(535)

and

$K_{+}^{*}=\\{\phi_{-}\phi_{-}\phi_{-}\phi_{-},\,\phi_{-}\phi_{-}\phi_{+}\phi_{+},\,\phi_{-}\phi_{+}\phi_{-}\phi_{+},\,\phi_{-}\phi_{+}\phi_{+}\phi_{-}\\}$
(536)

We can see that $K_{+}$ and $K_{+}^{*}$ are dual to each other in which
$K_{+}^{*}=*K_{+}$. For the negative partition, we have

$K_{-}=\\{-\phi_{+}\phi_{+}\phi_{+}\phi_{-},\,-\phi_{+}\phi_{+}\phi_{-}\phi_{+},\,-\phi_{+}\phi_{-}\phi_{+}\phi_{+},\,-\phi_{+}\phi_{-}\phi_{-}\phi_{-}\\}$
(537)

and

$K_{-}^{*}=\\{-\phi_{-}\phi_{-}\phi_{-}\phi_{+},\,-\phi_{-}\phi_{-}\phi_{+}\phi_{-},\,-\phi_{-}\phi_{+}\phi_{-}\phi_{-},\,-\phi_{-}\phi_{+}\phi_{+}\phi_{+}\\}$
(538)

We can see that $K_{-}$ and $K_{-}^{*}$ are dual to each other in which
$K_{-}^{*}=*K_{-}$. Therefore the full interaction term can be written as

$K^{(+)}\cup K^{(-)}=(K_{+}\cup K_{+}^{*})\cup(-K_{-}\cup-K_{-}^{*})\,.$ (539)

Now let’s reconsider for the $n=\mathrm{odd}$ case. The Lagrangian is given by

$\mathcal{L}=\frac{1}{2}\partial_{\mu}\phi_{+}\partial^{\mu}\phi_{+}+\frac{1}{2}\partial_{\mu}\phi_{-}\partial^{\mu}\phi_{-}-g(\pm\phi_{+}\pm\phi_{-})^{2k-1}\,,$
(540)

The interaction term $(\phi_{+}-\phi_{-})^{2k-1}$, upon expansion can be
separated into positive partition and negative partition that is dual to each
other. Let $P^{(+)}$ be the positive partition and $P^{(-)}$ be the negative
partition, we ill have $*P^{(+)}=P^{(-)}$. Mathematically

$(\phi_{+}-\phi_{-})^{2k-1}=\sum_{I\in
W}\prod_{l=1}^{2k-1}\phi_{i_{l}}=\sum_{I\in
P_{+}}\prod_{l=1}^{2k-1}\phi_{i_{l}}-\sum_{I\in
P_{-}}\prod_{l=1}^{2k-1}\phi_{i_{l}}\,,$ (541)

where

$\bigg{(}\sum_{I\in
P_{+}}\prod_{l=1}^{2k-1}\phi_{i_{l}}\bigg{)}^{*}=\sum_{I\notin
P_{+}}\prod_{l=1}^{2k-1}\phi_{i_{l}}^{*}=\sum_{I\in
P_{-}}\prod_{l=1}^{2k-1}\phi_{i_{l}}$ (542)

For example, this can be verified by $n=3$ case. Consider

$\displaystyle(\phi_{+}-\phi_{-})^{3}$
$\displaystyle=\phi_{+}\phi_{+}\phi_{+}-\phi_{+}\phi_{-}\phi_{+}-\phi_{-}\phi_{+}\phi_{+}+\phi_{-}\phi_{-}\phi_{+}$
(543)
$\displaystyle-\phi_{+}\phi_{+}\phi_{-}+\phi_{+}\phi_{-}\phi_{-}+\phi_{-}\phi_{+}\phi_{-}-\phi_{-}\phi_{-}\phi_{-}\,.$

We have for the positive partition,

$P^{(+)}=\\{\phi_{+}\phi_{+}\phi_{+},\,\phi_{+}\phi_{-}\phi_{-},\,\phi_{-}\phi_{+}\phi_{-},\,\phi_{-}\phi_{-}\phi_{+}\\}\,,$
(544)

and the negative partition,

$P^{(-)}=\\{-\phi_{-}\phi_{-}\phi_{-},\,-\phi_{-}\phi_{+}\phi_{+},\,-\phi_{+}\phi_{-}\phi_{+},\,-\phi_{+}\phi_{+}\phi_{-}\\}\,.$
(545)

We can clearly see that $P^{(-)}=*P^{(+)}$ with $**=1$, where the matrix
representation of $*$ is $\boldsymbol{\mathrm{M}}(*)$

$\boldsymbol{\mathrm{M}}(*)=\begin{pmatrix}0&-1\\\ -1&0\end{pmatrix}$ (546)

In conclusion, when $n=2k$ is even, $K^{(+)}$ and $K^{(-)}$ are non-dual;
while when $n=2k-1$ is odd, $P^{(+)}$ and $P^{(-)}$ are dual to each other.
Notice that dual and non-dual are a duality itself, this is reflects by the
odd and even power respectively. Define
$J_{\mathrm{odd}}=\\{(\phi_{+}-\phi_{-})^{2k-1}|\forall k\in\mathbb{N}\\}$ and
$J_{\mathrm{even}}=\\{(\phi_{+}-\phi_{-})^{2k}|\forall k\in\mathbb{N}\\}$.
They form a duality. This is because, consider the operator
$*=(\phi_{+}+\phi_{-})$,

$\displaystyle*J_{\mathrm{odd}}$
$\displaystyle=(\phi_{+}-\phi_{-})J_{\mathrm{odd}}=(\phi_{+}-\phi_{-})\\{(\phi_{+}+\phi_{-})^{2k-1}|\forall
k\in\mathbb{N}\\}$ (547) $\displaystyle=\\{(\phi_{+}-\phi_{-})^{2k}|\forall
k\in\mathbb{N}\\}$ $\displaystyle=J_{\mathrm{even}}$

Then

$\displaystyle**J_{\mathrm{odd}}$
$\displaystyle=*J_{\mathrm{even}}=(\phi_{+}-\phi_{-})J_{\mathrm{even}}$ (548)
$\displaystyle=(\phi_{+}-\phi_{-})\\{(\phi_{+}-\phi_{-})^{2k}|\forall
k\in\mathbb{N}\\}$ $\displaystyle=\\{(\phi_{+}-\phi_{-})^{2k+1}|\forall
k\in\mathbb{N}\\}$ $\displaystyle=J_{\mathrm{odd}}$

Therefore $**=1$ is an identity map. Since there exists an bijective map
between odd and even numbers, and also that $J_{\mathrm{odd}}\cap
J_{\mathrm{even}}=\emptyset$, therefore $J_{\mathrm{odd}}$ and
$J_{\mathrm{even}}$ is dual to each other. Thus the Lagrangians in 530 and 540
are dual to each other.

Next we would like to construct another possible classification for the terms
in 543. Notice that the following 4 terms are observational dual invariant,
which remains the same regardless the direction looking at it.

$(\phi_{+}\phi_{+}\phi_{+},\,-\phi_{-}\phi_{-}\phi_{-},\,-\phi_{+}\phi_{-}\phi_{+},\,\phi_{-}\phi_{+}\phi_{-}|S_{3})=(\phi_{+}\phi_{+}\phi_{+},\,-\phi_{-}\phi_{-}\phi_{-},\,-\phi_{+}\phi_{-}\phi_{+},\,\phi_{-}\phi_{+}\phi_{-}|S_{3}^{\star})$
(549)

Thus we have the $B$ partition as

$B=\\{\phi_{+}\phi_{+}\phi_{+},\,-\phi_{-}\phi_{-}\phi_{-},\,-\phi_{+}\phi_{-}\phi_{+},\,\phi_{-}\phi_{+}\phi_{-}\\}\,.$
(550)

The remaining $Q$ partition which does not have observation dual invariance
contains all the remaining terms

$Q=\\{-\phi_{+}\phi_{+}\phi_{-},\phi_{-}\phi_{-}\phi_{+},-\phi_{-}\phi_{+}\phi_{+},\phi_{+}\phi_{-}\phi_{-}\\}$
(551)

Mathematically, we can define a parity operator $\hat{P}=(-1)*$ acting on the
third therm or first term of the product, denoting $\hat{P}_{3}=(-1)*_{3}$ or
$\hat{P}_{1}=(-1)*_{1}$ respectively. We can see that

$\displaystyle\hat{P}_{3}B$
$\displaystyle=\\{\phi_{+}\phi_{+}[(-1)*\phi_{+}],\,-\phi_{-}\phi_{-}[(-1)*\phi_{-}],\,-\phi_{+}\phi_{-}[(-1)*\phi_{+}],\,\phi_{-}\phi_{+}[(-1)*\phi_{-}]\\}$
(552)
$\displaystyle=\\{-\phi_{+}\phi_{+}\phi_{-},\phi_{-}\phi_{-}\phi_{+},-\phi_{-}\phi_{+}\phi_{+},\phi_{+}\phi_{-}\phi_{-}\\}$
$\displaystyle=Q\,.$

One can see that $\hat{P}_{3}^{2}=((-1)*_{3})^{2}=(-1)(-1)*_{3}*_{3}=1$ which
is the identity. Next, we also see that

$\displaystyle\hat{P}_{1}B$
$\displaystyle=\\{[(-1)*\phi_{+}]\phi_{+}\phi_{+},\,-[(-1)*\phi_{-}]\phi_{-}\phi_{-},\,-[(-1)*\phi_{+}]\phi_{-}\phi_{+},\,[(-1)*\phi_{-}]\phi_{+}\phi_{-}\\}$
(553)
$\displaystyle=\\{-\phi_{-}\phi_{+}\phi_{+},\,\phi_{+}\phi_{-}\phi_{-},\,\phi_{-}\phi_{-}\phi_{+},\,-\phi_{+}\phi_{+}\phi_{-}\\}$
$\displaystyle=Q\,.$

And we have $\hat{P}_{1}^{2}=1$. Therefore, $B$ and $Q$ are dual to each other
under such parity maps. We interpret as, $B$ is a dual invariant while $Q$ is
a non-dual invariant under observation, such that $Q$ is dual to $B$.

The equation of motion for the $n-$th order interaction term is the following

$\Box\phi_{+}=n(\phi_{+}-\phi_{-})^{n-1}\quad\text{and}\quad\Box\phi_{-}=-n(\phi_{+}-\phi_{-})^{n-1}\,.$
(554)

Together we have

$\Box(\phi_{+}+\phi_{-})=0\,.$ (555)

To cover all the diagrams of different order for $g^{k}$, we demand a full
theory as follow

$\displaystyle\mathcal{L}$
$\displaystyle=\frac{1}{2}\partial_{\mu}\phi_{+}\partial^{\mu}\phi_{+}+\frac{1}{2}\partial_{\mu}\phi_{-}\partial^{\mu}\phi_{-}-\sum_{k=2}^{n}g_{k}(\phi_{+}+\phi_{-})^{k}$
(556)
$\displaystyle=\frac{1}{2}\partial_{\mu}\phi_{+}\partial^{\mu}\phi_{+}+\frac{1}{2}\partial_{\mu}\phi_{-}\partial^{\mu}\phi_{-}-\sum_{k=2}^{n}\sum_{i_{1},i_{2},\cdots,i_{k}\in
W}g_{k}\prod_{l=1}^{k}\phi_{i_{l}}\,.$

The the Lagrangian is invariant under the global degraded symmetry group of

$\mathbb{Z}_{2}\oplus(\mathbb{Z}_{2}\otimes\mathbb{Z}_{2})\oplus(\mathbb{Z}_{2}\otimes\mathbb{Z}_{2}\oplus\mathbb{Z}_{2})\otimes\cdots=\bigoplus_{k=1}^{n}\bigotimes_{l=1}^{k}\mathbb{Z}_{2}\,.$
(557)

Finally we would like to study the case of heterogeneous field case. We have
the following Lagrangian,

$\mathcal{L}=\frac{1}{2}\sum_{a=+,-}\sum_{i=1}^{n}\partial_{\mu}\phi_{ai}\partial^{\mu}\phi_{ai}-g\prod_{i=1}^{n}(\pm\phi_{+i}\pm\phi_{-i})\,,$
(558)

and

$\mathcal{L}=\frac{1}{2}\sum_{a=+,-}\sum_{i=1}^{n}\partial_{\mu}\phi_{ai}\partial^{\mu}\phi_{ai}-g\prod_{i=1}^{n}(\pm\phi_{+i}\mp\phi_{-i})\,,$
(559)

Using $n=2$ as an example, and the interaction term as
$(\phi_{+i}-\phi_{-i})^{2}$, we have four scalar fields in the Lagrangian

$\displaystyle\mathcal{L}$
$\displaystyle=\frac{1}{2}\partial_{\mu}\phi_{+1}\partial^{\mu}\phi_{+1}+\frac{1}{2}\partial_{\mu}\phi_{-1}\partial^{\mu}\phi_{-1}+\frac{1}{2}\partial_{\mu}\phi_{+2}\partial^{\mu}\phi_{+2}+\frac{1}{2}\partial_{\mu}\phi_{-2}\partial^{\mu}\phi_{-2}$
(560)
$\displaystyle-g(\phi_{+1}\phi_{+2}-\phi_{-1}\phi_{-2}-\phi_{+1}\phi_{-2}+\phi_{1-}\phi_{+2})$

The full Lagrangian for all orders will be, take the (+,+) for example,

$\mathcal{L}=\frac{1}{2}\sum_{a=+,-}\sum_{i=1}^{n}\partial_{\mu}\phi_{ai}\partial^{\mu}\phi_{ai}-g\sum_{k=1}^{n}\prod_{i=1}^{k}(\phi_{+i}+\phi_{-i})\,.$
(561)

##  9 Dual Invariant Topological Field Theory

The previous section investigates dual invariant interaction. In this chapter,
we would like to give a formal study of dual invariant action. Let’s suppose
we define the observation duality as left and right observation perspective,
$L$ and $R$. We want to construct an action which remains the same whenever
you look from the left or look from the right, i.e. mathematically

$(S|L)\equiv(S|R)\,,$ (562)

such that $S=S^{*}$.

The scalar field action takes the generic form of

$S=\frac{1}{2}\int d^{4}x\,\partial_{\mu}\phi\eta^{\mu\nu}\partial_{\nu}\phi$
(563)

is not dual invariant. First of all, $\partial_{\mu}\phi\partial^{\mu}\phi$,
when looking on the left is $-\phi\partial^{\mu}\phi\partial_{\mu}$, which is
an operator and not a scalar. If we carry out integration by parts,

$S=\frac{1}{2}\int d^{4}x(-\phi\Box\phi)$ (564)

It is tempted to think that $\phi\Box\phi$ is dual invariant, however bare in
mind that the $\Box$ operator when looking from the right, it is $-\Box$
instead of $\Box$ due to the minus signs in the $\Box$ operators. Also the
negative sign in front makes the action impossible to be dual invariant. Next
if we include a mass term of $-\frac{1}{2}m^{2}\phi^{2}$, this is not dual
invariant due to the negative sign. Similar reason goes to the action of gauge
field,

$S=-\frac{1}{4}\int d^{4}x\,F_{\mu\nu}F^{\mu\nu}=\frac{1}{2}\int
d^{4}xA^{\mu}(\Box\eta_{\mu\nu}-\partial_{\mu}\partial_{\nu})A^{\nu}=\frac{1}{2}\int
d^{4}x\,(\boldsymbol{\mathrm{E}}^{2}-\boldsymbol{\mathrm{B}}^{2})\,,$ (565)

which is non-dual invariant due to the minus sign and the $\Box$ operator. Or
we can see that the term
$\boldsymbol{\mathrm{E}}^{2}-\boldsymbol{\mathrm{B}}^{2}$ is non-dual
invariant due to the minus sign as in the form of $a-b$. If we add a mass term
$\frac{1}{2}m^{2}A_{\mu}A^{\mu}$, although we have a positive sign, the
$A_{\mu}A^{\mu}$ would induce a negative sign when we look from the right.

The Dirac action is also not dual invariant. The term
$i\bar{\psi}\gamma^{\mu}\partial_{\mu}\psi$ has a structure of row times
matrix times column, but when we look form the right we have something column
times matrix times a row, which is not possible. For the mass term
$-m\bar{\psi}\psi$, it has a minus sign it is immediately non-dual invariant.

In other words, we come into a conclusion that the standard model action is
not dual invariant. One of the major problems is due to the negative signs in
the metric, so the flat Minkowski metric tensor is the problematic cause of
non-dual invariance. This shows that the intrinsic structure of spacetime
itself is naturally non-dual invariant.

Generally speaking, a dual invariant action in the left and right observation
perspective must be positive, and it should possess $ada$ form if it contains
derivatives. It should also be independent of the metric tensor.

