# Connes spectral distance and nonlocality of generalized noncommutative phase
spaces

Bing-Sheng Lin1,2,†, Tai-Hua Heng 3
1School of Mathematics, South China University of Technology,
Guangzhou 510641, China
2Laboratory of Quantum Science and Engineering,
South China University of Technology, Guangzhou 510641, China
†Email<EMAIL_ADDRESS>
3School of Physics and Material Science, Anhui University, Hefei 230601, China

###### Abstract

We study the Connes spectral distance of quantum states and analyse the
nonlocality of the $4D$ generalized noncommutative phase space. By virtue of
the Hilbert-Schmidt operatorial formulation, we obtain the Dirac operator and
construct a spectral triple corresponding to the noncommutative phase space.
Based on the ball condition, we obtain some constraint relations about the
optimal elements, and then calculate the Connes spectral distance between two
Fock states. Due to the noncommutativity, the spectral distances between Fock
states in generalized noncommutative phase space are shorter than those in
normal phase space. This shortening of distances implies some kind of
nonlocality caused by the noncommutativity. We also find that these spectral
distances in $4D$ generalized noncommutative phase space are additive and
satisfy the normal Pythagoras theorem. When the noncommutative parameters
equal zero, the results return to those in normal quantum phase space.

##  1 Introduction

The ideas of noncommutative spacetime started in 1947 [1]. In the 1980’s,
Connes formulated the mathematically rigorous framework of noncommutative
geometry [2]. In the past decades, there has been much interest in the study
of physics in noncommutative spaces [3, 4, 5, 6, 7, 8, 9, 10, 11]. A
noncommutative spacetime also appeared in string theory, namely in the
quantization of open string [3]. The noncommutativity of spacetime also plays
an important role in quantum gravity [12, 13]. The concept of noncommutative
spacetime is also applied in condensed matter physics, such as the integer
quantum Hall effect [14]. Usually, only spatial noncommutativity is
considered. But many researchers have also studied models in which a
noncommutative geometry is defined on the whole phase space [15, 16, 17, 18,
19, 20, 21, 22, 23, 24]. Noncommutativity between momenta can be naturally
considered as a consequence of noncommutativity between positions, as momenta
are defined to be the partial derivatives of the action with respect to the
position coordinates.

Because of the noncommutativity, there are no traditional point and distance
in a noncommutative space. One can study states in noncommutative spaces. In a
noncommutative space, a pure state is the analog of a traditional point in a
normal commutative space. One can calculate some kinds of distance measures
between the states, such as the Connes spectral distance [25]. The Connes
spectral distances in some kinds of noncommutative spaces have already been
studied in the literatures [26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 37, 36].
For example, Cagnache _et. al._ have studied the Connes spectral distance in
the Moyal plane [27]. They explicitly computed Connes spectral distance
between the pure states which corresponding to eigenfunctions of the quantum
harmonic oscillators. Scholtz and his collaborators have developed the
Hilbert-Schmidt operatorial formulation [32, 33, 34], they studied the Connes
spectral distances of harmonic oscillator states and also coherent states in
Moyal plane and fuzzy space. Barrett _et. al._ also used Monte Carlo
simulation to study spectral distances in the fuzzy spaces [36].

In the present work, we study the Connes spectral distance of the quantum
states in $4D$ generalized noncommutative phase space (NCPS). We find that the
Connes spectral distances of the quantum states in noncommutative phase space
are shorter than those in normal phase space. This means that the
noncommutativity of the generalized noncommutative phase space can lead to
some kind of nonlocality. This nonlocality can equivalently make the distances
shorter.

This paper is organized as follows. In Sec. 2, we construct a spectral triple
corresponding to the $4D$ generalized noncommutative phase space. Using the
Hilbert-Schmidt operatorial formulation, we construct a boson Fock space and a
quantum Hilbert space, and obtain the Dirac operator. In Sec. 3, we review the
definition of Connes spectral distance. Based on the ball condition, we derive
some constraint relations about the optimal elements. The Connes spectral
distances between Fock states in generalized noncommutative phase space are
calculated in Sec. 4. Some discussions and conclusions are given in Sec. 5.
Some details of calculations of the Dirac operator and the Connes spectral
distances are given in Appendixes.

##  2 Spectral triple of generalized noncommutative phase space

Now let us consider the following $4D$ generalized noncommutative phase space
[38], the position operators $\hat{X}_{i}$ and the momentum operators
$\hat{P}_{i}$ satisfy the following commutation relations,

$[\hat{X}_{1},\,\hat{P}_{1}]=[\hat{X}_{2},\,\hat{P}_{2}]=\mathrm{i}\hbar\,,\qquad[\hat{X}_{1},\,\hat{X}_{2}]=\mathrm{i}\mu\,,\qquad[\hat{P}_{1},\,\hat{P}_{2}]=\mathrm{i}\nu\,,$
(1)

and others vanish. Here $\mu$, $\nu$ are some real parameters. In the present
work, we only consider the case $\mu,\nu>0$. For simplicity, we can use the
following transformations [39],

$\tilde{X}_{i}=\sqrt[4]{\frac{\nu}{\mu}}\hat{X}_{i},\qquad\tilde{P}_{i}=\sqrt[4]{\frac{\mu}{\nu}}\hat{P}_{i},$
(2)

and we have

$[\tilde{X}_{1},\,\tilde{P}_{1}]=[\tilde{X}_{2},\,\tilde{P}_{2}]=\mathrm{i}\hbar\,,\qquad[\tilde{X}_{1},\,\tilde{X}_{2}]=\mathrm{i}\theta\,,\qquad[\tilde{P}_{1},\,\tilde{P}_{2}]=\mathrm{i}\theta\,,$
(3)

and others vanish. Here $\theta=\sqrt{\mu\nu}$. Usually we also assume
$\theta\ll\hbar$.

In a normal $4D$ quantum phase space, the coordinate operators $\hat{x}_{i}$,
$\hat{p}_{i}$ satisfy the following commutation relations,

$[\hat{x}_{i},\,\hat{p}_{j}]=\mathrm{i}\delta_{ij}\hbar,\qquad i,j=1,2,$ (4)

and others vanish. Similarly, we can denote

$\tilde{x}_{i}=\sqrt[4]{\frac{\nu}{\mu}}\,\hat{x}_{i},\qquad\tilde{p}_{i}=\sqrt[4]{\frac{\mu}{\nu}}\,\hat{p}_{i},$
(5)

and there is the similar commutation relation
$[\tilde{x}_{i},\,\tilde{p}_{j}]=\mathrm{i}\delta_{ij}\hbar$. Define the
creation and annihilation operators

$\hat{a}_{i}=\frac{1}{\sqrt{2\hbar}}(\tilde{x}_{i}+\mathrm{i}\tilde{p}_{i}),\qquad\hat{a}_{i}^{{\dagger}}=\frac{1}{\sqrt{2\hbar}}(\tilde{x}_{i}-\mathrm{i}\tilde{p}_{i}),$
(6)

these operators satisfy the commutation relations
$[\hat{a}_{i},\hat{a}_{j}^{{\dagger}}]=\delta_{ij}$, and
$[\hat{a}_{i},\hat{a}_{j}]=[\hat{a}_{i}^{{\dagger}},\hat{a}_{j}^{{\dagger}}]=0$.
One can define the Fock states as follows,

$|n\rangle_{\\!i}=\frac{1}{\sqrt{n!}}(\hat{a}_{i}^{{\dagger}})^{n}|0\rangle,\quad
n=0,1,2,...\,,$ (7)

where $|0\rangle$ is the vacuum state, and $\hat{a}_{i}|0\rangle=0$. These
Fock states $|n\rangle_{\\!i}$ satisfy the relations

$\displaystyle\hat{a}_{i}|n\rangle_{\\!i}=\sqrt{n}|n-1\rangle_{\\!i},\qquad\hat{a}_{i}^{{\dagger}}|n\rangle_{\\!i}=\sqrt{n+1}|n+1\rangle_{\\!i}.$
(8)

We also have the resolution of the identity,

$\displaystyle\sum_{n=0}^{\infty}|n\rangle_{\\!i\,i\\!}\langle
n|=\mathds{I}_{i}.$ (9)

