}e^{-ip\cdot(x-y)}\,.$
(108)

#### Massive spin-1 gauge boson:

$D_{\mu\nu}^{(n)}(x-y)=\int\frac{d^{D}p}{(2\pi)^{D}}\frac{-i\cdot
4^{n}}{p^{4n}(p^{2}-M^{2})}\bigg{(}g_{\mu\nu}-\frac{p_{\mu}p_{\nu}}{M^{2}}\bigg{)}e^{-ip\cdot(x-y)}\,.$
(109)

#### Massive spin-1 gauge boson in Lorentz gauge

$D_{\mu\nu}^{(n)}(x-y)=\int\frac{d^{D}p}{(2\pi)^{D}}\frac{-i\cdot
4^{n}}{p^{4n}}\bigg{(}\frac{g_{\mu\nu}}{p^{2}-M^{2}}-\frac{1}{p^{2}-M^{2}}\bigg{(}1-\frac{1}{\xi}\bigg{)}\bigg{(}\frac{1}{M^{2}-\frac{p^{2}}{\xi}}\bigg{)}p_{\mu}p_{\nu}\bigg{)}e^{-ip\cdot(x-y)}\,.$
(110)

#### Dirac spin-1/2 fermion:

$S_{F}^{(n)}(x-y)=\int\frac{d^{D}p}{(2\pi)^{D}}\frac{i}{(\not{p}^{\dagger})^{n}(\not{p}-m)(\not{p})^{n}}e^{-ip\cdot(x-y)}\,.$
(111)

#### Massless spin-1 gluon:

$D^{ab\,(n)}_{\mu\nu}(x-y)=\delta^{ab}D^{(n)}_{\mu\nu}(x-y)=\int\frac{d^{D}p}{(2\pi)^{D}}\frac{-i\cdot
4^{n}\delta^{ab}g_{\mu\nu}}{p^{2+4n}}\bigg{(}g_{\mu\nu}-(1-\xi)\frac{p_{\mu}p_{\nu}}{p^{2}}\bigg{)}e^{-ip\cdot(x-y)}\,.$
(112)

#### Ghost field:

$G^{ab\,(n)}(x-y)=\int\frac{d^{D}p}{(2\pi)^{D}}\frac{i\delta^{ab}}{p^{2+4n}}e^{-ip\cdot(x-y)}\,.$
(113)

The corresponding vertices in rotored quantum field theory is

Figure 1:  Vertices in momentum space under rotor quantum field theory.

##  5 BRST Symmetry of the rotor model

It is known that the Faddeev-Popov ghost field arises from the gauge fixing
procedure of non-abelian gauge field, and so do the rotored ghost field
$\Box^{n}c$ appears in the generalized action (72) of the sourced partition
function. It is noted that the rotor propagators contains both physical
transverse polarized degree of freedom and the unpolarized time and
longitudinal-polarized degree of freedom, this potentially leads to problems
in loop diagrams, whicn contains non-physical degree of freedom. In order to
construct a valid quantum rotored Yang-Mills gauge field theory of general
$n$-th order, we need to check that if the Yang-Mills theory under rotor model
possess BRST (Becchi-Rouet-Stora-Tyutin) symmetry, such that the contributions
from the non-physical degree of freedom are cancelled by the rotored ghosts.
Since the cancellation extends to the full theory, the most convenient way is
to check if the rotored non-abelian gauge field and rotored ghost field has
BRST symmetry.

To see the BRST symmetry in the rotor mechanism, we need to slightly rewrite
the action of a set of rotored fields

$S_{\xi}[\Box^{n}T_{\mu},\Box^{n}c^{*},\Box^{n}c,\nabla_{\nu_{n}\cdots\nu_{1}}\bar{\psi}\,\gamma^{\dagger\nu_{1}\cdots\nu_{n}}\,,\,\gamma^{\mu_{n}\cdots\mu_{1}}\nabla_{\mu_{n}\cdots\mu_{1}}\psi]$
(114)

by using a Lagrange multiplier. We also introduce a set of new commuting rotor
auxiliary fields $\Box^{n}B=\Box^{n}B^{a}t^{a}\,,$ which has no independent
dynamics. Then

$\displaystyle\quad
S_{\xi}[\Box^{n}T_{\mu},\Box^{n}c^{*},\Box^{n}c,\nabla_{\nu_{n}\cdots\nu_{1}}\bar{\psi}\,\gamma^{\dagger\nu_{1}\cdots\nu_{n}}\,,\,\gamma^{\mu_{n}\cdots\mu_{1}}\nabla_{\mu_{n}\cdots\mu_{1}}\psi\,,\Box^{n}B]$
(115) $\displaystyle=\int
d^{4n+4}x\bigg{(}-\frac{1}{2}\mathrm{Tr}G_{n\,\mu\nu}G^{\mu\nu}_{n}+(\nabla_{\nu_{n}\cdots\nu_{1}}\bar{\psi})\,\gamma^{\dagger\nu_{1}\cdots\nu_{n}}(i\gamma^{\alpha}D_{n\,\alpha}-m)\,\gamma^{\mu_{n}\cdots\mu_{1}}\nabla_{\mu_{n}\cdots\mu_{1}}\psi$
$\displaystyle\quad\quad\quad\quad\quad\quad\quad+2\mathrm{Tr}(\partial^{\mu}\Box^{n}c^{*})(D_{n\,\mu}\Box^{n}c)+\frac{1}{4^{n}}\xi\mathrm{Tr}\,(\Box^{n}B\Box^{n}B)+\frac{2}{4^{n}}\mathrm{Tr}(\Box^{n}B\partial_{\mu}\Box^{n}T^{\mu})\bigg{)}\,.$

By ignoring an overall constant factor, the sourced partition function is
given by

$\displaystyle Z[\mathrm{source}]$
$\displaystyle=\int\mathcal{D}\Box^{n}T_{\mu}\mathcal{D}[(\nabla_{\nu_{n}\cdots\nu_{1}}\bar{\psi})\gamma^{\dagger\nu_{1}\cdots\nu_{n}}]\mathcal{D}[\gamma^{\mu_{n}\cdots\mu_{1}}\nabla_{\mu_{n}\cdots\mu_{1}}\psi]\mathcal{D}\Box^{n}c^{*}\mathcal{D}\Box^{n}c\mathcal{D}\Box^{n}B$
(116) $\displaystyle\quad\quad\quad\times\exp(iS_{\xi}+\text{source
terms})\,.$

We find that the action in (115) has a rotored BRST symmetry by the following
infinitesimal transformations. Let $\epsilon$ be a constant Grassmann number,

$\displaystyle\delta\Box^{n}T_{\mu}$ $\displaystyle=2^{n}\epsilon
D_{n\,\mu}\Box^{n}c$ (117)
$\displaystyle\delta(\gamma^{\mu_{n}\cdots\mu_{1}}\nabla_{\mu_{n}\cdots\mu_{1}}\psi)$
$\displaystyle=ig\epsilon
c\gamma^{\mu_{n}\cdots\mu_{1}}\nabla_{\mu_{n}\cdots\mu_{1}}\psi$
$\displaystyle\delta[(\nabla_{\nu_{n}\cdots\nu_{1}}\bar{\psi})\,\gamma^{\dagger\nu_{1}\cdots\nu_{n}}]$
$\displaystyle=-ig\epsilon
c(\nabla_{\nu_{n}\cdots\nu_{1}}\bar{\psi})\,\gamma^{\dagger\nu_{1}\cdots\nu_{n}}$
$\displaystyle\delta\Box^{n}c$ $\displaystyle=ig\epsilon\Box^{n}c\Box^{n}c$
$\displaystyle\delta\Box^{n}c^{*}$
$\displaystyle=\frac{1}{2^{n}}\epsilon\Box^{n}B$
$\displaystyle\delta\Box^{n}B$ $\displaystyle=0\,.$

Now we check it is a symmetry. If the action in 115 satisfies the BRST
symmetry, it satisfies $\delta S_{\xi}=0$ under the infinitesimal
transformation in 117. First notice that
$\delta\Box^{n}T_{\mu}=2^{n}D_{n\,\mu}(\epsilon\Box^{n}c)$ and
$\delta(\gamma^{\mu_{n}\cdots\mu_{1}}\nabla_{\mu_{n}\cdots\mu_{1}}\psi)=ig(\epsilon
c)\gamma^{\mu_{n}\cdots\mu_{1}}\nabla_{\mu_{n}\cdots\mu_{1}}\psi$, this is
just as if the gauge parameter $\Box^{n}\alpha=\epsilon\Box^{n}c$ where
$\Box^{n}\alpha\equiv\Box^{n}\alpha^{a}t^{a}$ is the phase matrix parameter in
the unitary transformation $U=\exp(i\Box^{n}\alpha)$. It follows that

$-\frac{1}{2}\delta(\mathrm{Tr}G_{n\,\mu\nu}G^{\mu\nu}_{n})=0\quad\text{and}\quad\delta[(\nabla_{\nu_{n}\cdots\nu_{1}}\bar{\psi})\,\gamma^{\dagger\nu_{1}\cdots\nu_{n}}(i\gamma^{\alpha}D_{n\,\alpha}-m)\,\gamma^{\mu_{n}\cdots\mu_{1}}\nabla_{\mu_{n}\cdots\mu_{1}}\psi]=0\,.$
(118)

Next we check

$\displaystyle\delta(D_{n\,\mu}\Box^{n}c)$
$\displaystyle=\delta\Big{(}\partial_{\mu}\Box^{n}c-ig\Big{[}\frac{1}{2}\Box^{n}T_{\mu}\,,\,\Box^{n}c\Big{]}\Big{)}$
(119)
$\displaystyle=\partial_{\mu}\delta\Box^{n}c-ig\Big{[}\frac{1}{2^{n}}\Box^{n}T_{\mu}\,,\,\delta\Box^{n}c\Big{]}-ig\Big{[}\frac{1}{2}\delta\Box^{n}T_{\mu}\,,\,\Box^{n}c\Big{]}$
$\displaystyle=D_{n\,\mu}\delta\Box^{n}c-ig[\epsilon
D_{n\,\mu}\Box^{n}c,\Box^{n}c]$
$\displaystyle=D_{n\,\mu}(ig\epsilon\Box^{n}c\Box^{n}c)-ig\epsilon(D_{n\,\mu}\Box^{n}c)\Box^{n}c+ig\Box^{n}c(\epsilon
D_{n\,\mu}\Box^{n}c)$
$\displaystyle=ig\epsilon(D_{n\,\mu}\Box^{n}c)\Box^{n}c+ig\epsilon\Box^{n}c(D_{n\,\mu}\Box^{n}c)-ig\epsilon(D_{n\,\mu}\Box^{n}c)\Box^{n}c+ig\Box^{n}c\epsilon(D_{n\,\mu}\Box^{n}c)$
$\displaystyle=-ig\Box^{n}c\epsilon(D_{n\,\mu}\Box^{n}c)+ig\Box^{n}c\epsilon(D_{n\,\mu}\Box^{n}c)$
$\displaystyle=0\,.$

And since $\delta\Box^{n}B=0$, then the remaining variation gives

$\displaystyle\quad\delta\Big{(}2\mathrm{Tr}(\partial^{\mu}\Box^{n}c^{*})(D_{n\,\mu}\Box^{n}c)+\frac{1}{4^{n}}\xi\mathrm{Tr}\,(\Box^{n}B\Box^{n}B)+\frac{2}{4^{n}}\mathrm{Tr}(\Box^{n}B\partial_{\mu}\Box^{n}T^{\mu})\Big{)}$
(120)
$\displaystyle=2\mathrm{Tr}(\partial^{\mu}\delta\Box^{n}c^{*})(D_{n\,\mu}\Box^{n}c)+\mathrm{Tr}(\partial^{\mu}\Box^{n}c^{*})(\delta
D_{n\,\mu}\Box^{n}c)+\frac{2}{4^{n}}\mathrm{Tr}(\Box^{n}B\partial_{\mu}\delta\Box^{n}T^{\mu})$
$\displaystyle=-2\mathrm{Tr}(\delta\Box^{n}c^{*}\partial^{\mu}D_{n\,\mu}\Box^{n}c)+\frac{2}{4^{n}}\mathrm{Tr}\Big{(}\Box^{n}B\partial^{\mu}(2^{n}\epsilon
D_{n\,\mu}\Box^{n}c)\Big{)}$
$\displaystyle=-2\mathrm{Tr}\Big{(}\frac{1}{2^{n}}\epsilon\Box^{n}B\partial_{\mu}D_{n\,\mu}\Box^{n}c\Big{)}+\frac{2}{2^{n}}\mathrm{Tr}\Big{(}\Box^{n}B\partial^{\mu}(2^{n}\epsilon
D_{n\,\mu}\Box^{n}c)\Big{)}$ $\displaystyle=0\,.$

Therefore, it satisfies $\delta S_{\xi}=0$. Thus, there exists a BRST symmetry
for the rotor mechanism.

Next, we can use a notation $Q$ to generalize the BRST symmetry,

$\delta_{\epsilon}\Box^{n}T_{\mu}=\epsilon Q\Box^{n}T_{\mu}\,,$ (121)

and

$\delta_{\epsilon}(\gamma^{\mu_{n}\cdots\mu_{1}}\nabla_{\mu_{n}\cdots\mu_{1}}\psi)=\epsilon
Q\gamma^{\mu_{n}\cdots\mu_{1}}\nabla_{\mu_{n}\cdots\mu_{1}}\psi\,.$ (122)

A key property is that

$Q^{2}=0$ (123)

so that BRST symmetry under rotor mechanism is nilpotent. For example

$\epsilon
Q\epsilon^{\prime}Q\Box^{n}T_{\mu}=\delta_{\epsilon}(\epsilon^{\prime}D_{n\,\mu}\Box^{n}c)=\epsilon^{\prime}\delta_{\epsilon}(D_{n\,\mu}\Box^{n}c)=0\,.$
(124)

And

$\displaystyle\epsilon Q\epsilon^{\prime}Q\Box^{n}c$
$\displaystyle=\delta_{\epsilon}(ig\epsilon^{\prime}\Box^{n}c\Box^{n}c)$ (125)
$\displaystyle=ig\epsilon^{\prime}[(\delta_{\epsilon}\Box^{n}c)\Box^{n}c+\Box^{n}c(\delta_{\epsilon}\Box^{n}c)]$
$\displaystyle=ig\epsilon^{\prime}[ig\epsilon(\Box^{n}c)^{2}\Box^{n}c+\Box^{n}cig\epsilon(\Box^{n}c)^{2}]$
$\displaystyle=-g^{2}\epsilon^{\prime}[\epsilon(\Box^{n}c)^{3}-\epsilon(\Box^{n}c)^{3}]$
$\displaystyle=0\,.$

Therefore in general, we have

$Q^{2}(\Box^{n}(\mathrm{field}))=0\,.$ (126)

###  5.1 Slavnov-Taylor Identities for the Rotor Model

The Ward identity imposes the condition of gauge symmetry in quantum
electrodynamics for abelian gauge fields. The generalization of Ward identity
to non-abelian gauge field is manifested in Slavnov-Taylor identities [37, 38,
39], which is far less trivial. And the generalization to rotor model is
complicated, but it is essential to check for gauge symmetry of rotored Yang
Mills theory. First consider the gauge-fixed sourced generating functional,

$\displaystyle Z[\Box^{n}J]$
$\displaystyle=\int\mathcal{D}\Box^{n}T^{\mu}\mathcal{D}\Box^{n}c^{*}\mathcal{D}\Box^{n}c\exp\bigg{(}i\int
d^{4n+4}x\big{(}\mathcal{L}^{(n)}_{\mathrm{YM}}+\Box^{n}J_{\mu}^{a}\Box^{n}B^{\mu\,a}$
(127) $\displaystyle\quad\quad\quad\quad+\frac{1}{2\cdot
4^{n}\xi}(\partial_{\mu}\Box^{n}T^{\mu})^{2}+\Box^{n}c^{*a}\partial^{\mu}D_{n\mu}^{ab}\Box^{n}c^{b}\big{)}\bigg{)}$
$\displaystyle=\int\mathcal{D}\Box^{n}T^{\mu}\mathcal{D}\Box^{n}c^{*}\mathcal{D}\Box^{n}c\exp\bigg{(}i\int
d^{4n+4}x\big{(}\mathcal{L}^{(n)}_{\mathrm{YM}}+\Box^{n}J_{\mu}^{a}\Box^{n}B^{\mu\,a}$
$\displaystyle\quad\quad\quad\quad+\frac{1}{2\cdot
4^{n}\xi}(\partial_{\mu}\Box^{n}T^{\mu})^{2}+\Box^{n}c^{*a}\Box^{n+1}c^{a}+\frac{g}{2^{n}}f^{abc}\Box^{n}c^{*a}\Box^{n}c^{b}\partial^{\mu}\Box^{n}T_{\mu}^{c}\big{)}\bigg{)}\,,$

where $\frac{1}{2\cdot 4^{n}\xi}(\partial_{\mu}\Box^{n}T^{\mu})^{2}$ is the
gauge-fixing term of rotored non-abelian gauge field. And we define the matrix
element

$M^{ab}=\partial^{\mu}D_{n\mu}^{ab}\,.$ (128)

To obtain the generalized Slavnov-Taylor identities of the rotor model, the
integral in (127) we make an infinitesimal rotored gauge transformation of the
field variables as in equation (17) and (18), where the phase satisfies

$M^{ab}\alpha^{b}(x)=\chi^{a}(x)$ (129)

for some arbitrary function $\chi^{a}(x)$. The inverse of $M^{ab}$,
$M^{-1\,ab}$ is the Green’s function of the rotor ghost-field under external
rotor gauge field $\Box^{n}T^{\mu}$. By carrying out variation for the
integral (127) and equating the coefficient $\chi^{a}(x)$ to zero, we obtain

$\displaystyle\int\mathcal{D}\Box^{n}T^{\mu}\mathcal{D}\Box^{n}c^{*}\mathcal{D}\Box^{n}c\exp\bigg{(}i\int
d^{4n+4}x\big{(}\mathcal{L}^{(n)}_{\mathrm{YM}}+\Box^{n}J_{\mu}^{a}\Box^{n}B^{\mu\,a}\frac{1}{2\cdot
4^{n}\xi}(\partial_{\mu}\Box^{n}T^{\mu})^{2}+\Box^{n}c^{*a})\bigg{)}$ (130)
$\displaystyle\times\bigg{\\{}-\frac{1}{2^{n}\xi}\partial_{\mu}\Box^{n}T_{\mu}^{a}(y)+\int
d^{4n+4}z\Box^{n}c^{*a}(y)\Box^{n}J^{\mu\,b}(z)[D_{n\mu}\Box^{n}c(z)]^{b}\bigg{\\}}=0\,.$

Equation (130) is the generalized Slavnov-Taylor identity for the rotor model.
By differentiating (130) with respect to external source $\Box^{n}J_{\mu}$ and
putting $\Box^{n}J_{\mu}=0$, we obtain the following identity

$\frac{1}{\alpha}\langle
0|\mathrm{T}\partial^{\mu}\Box^{n}A_{\mu}^{a}(x)\Box^{n}A^{b}_{\nu}(y)=-i\langle
0|\mathrm{T}\Box^{n}c^{*a}(x)(D_{n\nu}\Box^{n}c)^{b}(y)|0\rangle\,.$ (131)

##  6 Generalized Higgs mechanism under rotor mechanism

The Higgs mechanism is responsible for the explanation of mass acquisition of
gauge boson through the process of spontaneous symmetry breaking, it also
explains how the fermions couple with the Higgs field to main mass [40, 41,
42, 43, 44, 45]. In addition, it also predicts all possible Higgs interactions
and decays.

