# The resource theory of dephasing estimation in multiqubit systems

Zishi Chen  Xueyuan Hu<EMAIL_ADDRESS>School of Information Science and
Engineering, Shandong University, Qingdao 266237, China

###### Abstract

We present a resource theory to investigate the power of a multqubit system as
a probe in the task of dephasing estimation. Our approach employs the quantum
Fisher information about the dephasing parameter as the resource measure.
Based on the monotonicity of quantum Fisher information, we propose two sets
of free operations in our resource theory, the Hamming distance preserving
operations and the selectively Hamming distance preserving operations. We
derive a necessary condition for the state transformation under these free
operations and demonstrate that uniform superposition states are the golden
states in our resource theory. We further compare our resource theory with the
resource theory of coherence and thoroughly investigate the relation between
their free operations in both single-qubit and multiqubit cases. Additionally,
for multiqubit systems, we discover the incompatibility between the resource
theory of dephasing estimation and that of $U(1)$ asymmetry, which is
responsible for phase estimation. The condition for enhancing the performance
of a probe state in phase estimation while preserving its ability in dephasing
estimation is also discussed. Our results provide new insights into quantum
parameter estimation by the resource-theoretic approach.

###### pacs:

03.65.Ta, 03.65.Yz, 03.67.Mn

##  I INTRODUCTION

In quantum resource theories [1], certain quantum properties of physical
systems are treated as resources. By the resource-theoretic approach, the
advantages of these quantum resources are quantified for various quantum
information processing tasks such as quantum teleportation [2], channel
discrimination [3, 4] and quantum key distribution [5]. Moreover, the quantum
resource theory provides a general framework to characterize laws for state
transformations under certain classes of restricted operations called free
operations, and thus sheds new light on the research field of fundamental
physics, including thermodynamics [6].

Quantum parameter estimation is a task which employs quantum properties to
make precise estimation of given parameters [7, 8, 9, 10]. For instance, in
phase estimation, maximally entangled probe states can improve the error
scaling of the estimator[8, 11]. This improvement inspires researchers to
investigate the role of quantum properties in quantum parameter estimation
[12]. It has been found that there are quantum enhanced estimation tasks
without entanglement [13, 9], indicating that entanglement is not the unique
resource which underlies the quantum advantage in parameter estimation. In
particular, for the estimation of noise in teleportation-covariant channels,
including phase damping (PD) channels [14, 15], entanglement is not a
necessary condition for the optimal probe states [9, 16, 17, 18].

Recently, the notions of quantum parameter estimation have been introduced to
the framework of resource theories and made significant progress. The quantum
Fisher information (QFI), which represents the precision limit of the probe
states [19], can be used as a resource measure of coherence and asymmetry [20,
21, 22, 23], due to its nice mathematical properties such as convexity and
monotonicity [7, 24]. Furthermore, by the resource-theoretic approach, an
operational interpretation of the QFI is proposed in quantum thermodynamics
[25]. In this work, we propose and give answers to the following questions:
what is the quantum resource which underlies the precision of dephasing
estimation? How does this resource relate to some well-known resources, such
as quantum coherence and $U(1)$ asymmetry?

We establish a resource theory of dephasing estimation in multiqubit systems.
We first prove that incoherent states are free states in this present resource
theory. Due to the monotonicity of QFI, we define two sets of free operations
in the resource theory of dephasing estimation, the Hamming distance
preserving operations (HDP) and the selectively Hamming distance preserving
operations (SHP). Then we derive a necessary condition for the transformed
states under HDP. Furthermore, uniform superposition states can be transformed
into any state under SHP, that is, uniform superposition states are golden
probe states in the resource theory of dephasing estimation. Moreover, by
comparing our resource theory and the resource theory of quantum coherence, we
show the relation between their free operations in both single-qubit and
multiqubit cases. Finally, we consider the resource theory of $U(1)$
asymmetry. By employing SHP, it is possible to improve the performance of the
probe state in phase estimation, while also preserving the ability of phase
estimation.

##  II NOTIONS AND PRELIMINARIES

In this section, we review the definitions of the Fisher information (FI), the
QFI, and the PD channels [7, 14]. Before reviewing these definitions, let us
briefly recapitulate some notions.

We denote the Hilbert space associated with the qubits of our consideration by
$\mathcal{H}$. Let $\mathcal{L}(\mathcal{H})$ be the set of liner operators on
$\mathcal{H}$. Any quantum operation on a physical system can be
mathematically represented by a completely-positive and trace-preserving
(CPTP) map
$\mathcal{E}:\mathcal{L}(\mathcal{H})\rightarrow\mathcal{L}(\mathcal{H})$. In
this paper, we focus on the situations where the input and output spaces are
the same. We use $\mathbb{I}$ as the identity matrix of the input (output)
space.

###  II.1 FISHER INFORMATION (FI) AND QUANTUM FISHER INFORMATION (QFI)

In a general parameter estimation task, a probe system, initially in the state
$\rho$, is sent through a parameterized quantum channel
$\mathcal{E}_{\theta}$. The output state
$\rho(\theta)\equiv\mathcal{E}_{\theta}(\rho)$ is then measured by a positive-
operator-valued measurement (POVM) $\\{M_{x}\\}$, where $M_{x}$ is the
positive semi-definite operator satisfying $\sum_{x}M_{x}=\mathbb{I}$, and
each measurement result $x$ is obtained with the probability
$p(x|\theta)=\mathrm{tr}(\rho(\theta)M_{x})$. Finally, the parameter $\theta$
is estimated based on the probability distribution of measurement results.

The Cramér-Rao bound provides a lower bound for the variance of an unbiased
estimator [7, 9, 8]

${\rm
Var}(\theta)\geq\dfrac{1}{F(\theta)}\geq\dfrac{1}{\mathcal{F}(\rho(\theta))}.$
(1)

Here the Fisher information $F(\theta)$ is defined as [7, 9, 22]

$F(\theta)=\sum_{x}\dfrac{(\partial p(x|\theta))^{2}}{p(x|\theta)},$ (2)

where $\partial\coloneqq\partial/\partial\theta$. It depends on both the probe
state $\rho$ and the measurement $\\{M_{x}\\}$. The QFI
$\mathcal{F}(\rho(\theta))$ is defined as [7, 9, 8, 22]

$\mathcal{F}(\rho(\theta))=\mathrm{tr}(\rho(\theta)L^{2}),$ (3)

where $L$ is the symmetric logarithmic derivative (SLD) and defined by
$\frac{1}{2}(L\rho(\theta)+\rho(\theta)L)=\partial\rho(\theta)$. The QFI
satisfies the following properties [7, 9].
(P1) Additivity.
$\mathcal{F}(\rho(\theta)\otimes\rho^{\prime}(\theta))=\mathcal{F}(\rho(\theta))+\mathcal{F}(\rho^{\prime}(\theta))$.
It implies $\mathcal{F}(\rho(\theta)^{\otimes n})=n\mathcal{F}(\rho(\theta))$.
(P2) Convexity.
$\mathcal{F}(\sum_{j}p_{j}\rho_{j}(\theta))\leq\sum_{j}p_{j}\mathcal{F}(\rho_{j}(\theta))$.
(P3) Monotonicity.
$\mathcal{F}(\rho(\theta))\geq\mathcal{F}(\mathcal{E}(\rho(\theta)))$, for any
CPTP map $\mathcal{E}$ which is independent of $\theta$.

###  II.2 PHASE DAMPING CHANNELS

In this paper, we focus on the phase damping (PD) channel for single-qubit and
multiqubit states, which is denoted by $\mathcal{E}^{\mathrm{PD}}_{\theta}$.

A single-qubit state
$\sigma=\frac{1}{2}(\mathbb{I}+r\cos{\phi}\sigma_{x}+r\sin{\phi}\sigma_{y}+z\sigma_{z})$,
where $\sigma_{x}$, $\sigma_{y}$ and $\sigma_{z}$ are two-dimensional Pauli
matrices, can be represented as a Bloch vector $(r\cos{\phi},r\sin{\phi},z)$,
where $-1\leq z\leq 1$, $0\leq r\leq 1$ and $0\leq\phi<2\pi$. Then the density
matrix of the state $\sigma$ can be written as

$\sigma=\dfrac{1}{2}\begin{bmatrix}1+z&re^{-i\phi}\\\
re^{i\phi}&1-z\end{bmatrix}.$ (4)

After the action of a PD channel [14], the state becomes

$\mathcal{E}^{\mathrm{PD},1}_{\theta}(\sigma)=\dfrac{1}{2}\begin{bmatrix}1+z&re^{-\theta-i\phi}\\\
re^{-\theta+i\phi}&1-z\end{bmatrix},$ (5)

where $\theta$ is the parameter of the PD channel.

Now consider a general $n$-qubit state
$\rho=\sum_{x=0,y=0}^{2^{n}-1,2^{n}-1}\rho_{xy}|x_{b}\rangle\langle y_{b}|$,
where $x_{b}$ represents the binary representation of the non-negative integer
$x$. Let each qubit transmit through a PD channel
$\mathcal{E}^{\mathrm{PD},1}_{\theta}$, and the output $n$-qubit state reads

$\mathcal{E}^{\mathrm{PD}}_{\theta}(\rho)=\sum_{x=0,y=0}^{2^{n}-1,2^{n}-1}e^{-h(x,y)\theta}\rho_{xy}|x\rangle\langle
y|,\ n=1,2,...$ (6)

where $h(i,j)$ is the Hamming distance between two binary strings $i_{b}$ and
$j_{b}$. The mathematical definition of the Hamming distance is
$h(i,j)=\sum_{x=1}^{n}i^{x}_{b}\oplus j^{x}_{b}$, where $i^{x}_{b}(j^{x}_{b})$
represents the $x$th bit of the string $i_{b}(j_{b})$. The task, which is to
estimate $\theta$ by a probe state, is called dephasing estimation.

##  III STRUCTURE OF THE RESOURCE THEORY OF DEPHASING ESTIMATION

In the resource theory of dephasing estimation, we use the QFI as a resource
measure. Precisely, we define the power of a probe state $\rho$ in dephasing
estimation as follows

$\mathcal{F}_{\theta}^{\mathrm{PD}}(\rho):=\mathcal{F}(\mathcal{E}^{\mathrm{PD}}_{\theta}(\rho)).$
(7)

The free states are those which do not have the ability of dephasing
estimation, i.e., $\rho$ is a free state if and only if
$\mathcal{F}_{\theta}^{\mathrm{PD}}(\rho)=0$. Apparently, an incoherent state
$\rho_{\mathrm{inc}}$ on computational basis belongs to the set of free
states, because $\mathcal{E}^{\mathrm{PD}}_{\theta}(\rho_{\mathrm{inc}})$ is
independent of $\theta$. In the following, we prove that any state which is
not incoherent is a resourceful state.