Furthermore, one can define the boson Fock space as follows,

$\mathcal{F}=\mathrm{span}\left\\{|m,n\rangle:=|m\rangle_{\\!1}\otimes|n\rangle_{\\!2},\quad
m,n=0,1,2,...\right\\}.$ (10)

There are the relations,

$\displaystyle\sum_{n=0}^{\infty}|m,n\rangle\langle
m,n|=|m\rangle_{\\!1\,1\\!}\langle
m|\otimes\mathds{I}_{2},\qquad\sum_{m=0}^{\infty}|m,n\rangle\langle
m,n|=\mathds{I}_{1}\otimes|n\rangle_{\\!2\,2\\!}\langle n|,$
$\displaystyle\sum_{m,n=0}^{\infty}|m,n\rangle\langle
m,n|=\mathds{I}_{1}\otimes\mathds{I}_{2}.$ (11)

By virtue of the Hilbert-Schmidt operatorial formulation [40], one can
construct the quantum Hilbert space as follows,

$\mathcal{Q}=\mathrm{span}\left\\{|m_{1},m_{2}\rangle\langle
n_{1},n_{2}|\right\\}.$ (12)

The elements $\psi$ of the quantum Hilbert space $\mathcal{Q}$ are denoted by
$|\psi)$, and the inner product is defined as

$(\phi|\psi):=\mathrm{tr}_{\mathcal{F}}\left(\phi^{{\dagger}}\psi\right)=\sum_{m,n=0}^{\infty}\langle
m,n|\phi^{{\dagger}}\psi|m,n\rangle,$ (13)

where $\mathrm{tr}_{\mathcal{F}}(\cdot)$ denotes the trace over $\mathcal{F}$.

In general, a noncommutative space corresponds to a spectral triple
$(\mathcal{A},\mathcal{H},\mathcal{D})$ [2], where $\mathcal{A}$ is an
involutive algebra acting on a Hilbert space $\mathcal{H}$, and $\mathcal{D}$
is the Dirac operator on $\mathcal{H}$. One can construct a spectral triple
$(\mathcal{A},\mathcal{H},\mathcal{D})$ corresponding to the generalized
noncommutative phase space (3) as follows,

$\mathcal{A}=\mathcal{Q},\qquad\mathcal{H}=\mathcal{F}\otimes\mathbb{C}^{4},$
(14)

and an element $e\in\mathcal{A}$ acts on
$\Psi=\left(|\psi_{1}\rangle,|\psi_{2}\rangle,|\psi_{3}\rangle,|\psi_{4}\rangle\right)^{\prime}\in\mathcal{H}$
through the diagonal representation $\pi$ as

$\pi(e)\Psi=\pi(e)\left(\begin{array}[]{c}|\psi_{1}\rangle\\\
|\psi_{2}\rangle\\\ |\psi_{3}\rangle\\\ |\psi_{4}\rangle\\\
\end{array}\right)=\left(\begin{array}[]{cccc}e&0&0&0\\\ 0&e&0&0\\\ 0&0&e&0\\\
0&0&0&e\\\ \end{array}\right)\left(\begin{array}[]{c}|\psi_{1}\rangle\\\
|\psi_{2}\rangle\\\ |\psi_{3}\rangle\\\ |\psi_{4}\rangle\\\
\end{array}\right)=\left(\begin{array}[]{c}e|\psi_{1}\rangle\\\
e|\psi_{2}\rangle\\\ e|\psi_{3}\rangle\\\ e|\psi_{4}\rangle\\\
\end{array}\right).$ (15)

The Dirac operator for the $4D$ generalized noncommutative phase space (3) can
be defined as (see Appendix A for more details)

$\mathcal{D}=\beta\left(\begin{array}[]{cccc}0&0&-\hat{A}_{2}^{\dagger}&-\hat{A}_{1}^{\dagger}\\\
0&0&\hat{A}_{1}&-\hat{A}_{2}\\\ -\hat{A}_{2}&\hat{A}_{1}^{\dagger}&0&0\\\
-\hat{A}_{1}&-\hat{A}_{2}^{\dagger}&0&0\\\ \end{array}\right),$ (16)

where

$\beta=\frac{\sqrt{\hbar+\sqrt{\hbar^{2}-\theta^{2}}}}{\sqrt{\hbar^{2}-\theta^{2}}},\qquad\hat{A}_{i}=\hat{a}_{i}-\mathrm{i}t\varepsilon_{ij}\hat{a}_{j},$
(17)

and

$t=\frac{\theta}{\hbar+\sqrt{\hbar^{2}-\theta^{2}}},\qquad\varepsilon=\left(\begin{array}[]{cc}0&1\\\
-1&0\\\ \end{array}\right).$ (18)

Obviously, there is $0<t\ll 1$.

When the noncommutative parameters $\mu,\nu\to 0$, there is $\theta\to 0$, and
$t\to 0$. The operators $\hat{A}_{i}$ will return to the operators
$\hat{a}_{i}$, and the Dirac operator $\mathcal{D}$ (16) will return to the
Dirac operator of normal $4D$ quantum phase space.

##  3 Connes spectral distance and optimal elements

Let us consider the case where the quantum states $\omega$ are normal and
bounded, so they are representable by density matrices $\rho$ [32]. The action
of the state $\omega$ on an element $e\in\mathcal{A}$ can be written as

$\omega(e)=\mathrm{tr}_{\mathcal{F}}(\rho e).$ (19)

Suppose the quantum states $\omega$ and $\omega^{\prime}$ correspond to the
density matrices $\rho$ and $\rho^{\prime}$, respectively. The Connes spectral
distance between two states $\omega$ and $\omega^{\prime}$ is [25]

$d(\omega,\omega^{\prime})\equiv\sup_{e\in
B}|\omega(e)-\omega^{\prime}(e)|=\sup_{e\in B}|\mathrm{tr}_{\mathcal{F}}(\rho
e)-\mathrm{tr}_{\mathcal{F}}(\rho^{\prime}e)|,$ (20)

where

$B=\left\\{e\in\mathcal{A}:\big{\|}[\mathcal{D},\pi(e)]\big{\|}_{op}\leqslant
1\right\\},$ (21)

and the operator norm is defined as

$\|a\|_{op}\equiv\sup_{\psi\in\mathcal{H}}\frac{\|a\psi\|}{\|\psi\|},\qquad\|a\|^{2}\equiv(a,a)=\mathrm{tr}_{\mathcal{F}}(a^{{\dagger}}a).$
(22)

The inequality in (21) is the so-called ball condition. Using the Dirac
operator $\mathcal{D}$ (16), the commutator $[\mathcal{D},\pi(e)]$ for a
Hermitian element $e\in\mathcal{A}$ is

$[\mathcal{D},\pi(e)]=\beta\left(\begin{array}[]{cc}0&D_{1}\\\
-D_{1}^{\dagger}&0\\\ \end{array}\right),$ (23)

where

$D_{1}=\left(\begin{array}[]{cc}[\hat{A}_{2},e]^{\dagger}&[\hat{A}_{1},e]^{\dagger}\\\
{}[\hat{A}_{1},e]&-[\hat{A}_{2},e]\\\ \end{array}\right).$ (24)

After some straightforward calculations, one can obtain

$\left\|[\mathcal{D},\pi(e)]^{\dagger}[\mathcal{D},\pi(e)]\right\|_{op}=\beta^{2}\max\left\\{\|D_{1}^{\dagger}D_{1}\|_{op},\,\|D_{1}D_{1}^{\dagger}\|_{op}\right\\}.$
(25)

From the above expression (24), we have

$D_{1}^{\dagger}D_{1}=\left(\begin{array}[]{cc}[\hat{A}_{2},e][\hat{A}_{2},e]^{\dagger}+[\hat{A}_{1},e]^{\dagger}[\hat{A}_{1},e]&[\hat{A}_{2},e][\hat{A}_{1},e]^{\dagger}-[\hat{A}_{1},e]^{\dagger}[\hat{A}_{2},e]\\\
{}[\hat{A}_{1},e][\hat{A}_{2},e]^{\dagger}-[\hat{A}_{2},e]^{\dagger}[\hat{A}_{1},e]&[\hat{A}_{1},e][\hat{A}_{1},e]^{\dagger}+[\hat{A}_{2},e]^{\dagger}[\hat{A}_{2},e]\\\
\end{array}\right).$ (26)

