# Quantifying unsharpness of observables in an outcome-independent way

Arindam Mitra1,2 1Optics and Quantum Information Group, The Institute of
Mathematical Sciences, C. I. T. Campus, Taramani, Chennai 600113, India.
2Homi Bhabha National Institute, Anushakti Nagar, Mumbai 400094, India.

###### Abstract

Recently, a very beautiful measure of the unsharpness (fuzziness) of the
observables is discussed in the paper [Phys. Rev. A 104, 052227 (2021)]. The
measure which is defined in this paper is constructed via uncertainty and does
not depend on the values of the outcomes. There exist several properties of a
set of observables (e.g., incompatibility, non-disturbance) that do not depend
on the values of the outcomes. Therefore, the approach in the above-said paper
is consistent with the above-mentioned fact and is able to measure the
intrinsic unsharpness of the observables. In this work, we also quantify the
unsharpness of observables in an outcome-independent way. But our approach is
different than the approach of the above-said paper. In this work, at first,
we construct two Luder’s instrument-based unsharpness measures and provide the
tight upper bounds of those measures. Then we prove the monotonicity of the
above-said measures under a class of fuzzifying processes (processes that make
the observables more fuzzy). This is consistent with the resource-theoretic
framework. Then we relate our approach to the approach of the above-said
paper. Next, we try to construct two instrument-independent unsharpness
measures. In particular, we define two instrument-independent unsharpness
measures and provide the tight upper bounds of those measures and then we
derive the condition for the monotonicity of those measures under a class of
fuzzifying processes and prove the monotonicity for dichotomic qubit
observables. Then we show that for an unknown measurement, the values of all
of these measures can be determined experimentally. Finally, we present the
idea of the resource theory of the sharpness of the observables.

##  I Introduction

In quantum mechanics, the observables are mainly of two types-(i) sharp
observables and (ii) unsharp observables. Quantifying the unsharpness of
observables is an interesting research direction to look at. Few works in this
direction have been already done [1, 2, 3, 4, 5, 6, 7, 8, 9]. Recently, in the
Ref. [9], the unsharpness of observables is quantified using uncertainty. The
measure defined in the Ref. [9], is outcome-independent.

In this work, we also quantify the unsharpness of the observables in an
outcome-independent way. But our approach is different than the approach of
the Ref. [9]. We first define two Luder’s instrument-based measures. Then we
discuss the different properties of these measures. Then, we try to construct
two instrument-independent unsharpness measures. We provide a conjecture and
if that can be proven, those instrument-independent measures will be
consistent with the resource-theoretic framework for qubit observables. Then
we discuss that the values of all of these measures can be determined
experimentally. Then we make the justification for taking sharpness as a
resource and also present the idea of the resource theory which can be
completed in the future.

The rest of this paper is organized as follows. In Sec. II, we discuss the
preliminaries. From Sec. III we start discussing our main results. In
particular, in Sec. III.1, we construct two Luder’s instrument-based
unsharpness measures and provide the tight upper bounds of those measures. In
Sec. III.2, we prove the monotonicity of the above-said measures under a class
of fuzzifying processes. In Sec. III.3, we relate our approach to the approach
of the Ref. [9]. In the Sec. IV, we try to construct two instrument-
independent unsharpness measures. In particular, in Sec. IV.1, we define two
instrument-independent unsharpness measures and provide the tight upper bounds
of those measures. In Sec. IV.2, we derive the condition for the monotonicity
of those measures under a class of fuzzifying processes and prove the
monotonicity for dichotomic qubit observables. In Sec. V, we show that for an
unknown measurement, the values of all of these measures can be determined
experimentally. In Sec. VI, we present the idea of the resource theory of the
sharpness of the observables. Finally, in Sec. VII, we summarize our results
and discuss the future outlook.

##  II Preliminaries

In this section, we discuss the preliminaries.

###  II.1 Observables

An observable (positive operator-valued measure or POVM) $\mathcal{A}$ acting
on the Hilbert space $\mathcal{H}$ is defined as a set of positive Hermitian
matrices i.e., $\mathcal{A}=\\{A_{i}\\}^{n}_{i=1}$ such that
$\sum_{i}A_{i}=\mathbb{I}_{d\times d}$ where $d$ is the dimension of the
Hilbert space $\mathcal{H}$ [10, 11, 12]. The set $\\{1,....,n\\}$ is called
outcome set of $\mathcal{A}$ and is denoted by $\Omega_{\mathcal{A}}$. Clearly
$A_{i}\in\mathcal{L}^{+}(\mathcal{H})$ and $\mathbb{I}_{d\times d}\geq
A_{i}\geq 0$ for all $i\in\Omega_{\mathcal{A}}$ where
$\mathcal{L}^{+}(\mathcal{H})$ is the set of positive bounded linear operators
acting on the Hilbert space $\mathcal{H}$. Therefore, $A_{i}^{2}\leq A_{i}$
for all $i\in\Omega_{\mathcal{A}}$. If $A^{2}_{i}=A_{i}$ holds for all
$i\in\Omega_{\mathcal{A}}$, we call $A$ as a projection-valued measure (PVM).
PVMs are the sharp observable and clearly PVMs are the special cases of POVMs.
Clearly, one outcome trivial sharp observable is $\mathbb{I}_{d\times d}$. If
there exist at least one $j\in\Omega_{\mathcal{A}}$ such that
$A^{2}_{j}<A_{j}$ then the observable $\mathcal{A}$ is not a PVM. This type of
observables are unsharp observables [9].

###  II.2 Quantum Channels

A quantum channel
$\Gamma:\mathcal{S}(\mathcal{H})\rightarrow\mathcal{S}(\mathcal{K})$ is a
completely positive trace preserving (CPTP) map where
$\mathcal{S}(\mathcal{H})$ is the state space (i.e., the set of density
matrices on the Hilbert space $\mathcal{H}$) [10, 11]. For a quantum channel
$\Gamma$, $\Gamma(\rho)$ can always be written as
$\Gamma(\rho)=\sum_{i}K_{i}\rho K^{\dagger}_{i}$. This form of $\Gamma$ is
called as the Kraus representation of $\Gamma$ and $K_{i}$’s are called the
Kraus operators of $\Gamma$. The channel
$\Gamma^{*}:\mathcal{L}(\mathcal{K})\rightarrow\mathcal{L}(\mathcal{H})$ is
called as the dual channel (i.e., in Heisenberg picture) of
$\Gamma:\mathcal{S}(\mathcal{H})\rightarrow\mathcal{S}(\mathcal{K})$ if for
all $\rho\in\mathcal{S}(\mathcal{H})$ and $X\in\mathcal{L}(\mathcal{K})$,
$\text{Tr}[\Gamma(\rho)X]=\text{Tr}[\rho\Gamma^{*}(X)]$ holds, where
$\mathcal{L}(\mathcal{H})$ is the set of bounded linear operators on the
Hilbert space $\mathcal{H}$.

A special type of channel is the depolarising channel. A depolarising channel
$\Gamma^{t}_{d}$ is defined as
$\Gamma^{t}_{d}(\rho)=t\rho+(1-t)\frac{\mathbb{I}}{d}$ for all
$\rho\in\mathcal{S}(\mathcal{H})$ and $t\in[-\frac{1}{3},1]$. To simplify the
notation we have written $\mathbb{I}_{d\times d}$ as $\mathbb{I}$.

###  II.3 Quantum Instruments

A quantum instrument $\mathcal{I}$ is a set of completely positive (CP) maps
$\\{\Phi_{i}:\mathcal{S}(\mathcal{H})\rightarrow\mathcal{L}^{+}(\mathcal{K})\\}$
i.e., $\mathcal{I}=\\{\Phi_{i}\\}$ such that $\Phi=\sum_{i}\Phi_{i}$ is a
quantum channel where $\mathcal{L}^{+}(\mathcal{H})$ is the set of positive
bounded linear operators on the Hilbert space $\mathcal{H}$ [11]. Suppose
$\mathcal{A}=\\{A_{i}\\}$ is an observable. A quantum instrument
$\mathcal{I}=\\{\Phi_{i}\\}$ is called $\mathcal{A}$-compatible instrument if
$\text{Tr}[\Phi_{i}(\rho)]=\text{Tr}[\rho A_{i}]$ for all
$i\in\Omega_{\mathcal{A}}$. Therefore, the observable $\mathcal{A}$ can be
measured using the instrument $\mathcal{I}$.

There exist a special type of of instruments which are known as Luder’s
instruments. For an observable $\mathcal{A}=\\{A_{i}\\}$, the
$\mathcal{A}$-compatible Luder’s instrument is defined as
$\mathcal{I}^{L}_{\mathcal{A}}=\\{\Phi^{L}_{A_{i}}\\}^{n_{\mathcal{A}}}_{i=1}$
such that $\Phi^{L}_{A_{i}}(\rho)=\sqrt{A_{i}}\rho\sqrt{A_{i}}$ for all
$i\in\\{1,....,n_{\mathcal{A}}\\}$.

###  II.4 Quantifying unsharpness of observables via uncertainty

In this subsection, we briefly discuss the approach of the Ref. [9]. For the
complete discussion, readers can check the Ref. [9]. Suppose we have an
observable $\mathcal{A}=\\{A_{i}\\}^{n_{\mathcal{A}}}_{i=1}$ acting on the
Hilbert space $\mathcal{H}$. Suppose the exact value of the $i$th outcome is
$\alpha_{i}$. Let $\alpha$ be a row vector such that
$\alpha=(\alpha_{1},\alpha_{2},....,\alpha_{n_{\mathcal{A}}})$. Then
$K^{\mathcal{A}}_{\alpha}$ is defined as
$K^{\mathcal{A}}_{\alpha}=\sum_{i}\alpha_{i}A_{i}$. Similarly,
$K^{\mathcal{A}}_{\alpha^{2}}$ is defined as
$K^{\mathcal{A}}_{\alpha^{2}}=\sum_{i}\alpha^{2}_{i}A_{i}$. Next a noise
operator
$N^{\mathcal{A}}_{\alpha}=K^{\mathcal{A}}_{\alpha^{2}}-(K^{\mathcal{A}}_{\alpha})^{2}$
is introduced. Then a function $F_{\rho}(\mathcal{A},\alpha)$ is introduced
such that

$\displaystyle F_{\rho}(\mathcal{A},\alpha)=\text{Tr}[\rho
N^{\mathcal{A}}_{\alpha}].$ (1)

Clearly, $F_{\rho}(\mathcal{A},\alpha)\geq 0$.
$F_{\rho}(\mathcal{A},\alpha)=0$ for all $\rho\in\mathcal{S}(\mathcal{H})$ iff
$\mathcal{A}$ is a PVM. Now, it is shown in the Ref. [9] that

$\displaystyle F_{\rho}(\mathcal{A},\alpha)=\alpha
F_{\rho}(\mathcal{A})\alpha^{T}$ (2)

where $F_{\rho}(\mathcal{A})$ is a matrix such that

$\displaystyle[F_{\rho}(\mathcal{A})]_{ij}=\delta_{ij}\text{Tr}[\rho
A_{i}]-\text{Tr}[\rho(\frac{A_{i}A_{j}+A_{j}A_{i}}{2})].$ (3)

Here $\delta_{ij}$ is Kronecker delta. Since,
$F_{\rho}(\mathcal{A},\alpha)\geq 0$, $F_{\rho}(\mathcal{A})\geq 0$ and
$F_{\rho}(\mathcal{A})=0$ iff $\mathcal{A}$ is a PVM. _This
$F_{\rho}(\mathcal{A})$ matrix is independent of $\alpha$ and very important
to construct the the unsharpness measure of an observable $\mathcal{A}$._
Next, the matrix $\mathcal{F}(\mathcal{A})$ is defined as
$\mathcal{F}(\mathcal{A})=F_{\frac{\mathbb{I}}{d}}(\mathcal{A})$. Now it has
been mentioned in the Ref. [9] that _any unitarily invariant norm_ of
$\mathcal{F}(\mathcal{A})$ can quantify of the unsharpness of $\mathcal{A}$.
For simplicity they have taken $l^{1}$ norm $\|.\|_{1}$ which is defined as
$\|X\|_{1}=\sum_{ij}|[X]_{ij}|$ for a matrix $X$. Therefore, the unsharpness
measure of an observable $\mathcal{A}$ is

$\displaystyle f(\mathcal{A})=\|\mathcal{F}(\mathcal{A})\|_{1}.$ (4)

##  III Luder’s Instrument-based unsharpness measures of observables

From this section, we start to discuss our main results.