The most possible candidate so far we know is the Chern-Simon’s action. We
will show that in fact it is dual invariant. The Chern-Simons action is given
by

$S_{\mathrm{CS}}=\int_{M}d^{3}x\,\epsilon^{\mu\nu\rho}A_{\mu}\partial_{\nu}A_{\rho}\,,$
(566)

where $M$ is the topological manifold. We can check that it is left, right
observation dual invariant by expanding the terms in the action explicitly,

$\displaystyle S_{\mathrm{CS}}$
$\displaystyle=\int_{M}d^{3}x\,\Big{(}(\epsilon^{012}A_{0}\partial_{1}A_{2}+\epsilon^{210}A_{2}\partial_{1}A_{0})+(\epsilon^{120}A_{1}\partial_{2}A_{0}+\epsilon^{021}A_{0}\partial_{2}A_{1})$
(567)
$\displaystyle\quad\quad\quad\quad\quad\quad\quad\quad\quad+(\epsilon^{201}A_{2}\partial_{0}A_{1}+\epsilon^{102}A_{1}\partial_{0}A_{2})\Big{)}$
$\displaystyle=\int_{M}d^{3}x\,\Big{(}(A_{0}\partial_{1}A_{2}-A_{2}\partial_{1}A_{0})+(A_{1}\partial_{2}A_{0}-A_{0}\partial_{2}A_{1})$
$\displaystyle\quad\quad\quad\quad\quad\quad\quad\quad\quad+(A_{2}\partial_{0}A_{1}-A_{1}\partial_{0}A_{2})\Big{)}$

So we have terms of $A_{a}\partial_{b}A_{c}-A_{c}\partial_{b}A_{a}$, which is
dual invariant. Therefore the Chern-Simons action is dual invariant,
$(S_{\mathrm{CS}}|L)\equiv(S_{\mathrm{CS}}|R)$.

The Chern-Simons action is topological, which is independent of the metric, so
it perfectly fits our necessary criteria. The dual invariant nature of the
Chern-Simons action echoes the idea in section 2.1.6 (General Analysis and
Topological Duality) that, if a system is dual invariant, its corresponding
geometry is topological.

The most general Chern-Simons action takes the following form,

$S_{\mathrm{CS}}=\int_{M}d^{3}x\,\kappa\epsilon^{\mu\nu\rho}A_{\mu}\partial_{\nu}A_{\rho}\,,$
(568)

where $\kappa$ is some constant which can be positive or negative. It is noted
that if $\kappa$ is negative, let’s say $-1$, the Chern-Simons action becomes,

$\displaystyle S_{\mathrm{CS}}$
$\displaystyle=\int_{M}d^{3}x\,\Big{(}(A_{2}\partial_{1}A_{0}-A_{0}\partial_{1}A_{2})+(A_{0}\partial_{2}A_{1}-A_{1}\partial_{2}A_{0})$
(569)
$\displaystyle\quad\quad\quad\quad\quad\quad\quad\quad\quad+(A_{1}\partial_{0}A_{2}-A_{2}\partial_{0}A_{1})\Big{)}\,.$

Hence now we have terms of $A_{c}\partial_{b}A_{a}-A_{a}\partial_{b}A_{c}$,
which is another dual invariant. Therefore no matter $\kappa$ is positive or
negative, the Chern-Simons action is still dual invariant. Not that we cannot
add a mass term $\frac{1}{2}m^{2}A_{\mu}A^{\mu}$ nor add an external source
term coupling to to the gauge field $A_{\mu}J^{\mu}$ as both of these terms
depend on the metric, this will break the dual invariant symmetry. In other
words, in order to preserve dual invariance, the gauge field particle must be
massless. In order to gain mass, there must be involving some new mass
acquisition mechanisms.

The dual invariance of the Chern-Simons action can be proved by integration by
parts. Consider

$\displaystyle(S_{\mathrm{CS}}|L)$
$\displaystyle=\int_{M}d^{3}x\,\epsilon^{\mu\nu\rho}A_{\mu}\partial_{\nu}A_{\rho}$
(570)
$\displaystyle=\mathrm{B.C.}-\int_{M}d^{3}x\,\epsilon^{\mu\nu\rho}A_{\rho}\partial_{\nu}A_{\mu}$
$\displaystyle=-\int_{M}d^{3}x\,(-1)^{3}\epsilon^{\rho\nu\mu}A_{\rho}\partial_{\nu}A_{\mu}$
$\displaystyle=\int_{M}d^{3}x\,\epsilon^{\rho\nu\mu}A_{\rho}\partial_{\nu}A_{\mu}$
$\displaystyle=(S_{\mathrm{CS}}|R)\,,$

where we have dropped the boundary term. We can see that
$A_{\mu}\partial_{\nu}A_{\rho}\rightarrow-A_{\rho}\partial_{\nu}A_{\mu}$ and
$\epsilon^{\mu\nu\rho}\rightarrow-\epsilon^{\rho\nu\mu}$, so when we exchange
perspectives of the two terms from $L$ to $R$, each term will induce a
negative sign and the two negative signs become positive again. Thus we prove
that the Chern-Simons action is dual invariant under the left-right
perspective.

For non-Abelian case, where the gauge field are matrices and suppose
$A_{\mu}=A_{\mu}^{a}t^{a}$ where $t^{a}$ are the SU$(N)$ generators the dual
invariance still holds,

$S_{\mathrm{CS}}=\int_{M}d^{3}x\,\mathrm{Tr}(\kappa\epsilon^{\mu\nu\rho}A_{\mu}\partial_{\nu}A_{\rho})=\int_{M}d^{3}x\,\kappa\epsilon^{\mu\nu\rho}A^{a}_{\mu}\partial_{\nu}A^{a}_{\rho}$
(571)

If the gauge field is some generic matrix, for example Berry connection for
the degenerate case, we have

$A_{\mu mn}=\langle\psi_{m}|\partial_{\mu}\psi_{n}\rangle\,.$ (572)

Then the Chern-Simons action is

$\displaystyle
S_{\mathrm{CS}}=\int_{M}d^{3}x\,\kappa\sum_{m}\sum_{n}\epsilon_{\mu\nu\rho}A^{\mu}_{mn}\partial^{\nu}A^{\rho}_{nm}\,,$
(573)

which is obviously dual invariant. Explicitly, this is

$\displaystyle S_{\mathrm{CS}}$
$\displaystyle=\int_{M}d^{3}x\,\kappa\sum_{m}\sum_{n}\epsilon_{\mu\nu\rho}\langle\psi_{m}|\partial^{\mu}\psi_{n}\rangle\partial^{\nu}\langle\psi_{n}|\partial^{\mu}\psi_{m}\rangle$
(574)
$\displaystyle=\int_{M}d^{3}x\,\kappa\sum_{m}\sum_{n}\epsilon_{\mu\nu\rho}\langle\psi_{m}|\partial^{\mu}\psi_{n}\rangle(\langle\partial^{\nu}\psi|\partial^{\rho}\psi_{n}\rangle+\langle\psi_{n}|\partial^{\nu}\partial^{\rho}\psi_{m}\rangle)$
$\displaystyle=\int_{M}d^{3}x\,\kappa\sum_{m}\sum_{n}\epsilon_{\mu\nu\rho}\langle\psi_{m}|\partial^{\mu}\psi_{n}\rangle\langle\partial^{\nu}\psi|\partial^{\rho}\psi_{n}\rangle\,.$

The last term vanishes because $\partial_{\nu}\partial_{\rho}$ is symmetric.

The most generic Chern-Simons Three form endowed in SU(N) Lie algebra takes
the following action:

$S_{\mathrm{CS}}=\kappa\int_{M}\mathrm{Tr}(A\wedge dA+\frac{2}{3}A\wedge
A\wedge A)\,.$ (575)

The $A$ is gauge-field 1-form, $dA$ is the exterior on $A$ which is a two-
form. Explicitly we have,

$\displaystyle
S_{\mathrm{CS}}=\kappa\int_{M}\mathrm{Tr}\Big{(}(A^{a}_{\mu}t^{a}dx^{\mu})\wedge\partial_{\rho}A_{\nu}^{b}t^{b}dx^{\rho}\wedge
dx^{\nu}+\frac{2}{3}A_{\mu}^{a}t^{a}A_{\nu}^{b}t^{b}A_{\rho}^{c}t^{c}dx^{\mu}\wedge
dx^{\nu}\wedge dx^{\rho}\Big{)}$ (576)

By relabelling indices and grouping terms, we have

$S_{\mathrm{CS}}=\kappa\int_{M}\mathrm{Tr}\Big{(}A_{\mu}^{a}\partial_{\nu}A_{\rho}^{b}t^{a}t^{b}dx^{\mu}\wedge
dx^{\nu}\wedge
dx^{\rho}+\frac{2}{3}A_{\mu}^{a}A_{\nu}^{b}A_{\rho}^{c}t^{a}t^{b}t^{c}dx^{\mu}\wedge
dx^{\nu}\wedge dx^{\rho}\Big{)}\,.$ (577)

Next using using the the standard form of wedge product

$d^{n}x=\frac{1}{n!}\epsilon_{\mu_{1}\mu_{2}\cdots\mu_{n}}dx^{\mu_{1}}\wedge
dx^{\mu_{2}}\wedge\cdots\wedge dx^{\mu_{n}}\,.$ (578)

Then we use the fact that

$\epsilon^{\mu_{1}\mu_{2}\cdots\mu_{n}}\epsilon_{\mu_{1}\mu_{2}\cdots\mu_{n}}=n!\,.$
(579)

Therefore

$\epsilon^{\mu_{1}\mu_{2}\cdots\mu_{n}}d^{n}x=dx^{\mu_{1}}\wedge
dx^{\mu_{2}}\wedge\cdots\wedge dx^{\mu_{n}}\,.$ (580)

Therefore, 577 simplifies into

$S_{\mathrm{CS}}=\kappa\int_{M}d^{3}x\,\epsilon^{\mu\nu\rho}\mathrm{Tr}\Big{(}A_{\mu}^{a}\partial_{\nu}A_{\rho}^{b}t^{a}t^{b}+\frac{2}{3}A_{\mu}^{a}A_{\nu}^{b}A_{\rho}^{c}t^{a}t^{b}t^{c}\Big{)}$
(581)

Next, we use the trace identity,

$\mathrm{Tr}(t^{a}t^{b})=\frac{1}{2}\delta^{ab}\quad\text{and}\quad\mathrm{Tr}(t^{a}t^{b}t^{c})=\frac{1}{4}(d^{abc}+if^{abc}).$
(582)

Therefore finally we get,

$S_{\mathrm{CS}}=\frac{\kappa}{2}\int_{M}d^{3}x\,\epsilon^{\mu\nu\rho}\Big{(}A_{\mu}^{a}\partial_{\nu}A_{\rho}^{a}+\frac{1}{3}if^{abc}A_{\mu}^{a}A_{\nu}^{b}A_{\rho}^{c}\Big{)}\,.$
(583)

where $d^{abc}$ does not contribute as it is a totally symmetric tensor. For
the first term we have already shown that it is dual invariant, the remain
task to show is the $AAA$ term is dual invariant. First notice that

$\frac{1}{3}i\epsilon^{\mu\nu\rho}f^{abc}A_{\mu}^{a}A_{\nu}^{b}A_{\rho}^{c}=\frac{1}{3}i(-\epsilon^{\rho\nu\mu})(-f^{cba})A_{\rho}^{c}A_{\nu}^{b}A_{\mu}^{a}=\frac{1}{3}i\epsilon^{\rho\nu\mu}f^{cba}A_{\rho}^{c}A_{\nu}^{b}A_{\mu}^{a}$
(584)

Thus the last term is left-right dual invariant.

The dual invariant action is not unique to the Chern-Simons 3-form. It can
involve more derivatives, for example $adada$, which is also a left-right dual
invariant. The follow action is dual invariant:

$S=\int_{M}d^{5}x\,\epsilon^{\mu\nu\rho\alpha\beta}A_{\mu}\partial_{\nu}A_{\rho}\partial_{\alpha}A_{\beta}\,.$
(585)

The sum has a total of $5!=120$ terms. If we analyse it in detail, we can
write the action explicitly as follow:

$\displaystyle S$
$\displaystyle=\int_{M}d^{5}x\,\Big{(}\sum_{\mu\nu\alpha\beta}\epsilon^{\mu\nu
0\alpha\beta}A_{\mu}\partial_{\nu}A_{0}\partial_{\alpha}A_{\beta}+\sum_{\mu\nu\alpha\beta}\epsilon^{\mu\nu
1\alpha\beta}A_{\mu}\partial_{\nu}A_{1}\partial_{\alpha}A_{\beta}+\sum_{\mu\nu\alpha\beta}\epsilon^{\mu\nu
2\alpha\beta}A_{\mu}\partial_{\nu}A_{2}\partial_{\alpha}A_{\beta}$ (586)
$\displaystyle\quad\quad\quad\quad\quad\quad+\sum_{\mu\nu\alpha\beta}\epsilon^{\mu\nu
3\alpha\beta}A_{\mu}\partial_{\nu}A_{3}\partial_{\alpha}A_{\beta}+\sum_{\mu\nu\alpha\beta}\epsilon^{\mu\nu
4\alpha\beta}A_{\mu}\partial_{\nu}A_{4}\partial_{\alpha}A_{\beta}\Big{)}\,.$

Take the first term as an example,

$\sum_{\mu\nu\alpha\beta}\epsilon^{\mu\nu
0\alpha\beta}A_{\mu}\partial_{\nu}A_{0}\partial_{\alpha}A_{\beta}\,.$ (587)

There will be $5!/5=24$ terms, corresponding to $24/2=12$ pairs. We will list
our the 12 pairs as an example for the $\rho=0$ case, it will be similar for
the remaining $\rho=1,2,3,4$ case.

$L-$perspective |  $R-$perspective
---|---
$\epsilon^{12034}$ | $\epsilon^{43021}$
$\epsilon^{21034}$ | $\epsilon^{43012}$
$\epsilon^{12043}$ | $\epsilon^{34021}$
$\epsilon^{21043}$ | $\epsilon^{34012}$
$\epsilon^{13024}$ | $\epsilon^{42031}$
$\epsilon^{31024}$ | $\epsilon^{42013}$
$\epsilon^{13042}$ | $\epsilon^{24031}$
$\epsilon^{31042}$ | $\epsilon^{24013}$
$\epsilon^{14023}$ | $\epsilon^{32041}$
$\epsilon^{41023}$ | $\epsilon^{32014}$
$\epsilon^{14032}$ | $\epsilon^{23041}$
$\epsilon^{41032}$ | $\epsilon^{23014}$

Since noting that

$\epsilon^{\mu\nu\rho\alpha\beta}=(-1)^{4}(-1)^{3}(-1)^{2}(-1)^{1}\epsilon^{\beta\alpha\rho\nu\mu}=\epsilon^{\beta\alpha\rho\nu\mu}\,,$
(588)

so the terms in left perspective and right perspective are also +1. For
example, we will have

$\epsilon^{12034}A_{1}\partial_{2}A_{0}\partial_{3}A_{4}+\epsilon^{43021}A_{4}\partial_{3}A_{0}\partial_{2}A_{1}=A_{1}\partial_{2}A_{0}\partial_{3}A_{4}+A_{4}\partial_{3}A_{0}\partial_{2}A_{1}$
(589)

which is obviously an left-right dual invariant. For example,

$\epsilon^{21034}A_{2}\partial_{1}A_{0}\partial_{3}A_{4}+\epsilon^{43012}A_{4}\partial_{3}A_{0}\partial_{2}A_{1}=-A_{2}\partial_{1}A_{0}\partial_{3}A_{4}-A_{4}\partial_{3}A_{0}\partial_{1}A_{2}$
(590)

is also dual invariant.

We can in fact prove that this action is dual invariant using integration by
parts.

$\displaystyle(S|L)$
$\displaystyle=\int_{M}d^{5}x\,\epsilon^{\mu\nu\rho\alpha\beta}A_{\mu}\partial_{\nu}A_{\rho}\partial_{\alpha}A_{\beta}$
(591)
$\displaystyle=\mathrm{B.C.}-\int_{M}d^{5}x\,\epsilon^{\mu\nu\rho\alpha\beta}A_{\beta}\partial_{\alpha}(A_{\mu}\partial_{\nu}A_{\rho})$
$\displaystyle=-\int_{M}d^{5}x\,\epsilon^{\mu\nu\rho\alpha\beta}A_{\beta}(\partial_{\alpha}A_{\mu}\partial_{\nu}A_{\rho}+A_{\mu}\partial_{\alpha}\partial_{\nu}A_{\rho})$
$\displaystyle=-\int_{M}d^{5}x\,\epsilon^{\mu\nu\rho\alpha\beta}A_{\beta}\partial_{\alpha}A_{\mu}\partial_{\nu}A_{\rho}$
$\displaystyle=-\int_{M}d^{5}x\,(-1)^{4}\epsilon^{\beta\mu\nu\rho\alpha}A_{\beta}\partial_{\alpha}A_{\mu}\partial_{\nu}A_{\rho}$
$\displaystyle=-\int_{M}d^{5}x\,(-1)^{4}\epsilon^{\beta\rho\nu\mu\alpha}A_{\beta}\partial_{\alpha}A_{\rho}\partial_{\nu}A_{\mu}$
$\displaystyle=-\int_{M}d^{5}x\,(-1)^{4}(-1)^{3}\epsilon^{\beta\alpha\rho\nu\mu}A_{\beta}\partial_{\alpha}A_{\rho}\partial_{\nu}A_{\mu}$
$\displaystyle=\int_{M}d^{5}x\,\epsilon^{\beta\alpha\rho\nu\mu}A_{\beta}\partial_{\alpha}A_{\rho}\partial_{\nu}A_{\mu}$
$\displaystyle=(S|R)\,,$

where we have dropped the boundary term.