So there is 111For any operator $a$ with the matrix elements $a_{ij}$ in some
orthonormal bases, there is the following Bessel’s inequality [34],
$\|a\|_{op}^{2}\geqslant\sum_{i}|a_{ij}|^{2}\geqslant|a_{ij}|^{2}.$ (27)

$\|D_{1}^{\dagger}D_{1}\|_{op}\geqslant\\!\sup_{\phi\in\mathcal{F},\langle\phi|\phi\rangle=1}\langle\phi|[\hat{A}_{1},e]^{\dagger}[\hat{A}_{1},e]+[\hat{A}_{2},e][\hat{A}_{2},e]^{\dagger}|\phi\rangle.$
(28)

It is known that for any bounded operator $a$, there is

$\|a\|_{op}^{2}=\|a^{\dagger}\|_{op}^{2}=\|a^{\dagger}a\|_{op}.$ (29)

Using the ball condition in (21), there is the following inequality,

$1\geqslant\big{\|}[\mathcal{D},\pi(e)]\big{\|}_{op}^{2}=\left\|[\mathcal{D},\pi(e)]^{\dagger}[\mathcal{D},\pi(e)]\right\|_{op}=\beta^{2}\|D_{1}^{\dagger}D_{1}\|_{op}.$
(30)

So from (28) and (30), for any Hermitian element $e\in B$, one can obtain

$\sup_{\phi\in\mathcal{F},\langle\phi|\phi\rangle=1}\langle\phi|[\hat{A}_{1},e]^{\dagger}[\hat{A}_{1},e]+[\hat{A}_{2},e][\hat{A}_{2},e]^{\dagger}|\phi\rangle\leqslant\frac{1}{\beta^{2}}.$
(31)

Similarly, we also have the following inequalities,

$\displaystyle\sup_{\phi\in\mathcal{F},\langle\phi|\phi\rangle=1}\langle\phi|[\hat{A}_{1},e][\hat{A}_{1},e]^{\dagger}+[\hat{A}_{2},e]^{\dagger}[\hat{A}_{2},e]|\phi\rangle\leqslant\frac{1}{\beta^{2}},$
$\displaystyle\sup_{\phi\in\mathcal{F},\langle\phi|\phi\rangle=1}\langle\phi|[\hat{A}_{1},e]^{\dagger}[\hat{A}_{1},e]+[\hat{A}_{2},e]^{\dagger}[\hat{A}_{2},e]|\phi\rangle\leqslant\frac{1}{\beta^{2}},$
$\displaystyle\sup_{\phi\in\mathcal{F},\langle\phi|\phi\rangle=1}\langle\phi|[\hat{A}_{1},e][\hat{A}_{1},e]^{\dagger}+[\hat{A}_{2},e][\hat{A}_{2},e]^{\dagger}|\phi\rangle\leqslant\frac{1}{\beta^{2}}.$
(32)

Now consider an element $e\in B$ which can be expressed as

$e=\sum_{i,j,k,l=0}^{\infty}C^{i,j}_{k,l}|i,j\rangle\langle k,l|.$ (33)

Since Hermitian elements can give the supremum in the Connes spectral distance
function [41], we only need to consider the element $e$ being Hermitian. So
there are $C^{i,j}_{k,l}=(C^{k,l}_{i,j})^{*}$, and $C^{i,j}_{i,j}$ are real
numbers. Using the relations (8), after some straightforward calculations, we
have

$\displaystyle[\hat{A}_{1},e]$ $\displaystyle=$
$\displaystyle[\hat{a}_{1}-\mathrm{i}t\hat{a}_{2},e]$ (34) $\displaystyle=$
$\displaystyle\sum_{i,j,k,l=0}^{\infty}\bigg{(}\left(C^{i+1,j}_{k,l}\sqrt{i+1}-C^{i,j}_{k-1,l}\sqrt{k}\right)$
$\displaystyle\qquad-\mathrm{i}t\left(C^{i,j+1}_{k,l}\sqrt{j+1}-C^{i,j}_{k,l-1}\sqrt{l}\right)\bigg{)}|i,j\rangle\langle
k,l|,$

and then

$\displaystyle[\hat{A}_{1},e][\hat{A}_{1},e]^{\dagger}$ $\displaystyle=$
$\displaystyle\sum_{i,j=0}^{\infty}\Bigg{(}\sum_{k,l=0}^{\infty}\bigg{|}\left(C^{i+1,j}_{k,l}\sqrt{i+1}-C^{i,j}_{k-1,l}\sqrt{k}\right)$
(35)
$\displaystyle\quad-\mathrm{i}t\left(C^{i,j+1}_{k,l}\sqrt{j+1}-C^{i,j}_{k,l-1}\sqrt{l}\right)\bigg{|}^{2}\Bigg{)}|i,j\rangle\langle
i,j|+n.d.\,.$

Here “$n.d.$” denotes the sum of terms $|k,l\rangle\langle i,j|$ with $k\neq
i$ and/or $l\neq j$, and these terms will not affect the calculations in the
following content. So we can just simply ignore these terms.

Similar to Refs. [27, 34], one can just consider the optimal elements being
diagonal elements. So we can set $C^{i,j}_{k,l}=0$ if $k\neq i$ and/or $l\neq
j$, and denote $c_{i,j}\equiv C^{i,j}_{i,j}$. For convenience, we also denote

$E_{i,j}\equiv c_{i,j}-c_{i-1,j},\qquad F_{i,j}\equiv c_{i,j}-c_{i,j-1},$ (36)

and

$G_{i,j}\equiv iE_{i,j}^{2}=i(c_{i,j}-c_{i-1,j})^{2},\qquad H_{i,j}\equiv
jF_{i,j}^{2}=j(c_{i,j}-c_{i,j-1})^{2}.$ (37)

So the above equation (35) can be rewritten as

$\displaystyle[\hat{A}_{1},e][\hat{A}_{1},e]^{\dagger}$ $\displaystyle=$
$\displaystyle\sum_{i,j=0}^{\infty}\big{(}(i+1)(c_{i+1,j}-c_{i,j})^{2}$ (38)
$\displaystyle\qquad+t^{2}(j+1)(c_{i,j+1}-c_{i,j})^{2}\big{)}|i,j\rangle\langle
i,j|+n.d.$ $\displaystyle=$
$\displaystyle\sum_{i,j=0}^{\infty}\left((i+1)E_{i+1,j}^{2}+t^{2}(j+1)F_{i,j+1}^{2}\right)|i,j\rangle\langle
i,j|+n.d.$ $\displaystyle=$
$\displaystyle\sum_{i,j=0}^{\infty}\left(G_{i+1,j}+t^{2}H_{i,j+1}\right)|i,j\rangle\langle
i,j|+n.d.\,.$

Similarly, there is

$[\hat{A}_{1},e]^{\dagger}[\hat{A}_{1},e]=\sum_{i,j=0}^{\infty}\big{(}G_{i,j}+t^{2}H_{i,j}\big{)}|i,j\rangle\langle
i,j|+n.d.\,.$ (39)

For the operator $\hat{A}_{2}$, we also have

$\displaystyle[\hat{A}_{2},e]$ $\displaystyle=$
$\displaystyle\sum_{i,j,k,l=0}^{\infty}\bigg{(}\left(C^{i,j+1}_{k,l}\sqrt{j+1}-C^{i,j}_{k,l-1}\sqrt{l}\right)$
(40)
$\displaystyle\qquad+\mathrm{i}t\left(C^{i+1,j}_{k,l}\sqrt{i+1}-C^{i,j}_{k-1,l}\sqrt{k}\right)\bigg{)}|i,j\rangle\langle
k,l|,$

and

$\displaystyle[\hat{A}_{2},e][\hat{A}_{2},e]^{\dagger}$ $\displaystyle=$
$\displaystyle\sum_{i,j=0}^{\infty}\big{(}H_{i,j+1}+t^{2}G_{i+1,j}\big{)}|i,j\rangle\langle
i,j|+n.d.\,,$ $\displaystyle{}[\hat{A}_{2},e]^{\dagger}[\hat{A}_{2},e]$
$\displaystyle=$
$\displaystyle\sum_{i,j=0}^{\infty}\big{(}H_{i,j}+t^{2}G_{i,j}\big{)}|i,j\rangle\langle
i,j|+n.d.\,.$ (41)

From the inequalities (31) and (3), using the results (38), (39) and (3), one
can obtain the following relations,

