be generally applied for matrices that are not one dimensional (readers can
check that easily). We can take the basis as

$|V\rangle=|v_{1}\rangle\otimes|v_{2}\rangle\otimes\cdots\otimes|v_{N}\rangle=\bigoplus_{\eta_{i_{1}},\eta_{i_{2}}\cdots\eta_{i_{N}}=0,1}|\eta_{i_{1}}\rangle\otimes|\eta_{i_{2}}\rangle\otimes\cdots\otimes|\eta_{i_{N}}\rangle\,,$
(275)

such that each $\eta_{i_{j}}=0,1$. Then each
$|\eta_{i_{1}}\rangle\otimes|\eta_{i_{2}}\rangle\otimes|\eta_{i_{N}}\rangle$
would align with the $A_{i_{1}}(g_{i_{1}})\,A_{i_{2}}(g_{i_{2}})\cdots
A_{i_{N}}(g_{i_{N}})$. Then the set of all
$|\eta_{1}\eta_{2}\cdots\eta_{N}\rangle$ form the basis of irreducible
representation of
$\mathbb{Z}_{2}\otimes\mathbb{Z}_{2}\otimes\cdots\otimes\mathbb{Z}_{2}$.

The consequence of multiplicity for all $A_{i}(\mathrm{g})$ is equal to 1
follows directly from the fifth line. This is because there are $N=2^{n}$
terms for the direct sum therefore we can assign each $A_{i}$ by

$A_{i}(\mathrm{g})=\mathcal{A}_{i_{1}}(g_{i_{1}})\,\mathcal{A}_{i_{2}}(g_{i_{2}})\cdots\mathcal{A}_{i_{N}}(g_{i_{N}})$
(276)

and explicitly we have decomposed $D(\mathrm{g})$

$D(\mathrm{g})=D(g_{i_{1}})\otimes D(g_{i_{2}})\otimes\cdots\otimes
D(g_{i_{N}})=\bigoplus_{i_{1},i_{2},\cdots,i_{N}=1,2}\mathcal{A}_{i_{1}}(g_{i_{1}})\,\mathcal{A}_{i_{2}}(g_{i_{2}})\cdots\mathcal{A}_{i_{N}}(g_{i_{N}})=\bigoplus_{i=1}^{N_{\tau}=2^{n}}a_{i}\mathcal{A}_{i}(\mathrm{g})$
(277)

that $a_{i}=1$ for all $i$. Note that each $A_{i}(\mathrm{g})$ must be either
$1$ or $-1$ as it is the products of $1$s and $-1$s , and this comes from the
fact that character of the parity group can only be $1$ or $-1$.

With the basis defined, now we can construct vector. First consider the vector
for $\mathbb{Z}_{2}$, which is a qubit

$|\psi\rangle=c_{0}|0\rangle+c_{1}|1\rangle=c_{1}|\boldsymbol{\bm{\cdot\cdot}}\rangle+c_{2}|-\rangle\,,$
(278)

where $|c_{1}|^{2}+|c_{2}|^{2}=1$. This is demanded by the probability of
observing the $|0\rangle$ state being $|c_{1}|^{2}$ and that of $|c_{2}|^{2}$
for the $|1\rangle$ state. For example, the simplest case would be, having
half of the probabilities for getting each state,

$|\psi\rangle=\frac{1}{\sqrt{2}}(|0\rangle+|1\rangle)\,.$ (279)

In a more formal way to represent the $j^{\rm th}$ state of the parity group
we write

$|\psi_{j}\rangle=\sum_{\eta_{j}=0,1}c^{(j)}_{\eta_{j}}|\eta_{j}\rangle\,,$
(280)

with normalization of

$|c^{(j)}_{0}|^{2}+|c^{(j)}_{1}|^{2}=1\quad\text{for
all}\,\,j=1,2,\cdots,N\,.$ (281)

The general state vector with $N$ tensor product of individual vector is given
by

$|\Psi\rangle=|\psi_{1}\rangle\otimes|\psi_{2}\rangle\otimes\cdots\otimes|\psi_{N}\rangle=\sum_{\eta_{i_{1}},\eta_{i_{2}},\cdots,\eta_{i_{N}}=0,1}c_{\eta_{i_{1}}}^{(1)}c_{\eta_{i_{2}}}^{(2)}\cdots
c_{\eta_{i_{N}}}^{(N)}|\eta_{i_{1}}\rangle\otimes|\eta_{i_{2}}\rangle\otimes\cdots\otimes|\eta_{i_{N}}\rangle\,.$
(282)

Then we have the tensor component as

$T_{i_{1}\,i_{2}\,\cdots\,i_{N}}=c_{\eta_{i_{1}}}^{(1)}c_{\eta_{i_{2}}}^{(2)}\cdots
c_{\eta_{i_{N}}}^{(N)}\,.$ (283)

And we demand the completeness relation by the sum of the probability of each
tensor product state be unity,

$\sum_{\eta_{i_{1}},\eta_{i_{2}},\cdots,\eta_{i_{N}}=0,1}|c_{\eta_{i_{1}}}^{(1)}c_{\eta_{i_{2}}}^{(2)}\cdots
c_{\eta_{i_{N}}}^{(N)}|^{2}=1\,.$ (284)

Then we have to solve (281) and (284) simultaneously. The general solution is
given simply by

$c^{(1)}_{1}=\pm\sqrt{1-|c_{0}^{(1)}|^{2}}\,,\quad
c^{(2)}_{1}=\pm\sqrt{1-|c_{0}^{(2)}|^{2}}\,,\cdots\,,c^{(N)}_{1}=\pm\sqrt{1-|c_{0}^{(N)}|^{2}}\,,$
(285)

where the set of combinations of all possible sign orientation of $+$ and $-$
form the full solution set, where we can write it as
$\\{c^{(1)}_{1},c^{(2)}_{1},\cdots c^{(N)}_{1}\\}$. Since each $c^{(j)}_{1}$
can be $+$ve or $-$ve, then there would be a total of $2^{n}$ solutions, i.e.
the cardinality of the solution set is just $N=2^{n}$. This solution set
satisfies both (281) and (284).

The simplest case for the general $n$-level solution would be

$|\Psi\rangle=\frac{1}{\sqrt{N}}\sum_{\eta_{i_{1}},\eta_{i_{2}},\cdots,\eta_{i_{N}}=0,1}|\eta_{i_{1}}\rangle\otimes|\eta_{i_{2}}\rangle\otimes\cdots\otimes|\eta_{i_{N}}\rangle\,,$
(286)

so that the probability of getting each state is $1/N$.

####  4.0.1 Heterogeneous basis

The above study has illustrated the tensor product representation of the basis
of same representation of $\mathbb{Z}_{2}$, i.e.

$\Big{(}\,D(g)\otimes D(g)\otimes\cdots\otimes
D(g)\,\Big{)}\,(|v\rangle\otimes|v\rangle\otimes\cdots\otimes|v\rangle)\,.$
(287)

Now we would like to extend the study to the different representations of the
basis of the same vector basis, that means

$\Big{(}\,D_{1}(g)\otimes D_{2}(g)\otimes\cdots\otimes
D_{N}(g)\,\Big{)}\,(|v_{1}\rangle\otimes|v_{2}\rangle\otimes\cdots\otimes|v_{N}\rangle)\,,$
(288)

such that

$|v_{j}\rangle=\begin{pmatrix}|0_{j}\rangle\\\ |1_{j}\rangle\end{pmatrix}\,.$
(289)

where each $|v_{j}\rangle$ is not necessarily same as $|v_{k}\rangle$. For
example for $N=2^{6}=64$ case, we have a particular basis as

$|1_{1}0_{2}0_{3}1_{4}1_{5}0_{6}\rangle\,,$ (290)

which is the basis of $V_{1}^{*}\otimes V_{2}\otimes V_{3}\otimes
V_{4}^{*}\otimes V_{5}^{*}\otimes V_{6}$.

###  4.1 Binary and diagramatic representation of multi-duality

For simplicity, we can express all the above concept with the aid of diagrams.
And since multi-duality deals with basis of 0s and 1s, it is convenient to use
binary representation for the purpose. For two states $|0\rangle$ and
$|1\rangle$, consider the following diagramatic approach,

Decimal representation | 0 | 1
---|---|---
Diagram |  |
Binary representation | 0 | 1

Now for second order

$(V\oplus V^{*})\otimes(V\oplus V^{*})=(V\otimes V)\oplus(V\otimes
V^{*})\oplus(V^{*}\otimes V)\oplus(V^{*}\otimes V^{*})\,,$ (291)

which has basis $|00\rangle$, $|01\rangle$, $|10\rangle$ and $|11\rangle$
respectively. Diagramatically,

Decimal representation | 0 | 1 | 2 | 3
---|---|---|---|---
Diagram |  |  |  |
Binary representation | (00) | (01) | (10) | (11)

For third order, we have $(V\oplus V^{*})\otimes(V\oplus V^{*})\otimes(V\oplus
V^{*})$ and have 8 terms upon expansion. This corresponds to 8 basis,
$|000\rangle,|001\rangle,|010\rangle,|011\rangle,|100\rangle,|101\rangle,|110\rangle,|111\rangle$,
which can be expressed diagramatically. For convenience we also introduce
spectral terms for them.

Decimal representation | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7
---|---|---|---|---|---|---|---|---
Diagram |  |  |  |  |  |  |  |
Binary representation | (000) | (001) | (010) | (011) | (100) | (101) | (110) | (111)
Spectural term | K | z | k | d | g | l | x | q

For forth order, there are 16 basis and it can be formed by the tensor product
of two second ordered diagrams, which is shown in figure 19.

Figure 43: The 16 basis constructed from the tensor product of two second
ordered diagrams. Each number located under the basis diagram is the decimal
representation.

For higher order, it can be similarly constructed from tensor product of lower
ordered diagrams. For example, the 6-th order can be constructed from the
tensor product of two third ordered diagrams, which is presented in figure 20.

Figure 44:  The 64 basis constructed from the tensor product of two third
ordered diagrams. Each number located under the basis diagram is the decimal
representation

###  4.2 Chiral representation and matrix basis

We have previously introduced basis representation of the $\mathbb{Z}_{2}$ and
the $\mathbb{Z}_{2}\times\mathbb{Z}_{2}$ group. Now we will like to see in a
more advanced view that the basis are promoted to matrices. In particular we
would like to see how this can be related to the context of gamma matrices in
fermionic quantum field theory.

Consider that the parity group $\mathbb{Z}_{2}$ being represented in real
general linear space GL(4, $\mathbb{R}$) of 4-dimension that is aroused from
Clifford algebra,

$\\{\gamma^{\mu},\gamma^{\nu}\\}=2\eta^{\mu\nu}\boldsymbol{1}\,,$ (292)

where $\mu,\nu=0,1,2,3$ are the Lorentz indices. The identity element is the
$4\times 4$ identity matrix and the parity element is the Dirac $\gamma^{5}$
matrix, where

$\gamma^{5}=i\gamma^{0}\gamma^{1}\gamma^{2}\gamma^{3}\,.$ (293)

Then the parity group is $\mathbb{Z}_{2}=\\{\boldsymbol{1},\gamma^{5}\\}$, and
we know that $(\gamma^{5})^{2}=\boldsymbol{1}$ and is independent of
representation (Dirac, Weyl, etc).

Now recall that in QFT, the global axial (chiral) transformation for a spinor
is

$\psi^{\prime}(x)=e^{i\theta\gamma^{5}}\psi(x)\,,$ (294)

where $\theta$ is the global phase (independent of spacetime $x$) of the
chiral transformation. And it is easy to show that

$U(\theta)=e^{i\theta\gamma^{5}}=\cos\theta\,\boldsymbol{1}+i\sin\theta\,\gamma^{5}\,.$
(295)

Recall that a generic state vector for the duality group can be written as
666Note that the $|\psi\rangle$ state here has nothing to do with with the
spinor field $\psi(x)$ above, readers should be confused by the notation
ambiguity.

$|\psi\rangle=\cos\theta\,|0\rangle+\sin\theta\,|1\rangle=\cos\theta\,|--\rangle+\sin\theta\,|-\rangle,$
(296)

thus comparing to equation 295, we can identity the matrix group elements
$\boldsymbol{1},\gamma^{5}$ of $\mathbb{Z}_{2}$ as the the basis of the group
by

$|0\rangle=\boldsymbol{1}\quad\text{and}\quad|1\rangle=i\gamma^{5}\,.$ (297)

Thus the basis can be considered as the matrix group element of
$\mathbb{Z}_{2}$ itself in the chiral representation.

Therefore, under the basis matrix representation, the tensor product
transformation is,

$U(\theta_{1})\otimes U(\theta_{2})\otimes\cdots\otimes
U(\theta_{n})=e^{i\theta_{1}\gamma^{5}}\otimes
e^{i\theta_{2}\gamma^{5}}\otimes\cdots\otimes e^{i\theta_{n}\gamma^{5}}\,.$
(298)

In full expansion we have

$e^{i\theta_{1}\gamma^{5}}\otimes
e^{i\theta_{2}\gamma^{5}}\otimes\cdots\otimes
e^{i\theta_{n}\gamma^{5}}=\bigotimes_{j=1}^{n}(\cos\theta_{j}\,\boldsymbol{1}+i\sin\theta_{j}\,\gamma^{5})\,.$
(299)

For example if $n=5$ we can have a particular term like

$(i^{3}\cos\theta_{1}\sin\theta_{2}\cos\theta_{3}\sin\theta_{4}\sin\theta_{5})\,\boldsymbol{1}\otimes\gamma^{5}\otimes\boldsymbol{1}\otimes\gamma^{5}\otimes\gamma^{5}\,.$
(300)

An important case would be $n=2$, then we have the matrix basis for the
4-duality group $\mathbb{Z}_{2}\times\mathbb{Z}_{2}$, in which

$\\{\boldsymbol{1}\otimes\boldsymbol{1},\,\boldsymbol{1}\otimes\gamma^{5},\,\gamma^{5}\otimes\boldsymbol{1},\,\gamma^{5}\otimes\gamma^{5}\\}\mapsto\\{|00\rangle,|01\rangle\,,|10\rangle,|11\rangle\\}\,.$
(301)

The rank-2 tensor components are

$T_{ij}=\begin{pmatrix}\cos\theta_{1}\cos\theta_{2}&i\cos\theta_{1}\sin\theta_{2}\\\
i\sin\theta_{1}\cos\theta_{2}&-\sin\theta_{1}\sin\theta_{2}\\\
\end{pmatrix}\,.$ (302)

and $\mathrm{det}\,T_{ij}=0$, in which all states are unentangled.

Finally we would like to define the orthogonality relation of the matrix
basis. Recall that we have $\langle 0|0\rangle=\langle 1|1\rangle=1$ and
$\langle 0|1\rangle=\langle 1|0\rangle=0$. For matrix basis, we can define
such by trace. For $g_{i},g_{j}\in\mathbb{Z}_{2}$,

$\mathrm{Tr}\big{(}\,D(g_{i})D(g_{j})\,\big{)}=2\delta_{ij}\,.$ (303)

This follows nicely from that fact that $\mathrm{Tr}\,\gamma^{5}=0$. We can
explicitly check that

$\mathrm{Tr}(\boldsymbol{1}\cdot\boldsymbol{1})=\mathrm{Tr}(\gamma^{5}\cdot\gamma^{5})=2\quad\text{and}\quad\mathrm{Tr}(\boldsymbol{1}\cdot\gamma^{5})=\mathrm{Tr}(\gamma^{5}\cdot\boldsymbol{1})=0\,.$
(304)

The canonical form for the graded chiral algebra is

$U(\theta_{1})\oplus U(\theta_{1})\otimes U(\theta_{2})\oplus\cdots\oplus
U(\theta_{1})\otimes U(\theta_{2})\otimes\cdots\otimes U(\theta_{n})\,.$ (305)

###  4.3 Comparison Representation

In the above studies, we have shown how to represent the $\mathbb{Z}_{2}^{N}$
group in duality basis. We can further form duality basis by comparing two
diagrams. We will use the notation $(,::,)$. Let’s $V\times V$ be the
comparison vector space of the basis, we have

$(,\,::\,,)\,:V\times V\rightarrow V\,.$ (306)

We compare the two diagrams level by level. If the two levels are of the same
state, we assign it as $|0^{\prime}\rangle$, and $|1^{\prime}\rangle$ if not.
Here we use the primed subscript to indicate this is a new emergent duality
basis that is formed by comparison, but for simplicity one can drop such
indication as being understood. At the end they are still duality basis. The
most fundamental comparison begins from the $n=1$ level,

$\displaystyle(0::0)$ $\displaystyle=1\,,$ (307) $\displaystyle(0::1)$
$\displaystyle=0\,,$ $\displaystyle(1::0)$ $\displaystyle=0\,,$
$\displaystyle(1::1)$ $\displaystyle=1\,.$

This can also be referred as the same rule as the normal multiplication for
the signs, $--\rightarrow-$, $-+\rightarrow+$, $+-\rightarrow-$ and
$++\rightarrow+$.

For higher $n$, let’s take $n=3$ for work out some examples.

$(010::110)=(011)\,\,,\,\,(111::010)=(010)\,,$ (308)

or diagrammatically

$(\includegraphics[trim=0.0pt 0.0pt 0.0pt
0.0pt,clip,scale={0.5}]{2.png}::\includegraphics[trim=0.0pt 0.0pt 0.0pt
0.0pt,clip,scale={0.5}]{6.png})=\includegraphics[trim=0.0pt 0.0pt 0.0pt
0.0pt,clip,scale={0.5}]{3.png}\,\,\,\,,\,\,\,\,(\includegraphics[trim=0.0pt
0.0pt 0.0pt 0.0pt,clip,scale={0.5}]{7.png}::\includegraphics[trim=0.0pt 0.0pt
0.0pt 0.0pt,clip,scale={0.5}]{2.png})=\includegraphics[trim=0.0pt 0.0pt 0.0pt
0.0pt,clip,scale={0.5}]{2.png}\,,$ (309)

and in decimal places,

$(1::6)=3\,,\,\,(7::2)=2\,.$ (310)

The comparison map can work compositely and satisfies the following axioms.

######  Definition 4.5.

Let $a,b,c$ be states representation and $(,::,)$ be the comparison map, and
the composition

$(,\,::\,,\,::\,,\cdots,::,\,)\,:V\times V\times\cdots\times V\rightarrow
V\,.$ (311)

satisfies the the following axiom

  * •

$(a::b)::c=a::(b::c)$ (associativity)

  * •

$a::b=b::a$ (commutativity)

  * •

$a::1=a$

  * •

$a::a=1$

  * •

$a=a^{-1}$ (self inverse)

The last axiom directly comes from the second-last axiom. (Note that
$1=1\cdots 1$ and $0=00\cdots 0$). From this definition it follows that

$b::a::b=a\,.$ (312)

This is because

$\displaystyle a::b$ $\displaystyle=b::a$ (313) $\displaystyle b::a::b$
$\displaystyle=(b::b)::a$ $\displaystyle b::a::b$ $\displaystyle=1::a$
$\displaystyle b::a::b$ $\displaystyle=a$

For example

$(101::011::000::110)=(101::011::001)=(101::101)=111\,,$ (314)

or diagrammatically

$(\includegraphics[trim=0.0pt 0.0pt 0.0pt
0.0pt,clip,scale={0.5}]{5.png}::\includegraphics[trim=0.0pt 0.0pt 0.0pt
0.0pt,clip,scale={0.5}]{3.png}::\includegraphics[trim=0.0pt 0.0pt 0.0pt
0.0pt,clip,scale={0.5}]{0.png}::\includegraphics[trim=0.0pt 0.0pt 0.0pt
0.0pt,clip,scale={0.5}]{6.png})=(\includegraphics[trim=0.0pt 0.0pt 0.0pt
0.0pt,clip,scale={0.5}]{5.png}::\includegraphics[trim=0.0pt 0.0pt 0.0pt
0.0pt,clip,scale={0.5}]{3.png}::\includegraphics[trim=0.0pt 0.0pt 0.0pt
0.0pt,clip,scale={0.5}]{1.png})=(\includegraphics[trim=0.0pt 0.0pt 0.0pt
0.0pt,clip,scale={0.5}]{5.png}::\includegraphics[trim=0.0pt 0.0pt 0.0pt
0.0pt,clip,scale={0.5}]{5.png})=\includegraphics[trim=0.0pt 0.0pt 0.0pt
0.0pt,clip,scale={0.5}]{7.png}$ (315)

and in decimal places

$(5::3::0::6)=(5::3::1)=(5::5)=1\,.$ (316)

There are some general rules that apply to any $n$. First the comparison is
For $l<n-1$, we have

$(0::l)=n-l\,\,\,\,\text{and}\,\,\,(2^{n}::l)=l\,.$ (317)

We also have

$(0::1::2::\,,\cdots\,,::2^{n}-1)=2^{n-1}\,,$ (318)

that means of the comparison of all states always return to $(111\cdots 1)$.