Finally, we would like to investigate the action with chiral anomaly, which is
given by the following:

$S_{A}=\frac{\alpha}{4\pi}\int_{M}d^{4}x\,\theta\epsilon^{\mu\nu\alpha\beta}F_{\mu\nu}F_{\alpha\beta}=\frac{\alpha\theta}{\pi}\int_{M}d^{4}x\,\boldsymbol{\mathrm{E}}\cdot\boldsymbol{\mathrm{B}}\,,$
(592)

where $\alpha$ is the fine-structure constant, $\theta\in[0,2\pi]$ is a
constant. This action must be dual-invariant because

$(\boldsymbol{\mathrm{E}}\cdot\boldsymbol{\mathrm{B}}|L)\equiv(\boldsymbol{\mathrm{B}}\cdot\boldsymbol{\mathrm{E}}|R)\,,$
(593)

where
$\boldsymbol{\mathrm{E}}\cdot\boldsymbol{\mathrm{B}}=\boldsymbol{\mathrm{B}}\cdot\boldsymbol{\mathrm{E}}$.
At the first sight, the term
$\epsilon^{\mu\nu\alpha\beta}F_{\mu\nu}F_{\alpha\beta}$ is non-dual invariant,
but in fact we can prove that it is a dual invariant using divergence theorem.
First, we rewrite the action as follow:

$\displaystyle S_{A}$
$\displaystyle=\frac{\alpha}{4\pi}\int_{M}d^{4}x\,\theta\epsilon^{\mu\nu\alpha\beta}F_{\mu\nu}F_{\alpha\beta}$
(594)
$\displaystyle=\frac{\alpha\theta}{4\pi}\int_{M}d^{4}x\,\epsilon^{\mu\nu\alpha\beta}(\partial_{\mu}A_{\nu}-\partial_{\nu}A_{\mu})(\partial_{\rho}A_{\sigma}-\partial_{\sigma}A_{\rho})$
$\displaystyle=\frac{\alpha\theta}{4\pi}\int_{M}d^{4}x\,\epsilon^{\mu\nu\alpha\beta}(2\partial_{\mu}A_{\nu})(2\partial_{\rho}A_{\sigma})$
$\displaystyle=\frac{\alpha\theta}{\pi}\int_{M}d^{4}x\,\partial_{\mu}(\epsilon^{\mu\nu\rho\sigma}A_{\nu}\partial_{\rho}A_{\sigma})$
$\displaystyle=\frac{\alpha\theta}{\pi}\int_{\partial
M}d^{3}x\,n_{\mu}\epsilon^{\mu\nu\rho\sigma}A_{\nu}\partial_{\rho}A_{\sigma}$
$\displaystyle=\frac{\alpha\theta}{\pi}\int_{\partial
M}d^{3}x\,(n_{0}\epsilon^{0\nu\rho\sigma}A_{\nu}\partial_{\rho}A_{\sigma}+n_{1}\epsilon^{1\nu\rho\sigma}A_{\nu}\partial_{\rho}A_{\sigma}+n_{2}\epsilon^{2\nu\rho\sigma}A_{\nu}\partial_{\rho}A_{\sigma}+n_{3}\epsilon^{3\nu\rho\sigma}A_{\nu}\partial_{\rho}A_{\sigma})\,,$

where $n_{\mu}$ is the unit vector normal to the 3D volume surface. Let’s
investigate the first term. We have for the left perspective,

$\displaystyle\quad\frac{\alpha\theta}{\pi}\int_{\partial
M}d^{3}x\,\Big{(}(n_{0}\epsilon^{0123}A_{1}\partial_{2}A_{3}+n_{0}\epsilon^{0321}A_{3}\partial_{2}A_{1})+(n_{0}\epsilon^{0312}A_{3}\partial_{1}A_{2}+n_{0}\epsilon^{0213}A_{2}\partial_{1}A_{3})+$
(595)
$\displaystyle\quad\quad\quad\quad\quad\quad\quad\quad(n_{0}\epsilon^{0231}A_{2}\partial_{3}A_{1}+n_{0}\epsilon^{0132}A_{1}\partial_{3}A_{2})\Big{)}$
$\displaystyle=\frac{\alpha\theta}{\pi}\int_{\partial
M}d^{3}x\,\Big{(}(n_{0}A_{1}\partial_{2}A_{3}-n_{0}A_{3}\partial_{2}A_{1})+(n_{0}A_{3}\partial_{1}A_{2}-n_{0}A_{2}\partial_{1}A_{3})+$
$\displaystyle\quad\quad\quad\quad\quad\quad\quad\quad(n_{0}A_{2}\partial_{3}A_{1}-n_{0}A_{1}\partial_{3}A_{2})\Big{)}$
$\displaystyle=\frac{\alpha\theta}{\pi}\int_{\partial
M}d^{3}x\,n_{0}\Big{(}(A_{1}\partial_{2}A_{3}-A_{3}\partial_{2}A_{1})+(A_{3}\partial_{1}A_{2}-A_{2}\partial_{1}A_{3})+$
$\displaystyle\quad\quad\quad\quad\quad\quad\quad\quad(A_{2}\partial_{3}A_{1}-A_{1}\partial_{3}A_{2})\Big{)}$
$\displaystyle=\frac{\alpha\theta}{\pi}\int_{\partial
M}d^{3}x\,\Big{(}(A_{1}\partial_{2}A_{3}-A_{3}\partial_{2}A_{1})+(A_{3}\partial_{1}A_{2}-A_{2}\partial_{1}A_{3})$
$\displaystyle\quad\quad\quad\quad\quad\quad\quad\quad(A_{2}\partial_{3}A_{1}-A_{1}\partial_{3}A_{2})\Big{)}n_{0}\,,$

which is the same as looking from the right perspective. The same idea goes to
the second, third and fourth term. Therefore, the chiral anomaly action is
left-right dual invariant, and it is topological.

##  10 New exotic matter state-Dualiton

In this section, we will study the the duality wave oscillation, its promotion
to duality field, and how is second quantization give rise to the quasi
particle which we call the dualiton. Dualiton is a new quasi particle which
arises from the theory of duality, and is regarded as the quantization of the
states with $\mathbb{Z}_{2}\times\mathbb{Z}_{2}$ symmetry.

First, we will use the mathematics of 4-box tableaux we constructed in section
3. Let $U=\\{u\\}$ and $U^{*}=\\{u^{*}\\}$. We have the following

Figure 103:

The fullness is considered as $U\cup U^{*}$. By taking the negation,

$*(U\cup U^{*})=*U\cap*U^{*}=*U\cap U=U^{*}\cap U=\emptyset\,,$ (596)

where in the second step we have used the de-Morgan’s theorem. For example,
let $u=\text{creation}$, $u^{*}=\text{annihilation}$, the fullness is creation
or annihilation. Not creation or annihilation is nor creation and nor
annihilation, which we will prove here is indeed the empty set. Here we have
also excluded the middle of $\\{0\\}$ as we do not include it in our
definition, so the exclusion of middle is automatically satisfied.

First we would like investigate some their properties. We have respectively
the full set and the null set as

$W=U\cup U^{*}\quad\text{and}\quad\emptyset=U\cap U^{*}\,.$ (597)

Explicitly,

$W=\\{u,u^{*}\\}\quad\text{and}\quad\emptyset=\\{\\}\,.$ (598)

The respective cardinality is $|W|=2$ and $|\emptyset|=0$. The full set and
the empty set are dual to each other,

$\emptyset=*W\quad\text{and}\quad W=*\emptyset\,,$ (599)

which is shown by the de-Morgan’s law. In a detailed look, this can be
manifested as

$(1)\emptyset=(*)W\quad\text{and}\quad(1)W=(*)\emptyset\,,$ (600)

where $1$ is the identity operator element while $*$ is the dual operator
element. We can take $u=1$ and $u^{*}=*$ and the abstract perspective
$S_{k}=W$ and $S_{k}^{\star}=\emptyset$, then 599 means to be the standard
form,

$(1|\emptyset)\equiv(*|W)\quad\text{and}\quad(1|W)\equiv(*|\emptyset)\,.$
(601)

In a compact form, this is just the standard relation we had before,

$(1|0)\equiv(0|1)\quad\text{and}\quad(1|1)\equiv(0|0)\,.$ (602)

Similarly, for the half sets,

$U=\\{u\\}\quad\text{and}\quad U^{*}=\\{u^{*}\\}\,.$ (603)

The two half sets are dual to each other,

$U^{*}=*U\quad\text{and}\quad U=*U^{*}\,.$ (604)

Explicitly this means

$(1|U^{*})\equiv(*|U)\quad\text{and}\quad(1|U)\equiv(*|U^{*})\,.$ (605)

and the compact form shares the one in 602. Element-wise, figure 103 can be
represented as follows,

Figure 104:

###  10.1 The canonical 4-duality structure

Next, we define the 4-canonical structure as follows ,

$\psi=\sum_{i=0,1}^{\oplus}\sum_{j=0,1}^{\oplus}(\hat{P}_{i}\oplus\hat{P}_{j})(|u\rangle\oplus|u^{*}\rangle)\,,$
(606)

where $\hat{P}_{0}=\hat{e}$ and $\hat{P}_{1}=\star\hat{e}$. Explicitly, the
sum is

$\psi=(\hat{e}|u\rangle\oplus\hat{e}|u^{*}\rangle)\oplus(\hat{e}|u\rangle\oplus\star\hat{e}|u^{*}\rangle)\oplus(\star\hat{e}|u\rangle\oplus
e|u^{*}\rangle)\oplus(\star\hat{e}|u\rangle\oplus\star\hat{e}|u^{*}\rangle)\,.$
(607)

First consider the first statement of negation: “nor created and nor
annihilated”. In terms of state, we have $|u\rangle=|\text{created}\rangle$
and its dual state $|u^{*}\rangle=|\text{annihilated}\rangle$. The negation
nor operator is defined as $\hat{P}_{1}=\star\hat{e}=\hat{-}$. So the
statement of “nor created and nor annihilated” is mathematically represented
as

$\displaystyle\quad\star\hat{e}(|\text{created}\rangle\oplus|\text{annihilated}\rangle)$
(608)
$\displaystyle=(\star\hat{e}\oplus\star\hat{e})(|\text{created}\rangle\oplus|\text{annihilated}\rangle)$
$\displaystyle=\star\hat{e}|\text{created}\rangle\oplus\star\hat{e}|\text{annihilated}\rangle$
$\displaystyle=\hat{-}|\text{created}\rangle\oplus\hat{-}|\text{annihilated}\rangle\,.$

However, there also exists a state with the dual of it. The normal operator is
defined as $\hat{P}_{0}=\star\hat{e}\star\hat{e}=\hat{e}=\hat{+}$,

$\displaystyle\quad\star\hat{e}\big{(}\star\hat{e}(|\text{created}\rangle\oplus|\text{annihilated}\rangle)\big{)}$
(609)
$\displaystyle=(\star\hat{e}\oplus\star\hat{e})(\star\hat{e}\oplus\star\hat{e})(|\text{created}\rangle\oplus|\text{annihilated}\rangle)$
$\displaystyle=\hat{e}|\text{created}\rangle\oplus\hat{e}|\text{annihilated}\rangle$
$\displaystyle=\hat{+}|\text{created}\rangle\oplus\hat{+}|\text{annihilated}\rangle\,.$

This is the statement of “being created and being annihilated”, which is the
negation statement of “nor created and nor annihilated”.

Finally we have two more states, “being created and nor annihilated” and its
negation “nor created and being annihilated”, which are mathematically,
respectively,

$\hat{+}|\text{created}\rangle\oplus\hat{-}|\text{annihilated}\rangle\quad\text{and}\quad\hat{-}|\text{created}\rangle\oplus\hat{+}|\text{annihilated}\rangle$
(610)

Therefore, the full state consists of two dual pairs as follow,

$\displaystyle\psi$
$\displaystyle=(\hat{+}|\text{created}\rangle\oplus\hat{+}|\text{annihilated}\rangle)\oplus(\hat{-}|\text{created}\rangle\oplus\hat{-}|\text{annihilated}\rangle)$
(611)
$\displaystyle\quad\oplus(\hat{+}|\text{created}\rangle\oplus\hat{-}|\text{annihilated}\rangle)\oplus(\hat{-}|\text{created}\rangle\oplus\hat{+}|\text{annihilated}\rangle)$

Or neatly we write symbolically as a canonical 4-duality structure,

$\psi=(\hat{+}|u\rangle\oplus\hat{+}|u^{*}\rangle)\oplus(\hat{-}|u\rangle\oplus\hat{-}|u^{*}\rangle)\oplus(\hat{+}|u\rangle\oplus\hat{-}|u^{*}\rangle)\oplus(\hat{-}|u\rangle\oplus\hat{+}|u^{*}\rangle),$
(612)

In terms of 4-tableau, we have

Figure 105:

Therefore, $\hat{-}|u\rangle\oplus\hat{-}|u^{*}\rangle$ state corresponds to
the Null state, which is one of the four states of the canonical structure. On
the other hand, its dual is the illustration of the
$\hat{+}|u\rangle\oplus\hat{+}|u^{*}\rangle$ state. The
$(\hat{+}|\text{created}\rangle\oplus\hat{+}|\text{annihilated}\rangle)$ state
means that appearance is created and annihilated, Then
$(\hat{-}|\text{created}\rangle\oplus\hat{-}|\text{annihilated}\rangle)$ is
that state of no appearance. We remain to have two duals states left, these
are the half states.

Now, we evaluate the operations. First we can see that

$\hat{+}|i\rangle=|i\rangle\quad\text{and}\quad\hat{-}|i\rangle=\bar{\boldsymbol{0}}_{i}=\bar{\boldsymbol{0}}\,.$
(613)

Here, $\bar{\boldsymbol{0}}$ is defined as a empty vector which does not
contain any entries, it is a hollow column vector with zero dimension, which
we call the vacuum state,

$\bar{\boldsymbol{0}}=\begin{pmatrix}&\\\ &\\\ \end{pmatrix}\,,$ (614)

so that for all $|v\rangle\in{V}$,
$\bar{\boldsymbol{0}}\oplus|v\rangle=|v\rangle\oplus\bar{\boldsymbol{0}}$ and
$\bar{\boldsymbol{0}}\oplus\bar{\boldsymbol{0}}=\bar{\boldsymbol{0}}$.

This is because $\hat{+}$ is the being-operator, this gives the state itself;
while $\hat{-}$ is the nor-operator, the action on the state means the state
does not exist, so this gives the empty vector. Therefore, equation 612 is
evaluated to be

$\displaystyle\psi$
$\displaystyle=(|u\rangle\oplus|u^{*}\rangle)\oplus(\bar{\boldsymbol{0}}_{u}\oplus\bar{\boldsymbol{0}}_{u^{*}})\oplus(|u\rangle\oplus\bar{\boldsymbol{0}}_{u^{*}})\oplus(\bar{\boldsymbol{0}}_{u}\oplus|u^{*}\rangle)$
(615)
$\displaystyle=(|u\rangle\oplus|u^{*}\rangle)\oplus\bar{\boldsymbol{0}}\oplus|u\rangle\oplus|u^{*}\rangle)\,.$

This gives the result of figure 32. Therefore in terms of direct sum vector
space representation, this is figure 32,

Figure 106:

Therefore we have the following results

$\displaystyle\hat{-}|\text{created}\rangle\oplus\hat{-}|\text{annihilated}\rangle=\bar{\boldsymbol{0}}\,.$
(616)
$\displaystyle\hat{+}|\text{created}\rangle\oplus\hat{+}|\text{annihilated}\rangle=|\text{created}\rangle\oplus|\text{annihilated}\rangle\,.$
$\displaystyle\hat{+}|\text{created}\rangle\oplus\hat{-}|\text{annihilated}\rangle=|\text{created}\rangle\,.$
$\displaystyle\hat{-}|\text{created}\rangle\oplus\hat{+}|\text{annihilated}\rangle=|\text{annihilated}\rangle\,.$

Explicitly in terms of diagram,

Figure 107:

In terms of set, we have 104. We find the structure of the set is identical to
the case of direct-sum space. Therefore one has the isomorphism,

Figure 108:

The operation isomorphism for the two cases hold as follow,

$\displaystyle U\cup\emptyset=\emptyset\cup
U=U\quad,\quad|v\rangle\oplus\bar{\boldsymbol{0}}=\bar{\boldsymbol{0}}\oplus|v\rangle=|v\rangle\,.$
(617) $\displaystyle U^{*}\cup\emptyset=\emptyset\cup
U^{*}=U^{*}\quad,\quad|v^{*}\rangle\oplus\bar{\boldsymbol{0}}=\bar{\boldsymbol{0}}\oplus|v^{*}\rangle=|v^{*}\rangle\,.$

In addition, the power set of the full set $W$ is the discrete topology of
$W$,

$\tau=\\{\emptyset,\\{u\\},\\{u^{*}\\},\\{u,u^{*}\\}\\}\,.$ (618)

This is isomorphic to the binary number set of

$\tau=\\{00,10,01,11\\}\,.$ (619)

This topology set is dual invariant, i.e. it is the same no matter you look
from the left perspective or right perspective, i.e.