$\displaystyle(G_{i+1,j}+H_{i,j})+t^{2}(G_{i,j}+H_{i,j+1})\leqslant\frac{1}{\beta^{2}},\quad(1+t^{2})(G_{i+1,j}+H_{i,j+1})\leqslant\frac{1}{\beta^{2}},$
$\displaystyle(G_{i,j}+H_{i,j+1})+t^{2}(G_{i+1,j}+H_{i,j})\leqslant\frac{1}{\beta^{2}},\quad(1+t^{2})(G_{i,j}+H_{i,j})\leqslant\frac{1}{\beta^{2}}.\qquad\quad$
(42)

So in the calculation of Connes spectral distance (20), in general, one only
need to consider the so-called optimal elements,

$e_{o}=\sum_{i,j=0}^{\infty}c_{i,j}|i,j\rangle\langle i,j|,$ (43)

where the coefficients $c_{i,j}$ satisfy the above constraint relations (3).
The Connes spectral distance between the quantum states $\omega$ and
$\omega^{\prime}$ is obtained by

$d(\omega,\omega^{\prime})=|\mathrm{tr}_{\mathcal{F}}(\rho\,e_{o})-\mathrm{tr}_{\mathcal{F}}(\rho^{\prime}\,e_{o})|.$
(44)

##  4 Connes spectral distance between Fock states in NCPS

Using the spectral triple $(\mathcal{A},\mathcal{H},\mathcal{D})$ constructed
above, one can calculate the Connes spectral distances between Fock states
$|n_{1},n_{2}\rangle$. First, let us consider the adjacent Fock states
$|m+1,n\rangle$ and $|m,n\rangle$, and the corresponding density matrices are
$\rho_{m+1,n}=|m+1,n\rangle\langle m+1,n|$ and $\rho_{m,n}=|m,n\rangle\langle
m,n|$, respectively. The spectral distance is

$\displaystyle d(\omega_{m+1,n},\omega_{m,n})$ $\displaystyle=$
$\displaystyle\sup_{e\in
B}|\mathrm{tr}_{\mathcal{F}}(\rho_{m+1,n}e)-\mathrm{tr}_{\mathcal{F}}(\rho_{m,n}e)|$
(45) $\displaystyle=$ $\displaystyle\sup_{e\in B}|\langle
m+1,n|e|m+1,n\rangle-\langle m,n|e|m,n\rangle|$ $\displaystyle=$
$\displaystyle\sup_{e\in B}|c_{m+1,n}-c_{m,n}|=\sup_{e\in B}|E_{m+1,n}|.$

From (3), we have

$\displaystyle(1+t^{2})(G_{m+1,n}+H_{m,n+1})\leqslant\frac{1}{\beta^{2}},\qquad(1+t^{2})(G_{m+1,n}+H_{m+1,n})\leqslant\frac{1}{\beta^{2}},$
$\displaystyle(G_{m+1,n}+H_{m+1,n+1})+t^{2}(G_{m+2,n}+H_{m+1,n})\leqslant\frac{1}{\beta^{2}},$
$\displaystyle(G_{m+1,n}+H_{m,n})+t^{2}(G_{m,n}+H_{m,n+1})\leqslant\frac{1}{\beta^{2}}.$
(46)

So in order to attain the supremum of $|c_{m+1,n}-c_{m,n}|=|E_{m+1,n}|$, there
should be

$H_{m,n+1}=H_{m+1,n}=0,$ (47)

and

$(1+t^{2})G_{m+1,n}=(1+t^{2})(m+1)E_{m+1,n}^{2}=\frac{1}{\beta^{2}},$ (48)

or

$|c_{m+1,n}-c_{m,n}|=|E_{m+1,n}|=\frac{1}{\beta\sqrt{(1+t^{2})(m+1)}}.$ (49)

We can also set

$H_{m,n}=H_{m+1,n+1}=0,$ (50)

so there is

$F_{m,n+1}=F_{m+1,n}=F_{m,n}=F_{m+1,n+1}=0.$ (51)

Without loss of generality, we can assume $E_{m+1,n}\geqslant 0$. There is

$\displaystyle E_{m+1,n}$ $\displaystyle=$ $\displaystyle
F_{m+1,n+1}+E_{m+1,n}=c_{m+1,n+1}-c_{m,n}$ (52) $\displaystyle=$
$\displaystyle E_{m+1,n+1}+F_{m,n+1}=E_{m+1,n+1}.$

Similarly, one can obtain

$E_{m+1,i}=\frac{1}{\beta\sqrt{(1+t^{2})(m+1)}},\qquad
F_{m,i+1}=F_{m+1,i+1}=0,$ (53)

where $i=0,1,2,...$ .

For example, one can choose

$c_{m,i}=0\,,\qquad c_{m+1,i}=\frac{1}{\beta\sqrt{(1+t^{2})(m+1)}}\,,$ (54)

and the optimal element is

$\displaystyle e_{o}$ $\displaystyle=$
$\displaystyle\sum_{i=0}^{\infty}c_{m+1,i}|m+1,i\rangle\langle m+1,i|$ (55)
$\displaystyle=$
$\displaystyle\frac{1}{\beta\sqrt{(1+t^{2})(m+1)}}\sum_{i=0}^{\infty}|m+1,i\rangle\langle
m+1,i|$ $\displaystyle=$
$\displaystyle\frac{1}{\beta\sqrt{(1+t^{2})(m+1)}}|m+1\rangle_{\\!1\,1\\!}\langle
m+1|\otimes\mathds{I}_{2}.$

So the Connes distance between the states $|m+1,n\rangle$ and $|m,n\rangle$ is

$\displaystyle d(\omega_{m+1,n},\omega_{m,n})$ $\displaystyle=$
$\displaystyle|\mathrm{tr}_{\mathcal{F}}(\rho_{m+1,n}e_{o})-\mathrm{tr}_{\mathcal{F}}(\rho_{m,n}e_{o})|$
(56) $\displaystyle=$ $\displaystyle|\langle m+1,n|e_{o}|m+1,n\rangle-\langle
m,n|e_{o}|m,n\rangle|$ $\displaystyle=$
$\displaystyle\frac{1}{\beta\sqrt{(1+t^{2})(m+1)}}$ $\displaystyle=$
$\displaystyle\frac{\sqrt{\hbar^{2}-\theta^{2}}}{\sqrt{2\hbar}}\frac{1}{\sqrt{m+1}}.$

Similarly, the Connes spectral distance between Fock states $|m,n+1\rangle$
and $|m,n\rangle$ is

$d(\omega_{m,n+1},\omega_{m,n})=\frac{1}{\beta\sqrt{(1+t^{2})(n+1)}}=\frac{\sqrt{\hbar^{2}-\theta^{2}}}{\sqrt{2\hbar}}\frac{1}{\sqrt{n+1}}.$
(57)

Next, let us consider the Fock states $|m,n\rangle$ and $|m+k,n\rangle$,
$k>1$, and the corresponding Connes spectral distance is

$\displaystyle d(\omega_{m+k,n},\omega_{m,n})$ $\displaystyle=$
$\displaystyle\sup_{e\in
B}|\mathrm{tr}_{\mathcal{F}}(\rho_{m+k,n}e)-\mathrm{tr}_{\mathcal{F}}(\rho_{m,n}e)|=\sup_{e\in
B}|c_{m+k,n}-c_{m,n}|$ (58) $\displaystyle=$ $\displaystyle\sup_{e\in
B}|c_{m+k,n}-c_{m+k-1,n}+c_{m+k-1,n}-c_{m+k-2,n}$
$\displaystyle\qquad+...+c_{m+1,n}-c_{m,n}|$ $\displaystyle=$
$\displaystyle\sup_{e\in B}|E_{m+k,n}+E_{m+k-1,n}+...+E_{m+1,n}|.$

From (3), we have

$\displaystyle(1+t^{2})(G_{m+2,n}+H_{m+1,n+1})\leqslant\frac{1}{\beta^{2}},\qquad(1+t^{2})(G_{m+1,n}+H_{m+1,n})\leqslant\frac{1}{\beta^{2}},$
$\displaystyle(G_{m+2,n}+H_{m+1,n})+t^{2}(G_{m+1,n}+H_{m+1,n+1})\leqslant\frac{1}{\beta^{2}},$
$\displaystyle(G_{m+1,n}+H_{m+1,n+1})+t^{2}(G_{m+2,n}+H_{m+1,n})\leqslant\frac{1}{\beta^{2}}.$
(59)