###  III.1 Construction and the upper bound of the Luder’s Instrument-based
unsharpness measure $\mathcal{E}^{L}$

The sharp quantum observables (PVMs) have an interesting property that makes
those observables different from the unsharp observables. Next, we discuss
this property of PVMs which motivates us to quantify the unsharpness of the
observables in the following outcome independent way. Suppose Alice is
measuring an observable $\mathcal{A}=\\{A_{i}\\}$ on a quantum state
$\rho\in\mathcal{H}_{d}$ through the Luder’s instrument
$\mathcal{L}^{\mathcal{A}}=\\{\Phi^{L}_{A_{i}}(\rho)\\}$. After obtaining the
outcome $i$, the post measurement state will be
$\rho^{\prime}_{i}=\frac{\Phi^{L}_{A_{i}}(\rho)}{\text{Tr}[\Phi^{L}_{A_{i}}(\rho)]}=\frac{\sqrt{A_{i}}_{i}\rho\sqrt{A_{i}}}{\text{Tr}[\rho
A_{i}]}$. The probability of obtaining the outcome $i$ is
$p_{i}=\text{Tr}[\rho A_{i}]$. Now, after obtaining the outcome $i$ if one
more time the observable $\mathcal{A}$ is measured by Alice on this post
measurement state $\rho^{\prime}_{i}$, the probability of again obtaining the
same outcome $i$ is

$\displaystyle p_{ii}=\text{Tr}[\rho^{\prime}_{i}A_{i}]=\frac{\text{Tr}[\rho
A_{i}^{2}]}{\text{Tr}[\rho A_{i}]}.$ (5)

Now if $\mathcal{A}$ is PVM $A^{2}_{i}=A_{i}$ for all $i$. Therefore,
$p_{ii}=1$ if $\mathcal{A}$ is a PVM._Therefore, if $\mathcal{A}$ is a PVM, on
successive measurements of $\mathcal{A}$, the outcome will definitely repeat._
If $\mathcal{A}$ is not a PVM, there exist an outcome $j$ for which
$A_{j}<A^{2}_{j}$ and therefore, $p_{jj}<1$. Therefore, there is a non-zero
probability that an unsharp observable will not produce the same outcome on
immediate successive measurements of the same observable. _This is a feature
of an unsharp observable or equivalently is the evidence of the unsharpness of
an observable and is of course an essential difference between a PVM and
POVM_. This fact motivates us to quantify the unsharpness of the observables
in the following outcome-independent way.
In the above experiment, the average probability that any outcome will repeat
in the successive measurement of $\mathcal{A}$ is

$\displaystyle\mathcal{P}^{L}(\rho;\mathcal{A})$
$\displaystyle=\sum_{i}p_{i}p_{ii}=\sum_{i}\text{Tr}[\rho A^{2}_{i}]$
$\displaystyle=\text{Tr}[\rho\sum_{i}A^{2}_{i}]=\text{Tr}[\rho
E^{\mathcal{A}}]$ (6)

where $E^{\mathcal{A}}=\sum_{i}A^{2}_{i}$. We will call $E^{\mathcal{A}}$ as
$E$-matrix of $\mathcal{A}$. Clearly, $E^{\mathcal{A}}$ is a positive
Hermitian matrix and $E^{\mathcal{A}}\leq\mathbb{I}$. Now the average
probability that a outcome will never repeat is

$\displaystyle\mathcal{E}^{L}(\rho;\mathcal{A})$
$\displaystyle=1-\mathcal{P}^{L}(\rho;\mathcal{A})$
$\displaystyle=\text{Tr}[\rho(\mathbb{I}-E^{\mathcal{A}})]$ (7)
$\displaystyle\leq\|\rho\|_{tr}\|\mathbb{I}-E^{\mathcal{A}})\|$
$\displaystyle=\|(\mathbb{I}-E^{\mathcal{A}})\|$ (8)

where $\|X\|$ denotes the operator norm i.e., the highest eigen value of a
Hermitian matrix $X$ and $\|X\|_{tr}$ denotes the trace norm of a Hermitian
matrix $X$ i.e., $\|X\|_{tr}=\text{Tr}[\sqrt{X^{\dagger}X}]$. In the second
last inequality, we have used the fact that if $T\in\mathcal{L}(\mathcal{H})$
is a trace-class (i.e., has a finite trace norm) Hermitian operator and
$S\in\mathcal{L}(\mathcal{H})$ is a arbitray Hermitian operator, then
$\text{Tr}[ST]\leq\|T\|_{tr}\|S\|$ [11]. In the last equality, we have used
the fact that $\rho$ is Hermitian and $\rho\geq 0$ and therefore,
$\|\rho\|_{tr}=\text{Tr}[\rho]=1$. Now the bound written in equation (8), is
achievable. Suppose $\ket{e^{\prime}_{max}}$ is the eigen state (i.e.,
normalised eigen vector) corresponding to the maximum eigen value of
$(\mathbb{I}-E^{\mathcal{A}})$. Then,
$\bra{e^{\prime}_{max}}E^{\mathcal{A}}\ket{e^{\prime}_{max}}=\|(\mathbb{I}-E^{\mathcal{A}})\|$.
Taking maximization of the quantity $\mathcal{E}^{L}(\rho;\mathcal{A})$ over
all set density matrices $\rho$, we obtain

$\displaystyle\mathcal{E}^{L}(\mathcal{A})$
$\displaystyle=\max_{\rho}\mathcal{E}^{L}(\rho;\mathcal{A})$
$\displaystyle=\bra{e^{\prime}_{max}}E^{\mathcal{A}}\ket{e^{\prime}_{max}}$
$\displaystyle=\|(\mathbb{I}-E^{\mathcal{A}})\|.$ (9)

We define $\mathcal{E}^{L}(\mathcal{A})$ as the Luder’s instrument-based
unsharpness measure of the observable $\mathcal{A}$. Clearly, if $\mathcal{A}$
is a PVM, $E^{\mathcal{A}}=\mathbb{I}$ and therefore,
$\mathcal{E}^{L}(\mathcal{A})=0$. If $\mathcal{A}$ is not a PVM then there
exists at least one $i$ such that $A_{i}^{2}<A_{i}$ and therefore,
$E^{\mathcal{A}}<\mathbb{I}$ and therefore, $\mathcal{E}^{L}(\mathcal{A})>0$.
_Therefore, $\mathcal{E}^{L}$ is a faithful measure._ Clearly,
$\mathcal{E}^{L}$ measure is independent of the bijective relabeling of
outcomes and of the values of outcomes.

There exist a upper bound for this unsharpness measure $\mathcal{E}^{L}$. Our
following lemma states that-

######  Lemma 1.

For an observable $\mathcal{A}=\\{A_{i}\\}^{n_{\mathcal{A}}}_{i}$,
$\mathcal{E}^{L}(\mathcal{A})\leq 1-\frac{1}{n_{\mathcal{A}}}$. This bound is
achieved by the observable
$\mathcal{T}^{n_{\mathcal{A}}}=\\{I^{n_{\mathcal{A}}}_{i}=\frac{\mathbb{I}}{n_{\mathcal{A}}}\\}^{n_{\mathcal{A}}}_{i=1}$.

###### Proof.

Suppose, $a^{\prime_{max}}=1-a_{min}$ is the maximum eigen value of
$(\mathbb{I}-E^{\mathcal{A}})$ and $\ket{a^{\prime_{max}}}$ is the
corresponding eigen vector. Threfore,
$\|\mathbb{I}-E^{\mathcal{A}}\|=a^{\prime_{max}}$. This implies that $a_{min}$
is the minimum eigen value of $E^{\mathcal{A}}$ and
$\ket{a^{\prime_{max}}}=\ket{a_{min}}$ is the corresponding eigen vector. Now
suppose, $\\{\ket{n}\\}$ is the eigen basis of $E^{\mathcal{A}}$. Therefore,
for some $n=n^{\prime}$, $\ket{n^{\prime}}=\ket{a_{min}}$. Then

$\displaystyle a_{min}$
$\displaystyle=\bra{a_{min}}E^{\mathcal{A}}\ket{a_{min}}$
$\displaystyle=\bra{a_{min}}\sum_{i}A^{2}_{i}\ket{a_{min}}$
$\displaystyle=\sum_{i}\bra{a_{min}}A^{2}_{i}\ket{a_{min}}$
$\displaystyle=\sum_{i}\sum^{d}_{n=1}\bra{a_{min}}A_{i}\ket{n}\bra{n}E_{i}\ket{a_{min}}$
$\displaystyle=\sum_{i}\sum^{d}_{n=1}|\bra{a_{min}}A_{i}\ket{n}|^{2}$
$\displaystyle\geq\sum_{i}|\bra{a_{min}}A_{i}\ket{a_{min}}|^{2}$
$\displaystyle=\sum_{i}x_{i}^{2}$ (10)

where $x_{i}=|\bra{a_{min}}A_{i}\ket{a_{min}}|$. Now we know that

$\sum_{i}x_{i}=\sum_{i}|\bra{a_{min}}A_{i}\ket{a_{min}}|=1$ (11)

as $A_{i}\geq 0$ for all $i\in\\{1,....,n^{\mathcal{A}}\\}$ and
$\sum_{i}A_{i}=\mathbb{I}$. Now we know from the optimization method of
Lagrange’s undetermined multipliers that $\sum^{n}_{i=1}x^{2}_{i}$ takes the
minimum value subject to condition $\sum^{n}_{i}x_{i}=1$ for
$x_{1}=x_{2}=....=x_{n}=\frac{1}{n}$. Therefore, using this fact, the
inequality (10) becomes

$\displaystyle a_{min}$ $\displaystyle\geq\sum_{i}x_{i}^{2}$
$\displaystyle\geq\sum_{i}\frac{1}{n^{2}_{\mathcal{A}}}$
$\displaystyle=\frac{1}{n_{\mathcal{A}}}.$ (12)

This implies that

$\displaystyle\mathcal{E}^{L}(\mathcal{A})$
$\displaystyle=\|\mathbb{I}-E^{\mathcal{A}}\|$ $\displaystyle=1-a_{min}$
$\displaystyle\leq 1-\frac{1}{n_{\mathcal{A}}}.$ (13)

Now, for the observable $\mathcal{T}^{n_{\mathcal{A}}}$,

$\displaystyle\mathcal{E}^{L}(\mathcal{T}^{n_{\mathcal{A}}})$
$\displaystyle=\|\mathbb{I}-E^{\mathcal{T}^{n_{\mathcal{A}}}}\|$
$\displaystyle=\|\mathbb{I}-\sum_{i}\frac{\mathbb{I}}{n^{2}_{\mathcal{A}}}\|$
$\displaystyle=\|\mathbb{I}-\frac{\mathbb{I}}{n_{\mathcal{A}}}\|$
$\displaystyle=1-\frac{1}{n_{\mathcal{A}}}.$ (14)

∎

We can also define another measure of the unsharpness in a different way. It
is to be noted that equation (7) is linear in $\rho$. Now let
$\mathfrak{R}=\\{\rho_{1},....,\rho_{k}\\}$ is a set of $k$ states.

Then, the simple average (i.e., with same probability $\frac{1}{k}$) of
$\mathcal{E}^{L}(\rho;\mathcal{A})$ over this set $\mathfrak{R}$ is

$\displaystyle<\mathcal{E}^{L}(\rho;\mathcal{A})>_{\mathfrak{R}}=$
$\displaystyle\sum^{k}_{i=1}\frac{1}{k}\mathcal{E}^{L}(\rho_{i};\mathcal{A})$
$\displaystyle=$
$\displaystyle\sum^{k}_{i=1}\frac{1}{k}\text{Tr}[\rho_{i}(\mathbb{I}-E^{\mathcal{A}})]$
$\displaystyle=$
$\displaystyle\text{Tr}[(\sum^{k}_{i=1}\frac{1}{k}\rho_{i})(\mathbb{I}-E^{\mathcal{A}})]$
$\displaystyle=$
$\displaystyle\text{Tr}[<\rho>_{\mathfrak{R}}(\mathbb{I}-E^{\mathcal{A}})]$

where $<\rho>_{\mathfrak{R}}=(\sum^{k}_{i=1}\frac{1}{k}\rho_{i})$ is the
simple average (i.e., with same probability $\frac{1}{k}$) of the states over
the set $\mathfrak{R}$. Generalising equation (LABEL:eq:avg_r) for whole state
space $\mathcal{S}(\mathcal{H})$, we get that

$\displaystyle<\mathcal{E}^{L}(\rho;\mathcal{A})>_{\mathcal{S}(\mathcal{H})}=$
$\displaystyle\text{Tr}[<\rho>_{\mathcal{S}(\mathcal{H})}(\mathbb{I}-E^{\mathcal{A}})]$
$\displaystyle=$
$\displaystyle\text{Tr}[\frac{\mathbb{I}}{d}(\mathbb{I}-E^{\mathcal{A}})]$
$\displaystyle=$
$\displaystyle\mathcal{E}^{L}(\frac{\mathbb{I}}{d};\mathcal{A}).$ (16)

where $<\rho>_{\mathcal{S}(\mathcal{H})}$ is the simple average of the states
over the whole state space $\mathcal{S}(\mathcal{H})$ and in the second last
equality we have used the well-known fact that
$<\rho>_{\mathcal{S}(\mathcal{H})}=\frac{\mathbb{I}}{d}$. We define the
unsharpness measure of $\mathcal{A}$ as

$\displaystyle\mathcal{E}^{\prime L}(\mathcal{A})=$
$\displaystyle<\mathcal{E}^{L}(\rho;\mathcal{A})>_{\mathcal{S}(\mathcal{H})}$
$\displaystyle=\mathcal{E}^{L}(\frac{\mathbb{I}}{d};\mathcal{A}).$ (17)

Now the lemma below states the upper bound of $\mathcal{E}^{\prime
L}(\mathcal{A})$.

######  Lemma 2.

For an observable $\mathcal{A}=\\{A_{i}\\}^{n_{\mathcal{A}}}_{i}$,
$\mathcal{E}^{\prime L}(\mathcal{A})\leq 1-\frac{1}{n_{\mathcal{A}}}$. This
bound is achieved by the observable
$\mathcal{T}^{n_{\mathcal{A}}}=\\{I^{n_{\mathcal{A}}}_{i}=\frac{\mathbb{I}}{n_{\mathcal{A}}}\\}^{n_{\mathcal{A}}}_{i=1}$.

###### Proof.

From the equation (17), we get

$\displaystyle\mathcal{E}^{\prime L}(\mathcal{A})=$
$\displaystyle\mathcal{E}^{L}(\frac{\mathbb{I}}{d};\mathcal{A})$
$\displaystyle\leq$
$\displaystyle\max_{\rho}\mathcal{E}^{L}(\rho;\mathcal{A})$ $\displaystyle=$
$\displaystyle\mathcal{E}^{L}(\mathcal{A})$ $\displaystyle\leq$ $\displaystyle
1-\frac{1}{n_{\mathcal{A}}}.$ (18)

Now, it is easy to check that $\mathcal{E}^{\prime
L}(\mathcal{T}^{n_{\mathcal{A}}})=1-\frac{1}{n_{\mathcal{A}}}$. ∎

######  Remark 1.

For an observable $\mathcal{A}=\\{A_{i}\\}$, it is very easy to prove that
$\mathcal{E}^{L}(\mathcal{A})=\mathcal{E}^{L}(\mathcal{A}^{U})$ and
$\mathcal{E}^{\prime L}(\mathcal{A})=\mathcal{E}^{\prime L}(\mathcal{A}^{U})$
which $\mathcal{A}^{U}=\\{U^{\dagger}A_{i}U\\}$. Therefore, $\mathcal{E}^{L}$
and $\mathcal{E}^{\prime L}(\mathcal{A})$ does not change if an unitary is
acted on the observables in the Heisenberg picture.

