# From entanglement to discord: a perspective based on partial transposition

Liang-Liang Sun111email<EMAIL_ADDRESS>Xiang Zhou  Sixia Yu 222email:
<EMAIL_ADDRESS>Department of Modern Physics and National Laboratory for
Physical Sciences at Microscale, University of Science and Technology of
China, Hefei, Anhui 230026, China

(August 28, 2024)

###### Abstract

Here, we show that partial transposition, which is initially introduced to
study entanglement, can also inspire many results on quantum discord
including: (I) a discord criterion of spectrum invariant under partial
transposition, stating that one state must contain discord if its spectrum is
changed by the action of partial transposition, (II) an approach to estimate
the geometric quantum discord and the one-way deficit based on the change of
spectrum. To compare with entanglement theory, we also lower-bound the
geometric quantum entanglement and the entanglement of relative entropy. Thus,
on one hand, we illustrate an approach to specify and estimate discord based
on partial transposition. On the other hand, we show that, entanglement and
discord, two basic notions of nonclassical correlations, can be placed on the
same ground such that their interplay and distinction can be illustrated in
within a universal framework..

###### pacs:

98.80.-k, 98.70.Vc

##  I I. Introduction

One distinctive feature of quantum theory is that quantum systems can
demonstrate various forms of nonclassical correlations, which often find
applications in quantum information science Horodecki et al. (2009); Brunner
et al. (2014); Modi et al. (2012); Bera et al. (2017). Entanglement was the
first such notion to be known about and then appreciated as a key resource in
many quantum information tasks, such as quantum teleportation Bennett et al.
(1993), cryptography Ekert (1991), quantum algorithms Raussendorf and Briegel
(2001); Raussendorf et al. (2003); Briegel et al. (2009), metrology
Giovannetti et al. (2004). Entanglement was even believed to be responsible
for why quantum resources can outperform classical ones. This belief started
to change when quantum discord is discovered, which goes beyond entanglement
and exists in a wide range of quantum states that may be separable. Like
entanglement, discord can also demonstrate quantum advantages in diversified
tasks, such as mixed-state quantum computing Knill and Laflamme (1998); Datta
et al. (2008), bounding distributed entanglement Chuan et al. (2012), remote
state preparation Dakić et al. (2012), and quantum state merging Cavalcanti et
al. (2011); Madhok and Datta (2011). Although it is known that entanglement
must demonstrate quantum discord, the specification and quantification of
these two basic properties are commonly seen as distinct subjects and follow
different lines of research. One interesting question is whether or not these
two properties can be characterized and quantified within a universal
framework.

Much focus has been placed on entanglement and many powerful tools are
introduced Horodecki et al. (2009); Friis et al. (2018). One may concern
whether these tools can be borrowed to specify and estimate the relatively
less studied discord. For this purpose, we consider the primary tool of of
detecting entanglement, namely, the positive partial transpose (PPT) criterion
Peres (1996); Horodecki et al. (1996), with which, one state is certified to
be entangled if its transposed density matrix presents negative eigenvalue.
This criterion also induces one computable entanglement quantifier, referred
to as negativity Eisert (2006); Vidal and Werner (2002), as the sum of
negative eigenvalues in absolute values. Although the tool of PPT was related
to discord, $e.g.,$ in $2\times n$-dimensional system, all discord-free states
belong to a subclass of PPT states, called strong PPT states Chruściński et
al. (2008), quantum discord is mainly detected with discord witness Zhang et
al. (2011); Yu et al. (2011). To quantify discord, quite a few measures have
been introduced Modi et al. (2012); Bera et al. (2017). Unfortunately, most of
them are hard to compute. One exception is the geometric quantum discord (GQD)
Dakić et al. (2010); Luo and Fu (2011, 2010), which can be equivalently
defined as the minimal disturbance under a projective measurement performed on
one partite thus can be computed via an optimization on the measurement Luo
and Fu (2011, 2010); Miranowicz et al. (2012); Rana and Parashar (2012);
Hassan et al. (2012).

In this paper, we show that many results on quantum discord can be inspired by
the map of partial transposition. We first illustrate that, the spectrum of
discord-free states are invariant (up to a relabel of indexes) under partial
transposition. Therefore, the change of spectrum indicates the presence of
discord. We refer this criterion to as spectrum invariant under partial
transposition (SIPT) in contrast to the PPT in entanglement theory.
Quantitatively, we find that the spectrum change implies a lower bound on the
GQD and a discord-like quantity of one-way deficit. For the sake of comparing
discord and entanglement, we also provide lower bounds for the geometric
quantum entanglement and the entanglement of relative entropy. Thus, a
perspective based on partial transposition is provided, in which, entanglement
is specified and estimated based on the negative eigenvalues presented under
the map, in contrast, discord is specified and estimated based on the change
of the spectrum.

##  II II.Preliminaries

Let us first make a brief review on some basic notions. One quantum state
$\rho_{AB}$ is said to be entangled if it cannot be written as the mixture of
product states, namely,

$\displaystyle\rho_{AB}=\textstyle\sum_{i}f_{i}\cdot\rho_{i,A}\otimes\rho_{i,B}.$
(1)

One can expand a general bipartite state in $N\otimes M$ dimensional Hilbert’s
space, entangled or separable, in a chosen product basis as

$\displaystyle\textstyle\rho_{AB}=\sum_{i,j}^{N}\sum_{k,l}^{M}\rho_{ij,kl}|i\rangle\langle
j|_{A}\cdot|k\rangle\langle l|_{B}.$ (2)

Henceforth we shall omit the footnote $A,B$ where not introducing ambiguity.
Given this decomposition, performing a partial transposition on Alice’s side
leads to

$\displaystyle\rho^{\Gamma_{A}}=\sum_{i,j}^{N}\sum_{k,l}^{M}\rho_{ij,kl}|j\rangle\langle
i|\cdot|k\rangle\langle l|.$ (3)

The density matrix of any separable state is positive, and one state is
certified to be entangled if it violates PPT criterion. PPT criterion also
induces an entanglement quantifier of negativity as the sum of the negative
spectrum in the absolute value as

$\displaystyle\textstyle\mathcal{N}:=\frac{\|\rho^{\Gamma_{A}}\|_{\rm
tr}-1}{2},$ (4)

where $\|A\|_{\rm tr}={\rm Tr}\sqrt{AA^{\dagger}}$.

Quantum discord (QD) is a typical quantum correlation referring to the
phenomenon that a composite quantum system contains more information than the
subsystems taken separably. By definition, one discord-free state $\rho$ can
be written in the form of

$\displaystyle\rho=\sum_{i}f_{i}\cdot|i\rangle\langle i|\otimes\rho_{i}.$ (5)

where $\\{|i\rangle\\}$ is one set of orthogonal basis. It is clear that an
entangled state must contain discord, however, it is not necessary for the
converse.

##  III III. The criterion of spectrum invariant partial transposition

The operation of partial transposition depends on the partite and basis on
which it is performed. However, the spectrum of the partially transposed state
is independent on them. We specify the spectrum of $\rho$ by
$\lambda^{\downarrow}(\rho)$ shorten by $\lambda^{\downarrow}$, where the
elements are arranged in descending order, namely,
$\lambda^{\downarrow}=(\lambda^{\downarrow}_{1},\lambda^{\downarrow}_{2},\cdots\lambda^{\downarrow}_{M\cdot
N})$ with $\lambda^{\downarrow}_{i}\geq\lambda^{\downarrow}_{i+1}$, $\forall
i$. The fact that
$\lambda^{\downarrow}(\rho^{\Gamma_{A}})=\lambda^{\downarrow}(\rho^{\Gamma_{B}})$
can be illustrated via
$\lambda^{\downarrow}(\rho^{\Gamma_{A}})=\lambda^{\downarrow}([\rho^{\Gamma_{A}}]^{\Gamma})=\lambda^{\downarrow}(\rho^{\Gamma_{B}})$,
where $\Gamma$ specifies the usual transposition acting on the joint system,
which does not alter state’s spectrum. The different choices of basis to
perform partial transposition can be captured with a local unitary operation
${\rm U}_{A}$ acting on Alice’s side. The basis-independence feature can be
illustrated as
$\lambda^{\downarrow}[(\operatorname{U}_{A}\rho\operatorname{U}^{\dagger}_{A})^{\Gamma_{A}}]=\lambda^{\downarrow}[(\operatorname{U}_{A}\rho\operatorname{U}^{\dagger}_{A})^{\Gamma_{B}}]=\lambda^{\downarrow}[\operatorname{U}_{A}(\rho^{\Gamma_{B}})\operatorname{U}^{\dagger}_{A}]=\lambda^{\downarrow}[\rho^{\Gamma_{B}}]=\lambda^{\downarrow}[\rho^{\Gamma_{A}}]$.
Therefore, the spectrum of partially transposed state is determined solely by
state $\rho$.

Consider a discord-free state $\rho=\sum_{i}f_{i}\cdot|i\rangle\langle
i|_{A}\otimes\rho_{i,B}$, one can perform partial transposition with respect
to basis $\\{|i\rangle_{A}\\}$ without losing any generality. Definitely, the
state is invariant under the operation. An immediately consequence is our
criterion of SIPT

######  Theorem 1 (Spectrum invariant under partial transposition).

Spectrum of a discord-free state $\rho$ is invariant under partial
transposition:

$\displaystyle\lambda^{\downarrow}(\rho)=\lambda^{\downarrow}(\rho^{\Gamma_{A}}).$
(6)

Thus, from the perspective of partial transposition, a change of spectrum
under the map indicates a non-trivial discord. To certify quantum
entanglement, the change needs to be large enough to ensure the presence of
negative spectrum.

To justify discord via Theorem.1, one can compute the spectrum of $\rho$ and
$\rho^{\Gamma_{A}}$, then makes a statement after comparing them. Or
alteratively, using the method based on the moments of matrix defined as
$\Pi_{n}(\rho):={\rm Tr}(\rho^{n})$. Noting that,

$\displaystyle\Pi_{n}(\rho)=\Pi_{n}(\rho^{\Gamma_{A}}),\forall
n,if,\lambda^{\downarrow}(\rho)=\lambda^{\downarrow}(\rho^{\Gamma_{A}}).$ (7)

Such a justification begins with the third moment as
$\Pi_{1}(\rho)=\Pi_{1}(\rho^{\Gamma_{A}})$ and
$\Pi_{2}(\rho)=\Pi_{2}(\rho^{\Gamma_{A}})$ are trivially satisfied, which is
due to the fact partial transposition is trace-preserving and
$\Pi_{2}(\rho)=\Pi_{2}(\rho^{\Gamma_{A}})$ is trivially satisfied. To deal
with a state $\rho$ of $M\otimes N-$dimensional system, a sufficient judgement
of Eq.(7) stops at most in the $(M\cdot N+2)-$th moment.

