$\overline{\psi}=\psi^{\dagger}\gamma^{0}$ in the ordinary notation of QFT.

$\displaystyle\widetilde{\psi}_{f}$
$\displaystyle:=\psi^{\dagger}_{f}\Gamma_{11}.$ (204)

Thirdly, similar to left-handed and right-handed fields used in the SM, we
define

$\displaystyle\psi^{L}_{f}$ $\displaystyle:=\left(\begin{array}[]{cc}1&0\\\
0&0\end{array}\right)\psi_{f},\quad\psi^{R}_{f}:=\left(\begin{array}[]{cc}0&0\\\
0&1\end{array}\right)\psi_{f}.$ (209)

Combining the above two notations, we write

$\displaystyle\widetilde{\psi}^{L}_{f}$
$\displaystyle=\psi^{L{\dagger}}_{f}\Gamma_{11}=\psi^{{\dagger}}_{f}\left(\begin{array}[]{cc}1&0\\\
0&0\end{array}\right)\Gamma_{11},$ (210c)
$\displaystyle\widetilde{\psi}^{R}_{f}$
$\displaystyle=\psi^{R{\dagger}}_{f}\Gamma_{11}=\psi^{\dagger}_{f}\left(\begin{array}[]{cc}0&0\\\
0&1\end{array}\right)\Gamma_{11}.$ (210f)

Clearly,

$\displaystyle\psi_{f}=\psi^{L}_{f}+\psi^{R}_{f},\qquad\widetilde{\psi}_{f}=\widetilde{\psi}^{L}_{f}+\widetilde{\psi}^{R}_{f}.$
(211)

From Eq.(210), it is direct to see that

$\displaystyle\widetilde{\psi}^{L}_{f}$
$\displaystyle=\psi^{\dagger}_{f}\Gamma_{10},\quad\widetilde{\psi}^{R}_{f}=\psi^{\dagger}_{f}\Gamma_{01}.$
(212)

Now, we are ready to discuss interaction $H$-operators. We first discuss
${\mathscr{H}}^{f\overline{f}^{\prime}B}$ of those BPIPs that involve
$A$-mode. Fermionic modes, which are involved in these BPIPs, are
$f=f^{\prime}=e,u$, and $d$. Their $\Gamma$-matrices, given in Eqs.(179d),
(194b), and (194d), are real and proportional to $\Gamma_{11}$. Due to the
relation in Eq.(202), these $\Gamma$-matrices satisfy
$(\Gamma^{f\overline{f}^{\prime}\to
B}\gamma^{\mu})^{\dagger}=\Gamma^{f\overline{f}^{\prime}\to B}\gamma^{\mu}$.
Substituting these $\Gamma$-matrices into Eq.(163), one gets the following
$H$-operator densities,

$\displaystyle{\mathscr{H}}^{e\overline{e}A}$
$\displaystyle=\frac{1}{4}\widetilde{\psi}_{e}\gamma^{\mu}\psi_{e}\psi_{A\mu},$
(213a) $\displaystyle{\mathscr{H}}^{u_{i}\overline{u}_{i}A}$
$\displaystyle=\frac{1}{4}\widetilde{\psi}_{u_{i}}\gamma^{\mu}(\frac{2}{3})\psi_{u_{i}}\psi_{A\mu},$
(213b) $\displaystyle{\mathscr{H}}^{d_{i}\overline{d}_{i}A}$
$\displaystyle=\frac{1}{4}\widetilde{\psi}_{d_{i}}\gamma^{\mu}(-\frac{1}{3})\psi_{d_{i}}\psi_{A\mu},$
(213c)

where $\psi_{A\mu}({\bf x})$ is defined in Eq.(203).

Next, we discuss BPIPs that involve $Z$-mode. Fermionic modes in them are
$f=f^{\prime}=\nu,e,u$, and $d$. Their $\Gamma$-matrices, given in Eqs.(179a),
(179c), (194a), and (194c), are also real and, as a result, also satisfy
$(\Gamma^{f\overline{f}^{\prime}\to
B}\gamma^{\mu})^{\dagger}=\Gamma^{f\overline{f}^{\prime}\to B}\gamma^{\mu}$.
Moreover, one checks that

$\displaystyle\Gamma_{10}\gamma^{\mu}=\Gamma_{10}\gamma^{\mu}\left(\begin{array}[]{cc}1&0\\\
0&0\end{array}\right),$ (214c)
$\displaystyle\Gamma_{01}\gamma^{\mu}=\Gamma_{10}\gamma^{\mu}\left(\begin{array}[]{cc}0&0\\\
0&1\end{array}\right).$ (214f)

With these relations, substituting the above $\Gamma$-matrices into Eq.(163),
one gets that

$\displaystyle{\mathscr{H}}^{\nu\overline{\nu}Z}=\
\frac{1}{2}\widetilde{\psi}^{L}_{\nu}\gamma^{\mu}\psi_{\nu}^{L}\psi_{Z\mu},$
(215a)
$\displaystyle{\mathscr{H}}^{e\overline{e}Z}=\frac{1}{4}\left(\widetilde{\psi}^{L}_{e}\gamma^{\mu}\psi^{L}_{e}-\widetilde{\psi}^{R}_{e}\gamma^{\mu}\psi^{R}_{e}\right)\psi_{Z\mu},$
(215b)
$\displaystyle{\mathscr{H}}^{u_{i}\overline{u}_{i}Z}=\Big{(}-\frac{1}{2}\widetilde{\psi}^{L}_{u_{i}}\gamma^{\mu}\psi_{u_{i}}^{L}$
$\displaystyle\hskip
42.67912pt+\frac{1}{6}\widetilde{\psi}^{L}_{u_{i}}\gamma^{\mu}\psi^{L}_{u_{i}}-\frac{1}{6}\widetilde{\psi}^{R}_{u_{i}}\gamma^{\mu}\psi^{R}_{u_{i}}\Big{)}\psi_{Z\mu},$
(215c)
$\displaystyle{\mathscr{H}}^{d_{i}\overline{d}_{i}Z}=-\frac{1}{12}\left(\widetilde{\psi}^{L}_{d_{i}}\gamma^{\mu}\psi^{L}_{d_{i}}-\widetilde{\psi}^{R}_{d_{i}}\gamma^{\mu}\psi^{R}_{d_{i}}\right)\psi_{Z\mu}.$
(215d)

Similarly, for BPIPs that involve $W$-type modes, from the $\Gamma$-matrices
given in Eqs.(179e) and (193e), one gets that

$\displaystyle{\mathscr{H}}^{\nu\overline{e}W}=\frac{1}{2\sqrt{2}}\widetilde{\psi}^{L}_{e}\gamma^{\mu}\psi_{\nu}^{L}\psi_{W\mu},$
(216a)
$\displaystyle{\mathscr{H}}^{e\overline{\nu}W}=\frac{1}{2\sqrt{2}}\widetilde{\psi}^{L}_{\nu}\gamma^{\mu}\psi_{e}^{L}\psi_{\overline{W}\mu},$
(216b)
$\displaystyle{\mathscr{H}}^{u_{i}\overline{d}_{i}W}=-\frac{1}{2\sqrt{2}}\widetilde{\psi}^{L}_{d_{i}}\gamma^{\mu}\psi_{u_{i}}^{L}\psi_{W\mu},$
(216c)
$\displaystyle{\mathscr{H}}^{d_{i}\overline{u}_{i}W}=-\frac{1}{2\sqrt{2}}\widetilde{\psi}^{L}_{u_{i}}\gamma^{\mu}\psi_{d_{i}}^{L}\psi_{\overline{W}\mu}.$
(216d)

Finally, following a procedure similar to that given above, one may derive
expressions for the $H$-operator densities of those BPIPs that involve
$u/d$-type modes and $g$-modes. This can be done in a straightforward way, by
making use of the $\Gamma$-matrices given in Eqs.(199), (201), and (197), but,
we are not to write them explicitly.

##  IX Comparison with the SM

In this section, we give a comparison between the interaction $H$-operators of
the above-discussed model and interaction Hamiltonians given in the SM. We
mainly discuss electroweak interactions and give a brief discussion for strong
interactions at the end of this section.

We use ${\mathscr{H}}_{\rm gws}^{\rm int}$ to denote the part of the
Hamiltonian density given in the SM, which is for interactions between the
first-generation fermions (electron, electron neutrino, up quark, down quark,
and their anti particles) and vector bosons. In the ordinary notation (e.g.,
in Ref.[2]), ${\mathscr{H}}_{\rm gws}^{\rm int}$ is written as

$\displaystyle{\mathscr{H}}_{\rm gws}^{\rm
int}=-eA_{\mu}J^{\mu}_{EM}-g(W^{+}_{\mu}J^{\mu+}_{W}+W^{-}_{\mu}J^{\mu-}_{W}+Z^{0}_{\mu}J^{\mu}_{Z}),$
(217)

where the currents are given by

$\displaystyle
J^{\mu}_{EM}=\overline{e}\gamma^{\mu}(-1)e+\overline{u}\gamma^{\mu}(+\frac{2}{3})u+\overline{d}\gamma^{\mu}(-\frac{1}{3})d,$
(218a) $\displaystyle
J^{\mu}_{Z}=\frac{1}{\cos\theta_{W}}\Big{[}J^{\mu}_{Z-\nu}+J^{\mu}_{Z-e}+J^{\mu}_{Z-u}+J^{\mu}_{Z-d}\Big{]},$
(218b) $\displaystyle
J^{\mu+}_{W}=\frac{1}{\sqrt{2}}(\overline{\nu}_{L}\gamma^{\mu}e_{L}+\overline{u}_{L}\gamma^{\mu}d_{L}),$
(218c) $\displaystyle
J^{\mu-}_{W}=\frac{1}{\sqrt{2}}(\overline{e}_{L}\gamma^{\mu}\nu_{L}+\overline{d}_{L}\gamma^{\mu}u_{L}).$
(218d)

Current terms on the rhs of Eq.(218b) are given by

$\displaystyle
J^{\mu}_{Z-\nu}=\overline{\nu}_{L}\gamma^{\mu}(\frac{1}{2})\nu_{L},$ (219a)
$\displaystyle
J^{\mu}_{Z-e}=\overline{e}_{L}\gamma^{\mu}(-\frac{1}{2}+\sin^{2}\theta_{W})e_{L}+\overline{e}_{R}\gamma^{\mu}(\sin^{2}\theta_{W})e_{R},$
(219b) $\displaystyle
J^{\mu}_{Z-u}=\overline{u}_{L}\gamma^{\mu}(\frac{1}{2}-\frac{2}{3}\sin^{2}\theta_{W})u_{L}+\overline{u}_{R}\gamma^{\mu}(-\frac{2}{3}\sin^{2}\theta_{W})u_{R},$
(219c) $\displaystyle
J^{\mu}_{Z-d}=\overline{d}_{L}\gamma^{\mu}(-\frac{1}{2}+\frac{1}{3}\sin^{2}\theta_{W})d_{L}+\overline{d}_{R}\gamma^{\mu}(\frac{1}{3}\sin^{2}\theta_{W})d_{R},$
(219d)

where $\theta_{W}$ indicates the Weinberg angle. In Eqs.(218)-(219), the
overlines have the ordinary meaning; that is, e.g.,
$\overline{\psi}=\psi^{\dagger}\gamma^{0}$.

Below, we give a comparison between ${\mathscr{H}}_{\rm gws}^{\rm int}$ and
the $H$-operator densities given in Eqs.(213)-(216).

Firstly, we discuss the electromagnetism interaction. There is no essential
difference between the form of the term “$-eA_{\mu}J^{\mu}_{EM}$” with
$J^{\mu}_{EM}$ given in Eq.(218a) and that of the densities given in Eq.(213)
for the $A$-mode. In fact, there are only two differences in their forms:
(a) In the place where ${\mathscr{H}}_{\rm gws}^{\rm int}$ has a coupling
constant $e$, the $H$-operator densities have a number factor ($1/4$). This a
feature common to all terms and we are not to mention it below.
(b) The up-quark and down-quark terms have negative and positive signs,
respectively; meanwhile, the $u$-mode and $d$-mode terms have reverse signs,
i.e., positive and negative signs, respectively. Clearly, this difference in
sign has no physical significance.

Secondly, let us discuss interactions with $Z^{0}$-boson.
(i) The current $J^{\mu}_{Z-\nu}$ in Eq.(219a) for the neutrino field, if
added by the $Z^{0}$-boson field, has the same form as the $H$-operator
density ${\mathscr{H}}^{\nu\overline{\nu}Z}$ in Eq.(215a) for the $\nu$-mode
field.
(ii) The current $J^{\mu}_{Z-e}$ in Eq.(219b) and the $H$-operator density
${\mathscr{H}}^{e\overline{e}Z}$ in Eq.(215b) also have a similar formal
structure. Their main difference lies in coefficients contained; in
particular, the former contains the Weinberg angle, while, the latter does
not. Interestingly, they would have the same coefficients, if one replaces the
term $\sin^{2}\theta_{W}$ in the former by the number of $\frac{1}{4}$. We
observe that experimental results, interpreted within the framework of the SM,
give a close value, i.e., $\sin^{2}\theta_{W}\simeq 0.23$ [27].
(iii) The currents $J^{\mu}_{Z-u}$ and $J^{\mu}_{Z-d}$ in Eq.(219) also have a
structure formally similar to that of the $H$-operators
${\mathscr{H}}^{u_{i}\overline{u}_{i}Z}$ and
${\mathscr{H}}^{d_{i}\overline{d}_{i}Z}$ in Eq.(215). But, there exist some
further differences in the coupling coefficients, which are seen clearly in
Table 1. Among the differences, the most significant one is between the left-
handed part of $-J^{\mu}_{Z-d}$ and the corresponding part of
${\mathscr{H}}^{d_{i}\overline{d}_{i}Z}$.

| $-J^{\mu}_{Z-u}$ | ${\mathscr{H}}^{u_{i}\overline{u}_{i}Z}$ | $-J^{\mu}_{Z-d}$ | ${\mathscr{H}}^{d_{i}\overline{d}_{i}Z}$
---|---|---|---|---
left-handed | $-\frac{1}{2}+\frac{0.92}{6}$ | $-\frac{1}{2}+\frac{1}{6}$ | $\frac{1}{2}-\frac{0.23}{3}$ | $-\frac{1}{12}$
right-handed | $\frac{0.92}{6}$ | $-\frac{1}{6}$ | $-\frac{0.92}{12}$ | $\frac{1}{12}$
Table 1: Coupling coefficients of the left-handed and right-handed parts of
$-J^{\mu}_{Z-u}$ and $-J^{\mu}_{Z-d}$ in Eq.(219), with $\sin^{2}\theta_{W}$
taken as $0.23$, in comparison with the corresponding ones of the $H$-operator
densities ${\mathscr{H}}^{u_{i}\overline{u}_{i}Z}$ and
${\mathscr{H}}^{d_{i}\overline{d}_{i}Z}$ in Eq.(215).

Thirdly, let us discuss the currents $J^{\mu+}_{W}$ and $J^{\mu-}_{W}$ in
Eqs.(218c)-(218d), which are for interactions mediated by $W^{\pm}$-bosons
given in the GWS theory. One sees that there is no essential difference
between the form of this part of ${\mathscr{H}}_{\rm gws}^{\rm int}$ and that
of the $H$-operator densities for the $W$-type modes given in Eq.(216).

In addition, there is a difference related to interactions among bosons. In
fact, $W^{\pm}$-bosons and $Z^{0}$-bosons may interact directly in the SM,
while, bosonic modes do not interact directly in a theory of modes. This
difference does not necessarily imply a real confliction, because it is
possible for bosonic particles to interact effectively in a mode model, if the
particles are interpreted as eigenstates of certain (sums of) $H$-operators.

Finally, we give a brief discussion for strong interactions. In the SM, the
first generation has two quarks — the up and down quarks, each possessing a
color degree of freedom with three colors, and there are eight gluons that
mediate interactions of quarks. The studied model of modes has corresponding
properties. Specifically, it contains two types of six-layer fermionic modes —
$u$-type mode and $d$-type mode, each labelled by an index $i$ that takes
three values, and there are eight independent $g$-modes with which these two
types of fermionic modes interact. The index $i$ gives an interpretation to
the color degree of freedom.

Differences also exist between the two models. For example, interactions among
$u$-modes behave differently from those among $d$-modes [cf. Eq.(199) and
Eq.(201)], while, the $SU(3)$ group does not predict any difference between
interactions among up quarks and those among down quarks. However, detailed
comparison of exact expressions for the strong interaction predicted by the
two models is a complicated topic, which depends on the representations
employed for particle/mode states.

##  X Summary and discussions

In this section, we summarize results obtained in this paper and give some
further discussions. We discuss the framework of mode theory in Sec.X.1 and
discuss the studied model of modes in Sec.X.2. In Sec.X.3, we show that the
studied model contains a candidate for dark matter.

###  X.1 About the framework of mode theory

In this paper, a framework has been proposed for developing theories directly
on a quantum ground for elementary physical entities, which we call modes.
State spaces of modes are built by making use of representation spaces of the
Lorentz group, more exactly, of the $SL(2,C)$ group.

The framework is mainly based on five BAs (basic assumptions), which include
the following key points.
(i) Modes differ in their spinor-state spaces.
(ii) Spinor states of modes have a layer structure.
(iii) There are three types of fundamental processes: free process, VF (vacuum
fluctuation), and FIP (fundamental interaction process, as the change of two
fermionic modes into one bosonic mode, or the reverse).
(iv) Fermionic modes with negative $p^{0}$ may exist instantly only.
(v) Finite time evolution of states obeys a Schrödinger-type equation governed
by a time evolution operator.

In addition to the five BAs, three important assumptions have been used in the
proposed framework, as listed below:
(a) Eq.(29) for fermionicness of Weyl spinors within unit modes, as well as
for computation of $H$-operator amplitudes;
(b) assumptions stated around Eq.(48) about peculiar properties of
$s(\overline{s})$-modes; 101010 The peculiar properties of
$s(\overline{s})$-mode may be regarded as coming from Eq.(29).
(c) hermiticity of the time evolution operator.
Besides, relatively “minor” assumptions have been used, which look natural
from the viewpoint of the theory of spinors.

Associated with each fundamental process, an operator called $H$-operator is
considered, which maps the state space of the incoming mode(s) of the process
to that of its outgoing mode(s). BPIPs (basic positive-$p^{0}$ interaction
process), whose incoming and outgoing modes do not include any
negative-$p^{0}$ fermionic mode, are constructed from FIPs and VF (if needed),
with $H$-operators constructed accordingly. The time evolution operator is
given by the sum of $H$-operators of all free modes ($p^{0}>0$ for fermionic
modes) and of all BPIPs. It turns out that this operator is a function of
quantum fields, which are constructed from creation and annihilation operators
for free-mode states. Written with quantum fields, the interaction
$H$-operator automatically gets a local feature.

Compared with the framework supplied by QFT which is usually employed in the
field of particle physics, the proposed framework has the following merits.
That is, (i) it is directly founded on a quantum ground and, hence, no
procedure of quantizing classical field is needed. (ii) Certain unification in
the description of mode states is achieved at the fundamental level, in the
sense that spinor-state spaces of all modes are built from the two smallest
nontrivial representation spaces of the $SL(2,C)$ group. And, (iii) certain
unification in the description of interactions is also achieved at the
fundamental level, in the sense that interaction operators for fundamental
processes have the common feature of mapping the state space of incoming
mode(s) to that of outgoing mode(s).

###  X.2 About the model of modes

A model of modes has been constructed and studied within the above framework,
with three model assumptions addressing three things: (a) exact layer
structures of physical modes, (b) layer phases, and (c) exact forms of
$T_{k}$-matrices. These model assumptions are introduced, based on properties
of spinor states of modes, particularly on certain features of layer
structures of modes.

The studied model contains four types of fermionic modes — $\nu$-mode,
$e$-mode, $u$-mode, and $d$-mode, and four types of bosonic modes — $A$-mode,
$Z$-mode, $W$-mode, and $g$-mode (including their antimodes). These modes are
in parallel with elementary particles in the SM, more exactly, with the four
first-generation fermions of electron neutrino, electron, up quark, and down
quark, and with the four bosons of photon, $Z^{0}$-boson, $W$-boson, and
gluon, respectively.

The time evolution operator of the model consists of three parts. The first
part has a form similar to the electroweak interaction Hamiltonian (with
fermions) in the SM, with some differences in the coupling coefficients. The
second part shows some global similarities to the interaction Hamiltonian
given for strong interactions in the SM; for example, the number of bosonic
modes that mediate the interactions is also eight. But, a full comparison of
details of the second part with the strong-interaction part of the SM is not
available at the present stage for some technical reasons; in particular,
(future) research work is needed in order to find representations of
particle/mode states that are appropriate for such a comparison. 111111In
principle, by imposing sufficiently many model assumptions, it should be
possible to get a mode model whose time evolution operator has exactly the
same form as the corresponding part in the total Hamiltonian of the SM. But,
such a possibility is not of particular interest, because many assumptions
would be needed to achieve this goal.  The third part was not discussed
previously and is to be discussed in the next section.

The studied model gives interesting predictions, which are not available in
the SM. Rephrased in the terminology of the SM, some major ones are listed
below: (i) Up and down quarks should have $\frac{2}{3}$ and $\frac{1}{3}$
electronic charges, respectively, in unit of the charge of electron. (ii)
Neutrino should have zero electronic charge. (iii) The Weinberg angle may
correspond to $\sin^{2}\theta_{W}=0.25$. 121212This prediction is not
necessarily in confliction with the known result of $\sin^{2}\theta_{W}\simeq
0.23$, because this value was computed within the theoretical framework of the
SM.  And, (iv) the color degree of freedom of quarks may correspond to an
index $i=1,2,3$, which labels certain layer configuration of mode states.

At the present stage, it is too early to discuss direct comparison of
predictions of the studied model and experimental results of particles. In
fact, for this purpose, an interpretation is needed to experimentally observed
(free) particle states. In a natural interpretation, one may relate these
states to eigenstates of certain (sums of) $H$-operators. According to
experiences obtained in the study of experimental data with the SM, a long way
may be expected ahead to get a real comparison for a model of modes.