So in order to attain the supremum of
$|c_{m+k,n}-c_{m,n}|=|E_{m+k,n}+E_{m+k-1,n}+...+E_{m+1,n}|$, one must let
$|c_{m+2,n}-c_{m,n}|=|E_{m+2,n}+E_{m+1,n}|$ as large as possible. There should
be $H_{m+1,n}=H_{m+1,n+1}=0$, and

$\displaystyle G_{m+2,n}+t^{2}G_{m+1,n}=\frac{1}{\beta^{2}},\qquad
G_{m+1,n}+t^{2}G_{m+2,n}=\frac{1}{\beta^{2}},$
$\displaystyle(1+t^{2})G_{m+2,n}=\frac{1}{\beta^{2}},\qquad(1+t^{2})G_{m+1,n}=\frac{1}{\beta^{2}}.$
(60)

Therefore we have

$G_{m+1,n}=G_{m+2,n}=\frac{1}{\beta^{2}(1+t^{2})}.$ (61)

Similarly, in order to attain the supremum of $|c_{m+k,n}-c_{m,n}|$, there
must be

$H_{m+i,n}=H_{m+i,n+1}=0,\qquad G_{m+i,n}=\frac{1}{\beta^{2}(1+t^{2})},\qquad
i=1,2,...,k\,,$ (62)

so

$c_{m+i,n}-c_{m+i-1,n}=E_{m+i,n}=\frac{1}{\beta\sqrt{1+t^{2}}}\frac{1}{\sqrt{m+i}}.$
(63)

For example, one can choose

$c_{m+p,i}=\frac{1}{\beta\sqrt{1+t^{2}}}\sum_{j=1}^{m+p}\frac{1}{\sqrt{j}}=\zeta_{0;m+p}\frac{1}{\beta\sqrt{1+t^{2}}},$
(64)

where $i=0,1,2...$ , $p=0,1,...,k$, and the function $\zeta_{p;q}$ is

$\zeta_{p;q}\equiv\sum_{i=p+1}^{q}\frac{1}{\sqrt{i}}=\zeta\left(\frac{1}{2},p+1\right)-\zeta\left(\frac{1}{2},q+1\right),$
(65)

where $\zeta(s,q)$ is the Hurwitz zeta function

$\zeta(s,q)\equiv\sum_{i=0}^{\infty}\frac{1}{(q+i)^{s}}.$ (66)

Obviously, there is $\zeta_{i;j}=-\zeta_{j;i}$, and therefore $\zeta_{i;i}=0$.
We also have $\zeta_{i;j}-\zeta_{i;k}=\zeta_{k;j}$. It is easy to verify that
these coefficients $c_{i,j}$ (64) satisfy the relations (62).

Suppose $e_{o}$ to be the optimal element with the coefficients $c_{i,j}$ (64)
which can attain the supremum in (58). There is

$\displaystyle e_{o}$ $\displaystyle=$
$\displaystyle\sum_{p=0}^{k}\sum_{i=0}^{\infty}c_{m+p,i}|m+p,i\rangle\langle
m+p,i|$ (67) $\displaystyle=$
$\displaystyle\frac{1}{\beta\sqrt{1+t^{2}}}\sum_{p=0}^{k}\zeta_{0;m+p}|m+p\rangle\langle
m+p|\otimes\mathds{I}_{2},$

and the Connes spectral distance between the states $|m+k,n\rangle$ and
$|m,n\rangle$ is

$\displaystyle d(\omega_{m+k,n},\omega_{m,n})$ $\displaystyle=$
$\displaystyle|\mathrm{tr}_{\mathcal{F}}(\rho_{m+k,n}e_{o})-\mathrm{tr}_{\mathcal{F}}(\rho_{m,n}e_{o})|$
(68) $\displaystyle=$ $\displaystyle|\langle m+k,n|e_{o}|m+k,n\rangle-\langle
m,n|e_{o}|m,n\rangle|$ $\displaystyle=$
$\displaystyle\frac{1}{\beta\sqrt{1+t^{2}}}\sum_{i=1}^{k}\frac{1}{\sqrt{m+i}}=\frac{\sqrt{\hbar^{2}-\theta^{2}}}{\sqrt{2\hbar}}\sum_{i=1}^{k}\frac{1}{\sqrt{m+i}}$
$\displaystyle=$
$\displaystyle\zeta_{m;m+k}\frac{\sqrt{\hbar^{2}-\theta^{2}}}{\sqrt{2\hbar}}.$

Since these distances do not depend on $n$, for any integers $n\neq
n^{\prime}$, we always have
$d(\omega_{m+k,n},\omega_{m,n})=d(\omega_{m+k,n^{\prime}},\omega_{m,n^{\prime}})$.
Obviously, there is
$d(\omega_{m+k,n},\omega_{m,n})=d(\omega_{m,n},\omega_{m+k,n})$.

It is easy to see that, for $m,n,k,l\geqslant 0$, these distances satisfy

$d(\omega_{m+k+l,n},\omega_{m,n})=d(\omega_{m+k,n},\omega_{m,n})+d(\omega_{m+k+l,n},\omega_{m+k,n}).$
(69)

This means that these distances are additive.

Similarly, the Connes spectral distance between Fock states $|m,n+k\rangle$
and $|m,n\rangle$ is

$d(\omega_{m,n+k},\omega_{m,n})=\frac{\sqrt{\hbar^{2}-\theta^{2}}}{\sqrt{2\hbar}}\sum_{i=1}^{k}\frac{1}{\sqrt{n+i}}=\zeta_{n;n+k}\frac{\sqrt{\hbar^{2}-\theta^{2}}}{\sqrt{2\hbar}}.$
(70)

We also have
$d(\omega_{m,n+k},\omega_{m,n})=d(\omega_{m^{\prime},n+k},\omega_{m^{\prime},n})$
for $m\neq m^{\prime}$.

Compare with the result in Refs. [27, 32, 34], one can find that the Connes
spectral distances between Fock states in generalized noncommutative phase
space have one more factor
$\sqrt{1-\theta^{2}\\!/\hbar^{2}}=\sqrt{1-\mu\nu\\!/\hbar^{2}}$ than those in
normal quantum phase space. So in general, the Connes spectral distances in
generalized noncommutative phase space are shorter than those in normal
quantum phase space. This is intuitive and reasonable, because the
noncommutativity of the space will lead to some kind of nonlocality. The
partial nonlocality can equivalently make the distances shorter.

Obviously, when $\theta\to\hbar$, these Connes spectral distances will become
zero. This will cause some singularities.

Furthermore, using the above method, one can find that for any integers
$m,n,k,l\geqslant 0$, the Connes spectral distance between the Fock states
$|m,n\rangle$ and $|k,l\rangle$ is (see Appendix B for more details)

$\displaystyle d(\omega_{k,l},\omega_{m,n})$ $\displaystyle=$
$\displaystyle\frac{1}{\beta\sqrt{1+t^{2}}}\sqrt{\zeta_{m;k}^{2}+\zeta_{n;l}^{2}}$
(71) $\displaystyle=$
$\displaystyle\frac{\sqrt{\hbar^{2}-\theta^{2}}}{\sqrt{2\hbar}}\sqrt{\zeta_{m;k}^{2}+\zeta_{n;l}^{2}}\,.$

It is easy to verify that, these spectral distances between Fock states in
NCPS also satisfy the Pythagoras theorem,

$d(\omega_{m+k,n+l},\omega_{m,n})=\sqrt{d(\omega_{m+k,n},\omega_{m,n})^{2}+d(\omega_{m,n+l},\omega_{m,n})^{2}}.$
(72)

This also coincides with the result in Ref. [30].

##  5 Discussions and conclusions

In this paper, we study the Connes spectral distance of Fock states in $4D$
generalized noncommutative phase space. By virtue of the Hilbert-Schmidt
operatorial formulation, we obtain the Dirac operator and a spectral triple
corresponding to the generalized noncommutative phase space. Based on the ball
condition, we obtain some constraint relations about the optimal elements.
Using these constraint relations, we construct the corresponding optimal
elements and derive the Connes spectral distance between two Fock states.