###  4.4 2-Level, 4-level and General $N=2^{n}$ Quantum Dual systems

A dual system with opposite element can be expressed in terms of a qubit,

$|\psi\rangle=\alpha|u\rangle+\beta|u^{*}\rangle=\alpha|0\rangle+\beta|1\rangle=\alpha|--\rangle+\beta|-\rangle\,,$
(319)

satisfying $|\alpha|^{2}+|\beta|^{2}=1$ and in particular we are interested in
the case for which $\alpha$ and $\beta$ are real. The $|u^{*}\rangle$ state is
the dual of $|u\rangle$, in such $*|u\rangle=|u^{*}\rangle$ and
$*|u^{*}\rangle=|u\rangle$, with $**=1$ the identity map, also $\langle
u|u^{*}\rangle=0$. The $\alpha$ values and $\beta$ values are the weight for
each of the state respectively. A common form takes the following

$|\psi\rangle=\cos\frac{\theta}{2}|--\rangle+\sin\frac{\theta}{2}|-\rangle=\cos\frac{\theta}{2}|0\rangle+\sin\frac{\theta}{2}|1\rangle\,,$
(320)

where $\theta=\omega t$ such that the coefficient of the dual state is
oscillatory. The factor of $\frac{1}{2}$ is chosen by convenience so that the
probability is at extreme at $k\pi$. Note that we also infer
$|-\rangle\equiv|+\rangle$ as the $|1\rangle$ state and
$|--\rangle\equiv|-\rangle$ as the $|0\rangle$ state. Suppose the $|--\rangle$
state has energy $E_{--}$ (or $E_{0}$) and the $|-\rangle$ state has energy
$E_{-}$ (or $E_{1})$, then the expectation value of the energy of the system
is given by

$\langle E\rangle=\sum_{j=0,1}|\langle j|\psi\rangle|^{2}E_{j}\,,$ (321)

which is

$\langle
E(\theta)\rangle=\cos^{2}\frac{\theta}{2}\,E_{-}+\sin^{2}\frac{\theta}{2}\,E_{+}\,.$
(322)

Thus the oscillatory growing and decreasing of dual energy is described by the
probability

$P_{-}=\cos^{2}\frac{\omega
t}{2}\,\,\,\,\text{and}\,\,\,\,P_{+}=\sin^{2}\frac{\omega t}{2}\,.$ (323)

Figure 45:  The probability of $|0\rangle$ and $|1\rangle$ by a given phase.
Both probabilities are the same i.e. $P_{+}=P_{-}=\frac{1}{2}$ at phases
$\frac{\pi}{2},\frac{3\pi}{2}$ , etc.

This describes the philosophy of dualism in a mathematical way. The extremes
are points at integers of $\pi$, which represent the phases. When the
probability of the 0 state rises to its maximum it must fall again where the
probability of 1 state grows. When the probability of the 0 state drops to its
minimum it must rise again where the probability of 1 state drops. Similarly,
the probability of the 1 state rises to its maximum it must fall again where
the probability of yin 0 grows. When the probability of the 1 state drops to
its minimum it must rise again where the probability of 0 state drops. And 0
and 1 states are in opposition as they do not overlap each other given by the
fact that the 0 state and 1 state are orthogonal to each other, $\langle
0|1\rangle=\langle 1|0\rangle=0$.

At extreme phase values, this gives the full 0 state or full 1 state with
deterministic probability of 1. For example for the full 0 state, at $t=0$

$|\psi(0)\rangle=|--\rangle\,.$ (324)

Suppose now the phase evolve a little such that $t>0$, then we ought to have

$|\psi(0+\delta)\rangle=(1-\epsilon)|--\rangle+\eta|-\rangle\,,$ (325)

where $\delta,\epsilon\,\text{and}\,\eta$ are small. This means the portion of
0 state is decreasing and the portion of 1 state begins to grows. This is the
philosophy that “at the extreme 0 becomes 1”. To be exact, the probability of
0 state becomes less and the probability of 1 state grows a bit, so this gives
an illusion that some 0 state converts to 1 state but in fact it is just the
probability change. Using small angle approximation, for small $\delta$

$|\psi(\delta)\rangle=\cos\frac{\delta}{2}|--\rangle+\sin\frac{\delta}{2}|-\rangle\approx\bigg{(}1-\frac{\delta^{2}}{4}\bigg{)}|--\rangle+\frac{\delta}{2}|-\rangle\,.$
(326)

Therefore we identify $\epsilon=\frac{\delta^{2}}{4}$ and
$\eta=\frac{\delta}{2}$.

Similarly for the full 1 state,

$|\psi(\pi)\rangle=|-\rangle\,.$ (327)

Now suppose the phase evolve a little bit, let’s say before the phase $\pi$,
we ought to have

$|\psi(\pi-\delta)\rangle=\eta|--\rangle+\bigg{(}1-\frac{\delta^{2}}{4}\bigg{)}|-\rangle\,.$
(328)

This means the portion of 1 state is decreasing and the portion of 0 state
begins to grow. This is the philosophy that “at the extreme 1 becomes 0”. To
be exact, the probability of 1 state becomes less and the probability of 0
state grows a bit, so this gives an illusion that some 1 states converts to 0
state but in fact it is just the probability change. Again, using small angle
approximation,

$\displaystyle|\psi(\pi-\delta)\rangle$
$\displaystyle=\cos\bigg{(}\frac{\pi}{2}-\frac{\delta}{2}\bigg{)}|--\rangle+\sin\bigg{(}\frac{\pi}{2}-\frac{\delta}{2}\bigg{)}|-\rangle$
(329)
$\displaystyle=\sin\frac{\delta}{2}|--\rangle+\cos\frac{\delta}{2}|-\rangle$
$\displaystyle\approx\frac{\delta}{2}|--\rangle+\bigg{(}1-\frac{\delta^{2}}{4}\bigg{)}|-\rangle\,.$

Therefore again we identify $\epsilon=\frac{\delta^{2}}{4}$ and
$\eta=\frac{\delta}{2}$.

The first order derivative of the expectation energy is

$\frac{d\langle
E(\theta)\rangle}{d\theta}=\frac{1}{4}\sin\theta(E_{+}-E_{-})\,.$ (330)

Thus the expectation energy is at extreme when $\sin\theta=0$ or $\Delta
E=E_{+}-E_{-}=0$, i.e. $\theta=\pm k\pi$ or $E_{+}=E_{-}$. Whether it is the
minimum or maximum can be checked by the sign nature of the second order
derivative,

$\frac{d^{2}\langle
E(\theta)\rangle}{d\theta^{2}}=\frac{1}{4}\cos\theta(E_{+}-E_{-})\,.$ (331)

If $E_{+}>E_{-}$, then minimum occurs at $\theta=2k\pi$ and maximum occurs at
$\theta=\pm(2k+1)\pi$.

Suppose the energy of the $|1\rangle$ state $E_{+}>0$ is positive and the
energy if the $|0\rangle$ state $E_{-}<0$ is negative, and their magnitudes
are the same $|E_{+}|=|E_{-}|=E$, then we have

$\langle E(t)\rangle=E\cos\,\omega t\,.$ (332)

The energy of the dual system oscillate with time.

Next we would like to use the notion of heterogeneous bases for the
$\mathbb{Z}_{2}\times\mathbb{Z}_{2}$ group to describe probability relations
between the $|0\rangle$ and $|1\rangle$ states along the flow of time $t$. We
focuses on the study in one period, $0\leq\theta<2\pi$. Define the original
$|0\rangle$, $|1\rangle$ states as $|0_{1}\rangle=|0\rangle$ and
$|1_{1}\rangle=|1\rangle$, and the two dual states of tendency,
$|0_{2}\rangle=|\downarrow\rangle$ and $|1_{2}\rangle=|\uparrow\rangle$. The
$|\downarrow\rangle$ state describes the state whenever the probability of
$|0\rangle$ or $|1\rangle$ state is decreasing, while the $|\uparrow\rangle$
state describes the state whenever the probability of $|0\rangle$ or
$|1\rangle$ state is increasing. Then we have the heterogeneous basis for the
$\mathbb{Z}_{2}\times\mathbb{Z}_{2}$ group as
$\\{|0_{1}0_{2}\rangle,|0_{1}1_{2}\rangle,|1_{1}0_{2}\rangle,|1_{1}1_{2}\rangle\\}$.
For $0\leq\theta<\pi$, we have

$|0\downarrow\rangle\,\,\,\,\equiv\,\,\,\,|1\uparrow\rangle\,,$ (333)

for another half $\pi\leq\theta<2\pi$ we have

$|0\uparrow\rangle\,\,\,\,\equiv\,\,\,\,|1\downarrow\rangle\,.$ (334)

The heterogeneous basis of $\mathbb{Z}_{2}\times\mathbb{Z}_{2}$ can be also
interpreted as objects of element with observer frame. One can take
$\downarrow$ as the $S_{k}$ observer and $\uparrow$ as the $S_{k}^{\star}$
observer, such that we have

$(0|S_{k})\,\,\,\,\equiv\,\,\,\,(1|S_{k}^{\star})\quad\quad\text{and}\quad\quad(1|S_{k})\,\,\,\,\equiv\,\,\,\,(0|S_{k}^{\star})\,.$
(335)

So we can apply the operators $I$,$*$, $\star$ and $*\circ\star$ to these
objects. We can represent these by a 4-tableau,

Figure 46:  The 4-tableau representation of a dual system with heterogeneous
basis representation of the 4-duality group. The light green boxes represent
the phase regime of $0\leq\theta<\pi$, the light blue boxes represent the
phase regime of $\pi\leq\theta<2\pi$. Objects with same colour are equipped
with equivalent relations.

It is useful to study the entropy of the system. The Shannon entropy is given
by [19, 20]

$H=-\sum_{i}p_{i}\log p_{i}\,,$ (336)

where $p_{i}$ is the probability of each state. The entropy of the 2-level
quantum system is

$H(\theta)=-2\cos^{2}\frac{\theta}{2}\log\Big{|}\cos\frac{\theta}{2}\Big{|}-2\sin^{2}\frac{\theta}{2}\log\Big{|}\sin\frac{\theta}{2}\Big{|}\,.$
(337)

Graphically, the entropy is

Figure 47:  Entropy of a 2-level system. The plot is generated in the range of
$-4\pi\leq\theta_{1},\theta_{2}\leq 4\pi$.

The graph is continuous everywhere and is symmetric. Maximum entropy occurs at
$\pm\frac{k\pi}{2}$ and is 1 bit. Minimum entropy is zero at occurs at $\pm
k\pi$.

Next let’s study the case of the 4-level system. For heterogeneous basis we
have

$|\psi(\theta_{1},\theta_{2})\rangle=\cos\frac{\theta_{1}}{2}\cos\frac{\theta_{2}}{2}|0_{1}0_{2}\rangle+\cos\frac{\theta_{1}}{2}\sin\frac{\theta_{1}}{2}|0_{1}1_{2}\rangle+\sin\frac{\theta_{1}}{2}\cos\frac{\theta_{2}}{2}|1_{1}0_{2}\rangle+\sin\frac{\theta_{1}}{2}\sin\frac{\theta_{1}}{2}|1_{1}1_{2}\rangle\,.$
(338)

The probability tensor is given by the element-wise products of the rank-2
matrices,

$p_{ij}(\theta_{1},\theta_{2})=T_{ij}(\theta_{1})\bullet
T_{ij}(\theta_{2})\bullet T_{ij}(\theta_{1})\bullet T_{ij}(\theta_{2})\,.$
(339)

Thus the expectation value of the energy is

$\langle
E(\theta_{1},\theta_{2})\rangle=\cos^{2}\frac{\theta_{1}}{2}\cos^{2}\frac{\theta_{2}}{2}E_{-_{1}-_{2}}+\cos^{2}\frac{\theta_{1}}{2}\sin^{2}\frac{\theta_{2}}{2}E_{-_{1}+_{2}}+\sin^{2}\frac{\theta_{1}}{2}\cos^{2}\frac{\theta_{2}}{2}E_{+_{1}-_{1}}+\sin^{2}\frac{\theta_{1}}{2}\sin^{2}\frac{\theta_{2}}{2}E_{+_{1}+_{1}}\,.$
(340)

Here the random variable $\Theta$ is parametrized by $\theta_{1}$ and
$\theta_{2}$. We have the entropy as

$\displaystyle H(\Theta)$
$\displaystyle=-2\Big{(}\cos\frac{\theta_{1}}{2}\cos\frac{\theta_{2}}{2}\Big{)}^{2}\log\Big{|}\cos\frac{\theta_{1}}{2}\cos\frac{\theta_{2}}{2}\Big{|}-2\Big{(}\cos\frac{\theta_{1}}{2}\sin\frac{\theta_{2}}{2}\Big{)}^{2}\log\Big{|}\cos\frac{\theta_{1}}{2}\sin\frac{\theta_{2}}{2}\Big{|}$
(341)
$\displaystyle\quad-2\Big{(}\sin\frac{\theta_{1}}{2}\cos\frac{\theta_{2}}{2}\Big{)}^{2}\log\Big{|}\sin\frac{\theta_{1}}{2}\cos\frac{\theta_{2}}{2}\Big{|}-2\Big{(}\sin\frac{\theta_{1}}{2}\sin\frac{\theta_{2}}{2}\Big{)}^{2}\log\Big{|}\sin\frac{\theta_{1}}{2}\sin\frac{\theta_{2}}{2}\Big{|}\,.$

Graphically,

Figure 48:  Entropy of a 4-level system. The plot is generated in the range of
$-4\pi\leq\theta_{1},\theta_{2}\leq 4\pi$.

We can see that the entropy function is not continuous everywhere, it is
confined in local regions of $(\theta_{1},\theta_{2})$ which give discrete
comb-liked peaks. The remaining regions would give imaginary entropy values,
thus are not well-defined.

Next we would like to find out what what phases $(\theta_{1},\theta_{2})$ the
entropy is maximum,

$C=\underset{\Theta}{\mathrm{max}}\,H(\Theta)=\underset{\theta_{1},\theta_{2}}{\mathrm{max}}\,H(\theta_{1},\theta_{2})\,.$
(342)

There are two ways to do so. The first one is to just apply the theorem that
an even distribution has maximum entropy. In other words for our case this
happens when all the states
$|0_{1}0_{2}\rangle,|0_{1}1_{2}\rangle,|1_{1}0_{2}\rangle,|1_{1}1_{2}\rangle$
have the same probability. This takes place when

$\displaystyle|\psi\rangle$
$\displaystyle=\frac{1}{\sqrt{2}}(|0_{1}\rangle+|1_{1}\rangle)\,\otimes\,\frac{1}{\sqrt{2}}(|0_{2}\rangle+|1_{2}\rangle)$
(343)
$\displaystyle=\frac{1}{2}|0_{1}0_{2}\rangle+\frac{1}{2}|0_{1}1_{2}\rangle+\frac{1}{2}|1_{1}0_{2}\rangle+\frac{1}{2}|1_{1}1_{2}\rangle\,.$

Then we have an even probability distribution of

$p_{00}=p_{01}=p_{10}=p_{11}=\frac{1}{4}\,.$ (344)

Obviously this occurs at

$\cos\frac{\theta_{1}}{2}=\cos\frac{\theta_{2}}{2}=\sin\frac{\theta_{1}}{2}=\sin\frac{\theta_{2}}{2}=\frac{1}{\sqrt{2}}.$
(345)

The general solution $\big{(}\frac{\pi}{2}+2p\pi,\frac{\pi}{2}+2q\pi\big{)}$
for any positive integers $p$ and $q$.

The maximum entropy is

$H_{\mathrm{max}}=-\sum_{i=1}^{4}\frac{1}{4}\log\frac{1}{4}=\big{(}\frac{1}{4}\cdot
2\big{)}\cdot 4=2\,\,\text{bits}\,.$ (346)

For the second way, the local maxima can be found technically by solving
simultaneously

$\frac{\partial H}{\partial\theta_{1}}=0\,\,\text{and}\,\,\frac{\partial
H}{\partial\theta_{2}}=0$ (347)

The partial derivatives are very messy, and they are difficult to solve
algebraically but computationally with the aid of contour diagram we can read
the peak values.

Figure 49: Contour plot of the phase angles $(\theta_{1},\theta_{2})$ in the
range of $-4\pi\leq\theta_{1},\theta_{2}\leq 4\pi$

On the other hand, the lower limit of the entropy of the system tends to zero,
but it cannot be exactly zero. These takes place at $\pm p\pi,\pm q\pi$.
Therefore at the beginning, half-way, and at the end the entropy tends to
zero.

The results make sense, as the phase angles $\theta_{i}=\omega t_{i}$, when
$\theta=0,2\pi$ which are the initial time and final time, things are highly
ordered with no uncertainty. But as time goes disorderness increases and
reaches maximum.

Now let’s reduce the above problem to the condition of same phase. There are
two possible meanings for equal phase $\theta_{1}=\theta_{2}$. It can either
mean two different angular frequencies but at same time, or same angular
frequency but at different times. We will consider the latter case. Suppose
now the system we described above is synchronized with the same phase, such
that $\theta_{1}=\theta_{2}=\theta$. Then our original equation (338) reduces
to a 2 qubit state,

$|\psi(\theta)\rangle=\cos^{2}\frac{\theta}{2}|00\rangle+\frac{1}{2}\sin\theta|01\rangle+\frac{1}{2}\sin\theta|10\rangle+\sin^{2}\frac{\theta}{2}|11\rangle\,.$
(348)

Then the energy becomes

$\langle E(\theta)\rangle=\sum_{j=--,-+,+-,++}|\langle
j|\psi(\theta)\rangle|^{2}E_{j}\,,$ (349)

which is

$\langle
E(\theta)\rangle=\cos^{4}\frac{\theta}{2}E_{--}+\frac{1}{4}\sin^{2}\theta
E_{-+}+\frac{1}{4}\sin^{2}\theta E_{+-}+\sin^{4}\frac{\theta}{2}E_{++}\,.$
(350)

The probability of each state is plot as

Figure 50: Probability of each state in the range of $-4\pi\leq\theta\leq
4\pi$. At $\theta=\pm 2k\pi$, $|00\rangle$ is the fully dominating state. At
$\theta=\pm(2k+1)\pi$, $|11\rangle$ is the fully dominiating state. At
$\theta=\pm\frac{k\pi}{2}$, the system is evenly distributed in all the
$|00\rangle,|01\rangle,|10\rangle,|11\rangle$ states with equal probability of
$\frac{1}{4}$.

We can diagrammatically describe this with the phase evolution of 4-tableau.
In particular we are interested in the probability of the states at integer
and half-integer multiple of $\pi$.

Figure 51: Phase evolution of 4-tableau.

We assign the area of outer big square as 1. Then the area of each quarter
square as $\frac{1}{4}$. We can map the area of the sub-squares to the
probability of each state. Equivalently, we can represent the one big box of
full-0 state and one big box of full-1 state in (52) as

Figure 52:

We denote the full-0 state by odd integers $\\{\cdots-1,1,3,5,7\cdots\\}$, the
full-1 state by even integers $\\{\cdots-2,0,2,4,6\cdots\\}$, and the 4 states
with equal probability as half integers
$\\{\cdots-\frac{1}{2},0,\frac{1}{2},\frac{3}{2}\,\frac{5}{2}\,\cdots\\}$. We
say the 4 equally probable states at half integers of $\pi$ _spontaneously_
collapse to the full-0 state at odd multiples of $\pi$, and _spontaneously_
collapse to the full-1 state at even multiples of $\pi$. The quantum system is
said to be momentarily deterministic at $\pm k\pi$ as either the full-0 state
or full-1 state is at probability 1.

The first order derivative of the expectation energy is

$\frac{d\langle
E(\theta)\rangle}{d\theta}=\sin\theta\Big{(}\sin^{2}\frac{\theta}{2}E_{++}-\cos^{2}\frac{\theta}{2}E_{--}\Big{)}+\frac{1}{4}\sin
2\theta\,(E_{-+}+E_{+-})\,.$ (351)

Therefore the extreme of $\langle E(\theta)\rangle$ occurs when

$\sin\theta=0\quad\text{or}\quad\Big{(}\sin^{2}\frac{\theta}{2}E_{++}-\cos^{2}\frac{\theta}{2}E_{--}\Big{)}+\frac{1}{2}\cos\theta(E_{-+}+E_{+-})=0\,.$
(352)

For the first case again we have $\theta=\pm k\pi$. Thus at the integer
multiple of $\pi$, the energy expectation value is at maximum, this correspond
to the purely full-0 state of pure full-1 state which is at probability 1.
Thus when the system is at its pure state, it has the maximum energy. Since
the full-0 state also represents the state of nothing while the full-1 state
represent the state of All, then it means when the universe is at the state of
purely nothing or purely everything, its energy is the greatest (Note that if
we originally off-set by a phase factor of $\pi$, then the $|01\rangle$ and
$|10\rangle$ state would have the maximum energy). A special attention should
be given for the second case (352), in particular
$E_{--}=E_{++}=E_{-+}=E_{+-}=E$ give zero first order derivative regardless of
$\theta$. Thus when all energies for the four states are the same, the rate of
change of $\langle E(\theta)\rangle$ is always 0 for whatever $\theta$ values.
Thus $\langle E(\theta)\rangle$ is constant overall all $\theta$s. It is in
fact easy to check that in fact $\langle E(\theta)\rangle=E$ using simple
trigonometry.