$(\tau|L)\equiv(\tau|R)\,.$ (620)

Therefore, we arrive at a very important theorem that topology implies dual
invariance.

In conclusion, the “nor created and nor annihilated” state means empty space
(set) mathematically. Since the empty set does not contain any elements, this
implies such object has no appearance. The dual perspective is the view of the
full set (space). It contains both the appearance of creation and the
appearance of annihilation.

####  10.1.1 Cardinality representation

Finally, we would like to work on the cardinality representation of the above
representation. If there exists a $u$ or $|u^{\star}\rangle$, we label as 1;
while if it is $\boldsymbol{0}$ then we have 0. In other words we get,

Figure 109:

In the first diagram, we can actually see that the cardinality representation
is the representation of the $\mathbb{Z}_{2}\times\mathbb{Z}_{2}$ group. In
the right diagram, we calculate the cardinality of the set for each quadrant.
In the last diagram, the $1/2$ representation denotes the red and blue
quadrant are half quadrants , while the 0 representation is the empty quadrant
and the 1 representation is the full quadrant. Next, we would like to
construct the cardinality representation matrix by directing summing the
cardinality elements of the state,

Figure 110:

Now define the dual doublet as the full state

$\phi=\begin{pmatrix}|u\rangle\\\
|u^{*}\rangle\end{pmatrix}=|u\rangle\oplus|u^{*}\rangle\,.$ (621)

Each representation matrix can be defined by

$\boldsymbol{0}=\begin{pmatrix}0&0\\\
0&0\end{pmatrix}\quad,\quad\boldsymbol{1}=\begin{pmatrix}1&0\\\
0&1\end{pmatrix}\quad,\quad\hat{P}_{+}=\begin{pmatrix}1&0\\\
0&0\end{pmatrix}\quad,\quad\hat{P}_{-}=\begin{pmatrix}0&0\\\
0&1\end{pmatrix}\,.$ (622)

where $\hat{P}_{+}$ and $\hat{P}_{-}$ are projection matrices constructed by
the elements of the cardinality, which satisfies

$\hat{P}_{+}+\hat{P}_{-}=\boldsymbol{1}\quad,\quad\hat{P}_{+}\hat{P}_{-}=\hat{P}_{-}\hat{P}_{+}=\boldsymbol{0}\,.$
(623)

Define also

$\underline{\boldsymbol{0}}=\begin{pmatrix}0\\\ 0\end{pmatrix}=0\oplus
0\quad,\quad\phi_{+}=\begin{pmatrix}|u\rangle\\\
0\end{pmatrix}=|u\rangle\oplus 0\quad,\quad\phi_{-}=\begin{pmatrix}0\\\
|u^{*}\rangle\end{pmatrix}=0\oplus|u^{*}\rangle\,.$ (624)

Then we have

$\boldsymbol{1}\phi=\phi\quad,\quad\boldsymbol{0}\phi=\underline{\boldsymbol{0}}\quad,\quad\hat{P}_{+}\phi=\phi_{+}\quad,\quad\hat{P}_{-}\phi=\phi_{-}\,.$
(625)

Also we have

$\hat{P}_{+}\phi_{-}=\underline{\boldsymbol{0}}\quad,\quad\hat{P}_{-}\phi_{+}=\underline{\boldsymbol{0}}\,.$
(626)

Since $*\phi_{+}=\phi_{-}$ and $*\phi_{-}=\phi_{+}$ ;
$*\phi=\underline{\boldsymbol{0}}$ and $*\underline{\boldsymbol{0}}=\phi$, it
follows from equation 625 that for operators, we have

$*\hat{P}_{+}=\hat{P}_{-}\,\,,\,\,*\hat{P}_{-}=\hat{P}_{+}\quad;\quad*\boldsymbol{1}=\boldsymbol{0}\,\,,\,\,*\boldsymbol{0}=\boldsymbol{1}\,.$
(627)

We term the second dual equivalence as _associated duality_ given in 79, for
which such duality is not given by normal matrix operation that we introduced
before. In addition, for $\hat{P}_{+}$ and $\hat{P}_{-}$, the dual operator
can be regarded as off-diagonal transpose $S$, such that
$\hat{P}_{+}=(\hat{P}_{+})^{S}$. We can also check that the dual operators are
orthogonal.

$\hat{P}_{+}\cdot\hat{P}_{-}=\mathrm{Tr}(\hat{P}_{+}^{T}\hat{P}_{-})=0\quad,\quad\hat{P}_{-}\cdot\hat{P}_{+}=\mathrm{Tr}(\hat{P}_{-}^{T}\hat{P}_{+})=0\,,$
(628)

and

$\boldsymbol{0}\cdot\boldsymbol{1}=\mathrm{Tr}(\boldsymbol{0}^{T}\boldsymbol{1})=0\quad,\quad\boldsymbol{1}\cdot\boldsymbol{0}=\mathrm{Tr}(\boldsymbol{1}^{T}\boldsymbol{0})=0\,.$
(629)

The cardinality is given by the norm of the state vector,

$\displaystyle\text{card}(\underline{\boldsymbol{0}})=\langle\phi|\phi\rangle=\mathrm{Tr}(0\oplus
0)=0\,,$ (630)
$\displaystyle\text{card}(\phi_{+})=\langle\phi_{+}|\phi_{+}\rangle=\mathrm{Tr}(1\oplus
0)=1\,,$
$\displaystyle\text{card}(\phi_{-})=\langle\phi_{-}|\phi_{-}\rangle=\mathrm{Tr}(0\oplus
1)=1\,,$
$\displaystyle\text{card}(\phi)=\langle\phi|\phi\rangle=\mathrm{Tr}(1\oplus
1)=2\,.$

There is also one important property, note that

$(\hat{P}_{+}+\hat{P}_{-})^{2}=\hat{P}_{+}^{2}+\hat{P}_{-}^{2}\,,$ (631)

which takes the form of $(a+b)^{2}=a^{2}+b^{2}$, this is prohibited when $a,b$
are real numbers (except zero), but allowed when $a,b$ are matrices.
Similarly, we have

$(\hat{P}_{+}+\hat{P}_{-})^{2}=(\hat{P}_{+}-\hat{P}_{-})^{2}$ (632)

and this only holds for the matrix case.

###  10.2 The Duality Wave

With all the studies above, we now can proceed to study the duality wave given
by the $\mathbb{Z}_{2}\times\mathbb{Z}_{2}$ representation. For convenience
take

$|u\rangle=|0\rangle=|\text{creation}\rangle=\begin{pmatrix}1\\\
0\end{pmatrix}\quad,\quad|u^{*}\rangle=|1\rangle=|\text{annihilation}\rangle=\begin{pmatrix}0\\\
1\end{pmatrix}\,.$ (633)

In physical sense, the duality system, can also infer as system with two spins
which are dual to each other, the up-spin and down spin. By isomorphism, we
take an ansatz that we can construct the quantum states as follow,

$\bar{\boldsymbol{0}}\rightarrow|0\oplus
0\rangle\quad,\quad|u\rangle\oplus\bar{\boldsymbol{0}}\rightarrow|1\oplus
0\rangle\quad,\quad\bar{\boldsymbol{0}}\oplus|u^{*}\rangle\rightarrow|0\oplus
1\rangle\quad,\quad|u\rangle\oplus|u^{*}\rangle\rightarrow|1\oplus
1\rangle\,.$ (634)

where $|a\oplus b\rangle=|a\rangle\oplus|b\rangle$ for $a,b=0,1$. The general
form takes the following

$\hat{P}|0\rangle\oplus\hat{Q}|1\rangle=|a\rangle\oplus|b\rangle\,,$ (635)

where $\hat{P},\hat{Q}$ are $\hat{+}$ or $\hat{-}$.

In the original representation

$\hat{-}=\boldsymbol{0}=\begin{pmatrix}0&0\\\
0&0\end{pmatrix}\quad,\quad\hat{+}=\boldsymbol{1}=\begin{pmatrix}1&0\\\
0&1\end{pmatrix}$ (636)

where $\hat{-}$ and $\hat{+}$ are dual to each other as we have stated above.
Then we have the following computation,

$\hat{-}|0\rangle\oplus\hat{-}|1\rangle=\begin{pmatrix}0\\\ 0\\\ 0\\\
0\end{pmatrix}\quad,\quad\hat{+}|0\rangle\oplus\hat{+}|1\rangle=\begin{pmatrix}0\\\
1\\\ 1\\\ 0\end{pmatrix}\,,$ (637)
$\hat{+}|0\rangle\oplus\hat{-}|1\rangle=\begin{pmatrix}0\\\ 1\\\ 0\\\
0\end{pmatrix}\quad,\quad\hat{-}|0\rangle\oplus\hat{+}|1\rangle=\begin{pmatrix}0\\\
0\\\ 1\\\ 0\end{pmatrix}\,,$ (638)

We can see that
$*(\hat{-}|0\rangle\oplus\hat{-}|1\rangle)=\boldsymbol{M}(\hat{-}|0\rangle\oplus\hat{-}|1\rangle)=\hat{+}|0\rangle\oplus\hat{+}|1\rangle)$.
Therefore $\hat{-}|0\rangle\oplus\hat{-}|1\rangle)$ and
$\hat{+}|0\rangle\oplus\hat{+}|1\rangle)$ are dual and orthogonal to each
other. Also $\hat{-}|0\rangle\oplus\hat{-}|1\rangle$ and
$\hat{+}|0\rangle\oplus\hat{+}|1\rangle$ are of associated duality, and they
are orthogonal.

Now we would like to use another matrix representation of $\hat{-}$ and
$\hat{+}$, so that we can obtain the canonical duality in equation 634. We
take

$\hat{-}=\begin{pmatrix}1&1\\\
0&0\end{pmatrix}\quad,\quad\hat{+}=\begin{pmatrix}0&0\\\ 1&1\end{pmatrix}\,.$
(639)

Clearly that $*\hat{-}=\boldsymbol{M}\hat{+}$ and
$*\hat{+}=\boldsymbol{M}\hat{-}$.

$\begin{pmatrix}0&1\\\ 1&0\end{pmatrix}\begin{pmatrix}1&1\\\
0&0\end{pmatrix}=\begin{pmatrix}0&0\\\
1&1\end{pmatrix}\quad\text{and}\quad\begin{pmatrix}0&1\\\
1&0\end{pmatrix}\begin{pmatrix}0&0\\\ 1&1\end{pmatrix}=\begin{pmatrix}1&1\\\
0&0\end{pmatrix}\,.$ (640)

Using this canonical representation,

$\hat{-}|0\rangle\oplus\hat{-}|1\rangle=\begin{pmatrix}1\\\ 0\\\ 1\\\
0\end{pmatrix}=|0\rangle\oplus|0\rangle\quad,\quad\hat{+}|0\rangle\oplus\hat{+}|1\rangle=\begin{pmatrix}0\\\
1\\\ 0\\\ 1\end{pmatrix}=|1\rangle\oplus|1\rangle\,,$ (641)
$\hat{+}|0\rangle\oplus\hat{-}|1\rangle=\begin{pmatrix}0\\\ 1\\\ 1\\\
0\end{pmatrix}=|0\rangle\oplus|1\rangle\quad,\quad\hat{-}|0\rangle\oplus\hat{+}|1\rangle=\begin{pmatrix}1\\\
0\\\ 0\\\ 1\end{pmatrix}=|1\rangle\oplus|0\rangle\,,$ (642)

Now we can see that
$*(\hat{-}|0\rangle\oplus\hat{-}|1\rangle)=\boldsymbol{M}(\hat{-}|0\rangle\oplus\hat{-}|1\rangle)=(\hat{+}|0\rangle\oplus\hat{+}|1\rangle)$
is dual and orthogonal to each other $\langle 0\oplus 0|1\oplus
1\rangle=\langle 0|1\rangle+\langle 0|1\rangle=0$, while
$\hat{+}|0\rangle\oplus\hat{-}|1\rangle$ and
$\hat{-}|0\rangle\oplus\hat{+}|1\rangle$ are associated dual and orthogonal
$\langle 0\oplus 1|1\oplus 0\rangle=\langle 0|1\rangle+\langle 1|0\rangle=0$ .

Diagramatically,

Figure 111:

The states $\\{|0\oplus 0\rangle,|0\oplus 1\rangle,|1\oplus 0\rangle,|1\oplus
1\rangle\\}$ form the basis of $\mathbb{Z}_{2}\times\mathbb{Z}_{2}$. The
transformation of each of the basis is given by

Figure 112:

Explicitly, the all six transformation reads

$\displaystyle\boldsymbol{1}\oplus\boldsymbol{M}|0\oplus
0\rangle=(\boldsymbol{1}\oplus\boldsymbol{M})(|0\rangle\oplus
0\rangle)=|0\rangle\oplus|1\rangle\,,$ (643)
$\displaystyle\boldsymbol{M}\oplus\boldsymbol{1}|0\oplus
1\rangle=(\boldsymbol{M}\oplus\boldsymbol{1})(|0\rangle\oplus
1\rangle)=|1\rangle\oplus|1\rangle\,,$
$\displaystyle\boldsymbol{M}\oplus\boldsymbol{M}|0\oplus
0\rangle=(\boldsymbol{M}\oplus\boldsymbol{M})(|0\rangle\oplus
0\rangle)=|1\rangle\oplus|1\rangle\,,$
$\displaystyle\boldsymbol{M}\oplus\boldsymbol{1}|0\oplus
0\rangle=(\boldsymbol{M}\oplus\boldsymbol{1})(|0\rangle\oplus
0\rangle)=|1\rangle\oplus|0\rangle\,,$
$\displaystyle\boldsymbol{1}\oplus\boldsymbol{M}|1\oplus
0\rangle=(\boldsymbol{1}\oplus\boldsymbol{M})(|1\rangle\oplus
0\rangle)=|1\rangle\oplus|1\rangle\,,$
$\displaystyle\boldsymbol{M}\oplus\boldsymbol{M}|1\oplus
0\rangle=(\boldsymbol{M}\oplus\boldsymbol{M})(|1\rangle\oplus
0\rangle)=|0\rangle\oplus|1\rangle\,.$

It is noted that

$\boldsymbol{M}\oplus\boldsymbol{M}=(\boldsymbol{1}\oplus\boldsymbol{M})(\boldsymbol{M}\oplus\boldsymbol{1})=(\boldsymbol{M}\oplus\boldsymbol{1})(\boldsymbol{1}\oplus\boldsymbol{M})\,.$
(644)

We have the Klein-4 group as

$\mathbb{Z}_{2}\times\mathbb{Z}_{2}=\\{\boldsymbol{1}\oplus\boldsymbol{1},\boldsymbol{1}\oplus\boldsymbol{M},\boldsymbol{M}\oplus\boldsymbol{1},\boldsymbol{M}\oplus\boldsymbol{M}\\}\,.$
(645)

The identity operator is given by the completeness relation,

$\boldsymbol{1}=|00\rangle\langle 00|+|11\rangle\langle 11|+|01\rangle\langle
01|+|10\rangle\langle 01|\,,$ (646)

and the duality operator is given by

$\boldsymbol{M}=|00\rangle\langle 11|+|11\rangle\langle 00|+|01\rangle\langle
10|+|10\rangle\langle 01|\,.$ (647)

Now we define an qubit operator,

$\hat{\Psi}(\theta)=\cos\theta(\hat{-}\oplus\hat{-})+\sin\theta(\hat{+}\oplus\hat{+})\,.$
(648)

where the phase dependent qubit is

$|\Psi(\theta)\rangle=\hat{\Psi}(\theta)|u\oplus u^{*}\rangle\,.$ (649)

