Dimensionally sharp inequalities for the linear entropy

TL;DR

This paper establishes sharp bounds for the linear entropy of finite-dimensional quantum systems using generalized Bloch decompositions, extending entropy inequalities to finite regimes and aiding in entanglement detection and quantum state characterization.

Contribution

It introduces a new inequality for linear entropy that provides strict bounds for all finite-dimensional quantum states, improving upon previous asymptotic results.

Findings

Derived sharp bounds for linear entropy in finite dimensions

Extended entropy inequalities from asymptotic to finite regimes

Potential applications in entanglement detection and quantum state characterization

Abstract

We derive an inequality for the linear entropy, that gives sharp bounds for all finite dimensional systems. The derivation is based on generalised Bloch decompositions and provides a strict improvement for the possible distribution of purities for all finite dimensional quantum states. It thus extends the widely used concept of entropy inequalities from the asymptotic to the finite regime, and should also find applications in entanglement detection and efficient experimental characterisations of quantum states.

Equations

$$S(\rho ):=-\mathrm{Tr}\rho {\log }_2(\rho ).$$

$$\vec{\mathbf{v}}(\rho_{A_{[n]}}):=\big{(}S(\rho_{A_{1}}),...,S(\rho_{A_{n}}),S(\rho_{A_{1}A_{2}}),...,S(\rho_{A_{1}...A_{n}})\big{)}$$

$$C_n:=\overline{\{\vec{\mathbf{v}}(\rho )|\rho \text{ a density matrix}\}}$$

$$\text{non-negativity}:$$

$$\text{triangle inequality or Araki-Lieb (AL)}:$$

$$\text{subadditivity (SA)}:$$

$$\text{weak monotonicity (WM)}:$$

$$\text{strong subadditivity (SSA)}:$$

$$S^{\alpha }(\rho ):=\frac{1}{1-\alpha }{\log }_2\mathrm{Tr}({\rho }^{\alpha })$$

$$S_q(\rho ):=\frac{1}{q-1}(1-\mathrm{Tr}({\rho }^q))$$

$$\Vert {\rho }_A{\Vert }_q+\Vert {\rho }_B{\Vert }_q\le 1+\Vert {\rho }_{AB}{\Vert }_q$$

$$\displaystyle S^{q}(\rho)=\frac{1}{1-q}\log_{2}\big{(}1-(q-1)S_{q}(\rho)\big{)}$$

$$S_L(\rho ):=1-\mathrm{Tr}({\rho }^2)$$

$$\displaystyle S_{L}(\rho)=1-\frac{1}{d}\big{(}1+\|\vec{\mathbf{b}}\|^{2}\big{)}$$

$$0\le S_L(\rho )\le 1-\frac{1}{d}=:D$$

$$S_L({\rho }_{AB})$$

$$S_L({\rho }_A\otimes {\rho }_B)$$

$$S_L({\rho }_{AB})\ge |S_L({\rho }_A)-S_L({\rho }_B)|$$

$$S_L({\rho }_{ABC})+\frac{1}{d_A{d}_B}{S}_L({\rho }_C)\le \frac{1}{d_B}{S}_L({\rho }_{AC})+\frac{1}{d_A}{S}_L({\rho }_{BC})+\frac{d_A{d}_B+1-d_A-d_B}{d_A{d}_B}$$

$$1-\frac{d_A{d}_B}{4}(1-S_L({\rho }_{AB})+\frac{1}{d_A{d}_B}{)}^2\le S_L({\rho }_A)+S_L({\rho }_B)-S_L({\rho }_A)S_L({\rho }_B)$$

$$S_L({\rho }_{AB})\le$$

$$\sqrt{1-S_L({\rho }_A)}+\sqrt{1-S_L({\rho }_B)}\ge 1$$

$$\Vert {\rho }_A{\Vert }_2+\Vert {\rho }_B{\Vert }_2\le 1+\Vert {\rho }_{AB}{\Vert }_2$$

$$\sqrt{\mathrm{Tr}({\rho }_A^2)}+\sqrt{\mathrm{Tr}({\rho }_B^2)}-1\le \sqrt{\mathrm{Tr}({\rho }_{AB}^2)}$$

$$S_L({\rho }_{AB})\le$$

$$\sqrt{1-S_L({\rho }_A)}+\sqrt{1-S_L({\rho }_B)}\ge 1$$

$$S_L({\rho }_{AB})$$

$$S_L({\rho }_{ABC})+\frac{1}{d_A{d}_B}{S}_L({\rho }_C)\le \frac{1}{d_B}{S}_L({\rho }_{AC})+\frac{1}{d_A}{S}_L({\rho }_{BC})+\frac{d_A{d}_B+1-d_A-d_B}{d_A{d}_B}$$

$$\displaystyle\big{(}S_{L}(\rho_{AB})+S_{L}(\rho_{C})-S_{L}(\rho_{AB})S_{L}(\rho_{C})\big{)}+\frac{1}{d_{A}d_{B}}S_{L}(\rho_{C})\leq$$

$$\displaystyle\frac{1}{d_{B}}\big{(}S_{L}(\rho_{A})+S_{L}(\rho_{C})-S_{L}(\rho_{A})S_{L}(\rho_{C})\big{)}+\frac{1}{d_{A}}\big{(}S_{L}(\rho_{B})+S_{L}(\rho_{C})-S_{L}(\rho_{B})S_{L}(\rho_{C})\big{)}$$

$$+\frac{d_A{d}_B+1-d_A-d_B}{d_A{d}_B}$$

$$S_L({\rho }_{AB})$$

$$\displaystyle-2D_{A}D_{B}\big{(}1-\sqrt{1-\frac{S_{L}(\rho_{A})}{D_{A}}}\big{)}\big{(}1-\sqrt{1-\frac{S_{L}(\rho_{B})}{D_{B}}}\big{)}$$