###  III.2 Monotonicity of $\mathcal{E}^{L}$ and $\mathcal{E}^{\prime L}$
under a class of fuzzifying processes

If $\mathcal{E}^{L}$ is a useful measure of unsharpness (fuzziness), it should
be monotonically non-decreasing under the processes which fuzzify the
observables i.e., under the processes which make the observables more unsharp.
These processes are called _fuzzifying processes_. One may intuit that coarse-
graining (a process where two or more outcomes are treated as a single one) is
a fuzzifying process. We show through the next example that this is not true
in general.

######  Example 1.

Consider two observables $\mathcal{A}=\\{A_{i}\\}^{3}_{i=1}$ and
$\mathcal{B}=\\{B_{i}\\}^{2}_{i=1}$ acting on $\mathcal{H}_{3}$ where
$A_{1}=\frac{1}{2}\ket{1}\bra{1}+\frac{1}{4}\ket{2}\bra{2},~{}A_{2}=\frac{1}{2}\ket{1}\bra{1}+\frac{3}{4}\ket{2}\bra{2},~{}A_{3}=\ket{3}\bra{3}$
and $B_{1}=\ket{1}\bra{1}+\ket{2}\bra{2},~{}B_{2}=\ket{3}\bra{3}$. clearly,
$A_{1}+A_{2}=B_{1}$ and $A_{3}=B_{2}$ and therefore, $\mathcal{B}$ is a
coarse-graining of $\mathcal{A}$. But $\mathcal{B}$ is a PVM and $\mathcal{A}$
is not a PVM. Therefore, $\mathcal{E}^{L}(\mathcal{A})>0$ and
$\mathcal{E}^{L}(\mathcal{B})=0$. Therefore, under this kind of classical
post-processing of the outcomes $\mathcal{E}^{L}$ may be decreasing.

The above example shows that it is not possible to prove that
$\mathcal{E}^{\mathcal{L}}$ is monotonically non-decreasing under the
classical post-processing of outcomes as it is not a fuzzifying process, in
general. Furthermore, one may intuit that the convex combination of
observables is a fuzzifying process i.e. if an arbitrary observable
$\mathcal{A}$ is convexly combined with any other arbitrary observable, the
resulting observable will be more unsharp than $\mathcal{A}$. We will show
through the next example that this is also not true, in general.

######  Example 2.

Consider a pair of observables $\mathcal{A}=\\{A_{i}\\}$ and
$\mathcal{B}=\\{B_{i}\\}$ acting on $\mathcal{H}_{3}$ where
$A_{1}=\frac{1}{2}\ket{1}\bra{1}+\frac{1}{4}\ket{2}\bra{2},~{}A_{2}=\frac{1}{2}\ket{1}\bra{1}+\frac{3}{4}\ket{2}\bra{2},~{}A_{3}=\ket{3}\bra{3}$
and $B_{1}=\ket{1}\bra{1},~{}B_{2}=\ket{2}\bra{2},~{}B_{3}=\ket{3}\bra{3}$. We
define a observable $\mathcal{C}^{\lambda}=\\{C^{\lambda}_{i}\\}$ where
$C^{\lambda}_{i}=\lambda A_{i}+(1-\lambda)B_{i}$ and $0\leq\lambda\leq 1$.
Clearly,
$C^{\lambda}_{1}=[1-\frac{\lambda}{2}]\ket{1}\bra{1}+\frac{\lambda}{4}\ket{2}\bra{2},~{}C^{\lambda}_{2}=\frac{\lambda}{2}\ket{1}\bra{1}+[1-\frac{\lambda}{4}]\ket{2}\bra{2},~{}C^{\lambda}_{3}=\ket{3}\bra{3}$.
It can be observed that the sharpness of the observable
$\mathcal{C}^{\lambda}$ increases with with the decrement of $\lambda$ and for
$\lambda=0$, $\mathcal{C}^{0}=\mathcal{B}$ which is a PVM. Now since, for
$\lambda=1$, $\mathcal{C}^{1}=\mathcal{A}$, $\mathcal{C}^{\lambda}$ is always
sharper than $\mathcal{A}$ for all values of $\lambda$. It can be easily shown
that $\mathcal{E}^{L}(\mathcal{A})\geq\mathcal{E}^{L}(\mathcal{C}^{\lambda})$
for all values of $\lambda$’s.

The above example shows that it is also not possible to prove that
$\mathcal{E}^{\mathcal{L}}$ is monotonically non-decreasing under the convex
combination of observables as it is not a fuzzifying process, in general.
Example 1 and example 2 suggest that it is not an easy task to specify all
fuzzifying processes. But one can specify the special classes of fuzzifying
processes. One can easily understand that the addition of white noise in the
observables is a fuzzifying process. Therefore, we restrict ourselves to this
particular class of fuzzifying processes and show that
$\mathcal{E}^{\mathcal{L}}$ is monotonically non-decreasing under this class
of fuzzifying processes in the following theorem.

######  Theorem 1.

Suppose $\mathcal{A}^{\lambda}=\\{A^{\lambda}_{i}\\}_{i=1}^{n_{\mathcal{A}}}$
is an unsharp version of $\mathcal{A}=\\{A_{i}\\}_{i=1}^{n_{\mathcal{A}}}$
i.e., $A^{\lambda}_{i}=\lambda
A_{i}+(1-\lambda)\frac{\mathbb{I}}{n_{\mathcal{A}}}$ for all
$i\in\\{1,.....,n_{\mathcal{A}}\\}$ where $1\geq\lambda\geq 0$. Then
$\mathcal{E}^{L}(\mathcal{A}^{\lambda})\geq\mathcal{E}^{L}(\mathcal{A})$ for
all $1\geq\lambda\geq 0$.

###### Proof.

The $E$-matrix of $\mathcal{A}^{\lambda}$, is given by

$\displaystyle E^{\mathcal{A}^{\lambda}}$
$\displaystyle=\sum_{i}(A^{\lambda}_{i})^{2}$ $\displaystyle=\sum_{i}(\lambda
A_{i}+\frac{1-\lambda}{n_{\mathcal{A}}}\mathbb{I})^{2}$
$\displaystyle=\sum_{i}(\lambda^{2}A_{i}^{2}+\frac{2\lambda(1-\lambda)}{n_{\mathcal{A}}}A_{i}+\frac{(1-\lambda)^{2}}{n^{2}_{\mathcal{A}}}\mathbb{I})$
$\displaystyle=\lambda^{2}\sum_{i}A_{i}^{2}+\frac{2\lambda(1-\lambda)}{n_{\mathcal{A}}}\mathbb{I}+\frac{(1-\lambda)^{2}}{n_{\mathcal{A}}}\mathbb{I}$
$\displaystyle=\lambda^{2}\sum_{i}A_{i}^{2}+\frac{(1-\lambda^{2})}{n_{\mathcal{A}}}\mathbb{I}$
(19)

Therefore,

$\displaystyle\mathbb{I}-E^{\mathcal{A}^{\lambda}}=$
$\displaystyle\mathbb{I}-(\lambda^{2}\sum_{i}A_{i}^{2}+\frac{(1-\lambda^{2})}{n_{\mathcal{A}}}\mathbb{I})$
$\displaystyle=$
$\displaystyle\lambda^{2}(\mathbb{I}-E^{\mathcal{A}})+(1-\lambda^{2})(1-\frac{1}{n_{\mathcal{A}}})\mathbb{I}.$
(20)

Now, using the properties of the operator norm, we get

$\displaystyle\|\mathbb{I}-E^{\mathcal{A}^{\lambda}}\|\leq$
$\displaystyle\|\lambda^{2}(\mathbb{I}-E^{\mathcal{A}})+(1-\lambda^{2})(1-\frac{1}{n_{\mathcal{A}}})\mathbb{I}\|$
$\displaystyle=\lambda^{2}\|\mathbb{I}-E^{\mathcal{A}}\|+(1-\lambda^{2})(1-\frac{1}{n_{\mathcal{A}}})\|\mathbb{I}\|$
$\displaystyle=\lambda^{2}\|\mathbb{I}-E^{\mathcal{A}}\|+(1-\lambda^{2})(1-\frac{1}{n_{\mathcal{A}}}).$
(21)

Suppose, $\ket{e^{\prime}_{max}}$ is the eigen state (i.e., normalised
eigenvector) of $(\mathbb{I}-E^{\mathcal{A}})$ corresponding to the highest
eigenvalue of $(\mathbb{I}-E^{\mathcal{A}})$. Then
$\|\mathbb{I}-E^{\mathcal{A}}\|=\bra{e^{\prime}_{max}}(\mathbb{I}-E^{\mathcal{A}})\ket{e^{\prime}_{max}}$.
Then, using the properties of the operator norm and equation (20), we get

$\displaystyle\|\mathbb{I}-E^{\mathcal{A}^{\lambda}}\|\geq$
$\displaystyle\bra{e^{\prime}_{max}}(\mathbb{I}-E^{\mathcal{A}^{\lambda}})\ket{e^{\prime}_{max}}$
$\displaystyle=$
$\displaystyle\lambda^{2}\bra{e^{\prime}_{max}}(\mathbb{I}-E^{\mathcal{A}})\ket{e^{\prime}_{max}}$
$\displaystyle+(1-\lambda^{2})(1-\frac{1}{n_{\mathcal{A}}})\bra{e^{\prime}_{max}}\mathbb{I}\ket{e^{\prime}_{max}}$
$\displaystyle=\lambda^{2}\|(\mathbb{I}-E^{\mathcal{A}})\|+(1-\lambda^{2})(1-\frac{1}{n_{\mathcal{A}}}).$
(22)

From inequality (21) and inequality (22), we get

$\|\mathbb{I}-E^{\mathcal{A}^{\lambda}}\|=\lambda^{2}\|(\mathbb{I}-E^{\mathcal{A}})\|+(1-\lambda^{2})(1-\frac{1}{n_{\mathcal{A}}}).$
(23)

Therefore,

$\displaystyle\mathcal{E}^{L}(\mathcal{A}^{\lambda})-\mathcal{E}^{L}(\mathcal{A})=$
$\displaystyle(1-\lambda^{2})(1-\frac{1}{n_{\mathcal{A}}})-(1-\lambda^{2})\|(\mathbb{I}-E^{\mathcal{A}})\|$
$\displaystyle=$
$\displaystyle(1-\lambda^{2})[(1-\frac{1}{n_{\mathcal{A}}})-\mathcal{E}^{L}(\mathcal{A})]$
$\displaystyle\geq$ $\displaystyle 0.$ (24)

We have used Lemma 1 to obtain the last inequality. Hence the theorem is
proved. ∎

Next we have an immediate corollary-

######  Corollary 1.

For any observable $\mathcal{A}=\\{A_{i}\\}$,
$\mathcal{E}^{L}(\mathcal{A}^{\lambda_{2}})\geq\mathcal{E}^{L}(\mathcal{A}^{\lambda_{1}})$
for all $1\geq\lambda_{1}\geq\lambda_{2}\geq 0$.

###### Proof.

The observable
$\mathcal{A}^{\lambda_{1}}=\\{A^{\lambda_{1}}_{i}=\lambda_{1}A_{i}+(1-\lambda_{1})\frac{\mathbb{I}}{n_{\mathcal{A}}}\\}$.
For notational simplicity we denote all $A^{\lambda_{1}}_{i}$ as
$A^{\prime}_{i}$ i.e., $A^{\lambda_{1}}_{i}=A^{\prime}_{i}$ for all
$i\in\\{1,.....,n_{\mathcal{A}}\\}$ and we also denote the observable
$\mathcal{A}^{\lambda_{1}}$ as $\mathcal{A}^{\prime}$ i.e.,
$\mathcal{A}^{\lambda_{1}}=\mathcal{A}^{\prime}$. Now the observable
$\mathcal{A}^{\lambda_{2}}=\\{A^{\lambda_{2}}_{i}=\lambda_{2}A_{i}+(1-\lambda_{2})\frac{\mathbb{I}}{n_{\mathcal{A}}}\\}$.
Suppose $\gamma=\frac{\lambda_{2}}{\lambda_{1}}$. Clearly, $1\geq\gamma\geq 0$
as $\lambda_{2}\leq\lambda_{1}$ and both are positive. Then for all
$i\in\\{1,.....,n_{\mathcal{A}}\\}$,

$\displaystyle A^{\lambda_{2}}_{i}=$
$\displaystyle\lambda_{2}A_{i}+(1-\lambda_{2})\frac{\mathbb{I}}{n_{\mathcal{A}}}$
$\displaystyle=$
$\displaystyle\gamma\lambda_{1}A_{i}+(1-\gamma\lambda_{1}+\gamma-\gamma)\frac{\mathbb{I}}{n_{\mathcal{A}}}$
$\displaystyle=$
$\displaystyle\gamma\lambda_{1}A_{i}+[(1-\gamma)+\gamma(1-\lambda_{1})]\frac{\mathbb{I}}{n_{\mathcal{A}}}$
$\displaystyle=$
$\displaystyle\gamma[\lambda_{1}A_{i}+(1-\lambda_{1})\frac{\mathbb{I}}{n_{\mathcal{A}}}]+(1-\gamma)\frac{\mathbb{I}}{n_{\mathcal{A}}}$
$\displaystyle=$ $\displaystyle\gamma
A^{\prime}_{i}+(1-\gamma)\frac{\mathbb{I}}{n_{\mathcal{A}}}$ $\displaystyle=$
$\displaystyle A^{\prime\gamma}_{i}$ (25)

where $A^{\prime\gamma}_{i}=\gamma
A^{\prime}_{i}+(1-\gamma)\frac{\mathbb{I}}{n_{\mathcal{A}}}$ for all
$i\in\Omega_{\mathcal{A}^{\lambda_{2}}}$. Therefore,
$\mathcal{A}^{\lambda_{2}}=\mathcal{A}^{\prime\gamma}=\\{A^{\prime\gamma}_{i}\\}$.
Then using the fact that $\mathcal{A}^{\lambda_{1}}_{i}=\mathcal{A}^{\prime}$
and Theorem 1, we get that
$\mathcal{E}^{L}(\mathcal{A}^{\lambda_{2}})\geq\mathcal{E}^{L}(\mathcal{A}^{\lambda_{1}})$.
Hence the corollary is proved. ∎

Therefore, $\mathcal{E}^{L}(\mathcal{A}^{\lambda})$ is monotonically non-
decreasing with decreasing value of $\lambda$ or equivalently
$\mathcal{E}^{L}$ is monotonically non-increasing with increasing value of
$\lambda$.