The qubit operator acts on the full state $|u\oplus u^{*}\rangle$,

$\begin{aligned} \hat{\Psi}(\theta)|u\oplus
u^{*}\rangle&=\cos\theta(\hat{-}|u\rangle\oplus\hat{-}|u^{*}\rangle)+\sin\theta(\hat{+}|u\rangle\oplus\hat{+}|u^{*}\rangle)\\\
&=\cos\theta(\hat{-}|0\rangle\oplus\hat{-}|1\rangle)+\sin\theta(\hat{+}|0\rangle\oplus\hat{+}|1\rangle)\\\
&=\cos\theta|0\oplus 0\rangle+\sin\theta|1\oplus 1\rangle\end{aligned}\,,$
(650)

Thus we write

$|\Psi(\theta)\rangle=\cos\theta|\text{nothingness}\rangle+\sin\theta|\text{All}\rangle\,,$
(651)

where $|\text{nothingness}\rangle=*|\text{All}\rangle$ and
$*|\text{nothingness}\rangle=|\text{All}\rangle$. For the probabilities, we
have

$P_{\text{nothingness}}(\theta)=\langle 0\oplus
0|\Psi(\theta)\rangle=\cos^{2}\theta\,,$ (652)

and

$P_{\text{All}}(\theta)=\langle 1\oplus
1|\Psi(\theta)\rangle=\sin^{2}\theta\,.$ (653)

Therefore, the probability of nothingness and All transforms one-another. In
particular we have

$|\Psi(0)\rangle=|\text{nothingness}\rangle\,,$ (654)

with probability equal to 1 for the nothingness state. And

$|\Psi(\pi/2)\rangle=|\text{All}\rangle\,.$ (655)

with probability equal to 1 for the All state. The two states are orthogonal,

$\langle\Psi(0)|\Psi(\pi/2)\rangle=\langle\text{nothingness}|\text{All}\rangle=0\,.$
(656)

There is a special state when $\theta=\pi/2$, then we have the analogous Bell
state (which is a entangled EPR pair)

$|\Psi(\pi/4)\rangle=\frac{1}{\sqrt{2}}(|0\oplus 0\rangle+|1\oplus 1\rangle)$
(657)

which is equal to

$|\Psi(\pi/4)\rangle=\frac{1}{\sqrt{2}}(|\text{nothingness}\rangle+|\text{All}\rangle)$
(658)

This is same as the famous case of Schrodinger’s cat in physics, where the cat
is regarded as nor alive nor dead, or both alive or dead, unless measurement
is taken. Therefore, the ultimate truth here is a qubit, which is
indeterministic unless an observation is made, where observation means
measurement.

Next we will study some properties of the qubit. Define the orthogonal state
by differentiating the state with respect to $\theta$, then we have

$\langle\Psi(\theta)|\frac{d}{d\theta}|\Psi(\theta)\rangle=0\,.$ (659)

Consider the state where we first take the derivative and the the dual
operator,

$|\psi^{\prime}(\theta)\rangle=*\frac{d}{d\theta}|\Psi(\theta)\rangle\,.$
(660)

This new state will be probability invariant as compared to the original
state.

Finally, we are particularly interested in local phase for which $\theta$ is
dependent on space and time and is linear of the spacetime variables. In other
words, we are interested in the qubit wave operator as

$\hat{\Psi}(x,y,z,t)=\cos(\omega
t-k_{x}x-k_{y}y-k_{z}z)(\hat{+}\oplus\hat{+})+\sin(\omega
t-k_{x}x-k_{y}y-k_{z}z)(\hat{-}\oplus\hat{-})$ (661)

The qubit wave operator acts on the full state $|u\oplus u^{*}\rangle$,

$\displaystyle\hat{\Psi}|u\oplus u^{*}\rangle$ $\displaystyle=\cos(\omega
t-k_{x}x-k_{y}y-k_{z}z)(\hat{-}|u\rangle\oplus\hat{-}|u^{*}\rangle)+\sin(\omega
t-k_{x}x-k_{y}y-k_{z}z)(\hat{+}|u\rangle\oplus\hat{+}|u^{*}\rangle)$ (662)
$\displaystyle=\cos(\omega t-k_{x}x-k_{y}y-k_{z}z)|0\oplus
0\rangle+\sin(\omega t-k_{x}x-k_{y}y-k_{z}z)|1\oplus 1\rangle\,.$

The qubit operator satisfies the wave equation,

$\Box\hat{\Psi}|u\oplus
u^{*}\rangle=\bigg{(}\nabla^{2}\hat{\Psi}-\frac{1}{c^{2}}\frac{\partial^{2}}{\partial
t^{2}}\hat{\Psi}\bigg{)}|u\oplus u^{*}\rangle=0$ (663)

where $c$ is the speed of the qubit wave.

Next for the dual qubit wave, we have

$\hat{\Psi^{\star}}(\theta)=\cos\theta(\hat{+}\oplus\hat{-})+\sin\theta(\hat{-}\oplus\hat{+})$
(664)

By acting on the full state $|u\otimes u^{*}\rangle$,

$\begin{aligned} \hat{\Psi}^{\star}(\theta)|u\oplus
u^{*}\rangle&=\cos\theta(\hat{+}|u\rangle\oplus\hat{-}|u^{*}\rangle)+\sin\theta(\hat{-}|u\rangle\oplus\hat{+}_{3}|u^{*}\rangle)\\\
&=\cos\theta(\hat{+}|0\rangle\oplus\hat{-}|1\rangle)+\sin\theta(\hat{-}|0\rangle\oplus\hat{+}|1\rangle)\\\
&=\cos\theta|1\oplus 0\rangle+\sin\theta|0\oplus 1\rangle\end{aligned}\,,$
(665)

We have

$\hat{\Psi}^{\star}(\theta)=\cos\theta|\text{creation}\rangle+\sin\theta|\text{annihilation}\rangle$
(666)

where $|\text{creation}\rangle=*|\text{annihilation}\rangle$ and
$*|\text{creation}\rangle=|\text{annihilation}\rangle$.

$P_{\text{creation}}(\theta)=\langle 1\oplus
0|\Psi^{\star}(\theta)\rangle=\cos^{2}\theta\,,$ (667)

and

$P_{\text{annihilation}}(\theta)=\langle 0\oplus
1|\Psi^{\star}(\theta)\rangle=\sin^{2}\theta\,.$ (668)

In phase equal to zero, we have

$|\Psi^{\star}(0)\rangle=|\text{creation}\rangle$ (669)

with probability equal to 1 for creation state. And

$|\Psi^{\star}(\pi/2)\rangle=|\text{annihilation}\rangle$ (670)

with probability equal to 1 for the annihilation state. The two states are
orthogonal,

$\langle\Psi^{\star}(0)|\Psi^{\star}(\pi/2)\rangle=\langle\text{creation}|\text{annihilation}\rangle=0\,.$
(671)

There is a special state when $\theta=\pi/2$, then we have the entangled EPR
state as

$|\Psi(\pi/4)\rangle=\frac{1}{\sqrt{2}}(|1\oplus 0\rangle+|0\oplus 1\rangle)$
(672)

which is equal to

$|\Psi(\pi/4)\rangle=\frac{1}{\sqrt{2}}(|\text{creation}\rangle+|\text{annihilation}\rangle)$
(673)

Then the dual wave qubit operator is

$\hat{\Psi}^{\star}(x,y,z,t)=\cos(\omega
t-k_{x}x-k_{y}y-k_{z}z)(\hat{+}\oplus\hat{-})+\sin(\omega
t-k_{x}x-k_{y}y-k_{z}z)(\hat{-}\oplus\hat{+})\,.$ (674)

By acting on the full state

$\displaystyle\hat{\Psi}^{\star}|u\oplus u^{*}\rangle$
$\displaystyle=\cos(\omega
t-k_{x}x-k_{y}y-k_{z}z)(\hat{+}|u\rangle\oplus\hat{-}|u^{*}\rangle)+\sin(\omega
t-k_{x}x-k_{y}y-k_{z}z)(\hat{-}|u\rangle\oplus\hat{+}|u^{*}\rangle)$ (675)
$\displaystyle=\cos(\omega t-k_{x}x-k_{y}y-k_{z}z)|1\oplus
0\rangle+\sin(\omega t-k_{x}x-k_{y}y-k_{z}z)|0\oplus 1\rangle\,.$

The dual qubit also satisfies the wave equation,

$\Box\hat{\Psi}^{\star}|u\oplus
u^{*}\rangle=\bigg{(}\nabla^{2}\hat{\Psi}^{\star}-\frac{1}{c^{2}}\frac{\partial^{2}}{\partial
t^{2}}\hat{\Psi}^{\star}\bigg{)}|u\oplus u^{*}\rangle=0\,.$ (676)

The pair of states are orthogonal to each other

$\langle\Psi(\theta)|\Psi^{\star}(\theta)\rangle=0\,.$ (677)

Note that the linear combination of the two states is also the solution of the
wave equation,

$\phi=A\Psi+B\Psi^{\star}\,.$ (678)

Therefore, when the dual states oscillate, it generates a wave propagation.
For example, in equation 675, the periodic cycle of creation and annihilation
is a wave. Since mind creates and annihilates, this shows that mind is a wave.
And for equation 662, when the state of nothingness and All oscillate, it
creates a wave. So the periodic transformation of nothingness and All will
produce a wave signal, that maybe detected in experimental sense.

Next we would like to investigate what happens when the operators act on the
dual of $|u\oplus u^{*}\rangle$, i.e. $*|u\oplus u^{*}\rangle=|u^{*}\oplus
u\rangle$. Using the representation in 639 we find that the following holds,

$\hat{P}*|0\rangle\oplus\hat{Q}*|1\rangle=\hat{P}|1\rangle\oplus\hat{Q}|0\rangle\,.$
(679)

Explicitly, we can check that

$\displaystyle\hat{-}*|0\rangle\oplus\hat{-}*|1\rangle$
$\displaystyle=\hat{-}|1\rangle\oplus\hat{-}|0\rangle=|0\rangle\oplus|0\rangle=\hat{-}|0\rangle\oplus\hat{-}|1\rangle\,,$
(680) $\displaystyle\hat{+}*|0\rangle\oplus\hat{+}*|1\rangle$
$\displaystyle=\hat{+}|1\rangle\oplus\hat{+}|0\rangle=|1\rangle\oplus|1\rangle=\hat{+}|0\rangle\oplus\hat{+}|1\rangle\,,$
$\displaystyle\hat{+}*|0\rangle\oplus\hat{-}*|1\rangle$
$\displaystyle=\hat{+}|1\rangle\oplus\hat{-}|0\rangle=|0\rangle\oplus|1\rangle=\hat{+}|0\rangle\oplus\hat{-}|1\rangle\,,$
$\displaystyle\hat{-}*|0\rangle\oplus\hat{+}*|1\rangle$
$\displaystyle=\hat{-}|1\rangle\oplus\hat{+}|0\rangle=|1\rangle\oplus|0\rangle=\hat{-}|0\rangle\oplus\hat{+}|1\rangle\,.$

Therefore, we have

$\hat{-}*=\hat{-}\quad,\quad\hat{+}*=\hat{+}\,.$ (681)

We can explicitly check that

$\begin{pmatrix}1&1\\\ 0&0\end{pmatrix}\begin{pmatrix}0&1\\\
1&0\end{pmatrix}=\begin{pmatrix}1&1\\\
0&0\end{pmatrix}\quad,\quad\begin{pmatrix}0&0\\\
1&1\end{pmatrix}\begin{pmatrix}0&1\\\ 1&0\end{pmatrix}=\begin{pmatrix}0&0\\\
1&1\end{pmatrix}\,.$ (682)

Now we would like to give a more detailed study in the $\boldsymbol{M}$ dual
matrix operator. First notice that the action of $*=\boldsymbol{M}$ swap the
position of rows as a vertical reflection. For example, for a general matrix

$A=\begin{pmatrix}a&b\\\
c&d\end{pmatrix}\quad\text{then}\quad*A=\boldsymbol{M}A=A^{*}=\begin{pmatrix}c&d\\\
a&b\end{pmatrix}\,.$ (683)

If $A$ acts on $\boldsymbol{M}$, the action is to swap the position of the
columns as a horizontal reflection,

$A*=A\boldsymbol{M}=\,^{*}A=\begin{pmatrix}b&a\\\ d&c\end{pmatrix}\,.$ (684)

Also we have

$**A=A**=A\,.$ (685)

We have the following identity, recalling $T$ is the transpose action and $R$
is the off-transpose action, we have

$A^{(TR)}=A^{(RT)}=\boldsymbol{M}A\boldsymbol{M}^{-1}=\boldsymbol{M}A\boldsymbol{M}=*A*^{-1}=*A*\,,$
(686)

as $\boldsymbol{M}^{-1}=\boldsymbol{M}$. It follows that
$\mathrm{Tr}(A^{(TR)})=\mathrm{Tr}A$ and
$\mathrm{det}(A^{(TR)})=\mathrm{det}A$. According to equations 681,

$*\hat{-}*=\hat{+}\quad,\quad*\hat{+}*=\hat{-}\,.$ (687)

And as $*=*^{-1}$, thus $\hat{+}$ and $\hat{-}$ are related by similarity
transform

$*\hat{-}*^{-1}=\hat{+}\quad,\quad*\hat{+}*^{-1}=\hat{-}\,.$ (688)

Next we would like to show that the identity

$(*A*)^{-1}=*A^{-1}*\,.$ (689)

This is because $(*A*)^{-1}=*^{-1}(*A)^{-1}=*A^{-1}*$.

Based on the result of 680, if follows that $|u^{*}\oplus u\rangle$ also
satisfies the wave equation,

$\Box\hat{\Psi}|u^{*}\oplus
u\rangle=\bigg{(}\nabla^{2}\hat{\Psi}-\frac{1}{c^{2}}\frac{\partial^{2}}{\partial
t^{2}}\hat{\Psi}\bigg{)}|u^{*}\oplus u\rangle=0$ (690)

and

$\Box\hat{\Psi}^{\star}|u^{*}\oplus
u\rangle=\bigg{(}\nabla^{2}\hat{\Psi}-\frac{1}{c^{2}}\frac{\partial^{2}}{\partial
t^{2}}\hat{\Psi}^{\star}\bigg{)}|u^{*}\oplus u\rangle=0$ (691)

Finally we will investigate the entropy of the system. The entropy for the
both systems $|\Psi\rangle$ and $|\Psi^{\star}\rangle$ are

$H(\theta)=-2\cos^{2}\theta\log|\cos\theta|-2\sin^{2}\theta\log|\sin\theta|\,.$
(692)

When the probability is equal the entropy is maximized, thus the EPR state is
the equilibrium state, which is the most stable state. The system has 0.693
bits. When it is either at its designated state with probability equal to 1,
the entropy is zero, which is minimum and hence unstable.

###  10.3 Canonical Quantization

In this section, we will dive into the core part of the whole subject, the
quantization of duality wave. Upon quantization, this will give the quanta of
the duality field. The procedure is called second quantization, which is
originated from quantum field theory. Second quantization turns a wave into a
particle quanta.