Under rotor mechanism, the potential term is

$V=\frac{\mu^{2}}{4^{n}}\Box^{n}\phi^{\dagger}\Box^{n}\phi+\frac{\lambda}{16^{n}}(\Box^{n}\phi^{\dagger}\Box^{n}\phi)^{2}\,,$
(132)

where $\mu^{2}<0$ and $\lambda>0$ for spontaneous symmetry breaking. The Higgs
action is

$S=\int
d^{D}x\bigg{(}-\frac{1}{4^{n+1}}\Box^{n}G_{\mu\nu}\Box^{n}G^{\mu\nu}+\frac{1}{4^{n}}(D_{n\,\mu}\Box^{n}\phi)^{\dagger}(D_{n\,\mu}\Box^{n}\phi)-\frac{\mu^{2}}{4^{n}}\Box^{n}\phi^{\dagger}\Box^{n}\phi-\frac{\lambda}{16^{n}}(\Box^{n}\phi^{\dagger}\Box^{n}\phi)^{2}\bigg{)}\,,$
(133)

This Lagrangian is invariant under U(1) transformation,

$(\Box^{n}\phi)\rightarrow(\Box^{n}\phi)^{\prime}=e^{i\theta}(\Box^{n}\phi)\,.$
(134)

When expressed in terms of two real rotored scalar fields $\Box^{n}\phi_{1}$
and $\Box^{n}\phi_{2}$, action reads

$V(\Box^{n}\phi_{1},\Box^{n}\phi_{2})=\frac{\mu^{2}}{2\cdot
4^{n}}\big{(}(\Box^{n}\phi_{1})^{2}+(\Box^{n}\phi_{2})^{2}\big{)}+\frac{\lambda}{4\cdot
16^{n}}\big{(}(\Box^{n}\phi_{1})^{2}+(\Box^{n}\phi_{2})^{2}\big{)}^{2}\,.$
(135)

The potential has an infinite minimum when

$(\Box^{n}\phi_{1})^{2}+(\Box^{n}\phi_{2})^{2}=\frac{-4^{n}\cdot\mu^{2}}{\lambda}=v_{n}^{2}\,.$
(136)

Upon the elimination of Goldstone boson by Unitary gauge and by perturbation
of the non-zero vacuum expectation value, we have

$\Box^{n}\phi(x)=\frac{1}{\sqrt{2}}\big{(}v_{n}+\Box^{n}h(x)\big{)}\,.$ (137)

Then after spontaneously symmetry breaking, we have

$\displaystyle S$ $\displaystyle=\int
d^{D}x\bigg{(}-\frac{1}{4^{n+1}}\Box^{n}G_{\mu\nu}\Box^{n}G^{\mu\nu}$ (138)
$\displaystyle\,\,\,\,\,\,\,\,+\frac{1}{2\cdot
4^{n}}\Big{(}\partial_{\mu}-\frac{i}{2^{n}}g\Box^{n}T_{\mu}(v_{n}+\Box^{n}h)\Big{)}\Big{(}\partial^{\mu}+\frac{i}{2^{n}}g\Box^{n}T_{\mu}(v_{n}+\Box^{n}h)\Big{)}$
$\displaystyle\,\,\,\,\,\,\,\,-\frac{\mu^{2}}{2\cdot
4^{n}}(v_{n}+\Box^{n}h)^{2}-\frac{\lambda}{4\cdot
16^{n}}(v_{n}+\Box^{n}h)^{4}\bigg{)}$ $\displaystyle=\int
d^{D}x\bigg{(}-\frac{1}{4^{n+1}}\Box^{n}G_{\mu\nu}\Box^{n}G^{\mu\nu}+\frac{1}{2\cdot
4^{n}}g^{2}v_{n}^{2}\Box^{n}T_{\mu}\Box^{n}T^{\mu}$
$\displaystyle\,\,\,\,\,\,\,\,\quad\quad\quad\quad+\frac{1}{2\cdot
4^{n}}\partial_{\mu}\Box^{n}h\partial^{\mu}\Box^{n}h-\frac{1}{4^{n}}\lambda
v_{n}^{2}\Box^{n}h\Box^{n}h$
$\displaystyle\,\,\,\,\,\,\,\,\quad\quad\quad\quad+\frac{1}{8^{n}}g^{2}v_{n}\Box^{n}T_{\mu}\Box^{n}T^{\mu}\Box^{n}h+\frac{1}{2\cdot
8^{n}}g^{2}\Box^{n}T_{\mu}\Box^{n}T^{\mu}\Box^{n}h\Box^{n}h$
$\displaystyle\,\,\,\,\,\,\,\,\quad\quad\quad\quad-\frac{1}{8^{n}}\lambda
v_{n}\Box^{n}h\Box^{n}h\Box^{n}h-\frac{1}{4\cdot
16^{n}}\lambda\Box^{n}h\Box^{n}h\Box^{n}h\Box^{n}h\bigg{)}\,.$

The mass term of the rotored massive gauge boson is identified as
$M_{n}=gv_{n}$, and the mass term of the rotored Higgs boson is
$m_{n}=\sqrt{2\lambda}v_{n}$. It is noted that when $n=0$, the action (138)
will restore back to the normal Higgs action with spontaneous symmetry
breaking. From the third line in (138), this gives the interactions between
the rotored Higgs boson and rotored gauge field. We can rewrite them in terms
of the metric,

$\frac{1}{8^{n}}gM_{n}g_{\mu\nu}\Box^{n}T^{\mu}\Box^{n}T^{\nu}\Box^{n}h\,,\quad\frac{1}{2\cdot
8^{n}}g^{2}g_{\mu\nu}\Box^{n}T^{\mu}\Box^{n}T^{\nu}\Box^{n}h\Box^{n}h\,.$
(139)

In terms of vertex diagram, we have the following feynman rules,

Figure 2:  Feyman rules for rotored Higgs boson decaying into two rotored
bosons (left) and rotored scattering process (right).

From the fourth line in 138, this gives the rotored Higgs boson self
interactions. The Feynman rules are as follow:

Figure 3:  Feyman rules for third order and fourth order rotored Higgs boson
interactions.

##  7 A discussion of Hierarchy Problem and the partial solution provided by
the rotor mechanism

It has been known that high-order derivative quantum field theory makes a good
job in eliminating UV divergence [4, 5, 6, 7, 8]. In this section, we will
investigate how the high-order-derivative quantum field theory by the rotor
mechanism can suppress UV divergence in the quantum loop processes. Although
the standard model is renormalizable, i.e. the divergences in the Lagrangian
can be treated by adding finite number of counter terms, the Hierarchy Problem
which involves fine tunning of the Higgs mass remains a long-lasting problem
in high-energy particle physics. Recently, reference [1] generalizes Lee-Wick
electrodynamics with high-order field derivatives to the Standard Model and
offers a solution to the Hierarchy Problem to tame UV divergences in one-loop
level. Here, we will see how high-order derivative field theory by rotor
mechanism can eliminate the UV divergence in amplitudes.

Let us have a revisit to the Hierarchy problem. For example, taking the
4-fields self-interaction of the Higgs boson the one-particle irreducible (1
PI) function can be easily calculated by integrating through the loop momentum
$k$,

$\tilde{\Gamma}(p)=\frac{-i\lambda^{2}}{8}\int\frac{d^{4}k}{(2\pi)^{4}}\frac{i}{k^{2}-m_{\rm
H}^{2}}=\frac{-i\lambda^{2}}{128\pi^{2}}\big{[}\Lambda^{2}-2m_{\rm
H}^{2}\ln\left(\frac{\Lambda}{m_{\rm H}}\right)+\mathcal{O}\left(\frac{m_{\rm
H}^{2}}{\Lambda^{4}}\right)\big{]}+\mathcal{O}(\lambda^{4}),$ (140)

which has a dominant quadratic divergence of high energy scale $\Lambda$.
Hence, the renormalization of the bare Higgs boson’s mass for the 4-field self
interaction is,

$\tilde{m}_{\rm H}^{2}=m_{\rm H}^{2}+i\tilde{\Gamma}(p)=m_{\rm
H}^{2}+\frac{\lambda^{2}}{128\pi^{2}}\big{[}\Lambda^{2}-2m_{\rm
H}^{2}\ln\left(\frac{\Lambda}{m_{\rm H}}\right)+\mathcal{O}\left(\frac{m_{\rm
H}^{2}}{\Lambda^{2}}\right)\big{]}+\mathcal{O}(\lambda^{4}).$ (141)

It is noted that $m_{\rm H}=\mu$ above is the bare mass, while $\tilde{m}_{\rm
H}$ is the physical, observed mass-the renormalized mass. The full calculation
of all quantum loop corrections from electroweak and Yukawa interaction to the
$n^{\rm th}$-order loop takes the general form [46],

${\rm\tilde{m}_{H}^{2}}=m_{\rm
H}^{2}+\frac{\Lambda^{2}}{16\pi^{2}}C_{n}(\mu^{2}=m_{\rm H}^{2})\,,$ (142)

where $C_{n}$ is the a polynomial expansion of the bare Higgs mass scale and
it is a function of the Higgs self-coupling $\lambda$, electroweak couplings
$g_{W},g^{\prime}$ and the Yukawa coupling $Y^{f}_{ij}$. To the first-order
loop calculation [46],

$\mathrm{\tilde{m}_{H}^{2}=m_{\rm
H}^{2}}+\frac{\Lambda^{2}}{32\pi^{2}}(4\lambda^{2}+3g^{\prime
2}+9g_{W}^{2}-24Y^{f\,2}_{ij})+\mathcal{O}(\lambda^{2},g^{\prime
4},g_{W}^{4},Y^{f\,4}_{ij})\,.$ (143)

The physical Higgs boson mass is measured to be $\tilde{m}_{\rm H}\simeq
126\,\rm GeV$ and hence $\tilde{m}^{2}_{\rm H}\sim 10^{4},\rm GeV^{2}$, while
the quantum corrections lead to the quadratic divergence of the Planck scale
$\Lambda\sim 10^{19}\,\rm GeV$. From equation (143) we can see that the
physical Higgs mass depends strictly on the strength of each of the coupling.
The nature requires an extremely precise fine-tuning cancellation of the
couplings and the bare Higgs mass to each $n^{\rm th}$ order of quantum loop
divergence by $10^{17}$ GeV, so as to obtain the low-energy, EW-scale Standard
Model we observe today [47]. This is known as the Hierarchy Problem.

The Hierarchy Problem gives a strong motivation to search for New Physics (NP)
particles with mass scales $\Lambda_{\rm NP}>\Lambda_{\rm EW}$ such that they
can cancel these loop divergences. One important scheme is supersymmetry,
which introduces new particles between bosonic and fermonic symmetry, such
that each divergent bosonic loop has a fermionic counterpart, vice versa.
Since fermonic amplitude has an extra factor of minus sign, this can cancels
the bosonic loop.

Using the rotor mechanism, we will see that the divergent loop processes are
suppressed and this will lead to convergent result. The generic problematic
1-loop divergent process includes, for example, the self-correction of the
Higgs boson by the W-boson loop. The Feynman diagram is illustrated below.

Figure 4:  Feyman diagram of self-energy correction of the Higgs boson by a
W-boson loop.

The $n$-rotored amplitude is given by

$\mathcal{M}^{(n)}=\frac{g_{W}^{2}(16)^{n}M_{W}^{2}}{64^{n}(2\pi)^{D}}\int
d^{D}k\,g^{\alpha\beta}g^{\delta\gamma}\frac{g_{\beta\delta}-k_{\beta}k_{\delta}/M_{W}^{2}}{k^{4n}(k^{2}-M_{W}^{2})}\frac{g_{\gamma\alpha}-(p-k)_{\gamma}(p-k)_{\alpha})/M_{W}^{2}}{(p-k)^{4n}((p-k)^{2}-M_{W}^{2})}\,.$
(144)

By expanding, we obtain

$\displaystyle\mathcal{M}^{(n)}$
$\displaystyle=\frac{g_{W}^{2}M_{W}^{2}}{4^{n}(2\pi)^{D}}\bigg{(}\int
d^{D}k\frac{g^{\alpha\beta}g_{\beta\delta}g^{\delta\gamma}g_{\gamma\alpha}}{k^{4n}(p-k)^{4n}(k^{2}-M_{W}^{2})((p-k)^{2}-M_{W}^{2})}$
(145) $\displaystyle\quad\quad\quad\quad\quad-\frac{1}{M_{W}^{2}}\int
d^{D}k\frac{(p-k)^{\alpha}(p-k)_{\alpha}}{k^{4n}(p-k)^{4n}(k^{2}-M_{W}^{2})((p-k)^{2}-M_{W}^{2})}$
$\displaystyle\quad\quad\quad\quad\quad-\frac{1}{M_{W}^{2}}\int
d^{D}k\frac{k^{\alpha}k_{\alpha}}{k^{4n}(p-k)^{4n}(k^{2}-M_{W}^{2})((p-k)^{2}-M_{W}^{2})}$
$\displaystyle\quad\quad\quad\quad\quad+\frac{1}{M_{W}^{4}}\int
d^{D}k\frac{k_{\delta}(p-k)^{\delta}k^{\alpha}(p-k)_{\alpha}}{k^{4n}(p-k)^{4n}(k^{2}-M_{W}^{2})((p-k)^{2}-M_{W}^{2})}\bigg{)}\,.$

First, we can give a rough estimate of the amplitude by the leading term,

$\displaystyle\mathcal{M}^{(n)}_{1}$
$\displaystyle\sim\frac{g_{W}^{2}M_{W}^{2}}{4^{n}(2\pi)^{D}}\int
d^{D}k\frac{g^{\alpha\beta}g_{\beta\delta}g^{\delta\gamma}g_{\gamma\alpha}}{k^{4n}(p-k)^{4n}(k^{2}-M_{W}^{2})((p-k)^{2}-M_{W}^{2})}$
(146)
$\displaystyle\sim\frac{2^{1-D}Dg_{W}^{2}M_{W}^{2}}{4^{n}\sqrt{\pi^{D}}\Gamma\big{(}\frac{D}{2}\big{)}}\int^{\Lambda}\frac{k^{D-1}dk}{k^{8n+4}}\,,$

where have have used the fact that of the integration of D-dimensional solid
angle is given by $\int d\Omega_{D}=\frac{2\pi^{D/2}}{\Gamma(D/2)}$ with
$\Gamma(x)$ the Gamma function. When $n=0$ and in four spacetime dimension
$D=4$ which is the unrotored case in 4D spacetime (which is just the normal
case), we have logarithmic divergence

$\mathcal{M}^{(0)}_{1}\sim\frac{g_{W}^{2}M_{W}^{2}}{2\pi^{2}}\ln\Lambda\,.$
(147)

Hence in the normal case, the self-correction of the Higgs boson by the W
boson is logarithmic divergent. But under rotor mechanism, by the second line
of 154. we have

