On the two-copy distillability of Werner states and a new partial trace inequality

Introduction

Werner states

Let $(e_i)_{i\in[d]}$ be the standard basis of $\mathbb{C}^{d}$, and let $E_{ij}:=e_i e_j^{*}\in M_d(\mathbb{C})$ be the associated matrix units, so that $(E_{ij})_{i,j\in[d]}$ is an orthonormal basis of $M_d(\mathbb{C})$ for the inner product (6): $\langle E_{ij},E_{kl}\rangle_F=\delta_{ik}\delta_{jl}$. Let $F$ denote the flip on $\mathbb{C}^{d}\otimes\mathbb{C}^{d}$, defined by $F(x\otimes y)=y\otimes x$ and given by

The operator $F$ is Hermitian and satisfies $F^{2}=I$. For $\alpha\in[-1,1]$ the Werner state with parameter $\alpha$ is

following the convention of [7 Eq. (4)]; $\varrho_{\alpha}$ is written for $\varrho(d,\alpha)$ when $d$ is clear.

It is known that the partial transpose $\varrho(d,\alpha)^{T_{B}}$ is not positive semidefinite, i.e. has negative partial transpose (NPT), if and only if $\alpha<-\frac{1}{d}$ [1]. Thus, $\varrho(4,-\frac{1}{2})$ is NPT. For $\alpha\ge-\frac{1}{d}$ the state is separable [1].

Entanglement distillation

A bipartite state $\varrho$ is two-copy distillable if some vector $\psi$ of Schmidt rank at most two across $A_{1}A_{2}:B_{1}B_{2}$ satisfies $\langle\psi|(\varrho^{T_{B}})^{\otimes2}|\psi\rangle<0$, where $T_{B}$ is partial transposition on the second subsystem of each copy [2, 7], and two-copy undistillable otherwise.

A state with positive partial transpose cannot be distilled: if $\varrho^{T_{B}}\succeq0$, then $(\varrho^{T_{B}})^{\otimes n}\succeq0$ for every $n$, so $\langle\psi|(\varrho^{T_{B}})^{\otimes n}|\psi\rangle\ge0$ for every $\psi$ and $\varrho$ is undistillable at every number of copies. Thus non-positivity of the partial transpose (NPT)—the Peres criterion [3]—is necessary for distillability [2].

A sufficient condition is the reduction criterion: any state violating the inequality $\varrho_{A}\otimes I-\varrho\succeq0$ or $I\otimes\varrho_{B}-\varrho\succeq0$, where $\varrho_{A}=\operatorname{tr}_{2}(\varrho)$ and $\varrho_{B}=\operatorname{tr}_{1}(\varrho)$, is distillable [4].

Between the two lies a partial unresolved regime: whether every NPT state is distillable is a longstanding open question. Recall that $\varrho(d,\alpha)$ is NPT if and only if $\alpha<-\frac{1}{d}$ [1]. Furthermore,

Date: July 28, 2026.

Key words and phrases. Werner states, distillability, NPT bound entanglement, partial trace, singular values.

We thank Karol Życzkowski for fruitful discussions. TCF acknowledges financial support from VILLUM FONDEN via the QMATH Centre of Excellence (Grant No. 10059) and the Novo Nordisk Foundation (grant NNF20OC0059939 “Quantum for Life”). FH was funded in whole or in part by the National Science Centre, Poland 2024/54/E/ST2/00451 and by the Polish National Agency for Academic Exchange under the Strategic Partnership Programme grant BNI/PST/2023/1/00001. For the purpose of Open Access, the author has applied a CC-BY public copyright licence to any Author Accepted Manuscript (AAM) version arising from this submission. BP and IV were supported by the Hungarian National Research, Development and Innovation Office, NKFIH Grant No. K-145904. Werner states are one-copy undistillable if and only if $\alpha\ge-\frac{1}{2}$ [2, 5, 11]. Since one-copy distillability trivially implies $k$-copy for every $k \ge1$, $\varrho(d,\alpha)$ is in particular two-copy distillable for every $\alpha< -\frac{1}{2}$.