Assume that there is at least one non-diagonal element
$\rho_{xy}=\left|\rho_{xy}\right|e^{-i\beta}\neq 0$ in the probe state
$\rho_{\mathrm{coh}}$. After the action of
$\mathcal{E}^{\mathrm{PD}}_{\theta}$, one implements the following measurement

$\begin{split}&M_{1}=(|x\rangle\langle x|+e^{-i\beta}|x\rangle\langle
y|+e^{i\beta}|y\rangle\langle x|+|y\rangle\langle y|)/2,\\\
&M_{2}=(|x\rangle\langle x|-e^{-i\beta}|x\rangle\langle
y|-e^{i\beta}|y\rangle\langle x|+|y\rangle\langle y|)/2,\\\
&M_{3}=\mathbb{I}-|x\rangle\langle x|-|y\rangle\langle y|,\end{split}$ (8)

and obtains the outcome $a\in\\{1,2,3\\}$ with probability
$p(a|\theta)=\mathrm{tr}(\mathcal{E}^{\mathrm{PD}}_{\theta}(\rho_{\mathrm{coh}})M_{a})$.
Direct calculation leads to

$\begin{split}p(1|\theta)&=\dfrac{1}{2}(\rho_{xx}+\rho_{yy}+2\left|\rho_{xy}\right|e^{-h(x,y)\theta}),\\\
p(2|\theta)&=\dfrac{1}{2}(\rho_{xx}+\rho_{yy}-2\left|\rho_{xy}\right|e^{-h(x,y)\theta}),\\\
p(3|\theta)&=1-p(1|\theta)-p(2|\theta).\end{split}$ (9)

It follows that

$\begin{split}F(\theta)&=\sum_{a=1}^{3}\dfrac{(\partial
p(a|\theta))^{2}}{p(a|\theta)}\\\
&=\dfrac{[h(x,y)\left|\rho_{xy}\right|e^{-h(x,y)\theta}]^{2}}{p(1|\theta)}\\\
&+\dfrac{[h(x,y)\left|\rho_{xy}\right|e^{-h(x,y)\theta}]^{2}}{p(2|\theta)}>0.\end{split}$
(10)

Because the QFI is lower bounded by the FI, we have
$\mathcal{F}_{\theta}^{\mathrm{PD}}(\rho_{\mathrm{coh}})\geq F(\theta)>0$.
Therefore, a free state in the resource theory of dephasing estimation is of
the following form

$\rho_{\mathrm{free}}=\sum_{x=0}^{2^{n}-1}\rho_{x}|x\rangle\langle x|.$ (11)

As for free operations, the minimum requirement is that they do not increase
the resource of states. Nevertheless, considering the complexity in
calculating the QFI, defining the free operations to be the whole set of
$\mathcal{F}_{\theta}^{\mathrm{PD}}$-nonincreasing operations would cause
difficulties in analysing the properties of the present resource and
generalizing the method of studying the resources in dephasing estimation.
With this in mind, we propose two sets of free operations in the following.

The first set of free operations $\mathbb{F}_{1}$ is defined as the set of all
CPTP maps commutable with any PD channel,

$\mathcal{E}\circ\mathcal{E}^{\mathrm{PD}}_{\theta}=\mathcal{E}^{\mathrm{PD}}_{\theta}\circ\mathcal{E},\
\forall\mathcal{E}\in\mathbb{F}_{1},$ (12)

and will be thoroughly studied in Sec. III.1. The second set of free
operations $\mathbb{F}_{2}$, studied in Sec. III.2, is defined as those with a
Kraus decomposition $\mathcal{E}=\sum_{j}\mathcal{K}_{j}$ (with
$\mathcal{K}_{j}(\cdot)=K_{j}\cdot K_{j}^{\dagger}$ and $K_{j}$ being the
Kraus operators satisfying $\sum_{j}K_{j}^{\dagger}K_{j}=\mathbb{I}$), such
that each Kraus branch $\mathcal{K}_{j}$ can commute with PD channels, i.e.,

$\mathcal{K}_{j}\circ\mathcal{E}^{\mathrm{PD}}_{\theta}=\mathcal{E}^{\mathrm{PD}}_{\theta}\circ\mathcal{K}_{j},\
\forall j,\ \forall\mathcal{E}\in\mathbb{F}_{2}.$ (13)

By definition, $\mathbb{F}_{2}\subseteq\mathbb{F}_{1}$. The detailed
comparison between these two sets will be discussed in Sec. III.3.

The monotonicity of $\mathcal{F}_{\theta}^{\mathrm{PD}}$ under these free
operations can be proved by employing the monotonicity (P3) of the QFI.
Precisely, if $\mathcal{E}\in\mathbb{F}_{1}$, we have

$\mathcal{F}_{\theta}^{\mathrm{PD}}(\mathcal{E}(\rho))=\mathcal{F}(\mathcal{E}(\mathcal{E}^{\mathrm{PD}}_{\theta}(\rho)))\leq\mathcal{F}(\mathcal{E}^{\mathrm{PD}}_{\theta}(\rho))=\mathcal{F}_{\theta}^{\mathrm{PD}}(\rho),$
(14)

for any $n$-qubit state $\rho$.

Also, it is worth noting that, even for the single-qubit case,
$\mathbb{F}_{1}$ is a strict subset of the
$\mathcal{F}_{\theta}^{\mathrm{PD}}$-nonincreasing operations. See Appendix A
for detailed discussion.

###  III.1 HAMMING DISTANCE PRESERVING OPERATIONS (HDP)

In this subsetion, we first give the definition of Hamming distance preserving
operations (HDP) and then prove that HDP is equivalent to $\mathbb{F}_{1}$
defined in Eq. (12). Moreover, we derive a necessary condition for state
transformations under HDP.

######  Definition 1.

A CPTP map
$\mathcal{E}^{\textrm{HDP}}:\mathcal{L}(\mathcal{H}^{n})\mapsto\mathcal{L}(\mathcal{H}^{n})$
belongs to the set of Hamming distance preserving operations (HDP) if and only
if

$\langle i|\mathcal{E}^{\textrm{HDP}}(|x\rangle\langle y|)|j\rangle=0,$ (15)

for all $i,j,x,y\in\\{0,1,2,...,2^{n}-1\\}$ satisfying $h(i,j)\neq h(x,y)$.

The following proposition shows the equivalence between the Hamming distance
preserving condition and the condition of commutativity with PD channels for
completely positive maps.

######  Proposition 1.

Let $\tilde{\mathcal{E}}$ be a completely positive (CP) map. Then,

$\tilde{\mathcal{E}}\circ\mathcal{E}^{\mathrm{PD}}_{\theta}=\mathcal{E}^{\mathrm{PD}}_{\theta}\circ\tilde{\mathcal{E}},$
(16)

if and only if

$\langle i|\tilde{\mathcal{E}}(|x\rangle\langle y|)|j\rangle=0,$ (17)

for all $i,j,x,y\in\\{0,1,2,...,2^{n}-1\\}$ satisfying $h(i,j)\neq h(x,y)$.

###### Proof.

From Eq. (6), Eq. (16) can be rewritten as

$(e^{-h(i,j)\theta}-e^{-h(x,y)\theta})\langle
i|\tilde{\mathcal{E}}(|x\rangle\langle y|)|j\rangle=0\quad\forall i,j,x,y.$
(18)

Equivalently, $\langle i|\tilde{\mathcal{E}}(|x\rangle\langle y|)|j\rangle=0$
if $h(x,y)\neq h(i,j)$. ∎

By setting $\tilde{\mathcal{E}}$ in the above proposition to be a CPTP map
$\mathcal{E}$, we directly have $\mathbb{F}_{1}=\mathrm{HDP}$. Therefore, we
will label the first set of free operations defined in Eq. (12) as HDP.

The following proposition characterizes a necessary condition of state
transformations under HDP.

######  Proposition 2.

For any $n$-qubit state $\rho$, it holds that

$\lvert\langle
i|\mathcal{E}^{\textrm{HDP}}(\rho)|j\rangle\rvert\leq{\sum_{x,y}}^{\prime}\lvert\rho_{xy}\rvert\sqrt{p_{i|x}p_{j|y}},$
(19)

where $\mathcal{E}^{\textrm{HDP}}\in\mathrm{HDP}$, $\rho_{xy}=\langle
x|\rho|y\rangle$, $p_{i|x}=\langle
i|\mathcal{E}^{\textrm{HDP}}(|x\rangle\langle x|)|i\rangle$, and
$\sum^{\prime}_{x,y}$ stands for a summation over all $x,y$ which satisfy
$h(x,y)=h(i,j)$.

###### Proof.

Based on Eq. (15), each element of the output density matrix is given by

$\langle
i|\mathcal{E}^{\textrm{HDP}}(\rho)|j\rangle={\sum_{x,y}}^{\prime}\rho_{xy}\langle
i|\mathcal{E}^{\textrm{HDP}}(|x\rangle\langle y|)|j\rangle.$ (20)

Now write $\mathcal{E}^{\textrm{HDP}}$ in a Kraus decomposition form
$\mathcal{E}^{\textrm{HDP}}(\cdot)=\sum_{a}D_{a}(\cdot)D_{a}^{\dagger}$ and
define the vector $\vec{X}^{xy}=(X^{xy}_{1},X^{xy}_{2},\dots)$ with
$X^{xy}_{a}\equiv\langle x|D_{a}|y\rangle$. Then we have $\langle
i|\mathcal{E}^{\textrm{HDP}}(|x\rangle\langle
y|)|j\rangle=\vec{X}^{ix}\cdot\vec{X}^{jy*}$, which implies
$p_{i|x}=|\vec{X}^{ix}|^{2}$. It follows from the Cauchy-Schwarz inequality
that

$\left|\langle i|\mathcal{E}^{\textrm{HDP}}(|x\rangle\langle
y|)|j\rangle\right|\leq\sqrt{p_{i|x}p_{j|y}}.$ (21)

The equality sign holds for one of the two cases: (1) either $p_{i|x}$ or
$p_{j|y}$ vanishes; (2) a parameter $\xi$ exists such that
$\vec{X}^{ix}=\xi\vec{X}^{jy}$. Therefore,

$\displaystyle\left|\langle
i|\mathcal{E}^{\textrm{HDP}}(\rho)|j\rangle\right|$ $\displaystyle\leq$
$\displaystyle{\sum_{x,y}}^{\prime}|\rho_{xy}||\langle
i|\mathcal{E}^{\textrm{HDP}}(|x\rangle\langle y|)|j\rangle|$ (22)
$\displaystyle\leq$
$\displaystyle{\sum_{x,y}}^{\prime}|\rho_{xy}|\sqrt{p_{i|x}p_{j|y}}.$

The equality sign in the first line holds if the terms in the summation of Eq.
(20) have the same phase. ∎

If we focus on the diagonal elements of the output state, the condition
$h(x,y)=h(i,i)=0$ leads to $x=y$, and Eq. (20) reduces to

$\langle
i|\mathcal{E}^{\textrm{HDP}}(\rho)|i\rangle=\sum_{x}\rho_{xx}p_{i|x}.$ (23)

It means that the diagonal elements of the output state are independent of the
off-diagonal elements of the input state.