It is interesting to ask a question whether SIPT provides a sufficient and
necessary justification of discord. Unfortunately, it is not the case as there
are states containing discord while the spectrum is invariant under partial
transposition, such as $X$ state having equal anti-diagonal terms Ali et al.
(2010). This is similar to the case in entanglement theory, where there are
bounded entangled states standing positivity under partial transposition, for
which, PPT criterion fails.

##  IV IV. Quantitative estimation of discord and discord-like quantity

Motivated by the fact that the negative spectrum of a partially transposed
state can be used to quantify entanglement, we consider how the spectrum
change under partial transposition quantitatively relates to quantum discord.

###  IV.1 A. Lower-bound GQD

GQD is one of the most studied discord measures and defined as the minimum
Hilbert-Schmidt distance or $2-$norm between state of interest $\rho$ and the
set of discord-free states specified by $\mathcal{D}$ Dakić et al. (2010); Luo
and Fu (2011, 2010)

$\displaystyle D_{\rm
HS}=\min_{\varrho\in\mathcal{D}}\|\rho-\varrho\|^{2}_{2}.$ (8)

Now, we provide two analytic lower bounds on GQD. We first make a quite useful
observation: $2-$norm is invariant under partial transposition as it only
involves the sum of the square modulus of matrix elements which is invariant
under partial transposition. This observation implies
$\|\rho-\varrho\|^{2}_{2}=\|\rho^{\Gamma_{A}}-\varrho^{\Gamma_{A}}\|^{2}_{2}$.
For latter use, we provide an estimate of the quantity
$\min_{\lambda_{d}}\|\lambda^{\prime\downarrow}-\lambda^{\downarrow}_{d}\|^{2}_{2}$
with ${\lambda^{\prime}}^{\downarrow}$ specifying the spectrum of
$\rho^{\Gamma_{A}}$, and a general positive and normalized spectrum is
specified by $\lambda_{d}$ with $\lambda^{\downarrow}_{d,i}\geq 0$, $\forall$
$i$ and $\sum_{i}\lambda^{\downarrow}_{d,i}=1$.

######  Lemma 1.

An analytic lower bound on
$\min_{\lambda_{d}}\|{\lambda^{\prime}}^{\downarrow}-\lambda^{\downarrow}_{d}\|^{2}_{2}$
is given as (see Supplemental Materials (SM) for proof)

$\displaystyle\min_{\lambda_{d}}\|{\lambda^{\prime}}^{\downarrow}-\lambda^{\downarrow}_{d}\|^{2}_{2}\geq\|{\lambda^{\prime}}_{(2)}\|^{2}_{2}+n\tau^{2}:=\mathfrak{L}({\lambda^{\prime}}^{\downarrow}),$
(9)

where we have partitioned ${\lambda^{\prime}}^{\downarrow}$ into two parts,
namely,
${\lambda^{\prime}}^{\downarrow}={\lambda^{\prime}}^{\downarrow}_{(1)}\bigcup{\lambda^{\prime}}^{\downarrow}_{(2)}$
with
${\lambda^{\prime}}^{\downarrow}_{(1)}=\\{{\lambda^{\prime}}^{\downarrow}_{1},\cdots,{\lambda^{\prime}}^{\downarrow}_{n}\\}$
and
${\lambda^{\prime}}^{\downarrow}_{(2)}=\\{{\lambda^{\prime}}^{\downarrow}_{n+1},\cdots,{\lambda^{\prime}}^{\downarrow}_{MN}\\}$
with $n$ specifying the minimal number such that
$\tau:=\frac{{\sum^{n}_{i=1}{\lambda^{\prime}}^{\downarrow}_{i}}-1}{n}\geq{\lambda^{\prime}}^{\downarrow}_{n+1}$
and $MN$ are the number of eigenvalues. The $\lambda_{d}$ achieving the
minimum is specified by $\lambda_{\rm PPT}$

Consider that a discord-free state is PPT, we have a lower-bound on GQD as

$\displaystyle D_{\rm HS}$ $\displaystyle=$
$\displaystyle\min_{\varrho\in\mathcal{D}}\|\rho-\varrho\|^{2}_{2}=\|\rho^{\Gamma_{A}}-\varrho^{\Gamma_{A}}_{\min}\|^{2}_{2}$
(10) $\displaystyle\geq$
$\displaystyle\|{\lambda^{\prime}}^{\downarrow}-\lambda^{\downarrow}_{\min}\|^{2}_{2}$
$\displaystyle\geq$
$\displaystyle\textstyle\min_{\lambda^{\downarrow}_{d}}\|{\lambda^{\prime}}^{\downarrow}-\lambda^{\downarrow}_{d}\|^{2}_{2}$
$\displaystyle=$
$\displaystyle\textstyle\mathfrak{L}(\lambda^{\prime}):=L^{D}_{\rm PPT},$
$\displaystyle\geq$
$\displaystyle\textstyle\frac{\mathcal{N}^{2}}{N_{+}}+\frac{\mathcal{N}^{2}}{N_{-}}:={L^{\prime}}^{D}_{\rm
PPT},$ (11)

where we have specified the closest discord-free state to $\rho$ (with respect
to Hilbert-Schmidt distance) by $\rho_{\min}$ and $N_{+}$ and $N_{-}$ are the
numbers of positive and negative elements of $\lambda^{\prime}$, respectively.
The first inequality follows from the inequality that, for any two states
$\rho$ and $\varrho$, it holds
$\|\rho-\varrho\|^{2}_{2}\geq\|{\lambda^{\prime}}^{\downarrow}(\rho)-\lambda^{\downarrow}(\varrho)\|^{2}_{2}$
Bhatia (1986). Eq.(10) is due to the above Lemma, and the proof for Eq.(11) is
left in SM.

Note that the lower bounds ${L}^{D}_{\rm PPT}$ and ${L^{\prime}}^{D}_{\rm
PPT}$ are non-zero only if $\rho$ violates PPT. For a general case, one state
containing discord may not violate PPT criterion, namely, the corresponding
transposed density matrix is positive. Here, we give another lower bound,
which may be non-trivial for such states. Based on the fact that any discord-
free state is SIPT, we have

$\displaystyle 2D_{\rm HS}$ $\displaystyle=$
$\displaystyle\min_{\varrho\in\mathcal{D}}\\{\|\rho-\varrho\|^{2}_{2}+\|\rho^{\Gamma_{A}}-\varrho^{\Gamma_{A}}\|^{2}_{2}\\}$
(12) $\displaystyle\geq$
$\displaystyle\|\lambda^{\downarrow}-\lambda^{\downarrow}(\rho_{\min})\|^{2}_{2}+\|{\lambda^{\prime}}^{\downarrow}-\lambda^{\downarrow}(\rho_{\min})\|^{2}_{2}$
$\displaystyle=$
$\displaystyle\textstyle\sum_{i}[{\lambda_{i}^{\downarrow}}^{2}+{{\lambda^{\prime}}_{i}^{\downarrow}}^{2}+2{\lambda_{i,\min}^{\downarrow}}^{2}-2(\lambda_{i}^{\downarrow}+{\lambda^{\prime}}_{i}^{\downarrow})\lambda_{i,\min}^{\downarrow}]$
$\displaystyle=$
$\displaystyle\textstyle\sum_{i}[{\lambda_{i}^{\downarrow}}^{2}+{{\lambda^{\prime}}_{i}^{\downarrow}}^{2}-\frac{({{\lambda^{\prime}}_{i}^{\downarrow}}+{\lambda_{i}^{\downarrow}})^{2}}{2}+2(\frac{{{\lambda^{\prime}}_{i}^{\downarrow}}+{\lambda_{i}^{\downarrow}}}{2}-\lambda^{\downarrow}_{i\min})^{2}]$
$\displaystyle=$
$\displaystyle\textstyle\frac{\|\lambda^{\downarrow}-{\lambda^{\prime}}^{\downarrow}\|^{2}_{2}}{2}+2\|\frac{{{\lambda^{\prime}}^{\downarrow}}+{\lambda^{\downarrow}}}{2}-\lambda^{\downarrow}_{\min}\|^{2}_{2}$
$\displaystyle\geq$
$\displaystyle\textstyle\frac{\|\lambda^{\downarrow}-{\lambda^{\prime}}^{\downarrow}\|^{2}_{2}}{2}+2\min_{\lambda^{\downarrow}_{d}}\|\frac{{{\lambda^{\prime}}^{\downarrow}}+{\lambda^{\downarrow}}}{2}-\lambda^{\downarrow}_{d}\|^{2}_{2}$
$\displaystyle=$
$\displaystyle\textstyle\frac{\|\lambda^{\downarrow}-{\lambda^{\prime}}^{\downarrow}\|^{2}_{2}}{2}+\mathfrak{L}(\frac{{\lambda^{\prime}}^{\downarrow}+{\lambda}^{\downarrow}}{2})$
$\displaystyle:=$ $\displaystyle 2L^{D}_{\rm SIPT}.$

The $\lambda_{d}$ achieving the minimum is specified by $\lambda_{\rm SIPT}$.