Regarding undistillability, Vianna and Doherty showed by numerical computation that qutrit Werner states are two-copy undistillable on the whole range $-\frac{1}{2} \le\alpha< -\frac{1}{3}$ [6]. For general $d$, Costa Rico and Wolf showed that Werner states are two-copy undistillable for all $\alpha\ge-\frac{1}{3}$ [11].

	$\alpha< -\frac{1}{2}$	$-\frac{1}{2} \le\alpha< -\frac{1}{3}$	$-\frac{1}{3} \le\alpha< -\frac{1}{d}$	$\alpha\ge-\frac{1}{d}$
		$(d \ge3\text{ only})$	$(d \ge4\text{ only})$
PPT [3]	N	N	N	Y
NPT [3]	Y	Y	Y	N
1-copy distillable [2, 5, 11]	Y	N	N	N
2-copy distillable	Y	N [here]$^{\dagger}$	N [11]$^{\ddagger}$	N [1]

The open problem

The concrete question addressed here is the following, listed as Problem 5 in Horodecki, Rudnicki and Życzkowski’s Five Open Problems in Quantum Information Theory [9].

Problem 1.1 (Problem 5 of [9]). Decide whether the Werner state $\varrho(4,-\frac{1}{2})$ of two ququarts, $d = 4$, defined in Eq. (2), is two-copy distillable.

The answer given in this work is negative [shown in Cor. B]. More generally, we ask:

Problem 1.2. For which $d$ and $\alpha$ is $\varrho(d,\alpha)$ two-copy distillable?

We show that for every $d \ge2$, the Werner state $\varrho(d,\alpha)$ is two-copy undistillable if and only if $\alpha\ge-\frac{1}{2}$ [shown in Cor. C]. Thus every Werner state is two-copy undistillable if and only if it is one-copy undistillable.

The proof presented in this work was found with ChatGPT Sol-5.6 and refined with Claude Opus and Fable. All results have been carefully checked by the authors.

During preparation of this manuscript we became aware of a similar work by Fu et al, arXiv:2607.21367 and a manuscript in preparation by Kishor Bharti, Rishikesh Gajjala, and Tobias Haug. They seem to use a different proof strategy but also use a variant of Lemma 6.6.

Undistillability from a partial trace inequality

To answer Problems 1.1 and 1.2, we will make use of the following partial trace inequality.

Costa Rico [10] states the criterion for $n$ copies and in the equivalent quadratic form $q_\alpha(C) \ge0$, where $q_\alpha(C)=\lVert C\rVert_F^2+\alpha(\lVert\operatorname{tr}_1(C)\rVert_F^2+\lVert\operatorname{tr}_2(C)\rVert_F^2)+\alpha^2\lvert\operatorname{tr}(C)\rvert^2$; only $n=2$ is used here. Writing $\alpha=-t$ with $t>0$ and dividing $q_{-t}(C)\ge0$ by $t$ gives the norm form (3) stated above, which is the form recorded by Costa Rico and Wolf [11 Eq. (31)] and the one applied here.

Costa Rico and Wolf [11] prove that (3) holds for every rank-two $C$ when $\alpha\ge-\frac{1}{3}$, and thus Werner states are two-copy undistillable in this range. Our aim is to show that this range extends to $\alpha=-\frac{1}{2}$.

Contributions

We prove the required inequality, (3) at $\alpha=-\frac{1}{2}$, in the following stronger form which is valid for every rank and local dimensions $d_1,d_2$. Note that no assumption on positive-semidefiniteness, hermiticity, or normality is made on $C$.

Theorem A is proved in § 8. By Theorem 2.1 follows:

In fact, we can deduce from Theorem A also:

Overview

Dependency table

Below we list, for every item, its provenance and dependencies. We cite the original reference for items that predate this note (these are collected in § 6), and highlight with [here] for results established in this work.