$\mathcal{M}^{(n)}_{1}\sim\frac{2^{1-D}Dg_{W}^{2}M_{W}^{2}}{4^{n}\sqrt{\pi^{D}}\Gamma\big{(}\frac{D}{2}\big{)}(D-8n-4)}\Lambda^{D-8n-4}\,,$
(148)

In our 4D universe, we simply have

$\mathcal{M}^{(n)}_{1}\sim-\frac{g_{W}^{2}M^{2}_{W}}{2^{2n+4}\pi^{2}n}\frac{1}{\Lambda^{8n}}\,.$
(149)

As $\Lambda$ is a very large value, which can be up to Planck scale $10^{19}$
GeV, even for $n=1$ $\mathcal{M}^{(1)}_{1}\rightarrow 0$. For the second term,

$\displaystyle\mathcal{M}^{(n)}_{2}$
$\displaystyle=\frac{g_{W}^{2}}{4^{n}(2\pi)^{D}}\int
d^{D}k\frac{1}{k^{4n}(p-k)^{4n-2}(k^{2}-M_{W}^{2})((p-k)^{2}-M_{W}^{2})}$
(150)
$\displaystyle\sim\frac{2^{1-D}g_{W}^{2}}{4^{n}\sqrt{\pi^{D}}\Gamma\big{(}\frac{D}{2}\big{)}}\int^{\Lambda}\frac{k^{D-1}dk}{k^{8n+2}}$
$\displaystyle=\frac{2^{1-D}g_{W}^{2}}{4^{n}\sqrt{\pi^{D}}\Gamma\big{(}\frac{D}{2}\big{)}(D-8n-2)}\Lambda^{D-8n-2}\,.$

For the unrotored case $n=0$ and in $D=4$,

$\mathcal{M}^{(0)}_{2}\sim\frac{g_{W}^{2}}{16\pi^{2}}\Lambda^{2}\,,$ (151)

which contributes to a quadratic divergence. But under rotor mechanism,

$\mathcal{M}^{(n)}_{2}\sim\frac{g_{W}^{2}}{2^{(2n+4)}\pi^{2}(1-4n)}\Lambda^{2-8n}\,.$
(152)

For example for $n=1$, $\mathcal{M}^{(1)}_{2}\propto\frac{1}{\Lambda^{6}}$
which is convergent. For the third term,

$\displaystyle\mathcal{M}^{(n)}_{3}$
$\displaystyle=\frac{g_{W}^{2}}{4^{n}(2\pi)^{D}}\int
d^{D}k\frac{1}{k^{4n-2}(p-k)^{4n}(k^{2}-M_{W}^{2})((p-k)^{2}-M_{W}^{2})}$
(153)
$\displaystyle\sim\frac{2^{1-D}g_{W}^{2}}{4^{n}\sqrt{\pi^{D}}\Gamma\big{(}\frac{D}{2}\big{)}}\int^{\Lambda}\frac{k^{D-1}dk}{k^{8n+2}}$
$\displaystyle=\frac{2^{1-D}g_{W}^{2}}{4^{n}\sqrt{\pi^{D}}\Gamma\big{(}\frac{D}{2}\big{)}(D-8n-2)}\Lambda^{D-8n-2}\,,$

which has the same result as the $\mathcal{M}_{2}^{(n)}$ case. For the last
term,

$\displaystyle\mathcal{M}^{(n)}_{4}$
$\displaystyle\sim\frac{g_{W}^{2}}{4^{n}(2\pi)^{D}M_{W}^{2}}\int
d^{D}k\frac{(k\cdot
p-k^{2})^{2}}{k^{4n}(p-k)^{4n}(k^{2}-M_{W}^{2})((p-k)^{2}-M_{W}^{2})}$ (154)
$\displaystyle=\frac{g_{W}^{2}}{4^{n}(2\pi)^{D}M_{W}^{2}}\int
d^{D}k\frac{(k\cdot p)^{2}-2(k\cdot
p)k^{2}+k^{4}}{k^{4n}(p-k)^{4n}(k^{2}-M_{W}^{2})((p-k)^{2}-M_{W}^{2})}$
$\displaystyle\sim\frac{g_{W}^{2}}{4^{n}(2\pi)^{D}M_{W}^{2}}\int
d^{D}k\frac{k^{2}p^{2}-2(k\cdot
p)k^{2}+k^{4}}{[k^{2}(p-k)^{2}]^{2n}(k^{2}-M_{W}^{2})((p-k)^{2}-M_{W}^{2})}$
$\displaystyle=\frac{g_{W}^{2}}{4^{n}(2\pi)^{D}M_{W}^{2}}\int
d^{D}k\frac{1}{[k^{2}(p-k)^{2}]^{2n-1}(k^{2}-M_{W}^{2})((p-k)^{2}-M_{W}^{2})}\,.$

From the second line to the third line, we have made the order approximation.
Notice that $(k\cdot
p)^{2}=k_{\mu}p^{\mu}k_{\alpha}p^{\alpha}=(k_{\mu}p^{\alpha})p^{\mu}k_{\alpha}$.
Then notice that
$k^{2}p^{2}=k_{\mu}k^{\mu}p_{\alpha}p^{\alpha}=(k_{\mu}p^{\alpha})k^{\mu}p_{\alpha}$.
They are obviously different but they have the same order in $k$ and $p$. As
here we are interested in order calculation, to simplify the analysis we make
the above approximation. Then we have

$\mathcal{M}^{(n)}_{4}\sim\frac{2^{1-D}g_{W}^{2}}{4^{n}\sqrt{\pi^{D}}\Gamma\big{(}\frac{D}{2}\big{)}M_{W}^{2}}\int^{\Lambda}\frac{k^{D-1}dk}{k^{8n}}=\frac{2^{1-D}g_{W}^{2}}{4^{n}\sqrt{\pi^{D}}M_{W}^{2}\Gamma\big{(}\frac{D}{2}\big{)}(D-8n)}\Lambda^{D-8n}\,.$
(155)

For the unrotored case $n=0$ and in $D=4$,

$\mathcal{M}^{(0)}_{4}\sim\frac{g_{W}^{2}}{32\pi^{2}}\Lambda^{4}\,,$ (156)

which is a seriously 4-th order divergent term. But under rotor mechanism,

$\mathcal{M}^{(n)}_{4}\sim\frac{g_{W}^{2}}{2^{(2n+5)}\pi^{2}M_{W}^{2}(1-2n)}\Lambda^{4-8n}\,.$
(157)

For example for $n=1$, $\mathcal{M}^{(1)}_{2}\propto\frac{1}{\Lambda^{4}}$
which is convergent. Therefore, the amplitude is

$\mathcal{M}^{(n)}=\mathcal{M}^{(n)}_{1}+\mathcal{M}^{(n)}_{2}+\mathcal{M}^{(n)}_{3}+\mathcal{M}^{(n)}_{4}\sim\frac{g_{W}^{2}}{4^{n+2}\pi^{2}\Lambda^{8n}}\bigg{(}\frac{M_{W}^{2}}{n}+\frac{\Lambda^{2}}{1-4n}+\frac{\Lambda^{4}}{4M_{W}^{2}(1-2n)}\bigg{)}$
(158)

in $D=4$ spacetime dimension. We can see that when $n=1$ under rotor
mechanism, the amplitude is already convergent in the high-energy regime.
Therefore, we see how the rotor mechanism can remove infinities in loop
diagrams at high energies.

Next, consider the self-correction of Higgs Boson by a fermion loop. The
Feynman diagram is as follow:

Figure 5:  Feyman diagram of self-energy correction of the Higgs boson by a
fermionic loop.

Since the vertex factor is just $-i\frac{m_{f}}{v_{n}}$, the $n$-rotored
amplitude is given by

$\mathcal{M}^{(n)}=-\frac{m_{f}^{2}}{v_{n}^{2}(2\pi)^{D}}\int
d^{D}k\,\mathrm{Tr}\bigg{(}\frac{1}{(\not{k}^{\dagger})^{n}(\not{k}-m_{f})(\not{k})^{n}}\frac{1}{(\not{p}^{\dagger}-\not{k}^{\dagger})^{n}(\not{p}-\not{k}-m_{f})(\not{p}-\not{k})^{n}}\bigg{)}\,.$
(159)

The minus sign is due to the Feynman rule for the fermion loop. Now using the
following fact that

$\displaystyle\gamma_{\mu}\gamma^{\mu}$ $\displaystyle=4I_{D}$ (160)
$\displaystyle(\gamma_{\mu}\gamma^{\mu})^{\dagger}$
$\displaystyle=4I_{D}^{\dagger}$
$\displaystyle\gamma^{\mu\dagger}\gamma_{\mu}^{\dagger}$
$\displaystyle=4I_{D}\,.$

Equivalently, the amplitude in (159) reads

$\mathcal{M}^{(n)}=-\frac{m_{f}^{2}}{v_{n}^{2}(2\pi)^{D}}\int
d^{D}k\,\mathrm{Tr}\bigg{(}\frac{(\not{k})^{n}(\not{k}+m_{f})(\not{k}^{\dagger})^{n}}{k^{4n}(k^{2}-m_{f}^{2})}\frac{(\not{p}-\not{k})^{n}(\not{p}-\not{k}+m_{f})(\not{p}^{\dagger}-\not{k}^{\dagger})^{n}}{(p-k)^{4n}((p-k)^{2}-m_{f}^{2})}\bigg{)}\,.$
(161)

The rough estimation of (159) gives

$\mathcal{M}^{(n)}\sim-\frac{m_{f}^{2}}{v_{n}^{2}(2\pi)^{D}}\int
d^{D}k\,\mathrm{Tr}\bigg{(}\frac{1}{\not{k}^{2+4n}}\bigg{)}\,.$ (162)

For the unrotored case, $n=0$ and in 4D spacetime, we have

$\mathcal{M}^{(n)}\sim-\frac{m_{f}^{2}}{v^{2}(2\pi)^{D}}\Lambda^{2}$ (163)

which has a quadratic divergence. But under rotor mechanism we have

$\mathcal{M}^{(n)}\sim-\frac{m_{f}^{2}}{v_{n}^{2}(2\pi)^{D}}\Lambda^{2-4n}\,.$
(164)

When $n=1$, we can see that $\mathcal{M}^{(n)}\propto\frac{1}{\Lambda^{2}}$,
which is convergent. Thus the divergence is eliminated.

Next, we consider the self-energy correction of the Higgs boson by the 3rd
order self-interaction. The vertex factor is $\frac{i}{2\cdot
8^{n}}\frac{m_{\mathrm{H}}^{2}}{v_{n}}$. The Feynman diagram is

Figure 6:  Feyman diagram of self-energy correction of the Higgs boson by a
3rd-order self interaction.

The amplitude for such process is

$\mathcal{M}^{(n)}=\frac{(16)^{n}m_{\mathrm{H}}^{4}}{4\cdot
64^{n}(2\pi)^{D}v_{n}^{2}}\int
d^{D}k\frac{1}{k^{4n}(k^{2}-m_{\mathrm{H}}^{2})}\frac{1}{(p-k)^{4n}((p-k)^{2}-m_{\mathrm{H}}^{2})}\,.$
(165)

By rough estimation, we have

$\displaystyle\mathcal{M}^{(n)}$
$\displaystyle\sim\frac{m_{\mathrm{H}}^{4}}{2^{D+2n+1}\sqrt{\pi^{D}}v_{n}^{2}}\int^{\Lambda}\frac{k^{D-1}dk}{k^{4+8n}}$
(166)
$\displaystyle\sim\frac{m_{\mathrm{H}}^{4}}{2^{D+2n+1}\sqrt{\pi^{D}}(D-8n-4)v_{n}^{2}}\Lambda^{D-8n-4}\,.$

For the unrotored case in 4D spacetime, $n=0$, we have logarithmic divergence

$\mathcal{M}^{(0)}\sim\frac{m_{\mathrm{H}}^{4}}{32\pi^{2}v_{n}^{2}}\ln\Lambda$
(167)

For the general case, if we consider $D=4$ for our universe, we have

$\mathcal{M}^{(n)}\sim-\frac{m_{\mathrm{H}}^{4}}{2^{D+4n+4}\sqrt{\pi^{D}}nv_{n}^{2}}\frac{1}{\Lambda^{8n}}\,.$
(168)

For $n=1$, the matrix element goes for
$\mathcal{M}^{(n)}\propto\frac{1}{\Lambda^{8}}$, which drops very quickly to
zero for large $\Lambda$. Therefore, the problem of divergence vanishes in the
rotor mechanism.

In general, the above method of power counting used by the above example
applies to other diagrams with higher number of loops. The generic form takes
the following:

$\mathcal{M}^{(n)}\sim\int\cdots\int\frac{d^{D}k_{1}d^{D}k_{2}\cdots
d^{D}k_{L}}{k_{i}^{4n}\cdots k_{l}^{4n}k^{2}_{i}\cdots
k^{2}_{l}(\not{k}_{j}^{\dagger})^{n}(\not{k}_{j})^{n+1}\cdots(\not{k}_{p}^{\dagger})^{n}(\not{k}_{p})^{n+1}}\,.$
(169)

Let $S$ be the superficial degree of divergence. And let $L$ be the number of
loops, $N_{\mathrm{B}}$ be the number of internal spin-1 gauge boson
propagator(s), $N_{\mathrm{H}}$ be the number of internal Higgs boson
propagator(s) and $N_{f}$ be the number of internal fermion propagators. Then
we have

$\displaystyle S$ $\displaystyle=\text{(power of momentum in numerator)-
(power of momentum in demoninator )}$ (170)
$\displaystyle=DL-(2N_{\mathrm{B}}+4N_{\mathrm{B}}n)-(2N_{\mathrm{H}}+4N_{\mathrm{H}}n)-(N_{f}+2N_{f}n)$
$\displaystyle=DL-(2N_{\mathrm{B}}+2N_{\mathrm{H}}+N_{f})-2n(2N_{\mathrm{B}}+2N_{\mathrm{H}}+N_{f})$
$\displaystyle=DL-(2n+1)(2N_{\mathrm{B}}+2N_{\mathrm{H}}+N_{f})\,.$

Naively, we expect a diagram to have a divergence proportional to
$\Lambda^{S}$. We expect logarithmic divergence $\log\Lambda$ when $S=0$, and
no divergence when $S<0$. This takes place when the following inequality
holds,

$\displaystyle DL-(2n+1)(2N_{\mathrm{B}}+2N_{\mathrm{H}}+N_{f})$
$\displaystyle<0$ (171) $\displaystyle n$
$\displaystyle>\frac{DL}{(2N_{\mathrm{B}}+2N_{\mathrm{H}}+N_{f})}-\frac{1}{2}\,,$

and also we demand $n>0$. In addition, as the number of loops satisfy the
following equation,

$L=I-V+1\,,$ (172)

where $I$ is the number of internal lines, $V$ is the number of vertices, it
follows that

$L=N_{\mathrm{B}}+N_{\mathrm{H}}+N_{f}-V+1\,.$ (173)

Let us illustrate by an example. Consider the following 3rd-order self-energy
correction of the Higgs boson,

Figure 7:  Feyman diagram of self-energy correction of the Higgs boson by a
3-loop interaction.

The Feynman amplitude can be easily written down as

$\displaystyle\mathcal{M}^{(n)}$
$\displaystyle=-\frac{M_{W}^{3}m_{f}^{2}m_{\mathrm{H}}^{2}}{2(2\pi)^{3D}v_{n}^{3}}\iiint
d^{D}k\,d^{D}q\,d^{D}l\,g^{\beta\alpha}g^{\gamma\delta}g^{\mu\nu}$ (174)
$\displaystyle\,\,\,\,\times\frac{g_{\alpha\gamma}-(p-k)_{\alpha}(p-k)_{\gamma}/M_{W}^{2}}{(p-k)^{4n}((p-k)^{2}-M_{W}^{2})}\frac{g_{\delta\mu}-(p-k-q)_{\delta}(p-k-q)_{\mu}/M_{W}^{2}}{(p-k-q)^{4n}((p-k-q)^{2}-m_{W}^{2})}\frac{g_{\nu\beta}-k_{\nu}k_{\beta}/M_{W}^{2}}{k^{4n}(k^{2}-M_{W}^{2})}$
$\displaystyle\,\,\,\,\times\bigg{(}\frac{1}{q^{4n}(q^{2}-m_{\mathrm{H}}^{2})}\bigg{)}^{2}\frac{1}{(p-q)^{4n}(p-q)^{2}-m_{\mathrm{H}}^{2}}$
$\displaystyle\,\,\,\,\times\mathrm{Tr}\bigg{(}\frac{1}{(\not{l}^{*})^{n}(\not{l}-m)(\not{l}^{n})}\frac{1}{(\not{q}^{*}-\not{l}^{*})^{n}(\not{q}-\not{l}-m)(\not{q}-\not{l})^{n}}\bigg{)}\,.$

To compute the superficial degree of UV divergence, we utilize the power
counting formula in (170). First notice that
$N_{\mathrm{B}}=3,\,\,N_{\mathrm{H}}=3,\,\,N_{f}=2,\,\,V=6$, then $L=3$. Hence
it follows that $S=3D-14(2n+1)=3D-28n-14$. We can see that when $n=0$ for the
unrotored case and $D=4$ spacetime $S=-2$, thus this diagram is convergent. We
can also see that in $D=4$ spacetime $S=-2-28n<0$, so there is no UV
divergence.