Moreover, an $n$-qubit state $\rho$ can be written as

$\rho=\sum_{h=0}^{n}\rho^{(h)},\
\rho^{(h)}\equiv\sum_{x,y:h(x,y)=h}\rho_{xy}|x\rangle\langle y|.$ (24)

We call $\rho^{(h)}$ a Hamming mode of $\rho$. Then, Proposition 2 indicates
that each Hamming mode of the input state is independently mapped by HDP to
the corresponding Hamming mode of the output state, namely,

$\mathcal{E}^{\textrm{HDP}}(\rho^{(h)})=[\mathcal{E}^{\textrm{HDP}}(\rho)]^{(h)}.$
(25)

It follows that, if $\rho^{(h)}=0$ for some $h$, then
$[\mathcal{E}^{\textrm{HDP}}(\rho)]^{(h)}=0$. For example, the state
$|\psi_{1}\rangle=\frac{1}{\sqrt{2}}(|0\rangle+|1\rangle)$ cannot be
transformed by HDP to state
$|\psi_{2}\rangle=\frac{1}{\sqrt{2}}(|0\rangle+|3\rangle)$, because
$(|\psi_{1}\rangle\langle\psi_{1}|)^{(2)}=0$ but
$(|\psi_{2}\rangle\langle\psi_{2}|)^{(2)}=\frac{1}{2}(|0\rangle\langle
3|+|3\rangle\langle 0|)\neq 0$. This sets the essential difference between the
present resource theory and the resource theory of coherence, because
$|\psi_{1}\rangle$ and $|\psi_{2}\rangle$ can be transformed to each other by
incoherent unitary operators. More details of the comparison between the two
resource theories are to be illustrated in Sec. IV.

An immediate problem is then whether the bound in Eq. (19) can be reached. We
will prove the attainability for single-qubit case, and show that it cannot be
reached for general $n$-qubit input states. Further, we will characterize a
set of $n$-qubit input states, for which the bound in Eq. (19) can be reached.

The set of states, which can be obtained from a given state $\rho$ by HDP, is
called the HDP cone of $\rho$. By Propostion 2, the HDP cone can be obtained
for any single-qubit state.

######  Corollary 1.

For two single-qubit states with Bloch vectors
$\vec{r}=(r\cos{\phi},r\sin{\phi},z)$ and
$\vec{r}^{\prime}=(r^{\prime}\cos{\phi^{\prime}},r^{\prime}\sin{\phi^{\prime}},z^{\prime})$,
$\vec{r}$ can be transformed to $\vec{r}^{\prime}$ via HDP if and only if

$r^{\prime}\leq\left\\{\begin{aligned} &r\sqrt{\frac{1-z^{\prime
2}}{1-z^{2}}},&|z^{\prime}|\geq|z|\\\
&r,&|z^{\prime}|<|z|.\end{aligned}\right.$ (26)

###### Proof.

If $\vec{r}^{\prime}$ can be obtained from $\vec{r}$ via HDP, Eq. (19) gives

$\displaystyle r^{\prime}$ $\displaystyle\leq$ $\displaystyle
r(\sqrt{p_{0|0}p_{1|1}}+\sqrt{p_{0|1}p_{1|0}})$ (27) $\displaystyle=$
$\displaystyle r\cos(\theta_{0}-\theta_{1}),$

where we set $p_{0|0}=\cos^{2}\theta_{0}$, $p_{1|1}=\cos^{2}\theta_{1}$, and
$\theta_{0},\theta_{1}\in[0,\frac{\pi}{2}]$. Furthermore, from Eq. (23), we
have

$\dfrac{1+z^{\prime}}{2}=\cos^{2}\theta_{0}\dfrac{1+z}{2}+(1-\cos^{2}\theta_{1})\dfrac{1-z}{2}.$
(28)

It follows that

$\begin{split}z^{\prime 2}&=[\dfrac{1}{2}\cos
2\theta_{0}(1+z)-\dfrac{1}{2}\cos 2\theta_{1}(1-z)]^{2}\\\
&=\dfrac{1}{4}(\operatorname{Re}[e^{2i\theta_{0}}(1+z)-e^{2i\theta_{1}}(1-z)])^{2}\\\
&\leq\dfrac{1}{4}|e^{2i\theta_{0}}(1+z)-e^{2i\theta_{1}}(1-z)|^{2}\\\
&=1-(1-z^{2})\cos^{2}(\theta_{0}-\theta_{1}),\end{split}$ (29)

or equivalently,

$\cos^{2}(\theta_{0}-\theta_{1})\leq\min\\{1,\frac{1-z^{\prime
2}}{1-z^{2}}\\}.$ (30)

By combining Eqs. (27) and (30), we conclude that Eq. (26) is a necessary
condition for the transformation from $\vec{r}$ to $\vec{r}^{\prime}$.

The sufficiency of Eq. (26) is proved by constructing a channel
$\mathcal{E}^{\textrm{HDP}}_{*}\in\mathrm{HDP}$ such that the boundary states
satisfying the equality sign in Eq. (26) can be reached. Precisely, we have
$\mathcal{E}^{\textrm{HDP}}_{*}(\cdot)=K_{1}\cdot K_{1}^{\dagger}+K_{2}\cdot
K_{2}^{\dagger}$ with

$\displaystyle K_{1}=\left(\begin{array}[]{cc}\cos\theta_{0}&0\\\
0&e^{i(\phi-\phi^{\prime})}\cos\theta_{1}\end{array}\right),$ (33)
$\displaystyle
K_{2}=\left(\begin{array}[]{cc}0&e^{i(\phi-\phi^{\prime})}\sin\theta_{1}\\\
\sin\theta_{0}&0\end{array}\right),$ (36)

where $\theta_{0},\theta_{1}\in[0,\frac{\pi}{2}]$ are determined by the
parameters in the input and output states as follows.

If $|z^{\prime}|<|z|$, we choose
$\sin{2\theta_{0}}=\sin{2\theta_{1}}=\frac{z^{\prime}}{z}$. It can be checked
that the channel $\mathcal{E}^{\textrm{HDP}}_{*}$ transforms the state
$\vec{r}$ to $\vec{r}^{\prime}$ with $r^{\prime}=r$ and
$z^{\prime}=z\sin{2\theta_{0}}$.

If $|z^{\prime}|\geq|z|$, the equality sign in Eq.(29) holds when

$\sin 2\theta_{0}(1+z)-\sin 2\theta_{1}(1-z)=0.$ (37)

Combining this equation with Eq. (28), we obtain

$\cos 2\theta_{0}=\frac{z+z^{\prime 2}}{z^{\prime}(1+z)},\cos
2\theta_{1}=\frac{z-z^{\prime 2}}{z^{\prime}(1-z)}.$ (38)

The channel $\mathcal{E}^{\textrm{HDP}}_{*}$ with the above parameters gives
an output state with
$z^{\prime}=\sqrt{1-(1-z^{2})\cos^{2}(\theta_{0}-\theta_{1})}$ and
$r^{\prime}=r\cos(\theta_{0}-\theta_{1})$.

∎

Figure 1: The colored region is the projection of the SHP cone in the x-z
plane, when the initial Bloch vector is $(0.6,0,0.6)$. The initial state is
marked by the star. The area bounded by the dotted line contains all single-
qubit states.

The unitary operator $U(\phi)=\textrm{diag}(1,e^{-i\phi})$ belongs to HDP.
Based on these unitary operations, the HDP cone is rotationally symmetric
about the $z$ axis. That is, the Bloch vector $(r^{\prime},0,z^{\prime})$ can
be transformed to Bloch vector
$(r^{\prime}\cos{\phi^{\prime}},r^{\prime}\sin{\phi^{\prime}},z^{\prime})$. In
FIG.1, we plot the projection of the SHP cone in the x-z plane for a given
initial state, whose Bloch vector is $(0.6,0,0.6)$.

In the above proof, we show that the bound in Eq. (19) can be reached for
single-qubit input states. Nevertheless, we will show in the following that
this bound cannot be reached in general. A counterexample goes as follows.
Consider a $2$-qubit state in the following form

$\rho=\frac{1}{4}\left(\begin{array}[]{cccc}1&\frac{1}{\sqrt{2}}&\frac{1}{\sqrt{2}}&0\\\
\frac{1}{\sqrt{2}}&1&0&0\\\ \frac{1}{\sqrt{2}}&0&1&0\\\
0&0&0&1\end{array}\right).$ (39)

For $\mathcal{E}\in\mathrm{HDP}$, the bound in Eq. (19) gives

$\displaystyle|\langle 0|\mathcal{E}(\rho)|1\rangle|$ $\displaystyle\leq$
$\displaystyle\sqrt{p_{0|0}p_{1|1}}|\rho_{01}|+\sqrt{p_{0|1}p_{1|0}}|\rho_{10}|$
(40)
$\displaystyle+\sqrt{p_{0|0}p_{1|2}}|\rho_{02}|+\sqrt{p_{0|2}p_{1|0}}|\rho_{20}|$

We aim to maximise $|\langle 0|\mathcal{E}(\rho)|1\rangle|$ for
$\mathcal{E}\in\mathrm{HDP}$. The bound in Eq. (19) gives

$\displaystyle|\langle 0|\mathcal{E}(\rho)|1\rangle|$ $\displaystyle\leq$
$\displaystyle\sqrt{p_{0|0}p_{1|1}}|\rho_{01}|+\sqrt{p_{0|1}p_{1|0}}|\rho_{10}|$
(41)
$\displaystyle+\sqrt{p_{0|0}p_{1|2}}|\rho_{02}|+\sqrt{p_{0|2}p_{1|0}}|\rho_{20}|$
$\displaystyle\leq$ $\displaystyle|\rho_{01}|+|\rho_{02}|.$

The second equality sign holds if and only if
$p_{0|0}=p_{1|1}=p_{1|2}\equiv\cos^{2}\phi$ and
$p_{1|0}=p_{0|1}=p_{0|2}\equiv\sin^{2}\phi$.

Now suppose $\cos^{2}\phi\neq 0$. It follows that the first equality sign in
Eq. (41) holds only if
$\vec{X}^{00}=e^{i\eta_{1}}\vec{X}^{11}=e^{i\eta_{2}}\vec{X}^{12}$, which in
turn gives $\cos^{2}\phi=|\vec{X}^{11}\cdot\vec{X}^{12*}|=|\langle
1|\mathcal{E}(|1\rangle\langle 2|)|1\rangle|$. However, the rhs of the above
equation equals zero from the Hamming distance preserving condition. This
violates the assumption that $\cos^{2}\phi\neq 0$. Therefore, we have
$\sin^{2}\phi=1$. Then the first equality sign in Eq. (41) holds only if
$\vec{X}^{10}=e^{i\xi_{1}}\vec{X}^{01}=e^{i\xi_{2}}\vec{X}^{02}$, which in
turn gives $\sin^{2}\phi=|\vec{X}^{01}\cdot\vec{X}^{02*}|=\langle
0|\mathcal{E}(|1\rangle\langle 2|)|0\rangle$, but again the rhs equals zero
because of the Hamming distance preserving condition. Therefore, it is
impossible to transform $\rho$ into

$\rho^{\prime}=\frac{1}{4}\left(\begin{array}[]{cccc}1&\sqrt{2}&0&0\\\
\sqrt{2}&2&0&0\\\ 0&0&0&0\\\ 0&0&0&1\end{array}\right),$ (42)

while this transformation is allowed by the condition as Eq. (19).