Next we find the entropy of the system. Equation 341 is simplified to,

$H(\theta)=-4\cos^{4}\frac{\theta}{2}\log\Big{|}\cos\frac{\theta}{2}\Big{|}-4\sin^{4}\frac{\theta}{2}\log\Big{|}\sin\frac{\theta}{2}\Big{|}-\sin^{2}\theta\log\Big{|}\frac{\sin\theta}{2}\Big{|}\,.$
(353)

The entropy function is plotted,

Figure 53: $H(\theta)$ plot in the range of $-4\pi\leq\theta\leq 4\pi$

The maximum occurs at $\frac{\pi}{2}\pm k\pi$, which is 2 bits. The roots of
$H(\theta)=0$ occurs as the limits when $\theta\rightarrow\pm k\pi$.

Our study of $2$-level and $4$-level system can be generalized to $N=2^{n}$
levels. Since the probability of each state is just the product of sine and
cosine, let’s write

$p_{i_{1}i_{2}\cdots
i_{N}}=\prod_{j=1}^{N}\mathrm{Trig}^{2}_{\,\,i_{j}}(\theta_{j})\,,$ (354)

where we define the trigonometry function $\mathrm{Trig}$ by

$\mathrm{Trig}_{\,\,i_{j}}(\theta_{j})=\begin{cases}\cos\theta_{j}&\quad\text{if}\,\,i_{j}=0\\\
\sin\theta_{j}&\quad\text{if}\,\,i_{j}=1\end{cases}\,.$ (355)

The full quantum state in $2^{n}$ level is generalized to

$|\psi(\theta_{1},\theta_{2}\,\cdots\,\theta_{N})\rangle=\sum_{i_{1},i_{2},\cdots,i_{N}=0,1}\bigg{(}\,\prod_{j=1}^{N}\mathrm{Trig}_{\,\,i_{j}}(\theta_{j})\,\bigg{)}\,|\eta_{i_{1}}\eta_{i_{2}}\cdots\eta_{i_{N}}\rangle\,.$
(356)

The expectation of energy is

$\langle E(\theta_{1},\theta_{2}\,\cdots\,\theta_{N})\rangle=\sum_{j\in
W}|\langle
j|\psi(\theta_{1},\theta_{2}\,\cdots\,\theta_{N})\rangle|^{2}E_{j}\,,$ (357)

which is

$\langle
E(\theta_{1},\theta_{2}\,\cdots\,\theta_{N})\rangle=\sum_{i_{1},i_{2},\cdots,i_{N}=0,1}\bigg{(}\,\prod_{j=1}^{N}\mathrm{Trig}^{2}_{\,\,i_{j}}(\theta_{j})\,\bigg{)}\,E_{i_{1}i_{2}\cdots
i_{N}}\,,$ (358)

where $W$ is the dual set. The entropy is

$H(\theta_{1},\theta_{2}\,\cdots\,\theta_{N})=-\sum_{i_{1},i_{2},\cdots,i_{N}=0,1}\bigg{[}\,\bigg{(}\,\prod_{j=1}^{N}\mathrm{Trig}^{2}_{\,\,i_{j}}(\theta_{j})\,\bigg{)}\,\,\bigg{(}\,\log\prod_{j=1}^{N}\
\mathrm{Trig}^{2}_{\,\,i_{j}}(\theta_{j})\bigg{)}\,\bigg{]}\,.$ (359)

The entropy is maximized when all the $\theta_{j}$s are equal such that the
probability of each state is even, i.e. equal to $\frac{1}{N}$.

For synchronized phase, all $\theta_{j}$s equal to a single $\theta$ variable,
and the full state (356) is just an unentangled N-qubit in the form of

$|\psi\rangle=\sum_{j=0}^{2^{n}}c_{j}|j\rangle\,.$ (360)

The coefficient $c_{j}$ is in the form of
$\cos^{p}\frac{\theta}{2}\sin^{q}\frac{\theta}{2}$, where $p$ is the number of
0 and $q$ is the number of 1 in the particular state. Then some of the states
have equal probability. For example in $n=3$ case, the 3 states

$|100\rangle\,\,,\,\,|010\rangle\,\,\text{and}\,\,|001\rangle$ (361)

all have the same coefficient $\cos\frac{\theta}{2}\sin^{2}\frac{\theta}{2}$,
and the same probability $\sin^{4}\frac{\theta}{2}\cos^{2}\frac{\theta}{2}$.
The probability distribution follows the binomial distribution,

$1=\Big{(}\cos^{2}\frac{\theta}{2}+\sin^{2}\frac{\theta}{2}\Big{)}^{n}=\sum_{k=0}^{n}\frac{n!}{k!(n-k)!}\cos^{2k}\frac{\theta}{2}\sin^{2n-2k}\frac{\theta}{2}\,,$
(362)

Therefore the number of states have the same probability is given by the
binomial coefficient

$\frac{n!}{k!(n-k)!}\,.$ (363)

The number of distinct groups of states that have same probabilities is just
$n+1$. (For example, in $n=4$, we have binomial coefficients $1,4,6,4,1$ then
we have $4+1=5$ groups). The total number of states is given by the sum of all
binomial coefficients, and this is just the identity

$\sum_{k=0}^{n}\frac{n!}{k!(n-k)!}=2^{n}=N\,,$ (364)

which is just $N=2^{n}$ states as expected.

Consider a large number of levels. When $n\rightarrow$, $N\rightarrow\infty$,
we would obtain a normal distribution for the probabilities.

Let’s study a few $n$ examples from small $n$ to large $n$. We will plot all
the probabilities in a same graph for each $n$. Consider 4 cases
$n=3,10,50,800$, the plots are shown in figure 54.

(a) $n=3$ case, 8 states

(b) $n=10$ case, 1024 states

(c) $n=50$ case, $\sim 10^{15}$ states

(d) $n=800$ case, $\sim 10^{24}$ states

Figure 54:  The plots of $p_{n}$ probability for $n=3,10,50$ and $800$. Note
that for each $n$ the probability of the states are plotted with the same
colour. These plots are in comparison to the plot 50 of $n=2$ case.

We can see that when $n$ grows larger, the contribution of probability from
the mixed states of 0 and 1 get lesser, and eventually when
$n\rightarrow\infty$ (here $n=800$ is sufficiently large enough ), the
$p_{00\cdots 0}=p_{0}=\cos^{2n}\frac{\theta}{2}$ and $p_{11\cdots
1}=p_{N}=\sin^{2n}\frac{\theta}{2}$ fully dominate. The contribution from
other individual mixed state is extremely small. Thus basically, when
$n\rightarrow\infty$, the system just automatically collapse to the full zero
state $|00\cdots 0\rangle$ at odd multiple of $\pi$ (half cycles) and to full
one state $|11\cdots 1\rangle$ at even multiple of $\pi$ (complete cycles). We
call the full zero state $|00\cdots\rangle=|0\rangle$ the beginning state and
the full one state $|11\cdots 1\rangle=|N-1\rangle$ state. Diagrammatically,

Figure 55:

Therefore, in fact we can interpret the the state $|\psi_{\infty}\rangle$ as a
new dual state $|\infty\rangle$

$|\bowtie\rangle=\frac{1}{\sqrt{2}}(|00\cdots 0\rangle+|11\cdots
1\rangle)=\frac{1}{\sqrt{2}}(|\bar{0}\rangle+|\bar{1}\rangle)=\frac{1}{\sqrt{2}}(|0\rangle+|\infty\rangle)\,,$
(365)

where in the last expression we have used the numerical expression for the
state ($0$ is always 0 and $\infty$ is always $\infty$ in any number base ).
This is because the $|0\rangle$ and $|\bar{1}\rangle$ dominate over all other
states, and hence we can basically think that $|\psi_{\infty}\rangle$ simply
contains two dual state. The $|\bowtie\rangle$ is a EPR pair, which is an
entangled state.

Now recall that the zeroth-level state $|\psi_{0}\rangle=|\bigcirc\rangle$
with probability 1. Thus when we go from the beginning to the end
$0\rightarrow\infty$, the originally deterministic $|\bigcirc\rangle$ state
now becomes the $|\bowtie\rangle$ state with two dual basis. We denote this as

$0\rightarrow\infty\,,$ (366)

or in terms of the change in number of states

$1\rightarrow 2\,.$ (367)

The general entropy for synchronized phase is,

$H_{n}(\theta)=-\sum_{k=0}^{n}\frac{n!}{k!(n-k)!}\Big{(}\cos^{2k}\frac{\theta}{2}\sin^{2n-2k}\frac{\theta}{2}\Big{)}\log\Big{(}\cos^{2k}\frac{\theta}{2}\sin^{2n-2k}\frac{\theta}{2}\Big{)}\,.$
(368)

Equivalently,

$H_{n}(\theta)=-2\sum_{k=0}^{n}\frac{n!}{k!(n-k)!}\Big{(}\cos^{2k}\frac{\theta}{2}\sin^{2n-2k}\frac{\theta}{2}\Big{)}\log\Big{|}\cos^{k}\frac{\theta}{2}\sin^{n-k}\frac{\theta}{2}\Big{|}\,.$
(369)

In general the entropy function $H_{n}(\theta)$ for different $n$s have same
shapes but just different maximum amplitudes. The local maximum and minimum
entropies occur at the same $\pm k\pi$s for all different $n$. The cases for
$n=3,10$ and $50$ are illustrated in figure 56.

Figure 56:

For synchronized phase, if all $E_{i_{1},i_{2}\,\cdots\,i_{N}}=E$ are equal,
then it is a degenerate quantum system. The expectation energy is constant
independent of phase

$\langle E(\theta)\rangle=E\,.$ (370)

This is simply because

$\langle
E(\theta)\rangle=\sum_{k=0}^{n}\frac{n!}{k!(n-k)!}\cos^{2k}\frac{\theta}{2}\sin^{2n-2k}\frac{\theta}{2}\,E=E\cdot
1=E\,.$ (371)

Finally, for the general $n$-qubit case, we have the state vector as

$|\Psi(\theta_{1},\theta_{2},\cdots\theta_{n})\rangle=|\psi_{1}(\theta_{1})\rangle\otimes|\psi_{2}(\theta_{2})\rangle\otimes\cdots\otimes|\psi_{n}(\theta_{n})\rangle=\bigotimes_{j=1}^{n}(\cos\theta_{j}|0_{j}\rangle+\sin\theta_{j}|1_{j}\rangle)\,.$
(372)

For each $i_{j}\in W_{j}$ where $W_{j}$ is the jth dual set, the probability
of each state is

$\displaystyle P_{i_{1}i_{2}\cdots i_{n}}$ $\displaystyle=|\langle
i_{1}i_{2}\cdots
i_{n}|\Psi(\theta_{1},\theta_{2},\cdots\theta_{n})\rangle|^{2}$ (373)
$\displaystyle=\bigg{|}\bigg{(}\bigotimes_{j=1}^{n}\langle
i_{j}|\bigg{)}\bigg{(}\bigotimes_{j=1}^{n}|\phi_{j}(\theta_{j})\rangle\bigg{)}\bigg{|}^{2}$
$\displaystyle=|(\langle i_{1}|\otimes\langle
i_{2}|\otimes\cdots\otimes\langle
i_{n}|)(|\phi_{1}(\theta_{1})\rangle\otimes|\phi_{2}(\theta_{2})\rangle\otimes\cdots\otimes|\phi_{n}(\theta_{n})\rangle)|^{2}$
$\displaystyle=|\langle i_{1}|\phi_{1}(\theta_{1})\rangle\otimes\langle
i_{2}|\phi_{2}(\theta_{2})\rangle\otimes\cdots\otimes\langle
i_{n}|\phi_{n}(\theta_{n})\rangle|^{2}$
$\displaystyle=\bigg{|}\bigotimes_{j=1}^{n}\langle
i_{j}|\psi(\theta_{j})\rangle\bigg{|}^{2}=\bigg{|}\prod_{j=1}^{n}\langle
i_{j}|\psi(\theta_{j})\rangle\bigg{|}^{2}$
$\displaystyle=\prod_{j=1}^{n}|\langle i_{j}|\psi(\theta_{j})\rangle|^{2}$
$\displaystyle=\prod_{j=1}^{n}P_{i_{j}}=P_{i_{1}}P_{i_{2}}\cdots P_{i_{n}}$

And the sum of probability is equal to 1,

$\displaystyle P_{tot}$ $\displaystyle=\sum_{i_{1},i_{2},\cdots
i_{n}}P_{i_{1}i_{2}\cdots i_{n}}$ (374) $\displaystyle=\sum_{i_{1}\in
W_{1}}\sum_{i_{2}\in W_{2}}\cdots\sum_{i_{n}\in
W_{n}}\prod_{j=1}^{n}P_{i_{j}}$ $\displaystyle=\bigg{(}\sum_{i_{1}\in
W_{1}}P_{i_{1}}\bigg{)}\bigg{(}\sum_{i_{2}\in
W_{2}}P_{i_{2}}\bigg{)}\cdots\bigg{(}\sum_{i_{n}\in W_{n}}P_{i_{n}}\bigg{)}$
$\displaystyle=\prod_{j=1}^{n}\sum_{i_{j}\in W_{j}}P_{i_{j}}$
$\displaystyle=1$

The expectation energy of the $n$-qubit system is given by

$\displaystyle\langle E(\theta_{1},\theta_{2},\cdots,\theta_{n})\rangle$
$\displaystyle=\sum_{i_{1}\in W_{1}}\sum_{i_{2}\in W_{2}}\cdots\sum_{i_{n}\in
W_{n}}|\langle i_{1}i_{2}\cdots
i_{n}|\Psi(\theta_{1},\theta_{2},\cdots\theta_{n})\rangle|^{2}E_{i_{1}i_{2}\cdots
i_{n}}$ (375) $\displaystyle=\sum_{i_{1}\in W_{1}}\sum_{i_{2}\in
W_{2}}\cdots\sum_{i_{n}\in W_{n}}P_{i_{1}i_{2}\cdots
i_{n}}(\theta_{1},\theta_{2},\cdots,\theta_{n})E_{i_{1}i_{2}\cdots i_{n}}$
$\displaystyle=\sum_{i_{1}\in W_{1},\cdots,i_{n}\in
W_{n}}\prod_{j=1}^{n}P_{i_{j}}(\theta_{j})E_{i_{1}i_{2}\cdots i_{n}}$

The entropy for the general case would be

$\displaystyle H(\theta_{1},\theta_{2},\cdots\theta_{n})$
$\displaystyle=-\sum_{i_{2}\in W_{2}}\cdots\sum_{i_{n}\in
W_{n}}P_{i_{1}i_{2}\cdots i_{n}}(\theta_{1},\theta_{2},\cdots\theta_{n})\log
P_{i_{1}i_{2}\cdots i_{n}}(\theta_{1},\theta_{2},\cdots\theta_{n})$ (376)
$\displaystyle=-\sum_{i_{2}\in W_{2}}\cdots\sum_{i_{n}\in W_{n}}|\langle
i_{1}i_{2}\cdots
i_{n}|\Psi(\theta_{1},\theta_{2},\cdots\theta_{n})\rangle|^{2}\log|\langle
i_{1}i_{2}\cdots
i_{n}|\Psi(\theta_{1},\theta_{2},\cdots\theta_{n})\rangle|^{2}$
$\displaystyle=-\sum_{i_{1}\in W_{1},\cdots,i_{n}\in
W_{n}}\prod_{j=1}^{n}P_{i_{j}}(\theta_{i_{j}})\log\prod_{j=1}^{n}P_{i_{j}}(\theta_{i_{j}})\,.$

###  4.5 Dual Pairs

Next, we would introduce a very important concept of quantum states of dual
pair. A dual pair is basically a pair of diagrams of which each level is dual
to each other. It means if a level is at $|0\rangle$ state then the same level
of the dual counterpart is $|1\rangle$, vice versa. We will give the formal
mathematical definition below.

######  Definition 4.6.

Let $|\eta_{i_{1}}\eta_{i_{2}}\cdots\eta_{i_{N}}\rangle$ of a $n$-level with
$N=2^{n}$ be the general state for a diagram, where $\eta_{i_{j}}=0$ or $1$.
Define the dual operator $*$ such that $*|0\rangle=|0^{*}\rangle=|1\rangle$
and $*|1\rangle=|1^{*}\rangle=|0\rangle$ which satisfies $*^{2}=I$,

$*|\eta_{i_{1}}\eta_{i_{2}}\cdots\eta_{i_{N}}\rangle=|\eta_{i_{1}}^{*}\eta_{i_{2}}^{*}\cdots\eta_{i_{N}}^{*}\rangle\,.$
(377)

Let $k$ be the decimal number recovered from the binary representation and
$k^{*}$ be the dual counter part, then $k$ and $k^{*}$ is related by

$k^{*}=(N-1)-k\,.$ (378)

The $(k,k^{*})$ is defined as a dual pair, and is arranged for $k<k^{*}$. The
dual pair must contain one odd and one even number.

The proof is straight forward. Let $a_{j}=0,1$ be the coefficient and let
$a_{j}^{*}$ be dual to $a_{j}$. The duality operator constrains that
$a_{j}+a_{j}^{*}=1$. Let

$k=\sum_{j=1}^{n}a_{j}2^{j-1}\quad\text{and}\quad
k^{*}=\sum_{j=1}^{n}a_{j}^{*}2^{j-1}\,.$ (379)

Then consider $k+k^{*}$,

$k+k^{*}=\sum_{j=1}^{n}(a_{j}+a_{j}^{*})\,2^{j-1}=\sum_{j=1}^{n}2^{j-1}=\frac{(1)(2^{n}-1)}{2-1}=2^{n}-1=N-1\,,$
(380)

where in the third step we have used the geometric series. It follows that
$k^{*}=(N-1)-k$. Since $N=2^{n}$ must be even, then $N-1$ must be odd. Thus
$k+k^{*}$ is odd. Using the fact that an odd number is composed of an even
number and an odd number, then $k,k^{*}$ must contain one odd number and one
even number. This completes the proof. $\Box$

For example, for $n=6$ level, we have $*|010010\rangle=|101101\rangle$. Then
$k=18$ and $k^{*}=63-18=45$. The dual pair is $(18,45)$.

There is a special kind of dual pair which has a common property in all
$n$-levels. The difference between $k$ and $k^{*}$ for this special pair is 3.

######  Definition 4.7.

Let $(k_{b},k_{b}^{*})$ be the boundary dual pair. The respective binary
representation are $(011\cdots 110)$ and $(100\cdots 001)$; and the respective
decimal representation is $\frac{N}{2}-1$ and $\frac{N}{2}+2$ for
$N>2\,\,\,\,(\text{or}\,\,n>1)$. The two integers in between are called the
confined boundary dual pair $\kappa,\kappa^{*}$. The respective binary
representation is $(011\cdots 1)$ and $(100\cdots 0)$, and the respective
decimal representation is $\frac{N}{2}-1$ and $\frac{N}{2}$. The boundary dual
is formally defined by $(k_{b},k_{b}^{*})$ such that $|k_{b}^{*}-k_{b}|=3$ and
the confined boundary dual is defined by $|\kappa^{*}-\kappa|=1$.

These can be easily shown. First we have

$k_{b}=\sum_{j=1}^{n-2}2^{j}=2(1+2+\cdots
2^{n-3})=\frac{2(2^{n-1}-1)}{(2-1)}=2^{n-1}-2=\frac{N}{2}-2\,,$ (381)

and

$k_{b}^{*}=1+2^{n-1}=\frac{N}{2}+1\,.$ (382)

Then the $\kappa$ in the confined binary dual pair is simply adding $1$ from
$k_{b}^{*}$, thus $\kappa=k_{b}+1=\frac{N}{2}-1$, and $\kappa^{*}$ is obtained
by subtracting $k_{b}^{*}$ from $1$, $\kappa^{*}=k_{b}^{*}-1=\frac{N}{2}$.
These can be expressed in terms in a diagramatic way,

Figure 57: Boundar pair and confined boundary dual pair for general $n$-level.
The black middle line is the duality mirror.

Examples for these pairs are shown in table 1, e.g. for $n=5$ and $n=6$,
respectively, we have boundary pairs as $(14,17)$ and $(30,33)$, with boundary
confined pair as $(10,21)$ and $(18,45)$. We call it boundary dual pair
because the pair appears nearest to the dual mirror, where the dual mirror
separates the dual numbers into two halves, the even half and the odd half.