First for the $\hat{\Phi}(\theta)$ we define the conjugated momentum operator
by

$\Pi=\frac{h}{i}\frac{\partial}{\partial\theta}\Psi\,.$ (693)

Therefore the conjugated momentum of $\Psi$ is

$\Pi(\theta)=\frac{h}{i}\Big{(}-\sin\theta(\hat{-}\oplus\hat{-})+\cos\theta(\hat{+}\oplus\hat{+})\Big{)}\,.$
(694)

Now we compute the commutator $[\hat{\Psi}(\theta),\hat{\Pi}(\theta)]$,

$\displaystyle\left[\Psi(\theta),\Pi(\theta)\right]$
$\displaystyle=(-ih)[\cos\theta(\hat{-}\oplus\hat{-})+\sin\theta(\hat{+}\oplus\hat{+})\,\,,\,\,-\sin\theta(\hat{-}\oplus\hat{-})+\cos\theta(\hat{+}\oplus\hat{+})]$
(695)
$\displaystyle=-ih\Big{(}-\cos\theta\sin\theta[(\hat{-}\oplus\hat{-}),(\hat{-}\oplus\hat{-})]-\sin^{2}\theta[(\hat{+}\oplus\hat{+}),(\hat{-}\oplus\hat{-})]$
$\displaystyle\,\,\,\,\,\,+\cos^{2}\theta[(\hat{-}\oplus\hat{-}),(\hat{+}\oplus\hat{+})]+\sin\theta\cos\theta[(\hat{+}\oplus\hat{+}),(\hat{+}\oplus\hat{+})]\Big{)}$
$\displaystyle=-ih[(\hat{-}\oplus\hat{-}),(\hat{+}\oplus\hat{+})]$
$\displaystyle=-ih\Big{(}(\hat{-}\oplus\hat{-})(\hat{+}\oplus\hat{+})-(\hat{+}\oplus\hat{+})(\hat{-}\oplus\hat{-})\Big{)}$
$\displaystyle=-ih(\hat{-}\hat{+}\oplus\hat{-}\hat{+}-\hat{+}\hat{-}\oplus\hat{+}\hat{-})$

Now we take an ansatz for the replacement of $\hat{-},\hat{+}$ matrix
operators by bosonic creation and annihilation operators
$\hat{a},\hat{a}^{\dagger}$ with commutation relation of
$[\hat{a},\hat{a}^{\dagger}]=1$

$\hat{-}\rightarrow\hat{a}^{\dagger}\quad,\quad\hat{+}\rightarrow\hat{a}$
(696)

such that the quantum fields for duality are

$\hat{\Psi}(\theta)=\cos\theta(\hat{a}^{\dagger}\oplus\hat{a}^{\dagger})+\sin\theta(\hat{a}\oplus\hat{a})$
(697)

and

$\hat{\Pi}(\theta)=\frac{h}{i}\Big{(}-\sin\theta(\hat{a}^{\dagger}\oplus\hat{a}^{\dagger})+\cos\theta(\hat{a}\oplus\hat{a})\Big{)}\,.$
(698)

Then

$\displaystyle\left[\hat{\Psi(\theta)},\hat{\Pi(\theta)}\right]$
$\displaystyle=-ih(\hat{a}^{\dagger}\hat{a}\oplus\hat{a}^{\dagger}\hat{a}-\hat{a}\hat{a}^{\dagger}\oplus\hat{a}\hat{a}^{\dagger})$
(699) $\displaystyle=-ih\bigg{(}\begin{pmatrix}\hat{a}^{\dagger}\hat{a}&0\\\
0&\hat{a}^{\dagger}\hat{a}\end{pmatrix}-\begin{pmatrix}\hat{a}\hat{a}^{\dagger}&0\\\
0&\hat{a}\hat{a}^{\dagger}\end{pmatrix}\bigg{)}$
$\displaystyle=-ih\begin{pmatrix}-[\hat{a},\hat{a}^{\dagger}]&0\\\
0&-[\hat{a},\hat{a}^{\dagger}]\end{pmatrix}$
$\displaystyle=-ih\begin{pmatrix}-1&0\\\ 0&-1\end{pmatrix}$
$\displaystyle=ih\boldsymbol{1}\,.$

In the meantime, we have
$\left[\hat{\Psi}(\theta),\hat{\Psi}(\theta)\right]=0$ and
$\left[\hat{\Pi}(\theta),\hat{\Pi}(\theta)\right]$. Therefore we have the
canonical quantization condition as

$\left[\hat{\Psi}(\theta),\hat{\Pi}(\theta)\right]=ih\boldsymbol{1}\quad,\quad\left[\hat{\Psi}(\theta),\hat{\Psi}(\theta)\right]=\left[\hat{\Pi}(\theta),\hat{\Pi}(\theta)\right]=0\,.$
(700)

Next for the $\hat{\Phi}^{\star}(\theta)$ we define the conjugated momentum
operator by

$\hat{\Pi}^{\star}(\theta)=\frac{h}{i}\frac{\partial}{\partial\theta}\hat{\Psi}^{\star}\,.$
(701)

Therefore the conjugated momentum of $\Psi$ is

$\Pi^{\star}(\theta)=\frac{h}{i}\Big{(}-\sin\theta(\hat{+}\oplus\hat{-})+\cos\theta(\hat{-}\oplus\hat{+})\Big{)}\,.$
(702)

Now we compute the commutator $[\hat{\Psi}(\theta),\hat{\Pi}(\theta)]$,

$\displaystyle\left[\Psi^{\star}(\theta),\Pi^{\star}(\theta)\right]$
$\displaystyle=(-ih)\left[\cos\theta(\hat{+}\oplus\hat{-})+\sin\theta(\hat{-}\oplus\hat{+})\,\,,\,\,-\sin\theta(\hat{+}\oplus\hat{-})+\cos\theta(\hat{-}\oplus\hat{+})\right]$
(703)
$\displaystyle=-ih\Big{(}-\cos\theta\sin\theta\left[(\hat{+}\oplus\hat{-}),(\hat{+}\oplus\hat{-})\right]-\sin^{2}\theta\left[(\hat{-}\oplus\hat{+}),(\hat{+}\oplus\hat{-})\right]$
$\displaystyle\,\,\,\,\,\,+\cos^{2}\theta\left[(\hat{+}\oplus\hat{-}),(\hat{-}\oplus\hat{+})\right]+\sin\theta\cos\theta\left[(\hat{-}\oplus\hat{+}),(\hat{-}\oplus\hat{+})\right]\Big{)}$
$\displaystyle=-ih\left[(\hat{+}\oplus\hat{-}),(\hat{-}\oplus\hat{+})\right]$
$\displaystyle=-ih\Big{(}(\hat{+}\oplus\hat{-})(\hat{-}\oplus\hat{+})-(\hat{-}\oplus\hat{+})(\hat{+}\oplus\hat{-})\Big{)}$
$\displaystyle=-ih(\hat{+}\hat{-}\oplus\hat{-}\hat{+}-\hat{-}\hat{+}\oplus\hat{+}\hat{-})$

Then we have

$\displaystyle\left[\hat{\Psi}^{\star}(\theta),\hat{\Pi}^{\star}(\theta)\right]$
$\displaystyle=-ih(\hat{a}\hat{a}^{\dagger}\oplus\hat{a}^{\dagger}\hat{a}-\hat{a}^{\dagger}\hat{a}\oplus\hat{a}\hat{a}^{\dagger})$
(704) $\displaystyle=-ih\bigg{(}\begin{pmatrix}\hat{a}\hat{a}^{\dagger}&0\\\
0&\hat{a}^{\dagger}\hat{a}\end{pmatrix}-\begin{pmatrix}\hat{a}^{\dagger}\hat{a}&0\\\
0&\hat{a}\hat{a}^{\dagger}\end{pmatrix}\bigg{)}$
$\displaystyle=-ih\begin{pmatrix}[\hat{a},\hat{a}^{\dagger}]&0\\\
0&-[\hat{a},\hat{a}^{\dagger}]\end{pmatrix}$
$\displaystyle=ih\begin{pmatrix}-1&0\\\ 0&1\end{pmatrix}$
$\displaystyle=ihM\,,$

where $M$ is another representation matrix of the duality operator with
$M^{2}=\boldsymbol{1}$. In the meantime, we have
$[\hat{\Psi}^{\star}(\theta),\hat{\Psi}^{\star}(\theta)]=0$ and
$[\hat{\Pi}^{\star}(\theta),\hat{\Pi}^{\star}(\theta)]=0$. Therefore we have
the canonical quantization condition as

$[\hat{\Psi}^{\star}(\theta),\hat{\Pi}^{\star}(\theta)]=ihM\quad,\quad[\hat{\Psi}^{\star}(\theta),\hat{\Psi}^{\star}(\theta)]=[\hat{\Pi}^{\star}(\theta),\hat{\Pi}^{\star}(\theta)]=0\,.$
(705)

Notice that the cross terms are not zero, this is given by the fact that the
states are not independent of each other. For example

$[\hat{\Psi}(\theta),\hat{\Psi}^{\star}(\theta)]=\begin{pmatrix}-\cos
2\theta&0\\\ 0&0\end{pmatrix}\,.$ (706)

####  10.3.1 Hamiltonian

The canonical momentum operator was defined by
$\hat{\Pi}=\frac{1}{i}\frac{\partial}{\partial\theta}$, the $h$ is introduced
such that we will get the same form of commutation relation as quantum
mechanics does. From now on, to calculate the canonical Hamiltonian, let’s
define the dimensionless conjugated momentum, still denoted as $\hat{\Pi}$,

$\hat{\Pi}=\frac{1}{i}\frac{\partial}{\partial\theta}\hat{\Psi}\quad\text{and}\quad\hat{\Pi}^{\star}=\frac{1}{i}\frac{\partial}{\partial\theta}\hat{\Psi}^{\star}$
(707)

We ansatz that the Hamiltonian is defined as

$\hat{H}=\bigg{(}\frac{1}{2}\hat{\Pi}^{2}+\frac{1}{2}\hat{\Psi}^{2}\bigg{)}\hbar\omega\quad\text{and}\quad\hat{H}^{\star}=\bigg{(}\frac{1}{2}\hat{\Pi}^{\star
2}+\frac{1}{2}\hat{\Psi}^{\star 2}\bigg{)}\hbar\omega\,.$ (708)

Now first we evaluate $\hat{H}(\theta)$.

$\displaystyle\quad\hat{\Pi}^{2}(\theta)$ (709)
$\displaystyle=\frac{1}{2i^{2}}\Big{(}-\sin\theta(\hat{-}\oplus\hat{-})+\cos\theta(\hat{+}\oplus\hat{+})\Big{)}\Big{(}-\sin\theta(\hat{-}\oplus\hat{-})+\cos\theta(\hat{+}\oplus\hat{+})\Big{)}$
$\displaystyle=-\frac{1}{2}\Big{(}\sin^{2}\theta(\hat{-}\oplus\hat{-})(\hat{-}\oplus\hat{-})-\sin\theta\cos\theta(\hat{-}\oplus\hat{-})(\hat{+}\oplus\hat{+})$
$\displaystyle\quad\quad-\cos\theta\sin\theta(\hat{+}\oplus\hat{+})(\hat{-}\oplus\hat{-})+\cos^{2}\theta(\hat{+}\oplus\hat{+})(\hat{+}\oplus\hat{+})\Big{)}$
$\displaystyle=-\frac{1}{2}\Big{(}\sin^{2}\theta(\hat{-}\hat{-}\oplus\hat{-}\hat{-})-\sin\theta\cos\theta(\hat{-}\hat{+}\oplus\hat{-}\hat{+})-\sin\theta\cos\theta(\hat{+}\hat{-}\oplus\hat{+}\hat{-})+\cos^{2}\theta(\hat{+}\hat{+}\oplus\hat{+}\hat{+})\Big{)}$
$\displaystyle=-\frac{1}{2}\bigg{[}(\sin^{2}\theta\begin{pmatrix}\hat{-}\hat{-}&0\\\
0&\hat{-}\hat{-}\end{pmatrix}-\sin\theta\cos\theta\begin{pmatrix}\hat{-}\hat{+}&0\\\
0&\hat{-}\hat{+}\end{pmatrix}-\sin\theta\cos\theta\begin{pmatrix}\hat{+}\hat{-}&0\\\
0&\hat{+}\hat{-}\end{pmatrix}+\cos^{2}\theta\begin{pmatrix}\hat{+}\hat{+}&0\\\
0&\hat{+}\hat{+}\end{pmatrix}\bigg{]}$
$\displaystyle=-\frac{1}{2}\begin{pmatrix}\sin^{2}\theta\hat{-}\hat{-}+\cos^{2}\theta\hat{+}\hat{+}-\sin\theta\cos\theta(\hat{-}\hat{+}+\hat{+}\hat{-})&0\\\
0&\sin^{2}\theta\hat{-}\hat{-}+\cos^{2}\theta\hat{+}\hat{+}-\sin\theta\cos\theta(\hat{-}\hat{+}+\hat{+}\hat{-})\end{pmatrix}$
$\displaystyle=-\frac{1}{2}\Big{(}\sin^{2}\theta\hat{-}\hat{-}+\cos^{2}\theta\hat{+}\hat{+}-\sin\theta\cos\theta(\hat{-}\hat{+}+\hat{+}\hat{-})\Big{)}\begin{pmatrix}1&0\\\
0&1\end{pmatrix}$
$\displaystyle=-\frac{1}{2}\Big{(}\sin^{2}\theta\hat{a}^{\dagger}\hat{a}^{\dagger}+\cos^{2}\theta\hat{a}\hat{a}-\sin\theta\cos\theta(\hat{a}\hat{a}^{\dagger}+\hat{a}^{\dagger}\hat{a})\Big{)}\boldsymbol{1}$
$\displaystyle=-\frac{1}{2}\Big{(}\sin^{2}\theta\hat{a}^{\dagger}\hat{a}^{\dagger}+\cos^{2}\theta\hat{a}\hat{a}\Big{)}+\frac{1}{2}(2\sin\theta\cos\theta)\bigg{(}\hat{a}^{\dagger}\hat{a}+\frac{1}{2}\bigg{)}\boldsymbol{1}$
$\displaystyle=-\frac{1}{2}\Big{(}\sin^{2}\theta\hat{a}^{\dagger}\hat{a}^{\dagger}+\cos^{2}\theta\hat{a}\hat{a}\Big{)}+\sin\theta\cos\theta\bigg{(}\hat{a}^{\dagger}\hat{a}+\frac{1}{2}\bigg{)}\boldsymbol{1}\,.$

Then

$\displaystyle\quad\hat{\Psi}^{2}(\theta)$ (710)
$\displaystyle=\frac{1}{2}\Big{(}\cos\theta(\hat{-}\oplus\hat{-})+\sin\theta(\hat{+}\oplus\hat{+})\Big{)}\Big{(}\cos\theta(\hat{-}\oplus\hat{-})+\sin\theta(\hat{+}\oplus\hat{+})\Big{)}$
$\displaystyle=\frac{1}{2}\Big{(}\cos^{2}\theta(\hat{-}\oplus\hat{-})(\hat{-}\oplus\hat{-})+\cos\theta\sin\theta(\hat{-}\oplus\hat{-})(\hat{+}\oplus\hat{+})$
$\displaystyle\quad\quad+\sin\theta\cos\theta(\hat{+}\oplus\hat{+})(\hat{-}\oplus\hat{-})+\sin^{2}\theta(\hat{+}\oplus\hat{+})(\hat{+}\oplus\hat{+})\Big{)}$
$\displaystyle=\frac{1}{2}\Big{(}\cos^{2}\theta(\hat{-}\hat{-}\oplus\hat{-}\hat{-})+\cos\theta\sin\theta(\hat{-}\hat{+}\oplus\hat{-}\hat{+})+\cos\theta\sin\theta(\hat{+}\hat{-}\oplus\hat{+}\hat{-})+\sin^{2}\theta(\hat{+}\hat{+}\oplus\hat{+}\hat{+})\Big{)}$
$\displaystyle=\frac{1}{2}\bigg{[}(\cos^{2}\theta\begin{pmatrix}\hat{-}\hat{-}&0\\\
0&\hat{-}\hat{-}\end{pmatrix}+\cos\theta\sin\theta\begin{pmatrix}\hat{-}\hat{+}&0\\\
0&\hat{-}\hat{+}\end{pmatrix}+\cos\theta\sin\theta\begin{pmatrix}\hat{+}\hat{-}&0\\\
0&\hat{+}\hat{-}\end{pmatrix}+\sin^{2}\theta\begin{pmatrix}\hat{+}\hat{+}&0\\\
0&\hat{+}\hat{+}\end{pmatrix}\bigg{]}$
$\displaystyle=\frac{1}{2}\begin{pmatrix}\cos^{2}\theta\hat{-}\hat{-}+\sin^{2}\theta\hat{+}\hat{+}+\cos\theta\sin\theta(\hat{-}\hat{+}+\hat{+}\hat{-})&0\\\
0&\cos^{2}\theta\hat{-}\hat{-}+\sin^{2}\theta\hat{+}\hat{+}+\cos\theta\sin\theta(\hat{-}\hat{+}+\hat{+}\hat{-})\end{pmatrix}$
$\displaystyle=\frac{1}{2}\Big{(}\cos^{2}\theta\hat{-}\hat{-}+\sin^{2}\theta\hat{+}\hat{+}+\cos\theta\sin\theta(\hat{-}\hat{+}+\hat{+}\hat{-})\Big{)}\begin{pmatrix}1&0\\\
0&1\end{pmatrix}$
$\displaystyle=\frac{1}{2}\Big{(}\cos^{2}\theta\hat{a}^{\dagger}\hat{a}^{\dagger}+\sin^{2}\theta\hat{a}\hat{a}+\cos\theta\sin\theta(\hat{a}\hat{a}^{\dagger}+\hat{a}^{\dagger}\hat{a})\Big{)}\boldsymbol{1}$
$\displaystyle=\frac{1}{2}\Big{(}\cos^{2}\theta\hat{a}^{\dagger}\hat{a}^{\dagger}+\sin^{2}\theta\hat{a}\hat{a}\Big{)}\boldsymbol{1}+\frac{1}{2}(2\cos\theta\sin\theta)\bigg{(}\hat{a}^{\dagger}\hat{a}+\frac{1}{2}\bigg{)}\boldsymbol{1}$
$\displaystyle=\frac{1}{2}\Big{(}\cos^{2}\theta\hat{a}^{\dagger}\hat{a}^{\dagger}+\sin^{2}\theta\hat{a}\hat{a}\Big{)}\boldsymbol{1}+\sin\theta\cos\theta\bigg{(}\hat{a}^{\dagger}\hat{a}+\frac{1}{2}\bigg{)}\boldsymbol{1}\,.$