Full text

I Introduction

Entropy is a widely used concept in various fields. In information theory, entropy is a measure of uncertainty or lack of knowledge about a given system described by a random variable. In a quantum setting this random variable is a density operator, describing the state of a system, and the entropy is a function of this density operator. So the entropy is an intrinsic property of a state and not an observable. Entropy inequalities characterise the distribution of entropy, and thus the distribution of information, in the various parts of a multipartite system. Therefore entropy inequalities are also referred to as the laws of information theory. Let us illustrate this concept for the best understood entropy measure, the von Neumann entropy

[TABLE]

Assume we have an $n$-partite quantum system $A_{[n]}:=A_{1}...A_{n}$ in the state $\rho_{A_{1}...A_{n}}\in\mathcal{H}_{A_{1}}\otimes\dots\otimes\mathcal{H}_{A_{n}}$. The reduced density operator $\rho_{A_{I}}=\rho_{A_{i_{1}}...A_{i_{|I|}}}:=\operatorname{Tr}_{A_{I^{c}}}(\rho_{A_{[n]}})$ (where $\emptyset\neq I\subseteq[n]$) describes the state of the subsystem $A_{I}$. We call the vector

[TABLE]

an entropy vector. Which points in $\mathbb{R}^{2^{n}-1}$ are entropy vectors? Since the density operator is non-negative definite and normalised to $\operatorname{Tr}{\rho}=1$, the entropy is always non-negative and all entropy vectors lie in the non-negative orthant. Every additional inequality reduces the possible set of entropy vectors. The remaining entropy vectors form a set, whose closure is a cone, called the entropy cone.

[TABLE]

For a detailed discussion in the classical case see Chapter 13. -15. of Yeung (2008).

For the von Neumann entropy the best-known inequalities are

[TABLE]

This set of inequalities is not minimal, in fact they can all be derived from strong subadditivity (8), or equivalently weak monotonicity (7), see Nielsen and Chuang (2010) and Cadney et al. (2012). An inequality derived from this set (i.e. SSA) is called a von Neumann inequality. It turns out, that every inequality for bi- or tripartite quantum systems can by derived from strong subadditivity. For the classical analogue, the Shannon entropy, there are known counterexamples for n-partite systems with $n\geq 4$. Contrary to the Shannon entropy, no unconstrained non-von Neumann inequalities are known for n-partite systems $(n\geq 4)$ yet. See Pippenger (2003), Linden and Winter (2005) and Cadney et al. (2012).

If the axioms defining the von Neumann entropy uniquely (Klir (2006)) are weakened, they allow more general classes of entropies. The most prominent families of parametrised entropies are the Rényi entropy and the Tsallis entropy.

Definition 1**.**

For a density operator $\rho\in\mathcal{H}$ and a positive constant $\alpha$, the Rényi entropy $S^{\alpha}(\rho)$ is given by

[TABLE]

It has been shown in Linden et al. (2013), that the only possible inequalities for every n-partite system for the Rényi entropy with $0<\alpha<1$ are the non-negativity of the entropy of the system and all possible subsystems.

”Somewhat surprisingly, we find for $0<\alpha<1$, that the only inequality is non-negativity: In other words, any collection of non-negative numbers assigned to the nonempty subsets of $n$ parties can be arbitrarily well approximated by the $\alpha$-entropies of the $2^{n-1}$ marginals of a quantum state.” Linden et al. (2013)

For $\alpha>1$ it is shown in the same paper that there are no further linear (even homogeneous) inequalities. But since other inequalities are known, the set of entropy vectors is not a cone anymore for $\alpha>1$.

For the special case $\alpha=0$, the Rényi entropy becomes the quantum Hartley measure, also known as the Schmidt rank of the density operator. For this entropy there are again other inequalities besides non-negativity, see Cadney et al. (2014).

Definition 2**.**

The Quantum Tsallis q-entropy is defined as (see Raggio (1995))

[TABLE]

*for $\rho$ a density matrix and $q>0$ a real number. *

For $q>1$ the Tsallis entropy is bounded by $S_{q}(\rho)\leq\frac{1-d^{1-q}}{q-1}\leq\frac{1}{q-1}$, where $d$ is the dimension of the Hilbert space. Equality in the first inequality holds only for the completely mixed state.

Audenaert Audenaert (2007) proved 2007 that the Tsallis entropy for $q>1$ is subadditive. The subadditivity of the Tsallis entropy follows from the following lemma, but is a weaker statement.

Lemma 1**.**

Let $\rho_{AB}$ be a density operator of a bipartite state on a finite-dimensional Hilbert space $\mathcal{H}_{1}\otimes\mathcal{H}_{2}$ and $\|\rho\|_{q}:=(\operatorname{Tr}(\rho^{q}))^{\frac{1}{q}}$ the Schatten $q$-norm. Then for $q>1$ the following holds

[TABLE]

Further the Tsallis entropy is known to not satisfy strong subadditivity.

Both the Rényi and the Tsallis entropy, in the limit $q\rightarrow 1$, converge to the von Neumann entropy. The Rényi entropy can be calculated from the Tsallis entropy by

[TABLE]

so inequalities for the Tsallis entropy can be reformulated as inequalities for the Rényi entropy and vice versa.

In this paper we restrict ourselves to the Tsallis 2-entropy, also known as linear entropy

[TABLE]

The linear entropy is of remarkable interest for quantum information theory. It is obviously closely related to the purity ($\gamma(\rho)=\operatorname{Tr}(\rho^{2})=1-S_{L}(\rho)$) of a quantum state and is therefore also called the impurity. The purity as information measure was introduced in Fano (1957). Important entanglement witnesses arise from the linear entropy. If the linear entropy of a subsystem is greater then the entropy of the composite system, it follows that there is nonclassical correlation. Further it is easily represented in the Bloch representation. Using the Bloch decomposition (105) and Eq. (106) the linear entropy can be written as

[TABLE]

where $\vec{\mathbf{b}}$ denotes the Bloch vector.

We have already seen how the Tsallis entropy can be transformed into the Rényi entropy by Eq. (12). In the case of the linear entropy this gives the Rényi 2-entropy, also known as collision entropy. This is an important entropy for many applications, e.g. privacy amplification in quantum cryptography.