However, for $n$-qubit states satisfying the following two conditions, we
construct a quantum operation $\mathcal{E}\in\mathrm{HDP}$ such that the
equality sign in Eq. (19) is reached. Let $\\{|x\rangle\\}$ be the
computational basis of the $n$-qubit Hilbert space and $\rho_{xy}\equiv\langle
x|\rho|y\rangle$. The two condition can be stated as follows.
(C1) $\exists c\in\mathbb{Z}^{+}$, if $h(x,y)\neq c$, then $\rho_{xy}=0$.
(C2) $\forall\rho_{xy},\rho_{x^{\prime}y^{\prime}}\neq 0$, where $x\neq
x^{\prime}$, $\mathrm{span}\\{|x\rangle,|y\rangle\\}$ is orthogonal to
$\mathrm{span}\\{|x^{\prime}\rangle,|y^{\prime}\rangle\\}$.
Condition (C1) means that the Hamming distance between the supports of each
non-zero off-diagonal element equals to a constant. Condition (C2) means that
the supports of any two off-diagonal elements do not have overlap. Therefore,
for any non-zero off-diagonal element $\rho_{xy}$, $y$ uniquely depends on
$x$, and hence we write $y$ as $y(x)$.

Assume the above two conditions are satisfied by the input state. Let $i,j$ be
$n$-bit strings satisfying $h(i,j)=h(x,y(x))$. We construct
$\mathcal{E}\in\mathrm{HDP}$ as
$\mathcal{E}(\cdot)=K_{0}(\cdot)K_{0}^{\dagger}+\sum_{x}K_{x,1}(\cdot)K_{x,1}^{\dagger}+K_{x,2}(\cdot)K_{x,2}^{\dagger}$
with

$\displaystyle K_{x,1}$ $\displaystyle=$
$\displaystyle\sqrt{p_{i|x}}e^{-i\phi_{x}}|i\rangle\langle
x|+\sqrt{p_{j|y(x)}}|j\rangle\langle y(x)|,$ $\displaystyle K_{x,2}$
$\displaystyle=$ $\displaystyle\sqrt{p_{j|x}}e^{-i\phi_{x}}|j\rangle\langle
x|+\sqrt{p_{i|y(x)}}|i\rangle\langle y(x)|,$ $\displaystyle K_{0}$
$\displaystyle=$
$\displaystyle[\mathbb{I}-\sum_{x}(K_{x,1}^{\dagger}K_{x,1}+K_{x,2}^{\dagger}K_{x,2})]^{\frac{1}{2}},$
(43)

where $x$ is the $n$-bit string such that $\rho_{xy}\neq 0$ and $x<y(x)$, and
$\phi_{x}$ is the phase of $\rho_{xy}$. It follows that

$\langle
i|\mathcal{E}(\rho)|j\rangle=\sum_{x}(\sqrt{p_{i|x}p_{j|y(x)}}+\sqrt{p_{j|x}p_{i|y(x)}})|\rho_{xy}|,$
(44)

which means that the bound in Eq. (19) is reached by the channel
$\mathcal{E}$. Furthermore, if one aims to maximize $\langle
i|\mathcal{E}(\rho)|j\rangle$ over all $\mathrm{HDP}$, we choose
$p_{i|x}=p_{j|y(x)}$ and $p_{j|x}=p_{i|y(x)}=1-p_{i|x}$, and obtain

$\langle i|\mathcal{E}(\rho)|j\rangle_{\max}=\sum_{x}|\rho_{xy}|.$ (45)

It means that, if conditions (C1) and (C2) are satisfied by input state, then
the off-diagonal elements of the same Hamming mode can be merged together by
HDP.

###  III.2 SELECTIVELY HAMMING DISTANCE PRESERVING OPERATIONS (SHP)

We give the definition of selectively Hamming distance preserving opertions
(SHP) as follows.

######  Definition 2.

Let
$\mathcal{E}^{\textrm{SHP}}:\mathcal{L}(\mathcal{H}^{n})\mapsto\mathcal{L}(\mathcal{H}^{n})$
be a CPTP map. Then $\mathcal{E}^{\textrm{SHP}}$ belongs to selectively
Hamming distance preserving operations (SHP) if it has a Kraus decomposition
$\mathcal{E}^{\textrm{SHP}}(\cdot)=\sum_{l}K_{l}\cdot K_{l}^{\dagger}$ such
that

$\langle i|K_{l}|x\rangle\langle y|K_{l}^{\dagger}|j\rangle=0,\ \forall l,$
(46)

if $h(i,j)\neq h(x,y)$.

By Proposition 1, SHP is equivalent to the second set of free operations
$\mathbb{F}_{2}$ defined in Eq. (13). Hence, we denote $\mathbb{F}_{2}$ as SHP
here and after.

The following proposition gives the explicit form of the Kraus operators of a
selectively Hamming distance preserving operation.

######  Proposition 3.

If $\mathcal{E}^{\textrm{SHP}}\in\mathrm{SHP}$, then
$\mathcal{E}^{\textrm{SHP}}$ has a Kraus decomposition
$\mathcal{E}^{\textrm{SHP}}=\sum_{l}K_{l}(\cdot)K_{l}^{\dagger}$ such that
every Kraus operator can be written as

$K_{l}=\sum_{x=0}^{2^{n}-1}c_{lx}|\pi_{l}^{n}(x)\rangle\langle x|,$ (47)

where $\sum_{l}|c_{lx}|^{2}=1$, and
$\pi_{l}^{n}:\\{1,...,2^{n}-1\\}\mapsto\\{1,...,2^{n}-1\\}$ is a one-to-one
map such that

$h(x,y)=h(\pi_{l}^{n}(x),\pi_{l}^{n}(y)),\forall x,y.$ (48)

We define $\\{\pi_{j}^{n}\\}$ as Hamming distance preserving functions and
discuss them in Appendix B.

###### Proof.

For $x=y$, Eq. (46) reduces to

$\langle i|K_{l}|x\rangle\langle
x|K_{l}^{\dagger}|j\rangle=\delta_{ij}|\langle i|K_{l}|x\rangle|^{2}.$ (49)

Similarly, we have

$\langle i|K_{l}|x\rangle\langle
y|K_{l}^{\dagger}|i\rangle=\delta_{xy}|\langle i|K_{l}|x\rangle|^{2}.$ (50)

Hence, there is at most one non-zero element in each row and each column of
$K_{l}$. The Kraus operators can then be written as

$K_{l}=\sum_{x=0}^{2^{n}-1}c_{lx}|f_{l}(x)\rangle\langle x|,$ (51)

where $f_{l}:\\{1,...,2^{n}-1\\}\mapsto\\{1,...,2^{n}-1\\}$ is a one-to-one
map and the completion identity $\sum_{l}K_{l}^{\dagger}K_{l}=\mathbb{I}$
demands $\sum_{l}|c_{lk}|^{2}=1$.

In order to prove that $f_{l}(x)$ is a Hamming distance preserving function,
we substitute the above form of $K_{l}$ in Eq. (46). It follows that the
necessary condition for

$\langle i|f_{l}(x)\rangle\langle f_{l}(y)|j\rangle c_{lx}c_{ly}^{*}\neq 0$
(52)

is that $f_{l}(x)=i,\ f_{l}(y)=j$, and $h(x,y)=h(i,j)$. In other words, for
any two coefficients $c_{lx},c_{ly}\neq 0$, it holds that

$h(x,y)=h(f_{l}(x),f_{l}(y)),$ (53)

or equivalently, $f_{l}(x)$ is a Hamming distance preserving function. ∎

For a resource theory, the states, which can be transformed to any state of
the same dimension under free operations, are called golden states. To explore
the existence of golden states is one of the most notable problems for a
resource theory. The golden states of the present resource theory are the
optimal probe states for dephasing estimation. This resource-theoretic way can
convert the calculation of the QFI into a state transformation problem. Here
we will prove that the uniform superposition states, denoted as

$|+^{(n)}\rangle=\frac{1}{\sqrt{2^{n}}}\sum_{x=0}^{2^{n}-1}e^{-i\eta_{x}}|x\rangle,\
\eta_{x}\in[0,2\pi),$ (54)

are golden states in the present resource theory.

######  Proposition 4.

An $n$-qubit uniform superposition state can be transformed to any $n$-qubit
state by SHP.

###### Proof.

By definition, SHP is convex. Hence it is sufficient to prove that any
$n$-qubit pure state can be obtained from $|+^{(n)}\rangle$ by SHP. The
general form of an $n$-qubit pure state reads
$|\psi^{(n)}\rangle=\sum_{x=0}^{2^{n}-1}e^{-i\phi_{x}}\sqrt{\psi_{x}}|x\rangle$,
where $\phi_{x}\in[0,2\pi)$ and $\psi_{x}\geq 0$ satisfying
$\sum_{x}\psi_{x}=1$.

Here we design a selectively Hamming distance preserving operation
$\mathcal{S}=\sum_{z=0}^{2^{n}-1}K_{z}\cdot K_{z}^{\dagger}$, which transforms
$|+^{(n)}\rangle$ to $|\psi^{(n)}\rangle$. Each Kraus operator is in the
following form

$K_{z}=\sum_{x=0}^{2^{n}-1}e^{-i(\phi_{x}-\eta_{q(z,x)})}\sqrt{\psi_{x}}|x\rangle\langle
q(z,x)|,$ (55)

where $q(z,x)$ is an $n$-bit string with the $m$th bit defined as

$(q(z,x))^{m}_{b}=z^{m}_{b}\oplus x^{m}_{b}.$ (56)

Therefore, $\\{|q(z,x)\rangle\\}_{z}$ is a basis of $n$-qubit Hilbert space
for any $x$, or equivalently, $\sum_{z=0}^{2^{n}-1}|q(z,x)\rangle\langle
q(z,x)|=\mathbb{I}$. It is directly checked that
$\sum_{z}K_{z}^{\dagger}K_{z}=\mathbb{I}$, which ensures that $\mathcal{S}$ is
a CPTP map. Besides, $h(x,x^{\prime})=h(q(z,x),q(z,x^{\prime}))$, which
ensures that $\mathcal{S}\in\mathrm{SHP}$.

Then we calculate

$\begin{split}K_{z}|+^{(n)}\rangle&=\dfrac{1}{\sqrt{2^{n}}}\sum_{x=0}^{2^{n}-1}e^{-i(\phi_{x}-\eta_{q(z,x)})}\sqrt{\psi_{x}}|x\rangle\langle
q(z,x)|\sum_{y=0}^{2^{n}-1}e^{-i\eta_{y}}|y\rangle\\\
&=\dfrac{1}{\sqrt{2^{n}}}\sum_{x=0}^{2^{n}-1}e^{-i\phi_{x}}\sqrt{\psi_{x}}|x\rangle\\\
&=\dfrac{1}{\sqrt{2^{n}}}|\psi^{(n)}\rangle.\end{split}$ (57)

Consequently,
$\mathcal{S}(|+^{(n)}\rangle\langle+^{(n)}|)=|\psi^{(n)}\rangle\langle\psi^{(n)}|$.
∎

The set of golden states contains $n$-qubit product states. For such states,
the resource of entanglement does not provide any advantage in dephasing
estimation. In fact,
$\mathcal{F}_{\theta}^{\mathrm{PD}}(|+^{(n)}\rangle\langle+^{(n)}|)=n\mathcal{F}_{\theta}^{\mathrm{PD}}(|+^{(1)}\rangle\langle+^{(1)}|)$.
This is consistent with the results in [16, 17, 18], where it is found that
dephasing estimation cannot beat the standard quantum limit.