Anyway, it is possible for future study to show some need of modification to
the mode model. In this case, one possible direction of modification is to
find some other structural feature of spinor states involved and, based on it,
introduce some new model assumption, e.g., for a new $T_{k}$ matrix. Another
direction is to make use of some symmetry consideration, perhaps, like gauge
symmetries in the SM. But, presently, the situation with the second direction
is still obscure, because gauge symmetries in the SM are first introduced for
classical fields, while, the proposed framework is already quantum mechanical
in the beginning.

###  X.3 A candidate for dark matter?

The studied model of modes has a part that have not been discussed. This is
related to the rest two three-layer fermionic modes mentioned in Sec.VIII.3
(besides $t$-modes). In this section, we discuss six-layer modes that are
constructed from these two three-layer modes, and show that they are
candidates for dark matter.

We call the two three-layer fermionic modes mentioned above _$t_{\nu}$ -mode_
and _$\overline{t}_{\nu}$ -mode_. They have the following spinor states,

$\displaystyle|\mathcal{S}_{t_{\nu}}^{\alpha}({\bf
p})\rangle=\frac{1}{\sqrt{3}}\left(\begin{array}[]{c}|u^{\alpha}({\bf
p})\rangle\\\ |u^{\alpha}({\bf p})\rangle\\\ |u^{\alpha}({\bf
p})\rangle\end{array}\right),$ (220d)
$\displaystyle|\mathcal{S}_{\overline{t}_{\nu}}^{\alpha}({\bf
p})\rangle=\frac{1}{\sqrt{3}}\left(\begin{array}[]{c}|\overline{v}^{\alpha}({\bf
p})\rangle\\\ |\overline{v}^{\alpha}({\bf p})\rangle\\\
|\overline{v}^{\alpha}({\bf p})\rangle\end{array}\right).$ (220h)

According to the assumption ${\rm MA}_{\rm mode}^{\rm species}$, the above two
modes give rise to two six-layer fermionic modes, which we call $u_{\nu}$-mode
and $d_{\nu}$-mode, respectively, with the following spinor states,

$\displaystyle|\mathcal{S}_{{u}_{\nu}}^{\alpha}({\bf
p})\rangle=\frac{1}{\sqrt{2}}\left(\begin{array}[]{c}|\mathcal{S}_{t_{\nu}}^{\alpha}({\bf
p})\rangle\\\ |\mathcal{S}_{t_{\nu}}^{\alpha}({\bf
p})\rangle\end{array}\right),$ (221c)
$\displaystyle|\mathcal{S}_{{d}_{\nu}}^{\alpha}({\bf
p})\rangle=\frac{1}{\sqrt{2}}\left(\begin{array}[]{c}|\mathcal{S}_{t_{\nu}}^{\alpha}({\bf
p})\rangle\\\ i|\mathcal{S}_{\overline{t}_{\nu}}^{\alpha}({\bf
p})\rangle\end{array}\right).$ (221f)

Their antimodes are indicated as $\overline{u}_{\nu}$-mode and
$\overline{d}_{\nu}$-mode, respectively, with spinor states denoted by
$|\mathcal{S}_{{\overline{u}}_{\nu}}^{\alpha}({\bf p})\rangle$ and
$|\mathcal{S}_{{\overline{d}}_{\nu}}^{\alpha}({\bf p})\rangle$.

Here, we are interested in interactions of the above six-layer modes with the
$A$-mode only. It is the matrices $T_{2}$ and $T_{3}$ that may predict such
interactions. Let us first discuss the $u_{\nu}$-mode. The matrix $T_{2}$
effectively predicts a one-layer structure from
$\overrightarrow{D}|\mathcal{S}_{{u}_{\nu}}^{\alpha}({\bf
p})\rangle|\mathcal{S}_{{\overline{u}}_{\nu}}^{\alpha^{\prime}}({\bf
p}^{\prime})\rangle$, which does not correspond to any physical state.
Meanwhile, the action of $T_{3}$ on
$\overrightarrow{D}|\mathcal{S}_{{u}_{\nu}}^{\alpha}({\bf
p})\rangle|\mathcal{S}_{{\overline{u}}_{\nu}}^{\alpha^{\prime}}({\bf
p}^{\prime})\rangle$ produces the same result as that for
$\overrightarrow{D}|\mathcal{S}_{{u}_{1}}^{\alpha}({\bf
p})\rangle|\mathcal{S}_{{\overline{u}}_{1}}^{\alpha^{\prime}}({\bf
p}^{\prime})\rangle$ and gives a vanishing $\Gamma$-matrix as in Eq.(193b).
Hence, one has

$\displaystyle\Gamma^{u_{\nu}\overline{u}_{\nu}\to
A}_{(T_{2})}=\Gamma^{u_{\nu}\overline{u}_{\nu}\to A}_{(T_{3})}=0.$ (222)

Next, we discuss the $d_{\nu}$-mode. Making use of the explicit expressions of
spinor states given above, we note that the first three layers of
$\overrightarrow{D}|\mathcal{S}_{{d}_{\nu}}^{\alpha}({\bf
p})\rangle|\mathcal{S}_{{\overline{d}}_{\nu}}^{\alpha^{\prime}}({\bf
p}^{\prime})\rangle$, which are obtained from the two upper sections, are
identical; and similar for the last three layers from lower sections. This
implies that the action of $T_{2}$ on
$\overrightarrow{D}|\mathcal{S}_{{d}_{\nu}}^{\alpha}({\bf
p})\rangle|\mathcal{S}_{{\overline{d}}_{\nu}}^{\alpha^{\prime}}({\bf
p}^{\prime})\rangle$ should produce the same result as $T_{1}$ on a pair of
$e$-mode and $\overline{e}$-mode, namely on
$\overrightarrow{D}|\mathcal{S}_{{e}}^{\alpha}({\bf
p})\rangle|\mathcal{S}_{{\overline{e}}}^{\alpha^{\prime}}({\bf
p}^{\prime})\rangle$. Hence, according to Eq.(179d), one has
$\Gamma^{d_{\nu}\overline{d}_{\nu}\to A}_{(T_{2})}=\frac{1}{4}\Gamma_{11}$. As
for $T_{3}$, similar to the above-discussed case of $u_{\nu}$-mode, like
Eq.(193d), one has $\Gamma^{d_{\nu}\overline{d}_{\nu}\to
A}_{(T_{3})}=-\frac{1}{4}\Gamma_{11}$. As a result,

$\displaystyle\Gamma^{d_{\nu}\overline{d}_{\nu}\to
A}=\Gamma^{d_{\nu}\overline{d}_{\nu}\to
A}_{(T_{2})}+\Gamma^{d_{\nu}\overline{d}_{\nu}\to A}_{(T_{3})}=0.$ (223)

Therefore, neither the $u_{\nu}$-mode nor the $d_{\nu}$-mode interacts with
the $A$-mode. This is just a feature expected for dark matter.

###### Acknowledgements.

The author is grateful to Yan Gu for valuable discussions and suggestions.
This work was partially supported by the National Natural Science Foundation
of China under Grant Nos. 11535011, 11775210, and 12175222.

##  Appendix A Abstract notation for spinors

In this appendix, we first recall basic properties of two-component Weyl
spinors [20, 21, 22, 23], written in the abstract notation as discussed in
Ref.[24] (Sec.A.1). Then, we recall basic properties of four-component vectors
(Sec.A.2) and write them also in the abstract notation (Sec.A.3).

Concerning bras to be used in this appendix, we discuss only spinor bras for
scalar product, which are indicated as $\langle\langle\cdot|$ in the main
text. But, for brevity, within this appendix, we use the ordinary notation of
$\langle\cdot|$ to indicate them.

###  A.1 Basic properties of two-component spinors

In the spinor theory, there are two smallest nontrivial representation spaces
of the $SL(2,C)$ group, which are spanned by two types of two-component Weyl
spinors, respectively, with the relationship of complex conjugation. The two
spaces are denoted by ${\mathscr{W}}$ and $\overline{\mathscr{W}}$, with
overline indicating consequence of the operation of complex conjugation, as a
convention in the theory of spinors.

In terms of components, a Weyl spinor in ${\mathscr{W}}$ is written as, say,
$\kappa_{A}$ with an index $A=0,1$. In the abstract notation, a basis in the
space ${\mathscr{W}}$ is written as $|S^{A}\rangle$ and the above spinor is
written as $|\kappa\rangle$, with the expansion

$|\kappa\rangle=\kappa_{A}|S^{A}\rangle,$ (224)

with repeated index implying summation. 131313In Ref.[24], Eq.(224) is written
as $|\kappa\rangle=\kappa^{A}|S_{A}\rangle=-\kappa_{A}|S^{A}\rangle$, which
would introduce a minus sign to the rhs of the two equalities in Eq.(235).
One may introduce a space that is dual to ${\mathscr{W}}$, composed of bras
with a basis written as $\langle S^{A}|$. In order to construct a product that
is a scalar under $SL(2,C)$ transformations [see Eq.(305)], the bra dual to
the ket $|\kappa\rangle$ should be written as

$\langle\kappa|=\langle S^{A}|\kappa_{A},$ (225)

which has the same components as $|\kappa\rangle$ in Eq.(224), but not their
complex conjugates. (See Appendix B for a brief discussion of basic properties
of $SL(2,C)$ transformations.)

By definition, one may set scalar products of the basis as

$\langle S^{A}|S^{B}\rangle=\epsilon^{AB},$ (226)

where $\epsilon^{AB}$ is a symbol defined by

$\epsilon^{AB}=\left(\begin{array}[]{cc}0&1\\\ -1&0\end{array}\right).$ (227)

It proves convenient to introduce another matrix $\epsilon_{AB}$, which has
the same elements as $\epsilon^{AB}$. These two matrices can be used to raise
and lower indexes of components, say,

$\kappa^{A}=\epsilon^{AB}\kappa_{B},\quad\kappa_{A}=\kappa^{B}\epsilon_{BA},$
(228)

as well as for the basis spinors, namely,

$\displaystyle|S^{A}\rangle=\epsilon^{AB}|S_{B}\rangle,\quad|S_{A}\rangle=|S^{B}\rangle\epsilon_{BA}.$
(229)

It is not difficult to verify that (i) $\langle
S_{A}|S_{B}\rangle=\epsilon_{AB}$; (ii)

${f_{\ldots}^{\ \ \ A}\ (g)^{\cdots}}_{A}=-{f_{\ldots A}\ (g)^{\cdots A}},$
(230)

i.e., a minus sign is introduced by the exchange of the positions of a pair of
repeated two-component index; and (iii) the symbols $\epsilon_{C}^{\ \
A}=\epsilon^{BA}\epsilon_{BC}$ and $\epsilon^{A}_{\ \
C}=\epsilon^{AB}\epsilon_{BC}$ satisfy the relation:

$\displaystyle\epsilon_{C}^{\ \ A}=-\epsilon^{A}_{\ \ C}=\delta^{A}_{C},$
(231)

where $\delta^{A}_{B}=1$ for $A=B$ and $\delta^{A}_{B}=0$ for $A\neq B$.
($\delta$-symbols for other types of indices are defined in the same way.)

The scalar product of two generic spinors $|\chi\rangle$ and $|\kappa\rangle$,
written as $\langle\chi|\kappa\rangle$, has the expression of

$\langle\chi|\kappa\rangle=\chi_{A}\kappa^{A}.$ (232)

The anti-symmetry of $\epsilon^{AB}$ implies that

$\langle\chi|\kappa\rangle=-\langle\kappa|\chi\rangle$ (233)

and, as a consequence, $\langle\kappa|\kappa\rangle=0$ for all
$|\kappa\rangle$. It is not difficult to verify the following properties: (a)
The identity operator in the space ${\mathscr{W}}$, denoted by
$I_{{\mathscr{W}}}$, can be written as

$\displaystyle I_{{\mathscr{W}}}=|S^{A}\rangle\langle S_{A}|,$ (234)

satisfying $I_{{\mathscr{W}}}|\kappa\rangle=|\kappa\rangle$ for all
$|\kappa\rangle\in{\mathscr{W}}$; and (b) the components of $|\kappa\rangle$
have the following expressions,

$\kappa^{A}=\langle S^{A}|\kappa\rangle,\quad\kappa_{A}=\langle
S_{A}|\kappa\rangle.$ (235)

The operation of complex conjugation converts ${\mathscr{W}}$ to
$\overline{\mathscr{W}}$, and changes spinors $|\kappa\rangle\in{\mathscr{W}}$
to spinors $|\overline{\kappa}\rangle\in\overline{\mathscr{W}}$. Corresponding
to the basis $|S_{A}\rangle\in{\mathscr{W}}$, the space
$\overline{\mathscr{W}}$ has a basis denoted by
$|\overline{S}_{A^{\prime}}\rangle$ with a primed index
$A^{\prime}=0^{\prime},1^{\prime}$. On the basis of
$|\overline{S}_{A^{\prime}}\rangle$, $|\overline{\kappa}\rangle$ is written as

$|\overline{\kappa}\rangle={\overline{\kappa}}_{A^{\prime}}|\overline{S}^{A^{\prime}}\rangle,$
(236)

where

$\overline{\kappa}^{A^{\prime}}:=(\kappa^{A})^{*}.$ (237)

Similar to the $\epsilon$ symbols discussed above, one introduces symbols
$\epsilon^{A^{\prime}B^{\prime}}$ and $\epsilon_{A^{\prime}B^{\prime}}$, which
have the same matrix elements as $\epsilon^{AB}$ and are used to raise and
lower primed indices. The identity operator in the space
$\overline{\mathscr{W}}$, denoted by $I_{\overline{\mathscr{W}}}$, has the
following form

$\displaystyle
I_{\overline{\mathscr{W}}}=|\overline{S}^{A^{\prime}}\rangle\langle\overline{S}_{A^{\prime}}|.$
(238)

When a spinor $\kappa^{A}$ is transformed by an $SL(2,C)$ matrix, the spinor
$\overline{\kappa}^{A^{\prime}}$ is transformed by the complex-conjugate
matrix (see Appendix B).

###  A.2 Basic properties of four-component vectors

In this section, we recall basic properties of four-component vectors as
special cases of spinors [20, 22, 21]. We use the ordinary notation in this
section and will discuss the abstract notation in the next section.

A key point is a one-to-one mapping between a direct-product space
${\mathscr{W}}\otimes\overline{\mathscr{W}}$ and a four-dimensional space
denote by ${\mathscr{V}}$. The mapping is given by the so-called Enfeld-van
der Waerden symbols, in short _EW-symbols_ , denoted by $\sigma^{\mu
AB^{\prime}}$, with $\mu=0,1,2,3$, which are invariant under $SL(2,C)$
transformations. An often-used set of explicit expressions for these symbols
is given by 141414Compared with Pauli matrices, the EW-symbols have a common
prefactor $1/\sqrt{2}$. This leads to the relations in Eq.(264), otherwise,
the rhs of the two equalities in it should be multiplied by $2$. (This
prefactor $1/\sqrt{2}$ was not used in Ref.[24].)

$\displaystyle\sigma^{0AB^{\prime}}=\frac{1}{\sqrt{2}}\left(\begin{array}[]{cc}1&0\\\
0&1\\\
\end{array}\right),\sigma^{1AB^{\prime}}=\frac{1}{\sqrt{2}}\left(\begin{array}[]{cc}0&1\\\
1&0\\\ \end{array}\right),\ \ \ $ (243)
$\displaystyle\sigma^{2AB^{\prime}}=\frac{1}{\sqrt{2}}\left(\begin{array}[]{cc}0&-i\\\
i&0\\\
\end{array}\right),\sigma^{3AB^{\prime}}=\frac{1}{\sqrt{2}}\left(\begin{array}[]{cc}1&0\\\
0&-1\\\ \end{array}\right).\ \ $ (248)

Lowering the two-component indices like in Eq.(228), it is straightforward to
verify that

$\displaystyle\sigma^{0}_{AB^{\prime}}=\frac{1}{\sqrt{2}}\left(\begin{array}[]{cc}1&0\\\
0&1\\\
\end{array}\right),\sigma^{1}_{AB^{\prime}}=\frac{1}{\sqrt{2}}\left(\begin{array}[]{cc}0&-1\\\
-1&0\\\ \end{array}\right),$ (253)
$\displaystyle\sigma^{2}_{AB^{\prime}}=\frac{1}{\sqrt{2}}\left(\begin{array}[]{cc}0&-i\\\
i&0\\\
\end{array}\right),\sigma^{3}_{AB^{\prime}}=\frac{1}{\sqrt{2}}\left(\begin{array}[]{cc}-1&0\\\
0&1\\\ \end{array}\right).$ (258)

For example, a spinor $\phi_{AB^{\prime}}$ in the space
${\mathscr{W}}\otimes\overline{\mathscr{W}}$ can be mapped to a vector
$K^{\mu}$ in the space ${\mathscr{V}}$ by the EW-symbols,

$K^{\mu}=\sigma^{\mu AB^{\prime}}\phi_{AB^{\prime}}.$ (259)

In the space ${\mathscr{V}}$, of particular importance is a symbol
$g^{\mu\nu}$, which is defined by

$g^{\mu\nu}:=\sigma^{\mu AB^{\prime}}\sigma^{\nu
CD^{\prime}}\epsilon_{AC}\epsilon_{B^{\prime}D^{\prime}}.$ (260)

One may introduce a lower-indexed symbol $g_{\mu\nu}$, which by definition has
the same matrix elements as $g^{\mu\nu}$, namely, $[g^{\mu\nu}]=[g_{\mu\nu}]$.
Then, like the $\epsilon$-symbols in the space ${\mathscr{W}}$, these two
symbols of $g$ can be used to raise and lower indexes, e.g.,

$K_{\mu}=K^{\nu}g_{\nu\mu},\quad K^{\mu}=g^{\mu\nu}K_{\nu}.$ (261)

Making use of the anti-symmetry of the $\epsilon$ symbols, it is easy to
verify that $g^{\mu\nu}$ is symmetric, i.e.,

$g^{\mu\nu}=g^{\nu\mu}.$ (262)

Due to this symmetry, the upper-lower positions of a pair of repeated index
are exchangeable, namely

${F_{\ldots}^{\ \ \ \mu}(f)^{\cdots}}_{\mu}={F_{\ldots\mu}(f)^{\cdots\mu}}.$
(263)

The EW-symbols have the following properties,

$\sigma^{AB^{\prime}}_{\mu}\sigma_{CD^{\prime}}^{\mu}=\delta^{AB^{\prime}}_{CD^{\prime}},\quad\sigma_{AB^{\prime}}^{\mu}\sigma^{AB^{\prime}}_{\nu}=\delta_{\nu}^{\mu},$
(264)

where
$\delta^{AB^{\prime}}_{CD^{\prime}}:=\delta^{A}_{C}\delta^{B^{\prime}}_{D^{\prime}}$.
Using the relations in Eq.(264), it is not difficult to check that the map
from ${\mathscr{W}}\otimes\overline{\mathscr{W}}$ to ${\mathscr{V}}$ given in
Eq.(259) is reversible. Making use of Eqs.(264) and (231), it is not difficult
to verify the following relation,

$\displaystyle\sigma_{\mu
AB^{\prime}}\sigma_{CD^{\prime}}^{\mu}=\epsilon_{AC}\epsilon_{B^{\prime}D^{\prime}}.$
(265)

Then, substituting the definition of $g^{\mu\nu}$ in Eq.(260) into the product
of $g^{\mu\nu}g_{\nu\lambda}$, after simple algebra, one gets

$\displaystyle g^{\mu\nu}g_{\nu\lambda}=g^{\mu}_{\ \ \lambda}=g^{\ \
\mu}_{\lambda}=\delta^{\mu}_{\lambda}.$ (266)

When a $SL(2,C)$ transformation is carried out on the space ${\mathscr{W}}$, a
related transformation is applied to the space ${\mathscr{V}}$. Requiring
invariance of the EW-symbols, transformations on the space ${\mathscr{V}}$ can
be fixed and turn out to constitute a (restricted) Lorentz group (see Appendix
B for detailed discussions.). Therefore, the space ${\mathscr{V}}$ is a four-
component vector space. In fact, substituting the explicit expressions of the
EW-symbols in Eq.(248) into Eq.(260), one gets that

$\displaystyle g^{\mu\nu}=\sigma^{\mu}_{A^{\prime}B}\sigma^{\nu
A^{\prime}B}=\left(\begin{array}[]{cccc}1&0&0&0\\\ 0&-1&0&0\\\ 0&0&-1&0\\\
0&0&0&-1\end{array}\right),$ (271)

which is just the Minkovski’s metric.

As shown in Appendix B [Eq.(317)], the following product,

$\displaystyle J_{\nu}K^{\nu}=J^{\mu}g_{\mu\nu}K^{\nu},$ (272)

is a scalar under Lorentz transformations. Physically, of more interest is a
product, in which one of the two vectors takes a complex-conjugate form, say,

$\displaystyle J^{*}_{\nu}K^{\nu}=J^{\mu*}g_{\mu\nu}K^{\nu}.$ (273)

Similarly, one finds that this product is also a scalar.

###  A.3 Abstract notation for four-component vectors

In the abstract notation, a basis in the space ${\mathscr{V}}$ is written as
$|T^{\mu}\rangle$ with $\mu=0,1,2,3$. The index of the basis can be lowered by
$g_{\mu\nu}$, i.e., $|T_{\mu}\rangle=g_{\mu\nu}|T^{\nu}\rangle$, and similarly
$|T^{\mu}\rangle=g^{\mu\nu}|T_{\nu}\rangle$. A generic four-component vector
$|K\rangle$ in the space ${\mathscr{V}}$ is expanded as

$|K\rangle=K_{\mu}|T^{\mu}\rangle=K^{\mu}|T_{\mu}\rangle.$ (274)

In consistency with the expression of bra in Eq.(225) for two-component
spinors, the bra corresponding to $|K\rangle$ is written as

$\displaystyle\langle K|=\langle T_{\mu}|K^{\mu}.$ (275)

To write the scalar product in Eq.(272) in the form of $\langle J|K\rangle$,
namely,

$\displaystyle\langle J|K\rangle=J_{\nu}K^{\nu},$ (276)

the following requirement is needed,

$\displaystyle\langle T^{\mu}|T^{\nu}\rangle=g^{\mu\nu},$ (277)

in a way similar to the case of two-component spinors [cf. Eq.(226)]. Due to
the symmetry of $g^{\mu\nu}$ in Eq.(262), one sees that

$\displaystyle\langle J|K\rangle=\langle K|J\rangle.$ (278)

It is not difficult to verify the following properties. (i) Making use of
Eq.(266), one can verify that the identity operator in the space
${\mathscr{V}}$ are written as

$I_{{\mathscr{V}}}=|T_{\mu}\rangle\langle T^{\mu}|=|T^{\mu}\rangle\langle
T_{\mu}|.$ (279)

(ii) The components $K^{\mu}$ and $K_{\mu}$ have the following expressions,

$K^{\mu}=\langle T^{\mu}|K\rangle,\quad K_{\mu}=\langle T_{\mu}|K\rangle.$
(280)

And, (iii) the symmetry of $g^{\mu\nu}$ implies that $\langle
T_{\mu}|T_{\nu}\rangle=\langle T_{\nu}|T_{\mu}\rangle$, as a result,

$\displaystyle\langle K|J\rangle=\langle J|K\rangle$ (281)

for two arbitrary vectors $|K\rangle$ and $|J\rangle$ in the space
${\mathscr{V}}$.