Therefore, finally we obtain the Hamiltonian as follow

$\hat{H}(\theta)=\bigg{(}\frac{1}{2}\hat{\Pi}^{2}(\theta)+\frac{1}{2}\hat{\Psi}^{2}(\theta)\bigg{)}\hbar\omega=\frac{1}{2}\cos
2\theta(\hat{a}^{\dagger}\hat{a}^{\dagger}-\hat{a}\hat{a})\hbar\omega\boldsymbol{1}+\sin
2\theta\bigg{(}\hat{a}^{\dagger}\hat{a}+\frac{1}{2}\bigg{)}\hbar\omega\boldsymbol{1}$
(711)

The expectation energy is hence given by

$\displaystyle\langle E\rangle=\langle n|\hat{H}(\theta)|n\rangle$
$\displaystyle=\frac{1}{2}\cos 2\theta\langle
n|(\hat{a}^{\dagger}\hat{a}^{\dagger}-\hat{a}\hat{a})|n\rangle\hbar\omega\,\boldsymbol{1}+\sin
2\theta\langle
n|\bigg{(}\hat{a}^{\dagger}\hat{a}+\frac{1}{2}\bigg{)}|n\rangle\hbar\omega\boldsymbol{1}$
(712) $\displaystyle=\sin
2\theta\bigg{(}n+\frac{1}{2}\bigg{)}\hbar\omega\boldsymbol{1}$
$\displaystyle=H\sin 2\theta\,\boldsymbol{1}\,,$

where

$H=\bigg{(}n+\frac{1}{2}\bigg{)}\hbar\omega.$ (713)

This is because $\langle n|\hat{a}^{\dagger}\hat{a}^{\dagger}|n\rangle=\langle
n|\hat{a}\hat{a}|n\rangle=0$ and $\langle
n|\hat{a}^{\dagger}\hat{a}|n\rangle=n$. Thus this gives the oscillation of
expectation of energy which depends upon the phase, which defines the quanta
of duality wave, the dualiton. The vacuum expectation energy of the dualiton
is given by

$\langle 0|\hat{H}(\theta)|0\rangle=\frac{1}{2}\sin
2\theta\hbar\omega\,\boldsymbol{1}=\sin\theta\cos\theta\hbar\omega\,\boldsymbol{1}\,.$
(714)

Now let’s evaluate the vacuum expectation value for different important
phases. When $\theta=0$ or $\frac{\pi}{2}$, we have

$E_{0}=\langle 0|\hat{H}(\theta)|0\rangle=\boldsymbol{0}\,.$ (715)

Recalling that $\Psi(\theta)=\cos\theta|0\oplus 0\rangle+\sin\theta|1\oplus
1\rangle=\cos\theta|\text{nothingness}\rangle+\sin\theta|\text{All}\rangle$ ,
this means when the dual wave function is at a particular state, then it has
zero vacuum expectation energy value. On the other hand, when $\theta=\pi/4$,

$E_{0}=\langle
0|\hat{H}(\theta)|0\rangle=\frac{1}{2}\hbar\omega\,\boldsymbol{1}=\begin{pmatrix}\frac{1}{2}\hbar\omega&0\\\
0&\frac{1}{2}\hbar\omega\end{pmatrix}\,.$ (716)

This means that when $\Psi(\frac{\pi}{4})=\frac{1}{\sqrt{2}}|0\oplus
0\rangle+\frac{1}{\sqrt{2}}|1\oplus
1\rangle=\frac{1}{\sqrt{2}}|\text{nothingness}\rangle+\frac{1}{\sqrt{2}}|\text{All}\rangle$,
the state is in linear superposition of the two states has its greatest
entropy, it has the maximized energy. This shows that when the system is in
its determined state of $|0\oplus 0\rangle$ or $|1\oplus 1\rangle$ it has
lower ground state energy than the undetermined superpositioned state of
$\Psi=\frac{1}{\sqrt{2}}|0\oplus 0\rangle+\frac{1}{\sqrt{2}}|1\oplus
1\rangle$, thus is considered to be more stable. We say the superpositioned
state is spontaneously collapsed into one of the state of full probability,
this statement is equivalent to spontaneous symmetry breaking when a state is
completely determined.

Next we evaluate $\hat{H}^{\star}(\theta)$.

$\displaystyle\quad\hat{\Pi}^{\star 2}(\theta)$ (717)
$\displaystyle=\frac{1}{2i^{2}}\Big{(}-\sin\theta(\hat{+}\oplus\hat{-})+\cos\theta(\hat{-}\oplus\hat{+})\Big{)}\Big{(}-\sin\theta(\hat{+}\oplus\hat{-})+\cos\theta(\hat{-}\oplus\hat{+})\Big{)}$
$\displaystyle=-\frac{1}{2}\Big{(}\sin^{2}\theta(\hat{+}\oplus\hat{-})(\hat{+}\oplus\hat{-})-\sin\theta\cos\theta(\hat{+}\oplus\hat{-})(\hat{-}\oplus\hat{+})$
$\displaystyle\quad\quad-\cos\theta\sin\theta(\hat{-}\oplus\hat{+})(\hat{+}\oplus\hat{-})+\cos^{2}\theta(\hat{-}\oplus\hat{+})(\hat{-}\oplus\hat{+})\Big{)}$
$\displaystyle=-\frac{1}{2}\Big{(}\sin^{2}\theta(\hat{+}\hat{+}\oplus\hat{-}\hat{-})-\sin\theta\cos\theta(\hat{+}\hat{-}\oplus\hat{-}\hat{+})-\sin\theta\cos\theta(\hat{-}\hat{+}\oplus\hat{+}\hat{-})+\cos^{2}\theta(\hat{-}\hat{-}\oplus\hat{+}\hat{+})\Big{)}$
$\displaystyle=-\frac{1}{2}\bigg{[}(\sin^{2}\theta\begin{pmatrix}\hat{+}\hat{+}&0\\\
0&\hat{-}\hat{-}\end{pmatrix}-\sin\theta\cos\theta\begin{pmatrix}\hat{+}\hat{-}&0\\\
0&\hat{-}\hat{+}\end{pmatrix}-\sin\theta\cos\theta\begin{pmatrix}\hat{-}\hat{+}&0\\\
0&\hat{+}\hat{-}\end{pmatrix}+\cos^{2}\theta\begin{pmatrix}\hat{-}\hat{-}&0\\\
0&\hat{+}\hat{+}\end{pmatrix}\bigg{]}$
$\displaystyle=-\frac{1}{2}\begin{pmatrix}\sin^{2}\theta\hat{+}\hat{+}+\cos^{2}\theta\hat{-}\hat{-}-\sin\theta\cos\theta(\hat{+}\hat{-}+\hat{-}\hat{+})&0\\\
0&\sin^{2}\theta\hat{-}\hat{-}+\cos^{2}\theta\hat{+}\hat{+}-\sin\theta\cos\theta(\hat{-}\hat{+}+\hat{+}\hat{-})\end{pmatrix}$
$\displaystyle=-\frac{1}{2}\begin{pmatrix}\sin^{2}\theta\hat{a}\hat{a}+\cos^{2}\theta\hat{a}^{\dagger}\hat{a}^{\dagger}-\sin\theta\cos\theta(\hat{a}\hat{a}^{\dagger}+\hat{a}^{\dagger}\hat{a})&0\\\
0&\sin^{2}\theta\hat{a}^{\dagger}\hat{a}^{\dagger}+\cos^{2}\theta\hat{a}\hat{a}-\sin\theta\cos\theta(\hat{a}^{\dagger}\hat{a}+\hat{a}\hat{a}^{\dagger})\end{pmatrix}$
$\displaystyle=-\frac{1}{2}\begin{pmatrix}\sin^{2}\theta\hat{a}\hat{a}+\cos^{2}\theta\hat{a}^{\dagger}\hat{a}^{\dagger}-\sin
2\theta\Big{(}\hat{a}^{\dagger}\hat{a}+\frac{1}{2}\Big{)}&0\\\
0&\sin^{2}\theta\hat{a}^{\dagger}\hat{a}^{\dagger}+\cos^{2}\theta\hat{a}\hat{a}-\sin
2\theta\Big{(}\hat{a}^{\dagger}\hat{a}+\frac{1}{2}\Big{)}\end{pmatrix}\,.$

Then

$\displaystyle\quad\hat{\Psi}^{\star 2}(\theta)$ (718)
$\displaystyle=\frac{1}{2}\Big{(}\cos\theta(\hat{+}\oplus\hat{-})+\sin\theta(\hat{-}\oplus\hat{+})\Big{)}\Big{(}\cos\theta(\hat{+}\oplus\hat{-})+\sin\theta(\hat{-}\oplus\hat{+})\Big{)}$
$\displaystyle=\frac{1}{2}\Big{(}\cos^{2}\theta(\hat{+}\oplus\hat{-})(\hat{+}\oplus\hat{-})+\cos\theta\sin\theta(\hat{+}\oplus\hat{-})(\hat{-}\oplus\hat{+})$
$\displaystyle\quad\quad+\sin\theta\cos\theta(\hat{-}\oplus\hat{+})(\hat{+}\oplus\hat{-})+\sin^{2}\theta(\hat{-}\oplus\hat{+})(\hat{-}\oplus\hat{+})\Big{)}$
$\displaystyle=\frac{1}{2}\Big{(}\cos^{2}\theta(\hat{+}\hat{+}\oplus\hat{-}\hat{-})+\cos\theta\sin\theta(\hat{+}\hat{-}\oplus\hat{-}\hat{+})+\cos\theta\sin\theta(\hat{-}\hat{+}\oplus\hat{+}\hat{-})+\sin^{2}\theta(\hat{-}\hat{-}\oplus\hat{+}\hat{+})\Big{)}$
$\displaystyle=\frac{1}{2}\bigg{[}(\cos^{2}\theta\begin{pmatrix}\hat{+}\hat{+}&0\\\
0&\hat{-}\hat{-}\end{pmatrix}+\cos\theta\sin\theta\begin{pmatrix}\hat{+}\hat{-}&0\\\
0&\hat{-}\hat{+}\end{pmatrix}+\cos\theta\sin\theta\begin{pmatrix}\hat{-}\hat{+}&0\\\
0&\hat{+}\hat{-}\end{pmatrix}+\sin^{2}\theta\begin{pmatrix}\hat{-}\hat{-}&0\\\
0&\hat{+}\hat{+}\end{pmatrix}\bigg{]}$
$\displaystyle=\frac{1}{2}\begin{pmatrix}\cos^{2}\theta\hat{+}\hat{+}+\sin^{2}\theta\hat{-}\hat{-}+\cos\theta\sin\theta(\hat{+}\hat{-}+\hat{-}\hat{+})&0\\\
0&\cos^{2}\theta\hat{-}\hat{-}+\sin^{2}\theta\hat{+}\hat{+}+\cos\theta\sin\theta(\hat{-}\hat{+}+\hat{+}\hat{-})\end{pmatrix}$
$\displaystyle=\frac{1}{2}\begin{pmatrix}\cos^{2}\theta\hat{a}\hat{a}+\sin^{2}\theta\hat{a}^{\dagger}\hat{a}^{\dagger}+\cos\theta\sin\theta(\hat{a}\hat{a}^{\dagger}+\hat{a}^{\dagger}\hat{a})&0\\\
0&\cos^{2}\theta\hat{a}^{\dagger}\hat{a}^{\dagger}+\sin^{2}\theta\hat{a}\hat{a}+\cos\theta\sin\theta(\hat{a}^{\dagger}\hat{a}+\hat{a}\hat{a}^{\dagger})\end{pmatrix}$
$\displaystyle=\frac{1}{2}\begin{pmatrix}\cos^{2}\theta\hat{a}\hat{a}+\sin^{2}\theta\hat{a}^{\dagger}\hat{a}^{\dagger}+\sin
2\theta\Big{(}\hat{a}^{\dagger}\hat{a}+\frac{1}{2}\Big{)}&0\\\
0&\cos^{2}\theta\hat{a}^{\dagger}\hat{a}^{\dagger}+\sin^{2}\theta\hat{a}\hat{a}+\sin
2\theta\Big{(}\hat{a}^{\dagger}\hat{a}+\frac{1}{2}\Big{)}\end{pmatrix}\,.$

Therefore, finally we obtain the Hamiltonian as follow

$\displaystyle\quad\hat{H}^{\star}(\theta)=\bigg{(}\frac{1}{2}\hat{\Pi}^{\star
2}(\theta)+\frac{1}{2}\hat{\Psi}^{\star 2}(\theta)\bigg{)}\hbar\omega$ (719)
$\displaystyle=\begin{pmatrix}\frac{1}{2}\cos
2\theta(\hat{a}\hat{a}-\hat{a}^{\dagger}\hat{a}^{\dagger})+\sin
2\theta\Big{(}\hat{a}^{\dagger}\hat{a}+\frac{1}{2}\Big{)}&0\\\
0&\frac{1}{2}\cos
2\theta(\hat{a}^{\dagger}\hat{a}^{\dagger}-\hat{a}\hat{a})+\sin
2\theta\Big{(}\hat{a}^{\dagger}\hat{a}+\frac{1}{2}\Big{)}\end{pmatrix}\hbar\omega$
$\displaystyle=\frac{1}{2}\cos
2\theta(\hat{a}^{\dagger}\hat{a}^{\dagger}-\hat{a}\hat{a})\hbar\omega M+\sin
2\theta\bigg{(}\hat{a}^{\dagger}\hat{a}+\frac{1}{2}\bigg{)}\hbar\omega\boldsymbol{1}\,.$

The expectation energy is hence given by

$\displaystyle\langle E^{\star}\rangle=\langle
n|\hat{H}^{\star}(\theta)|n\rangle$ $\displaystyle=\frac{1}{2}\cos
2\theta\langle n|(\hat{a}^{\dagger}\hat{a}^{\dagger}-\hat{a}\hat{a})|n\rangle
h\omega M+\sin 2\theta\langle
n|\bigg{(}\hat{a}^{\dagger}\hat{a}+\frac{1}{2}\bigg{)}|n\rangle
h\omega\boldsymbol{1}$ (720) $\displaystyle=\sin
2\theta\bigg{(}n+\frac{1}{2}\bigg{)}\hbar\omega\boldsymbol{1}$
$\displaystyle=H\sin 2\theta\,\boldsymbol{1}\,,$

Therefore the expectation energy of $\Psi^{*}(\theta)$ is same as the case of
$\Psi(\theta)$. That means

$\langle E(\theta)\rangle=\langle E^{\star}(\theta)\rangle=H\sin
2\theta\boldsymbol{1}\,.$ (721)

Hence the phase analysis remains the same as the $\Psi(\theta)$ case.

Figure 113: For the left figure, we compute the ground state $n=0$ for phase
$\theta=\pi/4,0,\pi/2$, corresponding to
$|\Psi(\pi/4)\rangle=\frac{1}{\sqrt{2}}(|0\oplus 0\rangle+|1\oplus
1\rangle),|\Psi(0)\rangle=|0\oplus
0\rangle\,\,\text{and}\,\,|\Psi(\pi/2)\rangle=|1\oplus 1\rangle$ respectively.
The $|\Psi(\pi/4)\rangle$ is a quantum state with probability of each state of
$1/2$, it has the average ground state energy of $\frac{1}{2}\hbar\omega$,
while $|\Psi(0)\rangle$ and $|\Psi(\pi/2)\rangle$ are classical states, with
probability equal to 1 with average ground energy equal to zero. The idea is
same for the right figure.