Note that the term
$\|\lambda^{\downarrow}-{\lambda^{\prime}}^{\downarrow}\|^{2}_{2}$ is non-zero
as long as state $\rho$ violates SIPT. Definitely, for PPT states, $L^{D}_{\rm
SIPT}$ can be nontrivial while $L^{D}_{\rm PPT}=0$. One may concern whether
$L^{D}_{\rm SIPT}\geq L^{D}_{\rm PPT}$. It turns out to be not the case.
Although it is clear that
$\|{\lambda^{\prime}}^{\downarrow}-\lambda^{\downarrow}_{\rm
PPT}\|^{2}_{2}\leq\|{\lambda^{\prime}}^{\downarrow}-\lambda^{\downarrow}_{\rm
SIPT}\|^{2}_{2}$ (where $\lambda^{\downarrow}_{\rm PPT}$ specifies the
spectrum achieving the minimum of
$\min_{\lambda_{d}}\|\lambda^{\downarrow}-\lambda^{\downarrow}_{d}\|^{2}_{2}$),
it is not necessary that $\|{\lambda}^{\downarrow}-\lambda^{\downarrow}_{\rm
PPT}\|^{2}_{2}\leq\|{\lambda}^{\downarrow}-\lambda^{\downarrow}_{\rm
SIPT}\|^{2}_{2}$. Consequently, $L^{D}_{\rm PPT}$ can be larger than
$L^{D}_{\rm SIPT}$. As an illustration, let us consider the maximum entangled
state $|\phi\rangle=\sum_{i}\frac{1}{\sqrt{d}}|ii\rangle$, for which, the
exact value of $D_{\rm HS}=1-\frac{1}{d}$. By the above considerations, we
have
$\lambda^{\downarrow}(|\phi\rangle\langle\phi|^{\Gamma_{A}})=\\{{\lambda^{\prime}}^{\downarrow}_{i}=\frac{1}{d};1,\leq
i\leq\frac{d(d+1)}{2}\\}\cup\\{{\lambda^{\prime}}^{\downarrow}_{i}=-\frac{1}{d};\frac{d(d+1)}{2}<i\leq
d^{2}\\}.$ We have $L^{D}_{\rm PPT}=1-\frac{2}{d+1}$ and $L_{\rm
SIPT}=\frac{3}{4}-\frac{1}{2d}-\frac{1}{2(d+1)}.$ For maximal entangled state,
$L^{D}_{\rm PPT}>L^{D}_{\rm SIPT}$ when $d\geq 3$. Given this fact, we
summarize our lower bounds on $D_{\rm HS}$ as

######  Theorem 2.

The geometric quantum discord is lower-bounded as

$\displaystyle D_{\rm HS}\geq L^{\mathcal{D}}:=\max\\{L^{D}_{\rm
PPT},L^{D}_{\rm SIPT}\\}.$ (13)

By this theorem, one can estimate the GQD by the change of the spectrum of
partially transposed state or the negative spectrum arising from the operation
of partial transpose. In the following, we show that such a lower bound also
induces the estimate of another discord-like quantity, namely, one-way
deficit.

###  IV.2 B. Lower-bound one-way deficit

Quantum work deficit Horodecki et al. (2003); Bera et al. (2017) is introduced
to capture the connection between thermodynamics and information and defined
as the additional extractable information, or work from a bipartite quantum
state when the two parties are in the same place as compared to the cases they
are in distant locations. Denote de-phasing operation on Bob’s side as
$\\{\Pi^{i}_{B}\\}$ and $\varrho_{AB}=\sum_{i}{\rm
I}\otimes\Pi^{i}_{B}\rho_{AB}{\rm I}\otimes\Pi^{i}_{B}$, one-way deficit reads

$\displaystyle\mathcal{WD}_{B}:=\min_{\\{\Pi^{i}_{B}\\}}S(\rho_{AB}\|\varrho_{AB}),$
(14)

where $S(\rho\|\varrho):={\rm Tr}(\rho\ln\rho-\rho\ln\varrho)$ and the
minimization is taken over all projective measurements performed on Bob’s
side. Clearly, $\varrho_{AB}$ is a discord-free state for any
$\\{\Pi^{i}_{B}\\}$. We thus have

$\displaystyle\min_{\\{\Pi^{i}_{B}\\}}S(\rho_{AB}\|\varrho_{AB})$
$\displaystyle\geq$
$\displaystyle\min_{\varrho\in\Omega}S(\rho_{AB}\|\varrho)=S(\rho_{AB}\|\varrho_{\min,re})$
(15) $\displaystyle\geq$
$\displaystyle\frac{\|\rho-\varrho_{\min,re}\|^{2}_{\rm tr}}{2\ln
2}\geq\min_{\varrho^{\prime}\in\Omega}\frac{\|\rho-\varrho^{\prime}\|^{2}_{\rm
tr}}{2\ln 2}$ $\displaystyle=$
$\displaystyle\frac{\|\rho-\varrho_{\min,tr}\|^{2}_{\rm tr}}{2\ln
2}\geq\frac{\|\rho-\varrho_{\min,tr}\|^{2}_{2}}{2\ln 2}$ $\displaystyle\geq$
$\displaystyle\textstyle\frac{L^{\mathcal{D}}}{2\ln 2},$

where $\varrho_{\min,re}$ specifies the closest discord-free state of $\rho$
with respect to the relative entropy, and $\rho_{\min,\rm tr}$ specifies the
closest state with respect to trace-distance, and we have used the quantum
Pinsker inequality $S(\rho\|\varrho)\geq\textstyle\frac{1}{2\ln
2}\|\rho-\varrho\|^{2}_{\rm tr}$ in the second inequality and the norm
inequality $\|\rho-\varrho\|^{2}_{\rm tr}\geq 2\|\rho-\varrho\|^{2}_{2}$ Coles
et al. (2019) in the fourth inequality.

##  V IV. Quantitative estimation of some entanglement measures

Let us move to entanglement theory and estimate two entanglement measures that
can be seen as the counterparts of the above ones in entanglement theory.

The geometric of entanglement, which is defined as the minimal Hilbert-Schmidt
distance between state of interest and the set of separable states specified
by $\mathcal{S}$ Ozawa (2000) as

$\displaystyle E_{\rm
HS}:=\min_{\varrho\in\mathcal{S}}\|\rho-\varrho\|^{2}_{2}.$ (16)

Note that sparable states are PPT and $2-$norm is invariant under partial
transposition. By the same consideration in Eq.(10) as

$\displaystyle E_{\rm HS}$ $\displaystyle=$
$\displaystyle\min_{\varrho\in\mathcal{S}}\|\rho^{\Gamma_{A}}-\varrho^{\Gamma_{A}}\|^{2}_{2}$
(17) $\displaystyle\geq$ $\displaystyle L^{D}_{\rm
PPT}\geq\textstyle\frac{\mathcal{N}^{2}}{N_{+}}+\frac{\mathcal{N}^{2}}{N_{-}},$
$\displaystyle\geq$ $\displaystyle\frac{4\mathcal{N}^{2}}{M\cdot N},$ (18)

where we have used $N_{-}+N_{+}\leq M\cdot N$ in the third inequality. Our
lower bound can be compared with the one reported in Ref Debarba et al. (2012)

$E_{\rm HS}\geq\frac{\mathcal{N}^{2}}{\min\\{(M-1)^{2},(N-1)^{2}\\}}.$

Immediately, when $\frac{MN}{4}\leq\min\\{(M-1)^{2},(N-1)^{2}\\}$, our lower
bound is tighter. As $N_{-}\leq(M-1)(N-1)$ Rana (2013), one example in case is
when $M=N$, $N_{-}\leq(N-1)^{2}=\min\\{(M-1)^{2},(N-1)^{2}\\}$ follows.

For a comparison to the one-way deficit, we now consider the entanglement of
relative entropy. This measure serves as an upper bound for the entanglement
of distillation, namely, the minimal number of singlets that are needed to
build a single copy of the concerned state. By the similar derivation done in
Eq.(15), we have

$\displaystyle E_{\rm
re}:=\min_{\varrho\in\mathcal{S}}S(\rho\|\varrho)\geq\frac{L^{D}_{\rm
PPT}}{2\ln 2}.$ (19)

_Conclusion_ Entanglement and discord are two typical quantum correlations
that are generally studied separably. In this paper, we show that one can use
the primary tool of entanglement theory, namely, the map of partial
transposition, to study discord, and many aspects of the two notions can be
connected in one-to-one correspondence. In contrast to the PPT in entanglement
theory, we show discord can be specified by the change of spectrum of the
density matrix after partial transposition, which leads to a discord criterion
of SIPT. Analogously to the entanglement measure of negativity, the change of
spectrum is shown to imply the estimation of the GQD and one-way deficit. We
also estimate the geometric quantum entanglement and the entanglement of
relative entropy. In this way, we provide not only one perspective to
investigate discord but also, a hierarchical specification and quantitative
estimation of entanglement and discord.

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##  Appendix A Supplemental Material.

We partition the ${\lambda^{\prime}}^{\downarrow}$ into two parts as
${\lambda^{\prime}}^{\downarrow}_{(1)}\bigcup{\lambda^{\prime}}^{\downarrow}_{(2)}$
with
${\lambda^{\prime}}^{\downarrow}_{(1)}=\\{{\lambda^{\prime}}^{\downarrow}_{1},\cdots,{\lambda^{\prime}}^{\downarrow}_{n}\\}$
and
${\lambda^{\prime}}^{\downarrow}_{(2)}=\\{{\lambda^{\prime}}^{\downarrow}_{n+1},\cdots,{\lambda^{\prime}}^{\downarrow}_{N\cdot
M}\\}$ with $n$ specifying the minimal number such that
$\tau:=\frac{{\sum^{n}_{i=1}{\lambda^{\prime}}^{\downarrow}_{i}}-1}{n}\geq{\lambda^{\prime}}^{\downarrow}_{n+1}$.
The minimum value of
$\min_{\lambda_{d}}\|{\lambda^{\prime}}^{\downarrow}-\lambda^{\downarrow}_{d}\|$
is obtained by taking $\lambda^{\downarrow}_{d}=\lambda^{\downarrow}_{\rm
PPT}$ with the element $\lambda^{\downarrow}_{\rm
i,PPT}={\lambda^{\prime}}^{\downarrow}_{i}-\tau$ for $i\leq n$ and $0$
otherwise. Without loss of generality, we assume that
$\lambda^{\downarrow}_{i,PPT}=0$ for $i>N_{+}$ with $N_{+}$ specifying the
number of positive elements of ${\lambda^{\prime}}^{\downarrow}$ as when
$i>N_{+}$ and ${\lambda^{\prime}}^{\downarrow}_{i}<0$, one has
$|{\lambda^{\prime}}^{\downarrow}_{i}-\lambda_{i,PPT}|>|{\lambda^{\prime}}^{\downarrow}_{i}|$,
namely, subtracting a positive element leads to an increase in amplitude, and
assume that $\lambda^{\prime}_{i}\geq\lambda_{i,PPT}$ for $i\geq N_{+}$. The
other elements are given as
${\lambda^{\prime}}^{\downarrow}_{i}-\lambda^{\downarrow}_{i,PPT}=\frac{\sum^{n}_{i}{\lambda^{\prime}}^{\downarrow}_{i}-1}{n}=\tau$,
$\forall$, $1\leq i\leq n$. Otherwise, if another normalized and positive
distribution $\lambda_{\gamma}$ leads to a smaller minimum, for which, we
assume that $\lambda^{\downarrow}_{i,\gamma}=0$ for $i>N_{+}$ and
${\lambda^{\prime}}^{\downarrow}_{i}-\lambda^{\downarrow}_{i,\gamma}\geq 0$
for $i\leq N_{+}$ without losing of generality. As
$\|{\lambda^{\prime}}^{\downarrow}_{i}-{\lambda^{\prime\prime}}^{\downarrow}_{i}\|^{2}_{2}={{\lambda^{\prime}}^{\downarrow}}^{2}_{i}-2\int^{{\lambda^{\prime\prime}}^{\downarrow}_{i}}_{0}(\lambda^{\prime}_{i}-x_{i})dx_{i},$
we have