Let us write basis spinors in the product space
${\mathscr{W}}\otimes\overline{\mathscr{W}}$ as

$\displaystyle|S_{AB^{\prime}}\rangle:=|S_{A}\rangle|\overline{S}_{B^{\prime}}\rangle,\
\ \langle S_{B^{\prime}A}|:=\langle\overline{S}_{B^{\prime}}|\langle S_{A}|.$
(282)

For this product space, we assume that Weyl spinors satisfy the
anticommutability relations given in Eq.(29). It proves convenient to
introduce an operator, which maps the space
${\mathscr{W}}\otimes\overline{\mathscr{W}}$ to the space ${\mathscr{V}}$ by
the EM-symbols. It is just the operator $\sigma$ in Eq.(30), i.e.,

$\sigma=\sigma^{\mu AB^{\prime}}|T_{\mu}\rangle\langle
S_{B^{\prime}A}|\qquad\text{[Eq.(\ref{sigma}) of main text].}$ (283)

For an arbitrary spinor
$|\phi\rangle=\phi_{AB^{\prime}}|S^{AB^{\prime}}\rangle\in{\mathscr{W}}\otimes\overline{\mathscr{W}}$,
this operator $\sigma$ acts as

$\displaystyle\sigma|\phi\rangle=\sigma^{\mu
AB^{\prime}}\phi_{AB^{\prime}}|T_{\mu}\rangle.$ (284)

Clearly, for $|K\rangle=\sigma|\phi\rangle$, this result is in agreement with
Eq.(259). Under the specific expressions of $\sigma^{\mu AB^{\prime}}$ in
Eq.(248), making use of Eq.(284) one finds that

$\displaystyle|T^{0}\rangle$
$\displaystyle=\frac{1}{\sqrt{2}}\sigma(|S^{11^{\prime}}\rangle+|S^{00^{\prime}}\rangle),$
(285a) $\displaystyle|T^{1}\rangle$
$\displaystyle=-\frac{1}{\sqrt{2}}\sigma(|S^{10^{\prime}}\rangle+|S^{01^{\prime}}\rangle),$
(285b) $\displaystyle|T^{2}\rangle$
$\displaystyle=\frac{i}{\sqrt{2}}\sigma(|S^{10^{\prime}}\rangle-|S^{01^{\prime}}\rangle),$
(285c) $\displaystyle|T^{3}\rangle$
$\displaystyle=\frac{1}{\sqrt{2}}\sigma(|S^{11^{\prime}}\rangle-|S^{00^{\prime}}\rangle).$
(285d)

Since
$\overline{|S_{AB^{\prime}}\rangle}=|S_{A^{\prime}B}\rangle=-|S_{BA^{\prime}}\rangle$,
the operation of complex conjugation maps the space ${\mathscr{V}}$ into
itself. Hence, one may use the same index $\mu$ to index
$\overline{|T_{\mu}\rangle}$, the complex conjugates of $|T_{\mu}\rangle$, and
write it as $|\overline{T}_{\mu}\rangle$. Making use of Eq.(30), the complex
conjugate of the operator $\sigma$, namely $\overline{\sigma}$, is written as

$\overline{\sigma}=\overline{\sigma}^{\mu
A^{\prime}B}|\overline{T}_{\mu}\rangle\langle S_{BA^{\prime}}|,$ (286)

where $\overline{\sigma}^{\mu A^{\prime}B}\equiv(\sigma^{\mu
AB^{\prime}})^{*}$.

One sees that both $|T_{\mu}\rangle$ and $|\overline{T}_{\mu}\rangle$ give a
basis for the space ${\mathscr{V}}$. In fact, one has a freedom of imposing
some relation between these two bases, which would consequently implies
certain relation between the two operators of $\sigma$ and
$\overline{\sigma}$, and vice versa. We found it convenient to set equivalence
of $\sigma$ and $\overline{\sigma}$, i.e.,

$\overline{\sigma}=\sigma.$ (287)

Moreover, we note that the explicit expressions of the EM-symbols in Eq.(248)
imply hermiticity of the EW-symbols, i.e.,

$\overline{\sigma}^{\mu B^{\prime}A}=\sigma^{\mu AB^{\prime}}.$ (288)

Then, we find that $\overline{\sigma}$ in Eq.(286) is written as

$\overline{\sigma}=\sigma^{\mu BA^{\prime}}|\overline{T}_{\mu}\rangle\langle
S_{BA^{\prime}}|=-\sigma^{\mu AB^{\prime}}|\overline{T}_{\mu}\rangle\langle
S_{B^{\prime}A}|,$ (289)

where Eq.(29a) has been used in the derivation of the second equality. From
Eqs.(30), (287), and (289), one sees that

${|\overline{T}_{\mu}\rangle}=-|T_{\mu}\rangle\Longleftrightarrow\langle\overline{T}_{\mu}|=-\langle
T_{\mu}|.$ (290)

Let us write the scalar product in Eq.(273) in the abstract form. Making use
of Eq.(290), one finds that

$\displaystyle|\overline{K}\rangle=-K^{\mu*}|T_{\mu}\rangle,$ (291a)
$\displaystyle\langle\overline{J}|=-\langle T_{\mu}|J^{\mu*}.$ (291b)

Then, one gets that

$\displaystyle\langle\overline{J}|K\rangle=-J^{*}_{\mu}K^{\mu}=-J^{\mu*}g_{\mu\nu}K^{\nu}.$
(292)

Clearly, the minus signs on the rhs of Eq.(292) comes from those in Eq.(290).
We recall that the same minus sign appears in the symbol $\widehat{g}$ [see
Eq.(47)]. In fact, using the hat symbol, one writes that

$\displaystyle\langle\overline{J}|K\rangle=J^{\mu*}\widehat{g}_{\mu\nu}K^{\nu}.$
(293)

Moreover, it is easy to verify that

$\displaystyle\langle\overline{K}|J\rangle^{*}=\langle\overline{J}|K\rangle.$
(294)

It proves convenient to introduce an operation of _transposition_ for generic
kets and bras, indicated by a superscript $T$, which is defined by

$\displaystyle|\phi\psi\rangle^{T}:=\langle\psi\phi|,\
\langle\psi\phi|^{T}:=|\phi\psi\rangle,\
(|\phi\rangle\langle\psi|)^{T}:=|\psi\rangle\langle\phi|.\hskip 25.6073pt$
(295)

For example, $|S_{AB^{\prime}}\rangle^{T}=\langle S_{B^{\prime}A}|$ and

$\sigma^{T}=\sigma^{\mu AB^{\prime}}|S_{AB^{\prime}}\rangle\langle T_{\mu}|.$
(296)

Computing the product of $\sigma$ in Eq.(30) and $\sigma^{T}$ in Eq.(296),
noting Eqs.(277) and (264), it is easy to verify that

$\sigma^{T}\sigma=I_{{\mathscr{W}}}\otimes
I_{\overline{\mathscr{W}}},\quad\sigma\sigma^{T}=I_{{\mathscr{V}}},$ (297)

where the identity operators are given in Eqs.(234), (238), and (279). This
implies that $\sigma^{T}$ is the reverse of $\sigma$, denote by $\sigma^{-1}$,

$\sigma^{-1}=\sigma^{T}.$ (298)

We use $\sigma^{\dagger}$ to denote the complex conjugate of the transposition
of $\sigma$,

$\displaystyle\sigma^{\dagger}:=\overline{\left(\sigma^{T}\right)}.$ (299)

Direction derivation shows that

$\displaystyle\sigma^{\dagger}=\sigma^{T}.$ (300)

###  A.4 Some detailed derivations

In this section, we give detailed derivations of Eqs.(284), (285), (291a),
(297) and (300).

Derivation of Eq.(284):

$\displaystyle\sigma|\phi\rangle=\sigma^{\mu
AB^{\prime}}|T_{\mu}\rangle\langle
S_{B^{\prime}A}|\phi_{CD^{\prime}}|S^{CD^{\prime}}\rangle$
$\displaystyle=\sigma^{\mu
AB^{\prime}}|T_{\mu}\rangle\phi_{CD^{\prime}}\delta^{C}_{A}\delta^{D^{\prime}}_{B^{\prime}}=\sigma^{\mu
AB^{\prime}}\phi_{AB^{\prime}}|T_{\mu}\rangle.$

Derivation of Eq.(285):
Making use of the fact that
$|S^{AB^{\prime}}\rangle=\phi_{CD^{\prime}}|S^{CD^{\prime}}\rangle$, with
$\phi_{CD^{\prime}}=\delta^{A}_{C}\delta^{B^{\prime}}_{D^{\prime}}$, and that
$\sigma|\phi\rangle=\sigma^{\mu
AB^{\prime}}\phi_{AB^{\prime}}|T_{\mu}\rangle$, one finds that

$\displaystyle\sigma|S^{00^{\prime}}\rangle=\sigma^{\mu
00^{\prime}}|T_{\mu}\rangle=\frac{1}{\sqrt{2}}(|T_{0}\rangle+|T_{3}\rangle)$
$\displaystyle=\frac{1}{\sqrt{2}}(|T^{0}\rangle-|T^{3}\rangle),$
$\displaystyle\sigma|S^{01^{\prime}}\rangle=\sigma^{\mu
01^{\prime}}|T_{\mu}\rangle=\frac{1}{\sqrt{2}}(|T_{1}\rangle-i|T_{2}\rangle)$
$\displaystyle=\frac{1}{\sqrt{2}}(-|T^{1}\rangle+i|T^{2}\rangle),$
$\displaystyle\sigma|S^{10^{\prime}}\rangle=\sigma^{\mu
10^{\prime}}|T_{\mu}\rangle=\frac{1}{\sqrt{2}}(|T_{1}\rangle+i|T_{2}\rangle)$
$\displaystyle=\frac{1}{\sqrt{2}}(-|T^{1}\rangle-i|T^{2}\rangle),$
$\displaystyle\sigma|S^{11^{\prime}}\rangle=\sigma^{\mu
11^{\prime}}|T_{\mu}\rangle=\frac{1}{\sqrt{2}}(|T_{0}\rangle-|T_{3}\rangle)$
$\displaystyle=\frac{1}{\sqrt{2}}(|T^{0}\rangle+|T^{3}\rangle).$

Derivation of Eq.(291a):

$\displaystyle\langle\overline{J}|K\rangle=-\langle
T_{\mu}|J^{\mu*}K^{\nu}|T_{\nu}\rangle=-J^{\mu*}K^{\nu}g_{\mu\nu}=-J_{\nu}^{*}K^{\nu}.$

Derivation of Eq.(297):

$\displaystyle\sigma^{T}\sigma=\sigma^{\mu
AB^{\prime}}|S_{AB^{\prime}}\rangle\langle T_{\mu}|\sigma^{\nu
CD^{\prime}}|T_{\nu}\rangle\langle S_{D^{\prime}C}|$
$\displaystyle=\sigma^{\mu AB^{\prime}}\sigma^{\nu
CD^{\prime}}g_{\mu\nu}|S_{AB^{\prime}}\rangle\langle S_{D^{\prime}C}|$
$\displaystyle=\sigma^{\mu}_{AB^{\prime}}\sigma_{\mu}^{CD^{\prime}}|S^{AB^{\prime}}\rangle\langle
S_{D^{\prime}C}|=|S^{AB^{\prime}}\rangle\langle S_{B^{\prime}A}|$
$\displaystyle\sigma\sigma^{T}=\sigma^{\nu CD^{\prime}}|T_{\nu}\rangle\langle
S_{D^{\prime}C}|\sigma^{\mu AB^{\prime}}|S_{AB^{\prime}}\rangle\langle
T_{\mu}|$ $\displaystyle=\sigma^{\nu CD^{\prime}}\sigma^{\mu
AB^{\prime}}\epsilon_{CA}\epsilon_{D^{\prime}B^{\prime}}|T_{\nu}\rangle\langle
T_{\mu}|$
$\displaystyle=\sigma_{\nu}^{CD^{\prime}}\sigma^{\mu}_{AB^{\prime}}|T^{\nu}\rangle\langle
T_{\mu}|=|T^{\mu}\rangle\langle T_{\mu}|$

Derivation of Eq.(300):

$\displaystyle\sigma^{\dagger}=\overline{\left(\sigma^{\mu
AB^{\prime}}|S_{AB^{\prime}}\rangle\langle
T_{\mu}|\right)}=\overline{\sigma}^{\mu
A^{\prime}B}|S_{A^{\prime}B}\rangle\langle\overline{T}_{\mu}|$
$\displaystyle=-\sigma^{\mu BA^{\prime}}|S_{A^{\prime}B}\rangle\langle
T_{\mu}|=\sigma^{\mu BA^{\prime}}|S_{BA^{\prime}}\rangle\langle
T_{\mu}|=\sigma^{T}$

##  Appendix B SL(2,C) transformations

In this appendix, we recall basic properties of $SL(2,C)$ transformations,
particularly, their relation to Lorentz transformations. We use tilde to
indicate results of $SL(2,C)$ transformations.

The group $SL(2,C)$ is composed of $2\times 2$ complex matrices with unit
determinant [20, 21, 22, 24, 23], written as

$h^{A}_{\ \ B}=\left(\begin{array}[]{cc}a&b\\\
c&d\end{array}\right)\quad\text{with}\ ad-bc=1.$ (301)

Under a transformation given by $h^{A}_{\ \ B}$, a two-component spinor
$\kappa^{A}$ is transformed to

$\widetilde{\kappa}^{A}=h^{A}_{\ \ B}\kappa^{B}.$ (302)

It is straightforward to verify that $\epsilon^{AB}$ is invariant under
$SL(2,C)$ transformations, that is, $\widetilde{\epsilon}^{AB}=h^{A}_{\ \
C}h^{B}_{\ \ D}\epsilon^{CD}$ has the same matrix form as $\epsilon^{AB}$ in
Eq.(227). Direct derivation gives the following relations,

$\displaystyle h^{A}_{\ \ B}h_{C}^{\ \ B}=h_{B}^{\ \ A}h^{B}_{\ \
C}=-\epsilon_{C}^{\ \ A}.$ (303) $\displaystyle h_{AD}h^{A}_{\ \
C}=\epsilon_{DC},\quad h^{A}_{\ \ B}h^{CB}=\epsilon^{AC}.$ (304)

It is not difficult to verify that the product $\chi_{A}\kappa^{A}$ is a
scalar product, that is,

$\displaystyle\widetilde{\chi}_{A}\widetilde{\kappa}^{A}=\chi_{A}\kappa^{A}.$
(305)

Note that no complex conjugate is involved in the formation of this scalar
product and this is exactly the reason why the bra for Weyl spinor has the
expansion given in Eq.(225).

When $\kappa^{A}$ is transformed by a matrix $h^{A}_{\ \ B}$,
$\overline{\kappa}^{A^{\prime}}$ is transformed by its complex-conjugate
matrix, namely,

$\widetilde{\overline{\kappa}}^{A^{\prime}}=\overline{h}^{A^{\prime}}_{\ \
B^{\prime}}\overline{\kappa}^{B^{\prime}},$ (306)

where

$\overline{h}^{A^{\prime}}_{\ \ B^{\prime}}:=(h^{A}_{\ \
B})^{*}\quad\text{with}\ A=A^{\prime},B=B^{\prime}.$ (307)

Now, we discuss relation between $SL(2,C)$ transformations and Lorentz
transformations. Related to an $SL(2,C)$ transformation $h^{A}_{\ \ B}$, which
is performed on a space ${\mathscr{W}}$, we use $\Lambda^{\mu}_{\ \ \nu}$ to
denote the corresponding transformation on an space ${\mathscr{V}}$,

$\ \widetilde{K}^{\mu}=\Lambda^{\mu}_{\ \ \nu}K^{\nu}.$ (308)

It proves convenient to require invariance of the EM-symbols under $SL(2,C)$
transformations, namely,

$\widetilde{\sigma}^{\mu A^{\prime}B}=\sigma^{\mu A^{\prime}B},$ (309)

where

$\displaystyle\widetilde{\sigma}^{\mu A^{\prime}B}=\Lambda^{\mu}_{\ \
\nu}\overline{h}^{A^{\prime}}_{\ \ C^{\prime}}h^{B}_{\ \ D}\sigma^{\nu
C^{\prime}D}.$ (310)

This requirement imposes a restriction to $\Lambda^{\mu}_{\ \ \nu}$. In fact,
substituting Eq.(310) into Eq.(309) and rearranging positions of some indices,
one gets that

$\displaystyle\sigma^{\mu}_{\ A^{\prime}B}=\Lambda^{\mu}_{\ \
\nu}\overline{h}_{A^{\prime}C^{\prime}}h_{BD}\sigma^{\nu C^{\prime}D}.$ (311)

Multiplying both sides of Eq.(311) by $\overline{h}^{A^{\prime}}_{\ \
E^{\prime}}h^{B}_{\ \ F}\sigma_{\nu}^{E^{\prime}F}$, its rhs gives

$\displaystyle\Lambda^{\mu}_{\ \
\eta}\overline{h}_{A^{\prime}E^{\prime}}h_{BF}\sigma^{\eta
E^{\prime}F}\overline{h}^{A^{\prime}}_{\ \ C^{\prime}}h^{B}_{\ \
D}\sigma_{\nu}^{C^{\prime}D}$ $\displaystyle=\Lambda^{\mu}_{\ \
\eta}\epsilon_{E^{\prime}C^{\prime}}\epsilon_{FD}\sigma^{\eta
E^{\prime}F}\sigma_{\nu}^{C^{\prime}D}=\Lambda^{\mu}_{\ \ \nu},$ (312)

where Eq.(304) and Eq.(264) have been used. Then, one gets the following
expression of $\Lambda^{\mu}_{\ \ \nu}$,

$\Lambda^{\mu}_{\ \
\nu}=\sigma^{\mu}_{A^{\prime}B}\overline{h}^{A^{\prime}}_{\ \
C^{\prime}}h^{B}_{\ \ D}\sigma_{\nu}^{C^{\prime}D}.$ (313)

Substituting Eq.(313) into the product $\Lambda^{\mu}_{\ \
\eta}\Lambda^{\nu}_{\ \ \xi}g^{\eta\xi}$, one gets

$\displaystyle\sigma^{\mu}_{A^{\prime}B}\overline{h}^{A^{\prime}}_{\ \
C^{\prime}}h^{B}_{\ \
D}\sigma_{\eta}^{C^{\prime}D}\sigma^{\nu}_{E^{\prime}F}\overline{h}^{E^{\prime}}_{\
\ G^{\prime}}h^{F}_{\ \ H}\sigma_{\xi}^{G^{\prime}H}g^{\eta\xi}.$

Using Eq.(265), this gives

$\displaystyle\sigma^{\mu}_{A^{\prime}B}\overline{h}^{A^{\prime}}_{\ \
C^{\prime}}h^{B}_{\ \
D}\sigma^{\nu}_{E^{\prime}F}\overline{h}^{E^{\prime}C^{\prime}}h^{FD}.$

Then, noting Eqs.(304) and (260), one gets the first equality in the following
two relations,

$\displaystyle\Lambda^{\mu}_{\ \ \eta}\Lambda^{\nu}_{\ \
\xi}g^{\eta\xi}=g^{\mu\nu},\quad\Lambda^{\mu}_{\ \ \eta}\Lambda^{\nu}_{\ \
\xi}g_{\mu\nu}=g_{\eta\xi};$ (314)

and, the second equality may be proved in a similar way. Therefore, the
transformations $\Lambda^{\mu}_{\ \ \nu}$ constitute the (restricted) Lorentz
group and the space ${\mathscr{V}}$ is composed of four-component vectors.

The transformation $\Lambda$ and the matrix $g$ have the following properties.
(i) The inverse transformation of $\Lambda^{\mu}_{\ \ \nu}$, denoted by
$\Lambda^{-1}$ has the simple expression of

$(\Lambda^{-1})^{\nu}_{\ \ \mu}=\Lambda_{\mu}^{\ \
\nu}\Longleftrightarrow(\Lambda^{-1})_{\nu\mu}=\Lambda_{\mu\nu}.$ (315)

In fact, substituting Eq.(313) into the product $\Lambda^{\mu}_{\ \
\nu}\Lambda_{\lambda}^{\ \ \nu}$ and making use of Eqs.(303), (264), and
(230), it is straightforward to verify Eq.(315).

(ii) Equation (314) implies that the matrix $g^{\mu\nu}$ is invariant under
the transformation $\Lambda$, that is,

$\widetilde{g}^{\mu\nu}=g^{\mu\nu}.$ (316)

(iii) The product $K^{\mu}g_{\mu\nu}J^{\nu}=K_{\mu}J^{\mu}$ is a scalar under
the transformation $\Lambda$, i.e.,

$\widetilde{K}_{\mu}\widetilde{J}^{\mu}=K_{\mu}J^{\mu},$ (317)

which can be readily proved making use of Eq.(314).

(iv) Making use of Eq.(288), it is straightforward to show that the
transformation $\Lambda$ is real, namely,

$\Lambda^{\mu}_{\ \ \nu}=(\Lambda^{\mu}_{\ \ \nu})^{*}.$ (318)

Then, it is easy to check that $K^{*}_{\mu}J^{\mu}$ is also a scalar product.