Therefore, in this section, we have seen how infinities arise from high-energy
ends of loop diagrams are tamed by higher-order-derivatives fields under rotor
mechanism. In such way, the UV divergences of from the 1-loop self-correction
of Higgs boson propagator are eliminated.

##  8 Explicit calculation of 1-loop self-correction of Higgs boson under
rotor mechanism and a discussion on the Hierarchy Problem

###  8.1 Correction by the 3rd order Higgs vertex

In this section, we will demonstrate the explicit calculation of 1-loop self
correction of Higgs boson by third order Higgs vertex under rotor mechanism.
We will show that although the rotor mechanism can tame ultraviolet UV
divergence at high energies, it introduces Infra-red IR divergence at low
energies in 4-dimensional spacetime. However, we will demonstrate that the
rotor mechanism can remove all infinities in higher dimension using the
example of 1-loop self-correction of Higgs Boson by third order Higgs vertex.
According to equation (165), the n-th order rotored amplitude is proportional
to the following integral,

$\mathcal{M}^{(n)}\propto\int
d^{D}k\frac{1}{k^{4n}(k^{2}-m_{\mathrm{H}}^{2})}\frac{1}{(p-k)^{4n}((p-k)^{2}-m_{\mathrm{H}}^{2})}$
(175)

Define the integral as

$I^{(n)}=\int
d^{D}k\frac{1}{k^{4n}(k^{2}-m_{\mathrm{H}}^{2})}\frac{1}{(p-k)^{4n}((p-k)^{2}-m_{\mathrm{H}}^{2})}$
(176)

We will first demonstrate the calculation of $n=1$ case, then the general $n$
rotored case. Using the Feynman parameter method given in [36]

$\frac{1}{A_{1}A_{2}\cdots
A_{n}}=\int_{0}^{1}\cdots\int_{0}^{1}dx_{1}dx_{2}\cdots
dx_{n}\,\delta\bigg{(}\sum_{i=1}^{n}x_{i}-1\bigg{)}\frac{(n-1)!}{(x_{1}A_{1}+x_{2}A_{2}+\cdots
x_{n}A_{n})^{n}}\,.$ (177)

For the $n=1$ rotored case, we can rewrite the Feynman integral using equation
(177)

$\displaystyle\frac{1}{k^{4}(p-k)^{4}(k^{2}-m_{\mathrm{H}}^{2})((p-k)^{2}-m_{\mathrm{H}}^{2})}$
(178)
$\displaystyle=\int_{0}^{1}\int_{0}^{1}\int_{0}^{1}\int_{0}^{1}\int_{0}^{1}\int_{0}^{1}dxdydrdsdudv\delta(x+y+r+s+u+v-1)$
$\displaystyle\quad\times\frac{5!}{[xk^{2}+yk^{2}+r(p-k)^{2}+s(p-k)^{2}+u(k^{2}-m_{\mathrm{H}}^{2})+v((p-k)^{2}-m_{\mathrm{H}}^{2})]^{6}}$

Now let us evaluate the terms in the denominator.

$\displaystyle\quad
xk^{2}+yk^{2}+r(p-k)^{2}+s(p-k)^{2}+u(k^{2}-m_{\mathrm{H}}^{2})+v((p-k)^{2}-m_{\mathrm{H}}^{2})$
(179) $\displaystyle=xk^{2}+yk^{2}+r(p^{2}-2p\cdot k+k^{2})+s(p^{2}-2p\cdot
k+k^{2})+u(k^{2}-m_{\mathrm{H}}^{2})$ $\displaystyle\quad\quad+v(p^{2}-2p\cdot
k+k^{2}-m_{\mathrm{H}}^{2})$
$\displaystyle=(x+y+r+s+u+v)k^{2}+(r+s+v)p^{2}-2(r+s+v)(p\cdot
k)-(u+v)m_{\mathrm{H}}^{2}$
$\displaystyle=k^{2}-2(r+s+v)+(r+s+v)p^{2}-(u+v)m_{\mathrm{H}}^{2}$
$\displaystyle=(k-(r+s+v)p)^{2}-(u+v)m_{\mathrm{H}}^{2}\,,$

where in the fourth line we used the fact that $x+y+r+s+u+v=1$. We define new
variable $l=k-(r+s+v)p$ and the effective mass
$\Delta=(u+v)m_{\mathrm{H}}^{2}$. Also it is clearly that $d^{D}k=d^{D}l$.
Therefore, now the integral reads

$\displaystyle I^{(1)}$ $\displaystyle=\int
d^{D}k\int_{0}^{1}\int_{0}^{1}\int_{0}^{1}\int_{0}^{1}\int_{0}^{1}\int_{0}^{1}dxdydrdsdudv$
(180)
$\displaystyle\quad\times\delta(x+y+r+s+u+v-1)\frac{120}{[(k-(r+s+v)p)^{2})-(u+v)m_{\mathrm{H}}^{2}]^{6}}\,.$

Then we have

$I^{(1)}=\int
d^{D}l\int_{0}^{1}\int_{0}^{1}\int_{0}^{1}\int_{0}^{1}\int_{0}^{1}\int_{0}^{1}dxdydrdsdudv\delta(x+y+r+s+u+v-1)\frac{120}{(l^{2}-\Delta)^{6}}\,.$
(181)

Now we carry out Wick’s rotation. The Wick’s rotation simply amount to
substitute $l^{0}=il_{E}^{0}$ and $\boldsymbol{l}=\boldsymbol{l}_{E}$. Then
the integral now reads

$\displaystyle I^{(1)}$ $\displaystyle=120i(-1)^{6}\int
d\Omega_{4}\int_{0}^{\infty}dl_{E}\int_{0}^{1}\int_{0}^{1}\int_{0}^{1}\int_{0}^{1}\int_{0}^{1}\int_{0}^{1}dxdydrdsdudv$
(182)
$\displaystyle\quad\times\delta(x+y+r+s+u+v-1)\frac{l_{E}^{3}}{(l_{E}^{2}+\Delta)^{6}}$
$\displaystyle=\frac{120i(2\pi^{2})}{\Gamma(2)}\int_{0}^{\infty}dl_{E}\int_{0}^{1}\int_{0}^{1}\int_{0}^{1}\int_{0}^{1}\int_{0}^{1}\int_{0}^{1}dxdydrdsdudv$
$\displaystyle\quad\times\delta(x+y+r+s+u+v-1)\frac{l_{E}^{3}}{(l_{E}^{2}+\Delta)^{6}}\,.$

Next using the integral fact that

$\int_{0}^{\infty}\frac{x^{3}}{(x^{2}+a)^{6}}dx=-\frac{a+5x^{2}}{40(a+x^{2})^{5}}\bigg{|}^{\infty}_{0}=\frac{1}{40a^{4}}\,.$
(183)

Therefore, now the integral reads

$\displaystyle I^{(1)}$
$\displaystyle=\frac{240i\pi^{2}}{40m_{\mathrm{H}}^{8}}\int_{0}^{1}\int_{0}^{1}\int_{0}^{1}\int_{0}^{1}\int_{0}^{1}\int_{0}^{1}dxdydrdsdudv\delta(x+y+r+s+u+v-1)\frac{1}{(u+v)^{4}}$
(184)
$\displaystyle=\frac{6i\pi^{2}}{m_{\mathrm{H}}^{8}}\int_{0}^{1}du\int_{0}^{1-u}dv\int_{0}^{1-u-v}ds\int_{0}^{1-s-u-v}dr\int_{0}^{1-r-s-
u-v}dy\frac{1}{(u+v)^{4}}$
$\displaystyle=\frac{6i\pi^{2}}{m_{\mathrm{H}}^{8}}\lim_{u\rightarrow
0}\bigg{(}-\frac{1}{4}+\frac{1}{36u^{2}}-\frac{1}{4u}-\frac{1}{2}\ln
u\bigg{)}$
$\displaystyle=\frac{-3i\pi^{2}}{2m_{\mathrm{H}}^{8}}+\mathrm{infinity}\,.$

We can see that the integral diverges when $u\rightarrow 0$, this is referred
as the infra-red IR divergence. Finally, we obtain the amplitude as

$\mathcal{M}^{(1)}=\frac{3i}{512\pi^{2}m_{\mathrm{H}}^{4}v_{n}^{2}}+\mathrm{infinity}\,.$
(185)

Therefore, we can see that although the rotor mechanism resolves the high-
energy infinity problem, it introduces infra-red IR divergences at low energy
in $D=4$ spacetime.

#### The General $n$ case

To calculate the amplitude of general $n$-rotored case, we need to use the
following Feynman integral formula given in [36]

$\displaystyle\frac{1}{A_{1}^{m_{1}}A_{2}^{m_{2}}\cdots A_{n}^{m_{n}}}$
$\displaystyle=\int_{0}^{1}\cdots\int_{0}^{1}\bigg{(}\prod_{i=1}^{n}dx_{i}\bigg{)}\delta\bigg{(}\sum_{i=1}^{n}x_{i}-1\bigg{)}\frac{\prod_{i=1}^{n}x_{i}^{m_{i}-1}}{\big{(}\sum_{i=1}^{n}x_{i}A_{i}\big{)}^{\sum_{i=1}^{n}m_{i}}}$
(186)
$\displaystyle\quad\quad\quad\quad\quad\quad\quad\quad\quad\times\frac{\Gamma(m_{1}+m_{2}+\cdots+m_{n})}{\Gamma({m_{1}})\Gamma({m_{2}})\cdots\Gamma(m_{n})}\,.$

The n-rotored integral is calculated to be

$\begin{aligned} I^{(n)}&=\int
d^{D}k\frac{1}{(k^{2})^{2n}(k^{2}-m_{\mathrm{H}}^{2})}\frac{1}{[(p-k)^{2}]^{2n}((p-k)^{2}-m_{\mathrm{H}}^{2})}\\\
&=\int
d^{D}k\int_{0}^{1}\int_{0}^{1}\int_{0}^{1}\int_{0}^{1}dxdydrds\delta(x+y+r+s-1)\\\
&\quad\times\frac{x^{2n-1}y^{2n-1}r^{0}s^{0}}{[xk^{2}+y(p-k)^{2}+r(k^{2}-m_{\mathrm{H}}^{2})+s((p-k)^{2}-m_{\mathrm{H}}^{2})]^{4n+2}}\frac{\Gamma(4n+2)}{\Gamma(2n)\Gamma(2n)\Gamma(1)\Gamma(1)}\\\
&=\int
d^{D}k\int_{0}^{1}\int_{0}^{1}\int_{0}^{1}\int_{0}^{1}\delta(x+y+r+s-1)dxdydrds\\\
&\quad\times\frac{x^{2n-1}y^{2n-1}}{[(k-(y+s)p)^{2}-(r+s)m_{\mathrm{H}}^{2}]^{4n+2}}\frac{\Gamma(4n+2)}{\Gamma^{2}(2n)}\end{aligned}\,.$
(187)

This time take $l=k-(y+s)p$ and the effective mass
$\Delta(r+s)=m_{\mathrm{H}}^{2}$. Then we have

$\displaystyle I^{(n)}$ $\displaystyle=\frac{\Gamma(4n+2)}{\Gamma^{2}(2n)}\int
d^{D}l\int_{0}^{1}\int_{0}^{1}\int_{0}^{1}\int_{0}^{1}dxdydrds\delta(x+y+r+s-1)\frac{x^{2n-1}y^{2n-1}}{(l^{2}-\Delta)^{4n+2}}$
(188)
$\displaystyle=\frac{i(-1)^{4n+2}\Gamma(4n+2)}{\Gamma^{2}(2n)}\int_{0}^{\infty}dl_{E}\int
d\Omega_{D}\int_{0}^{1}\int_{0}^{1}\int_{0}^{1}\int_{0}^{1}dxdydrds\delta(x+y+r+s-1)$
$\displaystyle\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\times
x^{2n-1}y^{2n-1}\frac{l_{E}^{D-1}}{(l_{E}^{2}+\Delta)^{4n+2}}$
$\displaystyle=\frac{2i(-1)^{4n+2}\pi^{D/2}\Gamma(4n+2)}{\Gamma^{2}(2n)\Gamma\big{(}\frac{D}{2}\big{)}}\int_{0}^{\infty}dl_{E}\int_{0}^{1}\int_{0}^{1}\int_{0}^{1}\int_{0}^{1}dxdydrds\delta(x+y+r+s-1)$
$\displaystyle\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\times
x^{2n-1}y^{2n-1}\frac{l_{E}^{D-1}}{(l_{E}^{2}+\Delta)^{4n+2}}\,.$

Next we calculate the $l_{E}$ integral

$\int_{0}^{\infty}dl_{E}\frac{l_{E}^{D-1}}{(l_{E}^{2}+\Delta)^{4n+2}}=\frac{\Gamma\big{(}\frac{D}{2}\big{)}\Gamma\big{(}4n+2-\frac{D}{2}\big{)}}{2\Delta^{4n+2-\frac{D}{2}}\Gamma(4n+2)}\,\,\,\,\text{if}\,\,0<D<8n+4\,.$
(189)

Note that $D$ is out of the range, the integral does not converge. Therefore,
now the $I^{(n)}$ integral is

$\displaystyle I^{(n)}$
$\displaystyle=\frac{i(-1)^{4n+2}\pi^{D/2}\Gamma\big{(}4n+2-\frac{D}{2}\big{)}}{\Gamma^{2}(2n)}\int_{0}^{1}\int_{0}^{1}\int_{0}^{1}\int_{0}^{1}dxdydrds\delta(x+y+r+s-1)\frac{x^{2n-1}y^{2n-1}}{\Delta^{4n+2-\frac{D}{2}}}$
(190)
$\displaystyle=\frac{i(-1)^{4n+2}\pi^{D/2}\Gamma\big{(}4n+2-\frac{D}{2}\big{)}}{\Gamma^{2}(2n)m_{\mathrm{H}}^{8n+4-D}}\int_{0}^{1}\int_{0}^{1}\int_{0}^{1}\int_{0}^{1}dxdydrds\delta(x+y+r+s-1)\frac{x^{2n-1}y^{2n-1}}{(r+s)^{4n+2-\frac{D}{2}}}$
$\displaystyle=\frac{i(-1)^{4n+2}\pi^{D/2}\Gamma\big{(}4n+2-\frac{D}{2}\big{)}}{\Gamma^{2}(2n)m_{\mathrm{H}}^{8n+4-D}}\int_{0}^{1}dr\int_{0}^{1-r}ds\int_{0}^{1-r-s}dy\frac{(1-y-r-s)^{2n-1}y^{2n-1}}{(r+s)^{4n+2-\frac{D}{2}}}$
$\displaystyle=\frac{i(-1)^{4n+2}\pi^{D/2}\Gamma\big{(}4n+2-\frac{D}{2}\big{)}}{\Gamma^{2}(2n)m_{\mathrm{H}}^{8n+4-D}}\frac{\Gamma^{2}(2n)}{\Gamma(4n)}\int_{0}^{1}dr\int_{0}^{1-r}ds\frac{(1-r-s)^{4n-1}}{(r+s)^{4n+2-\frac{D}{2}}}$
$\displaystyle=\frac{i(-1)^{4n+2}\pi^{D/2}\Gamma\big{(}4n+2-\frac{D}{2}\big{)}}{(-1)^{4n}\Gamma(4n)m_{\mathrm{H}}^{8n+4-D}}\int_{0}^{1}drB\bigg{(}1-\frac{1}{r};4n,2-\frac{D}{2}\bigg{)}\,,$

where

$B\bigg{(}1-\frac{1}{r};4n,2-\frac{D}{2}\bigg{)}=\int_{0}^{1-\frac{1}{r}}t^{4n-1}(1-t)^{1-\frac{D}{2}}dt$
(191)

is the incomplete Beta function. Hence, we need to compute the following
integral

$I^{(n)}(n,D)=\frac{i\pi^{D/2}\Gamma\big{(}4n+2-\frac{D}{2}\big{)}}{\Gamma(4n)m_{\mathrm{H}}^{8n+4-D}}\int_{0}^{1}dr\int^{1-\frac{1}{r}}_{0}t^{4n-1}(1-t)^{1-\frac{D}{2}}dt\,.$
(192)

Let us define the integral

$J(n,D)=\int_{0}^{1}dr\int^{1-\frac{1}{r}}_{0}t^{4n-1}(1-t)^{1-\frac{D}{2}}dt\,.$
(193)

First we analyse the case when $n=1/4$, which is not an integer. Although this
corresponds to an unphysical rotor number, it is worth to study this case. It
is found that, when $n=1/4$, the integral is computed to be

$J\bigg{(}\frac{1}{4},D\bigg{)}=\frac{2}{D-4}\bigg{(}\frac{2}{D-2}-1\bigg{)}\,\,\mathrm{if}\,\,D>2\,.$
(194)

The plot is as follow:

Figure 8: The plot of $J\Big{(}\frac{1}{4},D\Big{)}$ against spacetime
dimension $D$.

When $n=1/4$, so the converged amplitude takes place only when

$-i\mathcal{M}^{(1/4)}\bigg{(}\frac{1}{4},D\bigg{)}=\frac{\Gamma\big{(}3-\frac{D}{2}\big{)}}{2^{D+2}\sqrt{2\pi^{D}}m_{\mathrm{H}}^{2-D}v_{\frac{1}{4}}^{2}}\bigg{[}\frac{2}{D-4}\bigg{(}\frac{2}{D-2}-1\bigg{)}\bigg{]}\,.$
(195)

The amplitude against spacetime dimension is plotted as follow:

Figure 9: The plot of $-i\mathcal{M}\Big{(}\frac{1}{4},D\Big{)}$ against
spacetime dimension $D$.