$\displaystyle\|{\lambda^{\prime}}^{\downarrow}-\lambda^{\downarrow}_{\rm
PPT}\|^{2}_{2}-\|{\lambda^{\prime}}^{\downarrow}-\lambda^{\downarrow}_{\gamma}\|^{2}_{2}$
$\displaystyle=$
$\displaystyle-2\sum_{j|\lambda^{\downarrow}_{j,\gamma}<\lambda^{\downarrow}_{j,PPT}}\int^{\lambda^{\downarrow}_{j,PPT}}_{\lambda^{\downarrow}_{j,PPT}-\kappa_{j}}({\lambda^{\prime}}^{\downarrow}_{j}-x_{j})dx_{j}+2\sum_{i|\lambda^{\downarrow}_{i,PPT}<\lambda^{\downarrow}_{i,\gamma}}\int^{{\lambda^{\downarrow}}_{i,PPT}+{\delta_{i}}}_{\lambda^{\downarrow}_{i,PPT}}({\lambda^{\prime}}^{\downarrow}_{i}-y_{i})dy_{i}.$
(20) $\displaystyle\leq$
$\displaystyle-\tau\sum_{j}\kappa_{j}+\tau\sum_{i}\delta_{i}=0$

where $\delta_{ij}\geq 0$ can always be defined such that, for the case
$\lambda^{\downarrow}_{i,PPT}\geq\lambda^{\downarrow}_{i,\gamma}$,
$\kappa_{i}=\lambda^{\downarrow}_{i,PPT}-\lambda^{\downarrow}_{i,\gamma},$ and
for the case
$\lambda^{\downarrow}_{i,PPT}\leq\lambda^{\downarrow}_{i,\gamma}$,
$\delta_{i}=\lambda^{\downarrow}_{i,\gamma}-\lambda^{\downarrow}_{i,PPT}.$ In
the interval of integration, we have
${\lambda^{\prime}}^{\downarrow}_{j}-x_{j}\geq{\lambda^{\prime}}^{\downarrow}_{j}-\lambda_{j,PPT}\geq\tau$
and
${\lambda^{\prime}}^{\downarrow}_{j}-y_{j}\leq{\lambda^{\prime}}^{\downarrow}_{j}-\lambda^{\downarrow}_{j,PPT}\leq\tau,$
and $\sum_{i}\kappa_{i}=\sum_{j}\delta_{j}$. We have completed the proof for
that $\lambda_{\rm PPT}$ is the spectrum achieving the minimum.

To show that $L_{\rm
PPT}\geq\frac{\mathcal{N}^{2}}{N_{+}}+\frac{\mathcal{N}^{2}}{N_{-}}$, we
consider that another kind of
$\lambda^{\downarrow}_{\gamma^{\prime}}=\lambda^{\downarrow}_{\gamma^{\prime},(1)}\bigcup\lambda^{\downarrow}_{\gamma^{\prime},(2)}$
with
$\lambda^{\downarrow}_{\gamma^{\prime},(1)}=\\{\lambda^{\downarrow}_{1,\gamma^{\prime}},\cdots,\lambda^{\downarrow}_{N_{+},\gamma^{\prime}}\\}$
and
$\lambda^{\downarrow}_{\gamma^{\prime},(2)}=\\{\lambda^{\downarrow}_{N_{+}+1,\gamma^{\prime}},\cdots,\lambda^{\downarrow}_{MN,\gamma^{\prime}}\\}$
with $N_{+}$ specifying the number of positive elements in $\lambda^{\prime}$.
We assume also that $\lambda^{\downarrow}_{i,\gamma^{\prime}}=0$ for
$i>N_{+}$, however, other elements are not necessary to be positive. This
setting, involving looser constraint than that for $\lambda_{d}$, should lead
to a smaller lower-bound. As
$\min_{\lambda_{\gamma^{\prime}}}\|{\lambda^{\prime}}^{\downarrow}_{(1)}-\lambda^{\downarrow}_{\gamma^{\prime},(1)}\|^{2}_{2}$
is only subject to the constraint that $\sum_{i\leq
N_{+}}({\lambda^{\prime}}^{\downarrow}_{i}-\lambda^{\downarrow}_{i,\gamma^{\prime}})=\mathcal{N}$
and $2-$norm is convex, the minimum is obtained as
$\frac{\mathcal{N}^{2}}{N_{+}}$ when
${\lambda^{\prime}}^{\downarrow}_{i}-\lambda^{\downarrow}_{i,\gamma^{\prime}}=\frac{\mathcal{N}}{N_{+}}$
for $1\leq i\leq N_{+}$. Consider the negative terms of
${\lambda^{\prime}}^{\downarrow}_{(2)}$, one has
$\sum_{i>N_{+}}{\lambda^{\prime}}^{\downarrow}_{i}=-\mathcal{N}$,
$\|{\lambda^{\prime}}^{\downarrow}_{(2)}\|\geq\frac{\mathcal{N}^{2}}{N_{-}}$
follows due to the convexity of $2-$ norm, which is saturated when
${\lambda^{\prime}}^{\downarrow}_{i}=\frac{\mathcal{N}}{N_{-}}$ for $i>N_{+}$
with $N_{-}$ specifying the number of negative terms of $\lambda^{\prime}$.
Therefore, we have

$L^{D}_{\rm
PPT}\geq\frac{\mathcal{N}^{2}}{N_{+}}+\frac{\mathcal{N}^{2}}{N_{-}}.$


# Introducing DASEP: the doubly asymmetric simple exclusion process

David W. Ash

(Date: July 30, 2022)

###### Abstract.

Research in combinatorics has often explored the asymmetric simple exclusion
process (ASEP). The ASEP, inspired by examples from statistical mechanics,
involves particles of various species moving around a lattice. With the
traditional ASEP particles of a given species can move but do not change
species. In this paper a new combinatorial formalism, the DASEP (doubly
asymmetric simple exclusion process), is explored. The DASEP is inspired by
biological processes where, unlike the ASEP, the particles can change from one
species to another. The combinatorics of the DASEP on a one dimensional
lattice are explored, including the associated generating function. The
stationary probabilities of the DASEP are explored, and results are proven
relating these stationary probabilities to those of the simpler ASEP.

The ASEP (asymmetric simple exclusion process) is a structure that has
frequently been referred to in the combinatorics literature. In its simplest
form, the ASEP consists of a one dimensional infinite lattice, with each point
on the lattice being populated with either a particle or a hole. At random
intervals, each particle attempts to move either to the left or the right with
different but fixed probabilites (hence the term ‘asymmetric’). The ASEP can
be thought of as a form of Markov process as noted in [4] by Corteel et al.
Multiline queues [5] were introduced by Ferrari et al. as a combinatorial
approach to the analysis of the ASEP. Originally the ASEP particles were
thought of as all belonging to a single species. More recent work by Cantini
et al. [3] generalized the concept to multiple species and uncovered a link
with Macdonald polynomials. Although we focus on the homogeneous ASEP
(transition probabilities do not depend on position in the lattice), several
researchers (Lam et al. [7], Ayyer et al. [1], Cantini [2], Mandelshtam [9],
and Kim et al. [6]) have explored the inhomogeneous ASEP in which transition
probabilities do depend on lattice position.

##  1\. Definitions

Following [4], a partition can be defined as follows:

######  Definition 1.1.

A partition $\lambda$ is a nonincreasing sequence of $n$ nonnegative integers
$\lambda=(\lambda_{1}\geq\lambda_{2}\geq...\geq\lambda_{n}\geq 0)$.

We will start by working through a simple example of the ASEP before
introducing the new concept of the DASEP. We will ordinarily write a partition
as defined above as an $n$-tuple:
$\lambda=(\lambda_{1},\lambda_{2},...,\lambda_{n})$.

######  Definition 1.2.

We write $S_{n}(\lambda)$ to mean the set of all permutations of $\lambda$.

######  Example 1.3.

So, for $\lambda=(2,2,1)$, $S_{3}(\lambda)=\\{(2,2,1),(2,1,2),(1,2,2)\\}$.

The multispecies asymmetric simple exclusion process ASEP($\lambda$) is then
defined to be a Markov process on $S_{n}(\lambda)$ with certain specific
transition probabilities:

######  Definition 1.4.

For all partitions $\lambda$ as defined in Definition 1.1, ASEP($\lambda$) is
a Markov process on $S_{n}(\lambda)$. We let $t$ be a constant with $0\leq
t\leq 1$. Suppose we have two permutations $\mu\in S_{n}(\lambda)$ and $\nu\in
S_{n}(\lambda)$. The transition probability, $P_{\mu,\nu}$, is then given by:

  * •

If $\mu=(\mu_{1},\mu_{2},...,\mu_{k},i,j,\mu_{k+2},\mu_{k+3},...,\mu_{n})$ and
$\nu=(\mu_{1},\mu_{2},...,\mu_{k},j,i,\mu_{k+2},\mu_{k+3},...,\mu_{n})$ with
$i\neq j$, then $P_{\mu,\nu}=\frac{t}{n}$ if $i>j$ and
$P_{\mu,\nu}=\frac{1}{n}$ if $j>i$.