The linear entropy can be seen as the linear approximation to the von Neumann entropy at pure states. Therefore the efficient calculation in contrast to the von Neumann entropy makes it also attractive for practical use in larger systems.

We already know that the linear entropy, as a special case of the quantum Tsallis entropy, is non-negative and bounded.

[TABLE]

where we have equality in the first inequality iff $\rho$ is a pure state and equality in the second iff $\rho=\frac{\mathds{1}_{d}}{d}$, i.e. completely mixed. The dimension dependent constant $D$ denotes the maximal attainable entropy for a state in a Hilbert space with dimension $d$.

Moreover, the linear entropy is subadditive and pseudo-additive for product states

[TABLE]

From subadditivity the Araki-Lieb inequality can be derived via purification, i.e.

[TABLE]

For an alternative proof see Zhang et al. (2008).

It is known that the linear entropy is not strongly subadditive, only a weaker version holds, Petz and Virosztek (2015).

Recently Appel et al. Appel et al. (2017) have discovered two new inequalities, the first is a dimension dependent version of the strong subadditivity

Theorem 1**.**

For a tripartite quantum system $\rho_{ABC}$ we find the following entropy inequality for the linear entropy $S_{L}(\rho_{ABC})=1-\operatorname{Tr}(\rho_{ABC}^{2})$:

[TABLE]

The second inequality is non-linear and dimension dependent

Theorem 2**.**

For all $\rho_{AB}$ and the linear entropy or Tsallis 2-entropy

[TABLE]

Furthermore from Lemma 1. one can derive

Lemma 2**.**

For the linear entropy the following inequality holds

[TABLE]

for

[TABLE]

*and is sharper than subadditivity. *

Proof**.**

In the case $q=2$, Lemma 1. states

[TABLE]

Using $\|\rho\|_{2}=\sqrt{\operatorname{Tr}(\rho^{2})}$ this becomes

[TABLE]

Under the assumption that the left hand side is positive, both sides can be squared. Using $S_{L}(\rho)=1-\operatorname{Tr}(\rho^{2})$ one gets

[TABLE]

But the left hand side of Equation (24) is positive exactly when

[TABLE]

This inequality is sharper than subadditivity, since $0\leq S_{L}(\rho_{A}),S_{L}(\rho_{B})\leq 1$.

II Inhomogeneous Subadditivity

The inequality in Theorem 1. from Appel et al. (2017) can be reduced to give an inequality for bipartite systems.

Lemma 3**.**

For any bipartite quantum system $\rho_{AB}\in\mathcal{H}_{A}\otimes\mathcal{H}_{B}$ with the two subsystems $\rho_{A}=\operatorname{Tr}_{B}(\rho_{AB})$ and $\rho_{B}=\operatorname{Tr}_{A}(\rho_{AB})$ we have

[TABLE]

*where $d_{A}=\dim(\mathcal{H}_{A})$ and $d_{B}=\dim(\mathcal{H}_{B})$ denote the dimensions of the two partitions. *

Since these equations have a constant term, the inequality in Lemma 3. will be called inhomogeneous subadditivity (ISA) and the inequality from Theorem 1. strong inhomogeneous subadditivity (SISA).

This result follows either from Theorem 1. of Appel et al. (2017) and pseudo-additivity or directly from the Bloch decomposition.

Proof**.**

If we take the inequality in Theorem 1.

[TABLE]

and now assume that we can write the state as a product state $\rho_{ABC}=\rho_{AB}\otimes\rho_{C}$, where $\rho_{C}=|\phi\rangle\langle\phi|$ is a pure state, we can rewrite the inequality using pseudo-additivity

[TABLE]

and since $\rho_{C}$ is pure we have $S_{L}(\rho_{C})=0$ and it follows

[TABLE]

For an alternative proof see the Appendix B.1.

It turns out that this inequality is tighter than the inequality from Theorem 2.

Proposition 1**.**

*The inequality from Lemma 3. is always sharper than the inequality in Theorem 2. from Appel et al. (2017). *

For a proof see the Appendix B.2.

III Dimensionally sharp subadditivity

We now introduce a new inequality for the linear entropy. It is a dimension dependent inequality, which is sharper than subadditivity and in fact turns out to be the sharpest possible for every dimensions of a bipartite system. Therefore we call it dimensionally sharp subadditivity (DSSA).

Theorem 1**.**

Let $\rho_{AB}\in\mathcal{H}_{A}\otimes\mathcal{H}_{B}$ be the state of a bipartite quantum system. Then the following inequality holds

[TABLE]

under the assumption that

[TABLE]

*where $D_{A}=\frac{d_{A}-1}{d_{A}}$ and $D_{B}=\frac{d_{B}-1}{d_{B}}$. *

To prove this theorem, we first need some intermediate results, where we would like to point out that they are interesting in their own right.

Lemma 4**.**

*Let \big{\{}X_{i}\big{\}}_{i=0}^{d^{2}-1} be a basis for the Hilbert space $\mathcal{H}$, with Hermitian traceless operators (except $X_{0}=\mathds{1}_{d}$) and $\operatorname{Tr}(X_{i}X_{j})=d\delta_{ij}$.

*Then $\sqrt{d-1}\ \mathds{1}_{d}\pm X$ is a positive semi-definite operator. *

Proof**.**

For $X_{0}=\mathds{1}_{d}$ the statement is obvious, since $d\geq 2$.

Let now $\mu_{k}$ be the eigenvalues of a given $X_{i}$, $i\neq 0$.

We know that $\sum\limits_{k=0}^{d-1}\mu_{k}=0=\operatorname{Tr}(X_{i})$ and $\sum\limits_{k=0}^{d-1}\mu_{k}^{2}=d=\operatorname{Tr}(X^{2}_{i})$.