Item	Name	Source	Depends on	Used in
Lem. 6.1	Parker–Fillmore constant diagonal form	[13, 12]	—	Lem. 7.1
Lem. 6.2	singular-value decomposition	[19, 20]	spectral theorem	Lem. 7.1
Lem. 6.3	Lagrange (variance) identity	[16]	—	Thm. A, Step 6
Lem. 6.4	rank-one partial-trace inequality	[10, 11]	(13), (14), (19), (20)	Thm. A, Steps 2 and 3
Lem. 6.5	Theorem A for positive matrices	[15]	(23)	not used; delimits the elementary regime
Lem. 6.6	four-state marginal identity	[8]	quoted; or (19), (20)	Cor. 6.7
Cor. 6.7	crossed polarization	[here]	Lem. 6.6	Thm. A, Step 3
Lem. 7.1	balanced rank-$r$ decomposition	[here]	Lems. 6.2, 6.1	Thm. A
Thm. A	rank-constrained partial-trace inequality	[here]	Lems. 7.1, 6.4, 6.3; Cor. 6.7	Cors. B and C
Cor. B	Problem 1.1: $\varrho(4,-\frac{1}{2})$ is two-copy undistillable	[here]	Thm. A at $d_1=d_2=4$, $r=2$; Thm. 2.1	— (the target)
Cor. C	Problem 1.2 target	[here]	Thm. A at $r=2$; (23); Thm. 2.1; separability/NPT criterion [1]

Dependency graph

The same information in graphical form; arrows point from an ingredient to the result that uses it. Shaded nodes are known, framed nodes are established here.

Conventions and definitions

Index sets and matrix spaces

For a positive integer $d$, write $[d] := \{1,2,\ldots,d\}$. For positive integers $m,n$, let $M_{m,n}(\mathbb{C})$ denote the vector space of $m \times n$ complex matrices. Write $M_n(\mathbb{C}) := M_{n,n}(\mathbb{C})$. The conjugate transpose of a matrix $T$ is denoted by $T^*$. The ordinary transpose is denoted by $T^\mathsf{T}$. Entrywise complex conjugation is denoted by $\overline{T}$.

Inner products

For column vectors $x,y \in\mathbb{C}^n$ the inner product and norm are

For matrices $S,T \in M_{m,n}(\mathbb{C})$ the Frobenius, or Hilbert–Schmidt, inner product and norm are

Both are conjugate-linear in the first argument and linear in the second. A third pairing on a direct sum of two matrix spaces, appears in (26); it is again the same one, since the pair $(R_1,R_2)$ may be read as the block-diagonal matrix $\operatorname{diag}(R_1,R_2)$, for which $\operatorname{tr}(\operatorname{diag}(R_1,R_2)^*\operatorname{diag}(S_1,S_2)) = \langle R_1,S_1\rangle_F + \langle R_2,S_2\rangle_F$. No new notation is therefore introduced for it.

Rank-one operators

For column vectors $x,y \in\mathbb{C}^N$, the matrix $xy^*$ is the rank-at-most-one operator defined by $(xy^*)z = x(y^*z)$. Its trace and Frobenius norm satisfy

Note that the left-hand side of (8) is a matrix norm and the right-hand side a product of vector norms; the subscript $F$ marks the difference. Indeed, $\operatorname{tr}(xy^*) = \sum_{k=1}^{N}x_k\overline{y_k} = y^*x$, and

Trace cyclicity, including rectangular products

If $R \in M_{m,n}(\mathbb{C})$ and $S \in M_{n,m}(\mathbb{C})$, then

This is standard; see Horn and Johnson [19 Sec. 0.3] or Bhatia [20 Sec. I.1].

Tensor-product coordinates

Let $\mathcal{H}_1=\mathbb{C}^{d_1}$ and $\mathcal{H}_2=\mathbb{C}^{d_2}$, with standard orthonormal bases $(e_a)_{a\in[d_1]}$ and $(f_b)_{b\in[d_2]}$. We identify $\mathcal{H}_1\otimes\mathcal{H}_2\cong\mathbb{C}^{d_1d_2}$ using the ordered basis