##  Appendix C Helicity states of basic modes

In this appendix, we give detailed discussions on properties of helicity
states of basic modes. Specifically, in Sec.C.1 we derive angular momentum
operators, then, in Sec.C.2 discuss helicity states, and, finally, in Sec.C.3
discuss antimode-counterpart states and ACC-bras.

###  C.1 Angular momentum of basic modes

In this section, we derive explicit expressions for the spinor angular-
momentum operators of the $b$-mode and $\overline{b}$-mode, denoted by ${\bf
s}_{b}$ and ${\bf s}_{\overline{b}}$, respectively, which act on the two
spaces of ${\mathscr{W}}$ and $\overline{\mathscr{W}}$. This enables one to
prove the relation in Eq.(13).

The operator ${\bf s}_{b}$ is written as ${\bf
s}_{b}=(s_{b}^{1},s_{b}^{2},s_{b}^{3})\equiv(s_{b}^{23},s_{b}^{31},s_{b}^{12})$,
and similar for ${\bf s}_{\overline{b}}$. In the ket-bra form, their
components are written as

$\displaystyle s_{b}^{ij}$
$\displaystyle=|S_{A}\rangle(s_{b}^{ij})^{AB}\langle\langle S_{B}|,$ (319a)
$\displaystyle s_{\overline{b}}^{ij}$
$\displaystyle=|S_{A^{\prime}}\rangle(s_{\overline{b}}^{ij})^{A^{\prime}B^{\prime}}\langle\langle
S_{B^{\prime}}|.$ (319b)

To find expressions of $(s_{b}^{ij})^{AB}$ and
$(s_{\overline{b}}^{ij})^{A^{\prime}B^{\prime}}$, one may make use of
knowledge about the angular-momentum operator for Dirac spinors, denoted by
$\bf S$ with

${\bf S}=(S^{1},S^{2},S^{3})=(S^{23},S^{31},S^{12}).$ (320)

The above-discussed three operators have the following relationship,

$\displaystyle S^{ij}=\left(\begin{array}[]{cc}s_{b}^{ij}&0\\\
0&{s_{\overline{b}}}^{ij}\end{array}\right).$ (323)

As is well known, in the chiral representation of $\gamma^{\mu}$-matrices, the
component form of $\bf S$ satisfies
$[S^{\mu\nu}]=\frac{i}{4}[\gamma^{\mu},\gamma^{\nu}]$. With the upper part of
Dirac spinor associated with the space ${\mathscr{W}}$ and the lower part with
$\overline{\mathscr{W}}$, the $\gamma^{\mu}$-matrices have the following
expression [20, 23, 21, 22, 24],

$\displaystyle\gamma^{\mu}=\sqrt{2}\left(\begin{array}[]{cc}0&\sigma^{\mu
AB^{\prime}}\\\ \overline{\sigma}^{\mu}_{A^{\prime}B}&0\end{array}\right).$
(326)

Then, it is straightforward to find that the matrix $[S^{\mu\nu}]$ is written
as

$\displaystyle[S^{\mu\nu}]=\frac{i}{2}\left(\begin{array}[]{cc}\sigma^{\mu\nu,A}_{\
\ \ \ \ B}&0\\\ 0&-\overline{\sigma}^{\mu\nu,\ \ B^{\prime}}_{\ \ \
A^{\prime}}\end{array}\right),$ (329)

where

$\displaystyle\sigma^{\mu\nu,AC}:=\sigma^{\mu
AB^{\prime}}\overline{\sigma}^{\nu\ C}_{B^{\prime}}-\sigma^{\nu
AB^{\prime}}\overline{\sigma}^{\mu\ C}_{B^{\prime}}.$ (330)

To translate the component expression in Eq.(329) to an operator form and get
expressions for the operators in Eq.(323), let us consider an arbitrary Weyl
spinor $|u\rangle\in{\mathscr{W}}$ and use $|\kappa\rangle$ to indicate the
action of $s_{b}^{ij}$ on it, i.e.,

$\displaystyle|\kappa\rangle=s_{b}^{ij}|u\rangle.$ (331)

Substituting Eq.(319a) into Eq.(331), then, multiplying the result from left
by $\langle S^{A}|$ and making use of Eqs.(230)-(231) and (235), one gets that

$\displaystyle\kappa^{A}$ $\displaystyle=\langle\langle
S^{A}|\left(|S_{C}\rangle(s_{b}^{ij})^{CB}\langle
S_{B}|\right)|u\rangle=\epsilon^{A}_{\ C}(s_{b}^{ij})^{CB}u_{B}$
$\displaystyle=-(s_{b}^{ij})^{AB}u_{B}=(s_{b}^{ij})^{A}_{\ B}u^{B}.$ (332)

Clearly, applying the abstract Dirac operator in Eq.(323) to an abstract
spinor, one should get a result equivalent to that obtained by applying the
matrix form in Eq.(329) to spinor components. With Eq.(332), this gives that

$\displaystyle(s_{b}^{ij})^{AB}=\frac{i}{2}\sigma^{ij,AB},$ (333)

where positions of the index $B$ have been raised on both sides of the
equation.

The operator $s_{\overline{b}}^{ij}$ may be gotten by a similar method. In
fact, similar to Eq.(332), one gets that

$\displaystyle\overline{\kappa}_{A^{\prime}}$
$\displaystyle=(s_{\overline{b}}^{ij})_{A^{\prime}B^{\prime}}u^{B^{\prime}}=-(s_{\overline{b}}^{ij})_{A^{\prime}}^{\
\ B^{\prime}}u_{B^{\prime}},$ (334)

where positions of indices have been arranged for an easy comparison with
Eq.(329). Then, one sees that

$\displaystyle(s_{\overline{b}}^{ij})^{A^{\prime}B^{\prime}}=\frac{i}{2}\overline{\sigma}^{ij,A^{\prime}B^{\prime}}.$
(335)

From Eqs.(333) and (335), one directly gets that

$\displaystyle(s_{\overline{b}}^{ij})^{A^{\prime}B^{\prime}}=-\overline{({s}_{b}^{ij})^{AB}}.$
(336)

This is just the component form of Eq.(13).

###  C.2 Helicity states

In this section, we derive properties and expressions of helicity states of
basic modes, particularly, those given in Eqs.(16)-(17).

We recall that helicity states of a $b$-mode are written as $|w^{r}({\bf
p})\rangle$, satisfying $H^{\rm cy}_{b}|w^{r}({\bf
p})\rangle=h^{r}_{b}|w^{r}({\bf p})\rangle$, with $H^{\rm cy}_{b}={\bf
s}_{b}\cdot{\bf p}/|{\bf p}|$. To compute $|w^{r}({\bf p})\rangle$, let us
consider the action of angular momentum on a generic spinor $|\chi\rangle$.
Making use of Eqs.(319a), (230), and (235), one finds that

$\displaystyle
s_{b}^{k}|\chi\rangle=(s_{b}^{k})_{AB}|S^{A}\rangle\langle\langle
S^{B}|\chi_{C}|S^{C}\rangle$
$\displaystyle=(s_{b}^{k})_{AB}\chi_{C}\epsilon^{BC}|S^{A}\rangle=-(s_{b}^{k})_{A}^{\
\ C}\chi_{C}|S^{A}\rangle,$

thus,

$\displaystyle s_{b}^{k}|\chi\rangle=-(s_{b}^{k})_{A}^{\ \
B}\chi_{B}|S^{A}\rangle.$ (337)

Then, writing $|\kappa\rangle=s_{b}^{k}|\chi\rangle$ and expanding it as
$|\kappa\rangle=\kappa_{A}|S^{A}\rangle$, one finds that

$\displaystyle\kappa_{A}=-(s_{b}^{k})_{A}^{\ \ B}\chi_{B},$ (338a)
$\displaystyle\kappa^{A}=(s_{b}^{k})^{A}_{\ \ B}\chi^{B}.$ (338b)

We employ the explicit expressions of EW-symbols in Eqs.(248)-(258) to compute
explicit expressions of angular-momentum operators. From Eq.(333), one finds
the following well-known matrix forms of $s_{b}^{k}$, given by Pauli matrices,

$\displaystyle(s_{b}^{1})^{A}_{\ \
B}=\frac{1}{2}\left(\begin{array}[]{cc}0&1\\\ 1&0\\\ \end{array}\right),$
(339c) $\displaystyle(s_{b}^{2})^{A}_{\ \
B}=\frac{1}{2}\left(\begin{array}[]{cc}0&-i\\\ i&0\\\ \end{array}\right),$
(339f) $\displaystyle(s_{b}^{3})^{A}_{\ \
B}=\frac{1}{2}\left(\begin{array}[]{cc}1&0\\\ 0&-1\\\ \end{array}\right).$
(339i)

For example, for the third component, Eq.(333) gives that $(s_{b}^{3})^{A}_{\
C}=\frac{i}{2}\sigma^{12,A}_{\ \ \ \ \ C}$ and, then, one gets Eq.(339i) by
the following derivation for $\sigma^{12,A}_{\ \ \ \ \ C}$,

$\displaystyle\sigma^{12,A}_{\ \ \ \ \
C}=\sigma^{1AB^{\prime}}\overline{\sigma}^{2}_{B^{\prime}C}-\sigma^{2AB^{\prime}}\overline{\sigma}^{1}_{B^{\prime}C}$
$\displaystyle=\frac{1}{2}\left(\begin{array}[]{cc}0&1\\\ 1&0\\\
\end{array}\right)\left(\begin{array}[]{cc}0&i\\\ -i&0\\\
\end{array}\right)-\frac{1}{2}\left(\begin{array}[]{cc}0&-i\\\ i&0\\\
\end{array}\right)\left(\begin{array}[]{cc}0&-1\\\ -1&0\\\ \end{array}\right)$
$\displaystyle=\frac{1}{2}\left(\begin{array}[]{cc}-i&0\\\ 0&i\\\
\end{array}\right)-\frac{1}{2}\left(\begin{array}[]{cc}i&0\\\ 0&-i\\\
\end{array}\right)=\left(\begin{array}[]{cc}-i&0\\\ 0&i\\\
\end{array}\right).$

Let us consider a momentum ${\bf p}_{0}=(0,0,|{\bf p}_{0}|)$ along the
$z$-direction. Making use of Eqs.(319a) and (339i), one finds that

$\displaystyle H^{\rm cy}_{b}=s_{b}^{3}=-|S_{A}\rangle(s_{b}^{ij})^{A}_{\ \
B}\langle\langle S^{B}|=|S_{1}\rangle\langle\langle
S^{1}|-|S_{0}\rangle\langle\langle S^{0}|.$ (340)

Using Eq.(231), one then gets that $s_{b}^{3}|S_{0}\rangle=|S_{0}\rangle$ and
$s_{b}^{3}|S_{1}\rangle=-|S_{1}\rangle$. Thus, under the well-known spin-
angular-momentum expression in Eq.(339), the helicity states of $|w^{r}({\bf
p}_{0})\rangle$ of $b$-mode are in fact given by the basis spinors
$|S_{A}\rangle$.

More exactly, one may set components of $|w^{r}({\bf p}_{0})\rangle$ on the
basis of $|S_{A}\rangle$ as

$\displaystyle[w^{rA}({\bf
p}_{0})]=\left\\{\begin{array}[]{ll}\left(\begin{array}[]{c}1\\\ 0\\\
\end{array}\right)&\hbox{for $r=0$},\\\ \left(\begin{array}[]{c}0\\\ 1\\\
\end{array}\right)&\hbox{for $r=1$.}\end{array}\right.$ (347)

Correspondingly, the helicities are written as

$\displaystyle h^{r}_{b}=\frac{(-1)^{r}}{2}\quad(r=0,1).$ (348)

Then, noting that $|w^{r}({\bf p}_{0})\rangle=-w^{rA}({\bf
p}_{0})|S_{A}\rangle$, one gets

$\displaystyle|w^{r}({\bf
p}_{0})\rangle=\left\\{\begin{array}[]{ll}-|S_{0}\rangle&\hbox{for $r=0$},\\\
-|S_{1}\rangle&\hbox{for $r=1$}.\end{array}\right.$ (351)

One directly gets Eq.(17) from Eq.(351).

Further, the helicity states $|w^{r}({\bf p})\rangle$ can be obtained by a
Lorentz transformation that brings ${\bf p}_{0}$ to ${\bf p}$. Clearly, the
helicity is equal to $\frac{1}{2}$ for the state $|w^{0}({\bf p})\rangle$,
while, equal to $-\frac{1}{2}$ for $|w^{1}({\bf p})\rangle$.

For an $\overline{b}$-mode, from Eq.(13), one sees that its helicity states
are given by the complex conjugates of $|w^{r}({\bf p})\rangle$, i.e., by
$|\overline{w}^{r}({\bf p})\rangle$, with opposite eigenvalues. That is,

$\displaystyle H^{\rm cy}_{\overline{b}}|\overline{w}^{r}({\bf
p})\rangle=h^{r}_{\overline{b}}|\overline{w}^{r}({\bf p})\rangle,$ (352)
$\displaystyle
h^{r}_{\overline{b}}=\frac{(-1)^{r+1}}{2}=\left\\{\begin{array}[]{ll}-\frac{1}{2},&\hbox{for
$|\overline{w}^{0}({\bf p})\rangle$,}\\\ \frac{1}{2},&\hbox{for
$|\overline{w}^{1}({\bf p})\rangle$.}\end{array}\right.$ (355)

###  C.3 Antimode-counterpart states and ACC-bras

In this section, we give detailed discussions on properties of antimode-
counterpart states, as well as on ACC-bras and ACC-scalar products for spinor
states of basic modes. In particular, we derive Eqs.(21) and (22)-(24).

Making use of Eq.(20), which gives states $|{w^{r}_{\varrho}}({\bf
p})\rangle_{\rm ac}$ of $\varrho=+$, one may compute $|{w_{r\varrho}}({\bf
p})\rangle_{\rm ac}$ of $\varrho=+$. To this end, we note that the relation of
$|w_{r\varrho}({\bf p})\rangle=|w^{s}_{\varrho}({\bf p})\rangle\epsilon_{sr}$
implies that $|w_{r\varrho}({\bf p})\rangle_{\rm ac}=|w^{s}_{\varrho}({\bf
p})\rangle_{\rm ac}\epsilon_{sr}$. Substituting Eq.(20a) into the rhs of this
result and making use of the relation of $|\overline{w}_{s\varrho}({\bf
p})\rangle=|\overline{w}^{t}_{\varrho}({\bf p})\rangle\epsilon_{ts}$, one gets
that

$\displaystyle|w_{r\varrho}({\bf p})\rangle_{\rm
ac}=\sum_{s}|\overline{w}^{t}_{\varrho}({\bf
p})\rangle\epsilon_{ts}\epsilon_{sr}.$ (356)

Further, we note that

$\displaystyle\sum_{s}\epsilon_{ts}\epsilon_{sr}=-\delta_{tr},$ (357)

which can be easily checked by direct computation. Then, one gets that
$|w_{r\varrho}({\bf p})\rangle_{\rm ac}=-|\overline{w}^{r}_{\varrho}({\bf
p})\rangle$. Similarly, one computes $|w^{r}_{\varrho}({\bf p})\rangle_{\rm
ac}$ of $\varrho=-$ and gets that

$\displaystyle|w^{r}_{\varrho}({\bf p})\rangle_{\rm
ac}=\epsilon^{rs}|w_{s\varrho}({\bf p})\rangle_{\rm
ac}=\epsilon^{rs}|\overline{w}^{s}_{\varrho}({\bf
p})\rangle=\epsilon^{rs}\epsilon^{st}|\overline{w}_{t\varrho}({\bf
p})\rangle.$

To summarize, one gets the following relations,

$\displaystyle|w_{r\varrho}({\bf p})\rangle_{\rm
ac}=-|\overline{w}^{r}_{\varrho}({\bf p})\rangle\qquad\text{$(\varrho=+)$, }$
(358a) $\displaystyle|w^{r}_{\varrho}({\bf p})\rangle_{\rm
ac}=-|\overline{w}_{r\varrho}({\bf p})\rangle\qquad\text{$(\varrho=-)$. }$
(358b)

which are just those in Eq.(21).

Then, making use of Eqs.(20) and (21), one finds that

$\displaystyle|{w^{r}_{\varrho}}({\bf p})\rangle_{\rm ACC}=|w_{r\varrho}({\bf
p})\rangle\qquad(\varrho=+),$ (359a) $\displaystyle|{w_{r\varrho}}({\bf
p})\rangle_{\rm ACC}=-|w^{r}_{\varrho}({\bf p})\rangle\qquad(\varrho=+),$
(359b)

and

$\displaystyle|{w^{r}_{\varrho}}({\bf p})\rangle_{\rm ACC}=-|w_{r\varrho}({\bf
p})\rangle\qquad(\varrho=-),$ (360a) $\displaystyle|{w_{r\varrho}}({\bf
p})\rangle_{\rm ACC}=|w^{r}_{\varrho}({\bf p})\rangle\qquad(\varrho=-).$
(360b)

Clearly, Eq.(22) is just a concise form of the above equations. Further,
making use of Eqs.(17), (231), and (359)-(360), it is direct to compute the
ACC-scalar products, getting that

$\displaystyle\langle\widehat{w^{r}_{\varrho}}({\bf p})|{w^{s}_{\varrho}}({\bf
p})\rangle=\delta^{rs}\qquad(\varrho=+),$ (361a)
$\displaystyle\langle\widehat{w_{r\varrho}}({\bf p})|{w_{s\varrho}}({\bf
p})\rangle=\delta_{rs}\qquad(\varrho=+),$ (361b)

and

$\displaystyle\langle\widehat{w^{r}_{\varrho}}({\bf p})|{w^{s}_{\varrho}}({\bf
p})\rangle=-\delta^{rs}\qquad(\varrho=-),$ (362a)
$\displaystyle\langle\widehat{w_{r\varrho}}({\bf p})|{w_{s\varrho}}({\bf
p})\rangle=-\delta_{rs}\qquad(\varrho=-),$ (362b)

as summarized in Eq.(24).

Finally, we discuss commutability of taking antimode-counterpart state and
taking complex conjugation within an ACC operation. For example, for a state
$|{w^{r}_{\varrho}}({\bf p})\rangle$ with $\varrho=+$, reversing the two
suboperations, one first gets $|\overline{w^{r}_{\varrho}}({\bf p})\rangle$,
then, gets $|\overline{w^{r}_{\varrho}}({\bf p})\rangle_{\rm
ac}=|w_{r\varrho}({\bf p})\rangle$ according to Eq.(20a); this is just what is
given to $|{w^{r}_{\varrho}}({\bf p})\rangle_{\rm ACC}$ in Eq.(22a).
Meanwhile, for a state $|{w^{r}_{\varrho}}({\bf p})\rangle$ with $\varrho=-$,
reversing the two suboperations, one first gets
$|\overline{w^{r}_{\varrho}}({\bf p})\rangle$, then, gets
$|\overline{w^{r}_{\varrho}}({\bf p})\rangle_{\rm ac}=-|w_{r\varrho}({\bf
p})\rangle$, where Eq.(21b) has been used; this result is just what is given
to $|{w^{r}_{\varrho}}({\bf p})\rangle_{\rm ACC}$ in Eq.(22b).

##  Appendix D ACC operation and ACC-scalar product of generic spinors

In this appendix, we give detailed discussions on some properties of ACC-
scalar products of generic spinors (Appendix D.1), and on some relations about
ACC operations (Appendix D.2).

###  D.1 Properties of ACC-scalar products

Firstly, we show that, for two arbitrary single-mode states $|\psi\rangle$ and
$|\phi\rangle$, one has

$\displaystyle\overline{\langle\widehat{\psi}|\phi\rangle}\equiv\left(\langle\widehat{\psi}|\phi\rangle\right)^{*}=\langle\widehat{\phi}|\psi\rangle.$
(363)

To this end, let us write

$\displaystyle|\psi\rangle=\int d\widetilde{p}C_{\alpha}({\bf
p})|M^{\alpha}_{{\bf p}}\rangle,$ (364a) $\displaystyle|\phi\rangle=\int
d\widetilde{p}D_{\beta}({\bf p})|M^{\beta}_{{\bf p}}\rangle,$ (364b)

where $C_{\alpha}({\bf p})$ and $D_{\alpha}({\bf p})$ are expanding
coefficients. The ACC-bra of $|\phi\rangle$ is written as

$\displaystyle\langle\widehat{\phi}|=\int d\widetilde{p}D_{\beta}^{*}({\bf
p})\langle\widehat{M}^{\beta}_{\bf p}|.$ (365)

It is then easy to get that

$\displaystyle\langle\widehat{\phi}|\psi\rangle=\int
d\widetilde{p}d\widetilde{q}D^{*}_{\beta}({\bf q})C_{\alpha}({\bf
p})\langle\widehat{M}^{\beta}_{{\bf q}}|M^{\alpha}_{{\bf p}}\rangle$
$\displaystyle=\int d\widetilde{p}d\widetilde{q}D^{*}_{\beta}({\bf
q})C_{\alpha}({\bf p})|{\bf p}|\delta^{3}({\bf p}-{\bf
q})\Upsilon^{\beta\alpha}.$ (366)

Similarly, one has

$\displaystyle\langle\widehat{\psi}|\phi\rangle=\int
d\widetilde{p}d\widetilde{q}D_{\beta}({\bf q})C^{*}_{\alpha}({\bf p})|{\bf
p}|\delta^{3}({\bf p}-{\bf q})\Upsilon^{\alpha\beta}.$

Then, one gets Eq.(363).