It is interesting to notice that the set of golden states also contains
maximally entangled states, e.g., the two-qubit state
$\frac{1}{2}(|00\rangle+|01\rangle+|10\rangle-|11\rangle)$. For such states,
the local coherence of each qubit vanishes but the entanglement reaches
maximum.

To end this subsection, we introduce the concept of Hamming distance
preserving unitary operations, which are free unitary operations in the
resource theory of dephasing estimation.

######  Definition 3.

Let $\mathcal{U}(\cdot)=U_{j}\cdot U_{j}^{\dagger}$ be a unitary operation,
where $U_{j}$ is a unitary operator. Then $\mathcal{U}$ is said to be a
Hamming distance preserving unitary operation if and only if

$\mathcal{E}^{\mathrm{PD}}_{\theta}(U_{j}\rho{U_{j}}^{\dagger})=U_{j}\mathcal{E}^{\mathrm{PD}}_{\theta}(\rho){U_{j}}^{\dagger},\
\forall\rho.$ (58)

Because a Hamming distance preserving unitary operation belongs to
$\mathrm{SHP}$, from Proposition 3, we arrive at following structure of these
unitary operators.

######  Corollary 2.

Any $n$-qubit unitary operator satisfies Eq. (58) if and only if it can be
written as the form

$U_{j}=\sum_{x=0}^{2^{n}-1}e^{-i\omega_{x}}|\pi_{j}^{n}(x)\rangle\langle
x|\quad\forall j,\rho,$ (59)

where $\pi_{j}^{n}(\cdot)$ is a Hamming distance preserving function.

It is worth noticing that the inverse of $U_{j}$ still belongs to free unitary
operations for dephasing estimation. It means that the probe state preserves
the ability of dephasing estimation under Hamming distance preserving unitary
operations.

###  III.3 THE RELATION BETWEEN SHP AND HDP

In this subsection, we investigate the relation of SHP and HDP for single-
qubit and multiqubit states.

######  Proposition 5.

HDP is equivalent to SHP in the single-qubit case.

###### Proof.

Let $\mathcal{D}(\cdot)=\sum_{a}D_{a}(\cdot)D_{a}^{\dagger}$ be a single-qubit
CPTP map which belongs to HDP. The Choi-Jamiołkowski matrix of the
$d$-dimension operation $\mathcal{E}$ is defined as [26, 27]

$J_{\mathcal{E}}=\begin{bmatrix}\mathcal{E}(|0\rangle\langle
0|)&\cdots&\mathcal{E}(|0\rangle\langle
j|)&\cdots&\mathcal{E}(|0\rangle\langle d|)\\\
\vdots&\ddots&\vdots&\ddots&\vdots\\\ \mathcal{E}(|i\rangle\langle
0|)&\cdots&\mathcal{E}(|i\rangle\langle
j|)&\cdots&\mathcal{E}(|i\rangle\langle d|)\\\
\vdots&\ddots&\vdots&\ddots&\vdots\\\ \mathcal{E}(|d\rangle\langle
0|)&\cdots&\mathcal{E}(|d\rangle\langle
j|)&\cdots&\mathcal{E}(|d\rangle\langle d|)\\\ \end{bmatrix}.$ (60)

When the operation $\mathcal{D}$ belongs to $\mathrm{HDP}$, the corresponding
Choi-Jamiołkowski matrix is written as

$J_{\mathcal{D}}=\begin{bmatrix}p_{0|0}&0&0&\gamma_{0}\\\
0&p_{1|0}&\gamma_{1}&0\\\ 0&\gamma_{1}^{*}&p_{0|1}&0\\\
\gamma_{0}^{*}&0&0&p_{1|1}\end{bmatrix},$ (61)

where $\gamma_{0}=\sum_{a}\langle 0|D_{a}|0\rangle\langle
1|D_{a}^{\dagger}|1\rangle$, $\gamma_{1}=\sum_{a}\langle
1|D_{a}|0\rangle\langle 1|D_{a}^{\dagger}|0\rangle$, and
$p_{x|y}=\sum_{a}\langle x|D_{a}|y\rangle\langle y|D_{a}^{\dagger}|x\rangle$
with $x,y=0,1$.

Now consider an operation $\mathcal{S}\in\mathrm{SHP}$, which is related to
$\mathcal{D}$ as
$\mathcal{S}(\cdot)=\sum_{a}S_{a}(\cdot)S_{a}^{\dagger}+\sum_{a}T_{a}(\cdot)T_{a}^{\dagger}$
with

$\begin{split}S_{a}&=\begin{bmatrix}\langle 0|D_{a}|0\rangle&0\\\ 0&\langle
1|D_{a}|1\rangle\end{bmatrix},\\\ T_{a}&=\begin{bmatrix}0&\langle
0|D_{a}|1\rangle\\\ \langle 1|D_{a}|0\rangle&0\end{bmatrix}.\end{split}$ (62)

It is directly checked that the Choi-Jamiołkowski matrix of $\mathcal{S}$ is
also Eq. (61), and therefore, $\mathcal{S}=\mathcal{D}$. ∎

Furthermore, in Appendix A we prove that, for single-qubit states,
$\mathrm{HDP}$ does not represent all the operations, which cannot increase
the QFI of dephasing estimation.

######  Proposition 6.

In the multiqubit case, SHP $\subsetneq$ HDP.

SHP is subset to HDP by their definitions. Then we give a channel which
belongs to HDP but not to SHP. The channel can be written as
$\mathcal{W}(\cdot)=\sum_{i=0}^{4}W_{i}(\cdot)W_{i}^{\dagger}$, where

$\begin{split}W_{0}&=\dfrac{1}{2}|0\rangle\langle
1|+\dfrac{1}{2\sqrt{2}}(|1\rangle\langle 0|+|2\rangle\langle 0|),\\\
W_{1}&=\dfrac{1}{2}|1\rangle\langle 1|+\dfrac{1}{2\sqrt{2}}(|0\rangle\langle
0|+|3\rangle\langle 0|),\\\ W_{2}&=\dfrac{1}{2}|2\rangle\langle
1|+\dfrac{1}{2\sqrt{2}}(|0\rangle\langle 0|-|3\rangle\langle 0|),\\\
W_{3}&=\dfrac{1}{2}|3\rangle\langle 1|+\dfrac{1}{2\sqrt{2}}(|1\rangle\langle
0|-|2\rangle\langle 0|),\\\ W_{4}&=\mathbb{I}-|0\rangle\langle
0|-|1\rangle\langle 1|.\end{split}$ (63)

By the definition of HDP Eq. (15), this channel $\mathcal{W}$ is a Hamming
distance preserving operation. The following lemma is used to prove that
$\mathcal{W}$ does not belong to SHP.

######  Lemma 1 (Theorem 8.2 of [14]).

Consider two given quantum channels
$\mathcal{E}(\cdot)=\sum_{i=1}^{m}E_{i}\cdot E_{i}^{\dagger}$ and
$\mathcal{F}(\cdot)=\sum_{j=1}^{n}F_{j}\cdot F_{j}^{\dagger}$. Then
$\mathcal{E}=\mathcal{F}$ if and only if there is a $m\times n$ linear
isometry $u$ such that $E_{i}=\sum_{j}u_{ij}F_{j}$.

We find that the linear combination of $\left\\{W_{i}\right\\}$ cannot be
expressed as Eq. (47). Based on Lemma 1., $\mathcal{W}$ is not a selective
Hamming distance preserving operation.

##  IV THE COMPARISON BETWEEN THE RESOURCE THEORIES OF DEPHASING ESTIMATION
AND QUANTUM COHERENCE

Although some similarities can be found between the resource theories of
quantum coherence and dephasing estimation, e.g., they share the same set of
free states, there are essential differences between these two resource
theories. Take two states $|\psi_{1}\rangle=(|0\rangle+|1\rangle)/\sqrt{2}$
and $|\psi_{2}\rangle=(|0\rangle+|3\rangle)/\sqrt{2}$ as an example. They have
same amount of quantum coherence, but are different in the ability of
dephasing estimation. In fact, we have
$\mathcal{F}_{\theta}^{\mathrm{PD}}(|\psi_{1}\rangle\langle\psi_{1}|)=e^{-2\theta}/(1-e^{-2\theta})$
but
$\mathcal{F}_{\theta}^{\mathrm{PD}}(|\psi_{2}\rangle\langle\psi_{2}|)=4e^{-4\theta}/(1-e^{-4\theta})$.
The origin of the difference between these two resource theories lies in the
sets of free operations. In this section, we compare their free operations in
both single-qubit and multiqubit cases.

###  IV.1 A BRIEF REVIEW OF THE RESOURCE THEORY OF QUANTUM COHERENCE

In the resource theory of quantum coherence [28], a computational basis
$\\{|i\rangle\\}_{i=0}^{2^{n}-1}$ is prefixed. The completely dephasing (CD)
channel is defined as

$\Delta(\cdot)=\sum_{i=0}^{2^{n}-1}|i\rangle\langle i|\cdot|i\rangle\langle
i|.$ (64)

Then a state $\rho$ is free if and only if $\Delta(\rho)=\rho$. These free
states are called incoherent states.

In the following, we briefly review two sets of free operations. The first set
is called dephasing-covariant incoherent operations (DIO) [29, 30], defined as
quantum operations which commute with the CD channel, i.e.,

$\mathcal{E}\in\mathrm{DIO}\Leftrightarrow\Delta\circ\mathcal{E}=\mathcal{E}\circ\Delta.$
(65)

The second set is called strictly incoherently operations (SIO) [31, 29],
defined as

$\mathcal{E}\in\mathrm{SIO}\Leftrightarrow\Delta\circ\mathcal{E}_{j}=\mathcal{E}_{j}\circ\Delta,\
\forall j,$ (66)

where $\mathcal{E}=\sum_{j}\mathcal{E}_{j},\ \mathcal{E}_{j}(\cdot)=E_{j}\cdot
E_{j}^{\dagger}$, and $E_{j}$ are Kraus operators satisfying
$\sum_{j}E_{j}^{\dagger}E_{j}=\mathbb{I}$. It is worth noting that, the form
of every Kraus operator $E_{j}$ can be written as

$E_{j}=\sum_{x=0}^{2^{n}-1}c_{jx}|e_{j}(x)\rangle\langle x|,$ (67)

where $e_{j}:\\{1,...,2^{n}-1\\}\mapsto\\{1,...,2^{n}-1\\}$ is one-to-one and
$\sum_{j}|c_{jx}|^{2}=1$.

###  IV.2 SINGLE QUBIT OPERATIONS

We first explore the relation between the sets of free operations for single-
qubit case.