If we optimise any $\mu_{k}$ (with Lagrange) we get $\mu_{k}=\pm\sqrt{d-1}$ and therefore in general $\pm\mu_{k}\leq\sqrt{d-1}$. So we know that $\sqrt{d-1}\ \mathds{1}_{d}\pm X$ is a positive semi-definite operator.

Lemma 5**.**

*Let $\mathcal{H}_{A}$ and $\mathcal{H}_{B}$ be Hilbert spaces with dimension $d_{A}$ and $d_{B}$ respectively. Define a basis for each Hilbert space by \big{\{}X_{i}\big{\}}_{i=0}^{d_{A}^{2}-1} for the Hilbert space $\mathcal{H}_{A}$ and \big{\{}Y_{j}\big{\}}_{j=0}^{d_{B}^{2}-1} the Hilbert space $\mathcal{H}_{B}$. We require all bases to consist of Hermitian traceless operators (except $X_{0}=\mathds{1}_{d_{A}},\ Y_{0}=\mathds{1}_{d_{B}}$) that fulfil $\operatorname{Tr}(X_{i}X_{j})=d_{A}\delta_{ij}$ and $\operatorname{Tr}(Y_{i}Y_{j})=d_{B}\delta_{ij}$ respectively.

Then for all such basis elements $X_{i}$ and $Y_{j}$ the inequality

[TABLE]

*holds. *

Proof**.**

Pick a basis element of each basis, denoted by $X$ and $Y$.

By the previous lemma $\sqrt{d_{A}-1}\ \mathds{1}_{d_{A}}\pm X$ and $\sqrt{d_{B}-1}\ \mathds{1}_{d_{B}}\pm Y$ are positive semidefinite, and so is their product. Hence we can write

[TABLE]

And we can derive four different inequalities

[TABLE]

Dividing by $\sqrt{d_{A}-1}\sqrt{d_{B}-1}$ and renaming $a=\frac{\langle X\otimes\mathds{1}\rangle}{\sqrt{d_{A}-1}}$, $b=\frac{\langle\mathds{1}\otimes Y\rangle}{\sqrt{d_{B}-1}}$ and $c=\frac{\langle X\otimes Y\rangle}{\sqrt{d_{A}-1}\sqrt{d_{B}-1}}$ we can rewrite these inequalities as

[TABLE]

Assume that $a,b,c$ are nonzero (if not, one can get the same result easily).Then there are eight possible combinations of the signs of $a,b,c$. If

[TABLE]

then we can use one of the equations (39)-(42) to get

[TABLE]

Now lets look at the cases where this is not fulfilled.

Case 1: $a<0$, $b>0$, $c>0$

[TABLE]

Case 2: $a>0$, $b<0$, $c>0$

[TABLE]

Case 3: $a>0$, $b>0$, $c<0$

[TABLE]

Case 4: $a<0$, $b<0$, $c<0$

[TABLE]

So inequality (43) always holds and therefore we get

[TABLE]

Lemma 6**.**

For every bipartite quantum state $\rho_{AB}\in\mathcal{H}_{A}\otimes\mathcal{H}_{B}$ there exists a basis \big{\{}X_{i}\otimes Y_{j}\big{\}}_{i,j}, such that

• •

\big{\{}X_{i}\big{\}}_{i}* and \big{\{}Y_{j}\big{\}}_{j} are local bases for $\mathcal{H}_{A}$ and $\mathcal{H}_{B}$ respectively*

• •

all basis elements consist of Hermitian traceless operators (except $X_{0}=\mathds{1}_{d_{A}},\ Y_{0}=\mathds{1}_{d_{B}}$)

• •

$\operatorname{Tr}(X_{i}X_{j})=d_{A}\delta_{ij}$* and $\operatorname{Tr}(Y_{i}Y_{j})=d_{B}\delta_{ij}$*

• •

$\rho_{AB}$* can be expressed by*

[TABLE]

where $b_{i0}=\langle X_{i}\otimes\mathds{1}\rangle$, $b_{0j}=\langle\mathds{1}\otimes Y_{j}\rangle$ and $b_{ij}=\langle X_{i}\otimes Y_{j}\rangle$.

*In particular, in such a basis the local Bloch vectors are expressible by single basis elements instead of linear combinations of basis elements. *

Proof**.**

By Theorem 2., each state can be expressed in the well-known Bloch decomposition by

[TABLE]

Changing the operator basis \big{\{}X_{i}\big{\}}_{i} corresponds to a rotation of the Bloch vector, which can be realised by an orthogonal matrix $Q\in SO(d^{2}-1)$. I.e. take $X_{1}=\rho-\frac{\mathds{1_{d}}}{d}$, which is traceless and Hermitian, normalise it and choose it as the first basis element and then complete the basis in an appropriate way.

Do this for both partitions, $X_{1}:=\rho_{A}-\frac{\mathds{1_{d_{A}}}}{d_{A}}$ and $Y_{1}:=\rho_{B}-\frac{\mathds{1_{d_{B}}}}{d_{B}}$, and denote the bases by \big{\{}X_{i}\big{\}}_{i} and \big{\{}Y_{j}\big{\}}_{j} respectively. Then \big{\{}X_{i}\otimes Y_{j}\big{\}}_{i,j} is a basis for the joint system and we can write the state as

[TABLE]

We changed the Bloch basis locally, such that the marginal state is expressible by a single basis element, and therefore

[TABLE]

Lemma 7**.**

For the local Bloch vectors q-norm $\|C_{A}\|_{q}:=\sqrt[q]{\sum_{i}|\langle X_{i}\otimes\mathds{1}\rangle|^{q}}$, $\|C_{B}\|_{q}:=\sqrt[q]{\sum_{j}|\langle\mathds{1}\otimes Y_{j}\rangle|^{q}}$ and the correlation tensor norm $\|C_{AB}\|_{q}:=\sqrt[q]{\sum_{i,j}|\langle X_{i}\otimes Y_{j}\rangle|^{q}}$ of a given bipartite state the following holds for $q\geq 1$

[TABLE]

Proof**.**

We choose an appropriate basis for the Hilbert space $\mathcal{H}_{A}\otimes\mathcal{H}_{B}$, according to Lemma 6. Then we can express the correlation tensors of the joint system as $\|C_{A}\|_{q}=|\langle X_{1}\otimes\mathds{1}\rangle|$ and $\|C_{B}\|_{q}=|\langle\mathds{1}\otimes Y_{1}\rangle|$.