In particular, in our living spacetime dimension $D=4$, using the L-Hospital
rule,

$J\bigg{(}\frac{1}{4},4\bigg{)}=\lim_{D\rightarrow
4}\frac{2}{D-4}\bigg{(}\frac{2}{D-2}-1\bigg{)}=-1\,.$ (196)

We thus obtain a convergent amplitude for $n=1/4$, $D=4$

$i\mathcal{M}^{(1/4)}\bigg{(}\frac{1}{4},4\bigg{)}=\frac{m_{\mathrm{H}}^{2}}{64\sqrt{2}\pi^{2}v_{\frac{1}{4}}^{2}}=\frac{m_{\mathrm{H}}^{2}}{128\pi^{2}v^{2}}=2.044\times
10^{-4}\,,$ (197)

where we have used the fact that $v_{n}^{2}=4^{n}v^{2}$ in equation (136) and
$v=246\,\mathrm{GeV}$ is the vacuum expectation value. The amplitude is
inversely proportional to the square of the Higgs boson mass. It is noted that
the amplitude is ill-defined when $D=2$, for which it diverges.

Now, we investigate the physical $n,D\in\mathbb{Z}^{+}$ cases. We find that
for each integer $n$, the minimum dimension $D$ that gives convergent $J$
integral and amplitude $i\mathcal{M}^{\mathrm{(n)}}$ is given by the formula

$D_{\mathrm{min}}(n)=8n+1\,.$ (198)

For example, for $n=1$ case, $D_{\mathrm{min}}(1)$ = 9. That means the lowest
spacetime dimension for which the rotor mechanism to give finite result is
nine. For $n=1$, if $D<9$, the result will be all infinite. For example, for
the $n=2$ case, $D_{\mathrm{min}}(2)$ = 17. That means the lowest spacetime
dimension for which the rotor mechanism to give finite result is seventeen.
For $n=2$, if $D<17$, this will be all infinite. This applies to higher
positive integer $n$ values. And we conclude that for the Higgs boson self-
correction by 3rd order Higgs-self interaction, the minimum physical dimension
for taming the divergences (for both IR and UV) of the lowest-order rotor
model $(n=1)$ is 9.

Now we numerically work out the pair of $(n,d)$ which contributes the finite
convergent $J(n,D)$ integral. The result is plotted in figure (10).

Figure 10: The plot of $J(n,D)$ against rotor number $n$ and spacetime
dimension $D$.

The full amplitude is given by

$-i\mathcal{M}^{(n)}(n,D)=\frac{\Gamma\big{(}4n+2-\frac{D}{2}\big{)}}{2^{D+4n+2}\Gamma(4n)\sqrt{\pi^{D}}m_{\mathrm{H}}^{8n-D}v^{2}}J(n,D)\,.$
(199)

In general, in the nominator, the Gamma function
$\Gamma\big{(}4n+2-\frac{D}{2}\big{)}$ imposes constraints such that the
amplitude is finite. First notice that $\Gamma(-m)$ is infinity for $m$ equal
to positive integers. Therefore, such condition occurs when $D=2k$ and
$4n+2-k\leq 0$, so it takes place when $k\geq 4n+2$. Together with $D\geq
8n+1$ and $D<8n+4$, the conditions for finite amplitude are

$\begin{cases}D&<8n+4\\\ D&\geq 8n+1\\\ k&\ngeq 4n+2\\\ D&\neq 2k\\\
\end{cases}\,.$ (200)

For example $n=1$, we have $k\geq 6$ , $D\geq 12$, this will give
$\Gamma(-m)$, which is divergent. So for $n=1$, the required $(n,D)$ pair for
finite amplitude is $(1,9),(1,10),(1,11)$. For example $n=2$, the required
$(n,D)$ pair for finite amplitude is $(2,17),(2,18),(2,19)$. For example,

$-i\mathcal{M}^{(1)}(1,9)=\frac{\Gamma\big{(}\frac{3}{2}\big{)}}{2^{15}\Gamma(4)\pi^{9/2}m_{\mathrm{H}}^{-1}v^{2}}J(1,9)=4.931\times
10^{-11}\,\mathrm{GeV}^{-1}\,,$ (201)

which is a very small value. Therefore, under the rotor mechanism with rotor
index $n=1$, at $D=9$, the divergence is removed in high dimension. Thus, this
demonstrates how high-order-derivative field theory under rotor mechanism can
remove both UV and IR divergences in high spacetime dimension. This shows that
the Hierarchy problem can be solved by rotor mechanism in high spacetime
dimension.

###  8.2 Correction by the $W$ boson

Next, we will study the self-energy correction of Higgs boson by massive $W$
boson. Using the result in (154), we have the amplitude as

$\displaystyle\mathcal{M}^{(n)}$
$\displaystyle=\frac{g_{W}^{2}M_{W}^{2}}{4^{n}(2\pi)^{D}}\bigg{(}\int
d^{D}k\frac{D}{k^{4n}(p-k)^{4n}(k^{2}-M_{W}^{2})((p-k)^{2}-M_{W}^{2})}$ (202)
$\displaystyle\quad\quad\quad\quad\quad-\frac{1}{M_{W}^{2}}\int
d^{D}k\frac{(p-k)^{2}}{k^{4n}(p-k)^{4n}(k^{2}-M_{W}^{2})((p-k)^{2}-M_{W}^{2})}$
$\displaystyle\quad\quad\quad\quad\quad-\frac{1}{M_{W}^{2}}\int
d^{D}k\frac{k^{2}}{k^{4n}(p-k)^{4n}(k^{2}-M_{W}^{2})((p-k)^{2}-M_{W}^{2})}$
$\displaystyle\quad\quad\quad\quad\quad+\frac{1}{M_{W}^{4}}\int
d^{D}k\frac{[k\cdot(p-k)]^{2}}{k^{4n}(p-k)^{4n}(k^{2}-M_{W}^{2})((p-k)^{2}-M_{W}^{2})}\bigg{)}\,.$

From this we define the following integrals. First, from the first line, we
have

$I^{(n)}_{1}=\int
d^{D}k\frac{D}{(k^{2})^{2n}[(p-k)^{2}]^{2n}(k^{2}-M_{W}^{2})((p-k)^{2}-M_{W}^{2})}\,.$
(203)

From the second line we have

$I^{(n)}_{2}=\frac{1}{M_{W}^{2}}\int
d^{D}k\frac{1}{(k^{2})^{2n}[(p-k)^{2}]^{2n-1}(k^{2}-M_{W}^{2})((p-k)^{2}-M_{W}^{2})}\,.$
(204)

From the third line we have

$I^{(n)}_{3}=\frac{1}{M_{W}^{2}}\int
d^{D}k\frac{1}{(k^{2})^{2n-1}[(p-k)^{2}]^{2n}(k^{2}-M_{W}^{2})((p-k)^{2}-M_{W}^{2})}\,.$
(205)

For the last line, we use the following standard trick to convert the inner
product of $k\cdot p$ to squared values. The following identity is used:

$a\cdot b=\frac{1}{2}[a^{2}+b^{2}-(a-b)^{2}]\,.$ (206)

Therefore,

$\displaystyle I^{(n)}_{4}$ $\displaystyle=\frac{1}{M_{W}^{4}}\int
d^{D}k\frac{(k\cdot
p-k^{2})^{2}}{(k^{2})^{2n}[(p-k)^{2}]^{2n}(k^{2}-M_{W}^{2})((p-k)^{2}-M_{W}^{2})}$
(207) $\displaystyle=\frac{1}{M_{W}^{4}}\int
d^{D}k\frac{[\frac{1}{2}\big{(}(k^{2}+p^{2})-(k-p)^{2}\big{)}-k^{2}]^{2}}{(k^{2})^{2n}[(p-k)^{2}]^{2n}(k^{2}-M_{W}^{2})((p-k)^{2}-M_{W}^{2})}$
$\displaystyle=\frac{1}{4M_{W}^{4}}\int
d^{D}k\frac{(p^{2}-k^{2})^{2}-2(p^{2}-k^{2})(p-k)^{2}+(p-k)^{4}}{(k^{2})^{2n}[(p-k)^{2}]^{2n}(k^{2}-M_{W}^{2})((p-k)^{2}-M_{W}^{2})}$
$\displaystyle=\frac{1}{4M_{W}^{4}}\int
d^{D}k\frac{(p^{4}-2p^{2}k^{2}+k^{4})-2(p^{2}-k^{2})(p-k)^{2}+(p-k)^{4}}{(k^{2})^{2n}[(p-k)^{2}]^{2n}(k^{2}-M_{W}^{2})((p-k)^{2}-M_{W}^{2})}\,.$

For each term in the last line, we define six more integrals:

$I^{(n)}_{4a}=\frac{p^{4}}{4M_{W}^{4}}\int
d^{D}k\frac{1}{(k^{2})^{2n}[(p-k)^{2}]^{2n}(k^{2}-M_{W}^{2})((p-k)^{2}-M_{W}^{2})}\,,$
(208) $I_{4b}^{(n)}=\frac{-2p^{2}}{4M_{W}^{4}}\int
d^{D}k\frac{1}{(k^{2})^{2n-1}[(p-k)^{2}]^{2n}(k^{2}-M_{W}^{2})((p-k)^{2}-M_{W}^{2})}\,,$
(209) $I_{4c}^{(n)}=\frac{1}{4M_{W}^{4}}\int
d^{D}k\frac{1}{(k^{2})^{2n-2}[(p-k)^{2}]^{2n}(k^{2}-M_{W}^{2})((p-k)^{2}-M_{W}^{2})}\,,$
(210) $I_{4d}^{(n)}=\frac{-2p^{2}}{4M_{W}^{4}}\int
d^{D}k\frac{1}{(k^{2})^{2n}[(p-k)^{2}]^{2n-1}(k^{2}-M_{W}^{2})((p-k)^{2}-M_{W}^{2})}\,,$
(211) $I_{4e}^{(n)}=\frac{2}{4M_{W}^{4}}\int
d^{D}k\frac{1}{(k^{2})^{2n-1}[(p-k)^{2}]^{2n-1}(k^{2}-M_{W}^{2})((p-k)^{2}-M_{W}^{2})}\,,$
(212) $I_{4f}^{(n)}=\frac{1}{4M_{W}^{4}}\int
d^{D}k\frac{1}{(k^{2})^{2n}[(p-k)^{2}]^{2n-2}(k^{2}-M_{W}^{2})((p-k)^{2}-M_{W}^{2})}\,,$
(213)

Then now the computation becomes straight forward. The computation is similar
to that of the 3rd order Higgs vertex case, therefore we will just show the
final result in case some special issues are noted. First of all,

$I^{(n)}_{1}=\frac{iD\pi^{D/2}\Gamma\big{(}4n+2-\frac{D}{2}\big{)}}{\Gamma(4n)m_{W}^{8n+4-D}}\int_{0}^{1}B\bigg{(}1-\frac{1}{r},4n,2-\frac{D}{2}\bigg{)}\,.$
(214)

if $8n+1\leq D<8n+4$. Secondly,

$I_{2}^{(n)}=\frac{i\pi^{D/2}\Gamma\big{(}4n+1-\frac{D}{2}\big{)}}{\Gamma(4n-1)m_{W}^{8n+4-D}}\int_{0}^{1}B\bigg{(}1-\frac{1}{r},4n-1,2-\frac{D}{2}\bigg{)}\,.$
(215)

if $8n-1\leq D<8n+2$. Thirdly,

$I_{3}^{(n)}=\frac{i\pi^{D/2}\sqrt{\pi}\Gamma\big{(}4n+1-\frac{D}{2}\big{)}}{4^{2n-1}\Gamma(2n)\Gamma\big{(}2n-\frac{1}{2}\big{)}m_{W}^{8n+4-D}}\int_{0}^{1}B\bigg{(}1-\frac{1}{r},4n-1,2-\frac{D}{2}\bigg{)}\,.$
(216)

if $8n-1\leq D<8n+2$. Then

$I^{(n)}_{4a}=\frac{p^{4}}{4M_{W}^{4}}\times\frac{I^{(n)}_{1}}{D}=\frac{i\pi^{D/2}\Gamma\big{(}4n+2-\frac{D}{2}\big{)}p^{4}}{4\Gamma(4n)m_{W}^{8n+8-D}}\int_{0}^{1}B\bigg{(}1-\frac{1}{r},4n,2-\frac{D}{2}\bigg{)}\,.$
(217)

if $8n+1\leq D<8n+4$. Then

$I^{(n)}_{4b}=-\frac{2p^{2}}{4M_{W}^{2}}I_{3}^{(n)}=-\frac{2i\pi^{D/2}\sqrt{\pi}\Gamma\big{(}4n+1-\frac{D}{2}\big{)}p^{2}}{4^{2n}\Gamma(2n)\Gamma\big{(}2n-\frac{1}{2}\big{)}m_{W}^{8n+6-D}}\int_{0}^{1}B\bigg{(}1-\frac{1}{r},4n-1,2-\frac{D}{2}\bigg{)}\,.$
(218)

For the $I_{4c}^{(n)}$ integral, the situation is more complicated. It it is
worth to go through the whole computation process.

$\displaystyle I_{4c}^{(n)}$ $\displaystyle=\frac{1}{4M_{W}^{4}}\int
d^{D}l\int_{0}^{1}\int_{0}^{1}\int_{0}^{1}\int_{0}^{1}dxdydrds\delta(x+y+r+s-1)\frac{x^{2n-3}y^{2n-1}}{(l^{2}-\Delta)^{4n}}\frac{\Gamma(4n)}{\Gamma(2n-2)\Gamma(2n)}$
(219)
$\displaystyle=\frac{i(-1)^{4n}\Gamma(4n)(2\pi^{D/2})}{4M_{W}^{4}\Gamma(2n-2)\Gamma(2n)\Gamma\big{(}\frac{D}{2}\big{)}}\int_{0}^{1}\int_{0}^{1}\int_{0}^{1}\int_{0}^{1}dxdydrds\delta(x+y+r+s-1)x^{2n-3}y^{2n-1}$
$\displaystyle\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\times\int
dl_{E}\frac{l_{E}^{D-1}}{(l_{E}^{2}+\Delta)^{4n}}$
$\displaystyle=\frac{i(-1)^{4n}\Gamma(4n)(2\pi^{D/2})}{4M_{W}^{4}\Gamma(2n-2)\Gamma(2n)\Gamma\big{(}\frac{D}{2}\big{)}}\int_{0}^{1}\int_{0}^{1}\int_{0}^{1}\int_{0}^{1}dxdydrds\delta(x+y+r+s-1)x^{2n-3}y^{2n-1}$
$\displaystyle\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\times\frac{2n\Gamma\big{(}\frac{D}{2}\big{)}\Gamma\big{(}4n-\frac{D}{2}\big{)}}{\Delta^{4n-\frac{D}{2}}\Gamma(4n+1)}$
$\displaystyle=\frac{i(-1)^{4n}n\pi^{D/2}\Gamma(4n)\Gamma\big{(}4n-\frac{D}{2}\big{)}}{\Gamma(2n-2)\Gamma(2n)\Gamma(4n+1)M_{W}^{8n+4-D}}\int_{0}^{1}\int_{0}^{1}\int_{0}^{1}\int_{0}^{1}dxdydrds\delta(x+y+r+s-1)\frac{x^{2n-3}y^{2n-1}}{(r+s)^{4n-\frac{D}{2}}}$
$\displaystyle=\frac{i(-1)^{4n}n\pi^{D/2}\Gamma(4n)\Gamma\big{(}4n-\frac{D}{2}\big{)}}{\Gamma(2n-2)\Gamma(2n)\Gamma(4n+1)M_{W}^{8n+4-D}}\int_{0}^{1}dr\int_{0}^{1-r}ds\int_{0}^{1-r-s}dy\frac{(1-y-r-s)^{2n-3}y^{2n-1}}{(r+s)^{4n-\frac{D}{2}}}$
$\displaystyle=\frac{i(-1)^{4n}n\pi^{D/2}\Gamma(4n)\Gamma\big{(}4n-\frac{D}{2}\big{)}}{\Gamma(2n-2)\Gamma(2n)\Gamma(4n+1)M_{W}^{8n+4-D}}\int_{0}^{1}dr\int_{0}^{1-r}ds\frac{(1-r-s)^{4n-3}}{(r+s)^{4n-\frac{D}{2}}}\frac{\Gamma(2n)\Gamma(2n-2)}{\Gamma(4n-2)}$
$\displaystyle=\frac{i(-1)^{4n}n\pi^{D/2}\Gamma(4n)\Gamma\big{(}4n-\frac{D}{2}\big{)}}{\Gamma(4n+1)\Gamma(4n-2)M_{W}^{8n+4-D}}\int_{0}^{1}dr\int_{0}^{1-r}ds\frac{(1-r-s)^{4n-3}}{(r+s)^{4n-\frac{D}{2}}}$

if $0<D<8$. Now we evaluate the last integral. This gives

$\int_{0}^{1-r}ds\frac{(1-r-s)^{4n-3}}{(r+s)^{4n-\frac{D}{2}}}=\frac{\Gamma(4n-2)}{(1-r)^{4n-2}r^{4n-\frac{D}{2}}}\,_{2}\tilde{F}_{1}\bigg{(}1,4n-\frac{D}{2},4n-1,1-\frac{1}{r}\bigg{)}\,,$
(220)

where $\,{}_{2}\tilde{F}_{1}$ is the regularized generalized hypergeometric
$(2,1)$ function. We will give the definition of generalized hypergeometric
function and regularized generalized hypergeometric $(p,q)$ function here,
which is formally denoted as $\,{}_{p}F_{q}$. It is given that