  * •

If $\mu=(i,\mu_{2},\mu_{3},...,\mu_{n-1},j)$ and
$\nu=(j,\mu_{2},\mu_{3},...,\mu_{n-1},i)$ with $i\neq j$, then
$P_{\mu,\nu}=\frac{t}{n}$ if $j>i$ and $P_{\mu,\nu}=\frac{1}{n}$ if $i>j$.

  * •

If neither of the above conditions apply but $\nu\neq\mu$ then
$P_{\mu,\nu}=0$. If $\nu=\mu$ then
$P_{\mu,\mu}=1-\sum_{\nu\neq\mu}P_{\mu,\nu}$.

It is possible to compute steady state probabilities for ASEP($\lambda$). For
the purposes of the example that we will develop as we introduce DASEP, we are
primarily interested in ASEP($\lambda$) for $\lambda=(2,2,0)$,
$\lambda=(2,1,0)$, and $\lambda=(1,1,0)$, so we will focus mostly on these
three processes as we work through the computation of the steady state
probabilities. Continuing to follow [4] as we develop this example, to compute
these probabilities we need to define the concept of a multiline queue.

######  Definition 1.5.

A ball system $B$ is an $L\times n$ matrix each element of which is either $0$
or $1$. Moreover for all $i$ the number of $1$’s in row $i+1$ is less than or
equal to the number of $1$’s in row $i$.

######  Definition 1.6.

Given a ball system $B$ a multiline queue $Q$ is obtained by augmenting $B$
with a labeling and matching system. Each cell in $B$ will be labelled with a
number from $0$ to $L$ inclusive, and each cell with a $1$ element in row
$i+1$, for $i\geq 1$, will be matched to a cell with a $1$ element in row $i$.
Such a matching must be obtained through an application of the following
algorithm:

  * •

Step 1: Find the highest numbered row with unlabelled $1$ elements. Label each
of those elements with the number of the row. If this is row $1$, or there are
no remaining unlabelled $1$ elements in the matrix, exit.

  * •

Step 2: Find the row with labelled but unmatched elements. If this is row $1$,
go back to step 1. If it is row $i+1$, for $i\geq 1$, first match each
labelled but unmatched element that can be matched to an unlabelled element
directly below it to that element. This is considered a trivial match. Then
proceed from right to left (highest to lowest numbered columns) matching each
remaining labelled but unmatched element to an unlabelled element in the row
below–these are the nontrivial matches. Give all newly matched elements in row
$i$ the same label as the element it has just been matched to. Repeat step 2.

A multiline queue is often visualized as a ball system with an element with a
$1$ value being shown as a ball and a $0$ value by the absence of a ball.
Matches between elements (balls) are drawn by lines between the matched balls.
The following shows a multiline queue associated with ASEP($\lambda$) where
$\lambda=(2,2,0)$. Note that the line matching the ball at upper right to the
one at the lower middle wraps around to the right.

2222

The labels in the bottom row determine the partition of the associated ASEP.
The above multiline queue has $\lambda=(2,2,0)$ since the bottom row includes
two $2$’s and a $0$–by convention an element without a ball is assumed to be
labeled with a $0$. Likewise the following would be a multiline queue with
$\lambda=(2,1,0)$:

122

Each multiline queue is also associated with a permutation $\alpha\in
S_{n}(\lambda)$ corresponding to the labels of its bottom row in unsorted
order. For example, for the above multiline queue, $\lambda=(2,1,0)$ but
$\alpha=(0,1,2)$. We will write $\lambda(Q)=\lambda$ and $\alpha(Q)=\alpha$.

##  2\. Steady state probabilities with example

To determine steady state probabilities–and continue with the example started
in the introduction–we next assign to each nontrivial matching $p$ in $Q$ two
values $f(p)$ and $s(p)$. $f(p)$ is the number of choices that were available
for the match when the match was made. $s(p)$ is the number of legal matches
that were skipped, if we imagine ourselves considering possible matches from
left to right and wrapping around the end if needed, before the actual choice
was made. We can then define a weight on $p$ as
$\textnormal{wt}(p)=\frac{(1-t)t^{s(p)}}{1-t^{f(p)}}$. Here we are proceeding
from [4] but with the simplifying assumption that $q=1$, since in the sequel
we will rely on a theorem that requires $q=1$. Next we can define a weight on
the entire multiline queue $\textnormal{wt}(Q)=\prod_{p\in
Q}\textnormal{wt}(p)$ where the product is taken over all nontrivial matches
$p$ in $Q$. A theorem due to Martin [10] then gives the required steady state
probabilities:

$\textnormal{Pr}(\alpha)=\frac{\sum_{\alpha(Q)=\alpha}\textnormal{wt}(Q)}{\sum_{\lambda(Q)=\lambda}\textnormal{wt}(Q)}$

Before moving on to the DASEP, we need to evaluate the steady state
probabilities for the examples that we will ultimately use to develop the
DASEP. For the above multiline queue, there is exactly one nontrivial pair
$p$. When this pair is matched, there are two available options so $f(p)=2$.
As we picked the second available option, $s(p)=1$. So
$\textnormal{wt}(Q)=\frac{(1-t)t}{1-t^{2}}$. As noted above, $\alpha=(0,1,2)$
and the only other multiline queue with $\alpha=(0,1,2)$ is as follows:

122

Here there is no nontrivial matching pair, so $\textnormal{wt}(Q)=1$. Hence:

$\sum_{\alpha(Q)=(0,1,2)}\textnormal{wt}(Q)=1+\frac{(1-t)t}{1-t^{2}}=\frac{1+2t}{1+t}$

For reasons of symmetry:

$\sum_{\alpha(Q)=(0,1,2)}\textnormal{wt}(Q)=\sum_{\alpha(Q)=(1,2,0)}\textnormal{wt}(Q)=\sum_{\alpha(Q)=(2,0,1)}\textnormal{wt}(Q)=\frac{1+2t}{1+t}$

Next we look at $\alpha=(0,1,2)$, for which there are also two multiline
queues. The first of these is as follows:

122

Here there is one nontrivial matching pair $p$. When this pair is matched,
there are two available options so $f(p)=2$. As we picked the first available
option, $s(p)=0$. So $\textnormal{wt}(Q)=\frac{1-t}{1-t^{2}}$. The other
multiline queue with $\alpha=(2,1,0)$ is as follows:

122

Again there is no nontrivial matching pair, so $\textnormal{wt}(Q)=1$. Hence:

$\sum_{\alpha(Q)=(2,1,0)}\textnormal{wt}(Q)=1+\frac{1-t}{1-t^{2}}=\frac{2+t}{1+t}$

For reasons of symmetry:

$\sum_{\alpha(Q)=(2,1,0)}\textnormal{wt}(Q)=\sum_{\alpha(Q)=(1,0,2)}\textnormal{wt}(Q)=\sum_{\alpha(Q)=(0,2,1)}\textnormal{wt}(Q)=\frac{2+t}{1+t}$

So:

$\sum_{\lambda(Q)=(2,1,0)}\textnormal{wt}(Q)=3(\frac{1+2t}{1+t})+3(\frac{2+t}{1+t})=9$

We are now ready to give the steady state probabilities:

$\textnormal{Pr}(0,1,2)=\textnormal{Pr}(1,2,0)=\textnormal{Pr}(2,0,1)=\frac{1+2t}{9(1+t)}$

and:

$\textnormal{Pr}(2,1,0)=\textnormal{Pr}(1,0,2)=\textnormal{Pr}(0,2,1)=\frac{2+t}{9(1+t)}$

More trivially–we won’t work through the details–we can also give:

$\textnormal{Pr}(0,1,1)=\textnormal{Pr}(1,1,0)=\textnormal{Pr}(1,0,1)=\frac{1}{3}$

and:

$\textnormal{Pr}(0,2,2)=\textnormal{Pr}(2,2,0)=\textnormal{Pr}(2,0,2)=\frac{1}{3}$

##  3\. Doubly asymmetric simple exclusion process

We are now ready to introduce the DASEP (doubly asymmetric simple exclusion
process). As previously noted, the ASEP is inspired by statistical mechanics
where particles do not change species. The DASEP, by contrast, is inspired by
biological processes where particles can change species. We will often refer
to DASEP$(n,p,q)$ where $n$ is the number of positions on the lattice, $p$ is
the number of types of species, and $q$ is the number of particles.

######  Definition 3.1.

For all positive integers $n$, $p$, and $q$ with $n>q$, DASEP$(n,p,q)$ is a
Markov process on the following set:

$\bigcup_{\begin{subarray}{c}\lambda_{1}\leq p\\\
\lambda^{\prime}_{1}=q\end{subarray}}S_{n}(\lambda)$

Here the lower condition, $\lambda^{\prime}_{1}=q$, refers to the dual
partition [8] of $\lambda$, namely $\lambda^{\prime}$, and uses the fact that
$\lambda^{\prime}_{1}$ gives the number of nonzero terms in the original
partition $\lambda$. The transition probability $P_{\mu,\nu}$ on two
permutations $\mu$ and $\nu$ is as follows:

  * •

If $\mu=(\mu_{1},\mu_{2},...,\mu_{k},i,j,\mu_{k+2},\mu_{k+3},...,\mu_{n})$ and
$\nu=(\mu_{1},\mu_{2},...,\mu_{k},j,i,\mu_{k+2},\mu_{k+3},...,\mu_{n})$ with
$i\neq j$, then $P_{\mu,\nu}=\frac{t}{3n}$ if $i>j$ and
$P_{\mu,\nu}=\frac{1}{3n}$ if $j>i$.

  * •

If $\mu=(i,\mu_{2},\mu_{3},...,\mu_{n-1},j)$ and
$\nu=(j,\mu_{2},\mu_{3},...,\mu_{n-1},i)$ with $i\neq j$, then
$P_{\mu,\nu}=\frac{t}{3n}$ if $j>i$ and $P_{\mu,\nu}=\frac{1}{3n}$ if $i>j$.

  * •

If $\mu=(\mu_{1},\mu_{2},...,\mu_{k},i,\mu_{k+2},\mu_{k+3},...,\mu_{n})$ and
$\nu=(\mu_{1},\mu_{2},...,\mu_{k},i+1,\mu_{k+2},\mu_{k+3},...,\mu_{n})$ with
$i\geq 1$, then $P_{\mu,\nu}=\frac{u}{3n}$.