It follows $\|C_{AB}\|_{q}=\sqrt[q]{\langle X_{1}\otimes Y_{1}\rangle^{q}+\sum\limits_{i,j\geq 2}\langle X_{i}\otimes Y_{j}\rangle^{q}}\geq|\langle X_{1}\otimes Y_{1}\rangle|$ and by using Lemma 5. we can therefore conclude

[TABLE]

Corollary 1**.**

The following inequality holds true

[TABLE]

Proof**.**

From (106) and Eq. (47) we conclude for the linear entropy of a bipartite state

[TABLE]

and together with Eq. (14) therefore we have

[TABLE]

Now just plug these in the inequality of the previous lemma for $q=2$and the result follows.

Remark 1**.**

*Unfortunately Eq. (51) does not hold for values other then $q=2$, therefore our result follows only for the linear entropy. *

With this last result the proof of Theorem 1. becomes straightforward:

Proof**.**

First we need to square both sides to get rid of the roots. Therefore we have to be sure both sides are non-negative. This obviously holds for the left hand-side.

Rewriting the right side of the equation (II) in Corollary 1. and dividing it by $\sqrt{d_{A}-1}\sqrt{d_{B}-1}$ we get

[TABLE]

where $D_{A}=1-\frac{1}{d_{A}}$ and $D_{B}=1-\frac{1}{d_{B}}$.

The condition that the right hand-side (II) is non-negative then becomes

[TABLE]

or equivalently

[TABLE]

Assume now, that this constraint holds and that both sides of the equation in Corollary 1. are positive. Now square both sides and divide by $d_{A}d_{B}$. This gives

[TABLE]

which can be reformulated as

[TABLE]

Remark 2**.**

*The domain where inequality (59) holds is constrained. This is not a consequence of this proof and indeed it is wrong outside of its assigned domain.

Assume $d_{A}=d_{B}=d$ and $\rho_{AB}=\frac{\mathds{1}_{d^{2}}}{d^{2}}$ is the completely mixed state and therefore $S_{L}(\rho_{AB})=1-\frac{1}{d^{2}}$. But we also have $\rho_{A}=\rho_{B}=\frac{\mathds{1}_{d}}{d}$ and $S_{L}(\rho_{A})=S_{L}(\rho_{B})=1-\frac{1}{d}$.

Inequality (59) becomes $1-\frac{1}{d^{2}}\leq 0$, which can not hold since $d\geq 2$. *

Remark 3**.**

By taking the limits $d_{A}=d_{B}\rightarrow\infty$ DSSA converges towards the inequality from Lemma 2.

[TABLE]

IV Sharpest bound

We want to combine inhomogeneous subadditivity with dimensionally sharp subadditivity and prove this ”combination” to give a tight upper bound for the linear entropy of a bipartite quantum system.

From now on, $S_{L}(\rho_{A})$, $S_{L}(\rho_{B})$ and $S_{L}(\rho_{AB})$ will be denoted by the coordinates $x,y,z$. We will also use the notation $D_{A}:=\frac{d_{A}-1}{d_{A}}$ and $D_{B}:=\frac{d_{B}-1}{d_{B}}$.

Definition 3**.**

Let’s define the functions of the inequalities. The inequality in Lemma 3. gives

[TABLE]

and inequality (31) in Theorem 1.

[TABLE]

The restriction (32) then becomes

[TABLE]

So we can rewrite the inhomogeneous subadditivity as

[TABLE]

and the dimensionally sharp subadditivity as

[TABLE]

This two inequalities can be combined by the following proposition.

Proposition 2**.**

The function

[TABLE]

*is continuously differentiable. *

For a proof see Appendix C.

Lemma 8**.**

For arbitrary $0\leq a\leq D_{A}$ and $0\leq b\leq D_{B}$ there exists a state such that

[TABLE]

For the proof see Appendix D.

This means that every point of the restricting surface is an entropy vector, which implies the following theorem.

Theorem 1**.**

The inequality defined by

[TABLE]

*holds for a bipartite quantum system and is tight.

V Rényi 2-entropy and Purity

The Rényi 2-entropy, sometimes also called the collision entropy, is strongly related to the linear entropy by Equation (12). With this, the above inequalities can be translated into inequalities for the Rényi 2-entropy. This is not in contrast to the results of Linden et al. (2013), since the inequality will not be homogeneous.

The linear entropy and the Rényi 2-entropy depend on each other by

[TABLE]

Therefore Theorem 1. can be reformulated to an inequality for the Rényi 2-entropy.

Proposition 3**.**

Define the following function

[TABLE]

where

[TABLE]

Then $f_{R}(x,y)$ is a continuously differentiable function and the following inequality

[TABLE]

*holds for a bipartite quantum system. *

For the proof see Appendix E.

These inequalities may look somewhat unhandy, but it can be more useful for certain applications, since the Rényi entropy is additive for product states.

The purity of a quantum state $\rho$ is defined as $\gamma(\rho)=\operatorname{Tr}(\rho^{2})$ and therefore we can express the inequalities for the linear entropy also in terms of the purity.

Corollary 1**.**

Define the function

[TABLE]

Then $f_{P}(x,y)$ is a continuously differentiable function and the following inequality

[TABLE]

*holds for the purities of a bipartite quantum system. *

Proof**.**

This follows immediately from Theorem 1.

VI Inverted inequality

It is well known how one can use purification to obtain the lower bound to a bipartite entropy (Araki-Lieb inequality) from its upper bound given by subadditivity. We therefore call this lower bound the inverted inequality of subadditivity.