$\,{}_{p}F_{q}(a_{1},\cdots a_{p};b_{1},\cdots
b_{q};z)=\sum_{n=0}^{\infty}\frac{\prod_{k=1}^{p}(a_{k})^{(n)}}{\prod_{k=1}^{q}(b_{k})^{(n)}}\frac{z^{n}}{n!}\,,$
(221)

where the Pochhammer symbol for rising factorial is used,

$(a)^{(0)}=1,\cdots,(a)^{(n)}=a(a+1)(a+2)\cdots(a+n-1)=\prod_{l=0}^{n-1}(a+l)\,.$
(222)

The corresponding regularized generalized hypergeometric $(p,q)$ function is
defined by

$\,{}_{p}\tilde{F}_{q}(a_{1},\cdots a_{p};b_{1},\cdots
b_{q}:z)=\frac{\,{}_{p}F_{q}(a_{1},\cdots a_{p};b_{1},\cdots
b_{q};z)}{\Gamma(b_{1})\cdots\Gamma(b_{q})}\,.$ (223)

For our case, we are concerned with the regularized generalized hypergeometric
$(2,1)$ function, which is

$\,{}_{2}\tilde{F}_{1}(a_{1},a_{2};b_{1};z)=\frac{\,{}_{2}F_{1}(a_{1},a_{2};b_{1};z)}{\Gamma(b_{1})}=\frac{1}{\Gamma(b_{1})}\sum_{n=0}^{\infty}\frac{(a_{1})^{(n)}(a_{2})^{(n)}}{(b_{1})^{(n)}}\frac{z^{n}}{n!}\,.$
(224)

Therefore, for our case in (220), we have

$\displaystyle\,{}_{2}\tilde{F}_{1}\bigg{(}1,4n-\frac{D}{2},4n-1,1-\frac{1}{r}\bigg{)}$
$\displaystyle=\frac{1}{\Gamma(4n-1)}\sum_{m=0}^{\infty}\frac{1^{(m)}\big{(}4n-\frac{D}{2}\big{)}^{(m)}}{m!(4n-1)^{(m)}}\bigg{(}1-\frac{1}{r}\bigg{)}^{m}$
(225)
$\displaystyle=\frac{1}{\Gamma(4n-1)}\sum_{m=0}^{\infty}\frac{\big{(}4n-\frac{D}{2}\big{)}^{(m)}}{(4n-1)^{(m)}}\bigg{(}1-\frac{1}{r}\bigg{)}^{m}\,.$

Therefore, the last line of (219) yields

$\displaystyle I^{(n)}_{4c}$
$\displaystyle=\frac{i(-1)^{4n}n\pi^{D/2}\Gamma(4n)\Gamma\big{(}4n-\frac{D}{2}\big{)}}{\Gamma(4n+1)\Gamma(4n-1)M_{W}^{8n+4-D}}\int_{0}^{1}dr\frac{1}{(1-r)^{4n-2}r^{4n-\frac{D}{2}}}\,_{2}\tilde{F}_{1}\bigg{(}1,4n-\frac{D}{2},4n-1,1-\frac{1}{r}\bigg{)}$
(226)
$\displaystyle=\frac{i(-1)^{4n}n\pi^{D/2}\Gamma(4n)\Gamma\big{(}4n-\frac{D}{2}\big{)}}{\Gamma(4n+1)\Gamma(4n-1)M_{W}^{8n+4-D}}\int_{0}^{1}dr\frac{1}{(1-r)^{4n-2}r^{4n-\frac{D}{2}}}\sum_{m=0}^{\infty}\frac{\big{(}4n-\frac{D}{2}\big{)}^{(m)}}{(4n-1)^{(m)}}\bigg{(}1-\frac{1}{r}\bigg{)}^{m}\,.$

Thus this completes the mathematical calculation of $I_{4c}^{(n)}$.

Then we continue to calculate the integral $I_{4d}^{(n)}$. This is easy,

$I_{4d}^{(n)}=\frac{-2p^{2}}{4M_{W}^{2}}I_{2}^{(n)}=\frac{-i\pi^{D/2}\Gamma\big{(}4n+1-\frac{D}{2}\big{)}p^{2}}{2\Gamma(4n-1)m_{W}^{8n+6-D}}\int_{0}^{1}B\bigg{(}1-\frac{1}{r},4n-1,2-\frac{D}{2}\bigg{)}\,.$
(227)

Next,

$I_{4e}^{(n)}=\frac{i(-1)^{4n}\pi^{D/2}\sqrt{\pi}\Gamma(4n)\Gamma(4n-2)\Gamma\big{(}4n-\frac{D}{2}\big{)}}{2^{4n-4}\Gamma(2n-1)\Gamma(4n+1)\Gamma\big{(}2n-\frac{1}{2}\big{)}M_{W}^{8n+4-D}}\int_{0}^{1}dr\frac{\,{}_{2}\tilde{F}_{1}\big{(}1,4n-\frac{D}{2},4n-1,1-\frac{1}{r}\big{)}}{(1-r)^{4n-2}r^{4n-\frac{D}{2}}}\,.$
(228)

Finally,

$I_{4f}^{(n)}=\frac{i(-1)^{4n}\Gamma(4n)\Gamma\big{(}4n-\frac{D}{2}\big{)}}{4\Gamma(4n+1)M_{W}^{8n+4-D}}\int_{0}^{1}dr\frac{\,{}_{2}\tilde{F}_{1}\big{(}1,4n-\frac{D}{2},4n-1,1-\frac{1}{r}\big{)}}{(1-r)^{4n-2}r^{4n-\frac{D}{2}}}\,.$
(229)

The amplitude of the Higgs-self energy correction by $W$ boson is hence

$\mathcal{M}^{(n)}=\frac{g^{2}M_{W}^{2}}{4^{n}(2\pi)^{D}}\big{(}I_{1}^{(n)}-I_{2}^{(n)}-I_{3}^{(n)}+I_{4a}^{(n)}+I_{4b}^{(n)}+I_{4c}^{(n)}+I_{4d}^{(n)}+I_{4e}^{(n)}+I_{4f}^{(n)}\big{)}\,.$
(230)

It is worth to analyse the properties of these integrals. We can see that
unlike the case of the self-energy correction for the Higgs-boson from the 3rd
order Higgs interaction, here the amplitude is ill-defined for the $n=1/4$
case, as we have the denominator of $\Gamma(4n-1)$ in
$I_{2}^{(n)},I_{4c}^{(n)},I_{4d}^{(n)},I_{4e}^{(n)},I_{4f}^{(n)}$, which
diverges when $n=1/4$. Now, in order to get an idea of what pairs of $(n,D)$
give us convergent result, first we would like to analyse the integral in
(226). Define

$p(n,D,r)=\frac{1}{(1-r)^{4n-2}r^{4n-\frac{D}{2}}}\,_{2}\tilde{F}_{1}\bigg{(}1,4n-\frac{D}{2},4n-1,1-\frac{1}{r}\bigg{)}\,.$
(231)

Then we plot $p(n,D,r)$ with different $n$ and $D$ values.

Figure 11: The plot of $p(n,D,r)$ with of $n=1$ and different values of $D$.
Figure 12: The plot of $p(n,D,r)$ with of $n=2$ and different values of $D$.
Figure 13: The plot of $p(n,D,r)$ with of $n=3$ and different values of $D$.
Figure 14: The plot of $p(n,D,r)$ with of $n=4$ and different values of $D$.
Figure 15: The plot of $p(n,D,r)$ with of $n=5$ and different values of $D$.

We study the general properties of the $p(n,D,r)$ function and the integration
of it from 0 to 1. Let’s first study the $(1,D)$ case in detail. For each set
of $(1,D)$, we compute $p(1,D,r)$ and $\int_{0}^{1}p(1,D,r)dr$.

$(n,D)$ | $p(n,D,r)$ | $\int_{0}^{1}p(1,D,r)dr$
---|---|---
$(1,1)$ | $-\frac{2}{3}r^{-3/2}+\frac{2}{5}r^{-5/2}+\frac{4}{15}$ | $\infty$
$(1,2)$ | $\frac{(r-1)^{2}}{2r^{2}}$ | $\infty$
$(1,3)$ | $-\frac{2}{\sqrt{r}}+\frac{2}{3}r^{-3/2}+4$ | $\infty$
$(1,4)$ | $\frac{1}{r}+\ln r-1$ | $\infty$
$(1,5)$ | $2\sqrt{r}\bigg{(}\sqrt{\frac{1}{r}}-1\bigg{)}^{2}$ | $1.333$
$(1,6)$ | $r-1-\ln r$ | $0.5$
$(1,7)$ | $\frac{2}{3}\sqrt{r}\bigg{(}r+2\sqrt{\frac{1}{r}}-3\bigg{)}$ | $0.267$
$(1,8)$ | $\frac{1}{2}(r-1)^{2}$ | $0.167$
$(1,9)$ | $\frac{2}{15}\bigg{(}2\sqrt{\frac{1}{r}}-5r+3r^{2}\bigg{)}$ | $0.1143$
$(1,10)$ | $\frac{1}{6}(r-1)^{2}(2r+1)$ | $0.0833$

We can see that for the $n=1$ case, the integral only converges when $D>4$.
And in general, for higher dimension $D$ the integral decreases. This pattern
also applies for other $n$. For each $n$, there exists a threshold dimension
$D_{\mathrm{min}}(n)$ such that below this dimension the integral diverges,
upon computation we find that

$D_{\mathrm{min}}(n)=8n-3\,.$ (232)

When $n=1$, the threshold is $D_{\mathrm{min}}=5$, which is confirmed by the
above table. And when $n=2$, $D_{\mathrm{min}}=13$, and so on. From all the
plots above, we can see that larger $n$ with larger $D$ results in smaller
values of the integral. This shows that we will obtain smaller amplitudes with
increasing $(n,D)$ values.

The full amplitude is now given by

$\displaystyle\mathcal{M}^{(n)}$
$\displaystyle=\frac{ig_{W}^{2}}{2^{2n+D}\sqrt{\pi^{D}}M_{W}^{8n+2-D}}\bigg{\\{}\frac{\Gamma\big{(}4n+2-\frac{D}{2}\big{)}}{\Gamma(4n)}\bigg{(}D+\frac{p^{4}}{4M_{W}^{4}}\bigg{)}\int_{0}^{1}B\bigg{(}1-\frac{1}{r},4n,2-\frac{D}{2}\bigg{)}$
(233)
$\displaystyle\,\,-\Gamma\bigg{(}4n+1-\frac{D}{2}\bigg{)}\bigg{[}\frac{1}{\Gamma(4n-1)}\bigg{(}1+\frac{p^{2}}{2M_{W}^{2}}\bigg{)}+\frac{\sqrt{\pi}}{4^{2n}\Gamma(2n)\Gamma\big{(}2n-\frac{1}{2}\big{)}}\bigg{(}4+\frac{2p^{2}}{M_{W}^{2}}\bigg{)}\bigg{]}$
$\displaystyle\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\times\int_{0}^{1}B\bigg{(}1-\frac{1}{r},4n-1,2-\frac{D}{2}\bigg{)}$
$\displaystyle+\frac{(-1)^{4n}\Gamma(4n)\Gamma\big{(}4n-\frac{D}{2}\big{)}}{\Gamma(4n+1)}\bigg{[}\frac{n}{\Gamma(4n-1)}+\frac{\sqrt{\pi}\Gamma(4n-2)}{\Gamma(2n-1)\Gamma\big{(}2n-\frac{1}{2}\big{)}}+\frac{1}{4}\bigg{]}$
$\displaystyle\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\times\int_{0}^{1}dr\frac{\,{}_{2}\tilde{F}_{1}\big{(}1,4n-\frac{D}{2},4n-1,1-\frac{1}{r}\big{)}}{(1-r)^{4n-2}r^{4n-\frac{D}{2}}}\bigg{\\}}\,.$

The full amplitude is a function of $n$,$D$ and incoming momentum $p$,
$\mathcal{M}^{(n)}\equiv\mathcal{M}^{(n)}(n,D,p)$. It is found that for the
unrotored case $(n=0)$ in our $D=4$ spacetime, this amplitude diverges to
infinity. This is obvious due to the term $\Gamma(4n)$ appearing in the
denominator. This result is expected as in the Standard Model, one-loop
correction of the Higgs boson by $W$ boson diverges.

However, as by the above results, we see that the integrals converge only in
different set of $(n,D)$ values. We want to find the common $(n,D)$ values for
all the integrals, which amount to solve the three inequalities:

$\begin{cases}&8n+1\leq D<4n+4\\\ &8n-1\leq D<8n+2\\\ &0\leq D<8\\\ &D\geq
8n-3\\\ \end{cases}\,.$ (234)

However, there is no solution for this set of inequality. Therefore, the
amplitude in (233) means to be diverged in any particular set of $(n,D)$
values.

###  8.3 Correction by the fermion loop

Finally, we remain to calculate the self-energy correction of Higgs Boson by
fermion loop under rotor mechanism. This is essentially to compute the
amplitude calculation in (161). We need to develop some Dirac algebra and
trace identities involving $\gamma^{\mu\dagger}$, as well as $\gamma^{\mu}$
and $\gamma^{\mu\dagger}$ together in the most general $D$ spacetime
dimension. First notice that the complex conjugate of the $\gamma^{\mu}$
matrix is defined by

$\gamma^{\mu\dagger}=\gamma^{0}\gamma^{\mu}\gamma^{0\dagger}=\gamma^{0}\gamma^{\mu}\gamma^{0}\,,$
(235)

where we have used the fact that $\gamma^{0\dagger}=\gamma^{0}$. Then we
immediately obtain our first identity :

$\mathrm{Tr}(\gamma^{\mu^{\dagger}})=\mathrm{Tr}(\gamma^{0}\gamma^{\mu}\gamma^{0})=\mathrm{Tr}(\gamma^{0}\gamma^{0}\gamma^{\mu})=\mathrm{Tr}(\gamma^{\mu})=0\,,$
(236)

where we have used the trace identity $\mathrm{Tr}(ABC)=\mathrm{Tr}(CAB)$ and
the fact that $(\gamma^{0})^{2}=I$. Next, we prove the Dirac algebra for
hermitian gamma matrices

$\\{\gamma^{\mu\dagger},\gamma^{\nu\dagger}\\}=2\eta^{\mu\nu}I_{D}\,.$ (237)

The proof is straight forward,

$\displaystyle\\{\gamma^{\mu\dagger},\gamma^{\nu\dagger}\\}$
$\displaystyle=\gamma^{\mu\dagger}\gamma^{\nu\dagger}+\gamma^{\nu\dagger}\gamma^{\mu\dagger}$
(238)
$\displaystyle=\gamma^{0}\gamma^{\mu}\gamma^{0}\gamma^{0}\gamma^{\nu}\gamma^{0}+\gamma^{0}\gamma^{\nu}\gamma^{0}\gamma^{0}\gamma^{\mu}\gamma^{0}$
$\displaystyle=\gamma^{0}(\gamma^{\mu}\gamma^{\nu}+\gamma^{\nu}\gamma^{\mu})\gamma^{0}$
$\displaystyle=\gamma^{0}(2\eta^{\mu\nu}I_{D\times D})\gamma^{0}$
$\displaystyle=2\eta^{\mu\nu}\gamma^{0}I_{D\times D}\gamma^{0}$
$\displaystyle=2\eta^{\mu\nu}I_{D\times D}\,,$

where we have used the fact that $(\gamma^{0})^{2}=I$. This result implies

$\not{k}\not{k}=\not{k}^{\dagger}\not{k}^{\dagger}=k^{2}I_{D\times
D}\quad\text{and}\quad\not{k}\not{p}=\not{k}^{\dagger}\not{p}^{\dagger}=(k\cdot
p)I_{D\times D}\,.$ (239)

Next we will prove the following identity:

$\mathrm{Tr}(\gamma^{\mu}\gamma^{\rho\dagger})=D(2\eta^{0\mu}\eta^{0\rho}-\eta^{\mu\rho})\,.$
(240)

By the definition (235), it is noticed that
$\mathrm{Tr}(\gamma^{\mu}\gamma^{\rho\dagger})=\mathrm{Tr}(\gamma^{\mu}\gamma^{0}\gamma^{\rho}\gamma^{0})$.
Then using the identity of
$\mathrm{Tr}(\gamma^{\mu}\gamma^{\nu}\gamma^{\rho}\gamma^{\sigma})=D(\eta^{\rho\sigma}\eta^{\mu\nu}-\eta^{\nu\sigma}\eta^{\mu\rho}+\eta^{\mu\sigma}\eta^{\nu\rho})$,
by putting $\nu$ and $\sigma$ equal to 0, then we get