The dual pair is a very important idea to describe the appearing and hidden
states of nature. Suppose we construct a
$|\eta_{i_{N}}\eta_{i_{N-1}}\cdots\eta_{i_{2}}\eta_{i_{1}}\rangle$ state along
a positive flow of time. Each state $|\eta_{i_{j}}\rangle$ is generated at
time $t_{j}$, where $t_{f}>t_{N}>\cdots>t_{1}>t_{i}$. Consider a surface, for
example the surface of the table, then we flip a 2-sided coin which is either
head or tail successively in the time interval $t_{i}\leq t\leq t_{f}$. This
is called the even time interval, in which there is a generation of
information in this time interval. Let’s define the head state as $|1\rangle$
and represent it as a black circle, and the tail state as $|0\rangle$ and
represent it as a white circle. There are in addition, two more states, which
are two perspectives. The direction that the coin shows up to the observer is
called the appearing state $|\uparrow\rangle$, while the direction that the
coin that is not shown is called the hidden state $|\downarrow\rangle$. The
idea is illustrated in 58.

H

Figure 58: Illustration of a dual pair of $n=6$ case.

The two sides of the illustration are equivalent, we have

$(010100|\uparrow)\,\,\equiv\,\,(101011|\downarrow)\,,$ (383)

where we have the observer frame (or perspective) identified as
$S_{3}=\uparrow$ and $S_{3}^{\star}=\downarrow$. In terms of decimal number we
can write

$(18|\uparrow)\,\,\equiv\,\,(45|\downarrow)\,.$ (384)

In general we have

$(\eta_{i_{N}}\eta_{i_{N-1}}\cdots\eta_{i_{2}}\eta_{i_{1}}|S_{k})\,\,\equiv\,\,(\eta_{i_{N}}^{*}\eta_{i_{N-1}}^{*}\cdots\eta_{i_{2}}^{*}\eta_{i_{1}}^{*}|S_{k}^{\star})\,.$
(385)

Thus although the diagrams of both sides are not the same in terms of element,
when we change from the appearing perspective to the hidden perspective, they
are equivalent.

We can map the two dual states to effective $|\bar{0}\rangle$ and
$|\bar{1}\rangle$ states. We add an extra bar so as to distinguish them form
the original $|0\rangle$ and $|1\rangle$ states. This means

$|\eta_{i_{N}}\eta_{i_{N-1}}\cdots\eta_{i_{2}}\eta_{i_{1}}\rangle\rightarrow|\bar{0}\rangle\,\,,\,\,|\eta_{i_{N}}^{*}\eta_{i_{N-1}}^{*}\cdots\eta_{i_{2}}^{*}\eta_{i_{1}}^{*}\rangle\rightarrow|\bar{1}\rangle$
(386)

as $|0^{*}\rangle=|1\rangle$. Therefore in the above example we have

$|010100\rangle\rightarrow|\bar{0}\rangle\,\,,\,\,|101011\rangle\rightarrow|\bar{1}\rangle\,.$
(387)

It is interested to study, in particular, the case where $n$ is extremely
large. When $n\rightarrow\infty$,

$(000\cdots 00|\uparrow)\equiv(111\cdots 11|\downarrow)\,,$ (388)

which is, in decimal representation,

$(0|\uparrow)\equiv(\infty|\downarrow)\,.$ (389)

Hence, zero is equivalent to infinity under the appearing-hidden perspective.
We interpret as follow, zero in the appearing frame is equivalent to infinity
in the hidden frame. Therefore zero can be viewed as everything in the dual
frame. And of course, we can also have

$(0|\downarrow)\equiv(\infty|\uparrow)\,.$ (390)

Then zero in the hidden frame is equivalent to infinity in the appearing
frame.

The dual pair can be promoted to two dual quantum states. The coefficients can
show how much information is appearing and how much information is hidden. The
amount of information is described by the probability. For each dual pair in
$N=2^{n}$ diagrams (we have $2^{n-1}$ dual pairs),

$|\psi_{l}\rangle=a_{l}|l\rangle+a_{l}^{*}|l^{*}\rangle=\cos\theta_{l}|l\rangle+\sin\theta_{l}|l\rangle^{*}=\cos\theta_{l}|l\rangle+\sin\theta_{l}|2^{n}-l-1\rangle\,.$
(391)

The full state is given by

$|\Psi\rangle=\sum_{l=0}^{2^{n-1}-1}|\psi_{l}\rangle=\sum_{l=0}^{2^{n-1}-1}(\,\cos\theta_{l}|l\rangle+\sin\theta_{l}|l^{*}\rangle\,)=\sum_{l=0}^{2^{n-1}-1}(\,\cos\theta_{l}|l\rangle+\sin\theta_{l}|2^{n}-l-1\rangle\,)\,.$
(392)

We can also write it as the sum of paired
$|\bar{0}_{j}\rangle,|\bar{1}_{j}\rangle$ states,

$|\Psi\rangle=\sum_{j=0}^{2^{n-1}-1}(\,\cos\theta_{j}|\bar{0}_{j}\rangle+\sin\theta_{j}|\bar{1}_{j}\rangle\,)\,.$
(393)

If the appearing state and hidden state evolve under time, we can write it as

$|\Psi(t)\rangle=\sum_{l=0}^{2^{n-1}-1}(\,\cos\omega_{l}t|l\rangle+\sin\omega_{l}t|2^{n}-l-1\rangle\,)$
(394)

In the general form, we have

$|\Psi(t)\rangle=\frac{1}{2}\sum_{p+q=2^{n}-1}(\,a_{p}(t)|p\rangle+a_{q}(t)|q\rangle\,)\,.$
(395)

The factor of $\frac{1}{2}$ is required due to double counting.

One interesting property arises from the above theory is dual invariance.
Suppose the state remains the same regardless of forward time flow or backward
time flow, for example $(110011)$ looks the same in either case. Geometrically
this is simply left-right invariant by observation, or up-down invariant if
diagrammatically. We have

$(110011|RL,t>0)\,\,\equiv\,\,(110011|LR,t<0)\,.$ (396)

Such state can also infer the property of time-reversal symmetry of the state.
We will discuss give the formal definition of dual invariant in the next
section.

###  4.6 Dual Invariant

We will define dual invariant, which is an invariant under a dual observation
frame or perspective. There can be different kinds of dual invariants
subjected to the interest of study. Mathematically,

######  Definition 4.8.

Let the observer be situating in a space of dimension $k\geq 3$. Define a dual
space $S$ with two elements $S_{k}$ and $S_{k}^{\star}$. An object $\xi$ is
dual invariant under the two dual observer frames if $\xi$ satisfies

$(\xi|S_{k})=(\xi|S_{k}^{\star})$ (397)

for $S_{k}\neq S_{k}^{\star}$ and $S_{k}\cap S_{k}^{\star}=\emptyset$.

Note for the condition of $k\geq 3$, this is because one at least has to
observe perpendicularly to a dual system that is situated in a 2-dimensional
plane. The most common ones we will use is the up-down dual observing frame
and left-right dual observing frame. If an object is up-down (UD) dual
invariant, then the object remains the same regardless of looking from above
or looking from below. The two dual frames are $S_{k}^{UD}$ and $S_{k}^{DU}$.
If an object is left-right dual invariant, then the object remains the same
regardless of looking from the left or looking from the right. The two dual
frames are $S_{k}^{LR}$ and $S_{k}^{RL}$. Like-wise, other dual invariants
follow similar idea. For example, in the 3rd order case, all $(000)$, $(010)$,
$(101)$ and $(111)$ are all dual invariant, this corresponds to the dual
invariant decimal number $0,2,5,7$ respectively.

Next we would construct useful diagrams to highlight the important symmetry
properties in the multiplication table, they are called the feature diagrams.
In these diagrams, only elements in the table which have considerable dual
structure that can be form the basis of $\mathbb{Z}_{2}$ or
$\mathbb{Z}_{2}\times\mathbb{Z}_{2}$ will be high-lighted. The feature
diagrams are essential to study symmetry patterns of the elements. There are
two main parts in the feature diagram. Firstly for even $n$, the numbers are
arranged in the a $2^{\frac{n}{2}}\times 2^{\frac{n}{2}}$ grids with each grid
follows exactly the same order as in the multiplication table,

Figure 59:

Secondly, coloured symbols represent different dual structures of the diagram.
We will define some standard coloured symbols used in the feature diagrams as
follow:

  * •

Blue circle: Represent the diagram(s) that are up-down dual invariant, or
equivalently left-right dual invariant for the binary representation
$(\eta_{i_{1}}\eta_{i_{2}}\cdots\eta_{i_{n}}|S_{k}^{LR})=(\eta_{i_{1}}\eta_{i_{2}}\cdots\eta_{i_{n}}|S_{k}^{RL})$
where $\eta_{i}=0,1$. This corresponds to the condition of
$\eta_{i_{j}}=\eta_{j_{n-j}}$. The number recovered is defined as dual
invariant number (details will be discussed in the next section).

  * •

Purple square: Represent the diagram(s) that are invariant under the exchange
for two blocks of $n/2$-diagram in the multiplication.

  * •

Orange hexagon: Represent the diagram(s) where the upper diagram block is dual
to the lower diagram block, i.e. the upper block can be formed by acting the
dual operator to the lock block, and vice versa.

  * •

Green triangle: Represent diagram(s) with alternative $0$s and $1$s. There are
only two of these for any levels $n>3$, and this is a global property. One of
it is the dual of the other.

  * •

Four colour corners: Represent the diagram(s) of full, half and null states.

Basically, the diagrams that fall into Purple square, Orange hexagon, or Green
triangle are different basis of the reducible representation of the 2-duality
group $\mathbb{Z}_{2}$ For example, the feature diagram of 6th order is,

Figure 60:  Feature diagram of 6-th order diagram in correspondence of figure
5.

###  4.7 General properties for general $n$-level

In this section, we will study the properties of dual invariant number. For
each $n$-level, we collect all dual invariant numbers for each $n$-level as a
set $D_{n}$.

$D_{1}$ | $0,1$
---|---
$D_{2}$ | $0,3$
$D_{3}$ | $0,2,5,7$
$D_{4}$ | $0,6,9,15$
$D_{5}$ | $0,4,10,14,17,21,27,31$
$D_{6}$ | $0,12,18,30,33,45,51,63$
$D_{7}$ | $0,8,20,28,34,42,54,62,65,73,85,93,99,107,119,127$
$D_{8}$ | $0,24,36,60,66,90,102,126,129,153,165,189,195,219,231,255$
Table 1: Dual invariant numbers for $n$.

######  Definition 4.9.

Define the set of all dual invariant numbers $D$. It can be considered as the
union of all individual $D_{n}$,

$D=\bigcup_{i=1}^{\infty}D_{i}=D_{1}\cup D_{2}\cup\cdots\cup D_{\infty}\,.$
(398)

Zero is the common element for all of the $D_{n}$ s,

$\bigcap_{i=1}^{\infty}D_{i}=D_{1}\cap D_{2}\cap\cdots\cap
D_{\infty}=\\{0\\}\,.$ (399)

The dual invariant number $<100$ is shown in figure 61.

Figure 61:  Dual invariant number $<100$ from $n=1$ to $n=12$ level. The
number $0$ is general to all levels.

It is noted that the distribution pattern of dual invariant number is unclear
in figure 61. The pattern is only clear in individual $n$-levels when placed
in the $n\times n$ grid.

######  Definition 4.10.

Let $m$ be even and $m>2$. Let the dual invariant number of level $m$ be
$d_{m}$, where $d_{m}\in D_{m}$ and is defined by

$d_{m}=\sum_{j=0}^{m/2}a_{j+1}(2^{j}+2^{m-j+1})\,,$ (400)

with $a_{j+1}=0\,\text{or}\,1$.

This can be easily proved by the following. Consider the general form for a
binary number in decimal representation,

$k(n)=\sum_{j=1}^{n}a_{j}2^{j-1}\,\,\text{for}\,\,n\in\mathbb{N}\,.$ (401)

We consider the case for even $n$, where $n=m=2l$. Then the series sum can be
explicitly written down as

$\displaystyle k(m)$ $\displaystyle=a_{1}+a_{2}\cdot 2+a_{3}\cdot
2^{2}+\cdots+a_{\frac{m}{2}-1}\cdot 2^{\frac{m}{2}-2}+a_{\frac{m}{2}}\cdot
2^{\frac{m}{2}-1}+a_{\frac{m}{2}}\cdot
2^{\frac{m}{2}-1}+a_{\frac{m}{2}+1}\cdot
2^{\frac{n}{2}}+a_{\frac{n}{2}+2}\cdot 2^{\frac{n}{2}+1}$
$\displaystyle\quad+\cdots+a_{m-3}\cdot 2^{m-4}+a_{m-2}\cdot
2^{m-3}+a_{m-1}\cdot 2^{m-2}+a_{m}\cdot 2^{m-1}$
$\displaystyle=\big{(}a_{1}+a_{m}\cdot 2^{m-1}\big{)}+\big{(}a_{2}\cdot
2+a_{m-1}\cdot 2^{m-2}\big{)}+\big{(}a_{3}\cdot 2^{2}+a_{m-2}\cdot
2^{m-3}\big{)}$ $\displaystyle\quad+\cdots+\big{(}a_{\frac{m}{2}-1}\cdot
2^{\frac{m}{2}-1}+a_{\frac{m}{2}+2}\cdot
2^{\frac{m}{2}+1}\big{)}+\big{(}a_{\frac{m}{2}}\cdot
2^{\frac{m}{2}-1}+a_{\frac{m}{2}+1}\cdot 2^{\frac{m}{2}}\big{)}\,.$

The dual invariance imposes the condition of

$a_{j}=a_{m+1-j}\,.$ (402)

Then we have, for examples

$\begin{aligned} a_{1}&=a_{m}\\\ a_{2}&=a_{m-1}\\\
a_{\frac{m}{2}-1}&=a_{\frac{m}{2}+2}\\\
a_{\frac{m}{2}}&=a_{\frac{m}{2}+1}\end{aligned}\,.$

Thus we can write the series as

$d_{m}=\sum_{j=0}^{m/2}a_{j+1}(2^{j}+2^{m-j+1})$

which completes the proof.

######  Definition 4.11.

The $d_{n}$ for even $n$ and $n>2$ is divisible by 3.

This amounts to prove that the term $2^{j}+2^{m-j+1}$ in the sum is divisible
by 3. We will prove by induction by showing first the propositions $P(m+1,j)$
and $P(m,j+1)$ are true, followed by $P(m+1,j+1)$ is also true. Let
$l\in\mathbb{N}$ and suppose

$P(l,j)=2^{j}+2^{2l-j+1}=3q\quad\text{for some even integers}\,q\,\text{ and
max}\,j=\frac{m}{2}.$

For $l=1$, $P(1,j)=2^{j}+2^{3-j}$. And for $j=1$, $P(1,1)=6$ which is
divisible by 3 for $q=2$, which is true. Then

$\displaystyle P(l+1,j)$ $\displaystyle=2^{k}+2^{2(l+1)-j+1}$
$\displaystyle=2^{j}+4\cdot 2^{2l-j+1}$
$\displaystyle=2^{j}+4(3q-2^{j})\quad\text{(by assumption)}$
$\displaystyle=2^{j}-2^{l+2}+6q$ $\displaystyle=2^{j}(1-4)+6q$
$\displaystyle=3(2q-2^{j})\,\,,$

thus $P(l+1,j)$ is true. Next

$\displaystyle P(l,j+1)$ $\displaystyle=2^{j+1}+2^{2l-{j+1}+1}$
$\displaystyle=2^{j+1}+2^{-1}\cdot 2^{2l-j+1}$
$\displaystyle=2^{j+1}+2^{-1}\cdot(3q-2^{j})\quad\text{(by assumption)}$
$\displaystyle=2^{j+1}-2^{j-1}+\frac{3}{2}q$
$\displaystyle=2^{j-1}(2^{2}-1)+\frac{3}{2}q$
$\displaystyle=3\big{(}2^{j+1}+\frac{q}{2}\big{)}\,\,$

thus $P(l+1,j)$ is also true. Finally

$\begin{aligned} P(l+1,j+1)&=2^{j+1}+2^{2(l+1)-(j+1)+1}\\\ &=2^{j+1}+2\cdot
2^{2l-j+1}\\\ &=2^{j+1}+2(3q-2^{j})\quad\text{(by assumption)}\\\
&=6q\end{aligned}\,$

thus $P(l+1,j+1)$ is also true. Hence the proof completes. $\Box$

######  Definition 4.12.

Let positive integers $n>1$ be the level and the number of dual invariant
numbers be $d(n)$, where $d(n)=|D_{n}|$ is the cardinality of the $n^{\rm th}$
level dual invariant number set $D_{n}$. We have

$f(n)=\begin{cases}2^{\frac{n}{2}}&\text{if}\,\,n\,\,\text{is even}\\\
2^{\frac{n+1}{2}}&\text{if}\,\,n\,\,\text{is odd}\\\ \end{cases}$ (403)

The initial case is $|D_{2}|=2$.

Consider an dual invariant number number set of an even $n$-level. For the
next $n+1$ level which is odd, we can construct new dual invariants by
inserting a state in the middle,

$|\eta_{i_{1}}\eta_{i_{2}}\cdots\eta_{i_{\frac{n}{2}}}\eta_{i_{\frac{n}{2}}}\cdots\eta_{i_{2}}\eta_{i_{1}}\rangle\rightarrow|\eta_{i_{1}}\eta_{i_{2}}\cdots\eta_{i_{\frac{n}{2}}}\eta^{\prime}\eta_{i_{\frac{n}{2}}}\cdots\eta_{i_{2}}\eta_{i_{1}}\rangle\,,$
(404)

where again $\eta^{\prime}=0\,\,\text{or}\,\,1$. Therefore we have
$|D_{n+1}|=2|D_{n}|$ for $n$ is even. Next to construct new dual invariants
from the $n$ level, we add two states one at the beginning and one with the
end,

$|\eta_{i_{1}}\eta_{i_{2}}\cdots\eta_{i_{\frac{n}{2}}}\eta_{i_{\frac{n}{2}}}\cdots\eta_{i_{2}}\eta_{i_{1}}\rangle\rightarrow|\eta_{i}^{\prime\prime}\eta_{i_{1}}\eta_{i_{2}}\cdots\eta_{i_{\frac{n}{2}}}\eta^{\prime}\eta_{i_{\frac{n}{2}}}\cdots\eta_{i_{2}}\eta_{i_{1}}\eta^{\prime\prime}\rangle\,,$
(405)

where $\eta^{\prime\prime}=0\,\,\text{or}\,\,1$. Therefore we have
$|D_{n+2}|=2|D_{n}|$ for $n$ is even. We would like to introduce two set of
notations. The center insertion is

$|\eta_{i_{1}}\eta_{i_{2}}\cdots\eta_{i_{\frac{n}{2}}}\,{}_{\wedge}\,\eta_{i_{\frac{n}{2}}}\cdots\eta_{i_{2}}\eta_{i_{1}}\rangle$
(406)

and the beginning-ending insertion is

$|\,_{\wedge}\eta_{i_{1}}\eta_{i_{2}}\cdots\eta_{i_{\frac{n}{2}}}\,\eta_{i_{\frac{n}{2}}}\cdots\eta_{i_{2}}\eta_{i_{1}}\,{}_{\wedge}\rangle$
(407)

Hence, alternative odd and even layers have the same $|D_{n}|$. Using this way
the dual invariant numbers can be constructed recursively. The following
diagramatic approach illustrate the process if the state is represented as
$n-$ diagram.

Figure 62:

This allows us to construct 1 and 61.

######  Definition 4.13.

Define the ratio of the number of dual invariant number to the number of non-
dual invariant as $R(n)$, which is given by

$R(n)=\frac{2^{m}}{2^{n}-2^{m}}=\begin{cases}\frac{1}{2^{\frac{n}{2}}-1}&\text{for}\,\,m=\frac{n}{2}\,\,\text{and}\,\,n\,\text{is
even}\\\
\frac{1}{2^{\frac{n-1}{2}}-1}&\text{for}\,\,m=\frac{n+1}{2}\,\,\text{and}\,\,n\,\text{is
odd}\\\ \end{cases}\,.$ (408)

Thus we can see that the ratio decreases with increasing $n$, and both even
and odd cases converges to $0$ when $n\rightarrow\infty$. It means that as the
$n$-level increases, we are getting few dual invariants per $n$-diagram, and
eventually they drop to none.

Figure 63: The ratio $R(n)$ against $n$.