Next, we have to prove the monotonicity of $\mathcal{E}^{\prime L}$ under the
addition of white noise. We start with our next theorem.

######  Theorem 2.

Suppose $\mathcal{A}^{\lambda}=\\{A^{\lambda}_{i}\\}_{i=1}^{n_{\mathcal{A}}}$
is an unsharp version of $\mathcal{A}=\\{A_{i}\\}_{i=1}^{n_{\mathcal{A}}}$
i.e., $A^{\lambda}_{i}=\lambda
A_{i}+(1-\lambda)\frac{\mathbb{I}}{n_{\mathcal{A}}}$ for all
$i\in\\{1,.....,n_{\mathcal{A}}\\}$ where $1\geq\lambda\geq 0$. Then
$\mathcal{E}^{\prime L}(\mathcal{A}^{\lambda})\geq\mathcal{E}^{\prime
L}(\mathcal{A})$ for all $1\geq\lambda\geq 0$.

###### Proof.

From equation (17) and equation (20), we get that

$\displaystyle\mathcal{E}^{\prime L}(\mathcal{A}^{\lambda})=$
$\displaystyle\mathcal{E}^{L}(\frac{\mathbb{I}}{d};\mathcal{A}^{\lambda})$
$\displaystyle=$
$\displaystyle\text{Tr}[\frac{\mathbb{I}}{d}(\mathbb{I}-E^{\mathcal{A}^{\lambda}})]$
$\displaystyle=$
$\displaystyle\text{Tr}[\frac{\mathbb{I}}{d}(\lambda^{2}(\mathbb{I}-E^{\mathcal{A}})+(1-\lambda^{2})(1-\frac{1}{n_{\mathcal{A}}})\mathbb{I})]$
$\displaystyle=$
$\displaystyle\lambda^{2}\text{Tr}[\frac{\mathbb{I}}{d}(\mathbb{I}-E^{\mathcal{A}})]+(1-\lambda^{2})(1-\frac{1}{n_{\mathcal{A}}})$
$\displaystyle=$ $\displaystyle\lambda^{2}\mathcal{E}^{\prime
L}(\mathcal{A})+(1-\lambda^{2})(1-\frac{1}{n_{\mathcal{A}}})$ (26)

Therefore,

$\displaystyle\mathcal{E}^{\prime
L}(\mathcal{A}^{\lambda})-\mathcal{E}^{\prime L}(\mathcal{A})=$
$\displaystyle(\lambda^{2}-1)\mathcal{E}^{\prime
L}(\mathcal{A})+(1-\lambda^{2})(1-\frac{1}{n_{\mathcal{A}}})$ $\displaystyle=$
$\displaystyle(1-\lambda^{2})[(1-\frac{1}{n_{\mathcal{A}}})-\mathcal{E}^{\prime
L}(\mathcal{A})]$ $\displaystyle\geq$ $\displaystyle 0.$ (27)

We have used Lemma 2 to obtain the last inequality. Hence, the theorem is
proved. ∎

Next, we have an immdiate corollary

######  Corollary 2.

For any observable $\mathcal{A}=\\{A_{i}\\}$, $\mathcal{E}^{\prime
L}(\mathcal{A}^{\lambda_{2}})\geq\mathcal{E}^{\prime
L}(\mathcal{A}^{\lambda_{1}})$ for all $1\geq\lambda_{1}\geq\lambda_{2}\geq
0$.

###### Proof.

The proof is similar to the proof of Corollary 1. From the equation (25), we
get that
$\mathcal{A}^{\lambda_{2}}=\mathcal{A}^{\prime\gamma}=\\{A^{\prime\gamma}_{i}\\}$.
Then using the fact that $\mathcal{A}^{\lambda_{1}}_{i}=\mathcal{A}^{\prime}$
and Theorem 2, we get that $\mathcal{E}^{\prime
L}(\mathcal{A}^{\lambda_{2}})\geq\mathcal{E}^{\prime
L}(\mathcal{A}^{\lambda_{1}})$. Hence the corollary is proved. ∎

Therefore, $\mathcal{E}^{\prime L}(\mathcal{A}^{\lambda})$ is monotonically
non-decreasing with decreasing value of $\lambda$ or equivalently
$\mathcal{E}^{\prime L}$ is monotonically non-increasing with increasing value
of $\lambda$.

###  III.3 Relation of $\mathcal{E}^{L}(\mathcal{A})$ and $\mathcal{E}^{\prime
L}(\mathcal{A})$ with $F_{\rho}(\mathcal{A})$

In this subsection, we relate the approach given in the Ref. [9] (also briefly
discussed in Sec. II.4) with our apporach. More specifically, we relate
$F_{\rho}(\mathcal{A})$ with $\mathcal{E}^{L}(\mathcal{A})$ and
$\mathcal{E}^{\prime L}(\mathcal{A})$. From the expression of
$F_{\rho}(\mathcal{A})$ i.e., from the equation (3), we get that

$\displaystyle\text{Tr}[F_{\rho}(\mathcal{A})]=$
$\displaystyle\sum_{i}[F_{\rho}(\mathcal{A})]_{ii}$ $\displaystyle=$
$\displaystyle\sum_{i}[\text{Tr}[\rho
A_{i}]-\text{Tr}[\rho\frac{(A_{i}A_{i}+A_{i}A_{i})}{2}]]$ $\displaystyle=$
$\displaystyle\sum_{i}[\text{Tr}[\rho A_{i}]-\text{Tr}[\rho A^{2}_{i}]]$
$\displaystyle=$ $\displaystyle\text{Tr}[\rho(\mathbb{I}-\sum_{i}A^{2}_{i})]$
$\displaystyle=$ $\displaystyle\text{Tr}[\rho(\mathbb{I}-E^{\mathcal{A}})]$
$\displaystyle=$ $\displaystyle\mathcal{E}^{L}(\rho;\mathcal{A})$ (28)

Therefore,

$\displaystyle\mathcal{E}^{L}(\mathcal{A})$
$\displaystyle=\max_{\rho}\mathcal{E}^{L}(\rho;\mathcal{A})$
$\displaystyle=\max_{\rho}\text{Tr}[F_{\rho}(\mathcal{A})].$ (29)

Now as it is mentioned in the Ref. [9] that $F_{\rho}(\mathcal{A})$ is
Hermitian and $F_{\rho}(\mathcal{A})\geq 0$ for any arbitrary observable
$\mathcal{A}$, we have
$\text{Tr}[F_{\rho}(\mathcal{A})]=\|F_{\rho}(\mathcal{A})\|_{tr}$.

$\displaystyle\mathcal{E}^{L}(\mathcal{A})$
$\displaystyle=\max_{\rho}\|F_{\rho}(\mathcal{A})\|_{tr}.$ (30)

_Therefore, through our approach one of the operational meanings of the matrix
$F_{\rho}(\mathcal{A})$ can be understood._

Now taking $\rho=\frac{\mathbb{I}}{d}$ and from equation (28), we get that

$\displaystyle\text{Tr}[F_{\frac{\mathbb{I}}{d}}(\mathcal{A})]=$
$\displaystyle\mathcal{E}^{L}(\frac{\mathbb{I}}{d};\mathcal{A})$
$\displaystyle=$ $\displaystyle\mathcal{E}^{\prime L}(\mathcal{A}).$ (31)

As
$\text{Tr}[F_{\frac{\mathbb{I}}{d}}(\mathcal{A})]=\|F_{\frac{\mathbb{I}}{d}}(\mathcal{A})\|_{tr}$,
we have

$\displaystyle\mathcal{E}^{\prime
L}(\mathcal{A})=\|F_{\frac{\mathbb{I}}{d}}(\mathcal{A})\|_{tr}.$ (32)

Now it has been mentioned in the Ref. [9] that _any unitarily invariant norm_
of $\mathcal{F}(\mathcal{A})=F_{\frac{\mathbb{I}}{d}}(\mathcal{A})$ can
quantify of the unsharpness of $\mathcal{A}$. Therefore, we can take trace
norm of $\mathcal{F}(\mathcal{A})$ as a quantifier of the unsharpness of
$\mathcal{A}$ [21]. Therefore, $\mathcal{E}^{\prime L}$ measure is consistent
with the Ref. [9].

##  IV An attempt to construct instrument-independent unsharpness measures

###  IV.1 Construction and the upper bound of the instrument-independent
unsharpness measure $\mathcal{E}$

In the previous section, we discussed two Luder’s instrument-based unsharpness
measures of observables. This discussion raises an immediate question: can one
construct an instrument-independent unsharpness measure of observables? We try
to answer this question in this section.

Suppose Alice is using a general $\mathcal{A}$-compatible quantum instrument
$\mathcal{I}^{\mathcal{A}}=\\{\Phi^{\mathcal{A}}_{i}\\}$ on the state $\rho$
to measure an observable $\mathcal{A}=\\{A_{i}\\}$. Then,
$q_{i}=\text{Tr}[\Phi^{\mathcal{A}}_{i}(\rho)]=\text{Tr}[\rho A_{i}]$ is the
probability of getting the outcome $i$ and
$\rho^{\prime}_{i}=\frac{\Phi^{\mathcal{A}}_{i}(\rho)}{\text{Tr}[\Phi^{\mathcal{A}}_{i}(\rho)]}$
is the post-measurement after obtaining the outcome $i$. Now, after obtaining
the outcome $i$ if one more time the observable $\mathcal{A}$ is measured by
Alice on this post measurement state $\rho^{\prime}_{i}$, the probability of
again obtaining the same outcome $i$ is

$\displaystyle q_{ii}=\text{Tr}[\rho^{\prime}_{i}A_{i}].$ (33)

The average probability that the outcome will repeat on successive
measurements of the observable $\mathcal{A}$ using the instrument
$\mathcal{I}^{\mathcal{A}}$ on the state $\rho$ is

$\displaystyle\mathcal{Q}(\rho;\mathcal{A};\mathcal{I}^{\mathcal{A}})$
$\displaystyle=\sum_{i}q_{i}q_{ii}$ $\displaystyle=\sum_{i}\text{Tr}[\rho
A_{i}]\text{Tr}[\rho^{\prime}_{i}A_{i}]$
$\displaystyle\leq\sum_{i}\text{Tr}[\rho A_{i}]\|A_{i}\|$
$\displaystyle=\text{Tr}[\rho\mathcal{X}^{\mathcal{A}}]$ (34)

where $\mathcal{X}^{\mathcal{A}}=\sum_{i}\|A_{i}\|A_{i}$. We will call
$\mathcal{X}^{\mathcal{A}}$ as the $X$-matrix of the observable $\mathcal{A}$.

Therefore,

$\displaystyle\mathcal{Q}(\rho;\mathcal{A})=\max_{\mathcal{I}^{\mathcal{A}}}\mathcal{Q}(\rho;\mathcal{A};\mathcal{I}^{\mathcal{A}})$
$\displaystyle\leq\text{Tr}[\rho\mathcal{X}^{\mathcal{A}}].$ (35)

Now the average probability that a outcome will never repeat is

$\displaystyle\mathcal{E}(\rho;\mathcal{A};\mathcal{I}^{\mathcal{A}})$
$\displaystyle=1-\mathcal{Q}(\rho;\mathcal{A};\mathcal{I}^{\mathcal{A}}).$
(36)

Now suppose, $a_{max}$ is the highest eigenvalue of the matrix $A_{i}$ and
$\ket{a_{max}}$ is the corresponding eigen vector. Therefore,
$\bra{a_{max}}A_{i}\ket{a_{max}}=\text{Tr}[\ket{a_{max}}\bra{a_{max}}A_{i}]=\|A_{i}\|$.
Now consider an instrument
$\mathcal{J}^{\mathcal{A}}=\\{\Theta^{\mathcal{A}}_{i}\\}$ where
$\Theta^{\mathcal{A}}_{i}(\rho)=\text{Tr}[\rho
A_{i}]\ket{a_{max}}\bra{a_{max}}$. Therefore, the post-measurement states
after obtaining the outcome $i$ is $\sigma_{i}=\ket{a_{max}}\bra{a_{max}}$.
Now,

$\displaystyle\mathcal{Q}(\rho;\mathcal{A};\mathcal{J}^{\mathcal{A}})$
$\displaystyle=\sum_{i}\text{Tr}[\rho
A_{i}]\text{Tr}[\ket{a_{max}}\bra{a_{max}}(A_{i})]$
$\displaystyle=\sum_{i}\text{Tr}[\rho A_{i}]\|A_{i}\|$
$\displaystyle=\text{Tr}[\rho\mathcal{X}^{\mathcal{A}}].$ (37)

Now,

$\displaystyle\mathcal{Q}(\rho;\mathcal{A})$
$\displaystyle=\max_{\mathcal{I}^{\mathcal{A}}}\mathcal{Q}(\rho;\mathcal{A};\mathcal{I}^{\mathcal{A}})$
$\displaystyle\geq\mathcal{Q}(\rho;\mathcal{A};\mathcal{J}^{\mathcal{A}})$
$\displaystyle=\text{Tr}[\rho\mathcal{X}^{\mathcal{A}}].$ (38)

From inequality (35), equation (37) and inequality (38) we get

$\displaystyle\mathcal{Q}(\rho;\mathcal{A})=\text{Tr}[\rho\mathcal{X}^{\mathcal{A}}]=\mathcal{Q}(\rho;\mathcal{A};\mathcal{J}^{\mathcal{A}}).$
(39)

Therefore, choosing the best instrument $\mathcal{J}^{\mathcal{A}}$, one can
maximize the average probability that the outcome will repeat on successive
measurements of the observable $\mathcal{A}$ on the state $\rho$.