######  Proposition 7.

In the single-qubit case, SHP, HDP, SIO, and DIO are equivalent to each other.

###### Proof.

It has been proved in Ref. [29] that DIO$=$SIO for single-qubit case. In
Proposition 5, we proved that $\mathrm{SHP}=\mathrm{HDP}$. Therefore, here we
only need to prove the equivalence between $\mathrm{SIO}$ and $\mathrm{SHP}$.

According to [32], the general form of a single-qubit strictly incoherent
operation is $\mathcal{E}^{\mathrm{SIO}}(\cdot)=\sum_{l=1}^{4}K_{l}\cdot
K_{l}^{\dagger}$, where

$\begin{split}K_{1}&=\begin{bmatrix}a_{1}&0\\\
0&b_{1}\end{bmatrix},K_{2}=\begin{bmatrix}0&b_{2}\\\ a_{2}&0\end{bmatrix},\\\
K_{3}&=\begin{bmatrix}a_{3}&0\\\ 0&0\end{bmatrix},K_{4}=\begin{bmatrix}0&0\\\
a_{4}&0\end{bmatrix},\end{split}$ (68)

where $a_{i}$ is real for $i=1,2,3,4$ and
$\sum_{i=1}^{4}a^{2}_{i}=\sum_{j=1}^{2}\left|b_{j}\right|^{2}=1$. These Kraus
operators satisfy Eq. (46), so SIO $\subset$ SHP for single-qubit states.
Moreover, Proposition 3 implies SHP is a subset of SIO because Hamming
distance preserving functions are one-to-one. Hence SHP$=$SIO. This completes
the proof. ∎

In fact, the HDP cone Eq. $(\ref{SIO_region})$ is also the SIO cone [32, 33],
which reveals the equivalence of SIO and HDP for single-qubit states.

###  IV.3 MULTIQUBIT OPERATIONS

In this subsection, we explore the relation between SHP, HDP, SIO, and DIO in
the multiqubit case. The hierarchy of these sets of operations is shown in
Fig. 2. It is inferred by the definitions that
$\mathrm{SIO}\subseteq\mathrm{DIO}$, $\mathrm{HDP}\subseteq\mathrm{DIO}$ and
$\mathrm{SHP}\subseteq\mathrm{SIO}\cap\mathrm{HDP}$.

Figure 2: The relation between SHP, HDP, SIO and DIO in the multiqubit case.
The boundary between $\mathrm{SHP}$ and the intersection of $\mathrm{SIO}$ and
$\mathrm{HDP}$ is marked with the chain line. This boundary is unclear in the
$2$-qubit case but clear in the $m$-qubit case, where $m\geq 3$.

In the following, we will first show that neither $\mathrm{SIO}$ nor
$\mathrm{HDP}$ contains the other quantum operations belonging to
$\mathrm{SIO}\setminus\mathrm{HDP}$ and $\mathrm{HDP}\setminus\mathrm{SIO}$.
Consider an $n$-qubit ($n\geq 2$) unitary operator

$U=\mathbb{I}-|2\rangle\langle 2|+|2\rangle\langle 3|+|3\rangle\langle
2|-|3\rangle\langle 3|.$ (69)

The corresponding unitary operation belongs to $\mathrm{SIO}$ because it
satisfies Eq. (67). However, because $U|0\rangle\langle
2|U^{\dagger}=|0\rangle\langle 3|$ and $h(0,2)=1\neq 2=h(0,3)$, this unitary
operation does not belong to $\mathrm{HDP}$.

Next consider the $n$-qubit ($n\geq 2$) operation $\mathcal{W}$ in Eq. (63).
It belongs to $\mathrm{HDP}$. It can be checked that any linear combination of
the five Kraus operators of $\mathcal{W}$ is not of the form as Eq. (67).
Based on Lemma 1, $\mathcal{W}\in\mathrm{HDP}\setminus\mathrm{SIO}$.

Now we will show that $\mathrm{SHP}$ is strictly subset to
$\mathrm{SIO}\cap\mathrm{HDP}$ in the $m$-qubit case, where $m\geq 3$. This is
achieved by constructing the following channel
$\mathcal{R}(\cdot)=\sum_{i=0}^{4}R_{i}\cdot R_{i}^{\dagger}$, where

$\begin{split}R_{0}&=\dfrac{1}{\sqrt{2}}|0\rangle\langle
0|+\dfrac{1}{2}|2\rangle\langle 1|+\dfrac{1}{2}|3\rangle\langle 6|,\\\
R_{1}&=\dfrac{1}{\sqrt{2}}|6\rangle\langle 0|+\dfrac{1}{2}|2\rangle\langle
1|-\dfrac{1}{2}|3\rangle\langle 6|,\\\ R_{2}&=\dfrac{1}{2}|1\rangle\langle
1|+\dfrac{1}{2}|6\rangle\langle 6|+\dfrac{1}{\sqrt{2}}|7\rangle\langle 7|,\\\
R_{3}&=\dfrac{1}{2}|4\rangle\langle 1|-\dfrac{1}{2}|3\rangle\langle
6|+\dfrac{1}{\sqrt{2}}|7\rangle\langle 7|,\\\
R_{4}{\Large}&=\mathbb{I}-|0\rangle\langle 0|-|1\rangle\langle
1|-|6\rangle\langle 6|-|7\rangle\langle 7|.\\\ \end{split}$ (70)

The channel $\mathcal{R}\in\mathrm{SIO}\cap\mathrm{HDP}$ belongs to the
intersection of SIO and HDP but not to SHP by lemma 1, that is,
$\mathcal{R}\in\mathrm{SIO}\cap\mathrm{HDP}$ for $m$-qubit states. However, it
is unclear whether $\mathrm{SHP}=\mathrm{SIO}\cap\mathrm{HDP}$ in the
$2$-qubit case.

We conclude this subsection by showing that $\mathrm{SIO}\cup\mathrm{HDP}$ is
the strictly subset to $\mathrm{DIO}$. As proved in Ref. [34], the quantum
operation $\mathcal{N}(\cdot)=\sum_{i=0}^{4}N_{i}(\cdot)N_{i}^{\dagger}$ with

$\begin{split}N_{0}&=\dfrac{1}{2}|0\rangle\langle
1|+\dfrac{1}{2\sqrt{3}}|1\rangle\langle
0|-\dfrac{1}{2\sqrt{3}}|2\rangle\langle
0|+\dfrac{1}{2\sqrt{3}}|3\rangle\langle 0|,\\\
N_{1}&=\dfrac{1}{2\sqrt{3}}|0\rangle\langle
0|+\dfrac{1}{\sqrt{2}}|0\rangle\langle 2|+\dfrac{1}{\sqrt{6}}|0\rangle\langle
3|+\dfrac{1}{2}|1\rangle\langle 1|\\\ &+\dfrac{1}{2\sqrt{3}}|2\rangle\langle
0|+\dfrac{1}{2\sqrt{3}}|3\rangle\langle 0|,\\\
N_{2}&=\dfrac{1}{2\sqrt{3}}|0\rangle\langle
0|-\dfrac{1}{\sqrt{2}}|0\rangle\langle 2|+\dfrac{1}{\sqrt{6}}|0\rangle\langle
3|+\dfrac{1}{2\sqrt{3}}|1\rangle\langle 0|\\\ &+\dfrac{1}{2}|2\rangle\langle
1|-\dfrac{1}{2\sqrt{3}}|3\rangle\langle 0|,\\\
N_{3}&=\dfrac{1}{2\sqrt{3}}|0\rangle\langle
0|-\dfrac{\sqrt{6}}{3}|0\rangle\langle 3|-\dfrac{1}{2\sqrt{3}}|1\rangle\langle
0|-\dfrac{1}{2\sqrt{3}}|2\rangle\langle 0|\\\ &+\dfrac{1}{2}|3\rangle\langle
1|,\\\ N_{4}&=\mathbb{I}-(|0\rangle\langle 0|+|1\rangle\langle
1|+|2\rangle\langle 2|+|3\rangle\langle 3|),\end{split}$ (71)

belongs to $\mathrm{DIO}\setminus\mathrm{SIO}$. Because
$\mathcal{N}(|1\rangle\langle 2|)=\dfrac{1}{2\sqrt{2}}(|1\rangle\langle
0|-|2\rangle\langle 0|)$, and $h(1,2)\neq h(0,1)$ (or $h(1,2)\neq h(0,2)$),
$\mathcal{N}$ is not in $\mathrm{HDP}$.

##  V THE COMPARISON BETWEEN THE RESOURCE THEORIES OF DEPHASING ESTIMATION AND
$U(1)$ ASYMMETRY

The QFI of phase estimation is a resource measure of $U(1)$ asymmetry [21, 35,
25]. It has been shown that there is a probe incompatibility between phase
estimation and dephasing estimation, namely, no optimal probe states exist for
detecting phase and dephasing parameters simultaneously [15, 36]. This
motivates us to compare the resource theories of $U(1)$ asymmetry and
dephasing estimation.

###  V.1 A BRIEF REVIEW OF QUANTUM FISHER INFORMATION OF PHASE ESTIMATION

In this subsection, we briefly review the QFI of phase estimation. Given
Hamiltonian $H$ and the probe state $\rho$, $\rho(t)\equiv e^{-iHt}\rho
e^{iHt}$ represents the state of system at time $t$. The task of phase
estimation is to estimate the parameter $t$ by quantum measurement. Let
$\rho=\sum_{i}p_{i}|\psi_{i}\rangle\langle\psi_{i}|$ be the spectral
decomposition of the probe state $\rho$. The QFI of phase estimation can be
calculated as [37, 19, 8]

${\mathcal{F}}_{H}^{\mathrm{PE}}(\rho)=2\sum_{i,k}\dfrac{(p_{i}-p_{k})^{2}}{p_{i}+p_{k}}\left|\langle\psi_{i}|H|\psi_{k}\rangle\right|^{2}.$
(72)

Furthermore, when the probe is in a pure state $|\psi\rangle\langle\psi|$, the
QFI reduces to

${\mathcal{F}}_{H}^{\mathrm{PE}}(|\psi\rangle\langle\psi|)=4\Delta H^{2},$
(73)

where $\Delta
H^{2}:=\langle\psi|H^{2}|\psi\rangle-\langle\psi|H|\psi\rangle^{2}$ is the
variance of Hamiltonian $H$.

In this section, we consider the setting of $n$ qubits with no interaction.
For each qubit, the ground state energy is zero and the excited state energy
is $\mathfrak{h}$. The form of $n$-qubit Hamiltonian is given by

$H=\sum_{i=0}^{2^{n}-1}N^{i}_{1}\mathfrak{h}|i\rangle\langle i|,$ (74)

where $N^{i}_{1}$ represents the number of $1$’s in the binary string $i_{b}$.