Secondly, we show that the ACC-scalar product is an inner product for a
generic fermionic mode $F$ with $\varrho=+$, while, it is a negative inner
product for $F$ with $\varrho=-$. Indeed, for $|\psi\rangle$ in Eq.(364a) with
$M=F$, noting that
$\Upsilon^{\alpha\beta}=\varrho(\alpha)\delta^{\alpha\beta}$, from Eq.(366)
with $|\psi\rangle=|\phi\rangle$ one gets that

$\displaystyle\langle\widehat{\psi}|\psi\rangle$
$\displaystyle=\varrho(\alpha)\int
d\widetilde{p}\sum_{r(\alpha)}|C_{\alpha}({\bf p})|^{2}.$ (367)

Finally, we discuss the ACC-scalar product $\langle\widehat{\Phi}|\Psi\rangle$
of two arbitrary vectors of $|\Psi\rangle$ and $|\Phi\rangle$. If the product
is a $c$-number, then, it is straightforward to check that the relation in
Eq.(363) still holds for $\langle\widehat{\Phi}|\Psi\rangle$, that is,

$\displaystyle\langle\widehat{\Phi}|\Psi\rangle^{*}=\langle\widehat{\Psi}|\Phi\rangle\quad\text{for
$\langle\widehat{\Phi}|\Psi\rangle$ as a $c$-number}.$ (368)

Otherwise, one may divide both $\langle\widehat{\Phi}|$ and $|\Psi\rangle$
into two parts, written as
$\langle\widehat{\Phi}|=\langle\widehat{\Phi}_{2}|\langle\widehat{\Phi}_{1}|$
and $|\Psi\rangle=|\Psi_{1}\rangle|\Psi_{2}\rangle$, such that
$\langle\widehat{\Phi}_{1}|$ forms an ACC-scalar product with
$|\Psi_{1}\rangle$, meanwhile, $\langle\widehat{\Phi}_{2}|$ and
$|\Psi_{2}\rangle$ remain purely vectors, i.e.,

$\displaystyle\langle\widehat{\Phi}|\Psi\rangle=\langle\widehat{\Phi}_{2}|\left(\langle\widehat{\Phi}_{1}|\Psi_{1}\rangle\right)|\Psi_{2}\rangle.$
(369)

In this case, making use of Eq.(368), one gets that

$\displaystyle\langle\widehat{\Phi}|\Psi\rangle^{*}=\overline{\langle\widehat{\Phi}_{2}|}\left(\langle\widehat{\Psi}|\Phi_{1}\rangle\right)\overline{|\Psi_{2}\rangle}.$
(370)

###  D.2 Some relations for ACC operations

In this section, we derive some relations for ACC operations, particularly,
those for double ACC operations. For brevity, for ket, we also use a hat to
indicate the result of an ACC operation, i.e.,

$\displaystyle|\widehat{\Psi}\rangle\equiv|\Psi\rangle_{\rm ACC}.$ (371)

By definition, for a state $|\Psi\rangle$, the ACC operation gives
$|\widehat{\Psi}\rangle$. Double ACC operations bring no change to momentum
states [see Eq.(27)]. It is natural to assume that
$|\widehat{\widehat{s}}\rangle=|s\rangle$. These two properties, together with
Eq.(44), imply that

$\displaystyle|\widehat{\widehat{B^{\lambda}_{\bf
k}}}\rangle=|B^{\lambda}_{\bf k}\rangle,$ (372)

i.e., double application of ACC operation does not change states of bosonic
modes.

The situation is a little more complex with spinor states of fermionic modes.
For such a state, making use of Eq.(50), one gets that

$\displaystyle|\widehat{\widehat{\mathcal{S}_{f}^{\alpha}}}({\bf
p})\rangle=-|{{\mathcal{S}}}_{f}^{\alpha}({\bf p})\rangle.$ (373)

i.e., double application of ACC operation generates a minus sign. Therefore,
for a generic vector $|\Psi\rangle$, one has

$\displaystyle|\widehat{\widehat{\Psi}}\rangle=(-1)^{N_{f}(\Psi)}|\Psi\rangle,$
(374)

where $N_{f}$ is the number of fermionic modes contained in $|\Psi\rangle$.

Below, we discuss ACC operation on a term of the form of
$\langle\widehat{\Phi}|\Psi\rangle$. Let us first consider single modes. In
this case, $\langle\widehat{\Phi}|\Psi\rangle$ is a $c$-number and, by
definition, the ACC operation gives its complex conjugate. For bosonic modes,
one gets that

$\displaystyle\widehat{\langle\widehat{B^{\alpha}_{{\bf p}}}|B^{\beta}_{{\bf
q}}\rangle}=\langle\widehat{B^{\alpha}_{{\bf p}}}|B^{\beta}_{{\bf
q}}\rangle^{*}=\langle\widehat{B^{\beta}_{{\bf q}}}|{B^{\alpha}_{{\bf
p}}}\rangle,$ (375)

where Eq.(363) has been used when deriving the last equality. Consistently,
one may directly compute with Eq.(372) and get

$\displaystyle\widehat{\langle\widehat{B^{\alpha}_{{\bf p}}}|B^{\beta}_{{\bf
q}}\rangle}=\langle{B^{\alpha}_{{\bf p}}}|\widehat{B^{\beta}_{{\bf
q}}}\rangle=\langle\widehat{B^{\beta}_{{\bf q}}}|{B^{\alpha}_{{\bf
p}}}\rangle,$ (376)

where Eq.(278) has been used when deriving the last equality.

For spinor states of fermionic modes, say, for
$\langle\widehat{\mathcal{S}_{f}^{\alpha}}|\mathcal{S}_{f}^{\beta}\rangle$,
one compute that

$\displaystyle\widehat{\langle\widehat{\mathcal{S}_{f}^{\alpha}}|\mathcal{S}_{f}^{\beta}\rangle}=\left(\langle\widehat{\mathcal{S}_{f}^{\alpha}}|\mathcal{S}_{f}^{\beta}\rangle\right)^{*}=\langle\widehat{\mathcal{S}_{f}^{\beta}}|{\mathcal{S}_{f}^{\alpha}}\rangle;$
(377)

and, consistently, with Eq.(373),

$\displaystyle\widehat{\langle\widehat{\mathcal{S}_{f}^{\alpha}}|\mathcal{S}_{f}^{\beta}\rangle}=\langle\widehat{\widehat{\mathcal{S}_{f}^{\alpha}}}|\widehat{\mathcal{S}_{f}^{\beta}}\rangle=-\langle\mathcal{S}_{f}^{\alpha}|\widehat{\mathcal{S}_{f}^{\beta}}\rangle=\langle\widehat{\mathcal{S}_{f}^{\beta}}|{\mathcal{S}_{f}^{\alpha}}\rangle,$
(378)

where the anti-symmetry of scalar product of Weyl spinors in Eq.(233) has been
used in the derivation of the last equality.

Next, it is straightforward to generalize the above arguments to a generic
$c$-number scalar product $\langle\widehat{\Phi}|\Psi\rangle$. As a
generalization of Eqs.(375) and (377), one gets that

$\displaystyle\widehat{\langle\widehat{\Phi}|\Psi\rangle}=\langle\widehat{\Psi}|\Phi\rangle\quad\text{for
$\langle\widehat{\Phi}|\Psi\rangle$ as a $c$-number}.$ (379)

Finally, we discuss $\langle\widehat{\Phi}|\Psi\rangle$ of two arbitrary
vectors, as written in Eq.(369). By definition, one writes

$\displaystyle\widehat{\langle\widehat{\Phi}|\Psi\rangle}=\langle\widehat{\widehat{\Phi}}_{2}|\left(\langle\widehat{\Phi}_{1}|\Psi_{1}\rangle\right)^{*}|\widehat{\Psi}_{2}\rangle.$
(380)

Then, making use of Eqs.(374) and (379), one gets that

$\displaystyle\widehat{\langle\widehat{\Phi}|\Psi\rangle}=(-1)^{N_{f}(\Phi_{2})}\langle\Phi_{2}|\left(\langle\widehat{\Psi}_{1}|\Phi_{1}\rangle\right)|\widehat{\Psi}_{2}\rangle.$
(381)

###  D.3 Adjoint operators for ACC-scalar products

In this appendix, we discuss adjoint operators, which are defined with respect
to the ACC-scalar product. In particular, we give detailed derivations for
Eqs.(153) and (154).

For this purpose, it proves convenient to use the notation of
$(|\Phi\rangle,|\Psi\rangle)$ for ACC-scalar product of two arbitrary states
$|\Psi\rangle$ and $|\Phi\rangle$, i.e.,

$\displaystyle(|\Phi\rangle,|\Psi\rangle)\equiv\langle\widehat{\Phi}|\Psi\rangle.$
(382)

The adjoint operator of an operator $A$, indicated as $A^{+}$, is defined by
the following relation,

$\displaystyle(|\Phi\rangle,A^{+}|\Psi\rangle)\equiv(A|\Phi\rangle,|\Psi\rangle).$
(383)

We first derive the relation in Eq.(153) for annihilation operators. Using
Eq.(382), one writes

$\displaystyle(|\Phi\rangle,b_{M}^{\alpha}({\bf
p})|\Psi\rangle)=\langle\widehat{\Phi}|b_{M}^{\alpha}({\bf p})|\Psi\rangle.$
(384)

Meanwhile, making use of the definition of $b_{M}^{\alpha}({\bf
p}):=\left[b_{M}^{\alpha{\dagger}}({\bf p})\right]^{+}$ and Eq.(151), one gets
that

$\displaystyle(|\Phi\rangle,b_{M}^{\alpha}({\bf p})|\Psi\rangle)$
$\displaystyle=(b_{M}^{\alpha{\dagger}}({\bf
p})|\Phi\rangle,|\Psi\rangle)=\left(\langle\widehat{\Phi}|\langle\widehat{M}^{\alpha}_{\bf
p}|\right)|\Psi\rangle.$ (385)

Then, one gets Eq.(153), i.e.,

$\displaystyle\langle\widehat{\Phi}|b_{M}^{\alpha}({\bf
p})|\Psi\rangle=\langle\widehat{\Phi}|\langle\widehat{M}^{\alpha}_{{\bf
p}}||\Psi\rangle.$ (386)

Next, we derive Eq.(154). To this end, let us consider a term
$(|\Phi\rangle,\left(b_{M}^{\alpha}({\bf p})\right)^{+}|\Psi\rangle)$. By
definition, $(|\Phi\rangle,\left(b_{M}^{\alpha}({\bf
p})\right)^{+}|\Psi\rangle)=(b_{M}^{\alpha}({\bf
p})|\Phi\rangle,|\Psi\rangle)$. This relation, together with the equivalence
of $b_{M}^{\alpha}({\bf p})$ and $\langle\widehat{M}^{\alpha}_{{\bf p}}|$ as
shown in Eq.(386), gives that

$\displaystyle(|\Phi\rangle,\left(b_{M}^{\alpha}({\bf
p})\right)^{+}|\Psi\rangle)=\left(\langle\widehat{M}^{\alpha}_{{\bf
p}}|\Phi\rangle,|\Psi\rangle\right).$ (387)

The rhs of Eq.(387) may be written in the form of $\langle\Omega|\Psi\rangle$,
where $\langle\Omega|$ is obtained from
$\widehat{\langle\widehat{M}^{\alpha}_{{\bf p}}|\Phi\rangle}$ by changing its
pure ket part to a bra, with the $c$-number part unchanged.

To get $\langle\Omega|$, we note that, when being used in the computation of
physical quantities, the final place where the term
$(|\Phi\rangle,\left(b_{M}^{\alpha}({\bf p})\right)^{+}|\Psi\rangle)$ appears
should be in the computation of some $c$-number quantity. We assume that the
term already has this $c$-number feature with respect to the mode $M$. This
requires that, for such a term not definitely vanishing, the state
$|\Phi\rangle$ should contain at least one state of the mode $M$. With this
property, making use of Eq.(381), one finds that
$\langle\Omega|=\langle\widehat{\Phi}|M^{\alpha}_{{\bf p}}\rangle$. This gives
that

$\displaystyle(|\Phi\rangle,\left(b_{M}^{\alpha}({\bf
p})\right)^{+}|\Psi\rangle)=\langle\widehat{\Phi}|b_{M}^{\alpha{\dagger}}({\bf
p})|\Psi\rangle,$ (388)

where the definition of creation operator in Eq.(151) has been used. This
justifies Eq.(154), i.e., $b_{M}^{\alpha{\dagger}}({\bf
p})=\left[b_{M}^{\alpha}({\bf p})\right]^{+}$.

Moreover, for an arbitrary operator $A$, it is easy to check that

$\displaystyle\left(A^{+}\right)^{+}=A.$ (389)

In fact,

$\displaystyle\langle\widehat{\Phi}|[(A^{+})^{+}|\Psi\rangle]=(|\Phi\rangle,(A^{+})^{+}|\Psi\rangle)$
$\displaystyle=(A^{+}|\Phi\rangle,|\Psi\rangle)=(|\Psi\rangle,A^{+}|\Phi\rangle)^{*}$
$\displaystyle=(A|\Psi\rangle,|\Phi\rangle)^{*}=(|\Phi\rangle,A|\Psi\rangle).$

Furthermore, for an operator $A=c|\psi\rangle\langle\widehat{\phi}|$, making
use of Eq.(381), one gets that

$\displaystyle(|\Phi\rangle,A^{+}|\Psi\rangle)=c^{*}\left(|\psi\rangle\langle\widehat{\phi}|\Phi_{1}\rangle|\Phi_{2}\rangle,|\Psi\rangle\right)$
$\displaystyle=c^{*}\langle\widehat{\Phi}|\left(|\phi\rangle\langle\widehat{\psi}|\right)|\Psi\rangle,$

which implies that

$\displaystyle A^{+}=c^{*}|\phi\rangle\langle\widehat{\psi}|.$ (390)

##  Appendix E Properties of z-mode and s-mode

In Appendix E.1, we discuss a method of computing the angular-momentum
operator ${\bf J}_{\mathscr{V}}$ of $z$-mode and, as an application, derive
Eq.(36) for its third component. In Appendix E.2, we show that the spinor
spaces of $s$-mode and $\overline{s}$-mode are one-dimensional scalar spaces.

###  E.1 Angular momentum of z-mode

Let us consider an operator $X^{k}$ of $k=1,2,3$, defined by

$\displaystyle X^{k}:=\sigma s_{b}^{k}I_{\overline{\mathscr{W}}}\sigma^{T},$
(391)

where $s_{b}^{k}$ are components of the angular-momentum operator ${\bf
s}_{b}$ of $b$-mode and $I_{\overline{\mathscr{W}}}$ is the identity operator
on the space $\overline{\mathscr{W}}$. Using the relation of
$\overline{\sigma}=\sigma$ [Eq.(287)], one finds that the complex conjugate of
$X^{k}$ is written as

$\displaystyle\overline{X}^{k}=\sigma\overline{s}_{b}^{k}I_{{\mathscr{W}}}\sigma^{T}.$
(392)

Noting that ${\bf s}_{\overline{b}}=-{\overline{\bf s}}_{b}$ [Eq.(13)], one
sees that components of ${\bf J}_{\mathscr{V}}$ in Eq.(31) are written in the
following form,

$\displaystyle J^{k}_{{\mathscr{V}}}=X^{k}-\overline{X}^{k}.$ (393)

Substituting Eq.(319a) and the identity operator
$I_{\overline{\mathscr{W}}}=|\overline{S}^{C^{\prime}}\rangle\langle\overline{S}_{C^{\prime}}|$
[Eq.(238)] into Eq.(391), one gets that

$\displaystyle X^{k}=\sigma(s_{b}^{k})^{AB}|S_{A}\rangle\langle\langle
S_{B}|(|\overline{S}^{C^{\prime}}\rangle\langle\langle\overline{S}_{C^{\prime}}|)\sigma^{T}.$
(394)

Then, substituting Eqs.(30) and (296) into Eq.(394), one finds that

$\displaystyle X^{k}=\sigma^{\mu EF^{\prime}}|T_{\mu}\rangle\langle\langle
S_{F^{\prime}E}|(s_{b}^{k})^{AB}|S_{A}\rangle|\overline{S}^{C^{\prime}}\rangle\langle\langle\overline{S}_{C^{\prime}}|$
$\displaystyle\times\langle\langle S_{B}|\sigma^{\nu
GH^{\prime}}|S_{GH^{\prime}}\rangle\langle\langle T_{\nu}|$
$\displaystyle=\sigma^{\mu EF^{\prime}}(s_{b}^{k})^{AB}\sigma^{\nu
GH^{\prime}}|T_{\mu}\rangle\langle\langle T_{\nu}|\langle\langle
S_{F^{\prime}E}||S_{A}\rangle|\overline{S}^{C^{\prime}}\rangle\langle\langle\overline{S}_{C^{\prime}}|$
$\displaystyle\times\langle\langle S_{B}||S_{GH^{\prime}}\rangle$
$\displaystyle=\sigma^{\mu EF^{\prime}}(s_{b}^{k})^{AB}\sigma^{\nu
GH^{\prime}}|T_{\mu}\rangle\langle\langle
T_{\nu}|\epsilon_{EA}\epsilon_{F^{\prime}}^{\ \
C^{\prime}}\epsilon_{BG}\epsilon_{C^{\prime}H^{\prime}}$
$\displaystyle=\sigma^{\mu EC^{\prime}}(s_{b}^{k})^{A}_{\ \ G}\sigma^{\nu
GH^{\prime}}|T_{\mu}\rangle\langle\langle
T_{\nu}|\epsilon_{EA}\epsilon_{C^{\prime}H^{\prime}}$
$\displaystyle=\sigma^{\mu}_{AH^{\prime}}(s_{b}^{k})^{A}_{\ \ G}\sigma^{\nu
GH^{\prime}}|T_{\mu}\rangle\langle\langle T_{\nu}|,$

where Eq.(231) has been used. Then, writing $X^{k}$ in the form of
$X^{k}\equiv(X^{k})^{\mu\nu}|T_{\mu}\rangle\langle\langle T_{\nu}|$, its
components are written as

$\displaystyle(X^{k})^{\mu\nu}=\sigma^{\mu}_{AH^{\prime}}(s_{b}^{k})^{A}_{\
G}\sigma^{\nu GH^{\prime}}.$ (395)

To go further, we make use of the explicit expression of $(s_{b}^{3})^{A}_{\
B}$ in Eq.(339i) and find that

$\displaystyle(X^{3})^{\mu\nu}=\frac{1}{2}\sigma^{\mu}_{0H^{\prime}}\sigma^{\nu
0H^{\prime}}-\frac{1}{2}\sigma^{\mu}_{1H^{\prime}}\sigma^{\nu 1H^{\prime}}.$
(396)

Making use of the explicit expressions of the EW-symbols in Eqs.(248) and
(258), direct computation shows that

$\displaystyle\sigma^{\mu}_{0H^{\prime}}\sigma^{\nu
0H^{\prime}}=\frac{1}{2}\left(\begin{array}[]{cccc}1&0&0&1\\\ 0&-1&i&0\\\
0&-i&-1&0\\\ -1&0&0&-1\end{array}\right),$ (401)
$\displaystyle\sigma^{\mu}_{1H^{\prime}}\sigma^{\nu
1H^{\prime}}=\frac{1}{2}\left(\begin{array}[]{cccc}1&0&0&-1\\\ 0&-1&-i&0\\\
0&i&-1&0\\\ 1&0&0&-1\end{array}\right).$ (406)

This gives that

$\displaystyle(X^{3})^{\mu\nu}=\frac{1}{2}\left(\begin{array}[]{cccc}0&0&0&1\\\
0&0&i&0\\\ 0&-i&0&0\\\ -1&0&0&0\end{array}\right).$ (411)

Finally, from Eqs.(393) and (411), one gets Eq.(36).

###  E.2 $s$-mode as a scalar mode

In this section, we show that the spinor space ${\mathscr{S}}$ for an $s$-mode
is a one-dimensional $SL(2,C)$-scalar space. Let us consider the product state
$|\kappa\rangle|\chi\rangle$, for two arbitrary spinors
$|\kappa\rangle=-\kappa^{A}|S_{A}\rangle$ and
$|\chi\rangle=-\chi^{A}|S_{A}\rangle$ in ${\mathscr{W}}$. From Eq.(29b), one
sees that

$\displaystyle|S^{0}\rangle|S^{0}\rangle=0,\ |S^{1}\rangle|S^{1}\rangle=0,$
(412a) $\displaystyle|S^{0}\rangle|S^{1}\rangle=-|S^{1}\rangle|S^{0}\rangle.$
(412b)

Then, simple derivation gives

$\displaystyle|\kappa\rangle|\chi\rangle=(\kappa^{0}\chi^{1}-\kappa^{1}\chi^{0})|S_{0}\rangle|S_{1}\rangle=\left|\begin{array}[]{cc}\kappa^{0}&\kappa^{1}\\\
\chi^{0}&\chi^{1}\end{array}\right||S_{01}\rangle.$ (415)

Similarly, the transformed-spinor product is written as
$|\widetilde{\kappa}\rangle|\widetilde{\chi}\rangle=(\widetilde{\kappa}^{0}\widetilde{\chi}^{1}-\widetilde{\kappa}^{1}\widetilde{\chi}^{0})|S_{01}\rangle$.

A generic $SL(2,C)$ transformation is written as $h^{A}_{\ \ B}$ [see
Eq.(301)],

$h^{A}_{\ \ B}=\left(\begin{array}[]{cc}a&b\\\
c&d\end{array}\right)\quad\text{with}\ ad-bc=1.$ (416)

It transforms the components $\kappa^{A}$ to

$\displaystyle\widetilde{\kappa}^{A}=h^{A}_{\ \
B}\kappa^{B}=\left(\begin{array}[]{c}a\kappa^{0}+b\kappa^{1}\\\
c\kappa^{0}+d\kappa^{1}\end{array}\right).$ (419)

It is ready to compute that

$\displaystyle\widetilde{\kappa}^{0}\widetilde{\chi}^{1}-\widetilde{\kappa}^{1}\widetilde{\chi}^{0}=(ad-
bc)\kappa^{0}\chi^{1}+(bc-ad)\kappa^{1}\chi^{0}$
$\displaystyle=\kappa^{0}\chi^{1}-\kappa^{1}\chi^{0}.$ (420)

Then, one sees that

$\displaystyle|\widetilde{\kappa}\rangle|\widetilde{\chi}\rangle=|\kappa\rangle|\chi\rangle.$
(421)

Hence, the product state $|\kappa\rangle|\chi\rangle$ is a scalar under
$SL(2,C)$ transformations. One gets the same result for the spinor space
$\overline{\mathscr{S}}$.

##  Appendix F Derivations for H-operators

In this appendix, we give detailed derivations for some expressions of
$H$-operators given in the main text.