We find that the spectral distances in noncommutative phase space are shorter
than those in normal phase space. This is intuitive and reasonable, because
the noncommutativity of the space will lead to some kind of nonlocality. This
nonlocality can make the distances shorter equivalently. We also find that
these spectral distances in $4D$ generalized noncommutative phase space are
additive and satisfy the normal Pythagoras theorem. When the noncommutative
parameter $\theta=\sqrt{\mu\nu}$ equals zero, the results return to those in
normal quantum phase space. But if $\theta\to\hbar$, these Connes spectral
distances will become zero. This will cause some singularities. Usually, we
assume $\theta\ll\hbar$.

Here we only study the Connes spectral distances between Fock states.
Obviously, these distances depend on the specific quantum states being
considered. One can also study the spectral distances between other kinds of
pure states and mixed states. In a noncommutative space, a pure state is the
analog of a traditional point in a normal commutative space, and the spectral
distance between pure states corresponds to the geodesic distance between
points. Different kinds of quantum states can form different kinds of abstract
spaces, and these spaces have different mathematical structures. So studies of
the Connes spectral distances between quantum states can help us to study the
mathematical structures of the spaces and also physical properties of the
quantum systems.

Our results are significant for the study of the Connes spectral distances of
physical systems in noncommutative spaces. Our methods can be used to study
other physical systems in other kinds of noncommutative spaces. Using our
method, one can also study the Connes spectral distance between quantum states
in higher-dimensional noncommutative spaces. But usually the calculations and
results will be much more complicated.

## Acknowledgements

This work is partly supported by Key Research and Development Project of
Guangdong Province (Grant No. 2020B0303300001), the National Natural Science
Foundation of China (Grant No. 11911530750), the Guangdong Basic and Applied
Basic Research Foundation (Grant No. 2019A1515011703), the Fundamental
Research Funds for the Central Universities (Grant No. 2019MS109) and the
Natural Science Foundation of Anhui Province (Grant No. 1908085MA16).

##  Appendix A Dirac operator for $4D$ generalized noncommutative phase space

Similar to Ref. [34], in order to construct the Dirac operator for the $4D$
generalized noncommutative phase space (3), one can consider the following
extended noncommutative phase space in which the coordinate operators
$\tilde{X}_{i}$, $\tilde{Y}_{i}$ and $\tilde{P}_{i}$, $\tilde{Q}_{i}$ satisfy
the following commutation relations,

$\displaystyle[\tilde{X}_{i},\,\tilde{P}_{j}]=[\tilde{Y}_{i},\,\tilde{Q}_{j}]=\mathrm{i}\delta_{ij}\hbar,\qquad[\tilde{X}_{i},\,\tilde{Y}_{j}]=\mathrm{i}\delta_{ij}\lambda,\qquad[\tilde{Q}_{i},\,\tilde{P}_{j}]=\mathrm{i}\delta_{ij}\lambda,$
$\displaystyle[\tilde{X}_{1},\,\tilde{X}_{2}]=[\tilde{Y}_{1},\,\tilde{Y}_{2}]=\mathrm{i}\theta,\qquad[\tilde{P}_{1},\,\tilde{P}_{2}]=[\tilde{Q}_{1},\,\tilde{Q}_{2}]=\mathrm{i}\theta,$
(73)

where $\lambda$ is some parameter.

It is easy to verify that, the following unitary representation acting on the
quantum Hilbert space $\mathcal{Q}$ satisfies the above noncommutative
relations (A),

$\displaystyle\tilde{X}_{i}|\phi)=\alpha\big{|}(\tilde{x}_{i}-t\varepsilon_{ij}\tilde{p}_{j})\phi\big{)}\,,\qquad\tilde{P}_{i}|\phi)=\alpha\big{|}(\tilde{p}_{i}+t\varepsilon_{ij}\tilde{x}_{j})\phi\big{)}\,,$
$\displaystyle\tilde{Y}_{i}|\phi)=\frac{\alpha}{\sqrt{\hbar^{2}-\theta^{2}}}\Big{[}\lambda\big{|}(\tilde{p}_{i}-t\varepsilon_{ij}\tilde{x}_{j})\phi\big{)}-\sqrt{\hbar^{2}+\lambda^{2}-\theta^{2}}\big{|}\phi(\tilde{x}_{i}+t\varepsilon_{ij}\tilde{p}_{j})\big{)}\Big{]},$
$\displaystyle\tilde{Q}_{i}|\phi)=\frac{\alpha}{\sqrt{\hbar^{2}-\theta^{2}}}\Big{[}\lambda\big{|}(\tilde{x}_{i}+t\varepsilon_{ij}\tilde{p}_{j})\phi\big{)}+\sqrt{\hbar^{2}+\lambda^{2}-\theta^{2}}\big{|}\phi(\tilde{p}_{i}-t\varepsilon_{ij}\tilde{x}_{j})\big{)}\Big{]},\qquad$
(74)

where

$\alpha=\frac{\sqrt{\hbar+\sqrt{\hbar^{2}-\theta^{2}}}}{\sqrt{2\hbar}}.$ (75)

The Dirac operator $\mathcal{D}$ for the generalized noncommutative phase
space (3) can be written as [5],

$\mathcal{D}=\frac{1}{\lambda}\gamma^{1}\tilde{Y}_{1}-\frac{1}{\lambda}\gamma^{2}\tilde{Q}_{1}+\frac{1}{\lambda}\gamma^{3}\tilde{Y}_{2}-\frac{1}{\lambda}\gamma^{4}\tilde{Q}_{2},$
(76)

where $\gamma^{k}$’s are the Euclidean Dirac matrices satisfying
$\gamma^{k}\gamma^{l}+\gamma^{l}\gamma^{k}=2\delta_{kl}\mathds{I}_{4}$, for
example,

$\displaystyle\gamma^{1}=\left(\begin{array}[]{cccc}0&0&0&\mathrm{i}\\\
0&0&\mathrm{i}&0\\\ 0&-\mathrm{i}&0&0\\\ -\mathrm{i}&0&0&0\\\
\end{array}\right),\qquad\gamma^{2}=\left(\begin{array}[]{cccc}0&0&0&1\\\
0&0&-1&0\\\ 0&-1&0&0\\\ 1&0&0&0\\\ \end{array}\right),$ (85)
$\displaystyle\gamma^{3}=\left(\begin{array}[]{cccc}0&0&\mathrm{i}&0\\\
0&0&0&-\mathrm{i}\\\ -\mathrm{i}&0&0&0\\\ 0&\mathrm{i}&0&0\\\
\end{array}\right),\qquad\gamma^{4}=\left(\begin{array}[]{cccc}0&0&1&0\\\
0&0&0&1\\\ 1&0&0&0\\\ 0&1&0&0\\\ \end{array}\right).$ (94)

So the Dirac operator (76) can be expressed as,

$\mathcal{D}=\frac{1}{\lambda}\left(\begin{array}[]{cccc}0&0&\mathrm{i}\tilde{Y}_{2}-\tilde{Q}_{2}&\mathrm{i}\tilde{Y}_{1}-\tilde{Q}_{1}\\\
0&0&\mathrm{i}\tilde{Y}_{1}+\tilde{Q}_{1}&-\mathrm{i}\tilde{Y}_{2}-\tilde{Q}_{2}\\\
-\mathrm{i}\tilde{Y}_{2}-\tilde{Q}_{2}&-\mathrm{i}\tilde{Y}_{1}+\tilde{Q}_{1}&0&0\\\
-\mathrm{i}\tilde{Y}_{1}-\tilde{Q}_{1}&\mathrm{i}\tilde{Y}_{2}-\tilde{Q}_{2}&0&0\\\
\end{array}\right).$ (95)

After some straightforward calculations, one can obtain the commutator
$[\mathcal{D},\pi(e)]$ acting on an element
$\Phi\in\mathcal{Q}\otimes\mathbb{C}^{4}$ as