By using the same argument, a lower bound for the linear entropy following from Theorem 1. can be constructed. By purification, for every state $\rho$ there exists a pure state $|\phi\rangle\langle\phi|$ with $\rho$ as the reduced density operator. Now assume that $\rho=\rho_{AB}$ is a bipartite state and therefore $|\phi\rangle\langle\phi|$ is a tripartite state. Since it is pure, any two partitions have the same entropy and we can use this property to transform every upper bound inequality in a lower bound inequality.

Proposition 2. states, that the function $f(S_{L}(\rho_{A}),S_{L}(\rho_{B})$ is continuously differentiable. Therefore, using the implicit function theorem and the inverse function theorem (see Appendix F), the existence of an inverted inequality can be shown.

Lemma 9**.**

For $d_{A},\ d_{B}<\infty$ the function $f(x,y)$ defined in Proposition 2,

[TABLE]

*is strictly monotonically increasing in both variables. *

Proof**.**

For $h(x,y)$ this holds, since it is linear in both variables.

The gradient of $g(x,y)$, (122) in the proof of Proposition 2., is $\nabla g(x,y)|_{(0,0)}=\begin{pmatrix}1\\ 1\end{pmatrix}$ at the origin and decreases for growing $x,y$. But, since on the boundary $\Omega$ the Gradient $\nabla g(x,y)|_{\Omega}=\begin{pmatrix}\frac{1}{d_{B}}\\ \frac{1}{d_{A}}\end{pmatrix}$ (123) is still positive, $g(x,y)$ is a strictly monotonically increasing function for $x\leq r(y)$.

Proposition 4**.**

*The inequality from Theorem 1. can be inverted to two new inequalities. *

Proof**.**

Assume we have a system in state $\rho_{AB}\in\mathcal{H}_{A}\otimes\mathcal{H}_{B}$. If we take a copy of this Hilbert space (denoted by $\mathcal{H}_{R}$), there exists a pure state $|\phi\rangle\langle\phi|\in(\mathcal{H}_{A}\otimes\mathcal{H}_{B})^{2}=\mathcal{H}_{A}\otimes\mathcal{H}_{B}\otimes\mathcal{H}_{R}$ such that $\operatorname{Tr}_{R}(|\phi\rangle\langle\phi|)=\rho_{AB}$. Denote by $\rho_{R}=\operatorname{Tr}_{AB}(|\phi\rangle\langle\phi|)$.

For a pure bipartite state the entropies of the reduced states from the two subsystems are equal, $S_{L}(\rho_{AR})=S_{L}(\rho_{B})$, $S_{L}(\rho_{BR})=S_{L}(\rho_{A})$ and $S_{L}(\rho_{R})=S_{L}(\rho_{AB})$.

Every inequality for the linear entropy can therefore be transformed into an inverted inequality.

Let $f(x,y)$ denote the continuously differentiable function from Proposition 2. For the system $AR$ in the state $\rho_{AR}\in\mathcal{H}_{A}\otimes\mathcal{H}_{R}$, Theorem 1. states

[TABLE]

and, by the previous discussion

[TABLE]

For the other inequality just take the system $BR$ in the state $\rho_{BR}\in\mathcal{H}_{B}\otimes\mathcal{H}_{R}$ and, by the same argument

[TABLE]

Proposition 5**.**

There exists continuously differentiable functions $G_{1}$ and $G_{2}$, such that the inequalities (86) and (87) from the proof of the previous proposition can be written in the form

[TABLE]

Proof**.**

For equality we can rewrite (86) in the form

[TABLE]

Since this is linear in $y$ and by Lemma 9, we know that all the partial derivatives are always positive. Therefore, the implicit function theorem can be used and $U$ can be chosen to be the whole domain of $f(x,y)$. Setting $\mathbf{v}=(x,y)$ and $\mathbf{w}=z$ the theorem states, that for every $(x_{0},y_{0})$ there exists a continuously differentiable function $G_{1}(x,y)$, such that $G_{1}(x_{0},y_{0})=z$ and F\big{(}x,y,G(x,y)\big{)}=0. Therefore we know that

[TABLE]

Now fix $x$ to $x_{0}$. Define

[TABLE]

and note that $\frac{\partial f(x,y)}{\partial y}\geq\frac{1}{d_{A}d_{R}}>0$. By the inverse function theorem we can invert $f_{x_{0}}(y)$ and therefore rewrite Eq. (91) as

[TABLE]

Since this holds for every $x_{0}$, $G_{1}(x,y)$ is the continuously differentiable function of Eq. (88).

By an analogous argument we show the same for $G_{2}(x,y)$ of Eq. (89).

Note, that this is not the same as to invert $f(x,y):\mathds{R}^{2}\rightarrow\mathds{R}$, which is not possible.

So we have proven the existence and some properties of the functions $G_{1}$ and $G_{2}$, which give lower bound inequalities from purification. Let us now derive them explicitly. For the inhomogeneous subadditivity they are easy to determine, the two inverted inequalities following from the inequality from Lemma 3. are

[TABLE]

In the case of the dimensionally sharp subadditivity, things get a bit more complicated and one always has to be careful with the restriction. The inverted inequalities following from inequality (31) are given by

[TABLE]

and

[TABLE]

and hold for

[TABLE]

and

[TABLE]

respectively.

Now define

[TABLE]

and

[TABLE]

By Proposition 2. and the implicit function theorem, we know that $\tilde{f}_{1}(x,y)$ and $\tilde{f}_{2}(x,y)$ are continuously differentiable functions.

Now define

[TABLE]

The above discussion leads to

Corollary 1**.**

The inequality

[TABLE]

*holds and is sharper than the Araki-Lieb inequality for the Tsallis 2-entropy. *

Proof**.**

The first claim follows from the previous discussion.

Since the inequality from Theorem 1. is sharper than subadditivity, the inverted inequality is sharper than Araki-Lieb.