Now,

$\displaystyle\mathcal{E}(\rho;\mathcal{A})$
$\displaystyle=\min_{\mathcal{I}^{\mathcal{A}}}\mathcal{E}(\rho;\mathcal{A};\mathcal{I}^{\mathcal{A}})$
$\displaystyle=1-\max_{\mathcal{I}^{\mathcal{A}}}\mathcal{Q}(\rho;\mathcal{A};\mathcal{I}^{\mathcal{A}})$
$\displaystyle=1-\text{Tr}[\rho\mathcal{X}^{\mathcal{A}}]$
$\displaystyle=\text{Tr}[\rho(\mathbb{I}-\mathcal{X}^{\mathcal{A}})]$ (40)
$\displaystyle\leq\|\mathbb{I}-\mathcal{X}^{\mathcal{A}}\|.$ (41)

Therefore,

$\displaystyle\mathcal{E}(\mathcal{A})$
$\displaystyle=\max_{\rho}\mathcal{E}(\rho;\mathcal{A})$
$\displaystyle\leq\|\mathbb{I}-\mathcal{X}^{\mathcal{A}}\|.$ (42)

Now suppose $x^{\prime}_{max}$ is the highest eigen value of
$\mathbb{I}-\mathcal{X}^{\mathcal{A}}$ and $\ket{x^{\prime}_{max}}$ is the
corresponding eigen vector. Therefore,
$\bra{x^{\prime}_{max}}\mathbb{I}-\mathcal{X}^{\mathcal{A}}\ket{x^{\prime}_{max}}=\text{Tr}[\ket{x^{\prime}_{max}}\bra{x^{\prime}_{max}}(\mathbb{I}-\mathcal{X}^{\mathcal{A}})]=\|\mathbb{I}-\mathcal{X}^{\mathcal{A}}\|$.
Then,

$\displaystyle\mathcal{E}(\mathcal{A})$
$\displaystyle=\max_{\rho}\mathcal{E}(\rho;\mathcal{A})$
$\displaystyle\geq\mathcal{E}(\ket{x^{\prime}_{max}}\bra{x^{\prime}_{max}};\mathcal{A})$
$\displaystyle=\|\mathbb{I}-\mathcal{X}^{\mathcal{A}}\|.$ (43)

From inequality (42) and inequality (43) we get

$\displaystyle\mathcal{E}(\mathcal{A})=\|\mathbb{I}-\mathcal{X}^{\mathcal{A}}\|.$
(44)

We define $\mathcal{E}^{\mathcal{A}}$ as the instrument-independent
unsharpness measure of observables. Clearly, $\mathcal{E}$ measure is
independent of the bijective relabeling of outcomes and of the values of
outcomes. If $\mathcal{A}$ is a PVM, $\|A_{i}\|=1$ for all
$i\in\\{1,....,n_{\mathcal{A}}\\}$ and $\mathcal{E}^{\mathcal{A}}=0$. Next, we
provide a remark on the faithfulness of $\mathcal{E}$.

######  Remark 2.

For an observable $\mathcal{A}=\\{A_{i}\\}^{n_{\mathcal{A}}}_{i=1}$ acting on
the $d$-dimensional Hilbert space $\mathcal{H}$, we know that
$\mathcal{E}(\mathcal{A})=0$ only if $\|A_{i}\|=1$ for all $i\in\Omega_{A}$.
Let $\mathcal{A}$ be an observable such that $\|A_{i}\|=1$ for all
$i\in\Omega_{\mathcal{A}}$ and $\ket{a^{max}_{i}}$ be the eigenstate (one of
the eigen states if the maximum eigen value $1$ is degenerate) corresponding
to the maximum eigen value $1$ for all $i\in\Omega_{\mathcal{A}}$. Then for
any two $i,j\in\Omega_{\mathcal{A}}$ and $i\neq j$, suppose that
$\braket{a^{max}_{j}}{a^{max}_{i}}\neq 0$. Then
$\bra{a^{max}_{i}}A_{i}+A_{j}\ket{a^{max}_{i}}\geq
1+\mid\braket{a^{max}_{j}}{a^{max}_{i}}\mid^{2}>1$. But Since,
$A_{i}+A_{j}-\mathbb{I}$, $\bra{\psi}A_{i}+A_{j}\ket{\psi}\leq 1$ for all
$\ket{\psi}\bra{\psi}\in\mathcal{S}(\mathcal{H})$. Hence,
$\braket{a^{max}_{j}}{a^{max}_{i}}=0$ for all $i\in\Omega_{\mathcal{A}}$. Now
as $\sum_{k}A_{k}=\mathbb{I}$ and $\bra{a^{max}_{i}}A_{i}\ket{a^{max}_{i}}=1$,
we have $\bra{a^{max}_{i}}A_{j}\ket{a^{max}_{i}}=0$ for for any two
$i,j\in\Omega_{\mathcal{A}}$ and $i\neq j$. Therefore, for all
$j\in\Omega_{\mathcal{A}}$, there exist a $n_{\mathcal{A}}-1$-dimensional
subspace $\mathcal{K}_{j}$ of $\mathcal{H}$ such that for all
$\ket{\psi}\bra{\psi}\in\mathcal{K}_{j}$, $\bra{\psi}A_{j}\ket{\psi}=0$ and
for $1$-dimensional subspace (i.e., for $\ket{a^{max}_{j}}$),
$\bra{a^{max}_{j}}A_{j}\ket{a^{max}_{j}}=1$. Clearly, such construction is not
possible for $n_{\mathcal{A}}>d$ and for $n_{\mathcal{A}}=d$, such
construction implies
$\mathcal{A}=\\{A_{i}=\ket{a^{max}_{j}}\bra{a^{max}_{j}}\\}$ is a rank-$1$
PVM. Therefore, for $n_{\mathcal{A}}\geq d$, $\mathcal{E}(\mathcal{A})=0$
implies $\mathcal{A}$ is a PVM (sharp observable). Hence, the measure is
faithful. _Above statement implies that $\mathcal{E}$ is a faithful measure
for all qubit observables (i.e., for $d=2$)_. For $n\leq d$,
$\mathcal{E}(\mathcal{A})>0$ implise $\mathcal{A}$ is an unsharp observable.
But in this case $\mathcal{E}(\mathcal{A})=0$ does not implies $\mathcal{A}$
is a PVM. For example- The qutrit observable
$\mathcal{A}^{\prime}=\\{(\ket{1}\bra{1}+\frac{1}{2}\ket{2}\bra{2}),(\frac{1}{2}\ket{2}\bra{2}+\ket{3}\bra{3})\\}$
is an unsharp observable. But $\mathcal{E}(\mathcal{A}^{\prime})=0$.

Next we calculate the upper bound of $\mathcal{E}$.

######  Lemma 3.

For an observable $\mathcal{A}=\\{A_{i}\\}^{n_{\mathcal{A}}}_{i}$,
$\mathcal{E}(\mathcal{A})\leq 1-\frac{1}{n_{\mathcal{A}}}$. This bound is
achieved by the observable
$\mathcal{T}^{n_{\mathcal{A}}}=\\{I^{n_{\mathcal{A}}}_{i}=\frac{\mathbb{I}}{n_{\mathcal{A}}}\\}^{n_{\mathcal{A}}}_{i=1}$.

###### Proof.

From the definition of $\mathcal{E}^{\rho,\mathcal{A}}$, we have

$\displaystyle\mathcal{E}(\rho;\mathcal{A})$
$\displaystyle=\min_{\mathcal{I}^{\mathcal{A}}}\mathcal{E}^{\rho;\mathcal{A};\mathcal{I}^{\mathcal{A}}}$
$\displaystyle\leq\mathcal{E}(\rho;\mathcal{A};\mathcal{L}^{\mathcal{A}})$
(45)

Taking maximization over $\rho$ in both side of inequality 45 and from Lemma
1, we get

$\displaystyle\mathcal{E}^{\mathcal{A}}\leq\mathcal{E}^{L}(\mathcal{A})\leq
1-\frac{1}{n_{\mathcal{A}}}.$ (46)

Now for the observable $\mathcal{T}^{n_{\mathcal{A}}}$,

$\displaystyle\mathcal{E}(\mathcal{T}^{n_{\mathcal{A}}})$
$\displaystyle=\|\mathbb{I}-\mathcal{X}^{\mathcal{T}^{n_{\mathcal{A}}}}\|$
$\displaystyle=\|\mathbb{I}-\sum_{i}\frac{\mathbb{I}}{n^{2}_{\mathcal{A}}}\|$
$\displaystyle=1-\frac{1}{n_{\mathcal{A}}}$ (47)

Hence, the lemma is proved. ∎

Similar to $\mathcal{E}^{\prime L}$, we can define another instrument-
independent unsharpness measure $\mathcal{E}^{\prime}$ by taking average of
$\mathcal{E}(\rho;\mathcal{A})$ over full state space
$\mathcal{S}(\mathcal{H})$. Then

$\displaystyle\mathcal{E}^{\prime}(\mathcal{A})=$
$\displaystyle<\mathcal{E}(\rho;\mathcal{A})>_{\mathcal{S}(\mathcal{H})}$
$\displaystyle=$
$\displaystyle<\text{Tr}[\rho(\mathbb{I}-\mathcal{X}^{\mathcal{A}})]>_{\mathcal{S}(\mathcal{H})}$
$\displaystyle=$
$\displaystyle\text{Tr}[<\rho>_{\mathcal{S}(\mathcal{H})}(\mathbb{I}-\mathcal{X}^{\mathcal{A}})]$
$\displaystyle=$
$\displaystyle\text{Tr}[\frac{\mathbb{I}}{d}(\mathbb{I}-\mathcal{X}^{\mathcal{A}})]$
$\displaystyle=$ $\displaystyle\mathcal{E}(\frac{\mathbb{I}}{d};\mathcal{A}).$
(48)

The statement similar to Remark 2 also holds $\mathcal{E}^{\prime}$.

Now the lemma below states the upper bound of
$\mathcal{E}^{\prime}(\mathcal{A})$.

######  Lemma 4.

For an observable $\mathcal{A}=\\{A_{i}\\}^{n_{\mathcal{A}}}_{i}$,
$\mathcal{E}^{\prime}(\mathcal{A})\leq 1-\frac{1}{n_{\mathcal{A}}}$. This
bound is achieved by the observable
$\mathcal{T}^{n_{\mathcal{A}}}=\\{I^{n_{\mathcal{A}}}_{i}=\frac{\mathbb{I}}{n_{\mathcal{A}}}\\}^{n_{\mathcal{A}}}_{i=1}$.

###### Proof.

From the equation (48), we get

$\displaystyle\mathcal{E}^{\prime}(\mathcal{A})=$
$\displaystyle\mathcal{E}(\frac{\mathbb{I}}{d};\mathcal{A})$
$\displaystyle\leq$ $\displaystyle\max_{\rho}\mathcal{E}(\rho;\mathcal{A})$
$\displaystyle=$ $\displaystyle\mathcal{E}(\mathcal{A})$ $\displaystyle\leq$
$\displaystyle 1-\frac{1}{n_{\mathcal{A}}}.$ (49)

Now, it is easy to check that
$\mathcal{E}^{\prime}(\mathcal{T}^{n_{\mathcal{A}}})=1-\frac{1}{n_{\mathcal{A}}}$.
∎

The statement similar to Remark 1 also holds for $\mathcal{E}$ and
$\mathcal{E}^{\prime}$.

###  IV.2 Monotonicity of $\mathcal{E}$ and $\mathcal{E}^{\prime}$ under a
class of fuzzifying processes

Since from Example 1 and Example 2, we get that the coarse-graining and the
convex combination of the observables are not the fuzzifying processes,
monotonicity of $\mathcal{E}$ can not be shown. Therefore, here we try to show
that under the addition of white noise $\mathcal{E}$ is monotonically non-
decreasing. But unfortunately, it appears that the proof is not so
straightforward. Therefore, at first, we derive the condition for the
monotonicity of $\mathcal{E}$ under the addition of white noise (i.e., Theorem
3).

######  Theorem 3.

Suppose $\mathcal{A}^{\lambda}=\\{A^{\lambda}_{i}\\}_{i=1}^{n_{\mathcal{A}}}$
is an unsharp version of $\mathcal{A}=\\{A_{i}\\}_{i=1}^{n_{\mathcal{A}}}$
i.e., $A^{\lambda}_{i}=\lambda
A_{i}+(1-\lambda)\frac{\mathbb{I}}{n_{\mathcal{A}}}$ for all
$i\in\\{1,.....,n_{\mathcal{A}}\\}$ where $1\geq\lambda\geq 0$. Then
$\mathcal{E}(\mathcal{A}^{\lambda})\geq\mathcal{E}(\mathcal{A})$ for all
$1\geq\lambda\geq 0$ iff

$\Sigma^{\mathcal{A}}_{1}\geq\Sigma^{\mathcal{A}}_{2}$ (50)

holds where
$\Sigma^{\mathcal{A}}_{1}=(\bra{x^{\mathcal{A}}_{min}}\mathcal{X}^{\mathcal{A}}\ket{x^{\mathcal{A}}_{min}}-\frac{1}{n_{\mathcal{A}}})$
and
$\Sigma^{\mathcal{A}}_{2}=(\frac{\sum_{i}\|A_{i}\|}{n_{\mathcal{A}}}-\bra{x^{\mathcal{A}}_{min}}\mathcal{X}^{\mathcal{A}}\ket{x^{\mathcal{A}}_{min}})$
where $x^{\mathcal{A}}_{min}$ is the lowest eigen value of
$\mathcal{X}^{\mathcal{A}}$ and $\ket{x^{\mathcal{A}}_{min}}$ is the eigen
state of $\mathcal{X}^{\mathcal{A}}$ corresponding to the eigen value
$x^{\mathcal{A}}_{min}$.