Next we would like to study the divisibility of weak dual invariant numbers.
The concept of dual invariant can also be applied to levels of larger group.
For $N=2^{n}$ of which $n$ is even, we can easily divide the $n-$level diagram
into two halves. If the two halves are equal, it is also a dual invariant,
however it is a weaker case that the above. In decimal representation, it
takes the form of

$|p\rangle|p\rangle\,.$ (409)

For example, $|100100\rangle=|100\rangle|100\rangle=|4\rangle|4\rangle$ is a
3-level dual invariant. The dual invariant we introduced is called the strong
dual invariant and is a level-1 dual invariant.

######  Definition 4.14.

A strong dual invariant is also a weak dual invariant state for even
$n$-level(s) .

######  Definition 4.15.

Let $A$ be the set of strong dual invariant states and $B$ be the set of weak
dual invariant states, $B\subseteq A$ for even $n$-levels.

These two definitions are equivalent. It is note that the converse is not
true, a weak dual invariant state cannot be a strong dual invariant state.
Only when $n=2$, we have $A=B$.

######  Definition 4.16.

Any weak dual invariant number is divisible by $N+1$ for even $n$ .

The proof is easy. First we write the number in decimal representation as

$x=a_{1}2^{0}+a_{2}2^{1}+\cdots a_{i}2^{i-1}+a_{i+1}2^{i}+\cdots
a_{\frac{n}{2}}2^{\frac{n}{2}-1}+a_{\frac{n}{2}+1}2^{\frac{n}{2}}+\cdots+a_{\frac{n}{2}+j}2^{\frac{n}{2}+j-1}+a_{\frac{n}{2}+j+1}2^{\frac{n}{2}+j}+\cdots+a_{n}2^{n-1}\,.$
(410)

Due to the condition of weak dual invariant, we must have

$a_{1}=a_{\frac{n}{2}+1},\,\,\,\,a_{2}=a_{\frac{n}{2}+2}\,\,\,\,\cdots\,\,\,\,a_{j}=a_{\frac{n}{2}+j}\text{and}\,\,\,\,a_{n}=a_{\frac{n}{2}}\,.$
(411)

Thus we have

$\displaystyle x$
$\displaystyle=\sum_{i=0}^{\frac{n}{2}-1}a_{i+1}(2^{i}+2^{\frac{n}{2}+i})$
(412)
$\displaystyle=\sum_{i=0}^{\frac{n}{2}-1}a_{i+1}2^{i}(1+2^{\frac{n}{2}})$
$\displaystyle=(2^{\frac{n}{2}}+1)\sum_{i=0}^{\frac{n}{2}-1}a_{i+1}2^{i}$
$\displaystyle=(N+1)\sum_{i=0}^{\frac{n}{2}-1}a_{i+1}2^{i}$

for $n=2k$. Hence, the weak dual invariant number is divisible by $N+1$ for
even $n$. It is represented by the magenta boxes along the right diagonal
positions in the feature diagram.

##  5 Detailed symmetry studies of individual $n$ level

In this section, we would study the symmetry properties of each $n$-level
individually by the concepts we have introduced previously. There are not much
new concepts introduced, but instead applying all the previous concepts to
give a detailed analysis for each $n$-level.

###  5.1 The n =0 level

The n=0 level is known as the then null level, this is also called the
absolute ground state. We represent it as the a state vector as follow

$|\Psi\rangle=|\bigcirc\rangle\,.$ (413)

This state is completely deterministic and we obtain it by probability equal
to 1. The energy of this ground state is given by $E_{0}$.

###  5.2 The n=1 level

Next we would like to ask how this null state can promote to the dual state,
which is the $n=1$ level. We will model it with spontaneous symmetry breaking.
When spontaneous symmetry breaking begins to take place, we call the origin
$|\circ\rangle$ state tai chi, which allows the degeneracy of ground states.
The process from null to tai chi is a phase transition, the system changes
from a complete deterministic state to a probabilistic dual state.

The process can be model by a potential of scalar field $\phi$ as

$V(\phi)=\mu^{2}\phi^{2}+\lambda\phi^{4}\,.$ (414)

for $\mu^{2}<0$ and $\lambda>0$. The minimum takes place at

$v=\pm\sqrt{\frac{-\mu^{2}}{\lambda}}$ (415)

We would promote the field into quantum state, so we have

$|\Psi\rangle=\frac{1}{\sqrt{2}}(|-v\rangle+|+v\rangle)\,.$ (416)

Then the probability of choosing either the positive or negative vacuum state
is $1/2$ and one can infer the $|-v\rangle$ as $|0\rangle$ state and the
$|+v\rangle$ as $|1\rangle$. We have equal probability to choose the either
state. Once the either state is chosen, the state becomes classical again. The
dual state processes complex rotational symmetry, the probability is U(1)
invariant, i.e. a local or global transformation

$|\Psi^{\prime}\rangle=e^{i\theta}|\Psi\rangle$ (417)

leaves the probability invariant.

Next we study the qubit state vector $|\Psi\rangle$. take the form

$|\psi\rangle=a_{0}|0\rangle+a_{1}|1\rangle$ (418)

Here first we are interested in the case when $a_{0}=b_{0}$ or $a_{0}=-b_{0}$
(which are $1/\sqrt{2}$ and $-1/\sqrt{2}$) for our study of duality.

$\displaystyle|\psi_{++}\rangle$
$\displaystyle=\frac{1}{\sqrt{2}}(|0\rangle+|1\rangle)$ (419)
$\displaystyle|\psi_{-+}\rangle$
$\displaystyle=\frac{1}{\sqrt{2}}(-|0\rangle+|1\rangle)$
$\displaystyle|\psi_{+-}\rangle$
$\displaystyle=\frac{1}{\sqrt{2}}(|0\rangle-|1\rangle)$
$\displaystyle|\psi_{--}\rangle$
$\displaystyle=\frac{1}{\sqrt{2}}(-|0\rangle-|1\rangle)$

The basis with four elements
$\\{|\psi_{++}\rangle,|\psi_{-+}\rangle,|\psi_{+-}\rangle,|\psi_{--}\rangle\\}$
is isomorphic to $\\{|00\rangle,|01\rangle,|10\rangle,|11\rangle\\}$. The
conserved quantity for such $4-dual$ system is the probability of obtaining
each of the state, both are $\frac{1}{2}$.

The factor of $1/\sqrt{2}$ is not important here for discussion so we just
take as $+1,-1$. Let $U={1,-1}$ be a dual set and $V={|0\rangle,|1\rangle}$ be
another dual set. We can rewrite the qubit as follow,

$\psi=a_{0}|0\rangle+a_{1}|1\rangle=(a_{0}\oplus
a_{1})(|0\rangle\oplus|1\rangle)\,,$ (420)

where $a_{0},a_{1}=\pm 1$ are constant operators. Now define two dual
operators $*$ and $\star$, which act on $U$ and $V$ respectively. Note that
the two dual operators commute. For example,

$\displaystyle*|\psi\rangle$ $\displaystyle=*(a_{0}\oplus
a_{1})(|0\rangle\oplus|1\rangle)$ (421)
$\displaystyle=(*a_{0}\oplus*a_{1})(|0\rangle\oplus|1\rangle)$
$\displaystyle=(a_{1}\oplus a_{0})(|0\rangle\oplus|1\rangle)\,,$

and

$\displaystyle\star|\psi\rangle$ $\displaystyle=\star\big{(}(a_{0}\oplus
a_{1})(|0\rangle\oplus|1\rangle)\big{)}$ (422) $\displaystyle=(a_{0}\oplus
a_{1})\,\star(|0\rangle\oplus|1\rangle)\big{)}$ $\displaystyle=(a_{0}\oplus
a_{1})\,(\star|0\rangle\oplus\star|1\rangle)$ $\displaystyle=(a_{0}\oplus
a_{1})(|1\rangle\oplus|0\rangle)\,.$

Together we have

$*\star|\psi\rangle=*(a_{0}\oplus
a_{1})\star(|0\rangle\oplus|1\rangle)=(a_{1}\oplus
a_{0})(|1\rangle\oplus|0\rangle)=a_{1}|1\rangle+a_{0}|0\rangle$ (423)

Thus for the case of $a_{0}=1,a_{1}=-1$ (or vice versa) which is the dual set
$U$, we have

$|\psi\rangle=*\star|\psi\rangle\,.$ (424)

####  5.2.1 Dual Partition evolution

We would introduce the concept of dual partition evolution for $n=1$ case,
this essentially describe the process of a _tai chi_ diagram. This allows us
to study how the state $|--\rangle$ and state $|-\rangle$ exchange one another
under time evolution.

Figure 64: A standard _tai chi_ diagram

We would depict the evolution process in a cycle evolution of the 4-tableau.
First consider the initial full set All as $V\cup V^{*}$ where $V\cap
V^{*}=\emptyset$ such that the full set naturally partitioned into two equal
halves $V$ and $V^{*}$ (represented as $|0\rangle$ and $|1\rangle$ with equal
probability ).

Figure 65: A standard _tai chi_ cycle as the phase evolution of 4-tableau. The
$|0\rangle$ is denoted as black and the $|1\rangle$ is denoted by white. We
suppose the flow is moving downwards during the exchange. The phase $\theta$
is related to time by $\theta=\omega t$.

Let $Q$ be the left partition and $Q^{*}$ be the dual partition. The general
_tai chi_ state $|\psi\rangle$ can be expressed as linear superposition of the
two states in the two partitions. And we define our observable frame as
$S_{k}$, then

$(|\psi(t)\rangle\,\,|S_{k})=\Big{(}c_{1}(t)|0\rangle_{Q}+c_{2}(t)|1\rangle_{Q^{*}}+c^{*}_{1}(t)|0^{*}\rangle_{Q}+c_{2}^{*}(t)|1^{*}\rangle_{Q^{*}}\,\,\Big{|}S_{k}\Big{)}\,.$
(425)

The four states satisfy the orthogonality relations,

${}_{P}\langle i|j\rangle_{P^{\prime}}=\delta_{ij}\delta_{PP^{\prime}}\,,$
(426)

where $i,j=0,1$ and $P,P^{\prime}=Q,Q^{*}$. Note that one can also infer the
state $|0\rangle_{Q}$ as $|0\bar{0}\rangle$ , $|1\rangle_{Q^{*}}$ as
$|1\bar{1}\rangle$, $|0\rangle_{Q^{*}}$ as $|0\bar{1}\rangle$ and
$|1\rangle_{Q}$ as $|1\bar{0}\rangle$ , which contribute to a 4-dual system
emerged from a 2-dual system. The total probability is
$P_{1}(t)+P_{2}(t)+P_{1}^{*}(t)+P_{2}^{*}(t)=1$, explicitly

$|c_{1}(t)|^{2}+|c_{2}(t)|^{2}+|c_{1}^{*}(t)|^{2}+|c_{2}^{*}(t)|^{2}=1\,$
(427)

and the probability of each partition is always $\frac{1}{2}$,

$|c_{1}(t)|^{2}+|c_{1}^{*}(t)|^{2}=\frac{1}{2}\quad\text{and}\quad|c_{2}(t)|^{2}+|c_{2}^{*}(t)|^{2}=\frac{1}{2}\,.$
(428)

and also we must have

$|c_{1}(t)|^{2}+|c_{2}^{*}(t)|^{2}=\frac{1}{2}\quad\text{and}\quad|c_{2}(t)|^{2}+|c_{1}^{*}(t)|^{2}=\frac{1}{2}\,.$
(429)

At $t=0$, we have

$c_{1}(0)=c_{2}(0)=\frac{1}{\sqrt{2}}\quad\text{and }\quad
c_{1}^{*}(0)=c_{2}^{*}(0)=0\,,$ (430)

thus the probabilities are
$P_{1}=|c_{1}(0)|^{2}=P_{2}=|c_{1}(0)|^{2}=\frac{1}{2}$ and
$P_{1}^{*}=|c_{1}^{*}(0)|^{2}=P_{2}^{*}=|c_{1}^{*}(0)|^{2}=0$. This describe
the the initial state, which is the first diagram at $\theta=0$ in 65, which
is just the qubit,

$|\psi(0)\rangle=\Big{(}\frac{1}{\sqrt{2}}|0\rangle_{Q}+\frac{1}{\sqrt{2}}|1\rangle_{Q^{*}}\,\,\Big{|}S_{k}\Big{)}\,.$
(431)

The time evoluting quantum state is

$|\psi(t)\rangle=\frac{1}{\sqrt{2}}\bigg{(}\cos\frac{\omega
t}{2}|0\rangle_{Q}+\cos\frac{\omega t}{2}|1\rangle_{Q^{*}}+\sin\frac{\omega
t}{2}|0^{*}\rangle_{Q}+\sin\frac{\omega t}{2}|1^{*}\rangle_{Q^{*}}\bigg{)}$
(432)

We can check that when $\theta=\pi$,

$|\psi(T/2)\rangle=\frac{1}{\sqrt{2}}|0^{*}\rangle_{Q}+\frac{1}{\sqrt{2}}|1^{*}\rangle_{Q^{*}}=\frac{1}{\sqrt{2}}|1\rangle_{Q}+\frac{1}{\sqrt{2}}|0\rangle_{Q^{*}}$
(433)

and when $\theta=\pi/2$,

$|\psi(T/2)\rangle=\frac{1}{2}\Big{(}|0\rangle_{Q}+|1\rangle_{Q^{*}}+|1\rangle_{Q}+|0\rangle_{Q^{*}}\Big{)}\,.$
(434)

When we look from the $S_{k}^{\star}$ dual frame, the phase would be offset by
$\pi$, we have

$(|\psi(\theta)\rangle\,\,|S_{k})=(|\psi(\theta-\pi)\rangle\,\,|S_{k}^{\star})\,,$
(435)

i.e.

$\displaystyle\quad\bigg{(}\frac{1}{\sqrt{2}}\Big{(}\cos\frac{\omega
t}{2}|0\rangle_{Q}+\cos\frac{\omega t}{2}|1\rangle_{Q^{*}}+\sin\frac{\omega
t}{2}|0^{*}\rangle_{Q}+\sin\frac{\omega
t}{2}|1^{*}\rangle_{Q^{*}}\Big{)}\,\,\bigg{|}S_{k}\bigg{)}$ (436)
$\displaystyle\equiv\bigg{(}\frac{1}{\sqrt{2}}\bigg{(}\sin\frac{\omega
t}{2}|0\rangle_{Q}+\sin\frac{\omega t}{2}|1\rangle_{Q^{*}}+\cos\frac{\omega
t}{2}|0^{*}\rangle_{Q}+\cos\frac{\omega
t}{2}|1^{*}\rangle_{Q^{*}}\Big{)}\,\,\bigg{|}S_{k}^{\star}\bigg{)}$

The _tai chi_ diagram and the 65 representation have the same topology. We can
define the tai-chi process formally by the following:

######  Definition 5.1.

Let $U$ be the full space and $Q$ , $Q^{*}$ the partition and dual partition
space where $U=Q\cup Q^{*}$, and $U\subset\mathbb{R}^{2}$. Let the area
function of the partitions be $A(U)=1$. The exists an isomorphism $f$ between
the probability space $P(X)$ with random variable
$X=\\{|0\rangle,|1\rangle\\}$ as the state space and the area space $A$, which
defines the boxed partition diagram,

$f:P\rightarrow A\,.$ (437)

######  Definition 5.2.

The _tai chi_ diagram is topological equivalent to the boxed _tai chi_ cycle
process with the full diagram that has the area of 1 unit.

We can interpret backwards, the _tai chi_ cycle process can be deformed to the
_tai chi_ diagram by homeomorphism.

Figure 66:  Continuous deformation from _tai chi_ cycle process to _tai chi_
diagram.

###  5.3 The $n=2$ level

The basis of $n=2$ level, i.e. $\mathbb{Z}_{2}\otimes\mathbb{Z}_{2}$ 4-duality
group is naturally a 4-fundamental tableau.

Figure 67:

The null state corresponds to $0$, the 01 state corresponds to $V$, the 10
corresponds to $V^{*}$, and the full 11 state corresponds to All. We can
interpret also as either the 01 and 10 state contributes half of the system,
and the joining of them represents the fullness All. Therefore, it naturally
forms the basis of $\mathbb{Z}_{2}\times\mathbb{Z}_{2}$ 4-duality group and is
represented by the 4-fundamental tableau. The basis is
$\\{|00\rangle,|01\rangle,|10\rangle,|11\rangle\\}$, or in decimal
representation $\\{|0\rangle,|1\rangle,|2\rangle,|3\rangle\\}$.

Now recalling the comparison representation, here we can interpret the 4
states as group elements. The is equivalent to turning the basis as operators.
We call it quantization of the 4-basis. We can construct Cayley table as
follow.

Figure 68:

Therefore we write

$|0\rangle\rightarrow\hat{0}\,\,,\,\,|1\rangle\rightarrow\hat{1}\,\,,\,\,|2\rangle\rightarrow\hat{2}\,\,,\,\,|1\rangle\rightarrow\hat{3}\,.$
(438)

Or

$|00\rangle\rightarrow\hat{00}\,\,,\,\,|01\rangle\rightarrow\hat{01}\,\,,\,\,|10\rangle\rightarrow\hat{10}\,\,,\,\,|11\rangle\rightarrow\hat{11}\,.$
(439)

The 68 is isomorphic to the point group
$C_{2v}=\\{e,C_{2},\sigma_{1},\sigma_{2}\\}$, with the element identification
as

$\hat{3}\rightarrow e\,\,,\,\,\hat{0}\rightarrow
C_{2}\,\,,\,\,\hat{1}\rightarrow\sigma_{1}\,\,,\,\,\hat{2}\rightarrow\sigma_{2}\,.$
(440)

The feature diagram is

Figure 69:  Feature diagram of 4-yi.

The representation matrix of the 4-duality group is

$D(\mathbb{Z}_{2}\otimes\mathbb{Z}_{2})=\begin{pmatrix}A_{1}&0&0&0\\\
0&A_{2}&0&0\\\ 0&0&A_{3}&0\\\ 0&0&0&A_{4}\\\
\end{pmatrix}=\begin{pmatrix}\mathcal{A}_{1}(g)\mathcal{A}_{1}(g^{\prime})&0&0&0\\\
0&\mathcal{A}_{2}(g)\mathcal{A}_{2}(g^{\prime})&0&0\\\
0&0&\mathcal{A}_{2}(g)\mathcal{A}_{1}(g^{\prime})&0\\\
0&0&0&\mathcal{A}_{1}(g)\mathcal{A}_{2}(g^{\prime})\\\ \end{pmatrix}$ (441)

for $\mathrm{g}=(g,g^{\prime})=(I,I),(P,P),(P,I)\,\,{\text{and}}\,\,(I,P)$.
Explicitly

$\displaystyle D\big{(}([0],[0])\big{)}$ $\displaystyle=D\otimes
D(I,I)={\text{diag}}(1,1,1,1)\,,$ (442) $\displaystyle
D\big{(}([1],[1])\big{)}$ $\displaystyle=D\otimes
D(P,P)={\text{diag}}(1,1,-1,-1)\,,$ $\displaystyle D\big{(}([0],[1])\big{)}$
$\displaystyle=D\otimes D(I,P)={\text{diag}}(1,-1,-1,1)\,,$ $\displaystyle
D\big{(}([1],[0])\big{)}$ $\displaystyle=D\otimes
D(P,I)={\text{diag}}(1,-1,1,-1)\,.$

Next we take the basis as $|00\rangle$ , $|01\rangle$ , $|10\rangle$ and
$|11\rangle$. They correspond to, respectively

$\displaystyle|00\rangle$ $\displaystyle=|--\rangle|--\rangle$
$\displaystyle\equiv|\includegraphics[trim=0.0pt 0.0pt 0.0pt
0.0pt,clip,scale={0.5}]{04.png}\rangle$ (443) $\displaystyle|01\rangle$
$\displaystyle=|--\rangle|-\rangle$
$\displaystyle\equiv|\includegraphics[trim=0.0pt 0.0pt 0.0pt
0.0pt,clip,scale={0.5}]{24.png}\rangle$ $\displaystyle|10\rangle$
$\displaystyle=|-\rangle|--\rangle$
$\displaystyle\equiv|\includegraphics[trim=0.0pt 0.0pt 0.0pt
0.0pt,clip,scale={0.5}]{14.png}\rangle$ $\displaystyle|11\rangle$
$\displaystyle=|-\rangle|-\rangle$
$\displaystyle\equiv|\includegraphics[trim=0.0pt 0.0pt 0.0pt
0.0pt,clip,scale={0.5}]{34.png}\rangle$ $\displaystyle\,.$

We can further define $A_{1}=U_{1},A_{2}=U_{2}$ and $A_{3}=W_{1},A_{4}=W_{2}$,

$\displaystyle\quad U_{1}|{\text{00}}\rangle\oplus
U_{2}|{\text{11}}\rangle\oplus W_{1}|{\text{01}}\rangle\oplus
W_{2}|{\text{10}}\rangle$ (444)
$\displaystyle=U\big{(}|{\text{00}}\rangle\oplus|{\text{11}}\rangle\big{)}\oplus
W\big{(}|{\text{01}}\rangle\oplus|{\text{10}}\rangle\big{)}$

for $U=U_{1}\oplus U_{2}$ and $W=W_{1}\oplus W_{2}$, so we separate them into
two large categories, the (CC and DD) one, the fully connected or disconnected
case, and the (DC and CD) one which are both halfly connected. Thus one can
write