$\displaystyle\mathrm{Tr}(\gamma^{\mu}\gamma^{\rho\dagger})$
$\displaystyle=\mathrm{Tr}(\gamma^{\mu}\gamma^{0}\gamma^{\rho}\gamma^{0})$
(241) $\displaystyle=D(\eta^{\rho 0}\eta^{\mu
0}-\eta^{00}\eta^{\mu\rho}+\eta^{\mu 0}\eta^{0\rho})$
$\displaystyle=D(2\eta^{0\mu}\eta^{0\rho}-\eta^{\mu\rho})\,,$

where $\eta^{00}=1$ in our $\mathrm{diag}(+---)$ metric convention. Next we
verify the following computationally,

$\mathrm{Tr}(\gamma^{\mu\dagger}\gamma^{\nu}\gamma^{\rho\dagger})=\mathrm{Tr}(\gamma^{0}\gamma^{\mu}\gamma^{0}\gamma^{\nu}\gamma^{0}\gamma^{\rho}\gamma^{0})=\mathrm{Tr}(\gamma^{\mu}\gamma^{\nu\dagger}\gamma^{\rho})=0$
(242)

that the trace of a mixture of odd number of $\gamma^{\mu}$ and
$\gamma^{\mu\dagger}$ matrices is zero. Next, another useful identity we need
to use later is

$\displaystyle\quad\,\,\mathrm{Tr}(\gamma^{\mu}\gamma^{\nu\dagger}\gamma^{\rho}\gamma^{\sigma^{\dagger}})=\mathrm{Tr}(\gamma^{\mu}\gamma^{0}\gamma^{\nu}\gamma^{0}\gamma^{\rho}\gamma^{0}\gamma^{\sigma}\gamma^{0})$
(243)
$\displaystyle=D(8\eta^{0\mu}\eta^{0\nu}\eta^{0\rho}\eta^{0\sigma}-2\eta^{0\rho}\eta^{0\sigma}\eta^{\mu\nu}-2\eta^{0\mu}\eta^{0\sigma}\eta^{\nu\rho}-2\eta^{0\nu}\eta^{0\rho}\eta^{\mu\sigma}-2\eta^{0\mu}\eta^{0\nu}\eta^{\rho\sigma}$
$\displaystyle\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad+\eta^{\mu\nu}\eta^{\rho\sigma}-\eta^{\mu\rho}\eta^{\nu\sigma}+\eta^{\mu\sigma}\eta^{\nu\rho})\,.$

For simplicity, we will consider the $n=1$ case first in (161). Now we can
evaluate the nominator of (161)

$\displaystyle\quad(\not{k})(\not{k}+m_{f})(\not{k}^{\dagger})(\not{p}-\not{k})(\not{p}-\not{k}+m_{f})(\not{p}^{\dagger}-\not{k}^{\dagger})$
(244)
$\displaystyle=(k^{2}\not{k}^{\dagger}+m_{f}\not{k}\not{k}^{\dagger})\,[(p-k)^{2}(\not{p}^{\dagger}-\not{k}^{\dagger})+m_{f}(\not{p}-\not{k})(\not{p}^{\dagger}-\not{k}^{\dagger})]$
$\displaystyle=k^{2}(p-k)^{2}(k\cdot p-k^{2})I_{D\times
D}+m_{f}(p-k)^{2}\not{k}\not{k}^{\dagger}(\not{p}^{\dagger}-\not{k}^{\dagger})$
$\displaystyle\quad+m_{f}k^{2}\not{k}^{\dagger}(\not{p}-\not{k})(\not{p}^{\dagger}-\not{k}^{\dagger})+m_{f}^{2}\not{k}\not{k}^{\dagger}(\not{p}-\not{k})(\not{p}^{\dagger}-\not{k}^{\dagger})$
$\displaystyle=k^{2}(p-k)^{2}(k\cdot p-k^{2})I_{D\times
D}+m_{f}(p-k)^{2}\not{k}(k\cdot p-k^{2})$
$\displaystyle\quad+m_{f}k^{2}\not{k}^{\dagger}(\not{p}\not{p}^{\dagger}-\not{p}^{\dagger}\not{k}^{\dagger}-\not{k}^{\dagger}\not{p}^{\dagger}+\not{k}\not{k}^{\dagger})+m_{f}^{2}\not{k}\not{k}^{\dagger}(\not{p}\not{p}^{\dagger}-\not{p}^{\dagger}\not{k}^{\dagger}-\not{k}^{\dagger}\not{p}^{\dagger}+\not{k}\not{k}^{\dagger})$

Now we need to take the trace of the above expression. First notice that,

$\mathrm{Tr}(\not{k})=k_{\mu}\mathrm{Tr}(\gamma^{\mu})=0\,.$ (245)

Then using the identity of 242,

$\mathrm{Tr}(\not{k}^{\dagger}\not{p}\not{q}^{\dagger})=k_{\mu}p_{\nu}q_{\rho}\mathrm{Tr}(\gamma^{\mu\dagger}\gamma^{\nu}\gamma^{\rho\dagger})=0\,.$
(246)

Therefore using these two results and

$\displaystyle\quad\mathrm{Tr}\Big{(}k^{2}(p-k)^{2}(k\cdot p-k^{2})I_{D\times
D}+m_{f}(p-k)^{2}\not{k}(k\cdot p-k^{2})$ (247)
$\displaystyle\quad+m_{f}k^{2}\not{k}^{\dagger}(\not{p}\not{p}^{\dagger}-\not{p}^{\dagger}\not{k}^{\dagger}-\not{k}^{\dagger}\not{p}^{\dagger}+\not{k}\not{k}^{\dagger})+m_{f}^{2}\not{k}\not{k}^{\dagger}(\not{p}\not{p}^{\dagger}-\not{p}^{\dagger}\not{k}^{\dagger}-\not{k}^{\dagger}\not{p}^{\dagger}+\not{k}\not{k}^{\dagger})\Big{)}$
$\displaystyle=Dk^{2}(p-k)^{2}(k\cdot
p-k^{2})+m_{f}^{2}\mathrm{Tr}(\not{k}\not{k}^{\dagger}\not{p}\not{p}^{\dagger}-\not{k}\not{k}^{\dagger}\not{p}\not{k}^{\dagger}-\not{k}\not{k}^{\dagger}\not{k}\not{p}^{\dagger})+\not{k}\not{k}^{\dagger}\not{k}\not{k}^{\dagger})$
$\displaystyle=Dk^{2}(p-k)^{2}(k\cdot
p-k^{2})+m_{f}^{2}(k_{\mu}k_{\nu}p_{\rho}p_{\sigma}-k_{\mu}k_{\nu}p_{\rho}k_{\sigma}-k_{\mu}k_{\nu}k_{\rho}p_{\sigma}+k_{\mu}k_{\nu}k_{\rho}k_{\sigma})\mathrm{Tr}(\gamma^{\mu}\gamma^{\nu\dagger}\gamma^{\rho}\gamma^{\sigma\dagger})$
$\displaystyle=Dk^{2}(p-k)^{2}(k\cdot
p-k^{2})+m_{f}^{2}(k_{\mu}k_{\nu}p_{\rho}p_{\sigma}-k_{\mu}k_{\nu}p_{\rho}k_{\sigma}-k_{\mu}k_{\nu}k_{\rho}p_{\sigma}+k_{\mu}k_{\nu}k_{\rho}k_{\sigma})$
$\displaystyle\quad\times
D(8\eta^{0\mu}\eta^{0\nu}\eta^{0\rho}\eta^{0\sigma}-2\eta^{0\rho}\eta^{0\sigma}\eta^{\mu\nu}-2\eta^{0\mu}\eta^{0\sigma}\eta^{\nu\rho}-2\eta^{0\nu}\eta^{0\rho}\eta^{\mu\sigma}-2\eta^{0\mu}\eta^{0\nu}\eta^{\rho\sigma}$
$\displaystyle\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad+\eta^{\mu\nu}\eta^{\rho\sigma}-\eta^{\mu\rho}\eta^{\nu\sigma}+\eta^{\mu\sigma}\eta^{\nu\rho})$
$\displaystyle=Dk^{2}(p-k)^{2}(k\cdot p-k^{2})$
$\displaystyle\quad-2D(k^{0})^{2}p^{2}+8Dk^{0}p^{0}k^{2}-2D(p^{0})^{2}k^{2}+Dk^{2}p^{2}+8D(k^{0})^{2}(k\cdot
p)-4Dk^{0}p^{0}(k\cdot p)$ $\displaystyle\quad-2Dk^{2}(k\cdot
p)-8D(k^{0})k^{2}+Dk^{4}-16D(k^{0})^{3}p^{0}+8D(k^{0})^{2}(p^{0})^{2}+8D(k^{0})^{4}\,.$

Therefore, we have at least 13 integrals. But some of them vanish due to the
odd parity, we will look into details when we come across them. The integrals
are much more technically difficult than the previous ones, as now here we
have to also integrate terms regarding $k^{0}$. With some observation, first
we define the generic integral

$\displaystyle I(a,b,D)$ $\displaystyle=\int
d^{D}k\frac{1}{(k^{2})^{a}[(p-k)^{2}]^{b}(k^{2}-m_{f}^{2})((p-k)^{2}-m_{f}^{2})}$
(248)
$\displaystyle=\frac{2i\pi^{D/2}\Gamma\big{(}a+b+2-\frac{D}{2}\big{)}}{\Gamma(a+b)m_{f}^{2(a+b+2)-D}}\int_{0}^{1}drB\bigg{(}1-\frac{1}{r};a+b,2-\frac{D}{2}\bigg{)}$

if $0<D<2a+2b+4$. In addition, the integral of the beta function is also
convergent for some threshold $D$ values. Define the first integral as

$\displaystyle I^{(1)}_{1}$ $\displaystyle=\int
d^{D}k\frac{Dk^{2}(p-k)^{2}(k\cdot
p-k^{2})}{k^{4}(p-k)^{4}(k^{2}-m_{f}^{2})((p-k)^{2}-m_{f}^{2})}$ (249)
$\displaystyle=\frac{D}{2}\int
d^{D}k\frac{(p^{2}-k^{2})-(p-k)^{2}}{k^{2}(p-k)^{2}(k^{2}-m_{f}^{2})((p-k)^{2}-m_{f}^{2})}$
$\displaystyle=\frac{D}{2}\Big{(}p^{2}I(1,1,D)-I(0,1,D)-I(1,0,D)\Big{)}\,.$

The range for each integral is: for $I(1,1,D),4<D<8$, for $I(1,0,D)$ and
$I(0,1,D)$, $2<D<6$. Therefore the overlap $D$ for the first integral is
$D=5$.

The second integral is defined to be

$I^{(1)}_{2}=-2Dp^{2}\int
d^{D}k\frac{(k^{0})^{2}}{k^{4}(p-k)^{4}(k^{2}-m_{f}^{2})((p-k)^{2}-m_{f}^{2})}\,.$
(250)

Then using the old trick of Feynman parameters, we have

$I^{(1)}_{2}=\frac{-2Dp^{2}\Gamma(6)}{\Gamma^{2}(2)\Gamma^{2}(1)}\int
d^{D}k\int_{0}^{1}\int_{0}^{1}\int_{0}^{1}\int_{0}^{1}dxdydrds\delta(x+y+r+s-1)\frac{(k^{0})^{2}xy}{[(k-(y+s)p)^{2}-(r+s)m_{f}^{2}]^{6}}\,.$
(251)

Note that again we have $l=k-(y+s)p$, $\Delta=(r+s)m_{f}^{2}$. And we have
$k^{0}=l^{0}+(y+s)p^{0}$, therefore the integral becomes,

$\displaystyle I^{(1)}_{2}$ $\displaystyle=-240Dp^{2}\int
d^{D}l\int_{0}^{1}\int_{0}^{1}\int_{0}^{1}\int_{0}^{1}dxdydrds\delta(x+y+r+s-1)\frac{(k^{0})^{2}xy}{(l^{2}-\Delta)^{6}}$
(252) $\displaystyle=-240Dp^{2}\int
d^{D}l\int_{0}^{1}\int_{0}^{1}\int_{0}^{1}\int_{0}^{1}dxdydrds\delta(x+y+r+s-1)\frac{xy(l^{0}+(y+s)p)^{2}}{(l^{2}-\Delta)^{6}}$
$\displaystyle=-240Dp^{2}\int
d^{D}l\int_{0}^{1}\int_{0}^{1}\int_{0}^{1}\int_{0}^{1}dxdydrds\delta(x+y+r+s-1)$
$\displaystyle\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\times\frac{xy((l^{0})^{2}+2l^{0}(y+s)p+(y+s)^{2}p^{2})}{(l^{2}-\Delta)^{6}}\,.$

Next we carry out Wick’s rotation and perform the substitution of
$l^{0}=il_{E}^{0}$,

$\displaystyle I^{(1)}_{2}$
$\displaystyle=-(-1)^{6}240iDp^{2}\int_{0}^{\infty}dl_{E}\int
d\Omega_{D}\int_{0}^{1}\int_{0}^{1}\int_{0}^{1}\int_{0}^{1}dxdydrds\delta(x+y+r+s-1)$
(253)
$\displaystyle\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\times\frac{xy[l_{E}^{D-1}\big{(}-(l_{E}^{0})^{2}+2il_{E}^{0}(y+s)p+(y+s)^{2}p^{2})]}{(l_{E}^{2}+\Delta)^{6}}\,.$

Now this integral involve the integrating on $l_{E}^{0}$. To tackle this
integral, we concern integration on $D-$sphere. We set the spherical
coordinates $(l_{E}^{0},l_{E}^{1},\cdots l_{E}^{D-1})$ in the Euclidean $l$
momentum space as follows:

$\begin{cases}l_{E}^{0}&=l_{E}\sin\varphi_{1}\sin\varphi_{2}\cdots\sin\varphi_{D-2}\cos\varphi_{D-1}\\\
l_{E}^{1}&=l_{E}\sin\varphi_{1}\sin\varphi_{2}\cdots\sin\varphi_{D-2}\sin\varphi_{D-1}\\\
\vdots&\vdots\\\
l_{E}^{n-3}&=l_{E}\sin\varphi_{1}\sin\varphi_{2}\cos\phi_{3}\\\
l_{E}^{n-2}&=l_{E}\sin\varphi_{1}\cos\varphi_{2}\\\
l_{E}^{n-1}&=l_{E}\cos\varphi_{1}\,,\\\ \end{cases}$ (254)

where

$l_{E}^{2}=(l_{E}^{0})^{2}+(l_{E}^{1})^{2}+\cdots+(l_{E}^{D-1})^{2}\,.$ (255)

The differential volume is given by

$d^{D}l_{E}=l_{E}^{D-1}\sin^{D-2}\varphi_{1}\sin^{D-3}\varphi_{2}\cdots\sin\varphi_{D-2}dl_{E}d\varphi_{1}d\varphi_{2}\cdots
d\varphi_{D-1}\,.$ (256)

And the differential solid angle is given by

$d\Omega_{D}=\sin^{D-2}\varphi_{1}\sin^{D-3}\varphi_{2}\cdots\sin\varphi_{D-2}d\varphi_{1}d\varphi_{2}\cdots
d\varphi_{D-1}\,,$ (257)

where $\varphi_{1},\varphi_{2},\cdots,\varphi_{D-2}\in[0,\pi]$ and
$\varphi_{D-1}\in[0,2\pi]$. We first carry out the integral for the first term
involving $(l_{E}^{0})^{2}$, we have

$\displaystyle I^{(1)}_{2a}$
$\displaystyle=240iDp^{2}\int_{0}^{\infty}dl_{E}\int
d\Omega_{D}\int_{0}^{1}\int_{0}^{1}\int_{0}^{1}\int_{0}^{1}dxdydrds\delta(x+y+r+s-1)\frac{xyl_{E}^{D-1}(l_{E}^{0})^{2}}{(l_{E}^{2}+\Delta)^{6}}$
(258) $\displaystyle=240iDp^{2}\int_{0}^{\infty}dl_{E}\int\cdots\int
d\varphi_{1}d\varphi_{2}\cdots
d\varphi_{D-1}\sin^{D-2}\varphi_{1}\sin^{D-3}\varphi_{2}\cdots\sin\varphi_{D-2}$
$\displaystyle\quad\times\int_{0}^{1}\int_{0}^{1}\int_{0}^{1}\int_{0}^{1}dxdydrds\delta(x+y+r+s-1)\frac{xyl_{E}^{D-1}(l_{E}\sin\varphi_{1}\sin\varphi_{2}\cdots\sin\varphi_{D-2}\cos\varphi_{D-1})^{2}}{(l_{E}^{2}+\Delta)^{6}}$
$\displaystyle=240iDp^{2}\int_{0}^{\pi}\sin^{D}\varphi_{1}d\varphi_{1}\int_{0}^{\pi}\sin^{D-1}\varphi_{2}d\varphi_{2}\int_{0}^{\pi}\sin^{D-2}\varphi_{3}d\varphi_{3}\cdots\int_{0}^{\pi}\sin^{3}\varphi_{D-2}\int_{0}^{2\pi}\cos^{2}\varphi_{D-1}d\varphi_{D-1}$
$\displaystyle\quad\times\int_{0}^{1}\int_{0}^{1}\int_{0}^{1}\int_{0}^{1}dxdydrds\delta(x+y+r+s-1)\int_{0}^{\infty}dl_{E}\frac{xyl_{E}^{D+1}}{(l_{E}^{2}+\Delta)^{6}}\,.$