###### Proof.

The $X$-matrix of $\mathcal{A}^{\lambda}$ is

$\displaystyle\mathcal{X}^{\mathcal{A}^{\lambda}}=$
$\displaystyle\sum_{i}\|\lambda
A_{i}+(1-\lambda)\frac{\mathbb{I}}{n_{\mathcal{A}}}\|[\lambda
A_{i}+(1-\lambda)\frac{\mathbb{I}}{n_{\mathcal{A}}}]$ $\displaystyle=$
$\displaystyle\sum_{i}[\lambda\|A_{i}\|+(1-\lambda)\frac{1}{n_{\mathcal{A}}}][\lambda
A_{i}+(1-\lambda)\frac{\mathbb{I}}{n_{\mathcal{A}}}]$ $\displaystyle=$
$\displaystyle\lambda^{2}\sum_{i}\|A_{i}\|A_{i}+\frac{\lambda(1-\lambda)}{n_{\mathcal{A}}}[(\sum_{i}\|A_{i}\|)\mathbb{I}+\sum_{i}A_{i}]$
$\displaystyle+\frac{(1-\lambda)^{2}}{n_{\mathcal{A}}}\mathbb{I}$
$\displaystyle=$
$\displaystyle\lambda^{2}\mathcal{X}^{\mathcal{A}}+\frac{(1-\lambda)}{n_{\mathcal{A}}}[\lambda(\sum_{i}\|A_{i}\|)+1]\mathbb{I}$
(51)

Therefore,

$\displaystyle\mathbb{I}-\mathcal{X}^{\mathcal{A}^{\lambda}}=$
$\displaystyle\lambda^{2}[\mathbb{I}-\mathcal{X}^{\mathcal{A}}]+(1-\lambda^{2})\mathbb{I}$
$\displaystyle-\frac{(1-\lambda)}{n_{\mathcal{A}}}[\lambda(\sum_{i}\|A_{i}\|)+1]\mathbb{I}$
$\displaystyle=$
$\displaystyle\frac{(1-\lambda)}{n_{\mathcal{A}}}[(n_{\mathcal{A}}-1)+\lambda(n_{\mathcal{A}}-\sum_{i}\|A_{i}\|)]\mathbb{I}$
$\displaystyle+\lambda^{2}[\mathbb{I}-\mathcal{X}^{\mathcal{A}}]$
$\displaystyle=$
$\displaystyle\lambda^{2}[\mathbb{I}-\mathcal{X}^{\mathcal{A}}]+\gamma\mathbb{I}$
(52)

where
$\gamma=\gamma(\mathcal{A},\lambda)=\frac{(1-\lambda)}{n_{\mathcal{A}}}[(n_{\mathcal{A}}-1)+\lambda(n_{\mathcal{A}}-\sum_{i}\|A_{i}\|)]$.
As $A_{i}\leq\mathbb{I}$ and therefore, $\sum_{i}\|A_{i}\|\leq
n_{\mathcal{A}}$, we have $\gamma\geq 0$. Therefore,

$\displaystyle\mathcal{E}(\mathcal{A}^{\lambda})=$
$\displaystyle\|\mathbb{I}-\mathcal{X}^{\mathcal{A}^{\lambda}}\|$
$\displaystyle=$ $\displaystyle\lambda^{2}\mathcal{E}(\mathcal{A})+\gamma.$
(53)

Now,

$\displaystyle\mathcal{E}(\mathcal{A}^{\lambda})-\mathcal{E}(\mathcal{A})=$
$\displaystyle\gamma-(1-\lambda^{2})\mathcal{E}(\mathcal{A})$ $\displaystyle=$
$\displaystyle\frac{(1-\lambda)}{n_{\mathcal{A}}}[(n_{\mathcal{A}}-1)+\lambda(n_{\mathcal{A}}-\sum_{i}\|A_{i}\|)]$
$\displaystyle-(1-\lambda^{2})\mathcal{E}(\mathcal{A})$ $\displaystyle=$
$\displaystyle(1-\lambda)[(1-\frac{1}{n_{\mathcal{A}}}-\mathcal{E}(\mathcal{A}))$
$\displaystyle+\lambda(1-\frac{\sum_{i}\|A_{i}\|}{n_{\mathcal{A}}}-\mathcal{E}(\mathcal{A}))]$
(54)

Now,
$\mathcal{E}(\mathcal{A})=\|\mathbb{I}-\mathcal{X}^{\mathcal{A}}\|=1-x^{\mathcal{A}}_{min}=1-\bra{x^{\mathcal{A}}_{min}}\mathcal{X}^{\mathcal{A}}\ket{x^{\mathcal{A}}_{min}}$
where $x^{\mathcal{A}}_{min}$ is the lowest eigen value of
$\mathcal{X}^{\mathcal{A}}$ and $\ket{x^{\mathcal{A}}_{min}}$ is the eigen
state of $\mathcal{X}^{\mathcal{A}}$ corresponding to the eigen value
$x^{\mathcal{A}}_{min}$. Then

Therefore,

$\displaystyle\mathcal{E}(\mathcal{A}^{\lambda})-\mathcal{E}(\mathcal{A})=$
$\displaystyle(1-\lambda)[(\bra{x^{\mathcal{A}}_{min}}\mathcal{X}^{\mathcal{A}}\ket{x^{\mathcal{A}}_{min}}-\frac{1}{n_{\mathcal{A}}})$
$\displaystyle+\lambda(\bra{x^{\mathcal{A}}_{min}}\mathcal{X}^{\mathcal{A}}\ket{x^{\mathcal{A}}_{min}}-\frac{\sum_{i}\|A_{i}\|}{n_{\mathcal{A}}})]$
$\displaystyle=(1-\lambda)[\Sigma^{\mathcal{A}}_{1}-\lambda\Sigma^{\mathcal{A}}_{2}]$
$\displaystyle=(1-\lambda)\Sigma^{\mathcal{A}}(\lambda)$ (55)

where
$\Sigma^{\mathcal{A}}_{1}=(\bra{x^{\mathcal{A}}_{min}}\mathcal{X}^{\mathcal{A}}\ket{x^{\mathcal{A}}_{min}}-\frac{1}{n_{\mathcal{A}}})=x^{\mathcal{A}}_{min}-\frac{1}{n_{\mathcal{A}}}$,
$\Sigma^{\mathcal{A}}_{2}=(\frac{\sum_{i}\|A_{i}\|}{n_{\mathcal{A}}}-\bra{x^{\mathcal{A}}_{min}}\mathcal{X}^{\mathcal{A}}\ket{x^{\mathcal{A}}_{min}})=\frac{\sum_{i}\|A_{i}\|}{n_{\mathcal{A}}}-x^{\mathcal{A}}_{min}$
and
$\Sigma^{\mathcal{A}}(\lambda)=[\Sigma^{\mathcal{A}}_{1}-\lambda\Sigma^{\mathcal{A}}_{2}]$.
Now, since $\mathcal{E}(\mathcal{A})\leq(1-\frac{1}{n_{\mathcal{A}}})$,
$\Sigma^{\mathcal{A}}_{1}\geq 0$. Now, There are two following cases-
(I) For $\Sigma^{\mathcal{A}}_{2}<0$ \-
In this case, $\mathcal{E}(\mathcal{A}^{\lambda})-\mathcal{E}(\mathcal{A})\geq
0$ always. In this case $\Sigma^{\mathcal{A}}_{1}\geq\Sigma^{\mathcal{A}}_{2}$
trivially holds.
(II) For $\Sigma^{\mathcal{A}}_{2}\geq 0$ \-
In this case, the minimum value of $\Sigma^{\mathcal{A}}(\lambda)$ (for
$\lambda=1$) is
$\Sigma^{\mathcal{A}}_{min}=[\Sigma^{\mathcal{A}}_{1}-\Sigma^{\mathcal{A}}_{2}]=2x^{\mathcal{A}}_{min}-\frac{\sum_{i}\|A_{i}\|}{n_{\mathcal{A}}}-\frac{1}{n_{\mathcal{A}}}$.
Clearly, the condition for
$\mathcal{E}(\mathcal{A}^{\lambda})-\mathcal{E}(\mathcal{A})\geq 0$ for all
$\lambda$ is $\Sigma^{\mathcal{A}}_{min}\geq 0$ or equivalently
$\Sigma^{\mathcal{A}}_{1}\geq\Sigma^{\mathcal{A}}_{2}$.
∎

It appears that the proof of the inequality (50) for arbitray observable
acting on an arbitrary dimensional Hibert space, is difficult and therefore
proof of the statement that under the addition of white noise $\mathcal{E}$ is
monotonically non-decreasing is difficult. Therefore, next we prove the
inequality (50) for the qubit dichotomic observables.

######  Proposition 1.

For any dichotomic observable $\mathcal{W}$,
$\Sigma^{\mathcal{W}}_{1}\geq\Sigma^{\mathcal{W}}_{2}$ and therefore,
$\mathcal{E}(\mathcal{W}^{\lambda})\geq\mathcal{E}(\mathcal{W})$ for all
$1\geq\lambda\geq 0$.

###### Proof.

Suppose $\mathcal{W}=\\{W_{1},W_{2}\\}$ are two qubit dichotomic observables.
Clearly $W_{2}=\mathbb{I}-W_{1}$. Let
$W_{1}=\omega_{1}\ket{\omega_{1}}\bra{\omega_{1}}+\omega_{2}\ket{\omega_{2}}\bra{\omega_{2}}$.
Without the loss of generality, we can choose $\omega_{1}\geq\omega_{2}$. Then
$\|W_{1}\|=\omega_{1}$. Now
$W_{2}=(1-\omega_{1})\ket{\omega_{1}}\bra{\omega_{1}}+(1-\omega_{2})\ket{\omega_{2}}\bra{\omega_{2}}$.
Clearly, $\|W_{2}\|=(1-\omega_{2})$. Therefore,

$\displaystyle\mathcal{X}^{\mathcal{W}}=$
$\displaystyle\|W_{1}\|W_{1}+\|W_{2}\|W_{2}$ $\displaystyle=$
$\displaystyle\omega_{1}[\omega_{1}\ket{\omega_{1}}\bra{\omega_{1}}+\omega_{2}\ket{\omega_{2}}\bra{\omega_{2}})]$
$\displaystyle+(1-\omega_{2})[(1-\omega_{1})\ket{\omega_{1}}\bra{\omega_{1}}+(1-\omega_{2})\ket{\omega_{2}}\bra{\omega_{2}}]$
$\displaystyle=$
$\displaystyle[\omega^{2}_{1}+(1-\omega_{1})(1-\omega_{2})]\ket{\omega_{1}}\bra{\omega_{1}}$
$\displaystyle+[\omega_{1}\omega_{2}+(1-\omega_{2})^{2}]\ket{\omega_{2}}\bra{\omega_{2}}$
$\displaystyle=$
$\displaystyle\omega^{\prime}_{1}\ket{\omega_{1}}\bra{\omega_{1}}+\omega^{\prime}_{2}\ket{\omega_{2}}\bra{\omega_{2}}$
(56)

where $\omega^{\prime}_{1}=[\omega^{2}_{1}+(1-\omega_{1})(1-\omega_{2})]$ and
$\omega^{\prime}_{2}=[\omega_{1}\omega_{2}+(1-\omega_{2})^{2}]$. Now

$\displaystyle\omega^{\prime}_{1}-\omega^{\prime}_{2}=$
$\displaystyle[\omega^{2}_{1}+(1-\omega_{1})(1-\omega_{2})]-[\omega_{1}\omega_{2}+(1-\omega_{2})^{2}]$
$\displaystyle=$
$\displaystyle\omega^{2}_{1}+1+\omega_{1}\omega_{2}-\omega_{1}-\omega_{2}-\omega_{1}\omega_{2}-1+2\omega_{2}-\omega^{2}_{2}$
$\displaystyle=$
$\displaystyle(\omega_{1}-\omega_{2})[(\omega_{1}+\omega_{2})-1].$ (57)

Therefore, as $\omega_{1}\geq\omega_{2}$, we have
$\omega^{\prime}_{1}\geq\omega^{\prime}_{2}$ for $(\omega_{1}+\omega_{2})\geq
1$ and we have $\omega^{\prime}_{1}\leq\omega^{\prime}_{2}$ for
$(\omega_{1}+\omega_{2})\leq 1$.