That inhomogeneous subadditivity and dimensionally sharp subadditivity are the sharpest possible inequalities, does however not imply, that the inverted inequality is the sharpest. Why is this so? Remember that in the proof we used our inequality for states of the form $\rho_{AR}$ (or $\rho_{BR}$). But this means that the inverted inequality is only the sharpest for states $\rho_{AR}\in\mathcal{H}_{A}\otimes\mathcal{H}_{R}$ (or $\rho_{BR}\in\mathcal{H}_{B}\otimes\mathcal{H}_{R}$) which can be purified in the space $\mathcal{H}_{A}\otimes\mathcal{H}_{B}\otimes\mathcal{H}_{R}$.

VII Conclusion

We have introduced a dimensionally sharp entropy inequality for the Tsallis 2-entropy (ISA+DSSA). It provides a sharp improvement for all finite dimensional quantum systems, extending entropy inequalities beyond the asymptotic domain. From a physical perspective, these inequalities provides stronger constraints for the distribution of purities in multipartite quantum states. As any state that is prepared experimentally, and used for information encoding, can be represented by a finite dimensional Hilbert space, the inequalities should be useful whenever one is interested in the distribution of purities or equivalently Bloch vector lengths. The lengths of Bloch vectors over different ’sectors’ Eltschka and Siewert (2015); Eltschka et al. (2018) (i.e. tensor products of $k$ local Bloch operators would correspond to a ’k’-sector) or correlation tensors de Vicente and Huber (2011); Laskowski et al. (2011); Klöckl and Huber (2015); Asadian et al. (2016); Schwemmer et al. (2015) are often used to quantify correlations and detect entanglement, so sharper inequalities will also be useful in this context. For the future we hope to find a sharp lower bound and extend the approach to multipartite systems, also finding an inequality akin to strong subadditivity.

Acknowledgements The authors would like to acknowledge fruitful discussions with Paul Appel, Otfried Gühne, Felix Huber and Milan Mosonyi. MH and SM acknowledge funding from the FWF (Y879-N27, I3053-N27 and P31339-N27). CK gratefully acknowledges support from the European Research Council (“reFUEL” ERC-2017-STG 758149). CE and JS acknowledge funding from the German ResearchFoundation Project EL710/2-1 and JS from the Basque Government grant IT986-16 and grant PGC2018-101355-B-I00 (MCIU/AEI/FEDER,UE).

Appendix A Bloch decomposition

Recall the Schmidt decomposition:

Theorem 1**.**

Suppose $|\psi\rangle$ is a pure state of a composite system, $AB$. Then there exist orthonormal bases \big{\{}|X_{i}\rangle\big{\}}_{i=1}^{d_{A}} and \big{\{}|Y_{i}\rangle\big{\}}_{i=1}^{d_{B}} for the systems $A$ and $B$ respectively, such that

[TABLE]

*where $N=min(d_{A},d_{B})$ and $c_{i}$ are non-negative real numbers satisfying $\sum\limits_{i=1}^{N}c_{i}^{2}=1$ known as Schmidt co-efficients. *

A consequence of the Schmidt decomposition is

Corollary 1**.**

*For every state $\rho_{A}\in\mathcal{H}$ of a quantum system there is a pure state $|\psi\rangle\langle\psi|_{AR}\in\mathcal{H}^{*}\otimes\mathcal{H}$ such that $\operatorname{Tr}_{R}(|\psi\rangle\langle\psi|_{AR})=\rho_{A}$. *

We already know that there exists a orthogonal operator basis for the Hilbert Schmidt space. According to Bertlmann and Krammer (2008), every state can be represented in a Bloch basis.

Theorem 2**.**

For the Hilbert space $\mathcal{H}^{*}\otimes\mathcal{H}$ with $dim(\mathcal{H})=d$ there exists a basis \big{\{}X_{i}\big{\}}_{i=0}^{d^{2}-1} such that

• •

The identity matrix is included, $X_{0}=\mathds{1}_{d}$.

• •

The other $d^{2}-1$ elements are traceless Hermitian matrices.

• •

The basis elements are mutually orthogonal, $\operatorname{Tr}(X_{i}^{\dagger}X_{j})=d\ \delta_{ij}$.

Every state $\rho$ can be represented in the Bloch picture

[TABLE]

*where $\vec{\mathbf{b}}\in\mathbb{R}^{d^{2}-1}$ is the Bloch vector, $\|\vec{\mathbf{b}}\|^{2}\leq d-1$. The entries are given by $b_{i}=\langle X_{i}\rangle=\operatorname{Tr}(\rho X_{i})$. *

It follows immediately that

[TABLE]

Appendix B Notes on inhomogeneous subadditivity

B.1 Alternative proof of Lemma 3.

Proof**.**

A direct way to show this result comes from the obvious inequality of the correlation tensor $\|C_{AB}\|_{2}^{2}\geq 0$, implying

[TABLE]

B.2 Proof of Proposition 1.

Proof**.**

The inequality from Theorem 2. can be rewritten to the equivalent form:

[TABLE]

In line (111) the square root can be taken, since both sides are positive. If we subtract the right hand side of the equation from Lemma 3. from the right hand side of Equation (113) we get

[TABLE]

which shows that the inequality in Lemma 3. is sharper than inequality (113).

Appendix C Proof of Proposition 2.

Definition C.1**.**

Let’s define the following sets by abuse of notation

[TABLE]

as the graphs of these functions. Let

[TABLE]

denote the set where equality holds for inequality (32) and

[TABLE]

*the extension of $\Omega$ along the z-axis. *

Claim 1**.**

The intersection between the sets $DSSA$ and $\overline{\Omega}$ is exactly the intersection between the sets $ISA$ and $\overline{\Omega}$. Let us denote this intersection by

[TABLE]

Proof**.**

Since $\Omega$ divides the domain \big{\{}(x,y)|\ 0\leq x\leq D_{A},\ 0\leq y\leq D_{B}\big{\}} of $g(x,y)$ and $h(x,y)$ in two parts, obviously $DSSA\cap\overline{\Omega}$ and $ISA\cap\overline{\Omega}$ are non-empty.