###  V.2 SINGLE QUBIT CASE

For $n=1$, the Hamiltonian reads $H=\mathfrak{h}|1\rangle\langle 1|$. For a
single-qubit probe state $\sigma$, whose Bloch vector is
$(r\cos{\phi},r\sin{\phi},z)$, its spectral decomposition is

$\sigma=\frac{1}{2}[(1+\sqrt{z^{2}+r^{2}})|+r\rangle\langle+r|+(1-\sqrt{z^{2}+r^{2}})|-r\rangle\langle-r|],$
(75)

where eigenstates are given by

$\displaystyle|+r\rangle$ $\displaystyle=$
$\displaystyle\dfrac{(z+\sqrt{z^{2}+r^{2}})|0\rangle+re^{i\phi}|1\rangle}{\sqrt{r^{2}+(z+\sqrt{z^{2}+r^{2}})^{2}}},$
$\displaystyle|-r\rangle$ $\displaystyle=$
$\displaystyle\dfrac{-r|0\rangle+(z+\sqrt{z^{2}+r^{2}})e^{i\phi}|1\rangle}{\sqrt{r^{2}+(z+\sqrt{z^{2}+r^{2}})^{2}}}.$
(76)

Then the phase estimation QFI of $\sigma$ is calculated as

${\mathcal{F}}_{H}^{\mathrm{PE}}(\sigma)=r^{2}\mathfrak{h}^{2}.$ (77)

Therefore, in the single-qubit case, $\mathcal{F}_{\theta}^{\mathrm{PD}}$
cannot be increased under HDP because Eq. (26) shows that $r$ is non-
increasing under HDP.

Nevertheless, by comparing Eqs. (77) and (83), we find that
${\mathcal{F}}_{H}^{\mathrm{PE}}$ and $\mathcal{F}_{\theta}^{\mathrm{PD}}$
give different ordering of states even in the single-qubit case. It means that
even though both ${\mathcal{F}}_{H}^{\mathrm{PE}}$ and
$\mathcal{F}_{\theta}^{\mathrm{PD}}$ reaches maximum for uniform superposition
states, there is a possibility that for two single-qubit states $\sigma_{1}$
and $\sigma_{2}$,
${\mathcal{F}}_{H}^{\mathrm{PE}}(\sigma_{1})>{\mathcal{F}}_{H}^{\mathrm{PE}}(\sigma_{2})$
but
$\mathcal{F}_{\theta}^{\mathrm{PD}}(\sigma_{1})<\mathcal{F}_{\theta}^{\mathrm{PD}}(\sigma_{2})$.

###  V.3 MULTIQUBIT CASE

In the multiqubit case, the golden states in the resource theory of dephasing
estimation are uniform superposition states $|+^{(n)}\rangle$, while
${\mathcal{F}}_{H}^{\mathrm{PE}}$ reaches its maximum for GHZ states defined
as $|GHZ\rangle=(|0\rangle+e^{-i\eta_{0}}|2^{n}-1\rangle)/\sqrt{2}$ [11]. This
discordance in the maximally resourceful states is a possible origin of the
probe incompatibility, which stands for the absence of an optimal probe state
for detecting phase and dephase parameters simultaneously [36, 15]. It is then
natural to ask the following question: in what circumstances, can one increase
${\mathcal{F}}_{H}^{\mathrm{PE}}$ of a probe state, and meanwhile preserve its
ability of dephasing estimation?

An illustrative example is as follows. Consider a two-qubit pure state
$|\psi_{1}\rangle=\frac{1}{\sqrt{2}}(|1\rangle+|2\rangle)$. Because it is an
eigenstate of the Hamiltonian, we have
${\mathcal{F}}_{H}^{\mathrm{PE}}(|\psi_{1}\rangle\langle\psi_{1}|)=0$.
However, by applying a Hamming distance preserving unitary

$V=\begin{bmatrix}0&1&0&0\\\ 1&0&0&0\\\ 0&0&0&1\\\ 0&0&1&0\\\ \end{bmatrix},$
(78)

$|\psi_{1}\rangle$ is converted to
$|\psi_{2}\rangle=V|\psi_{1}\rangle=(|0\rangle+|3\rangle)/\sqrt{2}$, which is
one of the optimal probe states for phase estimation. It means that a free
unitary operation in the resource theory of dephasing estimation can transform
a free state in the resource theory of $U(1)$ asymmetry to a maximally
resourceful state in the latter resource theory. This shows the
incompatibility of the two resource theories.

This example is generalized to general two-qubit pure states as follows.

######  Proposition 8.

Let $|\psi^{(2)}\rangle=\sum_{x=0}^{3}e^{-i\phi_{x}}\sqrt{\psi_{x}}|x\rangle$
be a two-qubit pure state. Hamming distance preserving unitary operations can
increase the QFI of $|\psi^{(2)}\rangle$ in phase estimation, if and only if
$g(\psi_{1},\psi_{2})>g(\psi_{0},\psi_{3})$, where $g(x,y)\equiv
x+y-(x-y)^{2}$.

###### Proof.

The QFI of the state $|\psi^{(2)}\rangle$ in phase estimation is calculated as

${\mathcal{F}}_{H}^{\mathrm{PE}}(|\psi^{(2)}\rangle\langle\psi^{(2)}|)=g(\psi_{0},\psi_{3})\mathfrak{h}^{2}.$
(79)

After the action of the unitary $V$, the QFI of phase estimation becomes

${\mathcal{F}}_{H}^{\mathrm{PE}}(V|\psi^{(2)}\rangle\langle\psi^{(2)}|V^{\dagger})=g(\psi_{1},\psi_{2})\mathfrak{h}^{2},$
(80)

which is larger than Eq. (79) if $g(\psi_{1},\psi_{2})>g(\psi_{0},\psi_{3})$.

Moreover, when other Hamming preserving distance unitary operations are
performed on the state $|\psi^{(2)}\rangle$, the resulted
${\mathcal{F}}_{H}^{\mathrm{PE}}$ is either in the form of Eq. (79) or Eq.
(80). Therefore, the QFI of $|\psi^{(2)}\rangle$ in phase estimation can be
increased only if $g(\psi_{1},\psi_{2})>g(\psi_{0},\psi_{3})$. ∎

Interestingly, some nonunitary operations in SHP also have the ability in
improving ${\mathcal{F}}_{H}^{\mathrm{PE}}$ while preserving
$\mathcal{F}_{\theta}^{\mathrm{PD}}$. Consider an $n$-qubit mixed state in the
following form

$\rho=\sum_{x:x<\bar{x}}p_{x}|\psi(x)\rangle\langle\psi(x)|,$ (81)

where $|\psi(x)\rangle=\cos\zeta|x\rangle+\sin\zeta
e^{i\phi_{x}}|\bar{x}\rangle$, $\bar{x}$ stands for the bitwise bit flip of
$x$, and $\zeta$ is independent of $x$. It satisfies conditions (C1) and (C2)
for coherence merge in the same Hamming mode. Therefore, we apply the merging
operation defined in Eq. (III.1) with $i=0$, $j=2^{n}-1$, $y(x)=\bar{x}$,
$p_{i|x}=p_{j|\bar{x}}=1$ and $p_{j|x}=p_{i|\bar{x}}=0$. This results in a
pure state of the form

$\rho^{\prime}=|\psi(0)\rangle\langle\psi(0)|.$ (82)

By the convexity of QFI, we have
${\mathcal{F}}_{H}^{\mathrm{PE}}(\rho^{\prime})>\sum_{x}p_{x}{\mathcal{F}}_{H}^{\mathrm{PE}}(|\psi(x)\rangle\langle\psi(x)|)\geq{\mathcal{F}}_{H}^{\mathrm{PE}}(\rho)$.
This means that ${\mathcal{F}}_{H}^{\mathrm{PE}}$ is increased. Meanwhile,
$\mathcal{F}_{\theta}^{\mathrm{PD}}(\rho)=\mathcal{F}_{\theta}^{\mathrm{PD}}(\rho^{\prime})$.
The reason is as follows. On the one hand, from the monotonicity of
$\mathcal{F}_{\theta}^{\mathrm{PD}}$ under SHP, we have
$\mathcal{F}_{\theta}^{\mathrm{PD}}(\rho)\geq\mathcal{F}_{\theta}^{\mathrm{PD}}(\rho^{\prime})$.
On the other hand, from the convexity of QFI, we have
$\mathcal{F}_{\theta}^{\mathrm{PD}}(\rho^{\prime})=\sum_{x}p_{x}\mathcal{F}_{\theta}^{\mathrm{PD}}(|\psi(x)\rangle\langle\psi(x)|)\geq\mathcal{F}_{\theta}^{\mathrm{PD}}(\rho)$.

To sum up, one can use coherence merge in the same Hamming mode to increase
the ability of a probe state in phase estimation, and meanwhile, preserve its
ability of dephasing estimation.

##  VI CONCLUSION

In this work, we study the resource theory of dephasing estimation in
multiqubit systems. Due to the monotonicity of QFI, we define two sets of free
operations in the resource theory of dephasing estimation, HDP and SHP. Under
these free operations, the problem of finding optimal probes can be converted
to the problem of state transformations. This resource-theoretic approach
reduces the complexity to calculate the QFI. In the present resource theory,
the uniform superposition states are golden states, that is, the uniform
superposition states are optimal probe states for dephasing estimation. This
implies that there exists the connection between our resource theory and the
resource theory of quantum coherence. Therefore, we investigate the relation
between free operations of these two resource theories. In the single-qubit
case, SIO and DIO, which are free operations of the resource theory of quantum
coherence, are equivalent to SHP (or HDP). Furthermore, for multiqubit states,
the relation of these operations is depicted in FIG.2. Finally, we also
compare the resource theories of $U(1)$ asymmetry and dephasing estimation. By
employing SHP, it is possible to enhance the performance of a probe state in
phase estimation, while preserving its ability of dephasing estimation, which
implies the incompatibility of these two resource theories.

###### Acknowledgements.

This work was Supported by National Natural Science Foundation of China under
Grant No. 11774205.