###  F.1 H-operator of free process

(i) Derivation of Eq.(106):
Substituting Eqs.(96) and (105) and that ${\mathcal{C}}^{M\to M}_{\rm sp}=1$
into Eq.(100), and making use of Eq.(95), one gets that

$\displaystyle H_{M}^{0}=\left(\int d\widetilde{p}|M^{\alpha}_{{\bf
p}}\rangle\langle\widehat{M}_{{\bf p}\alpha}|\right)\times\left(\int
d\widetilde{p}^{\prime}|{\bf p}^{\prime}\rangle|{\bf p}^{\prime}|\langle{\bf
p}^{\prime}|\right)\times\left(\int
d\widetilde{p}^{\prime\prime}|M^{\alpha^{\prime\prime}}_{{\bf
p}^{\prime\prime}}\rangle\langle\widehat{M}_{{\bf
p}^{\prime\prime}\alpha^{\prime\prime}}|\right)$ $\displaystyle=\int
d\widetilde{p}d\widetilde{p}^{\prime}d\widetilde{p}^{\prime\prime}|M^{\alpha}_{{\bf
p}}\rangle\delta_{\alpha}^{\alpha^{\prime\prime}}|{\bf p}|\delta^{3}({\bf
p}-{\bf p}^{\prime})|{\bf p}^{\prime}||{\bf p}^{\prime\prime}|\delta^{3}({\bf
p}^{\prime\prime}-{\bf p}^{\prime})\langle\widehat{M}_{{\bf
p}^{\prime\prime}\alpha^{\prime\prime}}|=\int d\widetilde{p}|M^{\alpha}_{{\bf
p}}\rangle|{\bf p}|\langle\widehat{M}_{{\bf p}\alpha}|.$

(ii) Derivation of Eq.(108):
Firstly, we compute that

$\displaystyle\langle\widehat{M}^{\beta}_{\bf q}|H^{0}_{M}|M^{\alpha}_{\bf
p}\rangle=\langle\widehat{M}^{\beta}_{\bf q}|\int d\widetilde{p}^{\prime}\
|M^{\alpha^{\prime}}_{{\bf p}^{\prime}}\rangle|{\bf
p}^{\prime}|\langle\widehat{M}_{{\bf
p}^{\prime}\alpha^{\prime}}||M^{\alpha}_{\bf p}\rangle$
$\displaystyle=\langle\widehat{M}^{\beta}_{\bf q}|\int
d\widetilde{p}^{\prime}|M^{\alpha^{\prime}}_{{\bf p}^{\prime}}\rangle|{\bf
p}^{\prime}||{\bf p}|\delta^{3}({\bf p}-{\bf
p}^{\prime})\delta_{\alpha^{\prime}}^{\alpha}=\Upsilon^{\beta\alpha}|{\bf
q}|\delta^{3}({\bf q}-{\bf p})|{\bf p}|.$

Then, substituting this result into Eq.(107), one gets Eq.(108), i.e.,

$\displaystyle p^{0}=\int d\widetilde{q}\Upsilon(\alpha,\alpha)|{\bf
q}|\delta^{3}({\bf q}-{\bf p})|{\bf p}|=\Upsilon(\alpha,\alpha)|{\bf p}|.$
(422)

###  F.2 Dirac-spinor expression of FIP amplitude

In this section, we derive the FIP amplitude given in Eq.(136), which are
written in terms of Dirac spinors. To this end, one may insert the expression
of the operator $\sigma$ in Eq.(30) into Eq.(134). Making use of Eq.(326) for
$\gamma^{\mu}$, one gets that

$\displaystyle C^{f\overline{f}^{\prime}\to
B}_{\alpha\alpha^{\prime}\lambda}=\frac{1}{n_{f}\sqrt{n_{V}}}\langle\langle\overline{\varepsilon}_{\lambda}({\bf
k})|\sigma^{\mu AB^{\prime}}|T_{\mu}\rangle\langle
S_{B^{\prime}A}|\Big{(}c|u^{f}_{\alpha}({\bf
p})\rangle|\overline{v}^{\overline{f}^{\prime}}_{{\alpha^{\prime}}}({\bf
p}^{\prime})\rangle+d|\overline{v}^{f}_{\alpha}({\bf
p})\rangle|u^{\overline{f}^{\prime}}_{\alpha^{\prime}}({\bf
p}^{\prime})\rangle\Big{)}$
$\displaystyle=\frac{1}{n_{f}\sqrt{n_{V}}}\sigma^{\mu
AB^{\prime}}{\overline{\varepsilon}}_{\lambda\mu}({\bf
k})\Big{(}cu^{f}_{\alpha A}({\bf
p})\overline{v}^{\overline{f}^{\prime}}_{\alpha^{\prime}B^{\prime}}({\bf
p}^{\prime})-d\overline{v}^{f}_{\alpha B^{\prime}}({\bf
p})u^{\overline{f}^{\prime}}_{\alpha^{\prime}A}({\bf p}^{\prime})\Big{)}.$
(423)

This amplitude may be expressed in terms of the Dirac spinors of
$U^{\alpha}({\bf p})$ and $V^{\alpha}({\bf p})$ defined in Eq.(141). Indeed,
it is direct to check that

$\displaystyle C^{f\overline{f}^{\prime}\to
B}_{\alpha{\alpha^{\prime}}\lambda}=\frac{1}{n_{f}\sqrt{n_{V}}}{\overline{\varepsilon}}_{\lambda\mu}({\bf
k})\left(\begin{array}[]{cc}\overline{v}^{\overline{f}^{\prime}}_{\alpha^{\prime}B^{\prime}}({\bf
p}^{\prime}),-u^{\overline{f}^{\prime}}_{\alpha^{\prime}A}({\bf
p}^{\prime})\end{array}\right)\left(\begin{array}[]{cc}c\sigma^{\mu
AB^{\prime}}&0\\\ 0&d\sigma^{\mu
AB^{\prime}}\end{array}\right)\left(\begin{array}[]{cc}u^{f}_{\alpha A}({\bf
p})\\\ \overline{v}^{f}_{\alpha B^{\prime}}({\bf p})\end{array}\right)$ (429)
$\displaystyle=\frac{1}{n_{f}\sqrt{n_{V}}}{\overline{\varepsilon}}_{\lambda\mu}({\bf
k})\left(\begin{array}[]{cc}\overline{v}^{\overline{f}^{\prime}B^{\prime}}_{\alpha^{\prime}}({\bf
p}^{\prime}),-u^{\overline{f}^{\prime}}_{\alpha^{\prime}A}({\bf
p}^{\prime})\end{array}\right)\left(\begin{array}[]{cc}c\overline{\sigma}^{\mu
B^{\prime}A}&0\\\ 0&d\sigma^{\mu
AB^{\prime}}\end{array}\right)\left(\begin{array}[]{cc}u^{fA}_{\alpha}({\bf
p})\\\ \overline{v}^{f}_{\alpha B^{\prime}}({\bf p})\end{array}\right),$ (435)

where Eqs.(230) and (288) have been used. The mode-species labels in the Weyl
spinors used above, e.g., the superscript $f$ in $u^{fA}_{\alpha}({\bf p})$,
indicate where the Weyl spinors come from. But, properties of these Weyl
spinors are in fact independent of the mode species. Hence, it is unnecessary
to use these mode-species labels in the final expression of the interaction
amplitude, as already done in Eq.(136). Finally, noting the explicit
expression of the $\gamma^{\mu}$ matrices in Eq.(326) and the following
relation,

$\displaystyle\left(\begin{array}[]{cc}c\overline{\sigma}^{\mu}_{B^{\prime}A}&0\\\
0&d\sigma^{\mu AB^{\prime}}\end{array}\right)=\left(\begin{array}[]{cc}0&c\\\
d&0\end{array}\right)\left(\begin{array}[]{cc}0&\sigma^{\mu AB^{\prime}}\\\
\overline{\sigma}^{\mu}_{B^{\prime}A}&0\end{array}\right)=\frac{1}{\sqrt{2}}\Gamma^{f\overline{f}^{\prime}\to
B}\gamma^{\mu},$ (442)

one is ready to get Eq.(136).

###  F.3 Derivations of H-operators of BPIPs

In this section, we give detailed derivations for expressions of $H$-operator
of BPIPs given in the main text, particularly, for Eqs.(156), (157), and
(159).

(i) Derivation of Eq.(156).
(a) For $H^{ff^{\prime}\to B}_{{\rm BPIP},1}$ in Eq.(156a), one has
$\varrho(\alpha)=\varrho(\alpha^{\prime})=+$. Hence, this operator is directly
obtained by replacing the ket and bras in Eq.(111) by creation operator and
annihilation operators, respectively.
(b) To compute $H^{ff^{\prime}\to B}_{{\rm BPIP},2}$ Eq.(156b), one
substitutes Eqs.(111) and (146a) into Eq.(148b). For $H^{ff^{\prime}\to
B}_{{\rm BPIP},2}$, $\varrho(\alpha)=+$ and $\varrho(\alpha^{\prime})=-$,
hence, the VF pair should be $(f^{\prime},\overline{f}^{\prime})$, with the
state of $\overline{f}^{\prime}$ appearing in a scalar product. Thus, we have

$\displaystyle H^{f\overline{f}^{\prime}\to B}_{{\rm BPIP},2}=\int
d\widetilde{q}|{f^{\prime}}_{\overline{\bf
q}\overline{\beta}}\rangle\left(H_{\rm FIP}^{f\overline{f}^{\prime}\to
B}\right)^{\curvearrowleft}|{\overline{f}^{\prime}}_{{\bf q}\beta}\rangle=\int
d\widetilde{q}|{f^{\prime}}_{\overline{\bf
q}\overline{\beta}}\rangle\left(\int d\Omega\ |B^{\lambda}_{\bf
k}\rangle\delta^{3}_{P}C^{f\overline{f}^{\prime}\to
B}_{\alpha\alpha^{\prime}\lambda}\langle\widehat{\overline{f}^{\prime}}_{{\bf
p}^{\prime}}^{\alpha^{\prime}}|\langle\widehat{f}_{{\bf
p}}^{\alpha}|\right)^{\curvearrowleft}|{\overline{f}^{\prime}}_{{\bf
q}\beta}\rangle$ $\displaystyle=\int d\Omega\ |B^{\lambda}_{\bf
k}\rangle\delta^{3}_{P}C^{f\overline{f}^{\prime}\to
B}_{\alpha\alpha^{\prime}\lambda}\int
d\widetilde{q}|{f^{\prime}}_{\overline{\bf
q}\overline{\beta}}\rangle\langle\widehat{\overline{f}^{\prime}}_{{\bf
p}^{\prime}}^{\alpha^{\prime}}|{\overline{f}^{\prime}}_{{\bf
q}\beta}\rangle\langle\widehat{f}_{{\bf p}}^{\alpha}|=\int d\Omega\
|B^{\lambda}_{\bf k}\rangle\delta^{3}_{P}C^{f\overline{f}^{\prime}\to
B}_{\alpha\alpha^{\prime}\lambda}|{f^{\prime}}_{\overline{\bf
p}^{\prime}\overline{\alpha}^{\prime}}\rangle\langle\widehat{f}_{{\bf
p}}^{\alpha}|.$ (443)

For $\varrho(\alpha^{\prime})=-$, one has
$\varrho(\overline{\alpha}^{\prime})=+$ and, as a result,
$|{f^{\prime}}_{\overline{\bf
p}^{\prime}\overline{\alpha}^{\prime}}\rangle=|{f^{\prime}}_{\overline{\bf
p}^{\prime}}^{\overline{\alpha}^{\prime}}\rangle$. Then, it is ready to get
Eq.(156b).
(c) Similarly, for $H^{f\overline{f}^{\prime}\to B}_{{\rm BPIP},3}$ with
$\varrho(\alpha)=-$ and $\varrho(\alpha^{\prime})=+$, one has

$\displaystyle H^{f\overline{f}^{\prime}\to B}_{{\rm BPIP},3}=\int
d\widetilde{q}|\overline{f}_{\overline{\bf
q}\overline{\beta}}\rangle\left(H_{\rm FIP}^{f\overline{f}^{\prime}\to
B}\right)^{\curvearrowleft}|f_{{\bf q}\beta}\rangle=\int
d\widetilde{q}|\overline{f}_{\overline{\bf
q}\overline{\beta}}\rangle\left(\int d\Omega\ |B^{\lambda}_{\bf
k}\rangle\delta^{3}_{P}C^{f\overline{f}^{\prime}\to
B}_{\alpha\alpha^{\prime}\lambda}\langle\widehat{\overline{f}^{\prime}}_{{\bf
p}^{\prime}}^{\alpha^{\prime}}|\langle\widehat{f}_{{\bf
p}}^{\alpha}|\right)^{\curvearrowleft}|f_{{\bf q}\beta}\rangle$
$\displaystyle=\int d\Omega\ |B^{\lambda}_{\bf
k}\rangle\delta^{3}_{P}C^{f\overline{f}^{\prime}\to
B}_{\alpha\alpha^{\prime}\lambda}\int
d\widetilde{q}|\overline{f}_{\overline{\bf
q}\overline{\beta}}\rangle\langle\widehat{\overline{f}^{\prime}}_{{\bf
p}^{\prime}}^{\alpha^{\prime}}|\langle\widehat{f}_{{\bf p}}^{\alpha}|f_{{\bf
q}\beta}\rangle=\int d\Omega\ |B^{\lambda}_{\bf
k}\rangle\delta^{3}_{P}C^{f\overline{f}^{\prime}\to
B}_{\alpha\alpha^{\prime}\lambda}|\overline{f}_{\overline{\bf
p}}^{\overline{\alpha}}\rangle\varrho(\overline{\alpha})\langle\widehat{\overline{f}^{\prime}}_{{\bf
p}^{\prime}}^{\alpha^{\prime}}|.$ (444)

This gives Eqs.(156c) with $\varrho(\overline{\alpha})=+$.
(d) For $H^{f\overline{f}^{\prime}\to B}_{{\rm BPIP},4}$ with
$\varrho(\alpha)=\varrho(\alpha^{\prime})=-$, one has

$\displaystyle H^{f\overline{f}^{\prime}\to B}_{{\rm BPIP},4}=\int
d\widetilde{q}d\widetilde{q}^{\prime}|\overline{f}_{\overline{\bf
q}\overline{\beta}}\rangle\ \Big{[}\ |{f^{\prime}}_{\overline{\bf
q}^{\prime}\overline{\beta}^{\prime}}\rangle\left(H_{\rm
FIP}^{f\overline{f}^{\prime}\to
B}\right)^{\curvearrowleft}|{\overline{f}^{\prime}}_{{\bf
q}^{\prime}\beta^{\prime}}\rangle\Big{]}^{\curvearrowleft}|f_{{\bf
q}\beta}\rangle$ $\displaystyle=\int
d\widetilde{q}d\widetilde{q}^{\prime}|\overline{f}_{\overline{\bf
q}\overline{\beta}}\rangle\ \Big{[}\ |{f^{\prime}}_{\overline{\bf
q}^{\prime}\overline{\beta}^{\prime}}\rangle\left(\int d\Omega\
|B^{\lambda}_{\bf k}\rangle\delta^{3}_{P}C^{f\overline{f}^{\prime}\to
B}_{\alpha\alpha^{\prime}\lambda}\langle\widehat{\overline{f}^{\prime}}_{{\bf
p}^{\prime}}^{\alpha^{\prime}}|\langle\widehat{f}_{{\bf
p}}^{\alpha}|\right)^{\curvearrowleft}|{\overline{f}^{\prime}}_{{\bf
q}^{\prime}\beta^{\prime}}\rangle\Big{]}^{\curvearrowleft}|f_{{\bf
q}\beta}\rangle$ $\displaystyle=\int d\Omega\ |B^{\lambda}_{\bf
k}\rangle\delta^{3}_{P}C^{f\overline{f}^{\prime}\to
B}_{\alpha\alpha^{\prime}\lambda}\int
d\widetilde{q}d\widetilde{q}^{\prime}|\overline{f}_{\overline{\bf
q}\overline{\beta}}\rangle\ |{f^{\prime}}_{\overline{\bf
q}^{\prime}\overline{\beta}^{\prime}}\rangle\langle\widehat{\overline{f}^{\prime}}_{{\bf
p}^{\prime}}^{\alpha^{\prime}}|{\overline{f}^{\prime}}_{{\bf
q}^{\prime}\beta^{\prime}}\rangle\langle\widehat{f}_{{\bf p}}^{\alpha}|f_{{\bf
q}\beta}\rangle$ $\displaystyle=\int d\Omega\ |B^{\lambda}_{\bf
k}\rangle\delta^{3}_{P}C^{f\overline{f}^{\prime}\to
B}_{\alpha\alpha^{\prime}\lambda}|\overline{f}_{\overline{\bf
p}}^{\overline{\alpha}}\rangle\ |{f^{\prime}}_{\overline{\bf
p}^{\prime}}^{\overline{\alpha}^{\prime}}\rangle\varrho(\overline{\alpha}^{\prime})\varrho(\overline{\alpha}).$
(445)

This gives Eq.(156d) with
$\varrho(\overline{\alpha}^{\prime})=\varrho(\overline{\alpha})=+$.

(ii) Derivation of Eq.(157).
(a) Equation (157a) can be obtained directly by substituting Eq.(136) into
Eq.(156a) with $(\varrho(\alpha),\varrho(\alpha^{\prime}))=(+,+)$. To be
consistent with expressions to be derived below for other $H$-operators, we
reindex $({\bf p}^{\prime},\alpha^{\prime})$ as $({\bf q},\beta)$.
(b) To prove Eq.(157b) from Eq.(156b) with
$(\varrho(\alpha),\varrho(\alpha^{\prime}))=(+,-)$, we note that Eq.(469)
implies that

$\displaystyle U^{\alpha}(\overline{\bf p})=\left(\begin{array}[]{c}u^{\alpha
A}(\overline{\bf p})\\\ \overline{v}_{B^{\prime}}^{\alpha}(\overline{\bf
p})\end{array}\right)=\left(\begin{array}[]{c}-v^{\alpha A}({\bf p})\\\
\overline{u}_{B^{\prime}}^{\alpha}({\bf p})\end{array}\right)=-V^{\alpha}({\bf
p}),\quad V^{\alpha}(\overline{\bf p})=\left(\begin{array}[]{c}v^{\alpha
B}(\overline{\bf p})\\\ -\overline{u}_{A^{\prime}}^{\alpha}(\overline{\bf
p})\end{array}\right)=\left(\begin{array}[]{c}u^{\alpha B}({\bf p})\\\
\overline{v}_{A^{\prime}}^{\alpha}({\bf p})\end{array}\right)=U^{\alpha}({\bf
p}),$

as a result,

$\displaystyle V^{\alpha}({\bf p})=-U^{\alpha}(\overline{\bf p}),\quad
U^{\alpha}({\bf p})=V^{\alpha}(\overline{\bf p}).$ (446)

Substituting Eq.(136) into Eq.(156b) and using the first equality in Eq.(446),
we get that

$\displaystyle H^{ff^{\prime}\to B}_{{\rm BPIP},2}=\int d\Omega
b^{\lambda{\dagger}}_{B}({\bf
k})\delta^{3}_{P}{\overline{\varepsilon}}_{\lambda\mu}({\bf
k})V_{\alpha^{\prime}}^{{\dagger}}({\bf
p}^{\prime})\Gamma^{f\overline{f}^{\prime}\to B}\gamma^{\mu}U_{\alpha}({\bf
p})b^{\overline{\alpha}^{\prime}{\dagger}}_{f^{\prime}}(\overline{\bf
p}^{\prime})b^{\alpha}_{f}({\bf p})$ $\displaystyle=-\int d\Omega
b^{\lambda{\dagger}}_{B}({\bf k})\delta^{3}({\bf k}-{\bf p}+\overline{\bf
p}^{\prime}){\overline{\varepsilon}}_{\lambda\mu}({\bf
k})U_{\alpha^{\prime}}^{{\dagger}}(\overline{\bf
p}^{\prime})\Gamma^{f\overline{f}^{\prime}\to B}\gamma^{\mu}U_{\alpha}({\bf
p})b^{\overline{\alpha}^{\prime}{\dagger}}_{f^{\prime}}(\overline{\bf
p}^{\prime})b^{\alpha}_{f}({\bf p}).$

Noting that $-U_{\alpha^{\prime}}^{{\dagger}}(\overline{\bf
p}^{\prime})=\varrho(\alpha^{\prime})U_{\alpha^{\prime}}^{{\dagger}}(\overline{\bf
p}^{\prime})$ and making use of Eqs.(90) and (92), we find that
$\varrho(\alpha^{\prime})U_{\alpha^{\prime}}^{{\dagger}}(\overline{\bf
p}^{\prime})=U^{\alpha^{\prime}{\dagger}}(\overline{\bf p}^{\prime})$.
Moreover, according to the definition in Eq.(49), an index $\alpha^{\prime}$
in the upper position is independent of the sign $\varrho$ in it; this gives
that $U^{\alpha^{\prime}{\dagger}}(\overline{\bf
p}^{\prime})=U^{\overline{\alpha}^{\prime}{\dagger}}(\overline{\bf
p}^{\prime})$. Furthermore, since $\varrho(\overline{\alpha}^{\prime})=+$, we
have $U^{\overline{\alpha}^{\prime}{\dagger}}(\overline{\bf
p}^{\prime})=\varrho(\overline{\alpha}^{\prime})U_{\overline{\alpha}^{\prime}}^{{\dagger}}(\overline{\bf
p}^{\prime})=U_{\overline{\alpha}^{\prime}}^{{\dagger}}(\overline{\bf
p}^{\prime})$. Making use of the above properties, we get that

$\displaystyle H^{ff^{\prime}\to B}_{{\rm BPIP},2}=\int
d\widetilde{p}d\widetilde{p^{\prime}}d\widetilde{k}b^{\lambda{\dagger}}_{B}({\bf
k})\delta^{3}({\bf k}-{\bf p}+\overline{\bf
p}^{\prime}){\overline{\varepsilon}}_{\lambda\mu}({\bf
k})U_{\overline{\alpha}^{\prime}}^{{\dagger}}(\overline{\bf
p}^{\prime})\Gamma^{f\overline{f}^{\prime}\to B}\gamma^{\mu}U_{\alpha}({\bf
p})b^{\overline{\alpha}^{\prime}{\dagger}}_{f^{\prime}}(\overline{\bf
p}^{\prime})b^{\alpha}_{f}({\bf p}).$ (447)