$\displaystyle[\mathcal{D},\pi(e)]\Phi=[\mathcal{D},\pi(e)]\left(\begin{array}[]{c}|\phi_{1})\\\
|\phi_{2})\\\ |\phi_{3})\\\ |\phi_{4})\\\ \end{array}\right)$ (109)
$\displaystyle\\!\\!\\!\\!\\!\\!\\!\\!=\beta\left(\begin{array}[]{cccc}\\!0&\\!0&\\![-\hat{a}_{2}^{{\dagger}}{+}\mathrm{i}t\hat{a}_{1}\\!{}^{{\dagger}},e]&\\![-\hat{a}_{1}^{{\dagger}}{-}\mathrm{i}t\hat{a}_{2}\\!{}^{{\dagger}},e]\\\
\\!0&\\!0&\\![\hat{a}_{1}{-}\mathrm{i}t\hat{a}_{2},e]&\\![-\hat{a}_{2}{-}\mathrm{i}t\hat{a}_{1},e]\\\
{}\\![-\hat{a}_{2}{-}\mathrm{i}t\hat{a}_{1},e]&\\![\hat{a}_{1}^{{\dagger}}{+}\mathrm{i}t\hat{a}_{2}\\!{}^{{\dagger}},e]&\\!0&\\!0\\\
{}\\![-\hat{a}_{1}{+}\mathrm{i}t\hat{a}_{2},e]&\\![-\hat{a}_{2}^{{\dagger}}{+}\mathrm{i}t\hat{a}_{1}\\!{}^{{\dagger}},e]&\\!0&\\!0\\\
\end{array}\\!\right)\\!\\!\left(\begin{array}[]{c}\\!|\phi_{1})\\\
\\!|\phi_{2})\\\ \\!|\phi_{3})\\\ \\!|\phi_{4})\\\ \end{array}\\!\right)$
$\displaystyle=$
$\displaystyle\beta\left(\begin{array}[]{cccc}0&0&[-\hat{A}_{2}^{\dagger},e]&[-\hat{A}_{1}^{\dagger},e]\\\
0&0&[\hat{A}_{1},e]&[-\hat{A}_{2},e]\\\
{}[-\hat{A}_{2},e]&[\hat{A}_{1}^{\dagger},e]&0&0\\\
{}[-\hat{A}_{1},e]&[-\hat{A}_{2}^{\dagger},e]&0&0\\\
\end{array}\right)\\!\\!\left(\begin{array}[]{c}\\!|\phi_{1})\\\
\\!|\phi_{2})\\\ \\!|\phi_{3})\\\ \\!|\phi_{4})\\\ \end{array}\\!\right).$
(118)

So regarding $\Phi$ as a test function, one can identify the Dirac operator
$\mathcal{D}$ as

$\mathcal{D}=\beta\left(\begin{array}[]{cccc}0&0&-\hat{A}_{2}^{\dagger}&-\hat{A}_{1}^{\dagger}\\\
0&0&\hat{A}_{1}&-\hat{A}_{2}\\\ -\hat{A}_{2}&\hat{A}_{1}^{\dagger}&0&0\\\
-\hat{A}_{1}&-\hat{A}_{2}^{\dagger}&0&0\\\ \end{array}\right).$ (119)

##  Appendix B Calculations of Connes distance between Fock states
$|m,n\rangle$ and $|m+k,n+l\rangle$

First, let us consider the distance between the states $|m+1,n+1\rangle$ and
$|m,n\rangle$,

$\displaystyle d(\omega_{m+1,n+1},\omega_{m,n})=\sup_{e\in
B}|\mathrm{tr}_{\mathcal{F}}(\rho_{m+1,n+1}e)-\mathrm{tr}_{\mathcal{F}}(\rho_{m,n}e)|$
(120) $\displaystyle=$ $\displaystyle\sup_{e\in
B}|c_{m+1,n+1}-c_{m,n}|=\sup_{e\in B}|E_{m+1,n+1}+F_{m,n+1}|.$

From (3), there is

$\displaystyle(1+t^{2})(G_{m+1,n}+H_{m,n+1})\leqslant\frac{1}{\beta^{2}},\qquad(1+t^{2})(G_{m+1,n+1}+H_{m+1,n+1})\leqslant\frac{1}{\beta^{2}},$
$\displaystyle G_{m+1,n+1}+H_{m,n+1}\leqslant\frac{1}{\beta^{2}},\qquad
G_{m+1,n}+H_{m+1,n+1}\leqslant\frac{1}{\beta^{2}}.$ (121)

In order to attain the supremum of $|E_{m+1,n+1}+F_{m,n+1}|$, one must let
$G_{m+1,n+1}+H_{m,n+1}$ as large as possible. But if

$G_{m+1,n+1}+H_{m,n+1}>\frac{1}{(1+t^{2})\beta^{2}},$ (122)

then compare with (B), there must be

$G_{m+1,n}<G_{m+1,n+1},\qquad H_{m+1,n+1}<H_{m,n+1}.$ (123)

Without loss of generality, we can assume that $E_{m+1,n+1},F_{m,n+1}\geqslant
0$. Therefore there are

$E_{m+1,n}<E_{m+1,n+1},\qquad F_{m+1,n+1}<F_{m,n+1}.$ (124)

But

$E_{m+1,n}+F_{m+1,n+1}=c_{m+1,n+1}-c_{m,n}=E_{m+1,n+1}+F_{m,n+1},$ (125)

this is in contradiction with the relations (124). This means that the
inequality (122) is not true.

So there is

$G_{m+1,n+1}+H_{m,n+1}\leqslant\frac{1}{(1+t^{2})\beta^{2}},$ (126)

In order to attain the supremum of $|E_{m+1,n+1}+F_{m,n+1}|$, there must be

$G_{m+1,n+1}+H_{m,n+1}=\frac{1}{(1+t^{2})\beta^{2}},$ (127)

and

$G_{m+1,n}=G_{m+1,n+1},\qquad H_{m+1,n+1}=H_{m,n+1},$ (128)

or

$E_{m+1,n}=E_{m+1,n+1},\qquad F_{m+1,n+1}=F_{m,n+1}.$ (129)

So the Connes spectral distance between the states $|m+1,n+1\rangle$ and
$|m,n\rangle$ is

$\displaystyle d(\omega_{m+1,n+1},\omega_{m,n})=\sup_{e\in
B}|E_{m+1,n+1}+F_{m,n+1}|$ (130) $\displaystyle=$ $\displaystyle\sup_{e\in
B}\left|\frac{1}{\sqrt{m+1}}E_{m+1,n+1}\sqrt{m+1}+\frac{1}{\sqrt{n+1}}F_{m,n+1}\sqrt{n+1}\right|$
$\displaystyle\leqslant$ $\displaystyle\sup_{e\in
B}\sqrt{\frac{1}{m+1}+\frac{1}{n+1}}\sqrt{(m+1)E_{m+1,n+1}^{2}+(n+1)F_{m,n+1}^{2}}$
$\displaystyle=$ $\displaystyle\sup_{e\in
B}\sqrt{\frac{1}{m+1}+\frac{1}{n+1}}\sqrt{G_{m+1,n+1}+H_{m,n+1}}$
$\displaystyle=$
$\displaystyle\sqrt{\frac{1}{m+1}+\frac{1}{n+1}}\sqrt{\frac{1}{(1+t^{2})\beta^{2}}}.$

In the above inequality, we have used the Cauchy-Schwartz inequality,

$(x_{1}y_{1}+x_{2}y_{2})^{2}\leqslant(x_{1}^{2}+x_{2}^{2})(y_{1}^{2}+y_{2}^{2}),$
(131)

where the equality holds if $x_{1}y_{2}=x_{2}y_{1}$.