  * •

If $\mu=(\mu_{1},\mu_{2},...,\mu_{k},i+1,\mu_{k+2},\mu_{k+3},...,\mu_{n})$ and
$\nu=(\mu_{1},\mu_{2},...,\mu_{k},i,\mu_{k+2},\mu_{k+3},...,\mu_{n})$ with
$i\geq 1$, then $P_{\mu,\nu}=\frac{1}{3n}$.

  * •

If none of the above conditions apply but $\nu\neq\mu$ then $P_{\mu,\nu}=0$.
If $\nu=\mu$ then $P_{\mu,\mu}=1-\sum_{\nu\neq\mu}P_{\mu,\nu}$.

Figure 1 shows the simple example of the DASEP that we are working through.
Each of the $12$ small triangles represents one of the possible states of the
DASEP in the circular lattice with $3$ sites with two balls each of which can
be of species $1$ or $2$. Each possible state corresponds to a permutation in
either ASEP$(2,2,0)$, ASEP$(2,1,0)$, or ASEP$(1,1,0)$. All possible
transitions within a single ASEP (the first and second bullet points in the
definition above) are shown with blue arrows on this diagram. To keep the
diagram relatively clean in appearance, only selected transitions between
different ASEPs (the third and fourth bullet points) are shown, and these are
shown with red arrows. Other ASEPs such as ASEP$(1,0,0)$ or ASEP$(2,0,0)$ are
not show because these are not part of DASEP$(3,2,2)$. This is because, per
Definition 3.1, for DASEP$(3,2,2)$ we always have $\lambda^{\prime}_{1}=2$,
whereas for ASEP$(1,0,0)$ and ASEP$(2,0,0)$, we have $\lambda^{\prime}_{1}=1$.

Similar to with the ASEP, with the DASEP we wish to compute steady state
probabilities for permutations $\alpha$ which we will call
$\textnormal{Pd}(\alpha)$. We will focus on continuing to develop the example
we’ve been working on which turns out to be DASEP$(3,2,2)$. Here $n=3$ means
that the particles move on the the circular lattice with $3$ sites, $p=2$
means that each particle is allowed to take on the value $0$, $1$, or $2$, and
$q=2$ means that each permutation $\alpha$ has exactly $2$ nonzero values. We
therefore find ourselves interested in the following $12$ steady state
probabilities:

$\displaystyle{\definecolor[named]{.}{rgb}{0,0,0}\color[rgb]{0,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@gray@stroke{0}\pgfsys@color@gray@fill{0}\textnormal{ASEP}(2,1,0)}$$\displaystyle{\definecolor[named]{.}{rgb}{0,0,0}\color[rgb]{0,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@gray@stroke{0}\pgfsys@color@gray@fill{0}\textnormal{ASEP}(1,1,0)}$$\displaystyle{\definecolor[named]{.}{rgb}{0,0,0}\color[rgb]{0,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@gray@stroke{0}\pgfsys@color@gray@fill{0}\textnormal{ASEP}(2,2,0)}$022202220011101110012021102201120210

Figure 1: An example of the DASEP: DASEP(3,2,2)

$\textnormal{Pd}(0,1,1),\textnormal{Pd}(0,1,2),\textnormal{Pd}(0,2,1),\textnormal{Pd}(0,2,2),\textnormal{Pd}(1,0,1),\textnormal{Pd}(1,0,2)$
$\textnormal{Pd}(1,1,0),\textnormal{Pd}(1,2,0),\textnormal{Pd}(2,0,1),\textnormal{Pd}(2,0,2),\textnormal{Pd}(2,1,0),\textnormal{Pd}(2,2,0)$

Note here that particles in the DASEP are allowed to switch back and forth
between species $1$ and $2$, but not back and forth from $0$ to anything else.
That is because a value of $0$ is understood to not so much be a species but
the absence of a species. Due to symmetries we can now focus on solving for
the following four probabilities:

$w=\textnormal{Pd}(0,1,1),x=\textnormal{Pd}(0,1,2),y=\textnormal{Pd}(0,2,1),z=\textnormal{Pd}(0,2,2)$

From the above transition probabilities, this reduces to solving the following
four equations:

$2uw=x+y$ $(2+t)x+x+ux=(1+2t)y+z+uw$ $(1+2t)y+y+uy=(2+t)x+uw+z$ $2z=u(x+y)$

Multiplying the second equation by $2$ gives:

$(6+2t+2u)x=(2+4t)y+2z+2uw$

We can then substitute in the first and fourth equations to give:

$(6+2t+2u)x=(2+4t)y+(u+1)(x+y)$

or

$(5+2t+u)x=(3+4t+u)y$

We can then ask ourselves the question of when the proportions of steady state
probabilities for the DASEP are the same as for the previous ASEP. Noting that
$\textnormal{Pr}(0,1,2)=\frac{1+2t}{9(1+t)}$ and
$\textnormal{Pr}(2,1,0)=\frac{2+t}{9(1+t)}$ such equality will happen if:

$(5+2t+u)(1+2t)=(3+4t+u)(2+t)$

or

$5+2t+u+10t+4t^{2}+2tu=6+8t+2u+3t+4t^{2}+tu$

or

$t(1+u)=1+u$

This will happen iff $t=1$. We have therefore proved the following theorem:

######  Theorem 3.2.

If $D=\textnormal{DASEP}(3,2,2)$ is parameterized as described above by $t$
and $u$, then the following two statements are equivalent:

  * •

$t=1$

  * •

For all partitions $\lambda$ with $S_{n}(\lambda)\subseteq D$ and all
permutations $\mu$, $\nu\in S_{n}(\lambda)$ the following equality holds:
$\frac{\textnormal{Pr}(\mu)}{\textnormal{Pr}(\nu)}=\frac{\textnormal{Pd}(\mu)}{\textnormal{Pd}(\nu)}$.
That is, the ratio between steady state probabilities does not change in
moving from the ASEP to the DASEP.

As a result we wish to conjecture the following more general result:

######  Conjecture 3.3.

If $D=\textnormal{DASEP}(n,p,q)$ is parameterized as described above by $t$
and $u$, then the following two statements are equivalent:

  * •

$t=1$

  * •

For all partitions $\lambda$ with $S_{n}(\lambda)\subseteq D$ and all
permutations $\mu$, $\nu\in S_{n}(\lambda)$ the following equality holds:
$\frac{\textnormal{Pr}(\mu)}{\textnormal{Pr}(\nu)}=\frac{\textnormal{Pd}(\mu)}{\textnormal{Pd}(\nu)}$.
That is, the ratio between steady state probabilities does not change in
moving from the ASEP to the DASEP.

Partial Proof: We will prove this only in the $\implies$ direction. If $t=1$
we can replace $\lambda$ with a similar partition but with species of the same
type being replaced by similar distinct species. For example, if
$\lambda=(3,3,3,2,1,0,...)$ we would map this to
$\hat{\lambda}=(3_{1},3_{2},3_{3},2,1,0,...)$ and allow adjacent species
originally of the same type to be exchanged with the same transition
probability. This will create a completely symmetric situation, so all steady
state probabilities are equal. As an equal number of $\hat{\lambda}$’s can be
derived from each $\lambda$ this means all original steady state probabilities
are equal as well, so
$\frac{\textnormal{Pr}(\mu)}{\textnormal{Pr}(\nu)}=\frac{\textnormal{Pd}(\mu)}{\textnormal{Pd}(\nu)}=1$.
This completes the proof in the $\implies$ direction.

##  4\. Another specific case

We focus first on proving another specific case of the theorem for
DASEP$(3,3,2)$. Following are the nine values we must solve for to prove the
theorem for DASEP$(3,3,2)$:

$a_{1}=\textnormal{Pd}(0,1,1)$ $a_{2}=\textnormal{Pd}(0,2,2)$
$a_{3}=\textnormal{Pd}(0,3,3)$ $b_{1}=\textnormal{Pd}(0,2,3)$
$c_{1}=\textnormal{Pd}(0,3,2)$ $b_{2}=\textnormal{Pd}(0,1,3)$
$c_{2}=\textnormal{Pd}(0,3,1)$ $b_{3}=\textnormal{Pd}(0,1,2)$
$c_{3}=\textnormal{Pd}(0,2,1)$

These values can be obtained by solving the following set of nine equations:

$2ua_{1}=b_{3}+c_{3}$
$(2+t)b_{3}+ub_{3}+ub_{3}+b_{3}=(1+2t)c_{3}+b_{2}+a_{2}+ua_{1}$
$(1+2t)c_{3}+uc_{3}+uc_{3}+c_{3}=(2+t)b_{3}+c_{2}+a_{2}+ua_{1}$
$a_{2}+a_{2}+ua_{2}+ua_{2}=b_{1}+c_{1}+ub_{3}+uc_{3}$
$(2+t)b_{2}+ub_{2}+b_{2}=(1+2t)c_{2}+b_{1}+ub_{3}$
$(1+2t)c_{2}+uc_{2}+c_{2}=(2+t)b_{2}+c_{1}+uc_{3}$ $2a_{3}=ub_{1}+uc_{1}$
$(2+t)b_{1}+ub_{1}+b_{1}+b_{1}=(1+2t)c_{1}+a_{3}+ub_{2}+ua_{2}$
$(1+2t)c_{1}+uc_{1}+c_{1}+c_{1}=(2+t)b_{1}+a_{3}+ua_{2}+uc_{2}$

We can simplify (4) as follows:

$2(u+1)a_{2}=b_{1}+c_{1}+u(b_{3}+c_{3})$

We then can multiply (2) by $2(u+1)$ and then substitute in (1) and (10):

$(3+t+2u)b_{3}=(1+2t)c_{3}+b_{2}+a_{2}+ua_{1}$

or

$2(u+1)(3+t+2u)b_{3}=2(u+1)(1+2t)c_{3}+2(u+1)b_{2}+2(u+1)a_{2}+2u(u+1)a_{1}$

or

(13)
$\begin{split}(4u^{2}+2tu+2t+10u+6)b_{3}&=(4tu+4t+2u+2)c_{3}+2(u+1)b_{2}\\\
&+b_{1}+c_{1}+u(b_{3}+c_{3})+(u+1)(b_{3}+c_{3})\end{split}$

or

$(4u^{2}+2tu+2t+8u+5)b_{3}=(4tu+4t+4u+3)c_{3}+2(u+1)b_{2}+b_{1}+c_{1}$

Next we can add (5) and (6) to give:

$(3+t+u)b_{2}+(2+2t+u)c_{2}=(2+t)b_{2}+(1+2t)c_{2}+(b_{1}+c_{1})+u(b_{3}+c_{3})$

Rearranging terms and simplifying gives:

$b_{1}+c_{1}=(1+u)(b_{2}+c_{2})-u(b_{3}+c_{3})$

We can now substitute (16) into (14) to give:

(17) $\begin{split}&(4u^{2}+2tu+2t+8u+5)b_{3}\\\
&=(4tu+4t+4u+3)c_{3}+2(u+1)b_{2}+(1+u)(b_{2}+c_{2})-u(b_{3}+c_{3})\end{split}$

Collecting terms and rearranging:

(18) $\begin{split}&(4u^{2}+2tu+2t+9u+5)b_{3}\\\
&=(4tu+4t+3u+3)c_{3}+3(u+1)b_{2}+(u+1)c_{2}\end{split}$

We can factor out the $u+1$ to give us:

(19) $\begin{split}&(4u+2t+5)b_{3}=(4t+3)c_{3}+3b_{2}+c_{2}\end{split}$

We next multiply (9) by $2(u+1)$ and simplify, giving:

(20) $\begin{split}&2(u+1)(u+2t+3)c_{1}\\\
&=2(u+1)(2+t)b_{1}+2(u+1)a_{3}+2u(u+1)a_{2}+2u(u+1)c_{2}\end{split}$

Next substitute in (4) and (7) to get:

(21) $\begin{split}&2(u+1)(u+2t+3)c_{1}\\\
&=2(u+1)(2+t)b_{1}+u(u+1)(b_{1}+c_{1})+u(b_{1}+c_{1}+ub_{3}+uc_{3})+2u(u+1)c_{2}\end{split}$

Expanding and rearranging terms gives us:

(22) $\begin{split}&(u^{2}+4tu+4t+6u+6)c_{1}\\\
&=(u^{2}+2tu+2t+6u+4)b_{1}+2u(u+1)c_{2}+u^{2}b_{3}+u^{2}c_{3}\end{split}$

We next observe that (5) can be rewritten as:

(23) $\begin{split}&b_{1}=(3+t+u)b_{2}-(1+2t)c_{2}-ub_{3}\end{split}$

Likewise (6) can be written as:

(24) $\begin{split}&c_{1}=(2+2t+u)c_{2}-(2+t)b_{2}-uc_{3}\end{split}$

Substitute (23) and (24) into (22) to give:

(25) $\begin{split}&(u^{2}+4tu+4t+6+6)([2+2t+u]c_{2}-[2+t]b_{2}-uc_{3})\\\
&=(u^{2}+2tu+2t+6u+4)([3+t+u]b_{2}-[1+2t]c_{2}-ub_{3})\\\
&+2u(u+1)c_{2}+u^{2}b_{3}+u^{2}c_{3}\end{split}$

This can be expanded as follows:

(26)
$\begin{split}&(u^{3}+8tu^{2}+12t^{2}u+7u^{2}+38tu+12t^{2}+22u+30t+16)c_{2}\\\
&=(u^{3}+4tu^{2}+6t^{2}u+11u^{2}+28tu+6t^{2}+34u+24t+24)b_{2}\\\
&+(u^{3}+4tu^{2}+4tu+7u^{2}+6u)c_{3}-(u^{3}+2tu^{2}+2tu+5u^{2}+4u)b_{3}\end{split}$

A $u+1$ can be factored out to give:

(27)
$\begin{split}(u^{2}+8tu+12t^{2}+6u+30t+16)c_{2}&=(u^{2}+4tu+6t^{2}+10u+24t+24)b_{2}\\\
&+(u^{2}+4tu+6u)c_{3}-(u^{2}+2tu+4u)b_{3}\end{split}$

Similar to what we did with (9), we next multiply (8) by $2(u+1)$ and
simplify, giving:

(28) $\begin{split}&2(u+1)(4+t+u)b_{1}\\\
&=2(u+1)(1+2t)c_{1}+2(u+1)a_{3}+2u(u+1)a_{2}+2u(u+1)b_{2}\end{split}$

Again substitute in (4) and (7) to get:

(29) $\begin{split}&2(u+1)(4+t+u)b_{1}\\\
&=2(u+1)(1+2t)c_{1}+u(u+1)(b_{1}+c_{1})\\\
&+u(b_{1}+c_{1}+ub_{3}+uc_{3})+2u(u+1)b_{2}\end{split}$

Expanding and rearranging terms gives us:

(30) $\begin{split}&(u^{2}+2tu+2t+8u+8)b_{1}\\\
&=(u^{2}+4tu+4t+4u+2)c_{1}+2u(u+1)b_{2}+u^{2}b_{3}+u^{2}c_{3}\end{split}$

Substitute (23) and (24) into (30) to give:

(31) $\begin{split}&(u^{2}+2tu+2t+8u+8)([3+t+u]b_{2}-[1+2t]c_{2}-ub_{3})\\\
&=(u^{2}+4tu+4t+4u+2)([2+2t+u]c_{2}-[2+t]b_{2}-uc_{3})\\\
&+2u(u+1)b_{2}+u^{2}b_{3}+u^{2}c_{3}\end{split}$

This can be expanded as follows:

(32)
$\begin{split}&(u^{3}+4tu^{2}+6t^{2}u+11u^{2}+28tu+6t^{2}+38u+24t+28)b_{2}\\\
&=(u^{3}+8tu^{2}+12t^{2}u+7u^{2}+38tu+12t^{2}+18u+30t+12)c_{2}\\\
&+(u^{3}+2tu^{2}+2tu+9u^{2}+8u)b_{3}-(u^{3}+4tu^{2}+4tu+3u^{2}+2u)c_{3}\end{split}$

A $u+1$ can be factored out to give:

(33)
$\begin{split}(u^{2}+4tu+6t^{2}+10u+24t+28)b_{2}&=(u^{2}+8tu+12t^{2}+6u+30t+12)c_{2}\\\
&+(u^{2}+2tu+8u)b_{3}-(u^{2}+4tu+2u)c_{3}\end{split}$

Taking the difference of (27) and (33) and simplifying gives:

(34) $\begin{split}&b_{2}+c_{2}=u(b_{3}+c_{3})\end{split}$

Next (19) can be rearranged to give:

(35) $\begin{split}&3b_{2}+c_{2}=(4u+2t+5)b_{3}-(4t+3)c_{3}\end{split}$

We can solve (34) and (35) to get:

(36) $\begin{split}&2c_{2}=(4t+3u+3)c_{3}-(u+2t+5)b_{3}\end{split}$

(37) $\begin{split}&2b_{2}=(3u+2t+5)b_{3}-(u+4t+3)c_{3}\end{split}$

We can multiply (27) by $2$ and then substitute in (36) and (37):

(38)
$\begin{split}&(u^{2}+8tu+12t^{2}+6u+30t+16)([4t+3u+3]c_{3}-[u+2t+5]b_{3})\\\
&=(u^{2}+4tu+6t^{2}+10u+24t+24)([3u+2t+5]b_{3}-[u+4t+3]c_{3})\\\
&+2(u^{2}+4tu+6u)c_{3}-2(u^{2}+2tu+4u)b_{3}\end{split}$

This can be expanded to give:

(39)
$\begin{split}&(4u^{3}+36u^{2}t+90ut^{2}+72t^{3}+32u^{2}+206ut+270t^{2}+108u+322t+120)c_{3}\\\
&=(4u^{3}+24u^{2}t+54ut^{2}+36t^{3}+44u^{2}+190ut+198t^{2}+160u+350t+200)b_{3}\end{split}$

As previously discussed, $\textnormal{Pr}(0,1,2)=\frac{1+2t}{9(1+t)}$ and
$\textnormal{Pr}(2,1,0)=\frac{2+t}{9(1+t)}$, so for
$b_{3}=\textnormal{Pd}(0,1,2)$ and $c_{3}=\textnormal{Pd}(2,1,0)$ to be in the
same ratio we would require $b_{3}=k(1+2t)$ and $c_{3}=k(2+t)$ for some $k$.
It follows, after also dividing through by $2$, that:

(40) $\begin{split}&(2u^{3}+18u^{2}t+45ut^{2}+36t^{3}+16u^{2}+103ut\\\
&+135t^{2}+54u+161t+60)(t+2)\\\
&=(2u^{3}+12u^{2}t+27ut^{2}+18t^{3}+22u^{2}+95ut\\\
&+99t^{2}+80u+175t+100)(2t+1)\end{split}$

This can be expanded to:

(41)
$\begin{split}&2u^{3}t+18u^{2}t^{2}+45ut^{3}+36t^{4}+4u^{3}+52u^{2}t+193ut^{2}\\\
&+207t^{3}+32u^{2}+260ut+431t^{2}+108u+382t+120\\\
&=4u^{3}t+24u^{2}t^{2}+54ut^{3}+36t^{4}+2u^{3}+56u^{2}t+217ut^{2}\\\
&+216t^{3}+22u^{2}+255ut+449t^{2}+80u+375t+100\end{split}$

This can be reduced to:

(42) $\begin{split}&2u^{3}t+6u^{2}t^{2}+9ut^{3}-2u^{3}+4u^{2}t+24ut^{2}\\\
&+9t^{3}-10u^{2}-5ut+18t^{2}-28u-7t-20=0\end{split}$

This can be factored as:

(43)
$\begin{split}&(t-1)(2u^{3}+6u^{2}t+9ut^{2}+10u^{2}+33ut+9t^{2}+28u+27t+20)=0\end{split}$

Since $u\geq 0$ and $t\geq 0$, it follows that $t=1$. This completes the proof
in the $\impliedby$ direction for the DASEP$(3,3,2)$ case.