To simplify the notation, we introduce a index $\beta$ with
$\beta=\overline{\alpha}^{\prime}=(r^{\prime},+)$ and a momentum index ${\bf
q}$ with ${\bf q}=\overline{\bf p}^{\prime}$. Finally, changing the
integration over ${\bf p}^{\prime}$ to that over ${\bf q}$, we get Eq.(157b).
(c) Equation (157c) can be proved in a similar way from Eq.(156c) with
$(\varrho(\alpha),\varrho(\alpha^{\prime}))=(-,+)$. In fact, substituting
Eq.(136) into Eq.(156c) and using the second equality in Eq.(446), we get that

$\displaystyle H^{ff^{\prime}\to B}_{{\rm BPIP},3}=\int d\Omega
b^{\lambda{\dagger}}_{B}({\bf k})\delta^{3}({\bf k}-{\bf p}-{\bf
p}^{\prime}){\overline{\varepsilon}}_{\lambda\mu}({\bf
k})V_{\alpha^{\prime}}^{{\dagger}}({\bf
p}^{\prime})\Gamma^{f\overline{f}^{\prime}\to B}\gamma^{\mu}U_{\alpha}({\bf
p})b^{\overline{\alpha}{\dagger}}_{\overline{f}}(\overline{\bf
p})b^{\alpha^{\prime}}_{\overline{f}^{\prime}}({\bf p}^{\prime})$
$\displaystyle=\int d\Omega b^{\lambda{\dagger}}_{B}({\bf k})\delta^{3}({\bf
k}+\overline{\bf p}-{\bf
p}^{\prime}){\overline{\varepsilon}}_{\lambda\mu}({\bf
k})V_{\alpha^{\prime}}^{{\dagger}}({\bf
p}^{\prime})\Gamma^{f\overline{f}^{\prime}\to
B}\gamma^{\mu}V_{\alpha}(\overline{\bf
p})b^{\overline{\alpha}{\dagger}}_{\overline{f}}(\overline{\bf
p})b^{\alpha^{\prime}}_{\overline{f}^{\prime}}({\bf p}^{\prime})$
$\displaystyle=-\int d\Omega b^{\lambda{\dagger}}_{B}({\bf k})\delta^{3}({\bf
k}+\overline{\bf p}-{\bf
p}^{\prime}){\overline{\varepsilon}}_{\lambda\mu}({\bf
k})V_{\alpha^{\prime}}^{{\dagger}}({\bf
p}^{\prime})\Gamma^{f\overline{f}^{\prime}\to
B}\gamma^{\mu}V_{\overline{\alpha}}(\overline{\bf
p})b^{\overline{\alpha}{\dagger}}_{\overline{f}}(\overline{\bf
p})b^{\alpha^{\prime}}_{\overline{f}^{\prime}}({\bf p}^{\prime}),$ (448)

where in the derivation of the last equality we have used that
$V_{\alpha}(\overline{\bf p})=-V^{\alpha}(\overline{\bf
p})=-V^{\overline{\alpha}}(\overline{\bf
p})=-V_{\overline{\alpha}}(\overline{\bf p})$. Finally, one gets Eq.(157c) by
changing indices in Eq.(448) as $(\overline{\bf p},\overline{\alpha})\to({\bf
p},\alpha)$ and $({\bf p}^{\prime},\alpha^{\prime})\to({\bf q},\beta)$.
(d) To prove Eq.(157d) from Eq.(156d) with
$(\varrho(\alpha),\varrho(\alpha^{\prime}))=(-,-)$, one may follow a procedure
similar to that used above. Let us substitute Eq.(136) into Eq.(156d), getting
that

$\displaystyle H^{ff^{\prime}\to B}_{{\rm BPIP},4}=\int d\Omega
b^{\lambda{\dagger}}_{B}({\bf
k})\delta^{3}_{P}{\overline{\varepsilon}}_{\lambda\mu}({\bf
k})V_{\alpha^{\prime}}^{{\dagger}}({\bf
p}^{\prime})\Gamma^{f\overline{f}^{\prime}\to B}\gamma^{\mu}U_{\alpha}({\bf
p})b^{\overline{\alpha}{\dagger}}_{\overline{f}}(\overline{\bf
p})b^{\overline{\alpha}^{\prime}{\dagger}}_{f^{\prime}}(\overline{\bf
p}^{\prime})$ $\displaystyle=-\int d\Omega b^{\lambda{\dagger}}_{B}({\bf
k})\delta^{3}_{P}{\overline{\varepsilon}}_{\lambda\mu}({\bf
k})U_{\alpha^{\prime}}^{{\dagger}}(\overline{\bf
p}^{\prime})\Gamma^{f\overline{f}^{\prime}\to
B}\gamma^{\mu}V_{\alpha}(\overline{\bf
p})b^{\overline{\alpha}{\dagger}}_{\overline{f}}(\overline{\bf
p})b^{\overline{\alpha}^{\prime}{\dagger}}_{f^{\prime}}(\overline{\bf
p}^{\prime})$ $\displaystyle=-\int d\Omega b^{\lambda{\dagger}}_{B}({\bf
k})\delta^{3}({\bf k}-{\bf p}-{\bf
p}^{\prime}){\overline{\varepsilon}}_{\lambda\mu}({\bf
k})U_{\overline{\alpha}^{\prime}}^{{\dagger}}(\overline{\bf
p}^{\prime})\Gamma^{f\overline{f}^{\prime}\to
B}\gamma^{\mu}V_{\overline{\alpha}}(\overline{\bf
p})b^{\overline{\alpha}{\dagger}}_{\overline{f}}(\overline{\bf
p})b^{\overline{\alpha}^{\prime}{\dagger}}_{f^{\prime}}(\overline{\bf
p}^{\prime}).$

Then, reindexing as $(\overline{\bf p},\overline{\alpha})\to({\bf p},\alpha)$
and $(\overline{\bf p}^{\prime},\overline{\alpha}^{\prime})\to({\bf
q},\beta)$, we get Eq.(157d).

(v) Derivation of Eq.(159).
Making use of Eqs.(158), one computes that

$\displaystyle\int d^{3}{\bf x}\
\psi^{\dagger}_{f^{\prime}}\Gamma^{f\overline{f}^{\prime}\to
B}\gamma^{\mu}\psi_{f}\phi_{B\mu}^{\dagger}=(2\pi)^{3}\int
d\widetilde{q}d\widetilde{p}d\widetilde{k}\delta^{3}({\bf k}+{\bf q}-{\bf
p}){\overline{\varepsilon}}_{\lambda\mu}({\bf k})b^{\lambda{\dagger}}_{B}({\bf
k})U_{\beta}^{{\dagger}}({\bf q})\Gamma^{f\overline{f}^{\prime}\to
B}\gamma^{\mu}U_{\alpha}({\bf p})b^{\beta{\dagger}}_{f^{\prime}}({\bf
q})b^{\alpha}_{f}({\bf p})$ $\displaystyle+\delta^{3}({\bf k}+{\bf q}+{\bf
p}){\overline{\varepsilon}}_{\lambda\mu}({\bf k})b^{\lambda{\dagger}}_{B}({\bf
k})U_{\beta}^{{\dagger}}({\bf q})\Gamma^{f\overline{f}^{\prime}\to
B}\gamma^{\mu}V_{\alpha}({\bf p})b^{\beta{\dagger}}_{f^{\prime}}({\bf
q})b^{\alpha{\dagger}}_{\overline{f}}({\bf p})$ $\displaystyle+\delta^{3}({\bf
k}-{\bf q}-{\bf p}){\overline{\varepsilon}}_{\lambda\mu}({\bf
k})b^{\lambda{\dagger}}_{B}({\bf k})V_{\beta}^{{\dagger}}({\bf
q})\Gamma^{f\overline{f}^{\prime}\to B}\gamma^{\mu}U_{\alpha}({\bf
p})b^{\beta}_{\overline{f}^{\prime}}({\bf q})b^{\alpha}_{f}({\bf p})$
$\displaystyle+\delta^{3}({\bf k}-{\bf q}+{\bf
p}){\overline{\varepsilon}}_{\lambda\mu}({\bf k})b^{\lambda{\dagger}}_{B}({\bf
k})V_{\beta}^{{\dagger}}({\bf q})\Gamma^{f\overline{f}^{\prime}\to
B}\gamma^{\mu}V_{\alpha}({\bf p})b^{\beta}_{\overline{f}^{\prime}}({\bf
q})b^{\alpha{\dagger}}_{\overline{f}}({\bf p}).$

Taking the normal product, this gives Eq.(159).

##  Appendix G Relations between helicity states of b-modes with opposite
momenta

In this appendix, we derive relations between basic-mode helicity states
$|u^{r}({\bf p})\rangle$ and those for the opposite momentum $\overline{\bf
p}\equiv-{\bf p}$. For the simplicity in discussion, we set the direction of
${\bf p}$ along the $z$-axis. (Generalization to a generic direction is
straightforward.)

As is known, corresponding to a rotation performed in the ordinary three-
dimensional space by an angle $\psi$ around the $(\theta,\phi)$ direction, one
has the following rotation in the spinor space ${\mathscr{W}}$ [23], i.e.,

$\displaystyle R_{\mathscr{W}}(\theta,\phi,\psi)=e^{-i\frac{\psi}{2}{\bf
n}\cdot{\boldsymbol{\sigma}}}=\left(I\cos\frac{\psi}{2}-i{\bf
n}\cdot{\boldsymbol{\sigma}}\sin\frac{\psi}{2}\right),$ (449)

where $I$ indicates the unit matrix,
${\boldsymbol{\sigma}}=(\sigma_{x},\sigma_{y},\sigma_{z})$ represents the
vector of Pauli matrices, and ${\bf n}$ is the unit vector along the
$(\theta,\phi)$ direction. Or, in the notation of Eq.(53), the Weyl-spinor
rotation is written as

$\displaystyle
R_{\mathscr{W}}(\theta,\phi,\psi)=\exp\left(-i\theta_{k}s_{b}^{k}\right),$
(450)

where $s_{b}^{k}$ are components of the angular momentum of basic mode, which
are given in Eq.(339).

Clearly, a rotation in the ordinary $(x,y,z)$ space around the $y$-axis by an
angle $\psi=\pi$ transforms ${\bf p}$ to $\overline{\bf p}$. This rotation
corresponds to $R_{\mathscr{W}}(\frac{\pi}{2},\frac{\pi}{2},\pi)$ in the space
${\mathscr{W}}$ and gives $[w^{rA}(\overline{\bf p})]$, the matrix form of
$w^{rA}(\overline{\bf p})$, i.e.,

$\displaystyle[w^{rA}(\overline{\bf
p})]=R(\frac{\pi}{2},\frac{\pi}{2},\pi)[w^{rA}({\bf p})].$ (451)

From Eqs.(449)-(450), one gets that

$\displaystyle R_{\mathscr{W}}(\frac{\pi}{2},\frac{\pi}{2},\pi)$
$\displaystyle=(I\cos\frac{\psi}{2}-i\sigma_{y}\sin\frac{\psi}{2}),$
$\displaystyle=\left(\begin{array}[]{cc}\cos\frac{\psi}{2}&-\sin\frac{\psi}{2}\\\
\sin\frac{\psi}{2}&\cos\frac{\psi}{2}\end{array}\right)=\left(\begin{array}[]{cc}0&-1\\\
1&0\end{array}\right).$ (456)

Then, with $[w^{rA}({\bf p})]$ in Eq.(347), this gives that

$\displaystyle[{w}^{r=0,A}(\overline{\bf p})]=\left(\begin{array}[]{c}0\\\
1\end{array}\right),\quad[{w}^{r=1,A}(\overline{\bf
p})]=-\left(\begin{array}[]{c}1\\\ 0\end{array}\right).$ (461)

Comparing Eq.(347) and Eq.(461), one sees that

$\displaystyle|{w}^{r}(\overline{\bf p})\rangle=(-1)^{r}|{w}^{r+1}({\bf
p})\rangle.$ (462)

Then, usig Eq.(19), one gets that

$\displaystyle|{w}^{r}(\overline{\bf p})\rangle=-|{w}_{r}({\bf p})\rangle.$
(463)

To find an expression of $|{w}_{r}(\overline{\bf p})\rangle$, one may
substitute Eq.(463) into the relation of $|{w}_{r}(\overline{\bf
p})\rangle=|{w}^{s}(\overline{\bf p})\rangle\epsilon_{sr}$, getting that

$\displaystyle|{w}_{r}(\overline{\bf p})\rangle=-|{w}_{s}({\bf
p})\rangle\epsilon_{sr}=-|{w}^{t}({\bf p})\rangle\epsilon_{ts}\epsilon_{sr}.$
(464)

Then, using Eq.(357), one gets that

$\displaystyle|{w}_{r}(\overline{\bf p})\rangle=|{w}^{r}({\bf p})\rangle.$
(465) 151515 One should note that, due to the well-known difference between
rotations in the real three-dimensional space and those in the Weyl spinor
space, the relation between $|{w}_{r}(\overline{\bf p})\rangle$ and
$|{w}^{r}({\bf p})\rangle$ can not be obtained by directly exchanging ${\bf
p}$ and $\overline{\bf p}$ in Eq.(463), which would give a wrong relation of
$|{w}_{r}(\overline{\bf p})\rangle=-|{w}^{r}({\bf p})\rangle$. In fact, this
latter relation corresponds to the rotation of  $\displaystyle
R(\frac{\pi}{2},\frac{\pi}{2},3\pi)=\left(\begin{array}[]{cc}0&1\\\
-1&0\end{array}\right).$ (468) It is easy to check that, under this rotation,
the rhs of Eq.(462) should be multiplied by a factor $(-1)$ and, as a
consequence, so do Eqs.(463)-(469).

Finally, for the spinors of $u$ and $v$ defined in Eq.(49), one gets that

$\displaystyle|{u}^{\alpha}(\overline{\bf p})\rangle$
$\displaystyle=-|{v}^{\alpha}({\bf p})\rangle,$ (469a)
$\displaystyle|{v}^{\alpha}(\overline{\bf p})\rangle$
$\displaystyle=|{u}^{\alpha}({\bf p})\rangle.$ (469b)

##  Appendix H The matrices $\Gamma^{f\overline{f}^{\prime}\to B}$

In this appendix, we give detailed derivations for matrices
$\Gamma^{f\overline{f}^{\prime}\to B}$ given in the main text, as well as for
related quantities.

###  H.1 Expressions of c and d

In this section, we derive expressions of the two quantities $c$ and $d$ [see
Eq.(135)], which are useful when computing the matrices
$\Gamma^{f\overline{f}^{\prime}\to B}$ for $T_{1}$ and $T_{2}$.

It proves convenient to write the layer phase $\vartheta^{M}_{m}$ of a mode
$M$ as a function of the global phase $\vartheta^{M}_{1}$ and the phase
difference $\delta^{M}_{\theta}$. For two-layer and three-layer modes, the
assumption ${\rm MA}^{\rm layer}_{\rm phase}$ gives that

$\displaystyle\vartheta^{M}_{m}=\vartheta^{M}_{1}+(m-1)\delta^{M}_{\theta}.$
(470)

For $u$-modes and $d$-modes, within each sections the layer phases satisfy
Eq.(470) for a corresponding $t$-mode. Written explicitly,

$\displaystyle\vartheta^{u_{i}}_{m}=\left\\{\begin{array}[]{ll}\vartheta^{t_{i}}_{1}+(m-1)\delta^{t_{i}}_{\theta},&\hbox{for
$m=1,2,3$;}\\\ \vartheta^{t_{i}}_{1}+(m-4)\delta^{t_{i}}_{\theta},&\hbox{for
$m=4,5,6$;}\end{array}\right.$ (473)
$\displaystyle\vartheta^{d_{i}}_{m}=\left\\{\begin{array}[]{ll}\vartheta^{t_{i}}_{1}+(m-1)\delta^{t_{i}}_{\theta},&\hbox{for
$m=1,2,3$;}\\\
\vartheta^{t_{i}}_{1}+(m-4)\delta^{t_{i}}_{\theta}+\frac{\pi}{2},&\hbox{for
$m=4,5,6$;}\end{array}\right.$ (476)

where the phase $\pi/2$ for $d$-modes’ layers with $m=4,5,6$ comes from the
factor $i$ in states of their second sections [see Eq.(182f)].

For $T_{1}$ in Eq.(174) with $n_{f}=2,3$, making use of Eq.(470) and noting
the term $\delta_{nm}$, from Eq.(135) one gets that

$\displaystyle
c=e^{i(\vartheta^{f}_{1}+\vartheta^{f^{\prime}}_{1})}\sum_{m\in\mathcal{M}_{z}}e^{i(m-1)(\delta^{f}_{\theta}+\delta^{f^{\prime}}_{\theta}-\delta^{B}_{\theta})},$
(477a) $\displaystyle
d=e^{i(\vartheta^{f}_{1}+\vartheta^{f^{\prime}}_{1})}\sum_{m\in\mathcal{M}_{\overline{z}}}e^{i(m-1)(\delta^{f}_{\theta}+\delta^{f^{\prime}}_{\theta}-\delta^{B}_{\theta})},$
(477b)

where the fact of $\vartheta^{B}_{1}=0$ has been used.

The matrix $T_{2}$ in Eq.(186) is for two cases. In the first case,
$f=f^{\prime}=u_{i}$. From Eq.(182c), one sees that the two sections of a
$u_{i}$-mode are identical. Then, noting that $\vartheta^{B}_{1}=0$ and
$\vartheta^{B}_{2}=\delta^{B}_{\theta}$, one writes Eq.(135) as

$\displaystyle c=2\exp(i2\vartheta^{t_{i}}_{1})\sum_{m\in\mathcal{M}_{z}(m\leq
3)}e^{i2(m-1)\delta^{t}_{\theta}},$ (478a) $\displaystyle
d=2\exp(i2\vartheta^{t_{i}}_{1}-i\delta^{B}_{\theta})\sum_{m\in\mathcal{M}_{\overline{z}}(m\leq
3)}e^{i2(m-1)\delta^{t}_{\theta}}.$ (478b)

In the second case, $f=f^{\prime}=d_{i}$. In this case, the two sections of a
$d_{i}$-mode are antimodes of each other with an additional section phase of
$\pi/2$ and, as a result, expressions of $c$ and $d$ like those in Eq.(478)
are quite complicated. Hence, we simply write them in terms of
$\vartheta^{d_{i}}_{m}$, i.e.,

$\displaystyle c=\sum_{m\in\mathcal{M}_{z}}\exp(i2\vartheta^{d_{i}}_{m}),$
(479a) $\displaystyle
d=\exp(-i\delta^{B}_{\theta})\sum_{m\in\mathcal{M}_{\overline{z}}}\exp(i2\vartheta^{d_{i}}_{m}).$
(479b)

###  H.2 Derivation of some equations

In this section, we give detailed derivations for Eqs.(179), (187), (197), and
(202).

####  H.2.1  Derivation of Eq.(179)

This equation is for two-layer modes with $T_{1}$. As discussed in the main
text, for two-layer modes, one has

$\displaystyle\vartheta^{e}_{1}=\vartheta^{\nu}_{1}=0,$ (480)
$\displaystyle\delta^{\nu}_{\theta}=\delta_{\theta}^{Z}=0,\ \
\delta^{e}_{\theta}=\pi/2,\ \ \delta_{\theta}^{A}=\delta_{\theta}^{W}=\pi.$
(481)

Below, we discuss the FIPs separately.
(a) For the pair of $e\overline{e}$. For an FIP of $f\overline{f}^{\prime}\to
B$ with $f=f^{\prime}=e$, from Eq.(176) one gets that
$|{\mathcal{L}}^{e}_{1}\rangle|{\mathcal{L}}^{\overline{e}}_{1}\rangle=|u^{e}_{\alpha}({\bf
p})\rangle|\overline{v}^{\overline{e}}_{{\alpha^{\prime}}}({\bf
p}^{\prime})\rangle$ and
$|{\mathcal{L}}^{e}_{2}\rangle|{\mathcal{L}}^{\overline{e}}_{2}\rangle=|\overline{v}^{e}_{\alpha}({\bf
p})\rangle|u^{\overline{e}}_{\alpha^{\prime}}({\bf p}^{\prime})\rangle$. This
gives that $\mathcal{M}_{z}=\\{1\\}$ and $\mathcal{M}_{\overline{z}}=\\{2\\}$
for $\mathcal{M}_{z}$ and $\mathcal{M}_{\overline{z}}$ defined in Eq.(133).
Then, from Eq.(477), one gets that

$\displaystyle\text{$e\overline{e}\to A$: \ }c=e^{0}=1,\quad\&\quad
d=e^{0}=1;$ (482) $\displaystyle\text{$e\overline{e}\to Z$: \
}c=e^{0}=1,\quad\&\quad d=e^{i\pi}=-1.$ (483)

(b) For the pair of $\nu\overline{\nu}$, using Eq.(175), one finds that
$\mathcal{M}_{z}=\\{1,2\\}$ and $\mathcal{M}_{\overline{z}}=\emptyset$. This
gives that

$\displaystyle\text{$\nu\overline{\nu}\to A$: \ }c=1+e^{-i\pi}=0,\quad\&\quad
d=0;$ (484) $\displaystyle\text{$\nu\overline{\nu}\to Z$: \
}c=1+e^{0}=2,\quad\&\quad d=0.$ (485)

(c) For both pairs of $\nu\overline{e}$ and $e\overline{\nu}$, one finds that
$\mathcal{M}_{z}=\\{1\\}$ and $\mathcal{M}_{\overline{z}}=\emptyset$ and this
gives that

$\displaystyle\text{$\nu\overline{e}\to W$: \ }c=e^{0}=1,\quad\&\quad d=0;$
(486) $\displaystyle\text{$e\overline{\nu}\to\overline{W}$: \
}c=e^{0}=1,\quad\&\quad d=0.$ (487)

Finally, substituting the above values of $c$ and $d$ into Eq.(142), and
noting that $n_{f}=2$, $n_{V}=2$ for $B=A$ or $Z$, and $n_{V}=1$ for $B=W$ or
$\overline{W}$, one gets Eq.(179).