A vector $x\in\mathcal{H}_1\otimes\mathcal{H}_2$ is written $x=\sum_{a=1}^{d_1}\sum_{b=1}^{d_2}x_{ab}e_a\otimes f_b$. A matrix $C\in M_{d_1d_2}(\mathbb{C})$ has entries $C_{(a,b),(c,d)}$ with $a,c\in[d_1]$ and $b,d\in[d_2]$. For $P\in M_{d_1}(\mathbb{C})$ and $Q\in M_{d_2}(\mathbb{C})$, their Kronecker product is defined by

Partial traces

For $C\in M_{d_1d_2}(\mathbb{C})$, define $\operatorname{tr}_1(C)\in M_{d_2}(\mathbb{C})$ and $\operatorname{tr}_2(C)\in M_{d_1}(\mathbb{C})$ by

Thus $\operatorname{tr}_1$ traces out the first tensor factor, while $\operatorname{tr}_2$ traces out the second tensor factor. The partial traces are adjoint to tensoring with an identity matrix: for every $P\in M_{d_1}(\mathbb{C})$ and $Q\in M_{d_2}(\mathbb{C})$,

Equations (15)–(16) can be seen as the coordinate-free definition of the partial trace: $\operatorname{tr}_2$ is the unique linear map $M_{d_1d_2}(\mathbb{C})\to M_{d_1}(\mathbb{C})$ satisfying (15) for all $P$, and $\operatorname{tr}_1$ the unique one satisfying (16) for all $Q$ [18 §1.1.2], [17 §2.4.3].

Vectorization

For $X\in M_{d_1,d_2}(\mathbb{C})$, define the row-major vectorization,

This provides a linear bijection $\operatorname{vec}:M_{d_1,d_2}(\mathbb{C})\longrightarrow\mathcal{H}_1\otimes\mathcal{H}_2$, and it is an isometry between the two inner products of Section 5.2:

If $x=\operatorname{vec}(X)$ and $y=\operatorname{vec}(Y)$, then the rank-one operator $xy^*$ satisfies

Indeed,

These are partial-trace forms of the identity $\operatorname{vec}(ABC)=(A^T\otimes C)\operatorname{vec}(B)$ [see [14, §2.3, Eq. (4)]] for the column majorized form.

Rank and trace

For every $C \in M_N(\mathbb{C})$,

Indeed, for a singular-value decomposition $C = \sum_{j=1}^{s} \mu_j u_jv_j^*$ with $s = \operatorname{rank}(C)$, Cauchy–Schwarz gives $|\operatorname{tr}(C)| = |\sum_{j=1}^{s} \mu_j u_jv_j^*| = \sum_{j=1}^{s} \mu_j |u_j^*v_j| \le\sum_j \mu_j \le\sqrt{s}\,\lVert C\rVert_F$.

The paired partial-trace map

Define the linear map

by

The direct sum is equipped with the inner product

The two summands are orthogonal, so this is again the pairing (6), read on block-diagonal matrices. Thus

Equations (25) and (26) are definitions, and (27) is the norm they induce. The displayed norm is the genuine norm on the target direct sum: $\operatorname{tr}_1(C)$ occupies its $M_{d_2}(\mathbb{C})$ slot and $\operatorname{tr}_2(C)$ its $M_{d_1}(\mathbb{C})$ slot.

Known ingredients

Everything in this section predates the manuscript, with the exception of Corollary 6.7 which is a straightforward consequence of Lemma 6.6.

Constant diagonals

Lemma 6.1 (Parker–Fillmore constant diagonal form [13, 12]). Let $K \in M_r(\mathbb{C})$, and define $\tau:= \frac{\operatorname{tr}(K)}{r}$. Then there exists a unitary matrix $Q \in M_r(\mathbb{C})$ such that

equivalently, there is an orthonormal basis $q_1,\ldots,q_r$ of $\mathbb{C}^r$ with $q_i^*Kq_i = \tau$ for every $i \in[r]$.

The statement used here is [13 Theorem 9]: for every square complex matrix there is a unitary $Q$ for which $Q^*KQ$ has all diagonal entries equal; since the trace is unitarily invariant, that common value is $\tau= \operatorname{tr}(K)/r$. The same statement is [12 Corollary 2], with the constant likewise fixed by the trace condition of [12 Theorem 2].