##  Appendix A SINGLE-QUBIT HDP DOES NOT REPRESENT ALL SINGLE-QUBIT OPERATIONS
THAT CANNOT INCREASE THE QFI OF DEPHASING ESTIMATION

For a single-qubit state $\sigma=(r\cos\phi,r\sin\phi,z)$, the QFI of
dephasing estimation is independent on the phase $\phi$ due to the rotational
symmetry. Therefore, we only consider the case $\phi=0$ in the following
discussion. For a single-qubit state $\sigma_{0}=(r,0,z)$, the QFI of
dephasing estimation is

$\mathcal{F}_{\theta}^{\mathrm{PD}}(\sigma_{0})=\dfrac{r^{2}e^{-2\theta}(1-z^{2})}{1-z^{2}-r^{2}e^{-2\theta}}.$
(83)

In the following, we will show a channel $\mathcal{Z}\notin\mathrm{HDP}$,
which does not increase the QFI of phase estimation. The Choi matrix of the
channel $\mathcal{Z}$ is given by

$J_{\mathcal{Z}}=\begin{bmatrix}\frac{1}{2}&0&\frac{1}{4}&\frac{1}{4}\\\
0&\frac{1}{2}&\frac{1}{4}&-\frac{1}{4}\\\
\frac{1}{4}&\frac{1}{4}&\frac{1}{2}&0\\\
\frac{1}{4}&-\frac{1}{4}&0&\frac{1}{2}\end{bmatrix}.$ (84)

Its Kraus operators are given by

$\begin{split}Z_{0}&=\dfrac{1}{2}\begin{bmatrix}1&0\\\
0&1\end{bmatrix},Z_{1}=\dfrac{1}{2}\begin{bmatrix}0&1\\\ 1&0\end{bmatrix},\\\
Z_{2}&=\dfrac{1}{2}\begin{bmatrix}1&1\\\
0&0\end{bmatrix},Z_{3}=\dfrac{1}{2}\begin{bmatrix}0&0\\\
1&-1\end{bmatrix}.\end{split}$ (85)

Then the Bloch vector of $\mathcal{Z}(\sigma_{0})$ can be written as
$(\frac{1}{2}r,0,\frac{1}{2}r)$. The corresponding QFI of dephasing estimation
is

$\mathcal{F}_{\theta}^{\mathrm{PD}}(\mathcal{Z}(\sigma_{0}))=\dfrac{\frac{1}{4}r^{2}e^{-2\theta}(1-\frac{1}{4}r^{2})}{1-\frac{1}{4}r^{2}-\frac{1}{4}r^{2}e^{-2\theta}}$
(86)

When $r=0$ (or $z^{2}=1$), we find
$\mathcal{F}_{\theta}^{\mathrm{PD}}(\sigma_{0})=\mathcal{F}_{\theta}^{\mathrm{PD}}(\mathcal{Z}(\sigma_{0}))=0$.
Then we discuss the case $r\neq 0$. The inequality,
$\mathcal{F}_{\theta}^{\mathrm{PD}}(\sigma_{0})\geq\mathcal{F}_{\theta}^{\mathrm{PD}}(\mathcal{Z}(\sigma_{0}))$,
is equivalent to

$\dfrac{3}{r^{2}e^{-2\theta}}-\dfrac{4}{4-r^{2}}+\dfrac{1}{1-z^{2}}\geq
0,\forall r,z.$ (87)

When $z^{2}=0$ and $r^{2}=1$, the left side of the inequality reaches its
minimum $3e^{2\theta}-1/3$, which is always greater than $0$. Therefore, the
channel $\mathcal{Z}$ cannot increase the QFI of dephasing estimation and does
not belong to $\mathrm{HDP}$.

##  Appendix B HAMMING DISTANCE PRESERVING FUNCTION

In this appendix, we first give the definition of a Hamming distance
preserving function and then derive the explicit forms of these functions.

######  Definition 4.

Let $S^{n}$ be the set of $n$-bit strings. The one-to-one function
$\pi^{n}_{i}:S^{n}\mapsto S^{n}$ is said to be a Hamming distance preserving
function if

$h(x,y)=h(\pi^{n}_{i}(x),\pi^{n}_{i}(y)),\forall x,y\in S^{n}.$ (88)

The set of $n$-bit Hamming distance preserving functions is denoted as
$M^{n}$.

Next we study the explicit form of Hamming distance preserving functions. We
start with the simplest case where the input/output string consists of a
single bit. Obviously, $M^{1}=\\{\pi^{1}_{0},\pi_{1}^{1}\\}$ with

$\displaystyle\pi^{1}_{0}:$ $\displaystyle 0\mapsto 0,1\mapsto 1,$
$\displaystyle\pi_{1}^{1}:$ $\displaystyle 0\mapsto 1,1\mapsto 0.$ (89)

For two-bit case, there are eight elements in $M^{2}$. Precisely, we denote by
$x_{m}$ the $m$-th bit of string $x$, and define two sets of two-bit
functions, $R^{2}=\\{r_{0}^{2},r_{1}^{2}\\}$ with
$r_{0}^{2}(x_{1}x_{0})=x_{1}x_{0}$, $r_{1}^{2}(x_{1}x_{0})=x_{0}x_{1}$, and
$Q^{2}=\\{q_{z}\\}_{z=0}^{3}$ with $(q_{z}(x))_{l}=z_{l}\oplus x_{l},\ l=0,1$.
Then the set of two-bit Hamming distance preserving functions can be written
as

$M^{2}=Q^{2}\circ R^{2}.$ (90)

Here by $\circ$ we mean $A\circ B=\\{m|m=ab,a\in A,b\in B\\}$. This two-bit
case can be generalized to the $n$-bit case.

######  Proposition 9.

The set of $n$-bit Hamming distance preserving functions can be written as

$M^{n}=Q^{n}\circ R^{n},$ (91)

where $Q^{n}:=\\{q_{z}|q_{z}:S^{n}\mapsto S^{n},(q_{z}(x))_{l}=z_{l}\oplus
x_{l},l\in\\{0,\dots,n-1\\}\\}_{z=0}^{2^{n}-1}$, and $R^{n}$ represents the
set of $n$-bit functions which reorder the bits in an input string.

###### Proof.

Firstly, both $r\in R^{n}$ and $q\in Q^{n}$ are Hamming distance preserving
functions, so their multiplications also preserve Hamming distance, i.e.,
$Q^{n}\circ R^{n}\subseteq M^{n}$. In order to prove Eq. (91), we only need to
prove that $|M^{n}|=|Q^{n}\circ R^{n}|$, where $|X|$ denotes the number of
elements in the set $X$. For this purpose, we check that $|Q_{n}|=2^{n}$,
$|R^{n}|=n!$, and $|Q^{n}\circ R^{n}|=|Q^{n}|\cdot|R^{n}|$. In the following,
we focus on proving that

$|M^{n}|=2^{n}n!.$ (92)

Let $M_{z}^{n}:=\\{m|m\in M^{n},m:0\mapsto z\\}$ with $z\in S^{n}$ be a subset
of $M^{n}$. Then $\left|M_{z}^{n}\right|=\left|M_{0}^{n}\right|$ for all
$z=0,\dots,2^{n}-1$, because $M_{z}=q_{z}^{n}M_{0}$. It follows that

$|M^{n}|=2^{n}\left|M_{0}^{n}\right|.$ (93)

Next we consider a function $m\in M_{0}^{n}$ which satisfies $m:2^{0}\mapsto
i$. From the Hamming distance preserving condition $h(0,1)=h(0,i)$, there is
exactly one bit in string $i$ which equals to $1$, and hence $i=2^{j}$ with
$j=0,\dots,n$. Now we define $M_{0|j}^{n}:=\\{m|m\in M_{0}^{n},m:2^{0}\mapsto
2^{j}\\}$, which is a subset of $M_{0}^{n}$. Then any element $m_{j}\in
M_{0|j}^{n}$ is related to an element $m_{0}\in M_{0|0}^{n}$ via
$m_{j}=c^{n}_{j0}m_{0}$, where $c^{n}_{j0}$ represents an $n$-bit function
which exchanges the position of the zeroth bit and $j$-th bit of its input
string. It follows that $M_{0|j}^{n}=c_{j0}M_{0|0}^{n}$, and consequently,

$\left|M_{0}^{n}\right|=n\left|M_{0|0}^{n}\right|.$ (94)

Further, consider a function $m\in M_{0|0}^{n}$, which satisfies
$m:2^{1}\mapsto i$. The Hamming distance preserving condition requires that
$h(0,2)=h(0,i)$ and $h(1,2)=h(1,i)$. This is equivalent to say that there is
exactly one bit in string $i$ which equals to 1, and moreover, $i^{0}=0$.
Hence, we have $i=2^{j}$ with $j=1,\cdots,n-1$. Let $M_{0|0j}^{n}:=\\{m|m\in
M_{0|0}^{n},m:2^{1}\mapsto 2^{j}\\}$, and then we have
$|M_{0|0}^{n}|=\sum_{j=1}^{n-1}|M_{0|0j}^{n}|$, and
$|M_{0|0j}^{n}|=|M_{0|01}^{n}|$, $\forall j=1,\cdots,n-1$, or equivalently,

$\left|M_{0|0}^{n}\right|=(n-1)\left|M_{0|01}^{n}\right|.$ (95)

We apply the above discussion $n$ times and arrive at

$\left|M_{0}^{n}\right|=n!\left|M_{0|0\cdots n-1}^{n}\right|,$ (96)

where $M_{0|0\cdots n-1}^{n}=\\{m|m\in M^{n},m:2^{j}\mapsto 2^{j},\forall
j=0,\cdots,n-1\\}$.

For $m\in M_{0|0\cdots n-1}^{n}$, the Hamming distance preserving condition
requires that $h(m(x),0)=h(x,0)$ and

$h(m(x),2^{j})=\big{\\{}\begin{array}[]{cc}h(x,0),&x_{j}=0,\\\
h(x,0)-1,&x_{j}=1,\end{array}$ (97)

for all $x\in S^{n}$ and $j=0,\cdots,n-1$. This immediately implies that
$m(x)=x$. Therefore, the only function in $M_{0|0\cdots n-1}^{n}$ is identity,
and

$\left|M_{0|0\cdots n-1}^{n}\right|=1.$ (98)

Putting eqs. (93), (96), and (98) together, we arrive at Eq. (92). ∎

##  Appendix C THE CHANNEL $\mathcal{W}$ BELONGS TO
$\mathrm{HDP}\setminus\mathrm{SIO}$

The channel $\mathcal{W}$ is defined as Eq. (63). Its Kraus operators are
given by

$\begin{split}W_{0}&=\dfrac{1}{2}|0\rangle\langle
1|+\dfrac{1}{2\sqrt{2}}(|1\rangle\langle 0|+|2\rangle\langle 0|),\\\
W_{1}&=\dfrac{1}{2}|1\rangle\langle 1|+\dfrac{1}{2\sqrt{2}}(|0\rangle\langle
0|+|3\rangle\langle 0|),\\\ W_{2}&=\dfrac{1}{2}|2\rangle\langle
1|+\dfrac{1}{2\sqrt{2}}(|0\rangle\langle 0|-|3\rangle\langle 0|),\\\
W_{3}&=\dfrac{1}{2}|3\rangle\langle 1|+\dfrac{1}{2\sqrt{2}}(|1\rangle\langle
0|-|2\rangle\langle 0|),\\\ W_{4}&=\mathbb{I}-|0\rangle\langle
0|-|1\rangle\langle 1|.\end{split}$ (99)

We first introduce the definition of the $l_{1}$ norm of coherence

$C_{l_{1}}(\rho)=\sum_{i,j:i\neq j}|\rho_{i,j}|.$ (100)

The $l_{1}$ norm of coherence is monotonic under $\mathrm{SIO}$ [38], that is,
$\mathcal{C}_{l_{1}}(\mathcal{S}(\rho))\geq\mathcal{C}_{l_{1}}(\rho)$, where
$\mathcal{S}\in\mathrm{SIO}$. Let us take the state $\rho$ as follows,

$\rho=\dfrac{1}{2}(|0\rangle\langle 0|+|0\rangle\langle 1|+|1\rangle\langle
0|+|1\rangle\langle 1|).$ (101)

We calculate $\mathcal{C}_{l_{1}}(\rho)=1$ and
$\mathcal{C}_{l_{1}}(\mathcal{W}(\rho))=\sqrt{2}$. Therefore, the HDP channel
$\mathcal{W}$ can increase the $l_{1}$ norm of coherence. It implies
$\mathcal{W}\in\mathrm{HDP}\setminus\mathrm{SIO}$.

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