So one can set

$(m+1)E_{m+1,n+1}=(n+1)F_{m,n+1},$ (132)

and there should be

$d(\omega_{m+1,n+1},\omega_{m,n})\leqslant\frac{1}{\beta\sqrt{1+t^{2}}}\sqrt{\frac{1}{m+1}+\frac{1}{n+1}}.$
(133)

Furthermore, let us consider the Connes spectral distance between two Fock
states $|m,n\rangle$ and $|m+k,n+l\rangle$. For simplicity, we assume $k,l>1$.
The corresponding Connes spectral distance is

$\displaystyle d(\omega_{m+k,n+l},\omega_{m,n})=\sup_{e\in
B}|\mathrm{tr}_{\mathcal{F}}(\rho_{m+k,n+l}\,e)-\mathrm{tr}_{\mathcal{F}}(\rho_{m,n}\,e)|$
(134) $\displaystyle=$ $\displaystyle\sup_{e\in
B}|c_{m+k,n+l}-c_{m,n}|=\sup_{e\in
B}|c_{m+k,n+l}-c_{m+k,n}+c_{m+k,n}-c_{m,n}|$ $\displaystyle=$
$\displaystyle\sup_{e\in
B}\left|\sum_{i=1}^{l}(c_{m+k,n+i}-c_{m+k,n+i-1})+\sum_{j=1}^{k}(c_{m+j,n}-c_{m+j-1,n})\right|$
$\displaystyle=$ $\displaystyle\sup_{e\in
B}\left|\sum_{i=1}^{l}F_{m+k,n+i}+\sum_{j=1}^{k}E_{m+j,n}\right|.$

From (3), in order to attain the supremum of $|c_{m+k,n+l}-c_{m,n}|$, there
must be

$\displaystyle(1+t^{2})(G_{m+2,n+1}+H_{m+1,n+2})=\frac{1}{\beta^{2}},\quad(1+t^{2})(G_{m+1,n+1}+H_{m+1,n+1})=\frac{1}{\beta^{2}},$
$\displaystyle(G_{m+2,n+1}+H_{m+1,n+1})+t^{2}(G_{m+1,n+1}+H_{m+1,n+2})=\frac{1}{\beta^{2}},$
$\displaystyle(G_{m+1,n+1}+H_{m+1,n+2})+t^{2}(G_{m+2,n+1}+H_{m+1,n+1})=\frac{1}{\beta^{2}}.$
(135)

So there is

$G_{m+2,n+1}=G_{m+1,n+1},\qquad H_{m+1,n+1}=H_{m+1,n+2}.$ (136)

Similarly, one can obtain

$\displaystyle G_{m+i,n+j}=G_{m+1,n},\qquad 1\leqslant i\leqslant k,0\leqslant
j\leqslant l,$ $\displaystyle H_{m+i,n+j}=H_{m,n+1},\qquad 0\leqslant
i\leqslant k,1\leqslant j\leqslant l,$ (137)

or

$E_{m+i,n+j}=\sqrt{\frac{m+1}{m+i}}E_{m+1,n},\qquad
F_{m+i,n+j}=\sqrt{\frac{n+1}{n+j}}F_{m,n+1}.$ (138)

We also have

$(1+t^{2})(G_{m+1,n}+H_{m,n+1})=\frac{1}{\beta^{2}},$ (139)

or

$(m+1)E_{m+1,n}^{2}+(n+1)F_{m,n+1}^{2}=\frac{1}{(1+t^{2})\beta^{2}}.$ (140)

Therefore the Connes spectral distance between the Fock states $|m,n\rangle$
and $|m+k,n+l\rangle$ is

$\displaystyle d(\omega_{m+k,n+l},\omega_{m,n})=\sup_{e\in
B}\left|\sum_{i=1}^{k}E_{m+i,n}+\sum_{j=1}^{l}F_{m+k,n+j}\right|$ (141)
$\displaystyle=$ $\displaystyle\sup_{e\in
B}\left|\sqrt{m+1}E_{m+1,n}\sum_{i=1}^{k}\frac{1}{\sqrt{m+i}}+\sqrt{n+1}F_{m,n+1}\sum_{j=1}^{l}\frac{1}{\sqrt{n+j}}\right|$
$\displaystyle=$ $\displaystyle\sup_{e\in
B}\left|\sqrt{m+1}E_{m+1,n}\zeta_{m;m+k}+\sqrt{n+1}F_{m,n+1}\zeta_{n;n+l}\right|$
$\displaystyle\leqslant$ $\displaystyle\sup_{e\in
B}\sqrt{(m+1)E_{m+1,n}^{2}+(n+1)F_{m,n+1}^{2}}\sqrt{\zeta_{m;m+k}^{2}+\zeta_{n;n+l}^{2}}$
$\displaystyle=$
$\displaystyle\frac{1}{\beta\sqrt{1+t^{2}}}\sqrt{\zeta_{m;m+k}^{2}+\zeta_{n;n+l}^{2}}.$

Here we have used the Cauchy-Schwartz inequality (131) in the above
inequality, and the equality holds if

$\sqrt{m+1}E_{m+1,n}\zeta_{n;n+l}=\sqrt{n+1}F_{m,n+1}\zeta_{m;m+k}.$ (142)

From (140) and (142), one can derive

$E_{m+1,n}=\frac{1}{\beta\sqrt{1+t^{2}}}\frac{1}{\sqrt{m+1}}\frac{\zeta_{m;m+k}}{\sqrt{\zeta_{m;m+k}^{2}+\zeta_{n;n+l}^{2}}}=c_{0}\frac{1}{\sqrt{m+1}},$
(143)
$F_{m,n+1}=\frac{1}{\beta\sqrt{1+t^{2}}}\frac{1}{\sqrt{n+1}}\frac{\zeta_{n;n+l}}{\sqrt{\zeta_{m;m+k}^{2}+\zeta_{n;n+l}^{2}}}=d_{0}\frac{1}{\sqrt{n+1}},$
(144)

where

$c_{0}=\frac{1}{\beta\sqrt{1+t^{2}}}\frac{\zeta_{m;m+k}}{\sqrt{\zeta_{m;m+k}^{2}+\zeta_{n;n+l}^{2}}},\qquad
d_{0}=\frac{1}{\beta\sqrt{1+t^{2}}}\frac{\zeta_{n;n+l}}{\sqrt{\zeta_{m;m+k}^{2}+\zeta_{n;n+l}^{2}}}.$
(145)

So in order to attain the supremum of $|c_{m+k,n+l}-c_{m,n}|$, for example,
one can choose

$\displaystyle c_{p,q}$ $\displaystyle=$ $\displaystyle
c_{0}\sum_{i=1}^{p}\frac{1}{\sqrt{i}}+d_{0}\sum_{j=1}^{q}\frac{1}{\sqrt{j}}=c_{0}\zeta_{0;p}+d_{0}\zeta_{0;q}$
(146) $\displaystyle=$
$\displaystyle\frac{1}{\beta\sqrt{1+t^{2}}}\frac{\zeta_{0;p}\zeta_{m;m+k}+\zeta_{0;q}\zeta_{n;n+l}}{\sqrt{\zeta_{m;m+k}^{2}+\zeta_{n;n+l}^{2}}},$

where $p=m,m+1,...,m+k$ and $q=n,n+1,...,n+l$ . Obviously, there is

$E_{i+1,j}=c_{i+1,j}-c_{i,j}=\frac{1}{\sqrt{i+1}}c_{0},\qquad
F_{i,j+1}=c_{i,j+1}-c_{i,j}=\frac{1}{\sqrt{j+1}}d_{0}.$ (147)

The corresponding optimal element $e_{o}$ is

$\displaystyle e_{o}$ $\displaystyle=$
$\displaystyle\sum_{p=m}^{m+k}\sum_{q=n}^{n+l}c_{p,q}|p,q\rangle\langle p,q|$
(148) $\displaystyle=$
$\displaystyle\frac{1}{\beta\sqrt{1+t^{2}}}\sum_{p=m}^{m+k}\sum_{q=n}^{n+l}\frac{\zeta_{0;p}\zeta_{m;m+k}+\zeta_{0;q}\zeta_{n;n+l}}{\sqrt{\zeta_{m;m+k}^{2}+\zeta_{n;n+l}^{2}}}|p,q\rangle\langle
p,q|,$

and the spectral distance between the Fock states $|m+k,n+l\rangle$ and
$|m,n\rangle$ is

$\displaystyle d(\omega_{m+k,n+l},\omega_{m,n})$ $\displaystyle=$
$\displaystyle|\mathrm{tr}_{\mathcal{F}}(\rho_{m+k,n+l}\,e_{o})-\mathrm{tr}_{\mathcal{F}}(\rho_{m,n}\,e_{o})|$
(149) $\displaystyle=$ $\displaystyle|\langle
m+k,n+l|e_{o}|m+k,n+l\rangle-\langle m,n|e_{o}|m,n\rangle|$ $\displaystyle=$
$\displaystyle c_{0}\zeta_{m;m+k}+d_{0}\zeta_{n;n+l}$ $\displaystyle=$
$\displaystyle\frac{1}{\beta\sqrt{1+t^{2}}}\sqrt{\zeta_{m;m+k}^{2}+\zeta_{n;n+l}^{2}}.$

It is easy to verify that, these results (148) and (149) are also true for
$k\leqslant 0$ and/or $l\leqslant 0$.

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