####  H.2.2  Derivation of Eq.(187)

Equation (187) is for six-layer modes with $T_{2}$. Let us first discuss
$u$-mode, whose two sections are identical [see Eq.(182c)]. To compute the
values of $c$ and $d$, one needs to find out the two sets of $\mathcal{M}_{z}$
and $\mathcal{M}_{\overline{z}}$. Making use of the explicit expressions of
the spinor states of $u$-modes, one finds that

$\displaystyle u_{1}\overline{u}_{1}:\qquad$ $\displaystyle\
\mathcal{M}_{z}=\\{1,4\\},\ \mathcal{M}_{\overline{z}}=\\{2,3,5,6\\};$ (488a)
$\displaystyle{u_{2}\overline{u}_{2}:}\qquad$
$\displaystyle\mathcal{M}_{z}=\\{1,2,4,5\\},\
\mathcal{M}_{\overline{z}}=\\{3,6\\};$ (488b)
$\displaystyle{u_{3}\overline{u}_{3}:}\qquad$
$\displaystyle\mathcal{M}_{z}=\\{1,3,4,6\\},\
\mathcal{M}_{\overline{z}}=\\{2,5\\}.$ (488c)

Then, it is ready to compute $c$ and $d$ for $u_{i}\overline{u}_{i}\to A,Z$ by
making use of Eq.(478), getting the following results.

$\displaystyle u_{1}$ $\displaystyle\overline{u}_{1}\to A:$ $\displaystyle
c=2\exp(i2\vartheta^{t_{1}}_{1}),$ $\displaystyle
d=2e^{i(2\vartheta^{t_{1}}_{1}-\delta_{\theta}^{A})}(e^{i2\delta_{\theta}^{t}}+e^{i4\delta_{\theta}^{t}})=2e^{i2\vartheta^{t_{1}}_{1}};$
(489a) $\displaystyle u_{1}$ $\displaystyle\overline{u}_{1}\to Z:$
$\displaystyle c=2\exp(i2\vartheta^{t_{1}}_{1}),$ $\displaystyle
d=2e^{i(2\vartheta^{t_{1}}_{1}-\delta_{\theta}^{Z})}(e^{i2\delta_{\theta}^{t}}+e^{i4\delta_{\theta}^{t}})=-2e^{i2\vartheta^{t_{1}}_{1}}.$
(489b) $\displaystyle u_{2}$ $\displaystyle\overline{u}_{2}\to A:$
$\displaystyle
c=2\exp(i2\vartheta^{t_{2}}_{1})(1+e^{i2\delta_{\theta}^{t}})=2\exp(i2\vartheta^{t_{2}}_{1})e^{i\pi/3},$
$\displaystyle
d=2e^{i(2\vartheta^{t_{2}}_{1}-\delta_{\theta}^{A})}e^{i4\delta_{\theta}^{t}}=2e^{i2\vartheta^{t_{2}}_{1}}e^{i\delta_{\theta}^{t}};$
(490a) $\displaystyle u_{2}$ $\displaystyle\overline{u}_{2}\to Z:$
$\displaystyle
c=2\exp(i2\vartheta^{t_{2}}_{1})(1+e^{i2\delta_{\theta}^{t}})=2\exp(i2\vartheta^{t_{2}}_{1})e^{i\pi/3},$
$\displaystyle
d=2e^{i(2\vartheta^{t_{2}}_{1}-\delta_{\theta}^{Z})}e^{i4\delta_{\theta}^{t}}=-2e^{i2\vartheta^{t_{2}}_{1}}e^{i\delta_{\theta}^{t}}.$
(490b) $\displaystyle u_{3}$ $\displaystyle\overline{u}_{3}\to A:$
$\displaystyle
c=2\exp(i2\vartheta^{t_{3}}_{1})(1+e^{i4\delta_{\theta}^{t}})=2\exp(i2\vartheta^{t_{3}}_{1})e^{-i\pi/3},$
$\displaystyle
d=2e^{i(2\vartheta^{t_{3}}_{1}-\delta_{\theta}^{A})}e^{i2\delta_{\theta}^{t}}=2e^{i2\vartheta^{t_{3}}_{1}}e^{-i\pi/3};$
(491a) $\displaystyle u_{3}$ $\displaystyle\overline{u}_{3}\to Z:$
$\displaystyle
c=2\exp(i2\vartheta^{t_{3}}_{1})(1+e^{i4\delta_{\theta}^{t}})=2\exp(i2\vartheta^{t_{3}}_{1})e^{-i\pi/3},$
$\displaystyle
d=2e^{i(2\vartheta^{t_{3}}_{1}-\delta_{\theta}^{Z})}e^{i2\delta_{\theta}^{t}}=-2e^{i2\vartheta^{t_{3}}_{1}}e^{-i\pi/3}.$
(491b)

For $d$-modes, making use of the explicit expressions of their spinor states,
one finds that

$\displaystyle d_{1}\overline{d}_{1}:\qquad$ $\displaystyle\
\mathcal{M}_{z}=\\{1,5,6\\},\ \mathcal{M}_{\overline{z}}=\\{2,3,4\\};$ (492a)
$\displaystyle{d_{2}\overline{d}_{2}:}\qquad$
$\displaystyle\mathcal{M}_{z}=\\{1,2,6\\},\mathcal{M}_{\overline{z}}=\\{3,4,5\\};$
(492b) $\displaystyle{d_{3}\overline{d}_{3}:}\qquad$
$\displaystyle\mathcal{M}_{z}=\\{1,3,5\\},\mathcal{M}_{\overline{z}}=\\{2,4,6\\}.$
(492c)

Then, substituting Eq.(476) into Eq.(479), one finds that

$\displaystyle d_{1}$ $\displaystyle\overline{d}_{1}\to A:$ $\displaystyle
c=\exp(i2\vartheta^{t_{1}}_{1})(1+e^{i2\delta_{\theta}^{t}+i\pi}+e^{i4\delta_{\theta}^{t}+i\pi})=2\exp(i2\vartheta^{t_{1}}_{1}),$
$\displaystyle
d=\exp(i2\vartheta^{t_{1}}_{1}-i\delta_{\theta}^{A})(e^{i2\delta_{\theta}^{t}}+e^{i4\delta_{\theta}^{t}}+e^{i\pi})=2\exp(i2\vartheta^{t_{1}}_{1});$
(493a) $\displaystyle d_{1}$ $\displaystyle\overline{d}_{1}\to Z:$
$\displaystyle
c=\exp(i2\vartheta^{t_{1}}_{1})(1+e^{i2\delta_{\theta}^{t}+i\pi}+e^{i4\delta_{\theta}^{t}+i\pi})=2\exp(i2\vartheta^{t_{1}}_{1}),$
$\displaystyle
d=\exp(i2\vartheta^{t_{1}}_{1}-i\delta_{\theta}^{Z})(e^{i2\delta_{\theta}^{t}}+e^{i4\delta_{\theta}^{t}}+e^{i\pi})=-2\exp(i2\vartheta^{t_{1}}_{1}).$
(493b) $\displaystyle d_{2}$ $\displaystyle\overline{d}_{2}\to A:$
$\displaystyle
c=\exp(i2\vartheta^{t_{2}}_{1})(1+e^{i2\delta_{\theta}^{t}}+e^{i4\delta_{\theta}^{t}+i\pi})=2\exp(i2\vartheta^{t_{2}}_{1}+i\delta_{\theta}^{t}),$
$\displaystyle
d=\exp(i2\vartheta^{t_{2}}_{1}-i\delta_{\theta}^{A})(e^{i4\delta_{\theta}^{t}}+e^{i\pi}+e^{i2\delta_{\theta}^{t}+i\pi})$
$\displaystyle=2\exp(i2\vartheta^{t_{2}}_{1}+i\delta_{\theta}^{t});$ (494a)
$\displaystyle d_{2}$ $\displaystyle\overline{d}_{2}\to Z:$ $\displaystyle
c=\exp(i2\vartheta^{t_{2}}_{1})(1+e^{i2\delta_{\theta}^{t}}+e^{i4\delta_{\theta}^{t}+i\pi})=2\exp(i2\vartheta^{t_{2}}_{1}+i\delta_{\theta}^{t}),$
$\displaystyle
d=\exp(i2\vartheta^{t_{2}}_{1}-i\delta_{\theta}^{Z})(e^{i4\delta_{\theta}^{t}}+e^{i\pi}+e^{i2\delta_{\theta}^{t}+i\pi})$
$\displaystyle=-2\exp(i2\vartheta^{t_{2}}_{1}+i\delta_{\theta}^{t});$ (494b)
$\displaystyle d_{3}$ $\displaystyle\overline{d}_{3}\to A:$ $\displaystyle
c=\exp(i2\vartheta^{t_{3}}_{1})(1+e^{i4\delta_{\theta}^{t}}+e^{i2\delta_{\theta}^{t}+i\pi})=2\exp(i2\vartheta^{t_{3}}_{1}-i\delta_{\theta}^{t}),$
$\displaystyle
d=\exp(i2\vartheta^{t_{3}}_{1}-i\delta_{\theta}^{A})(e^{i2\delta_{\theta}^{t}}+e^{i\pi}+e^{i4\delta_{\theta}^{t}+i\pi})$
$\displaystyle=2\exp(i2\vartheta^{t_{3}}_{1}-i\delta_{\theta}^{t});$ (495a)
$\displaystyle d_{3}$ $\displaystyle\overline{d}_{3}\to Z:$ $\displaystyle
c=\exp(i2\vartheta^{t_{3}}_{1})(1+e^{i4\delta_{\theta}^{t}}+e^{i2\delta_{\theta}^{t}+i\pi})=2\exp(i2\vartheta^{t_{3}}_{1}-i\delta_{\theta}^{t}),$
$\displaystyle
d=\exp(i2\vartheta^{t_{3}}_{1}-i\delta_{\theta}^{Z})(e^{i2\delta_{\theta}^{t}}+e^{i\pi}+e^{i4\delta_{\theta}^{t}+i\pi})$
$\displaystyle=-2\exp(i2\vartheta^{t_{3}}_{1}-i\delta_{\theta}^{t});$ (495b)

####  H.2.3  Derivation of Eq.(197)

Equation (197) is for three-layer modes with $T_{1}$. For three-layer modes,
we have

$\displaystyle\vartheta^{t_{1}}_{1}=0,\quad\vartheta^{t_{2}}_{1}=-\delta_{\theta}^{t}/2,\quad\vartheta^{t_{3}}_{1}=\delta_{\theta}^{t}/2,$
(496)
$\displaystyle\delta_{\theta}^{t}=\pi/3,\quad\delta_{\theta}^{g}=2\delta_{\theta}^{t}.$
(497)

For the $t$-modes and $g$-modes, one has
$\delta^{t}_{\theta}+\delta^{t}_{\theta}-\delta^{g}_{\theta}=0$ and, as a
result, Eq.(477) becomes that

$\displaystyle
c=e^{i(\vartheta^{f}_{1}+\vartheta^{f^{\prime}}_{1})}\sum_{m\in\mathcal{M}_{z}}1,$
(498) $\displaystyle
d=e^{i(\vartheta^{f}_{1}+\vartheta^{f^{\prime}}_{1})}\sum_{m\in\mathcal{M}_{\overline{z}}}1.$
(499)

Using these expressions of $c$ and $d$, one finds that

$\displaystyle t_{1}\overline{t}_{1}\to g_{1\overline{1}}:$ $\displaystyle\
\mathcal{M}_{z}=\\{1\\}\ \&\ \mathcal{M}_{\overline{z}}=\\{2,3\\},$
$\displaystyle\to c=1,\ \&\ d=2;$ (500a)
$\displaystyle{t_{2}\overline{t}_{2}\to g_{2\overline{2}}:}$ $\displaystyle\
\mathcal{M}_{z}=\\{1,2\\}\ \&\ \mathcal{M}_{\overline{z}}=\\{3\\},$
$\displaystyle\to c=2e^{-i\delta^{t}_{\theta}}\ \&\
d=e^{-i\delta^{t}_{\theta}};$ (500b) $\displaystyle{t_{3}\overline{t}_{3}\to
g_{3\overline{3}}:}$ $\displaystyle\ \mathcal{M}_{z}=\\{1,3\\}\ \&\
\mathcal{M}_{\overline{z}}=\\{2\\},$ $\displaystyle\to
c=2e^{i\delta^{t}_{\theta}}\ \&\ d=e^{i\delta^{t}_{\theta}};$ (500c)
$\displaystyle{t_{1}\overline{t}_{2}\to g_{1\overline{2}}:}$ $\displaystyle\
\mathcal{M}_{z}=\\{1\\}\ \&\ \mathcal{M}_{\overline{z}}=\\{3\\},$
$\displaystyle\to c=e^{-i\delta^{t}_{\theta}/2}\ \&\
d=e^{-i\delta^{t}_{\theta}/2};$ (500d) $\displaystyle{t_{1}\overline{t}_{3}\to
g_{1\overline{3}}:}$ $\displaystyle\ \mathcal{M}_{z}=\\{1\\}\ \&\
\mathcal{M}_{\overline{z}}=\\{2\\},$ $\displaystyle\to
c=e^{i\delta^{t}_{\theta}/2}\ \&\ d=e^{i\delta^{t}_{\theta}/2};$ (500e)
$\displaystyle{t_{2}\overline{t}_{3}\to g_{2\overline{3}}:}$ $\displaystyle\
\mathcal{M}_{z}=\\{1\\},\mathcal{M}_{\overline{z}}=\emptyset,$
$\displaystyle\to c=1\ \&\ d=0;$ (500f)

Finally, substituting the above values of $c$ and $d$ into Eq.(142) and making
use of Eq.(526) to be proved later, one gets Eq.(197).

####  H.2.4  Derivation of Eq.(202)

From the expression of

$\displaystyle\gamma^{\mu}=\sqrt{2}\left(\begin{array}[]{cc}0&\sigma^{\mu
AB^{\prime}}\\\ \overline{\sigma}^{\mu}_{A^{\prime}B}&0\end{array}\right),$
(503)

one sees that

$\displaystyle\gamma^{\mu{\dagger}}=\sqrt{2}\left(\begin{array}[]{cc}0&\sigma^{\mu}_{AB^{\prime}}\\\
\overline{\sigma}^{\mu A^{\prime}B}&0\end{array}\right).$ (506)

Changing indices on the rhs of Eq.(506) by $A\leftrightarrow B$, one gets that

$\displaystyle\gamma^{\mu{\dagger}}=\sqrt{2}\left(\begin{array}[]{cc}0&\sigma^{\mu}_{BA^{\prime}}\\\
\overline{\sigma}^{\mu
B^{\prime}A}&0\end{array}\right)=\sqrt{2}\left(\begin{array}[]{cc}0&\overline{\sigma}^{\mu}_{A^{\prime}B}\\\
\sigma^{\mu AB^{\prime}}&0\end{array}\right),$ (511)

where Eq.(288) has been used in the derivation of the second equality. Then,
one finds that

$\displaystyle\left(\Gamma^{f\overline{f}^{\prime}\to
B}\gamma^{\mu}\right)^{\dagger}=\left(\gamma^{\mu}\right)^{\dagger}\left(\Gamma^{f\overline{f}^{\prime}\to
B}\right)^{\dagger}$
$\displaystyle=\sqrt{2}\left(\begin{array}[]{cc}0&\overline{\sigma}^{\mu}_{A^{\prime}B}\\\
\sigma^{\mu
AB^{\prime}}&0\end{array}\right)\left(\begin{array}[]{cc}0&d^{*}\\\
c^{*}&0\end{array}\right)$ (516)
$\displaystyle=\sqrt{2}\left(\begin{array}[]{cc}c^{*}\overline{\sigma}^{\mu}_{A^{\prime}B}&0\\\
0&d^{*}\sigma^{\mu AB^{\prime}}\end{array}\right)$ (519)
$\displaystyle=\sqrt{2}\left(\begin{array}[]{cc}0&c^{*}\\\
d^{*}&0\end{array}\right)\left(\begin{array}[]{cc}0&\sigma^{\mu
AB^{\prime}}\\\ \overline{\sigma}^{\mu}_{A^{\prime}B}&0\end{array}\right)$
(524) $\displaystyle=\left(\Gamma^{f\overline{f}^{\prime}\to
B}\right)^{*}\gamma^{\mu}.$ (525)

This gives Eq.(202).

###  H.3 A relation between $\Gamma$-matrices of FIPs of antimodes

In this section, we prove the following relation used above,

$\displaystyle\Gamma^{f^{\prime}\overline{f}\to\overline{B}}=\Gamma^{f\overline{f}^{\prime}\to
B}.$ (526)

To be specific in discussion, let us consider the following expressions of the
spinor states of $f$ and $\overline{f}^{\prime}$, namely,

$\displaystyle|\mathcal{S}_{f}^{\alpha}({\bf
p})\rangle=\frac{1}{\sqrt{n_{f}}}\left(\begin{array}[]{c}e^{i\vartheta_{1}^{f}}|u^{\alpha}({\bf
p})\rangle\\\ e^{i\vartheta_{2}^{f}}|\overline{v}^{\alpha}({\bf p})\\\
\vdots\\\ e^{i\vartheta_{n}^{f}}|u^{\alpha}({\bf p})\rangle\end{array}\right)$
(531)

and

$\displaystyle|\mathcal{S}_{\overline{f}^{\prime}}^{\beta}({\bf
q})\rangle=\frac{1}{\sqrt{n_{f}}}\left(\begin{array}[]{c}e^{i\vartheta_{1}^{f^{\prime}}}|\overline{v}^{\beta}({\bf
q})\rangle\\\ e^{i\vartheta_{2}^{f^{\prime}}}|u^{\beta}({\bf q})\\\ \vdots\\\
e^{i\vartheta_{n}^{f^{\prime}}}|u^{\beta}({\bf q})\rangle\end{array}\right).$
(536)

(Other expressions may be treated in exactly the same way.) The corresponding
spinor states of $\overline{f}$ and $f^{\prime}$ are obtained by replacements
of, say, $|u_{\alpha}({\bf
p})\rangle\leftrightarrow|\overline{v}_{\alpha}({\bf p})\rangle$, for all the
layer states, resulting in

$\displaystyle|\mathcal{S}_{\overline{f}}^{\alpha}({\bf
p})\rangle=\frac{1}{\sqrt{n_{f}}}\left(\begin{array}[]{c}e^{i\vartheta_{1}^{f}}|\overline{v}^{\alpha}({\bf
p})\rangle\\\ e^{i\vartheta_{2}}|u^{\alpha}({\bf p})\rangle\\\ \vdots\\\
e^{i\vartheta_{n}^{f}}|\overline{v}^{\alpha}({\bf
p})\rangle\end{array}\right),$ (541)
$\displaystyle|\mathcal{S}_{f^{\prime}}^{\beta}({\bf
q})\rangle=\frac{1}{\sqrt{n_{f}}}\left(\begin{array}[]{c}e^{i\vartheta_{1}^{f^{\prime}}}|u^{\beta}({\bf
q})\rangle\\\ e^{i\vartheta_{2}^{f^{\prime}}}|\overline{v}^{\beta}({\bf q})\\\
\vdots\\\ e^{i\vartheta_{n}^{f^{\prime}}}|\overline{v}^{\beta}({\bf
q})\rangle\end{array}\right).$ (546)

For the sake of clearness in comparing
$C^{f^{\prime}\overline{f}\to\overline{B}}_{\alpha^{\prime}\alpha\lambda}$ and
$C^{f\overline{f}^{\prime}\to B}_{\alpha\alpha^{\prime}\lambda}$, let us
relabel the spinor states of $f^{\prime}$ and $\overline{f}$ as
$|\mathcal{S}_{f^{\prime}}^{\alpha}({\bf p})\rangle$ and
$|\mathcal{S}_{\overline{f}}^{\beta}({\bf q})\rangle$, respectively, i.e.,

$\displaystyle|\mathcal{S}_{f^{\prime}}^{\alpha}({\bf
p})\rangle=\frac{1}{\sqrt{n_{f}}}\left(\begin{array}[]{c}e^{i\vartheta_{1}^{f^{\prime}}}|u^{\alpha}({\bf
p})\rangle\\\ e^{i\vartheta_{2}^{f^{\prime}}}|\overline{v}^{\alpha}({\bf
p})\\\ \vdots\\\ e^{i\vartheta_{n}^{f^{\prime}}}|\overline{v}^{\alpha}({\bf
p})\rangle\end{array}\right).$ (551)
$\displaystyle|\mathcal{S}_{\overline{f}}^{\beta}({\bf
q})\rangle=\frac{1}{\sqrt{n_{f}}}\left(\begin{array}[]{c}e^{i\vartheta_{1}^{f}}|\overline{v}^{\beta}({\bf
q})\rangle\\\ e^{i\vartheta_{2}}|u^{\beta}({\bf q})\rangle\\\ \vdots\\\
e^{i\vartheta_{n}^{f}}|\overline{v}^{\beta}({\bf
q})\rangle\end{array}\right),$ (556)

Then, it is easy to see that $\overrightarrow{D}|\mathcal{S}_{f}^{\alpha}({\bf
p})\mathcal{S}_{\overline{f}^{\prime}}^{\beta}({\bf q})\rangle$ has the same
vector-type layers as
$\overrightarrow{D}|\mathcal{S}_{f^{\prime}}^{\alpha}({\bf
p})\mathcal{S}_{\overline{f}}^{\beta}({\bf q})\rangle$.

Moreover, we note that, for the matrices $T_{k}$ used in the studied model of
modes, ${T_{k}}$ for ${f^{\prime}\overline{f}\to\overline{B}}$ is the same as
that for ${f\overline{f}^{\prime}\to B}$. Then, one sees that the values of
$c$ and $d$ for
$C^{f^{\prime}\overline{f}\to\overline{B}}_{\alpha^{\prime}\alpha\lambda}$ are
the same as those given on the rhs of Eq.(134) for
$C^{f\overline{f}^{\prime}\to B}_{\alpha\alpha^{\prime}\lambda}$. This proves
Eq.(526).

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