Therefore, following two cases-
(I) For $(\omega_{1}+\omega_{2})\geq 1$ \-
In this case the minimum eigen value of $\mathcal{X}^{\mathcal{W}}$ is
$x^{\mathcal{W}}_{min}=\omega^{\prime}_{2}$. Therefore,

$\displaystyle\Sigma^{\mathcal{W}}_{min}=$
$\displaystyle\Sigma^{\mathcal{W}}_{1}-\Sigma^{\mathcal{W}}_{2}$
$\displaystyle=$ $\displaystyle
2x^{\mathcal{W}}_{min}-\frac{1}{2}-\frac{\|W_{1}\|+\|W_{2}\|}{2}$
$\displaystyle=$ $\displaystyle
2[\omega_{1}\omega_{2}+(1-\omega_{2})^{2}]-\frac{1}{2}-\frac{\omega_{1}+(1-\omega_{2})}{2}$
$\displaystyle=$ $\displaystyle
1-4\omega_{2}+2\omega^{2}_{2}+2\omega_{1}\omega_{2}-\frac{(\omega_{1}-\omega_{2})}{2}.$
(58)

Now from Fig. 1(a), we get that that $\Sigma^{\mathcal{W}}_{min}\geq 0$ for
all $\omega_{1}$ and $\omega_{2}$ satisfying the conditions
$\omega_{1}\geq\omega_{2}$ and $\omega_{1}+\omega_{2}\geq 1$.

(II) For $(\omega_{1}+\omega_{2})\leq 1$ \-
In this case the minimum eigen value of $\mathcal{X}^{\mathcal{W}}$ is
$x^{\mathcal{W}}_{min}=\omega^{\prime}_{1}$. Therefore,

$\displaystyle\Sigma^{\mathcal{W}}_{min}=$
$\displaystyle\Sigma^{\mathcal{W}}_{1}-\Sigma^{\mathcal{W}}_{2}$
$\displaystyle=$ $\displaystyle
2x^{\mathcal{W}}_{min}-\frac{1}{2}-\frac{\|W_{1}\|+\|W_{2}\|}{2}$
$\displaystyle=$ $\displaystyle
2[\omega^{2}_{1}+(1-\omega_{1})(1-\omega_{2})]-\frac{1}{2}-\frac{\omega_{1}+(1-\omega_{2})}{2}$
$\displaystyle=$ $\displaystyle
1+2\omega^{2}_{1}+2\omega_{1}\omega_{2}-2(\omega_{1}+\omega_{2})-\frac{(\omega_{1}-\omega_{2})}{2}.$
(59)

Now from Fig. 1(b), we get that that $\Sigma^{\mathcal{W}}_{min}\geq 0$ for
all $\omega_{1}$ and $\omega_{2}$ satisfying the conditions
$\omega_{1}\geq\omega_{2}$ and $\omega_{1}+\omega_{2}<1$.

(a) Plot of $\Sigma^{\mathcal{W}}_{min}$ w.r.t. $\omega_{1}$ and $\omega_{2}$
satisfying the conditions $\omega_{1}\geq\omega_{2}$ and
$\omega_{1}+\omega_{2}\geq 1$

(b) Plot of $\Sigma^{\mathcal{W}}_{min}$ w.r.t. $\omega_{1}$ and $\omega_{2}$
satisfying the conditions $\omega_{1}\geq\omega_{2}$ and
$\omega_{1}+\omega_{2}<1$

Figure 1: Plots of $\Sigma^{\mathcal{W}}_{min}$ w.r.t. $\omega_{1}$ and
$\omega_{2}$ for $\omega_{1}\geq\omega_{2}$. These plots show that
$\Sigma^{\mathcal{W}}_{min}\geq 0$ always.

∎

Now, we have to prove monotonicity of $\mathcal{E}^{\prime}$ under the
addition of white noise. We start with our next theorem.

######  Theorem 4.

Suppose $\mathcal{A}^{\lambda}=\\{A^{\lambda}_{i}\\}_{i=1}^{n_{\mathcal{A}}}$
is an unsharp version of $\mathcal{A}=\\{A_{i}\\}_{i=1}^{n_{\mathcal{A}}}$
i.e., $A^{\lambda}_{i}=\lambda
A_{i}+(1-\lambda)\frac{\mathbb{I}}{n_{\mathcal{A}}}$ for all
$i\in\\{1,.....,n_{\mathcal{A}}\\}$ where $1\geq\lambda\geq 0$. Then
$\mathcal{E}^{\prime}(\mathcal{A}^{\lambda})\geq\mathcal{E}^{\prime}(\mathcal{A})$
for all $1\geq\lambda\geq 0$ iff

$\Sigma^{\prime\mathcal{A}}_{1}\geq\Sigma^{\prime\mathcal{A}}_{2}$ (60)

holds where
$\Sigma^{\prime\mathcal{A}}_{1}=\frac{1}{d}\text{Tr}[\mathcal{X}^{\mathcal{A}}]-\frac{1}{n}$
and
$\Sigma^{\prime\mathcal{A}}_{2}=\frac{\sum_{i}\|A_{i}\|)}{n_{\mathcal{A}}}-\frac{1}{d}\text{Tr}[\mathcal{X}^{\mathcal{A}}]$.

###### Proof.

From equation (52), we get that
$(\mathbb{I}-\mathcal{X}^{\mathcal{A}^{\lambda}})=\lambda^{2}[\mathbb{I}-\mathcal{X}^{\mathcal{A}}]+\gamma\mathbb{I}$
where
$\gamma=\gamma(\mathcal{A},\lambda)=\frac{(1-\lambda)}{n_{\mathcal{A}}}[(n_{\mathcal{A}}-1)+\lambda(n_{\mathcal{A}}-\sum_{i}\|A_{i}\|)]$.
As $A_{i}\leq\mathbb{I}$ and therefore, $\sum_{i}\|A_{i}\|\leq
n_{\mathcal{A}}$, we have $\gamma\geq 0$. Therefore, from the equation (48),
we get that

$\displaystyle\mathcal{E}^{\prime}(\mathcal{A}^{\lambda})=\lambda^{2}\mathcal{E}^{\prime}(\mathcal{A})+\gamma.$
(61)

Therefore,

$\displaystyle\mathcal{E}^{\prime}(\mathcal{A}^{\lambda})-\mathcal{E}^{\prime}(\mathcal{A})=$
$\displaystyle\gamma-(1-\lambda^{2})\mathcal{E}(\mathcal{A})$ $\displaystyle=$
$\displaystyle(1-\lambda)[(1-\frac{1}{n_{\mathcal{A}}}-\mathcal{E}^{\prime}(\mathcal{A}))$
$\displaystyle+\lambda(1-\frac{\sum_{i}\|A_{i}\|}{n_{\mathcal{A}}}-\mathcal{E}^{\prime}(\mathcal{A}))]$
$\displaystyle=$
$\displaystyle(1-\lambda)[\Sigma^{\prime\mathcal{A}}_{1}-\lambda\Sigma^{\prime\mathcal{A}}_{2}]$
$\displaystyle=$ $\displaystyle\Sigma^{\prime\mathcal{A}}(\lambda)$ (62)

where
$\Sigma^{\prime\mathcal{A}}_{1}=(1-\frac{1}{n_{\mathcal{A}}}-\mathcal{E}^{\prime}(\mathcal{A}))=(\frac{1}{d}\text{Tr}[\mathcal{X}^{\mathcal{A}}]-\frac{1}{n})$,
$\Sigma^{\prime\mathcal{A}}_{2}=(\frac{\sum_{i}\|A_{i}\|}{n_{\mathcal{A}}}-\frac{1}{d}\text{Tr}[\mathcal{X}^{\mathcal{A}}])$
and
$\Sigma^{\prime\mathcal{A}}(\lambda)=[\Sigma^{\prime\mathcal{A}}_{1}-\lambda\Sigma^{\prime\mathcal{A}}_{2}]$.
Now, since
$\mathcal{E}^{\prime}(\mathcal{A})\leq(1-\frac{1}{n_{\mathcal{A}}})$,
$\Sigma^{\prime\mathcal{A}}_{1}\geq 0$. Now, There are two following cases-
(I) For $\Sigma^{\prime\mathcal{A}}_{2}<0$ \-
In this case,
$\mathcal{E}^{\prime}(\mathcal{A}^{\lambda})-\mathcal{E}^{\prime}(\mathcal{A})\geq
0$ always. In this case $\Sigma^{\mathcal{A}}_{1}\geq\Sigma^{\mathcal{A}}_{2}$
trivially holds.
(II) For $\Sigma^{\prime\mathcal{A}}_{2}\geq 0$ \-
In this case, the minimum value of $\Sigma^{\prime\mathcal{A}}(\lambda)$ (for
$\lambda=1$) is
$\Sigma^{\prime\mathcal{A}}_{min}=[\Sigma^{\prime\mathcal{A}}_{1}-\Sigma^{\prime\mathcal{A}}_{2}]=2\frac{1}{d}\text{Tr}[\mathcal{X}^{\mathcal{A}}]-\frac{\sum_{i}\|A_{i}\|}{n_{\mathcal{A}}}-\frac{1}{n_{\mathcal{A}}}$.
Clearly, the condition for
$\mathcal{E}^{\prime}(\mathcal{A}^{\lambda})-\mathcal{E}^{\prime}(\mathcal{A})\geq
0$ for all $\lambda$ is $\Sigma^{\prime\mathcal{A}}_{min}\geq 0$ or
equivalently
$\Sigma^{\prime\mathcal{A}}_{1}\geq\Sigma^{\prime\mathcal{A}}_{2}$.

∎

Since, it is difficult to prove inequality (60), we prove it for dichotomic
qubit observables. Therefore, our next proposition is

######  Proposition 2.

For any dichotomic observable $\mathcal{W}$,
$\Sigma^{\mathcal{W}}_{1}\geq\Sigma^{\mathcal{W}}_{2}$ and therefore,
$\mathcal{E}^{\prime}(\mathcal{W}^{\lambda})\geq\mathcal{E}^{\prime}(\mathcal{W})$
for all $1\geq\lambda\geq 0$.

###### Proof.

Suppose $\mathcal{W}=\\{W_{1},W_{2}\\}$ are two qubit dichotomic observables.
Clearly $W_{2}=\mathbb{I}-W_{1}$. Let
$W_{1}=\omega_{1}\ket{\omega_{1}}\bra{\omega_{1}}+\omega_{2}\ket{\omega_{2}}\bra{\omega_{2}}$.
Without the loss of generality, we can choose $\omega_{1}\geq\omega_{2}$. Then
$\|W_{1}\|=\omega_{1}$. Now
$W_{2}=(1-\omega_{1})\ket{\omega_{1}}\bra{\omega_{1}}+(1-\omega_{2})\ket{\omega_{2}}\bra{\omega_{2}}$.
Clearly, $\|W_{2}\|=(1-\omega_{2})$. Therefore, from equation (56), we get
that

$\displaystyle\mathcal{X}^{\mathcal{W}}=$
$\displaystyle\omega^{\prime}_{1}\ket{\omega_{1}}\bra{\omega_{1}}+\omega^{\prime}_{2}\ket{\omega_{2}}\bra{\omega_{2}}$
(63)

where $\omega^{\prime}_{1}=[\omega^{2}_{1}+(1-\omega_{1})(1-\omega_{2})]$ and
$\omega^{\prime}_{2}=[\omega_{1}\omega_{2}+(1-\omega_{2})^{2}]$. Therefore,

$\displaystyle\Sigma^{\prime\mathcal{W}}_{min}$
$\displaystyle=2(\frac{1}{2}\text{Tr}[\mathcal{X}^{\mathcal{W}}])-\frac{1}{2}-\frac{\|W_{1}\|+\|W_{2}\|}{2}$
$\displaystyle=\omega^{\prime}_{1}+\omega^{\prime}_{2}-\frac{1}{2}-\frac{\omega_{1}+(1-\omega_{2})}{2}$
$\displaystyle=\omega^{2}_{1}+\omega_{1}\omega_{2}+(1-\omega_{2})(2-\omega_{1}-\omega_{2})-1-\frac{\omega_{1}-\omega_{2}}{2}$
(64)

Figure 2: Plot of $\Sigma^{\prime\mathcal{W}}_{min}$ w.r.t. $\omega_{1}$ and
$\omega_{2}$ for $\omega_{1}\geq\omega_{2}$. This plot show that
$\Sigma^{\prime\mathcal{W}}_{min}\geq 0$ always.

Figure 2, says that $\Sigma^{\prime\mathcal{W}}_{min}\geq 0$ for
$\omega_{1}\geq\omega_{2}$. Hence,
$\mathcal{E}^{\prime}(\mathcal{W}^{\lambda})\geq\mathcal{E}^{\prime}(\mathcal{W})$
for all $1\geq\lambda\geq 0$. ∎

Therefore, inequality (50) and inequality (60) hold for qubit dichotomic
observables. We have searched for examples for which inequality (50)
inequality (60) do not hold. But we could not find any such example. Noting
these facts, we provide the following conjecture-

######  Conjecture 1.

For any qubit observable $\mathcal{A}$, inequality
$\Sigma^{\mathcal{A}}_{1}\geq\Sigma^{\mathcal{A}}_{2}$ and inequality
$\Sigma^{\prime\mathcal{A}}_{1}\geq\Sigma^{\prime\mathcal{A}}_{2}$ hold and
therefore, $\mathcal{E}(\mathcal{A}^{\lambda})\geq\mathcal{E}(\mathcal{A})$
and
$\mathcal{E}^{\prime}(\mathcal{A}^{\lambda})\geq\mathcal{E}^{\prime}(\mathcal{A})$
for all $1\geq\lambda\geq 0$.

If Conjecture 1 can be proven then two corollaries similar to Corollary 1 and
Corollary 2 can also be proven which establishes the monotonicity of
$\mathcal{E}$ and $\mathcal{E}^{\prime}$ under the addition of white noise.