##  5\. More general results

To approach the more general theorem listed in Section 3, we first explore
DASEP$(3,p,2)$ having proven the result above for $p=2$ and $p=3$. To solve
this case we essentially need to solve for each of $p^{2}$ prior probabilities
$p_{i,j}=\textnormal{Pd}(0,i,j)$ for $1\leq i,j\leq p$. The steady state
probabilities can be obtained by solving a set of $p^{2}$ linear equations
each of which essentially demands equilibrium for each of the possible states
of the process. The generic form of such an equation, for $i<j$, is given by:

$(4+t+2u)p_{i,j}=(1+2t)p_{j,i}+p_{i+1,j}+p_{i,j+1}+up_{i-1,j}+up_{i,j-1}$

For $i>j$ the equation is:

$(3+2t+2u)p_{i,j}=(2+t)p_{j,i}+p_{i+1,j}+p_{i,j+1}+up_{i-1,j}+up_{i,j-1}$

For $i=j$ the equation simplifies to:

$(2+2u)p_{i,i}=p_{i+1,i}+p_{i,i+1}+up_{i-1,i}+up_{i,i-1}$

For $i=1<j$ the equation simplifies to:

$(3+t+2u)p_{i,j}=(1+2t)p_{j,i}+p_{i+1,j}+p_{i,j+1}+up_{i,j-1}$

For $i<j=p$ the equation simplifies to:

$(4+t+u)p_{i,j}=(1+2t)p_{j,i}+p_{i+1,j}+up_{i-1,j}+up_{i,j-1}$

For $i=1<j=p$ the equation simplifies to:

$(3+t+u)p_{i,j}=(1+2t)p_{j,i}+p_{i+1,j}+up_{i,j-1}$

For $i>j=1$ the equation simplifies to:

$(2+2t+2u)p_{i,j}=(2+t)p_{j,i}+p_{i+1,j}+p_{i,j+1}+up_{i-1,j}$

For $i=p>j$ the equation simplifies to:

$(3+2t+u)p_{i,j}=(2+t)p_{j,i}+p_{i,j+1}+up_{i-1,j}+up_{i,j-1}$

For $i=p>j=1$ the equation simplifies to:

$(2+2t+u)p_{i,j}=(2+t)p_{j,i}+p_{i,j+1}+up_{i-1,j}$

For $i=j=1$ the equation simplifies to:

$(2u)p_{i,i}=p_{i+1,i}+p_{i,i+1}$

For $i=j=p$ the equation simplifies to:

$2p_{i,i}=up_{i-1,i}+up_{i,i-1}$

From equation (44) we can define a polynomial $A_{i,j}$ by gathering all terms
on the left:

$A_{i,j}=(4+t+2u)p_{i,j}-(1+2t)p_{j,i}-p_{i+1,j}-p_{i,j+1}-up_{i-1,j}-up_{i,j-1}$

We can similarly define $A_{i,j}$ under the conditions stated for equations
(45) through (54) inclusive. We next define a $p^{2}\times p^{2}$ matrix $B$
as follows:

$B_{p(i_{1}-1)+j_{1},p(i_{2}-1)+j_{2}}=[p_{i_{1},j_{1}}]A_{i_{2},j_{2}}$

The next step is to prove that the rank of $B$ is $p^{2}-1$. To see this, we
first observe that the sum of all rows of $B$ is identically zero, meaning
that the rank cannot be $p^{2}$. For the rank to then be $p^{2}-1$, we would
then need to show that no nontrivial linear combination of a proper subset of
the rows can be zero. If we let row $i,j$ be $R_{i,j}$ and for some
coefficients $c_{i,j}$ we have $\sum_{i,j}c_{i,j}R_{i,j}=0$, then we need to
show that if any $c_{i,j}=0$, then all $c_{i,j}=0$. The only rows with a $t$
term in column $i,j$ will be $R_{i,j}$ and $R_{j,i}$. Hence if $c_{i,j}=0$, it
follows that $c_{j,i}$=0.

We next show that if $c_{i,j}=0$ it follows that $c_{i-1,j-1}=0$. We can do
this by first showing that $c_{i-1,j}$ and $c_{i,j-1}$ must be negations of
one another. The only rows with a $u$ term in column $i,j$ will be $R_{i,j}$,
$R_{i-1,j}$, and $R_{i,j-1}$, with the latter two having the same coefficient.
Hence the following two statements are equivalent: $c_{i,j}=0$ and
$c_{i-1,j}+c_{i,j-1}=0$. We can similarly show that $c_{i,j}=0$ and
$c_{i+1,j}+c_{i,j+1}=0$ are equivalent. So from $c_{i,j}=0$ we can derive
$c_{i-1,j-1}=0$. By repeated application of the same argument we will get
$c_{k,1}=0$ or $c_{1,k}=0$ for some $k$.

Likewise, using equations (47) and (50), the only rows with a $u$ term in
column $1,k$ will be $R_{1,k}$ and $R_{1,k-1}$ and the only rows with a $u$
term in column $k,1$ will be $R_{k,1}$ and $R_{k-1,1}$. So from $c_{k,1}=0$ we
can derive $c_{k-1,1}=0$ and from $c_{1,k}=0$ we can derive $c_{1,k-1}=0$. By
repeated application of this we will get to $c_{1,1}=0$. By reversing the
above arguments it follows that $c_{i,j}=0$ for any $i,j$ and we have proven:

######  Lemma 5.1.

The rank of the matrix $B$ as defined above is $p^{2}-1$.

We next prove a result about the values of the $p_{i,j}$:

######  Theorem 5.2.

If $i\neq j$,

$p_{i,j}+p_{j,i}=\frac{2u^{i+j-2}}{(\displaystyle\sum_{k=0}^{n-1}u^{k})^{2}}$

Likewise,

$p_{i,i}=\frac{u^{2i-2}}{(\displaystyle\sum_{k=0}^{n-1}u^{k})^{2}}$

Proof: This can be proven by eliminating the variable $t$ from the set of
linear equations above. For example, if we add (45) and (46) we get the
following:

(59) $\begin{split}&(4+t+2u)p_{i,j}+(3+2t+2u)p_{j,i}\\\
&=(2+t)p_{i,j}+(1+2t)p_{j,i}+p_{i+1,j}+p_{j,i+1}+p_{i,j+1}+p_{j+1,i}\\\
&+up_{i-1,j}+up_{j,i-1}+up_{i,j-1}+up_{j-1,i}\end{split}$

If we let $q_{i,j}=p_{i,j}+p_{j,i}$ the above can be simplified to:

(60)
$\begin{split}&(2+2u)q_{i,j}=q_{i+1,j}+q_{i,j+1}+uq_{i-1,j}+uq_{i,j-1}\end{split}$

If we substitute in the values for $q_{i,j}$ from the theorem we are
attempting to prove to the above equation, we will see that it does satisfy
the above equation. Therefore the values of $q_{i,j}$ given in the theorem
represent one possible feasible solution to the set of equations. Moreover,
based on the earlier theorem about the rank of $B$, the solution must be
unique. This completes the proof.

To continue with the proof of Conjecture 3.3 in the $\impliedby$ direction, we
note that from
$\frac{\textnormal{Pr}(\mu)}{\textnormal{Pr}(\nu)}=\frac{\textnormal{Pd}(\mu)}{\textnormal{Pd}(\nu)}$
it follows that
$\frac{\textnormal{Pr}(0,2,1)}{\textnormal{Pr}(0,1,2)}=\frac{\textnormal{Pd}(0,2,1)}{\textnormal{Pd}(0,1,2)}$
or $\frac{2+t}{1+2t}=\frac{p_{2,1}}{p_{1,2}}$. This expands as
$(2+t)p_{1,2}=(1+2t)p_{2,1}$. From the above theorem we know that:

$p_{1,2}+p_{2,1}=\frac{2u}{(\displaystyle\sum_{k=0}^{n-1}u^{k})^{2}}$

We can then solve for $p_{1,2}$ giving:

$p_{1,2}=\frac{2(1+2t)u}{3(1+t)(\displaystyle\sum_{k=0}^{n-1}u^{k})^{2}}$

Equation (47) then gives us:

$(3+t+2u)p_{1,2}=(1+2t)p_{2,1}+p_{2,2}+p_{1,3}+up_{1,1}$

Substitute in to get:

(64)
$\begin{split}&\frac{2(3+t+2u)(1+2t)u}{3(1+t)(\displaystyle\sum_{k=0}^{n-1}u^{k})^{2}}\\\
&=\frac{2(2+t)(1+2t)u}{3(1+t)(\displaystyle\sum_{k=0}^{n-1}u^{k})^{2}}+\frac{3(1+t)(1+u)u}{3(1+t)(\displaystyle\sum_{k=0}^{n-1}u^{k})^{2}}+p_{1,3}\end{split}$

This simplifies to:

(65)
$\begin{split}&p_{1,3}=\frac{(5ut+u+t-1)u}{3(1+t)(\displaystyle\sum_{k=0}^{n-1}u^{k})^{2}}\end{split}$

A similar argument to that used to produce equation (62) will give us:

$p_{1,3}=\frac{2(1+2t)u^{2}}{3(1+t)(\displaystyle\sum_{k=0}^{n-1}u^{k})^{2}}$

Equating (65) and (66) and solving will give us $t=1$. This completes the
proof of:

######  Theorem 5.3.

Conjecture 3.3 is now proved more generally for $D=\textnormal{DASEP}(3,p,2)$.

##  6\. Future work

Three main potential directions for future work are indicated. One is that
further results should be obtained with a view to eventually proving
Conjecture 3.3. It is now proven for $D=\textnormal{DASEP}(3,p,2)$ and the
suggestion would be to prove it for $D=\textnormal{DASEP}(n,2,2)$ and
$D=\textnormal{DASEP}(n,2,q)$ before eventually proceeding to
$D=\textnormal{DASEP}(n,p,q)$. Similarly considering the case where $0$
represents a ball with species $0$ rather than the absence of a species is a
variant that should be explored. The other, and more ambitious, possible goal
for future research would be to come up with a complete combinatorial
characterization of the steady state probabilities for the DASEP. For the
ASEP, this has been done in [4] and [10] leading to a deep relationship being
discovered between the ASEP and Macdonald polynomials. To the extent
Conjecture 3.3 can be proven, the ratios between steady state probabilities is
different for the ASEP than for the DASEP, suggesting there is more
combinatorial structure to be discovered.

## References

  * [1] Arvind Ayyer and Svante Linusson  “An inhomogeneous multispecies TASEP on a ring”  In _Advances in Applied Mathematics_ 57, 2014, pp. 21–43  arXiv:1206.0316
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