Since $\Gamma\subset\overline{\Omega}$, all we have to do is to check that the functions $g(x,y)$ and $h(x,y)$ coincide on $\Omega$.

[TABLE]

We conclude that $g(x,y)$ and $h(x,y)$ can be combined to a continuous surface, if we put $DSSA$ (left of $\overline{\Omega}$) and $ISA$ (right of $\overline{\Omega}$) together.

Now we only have to check the differentiability.

Since $g(x,y)$ and $h(x,y)$ are continuously differentiable functions, it suffices to show that the transition is too.

Look at the gradient of both functions on the overlap $\Omega$.

[TABLE]

[TABLE]

This shows that $f(x,y)$ is continuously differentiable.

Appendix D Proof of Lemma 8.

We want to show that the inequality

[TABLE]

is the sharpest possible inequality for a general bipartite quantum state. This can be shown by simply constructing states for which equality holds. The entropy vectors of those states lie on the boundary of the entropy body.

Lemma D.1**.**

Let $\rho_{AB}(\alpha)$ be a family of states of the form

[TABLE]

*where $\mu_{AB}$ is an arbitrary state, $\mathds{1}_{d_{AB}}$ the identity operator and $0\leq\alpha\leq 1$.

Then their entropies \big{(}S_{L}(\rho_{A}(\alpha)),S_{L}(\rho_{B}(\alpha)),S_{L}(\rho_{AB}(\alpha))\big{)} lie on a line in $\mathbb{R}^{3}$. *

Proof**.**

We have

[TABLE]

Using

[TABLE]

we can write

[TABLE]

and analogously

[TABLE]

Let

[TABLE]

be the entropy vector as defined in Eq. (2).

We then have

[TABLE]

and we can write

[TABLE]

We therefore know that the entropy vector of any convex combination of an arbitrary state and the completely mixed state $\rho_{AB}(\alpha):=\alpha\mu_{AB}+(1-\alpha)\frac{\mathds{1}_{d_{AB}}}{d_{AB}}$ lies on the line between the entropy vector of the state and the entropy vector of the completely mixed state (upper right corner).

Note however, that this is not true for two arbitrary states.

The set $\Gamma$ can be described as a parametrised curve in the following way

[TABLE]

where

[TABLE]

This can be seen by simply substituting $x=t$. The second argument $\gamma_{2}(t)$ follows immediately from the definition of $\Gamma$. To obtain the third entry $\gamma_{3}(t)$, plug the first two into the definition of $ISA$.

Lemma D.2**.**

The set of states depending on $0\leq\alpha\leq 1$.

[TABLE]

where

[TABLE]

*form a family of states with entropies on $\Gamma$ and on $ISA$. *

Proof**.**

We can calculate

[TABLE]

Using $t(\alpha)=\frac{d_{A}-1}{d_{A}}-\frac{d_{A}-1}{d_{A}}\alpha^{2}$ we can now write

[TABLE]

Therefore, we have found states whose entropies lie on $\Gamma$. We know, that the entropies of the mixture of an arbitrary state and the completely mixed state lie on a line. Since $\Gamma$ lies on $ISA$, we can conclude that $ISA$ (since it is linear) is the sharpest possible inequality for states on the right hand side of $\overline{\Omega}$.

Lemma D.3**.**

The set of states depending on $0\leq\alpha\leq 1$ and $0\leq\beta\leq 1$ with $\alpha+\beta\leq 1$

[TABLE]

*where $\mu_{1}$ and $\mu_{2}$ are defined as before and $\mu_{3}=|00\rangle\langle 00|$, form a family of states with entropies on $DSSA$. *

Proof**.**

In an analogous way to the previous proof we can calculate

[TABLE]

The surface $DSSA$ is given by the equation

[TABLE]

Since

[TABLE]

we have found a set of states whose entropies lie on $DSSA$, proving $DSSA$ to be sharp.

Appendix E Proof of Proposition 3.

Proof**.**

Use Lemma 3. and Eq. (73) to conclude that the inequality

[TABLE]

holds and analogously Theorem 1. to conclude that the inequality

[TABLE]

holds under the assumption that

[TABLE]

Since $f(x,y)$, as defined in Proposition 2., is continously differentiable and $e(x):=1-2^{-x}$ is a diffeomorphism for $x\in\mathbb{R}^{+}$, also

[TABLE]

is continuously differentiable.

Appendix F Implicit function theorem and inverted function theorem

The implicit function theorem states:

Theorem 1**.**

*Let $U$ be an open set in $\mathbb{R}^{k+n}$ and $F:U\rightarrow\mathbb{R}^{n}$ be a continuously differentiable function. Let $\mathbf{v}\in\mathbb{R}^{k}$ and $\mathbf{w}\in\mathbb{R}^{n}$, such that the function can be written as $F(\mathbf{v},\mathbf{w})$.

Suppose $(\mathbf{v}_{0},\mathbf{w}_{0})\in U$ such that $F(\mathbf{v}_{0},\mathbf{w}_{0})=0$ and the Jacobian restricted to $\mathbf{w}$ is invertible, that is*

[TABLE]

*Then there exists a neighbourhood $V\subset\mathbb{R}^{k}$ of $\mathbf{v}_{0}$ and a unique function $G:V\rightarrow\mathbb{R}^{n}$, such that $G(\mathbf{v}_{0})=\mathbf{w}_{0}$ and $F(\mathbf{v},G(\mathbf{v}))=0$ for every $\mathbf{v}\in V$. *

The inverse function theorem states:

Theorem 2**.**

*Let $U$ be an open set in $\mathbb{R}^{n}$ and $F:U\rightarrow\mathbb{R}^{n}$ be a continuously differentiable function. Let $\mathbf{v}_{0}\in\mathbb{R}^{n}$ such that the Jacobian is invertible.

Then there exists a neighbourhood $V\subset\mathbb{R}^{n}$ of $\mathbf{v}_{0}$, such that $F$ is invertible and the inverse function $F^{-1}$ is continuously differentiable.