$D(\mathbb{Z}_{2}\otimes\mathbb{Z}_{2})=\begin{pmatrix}U&0\\\
0&V\end{pmatrix}$ (445)

which is reducible, and this neatly represents that the 4-duality group as two
big categories, with $|00\rangle,\,|11\rangle$ being one and
$|01\rangle,\,|10\rangle$ another one. Compactly in dimension notation we can
write the tensor product as

$2\otimes 2=2\oplus 2=1\oplus 1\oplus 1\oplus 1\,.$ (446)

###  5.4 Quantum state with embedded 4-duality group

In this section, we would like to promote the basis of irreps of
$\mathbb{Z}_{2}\times\mathbb{Z}_{2}$ to quantum states and study the linear
combination of it. The full linear combination of the basis of irreps over the
4-Klein group can be expressed as a rank-2 tensor,

$|\,\phi(g,g^{\prime})\,\rangle=\frac{1}{2}\sum_{i_{1},i_{2}=0,1}a_{i_{1}i_{2}}(g,g^{\prime})|\eta_{i_{1}}\rangle\otimes|\eta_{i_{2}}\rangle\,,$
(447)

where we sum over repeating indices. The factor of $\frac{1}{2}$ is for
normalization. Explicitly we write

$\displaystyle|\,\phi(g,g^{\prime})\,\rangle$
$\displaystyle=\frac{1}{2}\Big{(}a_{00}(g,g^{\prime})|00\rangle+a_{11}(g,g^{\prime})|11\rangle+a_{01}(g,g^{\prime})|01\rangle+a_{10}(g,g^{\prime})|10\rangle\Big{)}$
(448)
$\displaystyle=\frac{1}{2}\Big{(}a_{00}(g,g^{\prime})|--\rangle|--\rangle+a_{11}(g,g^{\prime})|-\rangle|-\rangle+a_{01}(g,g^{\prime})|--\rangle|-\rangle+a_{10}(g,g^{\prime})|-\rangle|--\rangle\Big{)}\,.$

Define the tensor by

$\mathrm{A}(g,g^{\prime})=\begin{pmatrix}a_{00}(g,g^{\prime})&a_{01}(g,g^{\prime})\\\
a_{\rm{10}}(g,g^{\prime})&a_{\rm{11}}(g,g^{\prime})\end{pmatrix}=\begin{pmatrix}\mathcal{A}_{1}(g)A_{1}(g^{\prime})&\mathcal{A}_{1}(g)\mathcal{A}_{2}(g^{\prime})\\\
\mathcal{A}_{2}(g)\mathcal{A}_{1}(g^{\prime})&\mathcal{A}_{2}(g)\mathcal{A}_{2}(g^{\prime})\end{pmatrix}\,,$
(449)

The $\mathcal{A}_{1},\mathcal{A}_{2}$ are 1D irreps we had in 441. Using the
result we had in 442, we have

$\mathrm{A}(I,I)=\begin{pmatrix}1&1\\\
1&1\end{pmatrix}\,\,,\,\,\mathrm{A}(P,P)=\begin{pmatrix}1&-1\\\
-1&1\end{pmatrix}\,\,,\,\,\mathrm{A}(I,P)=\begin{pmatrix}1&-1\\\
1&-1\end{pmatrix}\,\,,\,\,\mathrm{A}(P,I)=\begin{pmatrix}1&1\\\
-1&-1\end{pmatrix}\,.$ (450)

And since the tensor product of two parity group is just same as the 4-duality
group, thus we have written
$(g,g^{\prime})\in\mathbb{Z}_{2}\otimes\mathbb{Z}_{2}$ as
$g\in\mathbb{Z}_{2}\times\mathbb{Z}_{2}$, and therefore from above we find

${\rm{det}}\,A(\rm{g})=0\,\,{\text{for all
\rm{g}}}\in\mathbb{Z}_{2}\times\mathbb{Z}_{2}\,.$ (451)

That means the linear combination of the 4 reducible representation basis of
the 4-dual set with the tensor components as the transformation of the
4-duality Klein 4 group is untangled. In other words, the doubly parity
transformation of a 4-dual set corresponds to 4 untangled basis.

This shows that the full state $\phi$ can be written as two independent tensor
product of the same dual duplet. For $g\in\mathbb{Z}_{2}$

$|\,\varphi(g)\,\rangle=\frac{1}{\sqrt{2}}\big{(}\mathcal{A}_{1}(g)|{\text{0}}\rangle+\mathcal{A}_{2}(g)|{\text{1}}\rangle\big{)}.$
(452)

Since all $\mathrm{A}_{1},\mathrm{A}_{2}$ are either $1$ or $-1$, thus it is
automatically properly normalized by $\frac{1}{\sqrt{2}}$. And we have

$|\,\phi(g,g^{\prime})\,\rangle=|\,\varphi(g)\,\rangle\otimes|\,\varphi(g^{\prime})\,\rangle\,.$
(453)

We can see that we have a probability of $\frac{1}{2}$ of having $|0\rangle$
or $|1\rangle$ state. More properties will be discussed later.

From the results of 450 , intuitively we can see that $\mathrm{A}(I,I)$ and
$\mathrm{A}(P,P)$ form a dual pair, while $\mathrm{A}(I,P)$ and
$\mathrm{A}(P,I)$ for a dual pair, this is because we see that the character
(trace) of $\mathrm{A}(I,I)$ and $\mathrm{A}(P,P)$ are the same, while the
character of $\mathrm{A}(I,P)$ and $\mathrm{A}(P,I)$ are the same,

${\mathrm{Tr}}\,\mathrm{A}(I,I)={\mathrm{Tr}}\,\mathrm{A}(P,P)=2\quad{\text{and}}\quad{\mathrm{Tr}}\,\mathrm{A}(I,P)={\mathrm{Tr}}\,\mathrm{A}(P,I)=0\,.$
(454)

Therefore, one can obtain $\mathrm{A}(P,P)$ from $\mathrm{A}(I,I)$ by some
similarity transformation, i.e. $\mathrm{A}(P,P)=U^{-1}\mathrm{A}(I,I)U$ for
some matrix $U$ , and likewise $\mathrm{A}(I,P)=V^{-1}\mathrm{A}(P,I)V$ for
some matrix $V$. In fact, the 4 representation matrices in 450 form a
4-duality group
$\\{\mathrm{A}(I,I),\mathrm{A}(I,P),\mathrm{A}(P,I),\mathrm{A}(P,P)\\}$ under
the operation of element-wise matrix multiplication $\bullet$ (known as the
Hadamard product). Such operation is Abelian. The identity is
$\mathrm{A}(I,I)$, and each element is of its own inverse. For example it is
easy to check that

$\mathrm{A}(I,P)\bullet\mathrm{A}(P,I)=\mathrm{A}(P,P)\,,$ (455)
$\mathrm{A}(P,P)\bullet\mathrm{A}(P,I)=\mathrm{A}(I,P)\,,$ (456)
$\mathrm{A}(P,P)\bullet\mathrm{A}(I,P)=\mathrm{A}(P,I)\,,$ (457)
$\mathrm{A}(I,P)\bullet\mathrm{A}(I,P)=\mathrm{A}(I,I)\,,$ (458)

etc. Therefore $\mathrm{A}$ is actually a function of the group elements of
the parity group. The idea can be represented by the following diagram

Figure 70: Diagramatic representation.

###  5.5 4-Duality basis representation by order rearrangement

In this subsection, we would study a very important construction of the
4-Duality basis by the order permutation of 4-yi of $n=2$ level. This creates
a new way of forming basis representation of the 4-duality group, and would
extend the originally simple basis representation from
$|--\rangle|--\rangle,\,|-\rangle|-\rangle,\,|--\rangle|-\rangle\,\,\text{and}\,\,|-\rangle|--\rangle$.

Let’s first define some clear notations. At the $n=1$ level splitting, we can
start with either $-$ or $--$ first, then followed by $--$ and $-$
respectively. The former is said to be left 0 right 1, symbolized by $(--,-)$,
while the latter would be left 1 right 0 symbolized by $(-,--)$. Next denote
the stacking of a yin state on the left and a yang state on the right as LR
and the opposite RL, and this can be done along the upward direction (U) and
downward direction (D). Next we assign the Gua’s order parameter as I, II,
III, IV as usual. The duality mirror lies in the middle to separate I, II and
III, IV into two halves. When the stacking carries, if there is no crossing
between the two halves we call it is a normal diagram, otherwise a crossing
diagram. The crossing reference is labelled by two pairs of order parameter,
for example I-III II-IV crossing. This is called the symmetry crossing as one
will see that this is symmetric along the dual mirror plane. For each diagram,
we collect the possible outcomes as $(ijkl)$, where each of them is the binary
number of the outcome. The following illustrates how do these principles work

Figure 71: Two examples for the splitting processes along the upward
direction. Both of them are noraml diagrma.  Figure 72: Two examples for the
splitting processes along the downward direction. The left one is a noraml
diagram and the right one is a I-III II-IV crossing diagram.

There would a total of 32 outcomes. The results are tabulated as follow

Normal diagram: Left 0 Right 1 $(--,-)$
---
| LL | RR | LR | RL
U | $\color[rgb]{1,0,0}{(0213)}$ | $\color[rgb]{1,0,0}{(2031)}$ | $\color[rgb]{1,0,0}{(0231)}$ | $\color[rgb]{1,0,0}{(2013)}$
D | $\color[rgb]{1,0,0}{(0123)}$ | $\color[rgb]{1,0,0}{(1032)}$ | $\color[rgb]{1,0,0}{(0132)}$ | $\color[rgb]{1,0,0}{(1023)}$
Normal diagram: Left 1 Right 0 $(-,--)$
| LL | RR | LR | RL
U | $\color[rgb]{1,0,0}{(1302)}$ | $\color[rgb]{1,0,0}{(3120)}$ | $\color[rgb]{1,0,0}{(1320)}$ | $\color[rgb]{1,0,0}{(3102)}$
D | $\color[rgb]{1,0,0}{(2301)}$ | $\color[rgb]{1,0,0}{(3210)}$ | $\color[rgb]{1,0,0}{(2310)}$ | $\color[rgb]{1,0,0}{(3201)}$
Crossing diagram: Left 0 Right 1 $(--,-)$
| LL | RR | LR | RL
U | $\color[rgb]{0,0,1}{(0303)}$ | $\color[rgb]{0,0,1}{(2121)}$ | $\color[rgb]{0,0.69140625,0.3125}{(0321)}$ | $\color[rgb]{0,0.69140625,0.3125}{(2103)}$
D | $\color[rgb]{0,0,1}{(0303)}$ | $\color[rgb]{0,0,1}{(1212)}$ | $\color[rgb]{0,0.69140625,0.3125}{(0312)}$ | $\color[rgb]{0,0.69140625,0.3125}{(1203)}$
Crossing diagram: Left 1 Right 0 $(-,--)$
| LL | RR | LR | RL
U | $\color[rgb]{0,0,1}{(1212)}$ | $\color[rgb]{0,0,1}{(3030)}$ | $\color[rgb]{0,0.69140625,0.3125}{(1230)}$ | $\color[rgb]{0,0.69140625,0.3125}{(3012)}$
D | $\color[rgb]{0,0,1}{(2121)}$ | $\color[rgb]{0,0,1}{(3030)}$ | $\color[rgb]{0,0.69140625,0.3125}{(2130)}$ | $\color[rgb]{0,0.69140625,0.3125}{(3021)}$
Table 2: Tabulated results.

Hence, the full combination of normal diagrams and crossing diagrams give all
the possibilities of the $4!=24$ elements of permutation group $S_{4}$ and 8
elements that have two repeated numbers. We call those disconnected diagrams,
the reason for calling them ‘disconnected’ with be soon addressed. Now we
would represent all the $S_{4}$ elements diagramatically with categories of
closed loops and would map all the red and green $(ijlk)$ the corresponding
categories. The result is shown as follow.

|  |  |  |  |
---|---|---|---|---|---
(0123) | (0321) | (0132) | (0231) | (0312) | (0213)
(3012) | (1032) | (2013) | (1023) | (2031) | (3021)
(2301) | (2103) | (3201) | (3102) | (1203) | (1302)
(1230) | (3210) | (1320) | (2310) | (3120) | (2130)
Table 3:  The $S_{4}$ permutation group represented by diagrams, with red and
green values obtained in 2 mapped onto them. Each successive row is formed by
cyclic permutation of the previous one, i.e.
$(ijkl)\rightarrow(lijk)\rightarrow$ etc. These operations form a rotation
group of $\mathrm{C}_{4}=\\{I,C_{4},C_{4}^{2},C_{4}^{3}\\}$.

The 8 blue elements with two repeated numbers are not contained in the $S_{4}$
group representations, as they cannot be drawn as a closed loop. There are
redundancies in each of these elements and they do not give rise to $(ijkl)$
for all different $i\neq j\neq k\neq l$. We represent these as two disjoint
pieces.

Figure 73: Disconnected diagram for symmetric I-III II-IV crossing. There are
2 set of diagrams for each of them.

There is another choice of crossing , I-IV II-III. This is the asymmetric
crossing, as it is not symmetric under the dual mirror plane. This would
recover all the 8 green values in 2, but different blue values.

Figure 74: Disconnected diagram for symmetric I-IV II-III crossing. There are
2 set of diagrams for each of them.

Since for the four all orders I, II, III, IV, there are only 3 possible ways
for forming two pairs. The first way I-II , III-IV generates all 16 normal
diagrams, then the remaining two ways are the symmetric I-III, II-IV crossing
and the I-IV, II-III asymmetric crossing. Each type of crossing generates
another 16 diagrams, in which 8 of them join with the 16 I-II , III-IV
diagrams that form a representation of $S_{4}$ group. There are two sets of
disconnected diagrams formed by the two different crossing, and the
differences are in the direction of the connecting lines (see 73 and 74). Now
we can reorganize the result in 2 to represent in two sets of 4-tableau due to
different crossings as

Figure 75: Two sets of 4-tableau diagram. Figure (a) is for I-III, II,IV
symmetric crossing and figure (b) is for I-IV, II-III asymemtric crossing. The
big number 8 in each box means that there are 8 elements of $(ijkl)$ in it.
The $8(--,-)$ means that there are 8 elements coming from left 0 right 1
diagrams while $4(--,-)\,4(-,--)$ means the 8 elements come from half of each
$(--,-)$ and $(-,--)$.

Thus for (a) we can mathematically write

$32_{\rm{sym}}=(24\oplus 8)_{\rm{sym}}$ (459)

and for (b)

$32_{\rm{asym}}=(24\oplus 8)_{\rm{asym}}\,.$ (460)

where $24_{\rm{sym}}=24_{\rm{asym}}=24$. Thus we have $32_{\rm{sym}}\cap
32_{\rm{asym}}=24_{\rm{sym}}=24_{\rm{asym}}=24$. Note that here the ‘$\oplus$”
symbol does not refer to the direct sum but a notation for classifying objects
into different categories. In terms of the box representation we can simply
write

$4_{\rm{sym}}=(3\oplus 1)_{\rm{sym}}$ (461)

and

$4_{\rm{asym}}=(3\oplus 1)_{\rm{asym}}\,.$ (462)

And we have $4_{\rm{sym}}\cap 4_{\rm{asym}}=3$.

It is important to note that there are no horizontal disconnect diagrams like
$(0101)\,\,(2323)$ and diagonal disconnect diagrams like $(0202)\,\,(1313)$.
This is because the former one involve both $|0\rangle$ states or both
$|0\rangle$ states at $n=1$ level for the split, which violates our rule.
Similarly the latter one involve all 4 $|0\rangle$ states or all 4 $|1\rangle$
states for the splitting at $n=2$ level, again this violates our splitting
rules. It is always important to bear in mind that the splitting processes
must be carried out in emerging one $|0\rangle$ and $|1\rangle$ state each
time from the original state.

We can also represent them by

Figure 76: Diagramatic representation.

There are fruitful information in the arrangement table 2. It can be
represented by two separate 4-tableau, one for normal diagram and the other
for crossing diagram.

Figure 77: Left standard tableau: Normal diagrams; Right standard tableau:
Crossing diagrams. $Q$ with subscripts denote different quadrants.

For normal diagrams, we have

$Q_{\mathrm{N}_{0}}=\mathrm{N\cap\bar{N}},\quad
Q_{\mathrm{N}_{1}}=\mathrm{N\cap\bar{N}^{*}},\quad
Q_{\mathrm{N}_{2}}=\mathrm{N^{*}\cap\bar{N}},\quad
Q_{\mathrm{N}_{3}}=\mathrm{N^{*}\cap\bar{N}^{*}}\,.$ (463)

We can establish isomorphism by

$Q_{\mathrm{N}_{0}}\mapsto(0_{1}0_{2})\quad
Q_{\mathrm{N}_{1}}\mapsto(0_{1}1_{1}),\quad
Q_{\mathrm{N}_{2}}\mapsto(1_{1}0_{2}),\quad
Q_{\mathrm{N}_{3}}\mapsto(1_{1}1_{2})\,,$ (464)

for which unstarred set is represented by 0 and starred set is represented by
1.

For crossing diagrams, we have

$Q_{\mathrm{C}_{0}}=\mathrm{C\cap\bar{C}},\quad
Q_{\mathrm{C}_{1}}=\mathrm{C\cap\bar{C}^{*}},\quad
Q_{\mathrm{C}_{2}}=\mathrm{C^{*}\cap\bar{C}},\quad
Q_{\mathrm{C}_{3}}=\mathrm{C^{*}\cap\bar{C}^{*}}\,.$ (465)

We can establish isomorphism by

$Q_{\mathrm{C}_{0}}\mapsto(0_{1}0_{2})\quad
Q_{\mathrm{C}_{1}}\mapsto(0_{1}1_{2}),\quad
Q_{\mathrm{C}_{2}}\mapsto(1_{1}0_{2}),\quad
Q_{\mathrm{C}_{3}}\mapsto(1_{1}1_{2})\,.$ (466)

Therefore the table establishes an isomorphism of two separate standard
4-tableau, each each of them is isomorphic to the heterogeneous representation
of the $\mathbb{Z}_{2}\times\mathbb{Z}_{2}$ group.

Note also that the full sets and empty sets for the normal digram and crossing
diagram are given by respectively,

$\mathrm{X=N\cup N^{*}=\bar{N}\cup\bar{N}^{*},\quad Y=C\cup
C^{*}=\bar{C}\cup\bar{C}^{*}}$ (467)

and

$\mathrm{N\cup N^{*}=\bar{N}\cup\bar{N}^{*}=\emptyset,\quad C\cup
C^{*}=\bar{C}\cup\bar{C}^{*}=\emptyset}\,.$ (468)

###  5.6  $n=3$ level

The 3-level case can be studied through the feature diagram, due to the fact
that 3 is odd and generally odd levels lack the symmetry property as the even
counterpart, there are less interesting properties to study.

The full state is given by

$|\psi\rangle=\sum_{i=0}^{7}a_{i}|i\rangle\,.$ (469)

We can write the sum by grouping the terms as 4 dual pairs, in which 2 of them
are dual invariants. In binary representation

$\displaystyle|\psi\rangle$
$\displaystyle=\big{[}\,(a_{000}|000\rangle+a_{111}|111\rangle)+(a_{010}|101\rangle+a_{101}|101\rangle)\,\big{]}$
(470)
$\displaystyle\quad\quad\quad\quad\quad\quad+\big{[}\,(a_{001}|001\rangle+a_{100}|100\rangle)+(a_{011}|011\rangle+a_{110}|110\rangle)\,\big{]}$

and in decimal representation,

$|\psi\rangle=\big{[}\,(a_{0}|0\rangle+a_{7}|7\rangle)+(a_{2}|2\rangle+a_{5}|5\rangle)\,\big{]}+\big{[}\,(a_{1}|1\rangle+a_{6}|6\rangle)+(a_{3}|3\rangle+a_{4}|4\rangle)\,\big{]}.$
(471)

We can represent this with the feature diagram as $2\times 2\times 2$
hypercube, but what we are more interested is its duality property, thus we
will find a way, if possible, to represent it by a 2D 4-duality diagram. This
time we can we have two diagrams in a single box instead of 1 as before and
each dual pair is treated as one dimension, collectively, in line with the
representation theory. The two square brackets in the above equation in 470 or
471 is showing the classification of two parts: dual invariant and non-dual
invariant. The feature diagram and its interpretations are shown as follow.