##  V Experimental Determination of the value of the unsharpness measures

Here we show that experimentally, one can determine the value of
$\mathcal{E}^{L}(\mathcal{A})$ and $\mathcal{E}^{\prime L}(\mathcal{A})$ for
an unknown qubit observable $\mathcal{A}=\\{A_{i}\\}$. We show this for the
qubit case. Generalization for the higher dimensions is straightforward.
Let $E$-matrix of an unknown qubit observable $\mathcal{A}=\\{A_{i}\\}$ be
$E^{\mathcal{A}}=\begin{bmatrix}a&c^{*}\\\ c&d\end{bmatrix}$ where this matrix
is written in $\sigma_{z}$ basis. Suppose $\ket{\pm,i}$ are the eigen states
of $\sigma_{i}$ corresponding to the eigen values $\pm 1$ for all
$i\in\\{x,y,z\\}$. Suppose we have $n_{i,\pm}$ copies of such states are
available to us. On each of these copies, $\mathcal{A}$ has been measured
twice successively using Luder’s instrument. Suppose that for $f_{i,\pm}$
copies outcomes have repeated (i.e., the outcome of the first $\mathcal{A}$
meaurement and the outcome of the second $\mathcal{A}$ meaurement are same)
Then from equation (6), we get that the average probability that any outcome
will repeat, is

$\displaystyle\mathcal{P}^{L}(\ket{\pm,i}\bra{\pm,i},\mathcal{A})=\text{Tr}[\ket{\pm,i}\bra{\pm,i}E^{\mathcal{A}}].$
(65)

Now we know that for large $n_{i,\pm}$,
$\mathcal{P}^{L}(\ket{\pm,i}\bra{\pm,i},\mathcal{A})\approx\frac{f_{i,\pm}}{n_{i,\pm}}$.
Now
$\mathcal{P}^{L}(\ket{+,z}\bra{+,z},\mathcal{A})=a\approx\frac{f_{z,+}}{n_{z,+}}$,
$\mathcal{P}^{L}(\ket{-,z}\bra{-,z},\mathcal{A})=b\approx\frac{f_{z,-}}{n_{z,-}}$,
$\mathcal{P}^{L}(\ket{\pm,x}\bra{\pm,x},\mathcal{A})=\frac{a+b\pm
2Re(c)}{2}\approx\frac{f_{x,\pm}}{n_{x,\pm}}$,
$\mathcal{P}^{L}(\ket{\pm,y}\bra{\pm,y},\mathcal{A})=\frac{a+b\pm
2Im(c)}{2}\approx\frac{f_{y,\pm}}{n_{y,\pm}}$ where $Re(c)$ is the real part
of $c$ and $Im(c)$ is the imaginary part of $c$. From these approximate
equalities, we get the following set of approximate equalities-

$\displaystyle a\approx$
$\displaystyle\frac{f_{z,+}}{n_{z,+}};~{}b\approx\frac{f_{z,-}}{n_{z,-}}$ (66)
$\displaystyle c\approx$
$\displaystyle(\frac{f_{x,+}}{2n_{x,+}}-\frac{f_{z,+}}{4n_{z,+}}-\frac{f_{z,-}}{4n_{z,-}})$
$\displaystyle+i(\frac{f_{y,+}}{2n_{y,+}}-\frac{f_{z,+}}{4n_{z,+}}-\frac{f_{z,-}}{4n_{z,-}}).$
(67)

Clearly for $n_{i\pm}\rightarrow\infty$, for all $i\in\\{x,y,z\\}$, above
approximate equalities become exact equalities. In this way, if $a$, $b$ and
$c$ are known approximately then $E^{\mathcal{A}}$ is known approximately. The
lowest eigenvalue of $E^{\mathcal{A}}$ is
$\frac{a+b-\sqrt{(a+b)^{2}-4(ab-|c|^{2})}}{2}$. Therefore,
$\mathcal{E}^{L}(\mathcal{A})=\|\mathbb{I}-E^{\mathcal{A}}\|=1-\frac{a+b-\sqrt{(a+b)^{2}-4(ab-|c|^{2})}}{2}$.
Similarly, $\mathcal{E}^{\prime
L}(\mathcal{A})=1-\frac{1}{2}\text{Tr}[E^{\mathcal{A}}]=1-\frac{a+b}{2}$.
Therefore, in this way, it is possible to determine the values of
$\mathcal{E}^{L}(\mathcal{A})$ and $\mathcal{E}^{\prime L}(\mathcal{A})$
experimentally.

The experimental determination of the values of $\mathcal{E}(\mathcal{A})$ and
$\mathcal{E}^{\prime}(\mathcal{A})$ is similar as above.

##  VI An attempt to construct the resource theory of the sharpness of the
observables

Quantification of quantum resources and the construction of the resource
theory is very important and interesting direction of research [13]. Few
examples of different resource theories are (i) the resource theory of
entanglement [13, 14], (ii) the resource theory of coherence [15, 16], (iii)
the resource theory of incompatibility [17], (iv) the resource theory of
quantum channels [18], (v) the resource theory of quantum thermodynamics [19,
20] etc. We do not claim we construct the complete resource theory here. But
we present the idea of the resource theory of the sharpness of the observables
here. We take the sharpness of the observables as a resource here. We first
provide the following reasons behind taking sharpness of the observables as a
resource-

  1. 1.

The Ref. [22] suggests that an ideal PVM have infinite resource costs.
Therefore, with finite amount of resource, a PVM can not be performed with
arbitrary accuracy. Therefore, this fact suggests that the ability to perform
PVMs (i.e., sharp measurements) or equivalently sharpness of the observables
itself can be considered as a resource.

  2. 2.

In practice, it is very difficult to get rid of the interaction between the
system and the environment. The interaction between the system and the
environment disturbs the quantum state of a system or equivalently one can say
that due to the interaction between the system and the environment, an
effective channel $\Lambda$ acts on the system state. In Heisenberg picture,
this channel acts on the observable $\mathcal{A}$, which we want to measure,
as $\Lambda^{*}(\mathcal{A})=\\{\Lambda^{*}(A_{i})\\}$. Depending on the type
of the interaction $\Lambda^{*}$ can convert a sharp observable into an
unsharp observable. For an example- if $\Lambda=\Gamma^{t}_{d}$ is
depolarising channel i.e.,
$\Lambda(\rho)=\Gamma^{t}_{d}(\rho)=t\rho+(1-t)\frac{\mathbb{I}}{d}$ and
$\mathcal{A}=\\{\ket{a_{i}}\bra{a_{i}}\\}$ is a rank one PVM, then
$\Lambda^{*}(\mathcal{A})=\Gamma^{t*}_{d}(\mathcal{A})=\\{\Gamma^{t*}_{d}(A_{i})=tA_{i}+(1-t)\frac{\mathbb{I}}{d}\\}$.
Therefore, for a given value of $t<1$, it is impossible to perform a PVM
accurately. Therefore, given the type of interaction, it may not be possible
to perform a PVM with arbitrary accuracy. Therefore, to perform a PVM in a
lab, one needs to make proper arrangements in the lab to get rid of such
interactions between the system and the environment which prevents one to
perform the desired PVM with arbitrary accuracy. Therefore, this fact also
suggests that the ability to perform PVMs (i.e., sharp measurements) or
equivalently sharpness of the observables itself can be considered as a
resource.

  3. 3.

There exist several information-theoretic tasks which can not be performed
perfectly without the sharp observables. For example- a set of orthogonal
states can be distinguished _perfectly_ only with certain PVMs. Therefore,
this fact also suggests that the ability to perform PVMs (i.e., sharp
measurements) or equivalently sharpness of the observables itself can be
considered as a resource.

Now we state the different elements of the resource theory of the sharpness of
the observables below-

  1. 1.

_The resource-_ The sharpness of the observables.

  2. 2.

_The free operation-_ The fuzzifying processes. For example- a class of
fuzzifying processes is the addition of white noise.

  3. 3.

_The resource measure-_ We know that the unsharpness is opposite to the
sharpness. Therefore, as sharpness is monotonically non-increasing under
fuzzifying processes, the unsharpness is monotonically non-decreasing under
fuzzifying processes. Since, from Theorem 1 and Corollary 1, we get that
$\mathcal{E}^{L}$ is monotonically non-decreasing under the addition of white
noise, $\mathcal{E}^{L}$ can be a possible measure of unsharpness. The higher
value of $\mathcal{E}^{L}$ corresponds to less sharpness (i.e., less
resource). Similarly, from Theorem 2 and Corollary 2, we get that
$\mathcal{E}^{\prime L}$ can be a possible measure of unsharpness. It is to be
noted that if the Conjecture 1 can be proven then $\mathcal{E}$ and
$\mathcal{E}^{\prime}$ also can be an unsharpness measure for qubit
observables consistent with the resource-theoretic framework.

  4. 4.

_Most resourceful measurements-_ The sharp measurements (PVMs).

  5. 5.

_Free measurements-_ Given the number of outcomes $n$, the observable
$\mathcal{T}^{n}=\\{I^{n}_{i}=\frac{\mathbb{I}}{n}\\}^{n}_{i=1}$ is a free
measurement (most unsharp).

  6. 6.

_Example of an information-theoretic task which requires the resource-_ Sharp
measurements are required in the _perfect_ discrimination of the orthogonal
states.

Now a complete resource theory can be constructed only if all the fuzzifying
processes are specified which is out of the scope of the present work. One
point should be mentioned that the above-said resource theory is completely
different the resource theory of quantum uncomplexity which is presented in
the Ref. [23] and the fuzzy operations which are discussed in the Ref. [23] is
quite different than our idea of fuzzifying processes.

##  VII Conclusion

In this work, at first, we have constructed two Luder’s instrument-based
unsharpness measures and provided the tight upper bounds of those measures.
Then we have proved the monotonicity of the above-said measures under a class
of fuzzifying processes (i.e., the addition of white noise). This is
consistent with the resource-theoretic framework. We have also discussed the
fact that these measures does not change if a unitary is acted on the
observables in the Heisenberg picture. Then we have related our approach to
the approach of the Ref. [9]. Next, we have tried to construct tried
instrument-independent unsharpness measures. In particular, we have defined
two instrument-independent unsharpness measures and provided the tight upper
bounds of those measures and then we have derived the condition for the
monotonicity of those measures under a class of fuzzifying processes and
proved the monotonicity for dichotomic qubit observables. Then we have shown
that for an unknown measurement, the values of all of these measures can be
determined experimentally. Finally, we have presented the idea of the resource
theory of the sharpness of the observables.

It would be interesting to prove Conjecture 1 in the future. It would be also
interesting to construct a complete resource theory of the sharpness of the
observables in the future.

##  VIII Acknowledgements

I would like to thank my advisor Prof. S. Ghosh for his valuable comments on
this work.

## References

  * [1] P. Busch, T. Heinonen, and P. Lahti, Noise and disturbance in quantum measurement, Phys. Lett. A 320, 261 (2004).
  * [2] M. Ozawa, Uncertainty relations for joint measurements of noncommuting observables, Phys. Lett. A 320, 367 (2004).
  * [3] S. Massar, Uncertainty relations for positive-operator-valued measures, Phys. Rev. A 76, 042114 (2007).
  * [4] P. Busch, T. Heinonen, P. Lahti, Heisenberg’s uncertainty principle, Phys. Rep. 452, 155 (2007).
  * [5] P. Busch, P. Lahti, and R. F. Werner, Quantum root-mean-square error and measurement uncertainty relations, Rev. Mod. Phys. 86, 1261 (2014).
  * [6] C. Carmeli, T. Heinonen, and A. Toigo, Intrinsic unsharpness and approximate repeatability of quantum measurements, J. Phys. A 40, 1303 (2007).
  * [7] K. Baek and W. Son, Unsharpness of generalized measurement and its effects in entropic uncertainty relations, Sci. Rep. 6, 30228 (2016).
  * [8] P. Busch, P. Lahti, J.-P. Pellonpää, and K. Ylinen, _Quantum Measurement_ (Springer, Berlin, 2016).
  * [9] Y. Liu and S. Luo, Quantifying unsharpness of measurements via uncertainty, Phys. Rev. A 104, 052227 (2021).
  * [10] M. A. Nielsen and I. L. Chuang, _Quantum Computation and Quantum Information: 10th Anniversary Edition_ (Cambridge University Press, Cambridge, 2010)
  * [11] T. Heinosaari and M. Ziman, _The Mathematical Language of Quantum Theory: From Uncertainty to Entanglement_ (Cambridge University Press, Cambridge, UK, 2012)
  * [12] M. M. Wilde, Quantum Information Theory (Cambridge University Press, Cambridge, 2013).
  * [13] E. Chitambar and G. Gour, Quantum resource theories, Rev. Mod. Phys. 91, 025001 (2019).
  * [14] F Shahandeh, The resource theory of entanglement, in: _Quantum correlations_ (Springer, 2019) pp. 61–109
  * [15] T. Baumgratz, M. Cramer, and M. B. Plenio, Quantifying Coherence, Phys. Rev. Lett. 113, 140401 (2014).
  * [16] A. Winter and D. Yang, Operational Resource Theory of Coherence, Phys. Rev. Lett. 116, 120404 (2016).
  * [17] F. Buscemi, E. Chitambar, and W. Zhou, Complete Resource Theory of Quantum Incompatibility as Quantum Programmability, Phys. Rev. Lett. 124, 120401 (2020).
  * [18] Y. Liu and X. Yuan, Operational resource theory of quantum channels, Phys. Rev. Research 2, 012035(R) (2020).
  * [19] Felix Binder, Luis A Correa, Christian Gogolin, Janet Anders, and Gerardo Adesso. _Thermodynamics in the quantum regime._ Fundamental Theories of Physics (Springer, 2018), 2019.
  * [20] J. Goold, M. Huber, A. Riera, L. d. Rio, P. Skrzypczyk, The role of quantum information in thermodynamics—a topical review, J. Phys. A: Math. Theor. 49, 143001 (2016).
  * [21] J.-T. Chan, C.-K. Li, N.-S. Sze, Isometries for unitarily invariant norms, Linear Algebra And Its Applications, 399 53-70 (2005).
  * [22] Y. Guryanova, N. Friis, M. Huber, Ideal Projective Measurements Have Infinite Resource Costs, Quantum 4, 222 (2020).
  * [23] N. Y. Halpern, N. B. T. Kothakonda, J. Haferkamp, A. Munson, J. Eisert, P. Faist, Resource theory of quantum uncomplexity, arXiv:2110.11371 [quant-ph].