In general,

$\int_{0}^{\pi}\sin^{m}x\,dx=\frac{\sqrt{\pi}\Gamma\big{(}\frac{m+1}{2}\big{)}}{\Gamma\big{(}\frac{m}{2}+1\big{)}}\quad\text{if}\,\,m>-1.$
(259)

Also,

$\int_{0}^{\infty}dl_{E}\frac{l_{E}^{D+1}}{(l_{E}^{2}+\Delta)^{6}}=\frac{\pi(D-8)(D-6)(D-4)(D-2)D\csc\big{(}\frac{\pi
D}{2}\big{)}}{7680\Delta^{5-\frac{D}{2}}}\,.$ (260)

Then we get

$\displaystyle I^{(1)}_{2a}$
$\displaystyle=\frac{iDp^{2}\pi^{2}}{32m_{f}^{10-D}}\bigg{(}\prod_{k=1}^{D-2}\int_{0}^{\pi}\sin^{D-k+1}\varphi_{k}d\varphi_{k}\bigg{)}\int_{0}^{1}\int_{0}^{1}\int_{0}^{1}\int_{0}^{1}dxdydrds\delta(x+y+r+s-1)\frac{xy}{(r+s)^{5-\frac{D}{2}}}$
(261)
$\displaystyle\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\times(D-8)(D-6)(D-4)(D-2)D\csc\bigg{(}\frac{\pi
D}{2}\bigg{)}$
$\displaystyle=\frac{D^{2}(D-8)(D-6)(D-4)(D-2)p^{2}\pi^{D/2+1}}{32\sin\big{(}\frac{\pi
D}{2}\big{)}m_{f}^{10-D}}\bigg{(}\prod_{k=1}^{D-2}\frac{\Gamma\big{(}\frac{D-k+2}{2}\big{)}}{\Gamma\big{(}\frac{D-k+3}{2}\big{)}}\bigg{)}$
$\displaystyle\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\times\int_{0}^{1}dr\int_{0}^{1-r}ds\int_{0}^{1-r-s}dy\frac{(1-y-r-s)y}{(r+s)^{5-\frac{D}{2}}}$
$\displaystyle=\frac{D^{2}(D-8)(D-6)(D-4)(D-2)p^{2}\pi^{D/2+1}}{32\sin\big{(}\frac{\pi
D}{2}\big{)}m_{f}^{10-D}}\frac{\Gamma(2)}{\Gamma\big{(}\frac{D+2}{2}\big{)}}\frac{1}{6}\int_{0}^{1}dr\int_{0}^{1-r}ds\frac{(1-r-s)^{3}}{(r+s)^{5-\frac{D}{2}}}$

if $-2<D<10$. Out of this range the integral will diverge. Now we calculate
the last integral

$\displaystyle\int_{0}^{1-r}ds\frac{(1-r-s)^{3}}{(r+s)^{5-\frac{D}{2}}}$
$\displaystyle=2\bigg{(}\frac{r^{3}}{D-2}-\frac{3r^{2}}{D-4}+\frac{3r}{D-6}+\frac{1}{8-D}\bigg{)}r^{\frac{D}{2}-4}$
(262) $\displaystyle\quad\quad\quad\quad+\frac{96}{(D-8)(D-6)(D-4)(D-2)}\,.$

And finally

$\int_{0}^{1}dr\int_{0}^{1-r}ds\frac{(1-r-s)^{3}}{(r+s)^{5-\frac{D}{2}}}=\frac{96}{(D-6)(D-4)(D-2)D}$
(263)

if $D>6$. It is noted that if $D\leq 6,$ the integral does not converge.
Therefore we get

$I_{2a}^{(1)}=\frac{4ip^{2}\pi^{D/2+1}D(D-8)}{\sin\big{(}\frac{\pi
D}{2}\big{)}\Gamma\big{(}\frac{D+2}{2}\big{)}m_{f}^{10-D}}$ (264)

in the overall range of $6<D<10$.

The next term to integrate will be

$I_{2b}^{(1)}=(-2i)(240)Dp^{2}\int_{0}^{\infty}dl_{E}\int
d\Omega_{D}\int_{0}^{1}\int_{0}^{1}\int_{0}^{1}\int_{0}^{1}dxdydrds\delta(x+y+r+s-1)\frac{l_{E}^{0}(y+s)}{(l_{E}^{2}+\Delta)^{6}}\,.$
(265)

Since
$l_{E}^{0}=l_{E}\sin\varphi_{1}\sin\varphi_{2}\cdots\sin\varphi_{D-2}\cos\varphi_{D-1}$,
when we integrate over $d\Omega_{D}$ where will be a vanishing term of

$\int_{0}^{2\pi}\cos\varphi_{D-1}d\varphi_{D-1}=0\,.$ (266)

Therefore the whole integral vanishes. In general, since if the integral is
odd

$\int\frac{d^{D}l}{(2\pi)^{D}}l^{m}(l^{0})=0$ (267)

for any integer $m$. Thus when we have odd order of $k^{0}$ in the integral,
the integral must vanish. Next we have the $I_{2c}^{(1)}$ integral, which is

$\displaystyle I_{2c}^{(1)}$
$\displaystyle=-240iDp^{4}\int_{0}^{\infty}dl_{E}\int
d\Omega_{D}\int_{0}^{1}\int_{0}^{1}\int_{0}^{1}\int_{0}^{1}dxdydrds\delta(x+y+r+s-1)\frac{l_{E}^{D-1}xy(y+s)^{2}}{(l_{E}^{2}+\Delta)^{6}}$
(268) $\displaystyle=-240iDp^{2}\int_{0}^{\infty}dl_{E}\int\cdots\int
d\varphi_{1}d\varphi_{2}\cdots
d\varphi_{D-1}\sin^{D-2}\varphi_{1}\sin^{D-3}\varphi_{2}\cdots\sin\varphi_{D-2}$
$\displaystyle\quad\times\int_{0}^{1}\int_{0}^{1}\int_{0}^{1}\int_{0}^{1}dxdydrds\delta(x+y+r+s-1)\frac{xy(y+s)^{2}l_{E}^{D-1}}{(l_{E}^{2}+\Delta)^{6}}\,.$

And as

$\int_{0}^{\infty}dl_{E}\frac{l_{E}^{D-1}}{(l_{E}^{2}+\Delta)^{6}}=-\frac{\pi(D-10)(D-8)(D-6)(D-4)(D-2)\csc\big{(}\frac{\pi
D}{2}\big{)}}{7680\Delta^{6-\frac{D}{2}}}$ (269)

if $0<D<12$. Then

$\displaystyle I_{2c}^{(1)}$
$\displaystyle=\frac{ip^{4}(D-10)(D-8)(D-6)(D-4)(D-2)D\pi^{D/2+1}}{32\sin\big{(}\frac{\pi
D}{2}\big{)}m_{f}^{12-D}}\bigg{(}\prod_{k=1}^{D-2}\frac{\Gamma\big{(}\frac{D-k}{2}\big{)}}{\Gamma\big{(}\frac{D-k+1}{2}\big{)}}\bigg{)}$
(270)
$\displaystyle\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\times\int_{0}^{1}dr\int_{0}^{1-r}ds\int_{0}^{1-r-s}dy\frac{(1-y-r-s)y(y+s)^{2}}{(r+s)^{6-\frac{D}{2}}}$
$\displaystyle=\frac{ip^{4}(D-10)(D-8)(D-6)(D-4)(D-2)D\cdot
2\pi^{D/2+1}}{32\sin\big{(}\frac{\pi
D}{2}\big{)}m_{f}^{12-D}}\frac{\Gamma(1)}{\Gamma\big{(}\frac{D}{2}\big{)}}$
$\displaystyle\quad\quad\quad\quad\quad\quad\quad\quad\quad\times\frac{1}{60}\int_{0}^{1}dr\int_{0}^{1-r}ds\frac{(1-r-s)^{3}(3(r-1)^{2}-4(r-1)s+3s^{2})}{(r+s)^{6-\frac{D}{2}}}$
$\displaystyle=\frac{ip^{4}\pi^{\frac{D}{2}+1}(D-10)(D^{2}-2D+24)}{6(D+2)\sin\big{(}\frac{\pi
D}{2}\big{)}\Gamma\big{(}\frac{D}{2}\big{)}m_{f}^{12-D}}$

if $D>8$. When $D<8$ the integral does not converge. So the $I_{2c}^{(1)}$
integral is converged in $8<D<12$. Therefore, for the $I^{(1)}$ integral to be
convergent, the spacetime dimension constraint is $6<D<10$ and $8<D<12$. The
solution is $D=9$ which is unique for the convergence.

The third integral is defined to be

$I^{(1)}_{3}=8Dp^{0}\int
d^{D}k\frac{(k^{0})k^{2}}{k^{4}(p-k)^{4}(k^{2}-m_{f}^{2})((p-k)^{2}-m_{f}^{2})}=0$
(271)

as the integral is odd.

The fourth integral is defined to be

$I_{4}^{(1)}=-2D(p^{0})^{2}\int
d^{D}k\frac{k^{2}}{k^{4}(p-k)^{4}(k^{2}-m_{f}^{2})((p-k)^{2}-m_{f}^{2})}=-2D(p^{0})^{2}I(1,2)$
(272)

which only converges when $6<D<10$.

The fifth integral is defined to be

$I_{5}^{(1)}=Dp^{2}\int
d^{D}k\frac{k^{2}}{k^{4}(p-k)^{4}(k^{2}-m_{f}^{2})((p-k)^{2}-m_{f}^{2})}=Dp^{2}I(1,2)$
(273)

which only converges when $6<D<10$.

The sixth integral is defined to be

$\displaystyle I_{6}^{(1)}$ $\displaystyle=8D\int
d^{D}k\frac{(k^{0})^{2}(k\cdot
p)}{k^{4}(p-k)^{4}(k^{2}-m_{f}^{2})((p-k)^{2}-m_{f}^{2})}$ (274)
$\displaystyle=4D\int
d^{D}k\frac{(k^{0})^{2}[(p^{2}-k^{2})-(p-k)^{2}]}{k^{4}(p-k)^{4}(k^{2}-m_{f}^{2})((p-k)^{2}-m_{f}^{2})}\,.$

Then we further define three more integrals,

$I^{(1)}_{6a}=4Dp^{2}\int
d^{D}k\frac{(k^{0})^{2}}{k^{4}(p-k)^{4}(k^{2}-m_{f}^{2})((p-k)^{2}-m_{f}^{2})}$
(275)

which takes the same form as $I_{2a}^{(1)}$ except the front constant. We have
$I^{(1)}_{6a}=-2I_{2a}^{(1)}$, which is defined in the overall range of
$6<D<10$. Next we have,

$I^{(1)}_{6b}=-4D\int
d^{D}k\frac{(k^{0})^{2}}{k^{2}(p-k)^{4}(k^{2}-m_{f}^{2})((p-k)^{2}-m_{f}^{2})}\,.$
(276)

Using the similar technique as above, this amounts to give

$\displaystyle I^{(1)}_{6b}$
$\displaystyle=-(-1)^{5}96iD\int_{0}^{\infty}dl_{E}\int
d\Omega_{D}\int_{0}^{1}\int_{0}^{1}\int_{0}^{1}\int_{0}^{1}dxdydrds\delta(x+y+r+s-1)$
(277)
$\displaystyle\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\times\frac{y[l_{E}^{D-1}\big{(}-(l_{E}^{0})^{2}+2il_{E}^{0}(y+s)p+(y+s)^{2}p^{2})]}{(l_{E}^{2}+\Delta)^{5}}$
$\displaystyle=\frac{i\pi^{D/2+1}D(D-6)}{8\sin\big{(}\frac{\pi
D}{2}\big{)}\Gamma\big{(}\frac{D+2}{2}\big{)}m_{f}^{8-D}}+\frac{2i\pi^{D/2+1}p^{2}(D-8)(D^{2}+2D+24)}{3(D+2)\sin\big{(}\frac{\pi
D}{2}\big{)}\Gamma\big{(}\frac{D}{2}\big{)}m_{f}^{10-D}}$

for $6<D<8$. Hence only when $D=7$ this integral is finite.

Finally, we have the integral of $I^{(1)}_{6c}$,

$I^{(1)}_{6c}=-4D\int
d^{D}k\frac{(k^{0})^{2}}{k^{4}(p-k)^{2}(k^{2}-m_{f}^{2})((p-k)^{2}-m_{f}^{2})}\,.$
(278)

Using the similar technique as above, this amounts to give,

$\displaystyle I^{(1)}_{6c}$
$\displaystyle=-(-1)^{5}24iD\int_{0}^{\infty}dl_{E}\int
d\Omega_{D}\int_{0}^{1}\int_{0}^{1}\int_{0}^{1}\int_{0}^{1}dxdydrds\delta(x+y+r+s-1)$
(279)
$\displaystyle\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\times\frac{x[l_{E}^{D-1}\big{(}-(l_{E}^{0})^{2}+2il_{E}^{0}(y+s)p+(y+s)^{2}p^{2})]}{(l_{E}^{2}+\Delta)^{5}}$
$\displaystyle=\frac{2i\pi^{D/2+1}(D-8)(D-6)}{\sin\big{(}\frac{\pi
D}{2}\big{)}\Gamma\big{(}\frac{D+2}{2}\big{)}m_{f}^{8-D}}-\frac{2i\pi^{D/2+1}(D-6)(D-8)(D^{2}+8)}{(D+2)(D+4)\sin\big{(}\frac{\pi
D}{2}\big{)}\Gamma\big{(}\frac{D}{2}\big{)}m_{f}^{8-D}}$

if $4<D<8$.

The seventh integral vanishes as it is an odd integral,

$I_{7}^{(1)}=-4Dp^{0}\int d^{D}k\frac{k^{0}(k\cdot
p)}{k^{4}(p-k)^{4}(k^{2}-m_{f}^{2})((p-k)^{2}-m_{f}^{2})}=0\,.$ (280)

The eighth integral is defined to be

$\displaystyle I_{8}^{(1)}$ $\displaystyle=-2D\int d^{D}k\frac{k^{2}(k\cdot
p)}{k^{4}(p-k)^{4}(k^{2}-m_{f}^{2})((p-k)^{2}-m_{f}^{2})}$ (281)
$\displaystyle=-D\int
d^{D}k\frac{(p^{2}-k^{2})-(p-k)^{2}}{k^{2}(p-k)^{4}(k^{2}-m_{f}^{2})((p-k)^{2}-m_{f}^{2})}$
$\displaystyle=-D(p^{2}I(1,2,D)-I(0,2,D)-I(1,1,D))\,.$

The range for each integral to be finite is: for $I(1,2,D)$, $6<D<10$. For
$I(0,2,D)$, $4<D<8$. For $I(1,1,D)$, $4<D<8$. Therefore the overall range is
$6<D<8$, which is $D=7$.

The ninth integral vanishes as it is an odd integral,

$I_{9}^{(1)}=-8D\int
d^{D}k\frac{(k^{0})k^{2}}{k^{4}(p-k)^{4}(k^{2}-m_{f}^{2})((p-k)^{2}-m_{f}^{2})}=0\,.$
(282)

The tenth integral is defined to be

$I_{10}^{(1)}=D\int
d^{D}k\frac{k^{4}}{k^{4}(p-k)^{4}(k^{2}-m_{f}^{2})((p-k)^{2}-m_{f}^{2})}=DI(0,2,D)\,,$
(283)

which is in the range of $4<D<8$.

The eleventh integral vanishes as it is an odd integral,

$I_{11}^{(1)}=-16Dp^{0}\int
d^{D}k\frac{(k^{0})^{3}}{k^{4}(p-k)^{4}(k^{2}-m_{f}^{2})((p-k)^{2}-m_{f}^{2})}=0\,.$
(284)

The twelfth integral is defined to be

$I_{12}^{(1)}=8D(p^{0})^{2}\int
d^{D}k\frac{(k^{0})^{2}}{k^{4}(p-k)^{4}(k^{2}-m_{f}^{2})((p-k)^{2}-m_{f}^{2})}=-\frac{4I_{2}^{(1)}(p^{0})^{2}}{p^{2}}$
(285)

with $D=9$.

The thirteenth integral is defined to be

$\displaystyle I_{13}^{(1)}$ $\displaystyle=8D\int
d^{D}k\frac{(k^{0})^{4}}{k^{4}(p-k)^{4}(k^{2}-m_{f}^{2})((p-k)^{2}-m_{f}^{2})}$
(286) $\displaystyle=960D\int
d^{D}l\int_{0}^{1}\int_{0}^{1}\int_{0}^{1}\int_{0}^{1}dxdydrds\delta(x+y+r+s-1)\frac{xy((l^{0})+(y+s)p)^{4}}{(l^{2}-\Delta)^{6}}$
$\displaystyle=960D\int
d^{D}l\int_{0}^{1}\int_{0}^{1}\int_{0}^{1}\int_{0}^{1}dxdydrds\delta(x+y+r+s-1)$
$\displaystyle\quad\quad\quad\times\frac{xy((l^{0})^{4}+4(l^{0})^{3}(y+s)p+6(l^{0})^{2}(y+s)^{2}p^{2}+6(l^{0})(y+s)^{3}p^{3}+(y+s)^{4}p^{4})}{(l^{2}-\Delta)^{6}}\,.$