Figure 78: 4-tableau of 3-level gua. (a) Dual structure (2+2) of dual
invariant states and non-dual invariant states; (b) (1+3) sturcture of pure
states and mixed states

There are two ways of interpretation here. The left one classifies the full
state into dual-invariant (indicated by blue and non-dual invariant (indicated
by green), giving a $(2\oplus 2)$ structure and hence further a $(1\oplus 1)$
dual structure. Thus there is the dual classification. Note that dual and non-
dual is itself a duality. The right one classifies the full states into pure-
states and mixed states, giving a $(1\oplus 3)$ structure. Pure states are
states of which all states are equal in each level, while mixed states are
states of which there exist one state that are different.

###  5.7  $n=4$ level

For $n=4$ level, the full state is given by

$|\psi\rangle=\sum_{i=0}^{15}a_{i}|i\rangle\,.$ (472)

And explicitly we write,

$|\psi\rangle=\sum_{i_{1},i_{2},i_{3},i_{4}=0,1}a_{i_{1}i_{2}i_{3}i_{4}}|\eta_{i_{1}}\eta_{i_{2}}\eta_{i_{3}}\eta_{i_{4}}\rangle\,.$
(473)

with the normalization

$\sum_{i_{1},i_{2},i_{3},i_{4}=0,1}|a_{i_{1}i_{2}i_{3}i_{4}}|^{2}=1\,.$ (474)

Next we would study the feature diagram,

Figure 79: Feature diagram of $16$-Gua.

We can link up the four spots of dual invariant numbers as a diamond. As we
will see later, for higher n-Gua, we can join the spots with different
interesting patterns, which can be easily recognized.

###  5.8 The study of $n=6$ level

For $n=6$ level, the full state is given by

$|\psi\rangle=\sum_{i=0}^{63}a_{i}|i\rangle\,.$ (475)

And explicitly we write,

$|\psi\rangle=\sum_{i_{1},i_{2},i_{3},i_{4},i_{5},i_{6}=0,1}a_{i_{1}i_{2}i_{3}i_{4}i_{5}i_{6}}|\eta_{i_{1}}\eta_{i_{2}}\eta_{i_{3}}\eta_{i_{4}}\eta_{i_{5}}\eta_{i_{6}}\rangle\,.$
(476)

with the normalization

$\sum_{i_{1},i_{2},i_{3},i_{4},i_{5},i_{6}=0,1}|a_{i_{1}i_{2}i_{3}i_{4}i_{5}i_{6}}|^{2}=1\,.$
(477)

The feature diagram is

Figure 80:  Feature diagram for $n=6$ level.

There are two diamonds. Alternatively, we can join it as a hexagon with two
external lines.

###  5.9 The study of higher n level

In general for higher even $n$ which can be square rooted, it shows
interesting patterns. For example, for $n=8$ case, which is $256$-diagram,
which can be represented by a $16\times 16$ feature diagram.

Figure 81:  Feature diagram.

And for convenience we can just show the spots for dual invariant numbers.

Figure 82:  Feature diagram.

For example, for $n=10$ case, which is $1024$-diagram, which can be
represented by a $32\times 32$ feature diagram. For convenience here we only
show the spots for dual invariant number.

Figure 83:  Feature diagram: diamond pattern  Figure 84:  Feature diagram:
second diamond pattern  Figure 85:  Feature diagram: polygon pattern  Figure
86:  Feature diagram: hexagon pattern

##  6 Duality in Rotor Mechanism

In the section, we would like to introduce duality in a new fashion by rotor
mechanism. We will study how duality arises from rotation of circulation of
states.

We will define two disks of states, the external layer and internal layer of
states. By rotating the internal or external layer of states, this can general
new state with duality structure.

###  6.1 Rotor mechanism in level 2

Consider an external disk and internal disk, associated with two states in
each layer. We rotate the internal disk, keeping the external disk unchanged.
The internal rotation introduces a new state which is dual to the original
one.

Figure 87: Rotor mechanism of 2 level.

First notice in the original diagram, we have the entangled state as

$|\psi\rangle=\frac{1}{\sqrt{2}}(|00\rangle+|11\rangle)\,.$ (478)

After the internal disk rotation for $\pi$ radian, this generates a new
entangled state,

$|\psi^{*}\rangle=\frac{1}{\sqrt{2}}(|01\rangle+|10\rangle)\,.$ (479)

This state is dual to the original state, this is because as shown in chapter
2. Moreover, it can be seen that the first diagram is for
$\\{-2n,\cdots-2,0,2,\cdots 2n\\}$ rotations and the second diagram is for
$\\{-2n+1,\cdots-1,1,\cdots 2n+1\\}$ rotations, and we know that the set of
even numbers and the set of odd numbers are dual to each other. Therefore, in
this way, we say $|\psi\rangle=\frac{1}{\sqrt{2}}(|00\rangle+|11\rangle)$ the
even state, while $|\psi^{*}\rangle=\frac{1}{\sqrt{2}}(|01\rangle+|10\rangle)$
the odd state. The even parity and odd parity can be seen also in the tensor
product states. Recalling in equation (348),

$|\psi(\theta_{1},\theta_{2})\rangle=\cos\frac{\theta_{1}}{2}\cos\frac{\theta_{2}}{2}|00\rangle+\cos\frac{\theta_{1}}{2}\sin\frac{\theta_{2}}{2}|01\rangle+\sin\frac{\theta_{1}}{2}\cos\frac{\theta_{2}}{2}|10\rangle+\sin\frac{\theta_{1}}{2}\sin\frac{\theta_{1}}{2}|11\rangle\,.$
(480)

So we see that the $|00\rangle$ and $|11\rangle$ states are associated with
even coefficients as $\cos\phi_{1}\cos\phi_{2}$ and $\sin\phi_{1}\sin\phi_{2}$
are even functions; while $|01\rangle$ and $|10\rangle$ states are associated
with odd coefficients as $\cos\phi_{1}\sin\phi_{2}$ and
$\sin\phi_{1}\cos\phi_{2}$ are odd functions. In addition, $|00\rangle$ and
$|00\rangle$ states are left-right observation invariant, while $|01\rangle$
and $|10\rangle$ states are not left-right observation invariant. This is an
example of, dual invariance and non-dual invariant are itself a duality.

Figure 88:

###  6.2 Internal and external observation duality

In this section we investigate the duality between internal and external
observer by constructing a two-layer rotation disk. Each layer consists of 4
elements with 0 or 1. The two-layer disk can be viewed from the internal
perspective or external perspective, which would give us different decimal
numbers respectively. The construction and the operation of the disk is shown
below.

Figure 89: Periodic 4-system from internal observer’s frame. The outter layer
shuffles in the clockwise direction while the internal layer shufffles in the
anticlockwise direction. Figure 90: Periodic 4-system from external observer’s
perspetive. The outter layer shuffles in the clockwise direction while the
internal layer shufffles in the anticlockwise direction.

The maps are given by the cyclic group $\mathbb{Z}_{4}$. We have the group
elements as

$G=\\{1\otimes 1,C_{4}\otimes C_{4}^{-1},C_{2}\otimes
C_{2}^{-1},C^{3}_{4}\otimes C_{4}^{-3}\\}$ (481)

The group elements are just isomorphic to $\mathbb{Z}_{4}$ because we have the
tensor product identity of $(A\otimes B)(C\otimes D)=(AC)\otimes(CD)$.

Figure 91: Periodic 4-system from external and internal observer’s frame

Therefore we can see that a square is dual to the twisted square, and the
hidden meaning is such duality. It is very difficult to see that just from
sretch that such two geometrical objects would have such meaning in
observation duality. And the difference between these object is that the
twisted one contains a node (or intersection) while the plain one does not.

Next let’s compute it in a table by unwrapping the periodicity.

Position | 1 | 2 | 3 | 4
---|---|---|---|---
External layer | 0101 | 1010 | 0101 | 1010
Internal layer | 0011 | 1001 | 1100 | 0110

The position is the label of each quadrant (for both internal and external
layer) and must be fixed. The property of periodicity also holds for the
comparison space.

Position $\displaystyle\,\,1:\,\,0101::0011=1001$ (482) Position
$\displaystyle\,\,2:\,\,1010::1001=1100$ Position
$\displaystyle\,\,3:\,\,0101::1100=0110$ Position
$\displaystyle\,\,4:\,\,1010::0110=0011$

We have two dual invariants after the $::$ computation, 1001 and its dual 0110
; and two non-dual invariants, 1100 and its dual 0011. And thus we have 1,3
dual and 2,4 dual respectively in terms of the position. Note that before, for
internal layer, we have 1,3 identical, 2,4 identical; for external layer, we
have non-dual invariants 1,3 dual and dual invariants 2,4 dual. The comparison
computation $::$ retains the dual structure followed by the external layer,
but the role of dual invariant and non-dual invariant has swapped.

###  6.3 Rotor mechanism in level 4

Next we promote to study the duality properties of rotor mechanism in level 4.
There are more complications than the case in level 2. Consider each layer
with four states $|00\rangle$, $|01\rangle$, $|10\rangle$ and $|11\rangle$, in
decimal place which are $|0\rangle$, $|1\rangle$, $|2\rangle$ and $|3\rangle$.
First consider that we rotate the internal disk in the anti-clockwise
direction, while keeping the external disk constant. Here first we employ the
internal observation perspective. We can see that in even number of rotation
of $\pi/4$, the structure of duality is remained.

Figure 92: Rotor mechanism in level 4. The blue ovals represent the pairs of
dual diagrams

For the unrotated diagram, we have the state

$|\psi_{0}\rangle=\frac{1}{2}|00\rangle\otimes|00\rangle+\frac{1}{2}|01\rangle\otimes|01\rangle+\frac{1}{2}|10\rangle\otimes|10\rangle+\frac{1}{2}|11\rangle\otimes|11\rangle$
(483)

and similar for the others.

In the original diagram (diagram of the upper-left corner), we have two sets
of dual diagrams, $\\{5,10\\}$ and $\\{0,15\\}$. The numerical difference of
the two sets is 5. When the internal layer is rotated by first even number of
$\pi/2$, such duality set structure is remained. We have $\\{7,8\\}$ and
$\\{2,13\\}$ two dual sets and the numerial difference of the two dual sets is
also 5. Therefore $n=0$, $n=2$ even $n$ (where $n$ is the number of $\pi/4$
phase rotation of the internal disk), there is an isomorphism of the two
structures. However, for $n=1$, $n=3$ odd $n$, there are no dual sets.

The following table illustrates the result of the rotor mechanism in level 4.

Figure 93:

The full space can be decomposed into four partitions due to four rotations,
mathematically we write,

$16=4\oplus 4\oplus 4\oplus 4\,.$ (484)

The collection of elements of each rotor diagram defines a vector space
$A_{i}$, where we have four of them. For $A_{0}$ and $A_{2}$, the space can be
decomposed into two dual spaces, which is

$4=2\oplus 2=1\oplus\bar{1}\oplus 1\oplus\bar{1}\,.$ (485)

For $A_{1}$ and $A_{3}$, there do not exist any decomposition into dual
spaces, but four distinctive unrelated spaces. Mathematically we write,

$4=1\oplus 1\oplus 1\oplus 1\,.$ (486)

One also finds that for each partition, the sum of all elements is always a
constant,

$\sum_{a\in A_{i}}a=30\,.$ (487)

Next we want to show that the total (net) change in numerical diagram is zero.
The proof is very easy. First define the change

$\Delta a_{i,j}=a_{A_{i+1,j}}-a_{A_{i},j}\,,$ (488)

where $j$ is the $j$-th element of the $i$-th partition Now it follows that

$\sum_{j}\Delta a_{i,j}=\sum_{a_{j}\in A_{i+1}}a_{j}-\sum_{a_{j}\in
A_{i}}a_{j}=15-15=0\,.$ (489)

Next we would like to study the conservation constants among the rotor
diagrams. In general, we see that properties within the same partition will
remain the same. We see that for the even partition (even number of $\pi/2$)

$a_{A_{0}\,1}+a_{A_{0}\,3}=0+10=a_{A_{2}\,1}+a_{A_{2}\,3}=2+8=10\,.$ (490)
$a_{A_{0}\,2}+a_{A_{0}\,4}=5+15=a_{A_{2}\,2}+a_{A_{2}\,4}=7+13=20\,.$ (491)

Also we see that for the odd partition (odd number of $\pi/2$)

$a_{A_{1}\,1}+a_{A_{1}\,3}=3+9=a_{A_{3}\,1}+a_{A_{3}\,3}=1+11=12\,.$ (492)
$a_{A_{1}\,2}+a_{A_{1}\,4}=4+14=a_{A_{3}\,2}+a_{A_{3}\,4}=6+12=18\,.$ (493)

And we also see that

$a_{A_{1}\,1}+a_{A_{1}\,2}=3+4=a_{A_{3}\,1}+a_{A_{3}\,2}=1+6=7\,.$ (494)
$a_{A_{1}\,3}+a_{A_{1}\,4}=9+14=a_{A_{3}\,3}+a_{A_{3}\,4}=11+12=23\,.$ (495)

Finally we will study the idea of sub-duality. We have already seen that the
even $n$ and odd $n$ are dual partition as shown in table (93). Now we will
show that in fact the $n=0$ partition is dual to the $n=2$ partition, and
$n=1$ partition is dual to the $n=3$ partition.

For the $n=1$ partition, the sum of difference is $(+3)+(-1)+(-1)+(-1)=0$. For
the $n=3$ partition, the sum of difference is $(-3)+(+1)+(+1)+(+1)=0$. Thus we
have

$\sum_{j}\Delta a_{1,j}=-\sum_{j}\Delta a_{3,j}\,.$ (496)

Therefore $n=1$ partition is dual to the $n=3$ partition. And the difference
in the odd partition is asymmetric.

For the $n=0$ partition, the sum of difference is trivial $0+0+0+0=0$. For the
$n=2$ partition, the sum of difference is $(+2)+(+2)+(-2)+(-2)=0$, which is
symmetric.

Now we consider the external perspective. We have the following result in 94
and 95. The properties from the internal perspective are basically inherited
to the external perspective, but just the numerical values are different. For
example, for the even partition
$a_{A_{0}\,1}+a_{A_{0}\,3}=0+5=a_{A_{2}\,1}+a_{A_{2}\,3}=4+1=5$ ;
$a_{A_{0}\,2}+a_{A_{0}\,4}=10+15=a_{A_{2}\,2}+a_{A_{2}\,4}=14+11=25.$. For the
odd partition,
$a_{A_{1}\,1}+a_{A_{1}\,3}=12+9=a_{A_{3}\,1}+a_{A_{3}\,3}=8+13=21.$ ;
$a_{A_{1}\,2}+a_{A_{1}\,4}=2+7=a_{A_{3}\,2}+a_{A_{3}\,4}=6+3=9\,.$ Also,
$a_{A_{1}\,1}+a_{A_{1}\,2}=12+2=a_{A_{3}\,1}+a_{A_{3}\,2}=8+6=14$ and
$a_{A_{1}\,3}+a_{A_{1}\,4}=9+7=a_{A_{3}\,3}+a_{A_{3}\,4}=13+3=26\,.$ For the
$n=1$ partition, the sum of difference is $(+12)+(-8)+(+4)+(-8)=0$. For the
$n=3$ partition, the sum of difference is $(-12)+(+8)+(-4)+(+8)=0$. Thus all
the properties are preserved.

Figure 94:  Figure 95:

Now we consider anti-clockwise rotation of the external layer, while keeping
the internal layer constant. We have the following result.

Figure 96:

Then we can establish mathematical relationships between the case of internal
disk rotation and external disk rotation. Denote the diagram function, which
is dependent on the rotation angle $\theta$ as $\mathcal{D}(\theta)$ being
considered under the internal rotation perspective or the external rotation
perspective. For example, the upper left diagram in figure 92 is denoted as
$(\mathcal{D}(0)|\text{internal rotation})$; the lower left diagram of figure
92 is denoted as $(\mathcal{D}(-\pi/2)|\text{internal rotation})$ and so on.
The same notion goes for the external rotation case in figure 94. We find the
following dual equivalent relationships:

$(\mathcal{D}(\pi/2)|\text{internal
rotation})\equiv(\mathcal{D}(-\pi/2)|\text{external rotation})$ (497)

with an offset of phase of $+\pi/2$ in the
$(\mathcal{D}(-\pi/2)|\text{external rotation})$ diagram. Similarly,

$(\mathcal{D}(-\pi/2)|\text{internal
rotation})\equiv(\mathcal{D}(\pi/2)|\text{external rotation})$ (498)

with an offset of phase of $-\pi/2$ in the $(\mathcal{D}(\pi/2)|\text{external
rotation})$ diagram. This is simply noted as

$(\text{anti-clockwise}|\text{internal
rotation})\equiv(\text{clockwise}|\text{external rotation})$ (499)

and

$(\text{clockwise}|\text{internal rotation})\equiv(\text{anti-
clockwise}|\text{external rotation})$ (500)

The above relationship just satisfies

$(u|S_{k})\equiv(u^{*}|S_{k}^{\star})\quad\text{and}\quad(u^{*}|S_{k})\equiv(u|S_{k}^{\star})$
(501)

by identifying $u(\theta)=\pi/2$, $u^{*}(\theta)=-\pi/2$,
$S_{k}=\text{internal rotation}$ and $S_{k}^{\star}=\text{external rotation}$.

The meaning of offset explicitly refers to the following. Let’s compare the
diagram for internal rotation and external rotation.

Figure 97:

We can see that for the external rotation, although the diagram is the same,
it is rotated by $+\pi/2$. And similarly

Figure 98:

for which the external diagram is rotated by $-\pi/2$.

Finally we have the dual invariant diagram for $\theta=\pi$ and $\theta=-\pi$.
We can see that

$(\mathcal{D}(\pi)|\text{internal
rotation})\equiv(\mathcal{D}(\pi)|\text{external rotation})$ (502)

and

$(\mathcal{D}(-\pi)|\text{internal
rotation})\equiv(\mathcal{D}(-\pi)|\text{external rotation})\,,$ (503)

with an offset of phase of $\pi$ (or $-\pi$) in the external diagram.

Figure 99:

##  7 Complementary Theorem in Duality

In this section, we will investigate a very important theorem in duality-The
Complementary Theorem. The theorem states a very deep principle: when a system
is dual invariant in a specific way, it has to be non-dual invariant in
another way, vice versa.

The theorem will be proved as the follows. Consider a duality of filled (full)
and unfilled (empty) states. The full state is denoted as $|0\rangle$ and the
empty state is denoted as $|1\rangle$. Then we have the dual element set as
$\\{\text{empty},\text{full}\\}=\\{|0\rangle,|1\rangle\\}$. A qubit
constructed will be
$|\psi(\theta)\rangle=\cos\theta|0\rangle+\sin\theta|1\rangle$. Now consider
the dual observer perspective. Consider $S_{k}=\text{matter}$ and
$S_{k}^{\star}=\text{empty space}$. Then we have the dual observer set as
$\\{\text{matter},\text{empty space}\\}=\\{S_{k},S_{k}^{\star}\\}$. Now first
consider a fully filled system, a glass of fully filled water as an example.
This is the full state $\psi(\pi/2)=|1\rangle$ with probability equal to 1. We
can say it is full of matter, or in another way we can say it is empty in
empty space. They are equivalent statements. Mathematically we write
$(|1\rangle|S_{k})\equiv(|0\rangle|S_{k}^{\star})$, or simply
$(1|S_{k})\equiv(0|S_{k}^{\star})$. Next consider its dual counter part, an
empty glass. This is the empty state $\psi(0)=|0\rangle$ with probability
equal to 1. We say it is empty in matter, but equivalently we can say it is
full (filled) of empty space. Mathematically, we have
$(0|S_{k})\equiv(1|S_{k}^{\star})$. Next consider a halfly filled glass. This
is the state of
$|\psi(\pi/4)\rangle\equiv|\frac{1}{2}\rangle=\frac{1}{\sqrt{2}}(|0\rangle+|1\rangle)$
We can say it is halfly filled with matter or equivalently halfly filled with
empty space. Mathematically,
$(\frac{1}{2}|S_{k})\equiv(\frac{1}{2}|S_{k}^{\star})$. It’s dual is also a
halfly filled glass, but the position of matter and empty space is
interchanged. It has a state of
$|\psi(\pi/4)\rangle\equiv|\frac{1}{2}\rangle=\frac{1}{\sqrt{2}}(|1\rangle+|0\rangle)=\frac{1}{\sqrt{2}}(|0\rangle+|1\rangle)$.
Mathematically $(\frac{1}{2}|S_{k}^{\star})\equiv(\frac{1}{2}|S_{k})$.
Diagramatically, we have the following:

Figure 100:

We can see that for dual invariant representation states $|11\rangle$ and
$|00\rangle$, we have non-dual relationships of
$(1|S_{k})\equiv(0|S_{k}^{\star})$ and $(0|S_{k})\equiv(1|S_{k}^{\star})$
respectively. Or we can think in terms of the states, both
$|\psi(\pi/2)\rangle=|1\rangle$ and $|\psi(0)\rangle=|0\rangle$ are non-dual
invarinat states ; while for non-dual representation states $|01\rangle$ and
$|10\rangle$, we have dual relationships of
$(\frac{1}{2}|S_{k})\equiv(\frac{1}{2}|S_{k}^{\star})$ (or
$(\frac{1}{2}|S_{k}^{\star})\equiv(\frac{1}{2}|S_{k})$ ). Or we can think in
terms of the states that $*|\psi(\pi/4)\rangle=|\psi(\pi/4)\rangle$. In a
summary, we can put the results in a 4-tableau,

Figure 101:

This marks the complementary theorem of duality. A dual invariant state adopts
a non-dual relationship establishment and a non-dual invariant state adopts a
dual relationship establishment.