In the Hamiltonian, we have the extra term of
$\hat{a}^{\dagger}\hat{a}^{\dagger}-\hat{a}\hat{a}$ in addition to the
standard Hamiltonian of simple harmonic oscillator. We would like to
investigate its meaning here. First consider

$\displaystyle\quad(\hat{a}^{\dagger}+\hat{a})(\hat{a}^{\dagger}-\hat{a})$
(722)
$\displaystyle=\hat{a}^{\dagger}\hat{a}^{\dagger}-\hat{a}^{\dagger}\hat{a}+\hat{a}\hat{a}^{\dagger}-\hat{a}\hat{a}$
$\displaystyle=\hat{a}^{\dagger}\hat{a}^{\dagger}-\hat{a}\hat{a}+[\hat{a},\hat{a}^{\dagger}]$
$\displaystyle=\hat{a}^{\dagger}\hat{a}^{\dagger}-\hat{a}\hat{a}+1$

Therefore

$\hat{a}^{\dagger}\hat{a}^{\dagger}-\hat{a}\hat{a}=(\hat{a}^{\dagger}+\hat{a})(\hat{a}^{\dagger}-\hat{a})-1\,.$
(723)

But recall that in quantum mechanics

$\hat{x}=\sqrt{\frac{\hbar}{2m\omega}}(\hat{a}^{\dagger}+\hat{a})\quad\text{and}\quad\hat{p}=i\sqrt{\frac{\hbar
m\omega}{2}}(\hat{a}^{\dagger}-\hat{a})\,.$ (724)

It follows that

$\hat{a}^{\dagger}\hat{a}^{\dagger}-\hat{a}\hat{a}=\sqrt{\frac{2m\omega}{\hbar}}\hat{x}\cdot\frac{1}{i}\sqrt{\frac{2}{\hbar
m\omega}}\hat{p}-1=\frac{2}{i\hbar}\hat{x}\hat{p}-1\,.$ (725)

Therefore

$\displaystyle\hat{H}(\theta)$ $\displaystyle=\cos
2\theta\bigg{(}\frac{1}{ih}\hat{x}\hat{p}-\frac{1}{2}\bigg{)}\hbar\omega\boldsymbol{1}+\sin
2\theta\bigg{(}\hat{a}^{\dagger}\hat{a}+\frac{1}{2}\bigg{)}\hbar\omega\boldsymbol{1}$
(726) $\displaystyle=\bigg{(}\frac{1}{i\hbar}\cos 2\theta\,\hat{x}\hat{p}+\sin
2\theta\hat{a}^{\dagger}\hat{a}\bigg{)}\hbar\omega\,\boldsymbol{1}+\frac{1}{2}(\sin
2\theta-\cos 2\theta)\hbar\omega\,\boldsymbol{1}$ $\displaystyle=(\sin
2\theta\hat{n}-i\cos
2\theta\hat{\theta})\hbar\omega\,\boldsymbol{1}+\frac{1}{2}(\sin 2\theta-\cos
2\theta)\hbar\omega\,\boldsymbol{1}\,,$

where

$\hat{\theta}=\frac{\hat{x}\hat{p}}{\hbar}=\hat{x}\frac{\partial}{\partial x}$
(727)

is the phase operator.

However, instead we can have

$\displaystyle\quad(\hat{a}^{\dagger}-\hat{a})(\hat{a}^{\dagger}+\hat{a})$
(728)
$\displaystyle=\hat{a}^{\dagger}\hat{a}^{\dagger}+\hat{a}^{\dagger}\hat{a}-\hat{a}\hat{a}^{\dagger}-\hat{a}\hat{a}$
$\displaystyle=\hat{a}^{\dagger}\hat{a}^{\dagger}-\hat{a}\hat{a}+[\hat{a}^{\dagger},\hat{a}]$
$\displaystyle=\hat{a}^{\dagger}\hat{a}^{\dagger}-\hat{a}\hat{a}-1$

Therefore

$\hat{a}^{\dagger}\hat{a}^{\dagger}-\hat{a}\hat{a}=(\hat{a}^{\dagger}-\hat{a})(\hat{a}^{\dagger}+\hat{a})+1=\frac{2}{i\hbar}\hat{p}\hat{x}+1\,.$
(729)

Then

$\displaystyle\hat{H}(\theta)$ $\displaystyle=\cos
2\theta\bigg{(}\frac{1}{ih}\hat{p}\hat{x}+\frac{1}{2}\bigg{)}\hbar\omega\boldsymbol{1}+\sin
2\theta\bigg{(}\hat{a}^{\dagger}\hat{a}+\frac{1}{2}\bigg{)}\hbar\omega\boldsymbol{1}$
(730) $\displaystyle=\bigg{(}\frac{1}{i\hbar}\cos 2\theta\,\hat{p}\hat{x}+\sin
2\theta\hat{a}^{\dagger}\hat{a}\bigg{)}\hbar\omega\,\boldsymbol{1}+\frac{1}{2}(\sin
2\theta+\cos 2\theta)\hbar\omega\,\boldsymbol{1}$ $\displaystyle=(\sin
2\theta\hat{n}-i\cos
2\theta\hat{\theta}^{*})\hbar\omega\,\boldsymbol{1}+\frac{1}{2}(\sin
2\theta+\cos 2\theta)\hbar\omega\,\boldsymbol{1}\,,$

where

$\hat{\theta}^{*}=\frac{\hat{p}\hat{x}}{\hbar}=\frac{\partial}{\partial
x}\hat{x}$ (731)

is the dual phase operator.

Next we evaluate the phase expectation value. Consider

$\langle
n|\hat{a}^{\dagger}\hat{a}^{\dagger}-\hat{a}\hat{a}|n\rangle=0=\langle
n|\frac{2}{i\hbar}\hat{x}\hat{p}-1|n\rangle=\frac{2}{i\hbar}\langle
n|\hat{x}\hat{p}|n\rangle-\langle n|n\rangle=\frac{2}{i\hbar}\langle
n|\hat{x}\hat{p}|n\rangle-1\\\ $ (732)

Therefore we have

$\langle n|\hat{\theta}|n\rangle=\frac{i}{2}\,.$ (733)

On the other hand

$\langle
n|\hat{a}^{\dagger}\hat{a}^{\dagger}-\hat{a}\hat{a}|n\rangle=0=\langle
n|\frac{2}{i\hbar}\hat{p}\hat{x}+1|n\rangle=\frac{2}{i\hbar}\langle
n|\hat{x}\hat{p}|n\rangle+\langle n|n\rangle=\frac{2}{i\hbar}\langle
n|\hat{x}\hat{p}|n\rangle+1\\\ $ (734)

Therefore we have

$\langle n|\hat{\theta}^{*}|n\rangle=\frac{-i}{2}\,.$ (735)

Then we have

$\langle n|\hat{\theta}+\hat{\theta}^{*}|n\rangle=0\,.$ (736)

Thus we have the dual invariant operator as

$\hat{\theta}+\hat{\theta}^{*}=\hat{x}\frac{\partial}{\partial
x}+\frac{\partial}{\partial x}\hat{x}\,.$ (737)

Therefore, we can see that $\hat{a}^{\dagger}\hat{a}^{\dagger}-\hat{a}\hat{a}$
relates to the dimensionless phase term, contrasting to then occupation number
term $\hat{n}$.

From equation 726 , in particular at phase $\theta=\pi/8$, we have equal
portion of $\hat{n}$ and $\hat{\theta}$,

$\hat{H}\Big{(}\frac{\pi}{8}\Big{)}=\frac{1}{\sqrt{2}}(\hat{n}-i\hat{\theta})\hbar\omega\,\boldsymbol{1}\,.$
(738)

We can see that at different phase, the Hamiltonian is dominated by different
behaviour. At $\theta=\pi/4$, it is energy dominated,

$H\Big{(}\frac{\pi}{4}\Big{)}=\bigg{(}\hat{a}^{\dagger}\hat{a}+\frac{1}{2}\bigg{)}\hbar\omega\boldsymbol{1}\,.$
(739)

At $\theta=0$, it is phase dominated,

$H(0)=\frac{1}{2}(\hat{a}^{\dagger}\hat{a}^{\dagger}-\hat{a}\hat{a})\hbar\omega\boldsymbol{1}=\bigg{(}-i\hat{\theta}-\frac{1}{2}\bigg{)}\hbar\omega\boldsymbol{1}\,.$
(740)

###  10.4 Partition function

In this section, we would study the partition function of the duality wave
system. First, the phase dependent partition function is given by

$Z(\theta)=\mathrm{Tr}\Big{[}\exp{\Big{(}-\beta\hat{H}(\theta)}\Big{)}\Big{]}\,.$
(741)

For the $\hat{\Psi}(\theta)$ field case, according to the Hamiltonian in 711,
it follows that

$\displaystyle Z(\theta)$
$\displaystyle=\mathrm{Tr}\Big{[}\exp\Big{(}-\frac{\beta}{2}\cos
2\theta(\hat{a}^{\dagger}\hat{a}^{\dagger}-\hat{a}\hat{a})\hbar\omega\boldsymbol{1}-\beta\sin
2\theta\Big{(}\hat{a}^{\dagger}\hat{a}+\frac{1}{2}\Big{)}\hbar\omega\boldsymbol{1}\Big{)}\Big{]}$
(742) $\displaystyle=\sum_{n}\langle n|\exp\Big{(}-\frac{\beta}{2}\cos
2\theta(\hat{a}^{\dagger}\hat{a}^{\dagger}-\hat{a}\hat{a})\hbar\omega\boldsymbol{1}-\beta\sin
2\theta\bigg{(}\hat{a}^{\dagger}\hat{a}+\frac{1}{2}\bigg{)}\hbar\omega\boldsymbol{1}\Big{)}|n\rangle$
$\displaystyle=\sum_{n}\exp\Big{(}-\sin
2\theta\Big{(}n+\frac{1}{2}\Big{)}\beta\hbar\omega\boldsymbol{1}\Big{)}$

The expectation energy is

$\displaystyle\quad\langle E(\theta)\rangle=-\frac{\partial}{\partial\beta}\ln
Z(\theta)$ (743)
$\displaystyle=-\frac{\partial}{\partial\beta}\ln\Big{(}\sum_{n}\exp\Big{(}-\sin
2\theta\Big{(}n+\frac{1}{2}\Big{)}\beta\hbar\omega\boldsymbol{1}\Big{)}\Big{)}$
$\displaystyle=\frac{\sum_{n}\sin
2\theta\big{(}n+\frac{1}{2}\big{)}\hbar\omega\exp\Big{(}-\sin
2\theta\Big{(}n+\frac{1}{2}\Big{)}\beta\hbar\omega\boldsymbol{1}\Big{)}}{\sum_{n}\exp\Big{(}-\sin
2\theta\Big{(}n+\frac{1}{2}\Big{)}\beta\hbar\omega\boldsymbol{1}\Big{)}}$
$\displaystyle=\frac{\exp\big{(}-\frac{\sin
2\theta}{2}\beta\hbar\omega\boldsymbol{1}\big{)}\big{[}\sum_{n}n\hbar\omega\sin
2\theta\exp\big{(}-\sin 2\theta
n\beta\hbar\omega\boldsymbol{1}\big{)}+\sum_{n}\frac{1}{2}\hbar\omega\sin
2\theta\exp\big{(}-\sin 2\theta\beta
n\hbar\omega\boldsymbol{1}\big{)}\big{]}}{\exp\big{(}-\frac{\sin
2\theta}{2}\beta\hbar\omega\boldsymbol{1}\big{)}\sum_{n}\exp\big{(}-\sin
2\theta n\beta\hbar\omega\boldsymbol{1}\big{)}}$
$\displaystyle=\frac{\sum_{n}n\hbar\omega\sin 2\theta\exp\big{(}-\sin 2\theta
n\beta\hbar\omega\boldsymbol{1}\big{)}}{\sum_{n}\exp\big{(}-\sin 2\theta
n\beta\hbar\omega\boldsymbol{1}\big{)}}+\frac{1}{2}\sin
2\theta\hbar\omega\boldsymbol{1}$ $\displaystyle=\hbar\omega\sin
2\theta\frac{\sum_{n}n\exp\big{(}-\sin 2\theta
n\beta\hbar\omega\boldsymbol{1}\big{)}}{\sum_{n}\exp\big{(}-\sin 2\theta
n\beta\hbar\omega\boldsymbol{1}\big{)}}+\frac{1}{2}\sin
2\theta\hbar\omega\boldsymbol{1}\,.$

Then let $X=\exp\big{(}-\sin 2\theta\hbar\omega\boldsymbol{1}\big{)}$ , and
using the identity,

$\sum_{n=0}^{\infty}X^{n}=\frac{\boldsymbol{1}}{\boldsymbol{1}-X}\,.$ (744)

It follows that

$\sum_{n=0}^{\infty}\exp\big{(}-\sin 2\theta
n\beta\hbar\omega\boldsymbol{1}\big{)}=\frac{\boldsymbol{1}}{\boldsymbol{1}-\exp\big{(}-\sin
2\theta\hbar\omega\boldsymbol{1}\big{)}}\,.$ (745)

Then using the identity of

$\displaystyle\sum_{n=0}^{\infty}\frac{d}{dX}X^{n}$
$\displaystyle=\frac{\boldsymbol{1}}{(\boldsymbol{1}-X)^{2}}$ (746)
$\displaystyle\lim_{n\rightarrow\infty}(X+2X^{2}+3X^{3}+\cdots+nX^{n})$
$\displaystyle=\frac{X}{(\boldsymbol{1}-X)^{2}}\,.$

Therefore,

$\sum_{n=0}^{\infty}n\exp\big{(}-\sin 2\theta
n\beta\hbar\omega\boldsymbol{1}\big{)}=\frac{\exp\big{(}-\sin 2\theta
n\beta\hbar\omega\boldsymbol{1}\big{)}}{\boldsymbol{1}-\exp\big{(}-\sin
2\theta n\beta\hbar\omega\boldsymbol{1}\big{)}}\,.$ (747)

Next

$\lim_{N\rightarrow\infty}\frac{\sum^{N}_{n=0}nX^{n}}{\sum_{n=0}^{N}X^{n}}=\frac{X}{1-X}\,.$
(748)

Then

$\lim_{N\rightarrow\infty}\frac{\sum_{n=0}^{N}n\exp\big{(}-\sin 2\theta
n\beta\hbar\omega\boldsymbol{1}\big{)}}{\sum_{n=0}^{N}\exp\big{(}-\sin 2\theta
n\beta\hbar\omega\boldsymbol{1}\big{)}}=\frac{\exp\big{(}-\sin 2\theta
n\beta\hbar\omega\boldsymbol{1}\big{)}}{\boldsymbol{1}-\exp\big{(}-\sin
2\theta
n\beta\hbar\omega\boldsymbol{1}\big{)}}=\frac{\boldsymbol{1}}{\exp\big{(}\sin
2\theta n\beta\hbar\omega\boldsymbol{1}\big{)}-\boldsymbol{1}}\,.$ (749)

Hence finally we obtain

$\langle E(\theta)\rangle=\frac{\hbar\omega\sin
2\theta\boldsymbol{1}}{\exp\big{(}\sin 2\theta
n\beta\hbar\omega\boldsymbol{1}\big{)}-\boldsymbol{1}}+\frac{1}{2}\sin
2\theta\hbar\omega\boldsymbol{1}\,.$ (750)

Now we evaluate $\langle E(\theta)\rangle$ in both high and low temperature
limit. In high temperature limit $k_{B}T>>\hbar\omega$, we have

$\langle E(\theta)\rangle\approx\frac{\hbar\omega\sin
2\theta\boldsymbol{1}}{\boldsymbol{1}+\sin 2\theta
n\beta\hbar{\omega}\boldsymbol{1}-\boldsymbol{1}}+\frac{1}{2}\sin
2\theta\hbar\omega\boldsymbol{1}=\frac{1}{\beta}=k_{B}T=\text{constant}\,,$
(751)

which is independent of the phase $\theta$.

In low temperature limit $\hbar{\omega}>>k_{B}T$ , or $T\rightarrow 0$,
$\exp\big{(}\sin 2\theta
n\beta\hbar\omega\boldsymbol{1}\big{)}\rightarrow\infty$, hence

$\langle E(\theta)\rangle=\frac{1}{2}\sin 2\theta\hbar\omega\boldsymbol{1}\,.$
(752)

This is just the same as the average ground state energy in 714 for $n=0$
case. That means at $T=0K$, all states are at their ground state.

##  11 Conclusion

In this paper, we demonstrate how the fundamental concept of duality can be
represented by rigorous mathematical and quantum physics formalism. The
symmetry of duality and multi-duality are represented by the groups
$\mathbb{Z}_{2}$ and $\mathbb{Z}_{2}\otimes\cdots\otimes\mathbb{Z}_{2}$. The
4-duality group $\mathbb{Z}_{2}\otimes\mathbb{Z}_{2}$ plays an important role
in various dual structures, and can be represented by a 4-tableau. The
concepts of dual pairs and dual invariant numbers are also introduced. We have
developed quantum field theory with duality and multi-duality symmetry and
demonstrated how duality is embedded in the interaction terms. We have
discovered that dual invariance implies topological property, for which the
Chern-Simons action and the chiral anomaly action is a topological invariant.
This profound result establishes a relation between dual invariance and
topological invariance. Finally and most importantly, we exhibit how the
oscillation of two dual states would produce a duality wave, and upon field
quantization this produces a quasi-particle quanta called the dualiton. This
is a new exotic particle produced by duality. A physical practice would be
local continuous spin flip of an electron that produces a wave signal, and
such wave is a particle quantum mechanically. This new particle is expected to
have various contributions in particle physics and condensed matter physics.

### Conflict of interest

There is no conflict of interest for this paper.

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