#### Implication

From the above analysis, in certain extent, it implies the incompleteness of
information carried by each state. When we are in the $|11\rangle$ state with
exact probability equal to 1, we basically know everything about the matter
realm but know nothing about the empty space realm. Vice versa, when we are
situated in the $|00\rangle$ state with exact probability equal to 1, we know
everything about the empty space realm but know nothing about the matter
world. When we are in the $|10\rangle$ state or $|01\rangle$ state, we only
know at most know half of each realm, therefore the information obtained is
still incomplete. We adopt the view that the full state, which is the linear
combination of four states contain the complete information of the universe,

$\psi=a_{00}|00\rangle+a_{11}|11\rangle+a_{01}|01\rangle+a_{10}|10\rangle\,.$
(504)

The complementary theorem will be best demonstrated through the following
identities. Upon direct computation, it is found that

$\displaystyle|0\rangle\oplus|1\rangle$
$\displaystyle=|0\rangle\otimes|0\rangle+|1\rangle\otimes|1\rangle$ (505)
$\displaystyle|1\rangle\oplus|0\rangle$
$\displaystyle=|0\rangle\otimes|1\rangle+|1\rangle\otimes|0\rangle\,.$

The first line is dual to the second line. From the left hand side, it is
clearly that it is is non-left,right invariant. However, for the right hand
side, it is $0\leftrightarrow 1$ dual invariant. Therefore, we have non-dual
invariant in one side and dual invariant in another, but the two expressions
are equivalent. Next we have

$\displaystyle|0\rangle\oplus|0\rangle$
$\displaystyle=|0\rangle\otimes|0\rangle+|1\rangle\otimes|0\rangle$ (506)
$\displaystyle|1\rangle\oplus|1\rangle$
$\displaystyle=|1\rangle\otimes|1\rangle+|0\rangle\otimes|1\rangle\,.$

Again, the first line is dual to the second line. From the left hand side, it
is clearly that it is left, right dual invariant. However, for the right hand
side, it is non $0\leftrightarrow 1$ dual invariant. Therefore, we have dual
invariant in one side and non-dual invariant in another, but the two
expressions are equivalent. Therefore, from 505 and 506, we have demonstrated
the complementary theorem in duality.

We can further demonstrate the complementary theorem of duality in the
following illustration:

Figure 102:

When we express the direct sum of the vectors explicitly in column vectors, we
can see the following property. For example, in the upper left box,
$|0\rangle\oplus|0\rangle$ is dual invariant under left and right observation
perspective, but the explicit column vector is not invariant under up and down
observation perspective. The same idea goes for the lower right box, which is
dual to the upper left box. Then in the upper right box,
$|0\rangle\oplus|1\rangle$ is not dual invariant under left and right
observation perspective, but the explicit column vector is dual invariant
under up and down observation perspective. The same idea goes for the lower
left box, which is dual to the upper right box.

And we have the identity

$|0\rangle\oplus|1\rangle+|1\rangle\oplus|0\rangle=|0\rangle\oplus|0\rangle+|1\rangle\oplus|1\rangle=\sum_{i,j=0,1}|i\rangle\otimes|j\rangle\,.$
(507)

The full dual invariant state, for which here is $(1\,\,1\,\,1\,\,1)^{T}$ can
be always decomposed into a sum of two non-dual invariant states or sum of two
dual invariant states.

##  8 Quantum field theory with duality symmetry

In this chapter, we would like apply the idea of duality to study Lagrangian
that process the duality $\mathbb{Z}_{2}$ symmetry. Consider a particle that
is described by the scalar field $\phi_{+}$ field, and its dual particle being
described by the $(\phi_{+})^{*}=\phi_{-}$ field. Next we want to construct an
interaction term that preserve $\mathbb{Z}_{2}$ symmetry, i.e. the Lagrangian
is invariant under the transformation of fields under duality,

$\phi_{+}\rightarrow\phi_{-}\quad\text{and}\quad\phi_{-}\rightarrow\phi_{+}\,.$
(508)

This means a field transformation as

$\begin{pmatrix}\phi_{-}\\\
\phi_{+}\end{pmatrix}=\boldsymbol{\mathrm{M}}\begin{pmatrix}\phi_{+}\\\
\phi_{-}\end{pmatrix}\,,$ (509)

where

$\boldsymbol{\mathrm{M}}=\begin{pmatrix}0&1\\\ 1&0\end{pmatrix}$ (510)

is the matrix representation of the duality group $\mathbb{Z}_{2}$. A
Lagrangian with dual symmetry can always be written as two separations of dual
terms,

$\mathcal{L}=\,\,^{*}\mathcal{L}+\mathcal{L^{*}}\,,$ (511)

where $*^{*}\mathcal{L}=\mathcal{L^{*}}$ and
$*\mathcal{L^{*}}=\,\,^{*}\mathcal{L}$. The simplest dual invariant Lagrangian
for scalar field is

$\mathcal{L}=\frac{1}{2}\partial_{\mu}\phi_{+}\partial^{\mu}\phi_{+}+\frac{1}{2}\partial_{\mu}\phi_{-}\partial^{\mu}\phi_{-}-g\phi_{+}\phi_{-}\,,$
(512)

where $g$ is the constant coupling. Upon symmetrization, we can write the
above as

$\mathcal{L}=\frac{1}{2}\partial_{\mu}\phi_{+}\partial^{\mu}\phi_{+}+\frac{1}{2}\partial_{\mu}\phi_{-}\partial^{\mu}\phi_{-}-\frac{g}{2}(\phi_{+}\phi_{-}+\phi_{-}\phi_{+})\,,$
(513)

which is invariant under the $\mathbb{Z}_{2}$ transformation in 509. We can
identify the two dual Lagrangians as,

${}^{*}\mathcal{L}=\frac{1}{2}\partial_{\mu}\phi_{+}\partial^{\mu}\phi_{+}-\frac{g}{2}\phi_{+}\phi_{-}$
(514)

and

$\mathcal{L^{*}}=\frac{1}{2}\partial_{\mu}\phi_{-}\partial^{\mu}\phi_{-}-\frac{g}{2}\phi_{-}\phi_{+}\,.$
(515)

The equation of motions for the $\phi^{+}$ field and the $\phi_{-}$ field are,

$\Box\phi_{+}=-g\phi_{-}\quad\text{and}\quad\Box\phi_{-}=-g\phi_{+}\,.$ (516)

Therefore, the dynamics of $\phi_{+}$ is sourced by $\phi_{-}$ and the
dynamics of $\phi_{-}$ is sourced by $\phi_{+}$. It is noted that we cannot
add heterogeneous mass terms to the above Lagrangian as this violates
$\mathbb{Z}_{2}$ duality symmetry. Consider,

$\mathcal{L}=\frac{1}{2}\partial_{\mu}\phi_{+}\partial^{\mu}\phi_{+}+\frac{1}{2}\partial_{\mu}\phi_{-}\partial^{\mu}\phi_{-}+\frac{1}{2}m_{+}^{2}\phi_{+}\phi_{+}+\frac{1}{2}m_{-}^{2}\phi_{-}\phi_{-}-\frac{g}{2}(\phi_{+}\phi_{-}+\phi_{-}\phi_{+})\,,$
(517)

Under the duality field transformation, we obtain

$\mathcal{L}^{\prime}=\frac{1}{2}\partial_{\mu}\phi_{-}\partial^{\mu}\phi_{-}+\frac{1}{2}\partial_{\mu}\phi_{+}\partial_{\mu}\phi_{+}+\frac{1}{2}m_{+}^{2}\phi_{-}\phi_{-}+\frac{1}{2}m_{-}^{2}\phi_{+}\phi_{+}-\frac{g}{2}(\phi_{-}\phi_{+}+\phi_{+}\phi_{-})\,,$
(518)

which is not equal to $\mathcal{L}$. Thus the mass terms break the
$\mathbb{Z}_{2}$ invariance. The dual symmetry is only preserved if
$m_{+}=m_{-}$. Therefore, for a dual field theory, both fields have to be of
same mass or massless.

Another example for $\mathbb{Z}_{2}$ invariant Lagrangian is

$\mathcal{L}=\frac{1}{2}\partial_{\mu}\phi_{+}\partial^{\mu}\phi_{+}+\frac{1}{2}\partial_{\mu}\phi_{-}\partial^{\mu}\phi_{-}-\lambda(\phi_{+}\phi_{+}+\phi_{-}\phi_{-})\,,$
(519)

In fact, the Lagrangian that possess $\mathbb{Z}_{2}$ symmetry is not unique.
The combination of the Lagrangians in 513 and 519 is also $\mathbb{Z}_{2}$
invariant,

$\displaystyle\mathcal{L}$
$\displaystyle=\frac{1}{2}\partial_{\mu}\phi_{+}\partial^{\mu}\phi_{+}+\frac{1}{2}\partial_{\mu}\phi_{-}\partial^{\mu}\phi_{-}-g(\phi_{+}\phi_{+}+\phi_{-}\phi_{-}+\phi_{+}\phi_{-}+\phi_{-}\phi_{+})$
(520)
$\displaystyle=\frac{1}{2}\partial_{\mu}\phi_{+}\partial^{\mu}\phi_{+}+\frac{1}{2}\partial_{\mu}\phi_{-}\partial^{\mu}\phi_{-}-g(\phi_{+}+\phi_{-})^{2}\,.$

The interaction term is invariant under $\mathbb{Z}_{2}\times\mathbb{Z}_{2}$
double duality transformation. We can construct representation matrices in
$\mathbb{R}^{4}$ for our purpose. The four elements are

$\displaystyle
I=D([0],[0])=\mathbb{I}\otimes\mathbb{I}=\begin{pmatrix}1&0&0&0\\\ 0&1&0&0\\\
0&0&1&0\\\
0&0&0&1\end{pmatrix}\,\,,\,\,a=D([0],[1])=\mathbb{I}\otimes\boldsymbol{\mathrm{M}}=\begin{pmatrix}0&0&1&0\\\
0&0&0&1\\\ 1&0&0&0\\\ 0&1&0&0\end{pmatrix}\,\,,\,\,$ (521) $\displaystyle
b=D([1],[0])=\boldsymbol{\mathrm{M}}\otimes\mathbb{I}=\begin{pmatrix}0&1&0&0\\\
1&0&0&0\\\ 0&0&0&1\\\
0&0&1&0\end{pmatrix}\,\,,\,\,c=D([1],[1])=\boldsymbol{\mathrm{M}}\otimes\boldsymbol{\mathrm{M}}=\begin{pmatrix}0&0&0&1\\\
0&0&1&0\\\ 0&1&0&0\\\ 1&0&0&0\end{pmatrix}\,\,.\,\,$

Note that the identity element $\mathbb{I}\otimes\mathbb{I}$ has all positive
eigenvalue of $+1$, while the remaining elements
$\boldsymbol{\mathrm{M}}\otimes\mathbb{I},\mathbb{I}\otimes\boldsymbol{\mathrm{M}}\,\,\text{and}\,\,\boldsymbol{\mathrm{M}}\otimes\boldsymbol{\mathrm{M}}$
would give two positive eigenvalues of $+1$ and two negative eigenvalues of
$-1$. Upon diagonalization, this will recover the result in 442. These
$4\times 4$ representation matrices act on the basis
$\\{|00\rangle,|01\rangle,|10\rangle,|11\rangle\\}$, which is
$\\{\phi_{+}\phi_{+},\phi_{+}\phi_{-},\phi_{-}\phi_{+},\phi_{-}\phi_{-}\\}$.

For each term in the interaction, we can carry out further analysis. First
notice that both $\phi_{+}\phi_{+}$ and $\phi_{-}\phi_{-}$ are observer dual
invariant, i.e. looking from left makes no difference to looking at right.
However, this is not the case for $\phi_{+}\phi_{-}$ and $\phi_{-}\phi_{+}$,
which are non-dual invariant under observer. We notice that

$(\phi_{\pm}\phi_{\pm}|S_{3})=(\phi_{\pm}\phi_{\pm}|S_{3}^{\star})\,.$ (522)

On the contrary,

$(\phi_{\pm}\phi_{\mp}|S_{3})=(\phi_{\mp}\phi_{\pm}|S_{3}^{\star})$ (523)

Therefore we introduce two partitions $B$ and $Q$

$B=\\{\phi_{+}\phi_{+},\phi_{-}\phi_{-}\,|\,(\phi_{\pm}\phi_{\pm}|S_{3})=(\phi_{\pm}\phi_{\pm}|S_{3}^{\star})\\}$
(524)
$Q=\\{\phi_{+}\phi_{-},\phi_{-}\phi_{+}\,|\,(\phi_{\pm}\phi_{\mp}|S_{3})\neq(\phi_{\pm}\phi_{\mp}|S_{3}^{\star})\\}$
(525)

Therefore, $B$ and $Q$ are dual partitions.

In general, the following Lagrangian is dual invariant

$\mathcal{L}=\frac{1}{2}\partial_{\mu}\phi_{+}\partial^{\mu}\phi_{+}+\frac{1}{2}\partial_{\mu}\phi^{-}\partial^{\mu}\phi^{-}-g(\phi_{+}+\phi_{-})^{n}\,.$
(526)

The interaction term $g(\phi_{+}+\phi_{-})^{n}$ is invariant under
$\mathbb{Z}_{2}\times\cdots\times\mathbb{Z}_{2}$
$(\mathrm{or}\,\mathbb{Z}_{2}\otimes\cdots\otimes\mathbb{Z}_{2})$ multi-
duality transformation.

Let $U$ be $\\{+\\}$ and its dual $U^{*}$ be $\\{-\\}$ and the complete set be
$W=U\cup U^{*}=\\{+,-\\}$. Since the interaction term can be written as the
following

$(\phi_{+}+\phi_{-})^{n}=\sum_{i_{1},i_{2},\cdots,i_{n}\in
W}\phi_{i_{1}}\phi_{i_{2}}\cdots\phi_{i_{N}}=\sum_{i_{1},i_{2},\cdots,i_{n}\in
W}\prod_{k=1}^{n}\phi_{i_{k}}\,,$ (527)

Therefore the Lagrangian can be written as

$\mathcal{L}=\frac{1}{2}\partial_{\mu}\phi_{+}\partial^{\mu}\phi_{+}+\frac{1}{2}\partial_{\mu}\phi_{-}\partial^{\mu}\phi_{-}-g\sum_{i_{1},i_{2},\cdots,i_{n}\in
W}\prod_{k=1}^{n}\phi_{i_{k}}\,.$ (528)

It is noted that such interaction term form the basis of representation of
homogeneous

In addition, for $n$ is even, the following four Lagrangians are also dual
invariant,

$\mathcal{L}=\frac{1}{2}\partial_{\mu}\phi_{+}\partial^{\mu}\phi_{+}+\frac{1}{2}\partial_{\mu}\phi_{-}\partial^{\mu}\phi_{-}-g(\pm\phi_{+}\pm\phi_{-})^{2k}\,,$
(529)

and

$\mathcal{L}=\frac{1}{2}\partial_{\mu}\phi_{+}\partial^{\mu}\phi_{+}+\frac{1}{2}\partial_{\mu}\phi_{-}\partial^{\mu}\phi_{-}-g(\pm\phi_{+}\mp\phi_{-})^{2k}\,,$
(530)

Note that for the second case in 530, if $n$ is odd, we have anti-dual
invariant symmetry, i.e. the Lagrangian in invariant under
$\phi_{+}\rightarrow-\phi_{-}$ and $\phi_{-}\rightarrow-\phi_{+}$. For 530,
the interaction can be separated further into the positive part and the
negative part. With some simple algebra, it can be shown that the positive
part is where the number of $+$ and the number of $-$ are even, and the
negative part is where the number of $+$ and the number of $-$ are odd. For
example,

$\displaystyle(\phi_{+}-\phi_{-})^{2k}$
$\displaystyle=\sum_{\begin{subarray}{c}i_{1},i_{2}\cdots,i_{n}\in W\\\
\\#+,\\#-=\mathrm{even}\end{subarray}}\phi_{i_{1}}\cdots\phi_{i2k}-\sum_{\begin{subarray}{c}i_{1},i_{2}\cdots,i_{n}\in
W\\\ \\#+,\\#-=\mathrm{odd}\end{subarray}}\phi_{i_{1}}\cdots\phi_{i_{k}}$
(531) $\displaystyle=\sum_{\begin{subarray}{c}i_{1},i_{2}\cdots,i_{n}\in W\\\
\\#+,\\#-=\mathrm{even}\end{subarray}}\prod_{l=1}^{2k}\phi_{i_{l}}-\sum_{\begin{subarray}{c}i_{1},i_{2}\cdots,i_{n}\in
W\\\ \\#+,\\#-=\mathrm{odd}\end{subarray}}\prod_{l=1}^{2k}\phi_{i_{l}}\,.$

The positive part and negative part, respectively, for each can be further
split into two dual partitions with $2^{n-1}$ elements. Let $K_{+}$ be the
partition and its dual $K_{+}^{*}$ for the positive partition, and the
complete positive partition is given by $K^{(+)}=K_{+}\cup K_{+}^{*}$ and
$K_{+}\cap K_{+}^{*}=\emptyset$; while $K_{-}$ be the partition and its dual
$K_{-}^{*}$ for the negative partition, and the complete negative partition is
given by $K^{(-)}=K_{-}\cup K_{-}^{*}$ and $K_{-}\cap K_{-}^{*}=\emptyset$. It
is important to note that the positive positive and negative partition are not
dual to each other. For simplicity denote $I=\\{i_{1},i_{2}\cdots i_{n}\\}$,
then we have

$\sum_{\begin{subarray}{c}I\in W\\\
\\#+,\\#-=\mathrm{even}\end{subarray}}\prod_{l=1}^{2k}\phi_{i_{l}}=\sum_{I\in
K_{+}}\prod_{l=1}^{2k}\phi_{i_{l}}+\sum_{I\in
K_{+}^{*}}\prod_{l=1}^{2k}\phi_{i_{l}}\,,$ (532)

and

$\sum_{\begin{subarray}{c}I\in W\\\
\\#+,\\#-=\mathrm{odd}\end{subarray}}\prod_{l=1}^{2k}\phi_{i_{l}}=\sum_{I\in
K_{-}}\prod_{l=1}^{2k}\phi_{i_{l}}+\sum_{I\in
K_{-}^{*}}\prod_{l=1}^{2k}\phi_{i_{l}}\,,$ (533)

This can be checked, for example $n=4$,

$\displaystyle(\phi_{+}-\phi_{-})^{4}$
$\displaystyle=\phi_{+}\phi_{+}\phi_{+}\phi_{+}-\phi_{+}\phi_{+}\phi_{+}\phi_{-}-\phi_{+}\phi_{+}\phi_{-}\phi_{+}+\phi_{+}\phi_{-}\phi_{+}\phi_{-}$
(534)
$\displaystyle-\phi_{+}\phi_{-}\phi_{+}\phi_{+}+\phi_{+}\phi_{-}\phi_{+}\phi_{-}+\phi_{+}\phi_{-}\phi_{-}\phi_{+}-\phi_{+}\phi_{-}\phi_{-}\phi_{-}$
$\displaystyle-\phi_{-}\phi_{+}\phi_{+}\phi_{+}+\phi_{-}\phi_{+}\phi_{+}\phi_{-}+\phi_{-}\phi_{+}\phi_{-}\phi_{+}-\phi_{-}\phi_{+}\phi_{-}\phi_{-}$
$\displaystyle+\phi_{-}\phi_{-}\phi_{+}\phi_{+}-\phi_{-}\phi_{-}\phi_{+}\phi_{-}-\phi_{-}\phi_{-}\phi_{-}\phi_{+}+\phi_{-}\phi_{-}\phi_{-}\phi_{-}$

There are $2^{4-1}=8$ positive terms and 8 negative terms respectively. For
the positive partition, we have

$K_{+}=\\{\phi_{+}\phi_{+}\phi_{+}\phi_{+},\phi_{+}\phi_{+}\phi_{-}\phi_{-},\phi_{+}\phi_{-}\phi_{+}\phi_{-},\phi_{+}\phi_{-}\phi_{-}\phi_{+}\\}$