Singular-value decomposition

Lemma 6.2 (Singular-value decomposition [19, 20]). Let $C \in M_N(\mathbb{C})$ have rank $s$. The case $s = 0$ is interpreted using empty families and an empty sum. Then there exist positive numbers $\mu_1,\ldots,\mu_s > 0$ and orthonormal families

such that $C = \sum_{i=1}^{s} \mu_i u_i v_i^*$.

Lagrange variance identity

Lemma 6.3 (Lagrange variance identity [16 Theorem 7]). For real numbers $\delta_1,\ldots,\delta_r$,

Equality in (30) is reached for $\delta_1=\cdots=\delta_r$.

To see this, note that both sides of (30) equal $(r-1)\sum_i\delta_i^2-2\sum_{i<j}\delta_i\delta_j$. On the left, $r\sum_i\delta_i^2-\left(\sum_i\delta_i\right)^2=\sum_i\delta_i^2+2\sum_{i<j}\delta_i\delta_j$; on the right, expanding each square gives $\sum_{i<j}(\delta_i-\delta_j)^2=(r-1)\sum_i\delta_i^2-2\sum_{i<j}\delta_i\delta_j$, since each $\delta_i^2$ occurs in exactly $r-1$ of the $\binom{r}{2}$ terms.

Rank-one partial-trace identities

Lemma 6.4 can be seen as follows: Let $F_A$ and $F_B$ denote the swap operators acting on the pairs of $\mathbb{C}^{d_1}$ and $\mathbb{C}^{d_2}$ factors respectively. Then,

where $P_A^- = \frac{1}{2}(I-F_A)$ and $P_B^- = \frac{1}{2}(I-F_B)$ are the antisymmetric projections.

Inequality (34) is the $q = 2$ case of positive-matrix entropy subadditivity [15]. We note that Lemma 6.5 is not used in the proof of Theorem A.

A four-state identity

The identity of Eltschka and Siewert [8] below converts a trace inner product of the marginals of two rank-one operators into a complementary expression with terms exchanged. What Theorem A eventually uses is its direct-sum consequence, Corollary 6.7.

Proof. By the definition (26) of the direct-sum inner product, the left-hand side of (39) is the sum of the left-hand sides of (37) and (38), and the right-hand side is the sum of their right-hand sides. Adding the two therefore gives the claim. This ends the proof. □

Remark 6.8. Setting $(u,v,s,t)=(x_i,y_i,x_j,y_j)$, Corollary 6.7 reads

the form used in the proof of Theorem A.

The balanced rank decomposition

We now state a main ingredient of the proof, a decomposition of $C$ such that the diagonal is constant and the Gram matrices corresponding to the left and right factors coincide.

Proof. Let $s:=\operatorname{rank}(C)\le r$. Take a singular-value decomposition from Lemma 6.2: $C=\sum_{i=1}^{s}\mu_i u_i v_i^*$. Extend $u_1,\ldots,u_s$ to an orthonormal family $u_1,\ldots,u_r$, and extend $v_1,\ldots,v_s$ to an orthonormal family $v_1,\ldots,v_r$. Define $\mu_{s+1}=\cdots=\mu_r=0$. Let

be the corresponding $N\times r$ isometries, and let $\Sigma:=\operatorname{diag}(\mu_1,\ldots,\mu_r)$. Then $C=U\Sigma V^*$. Define

Then $C=XY^*$ and $X^*X=\Sigma=Y^*Y$.

Set $K:=Y^*X\in M_r(\mathbb{C})$. By rectangular trace cyclicity (10) it holds that

By Lemma 6.1, there exists a unitary $Q\in M_r(\mathbb{C})$ such that every diagonal entry of $Q^*KQ$ is

Now let $\widetilde{X}:=XQ$ and $\widetilde{Y}:=YQ$ and note that

using $QQ^*=I$. Also define

using $X^*X=\Sigma=Y^*Y$. Note that $G\succeq0$, being a Gram matrix.

Let $x_i$ be the $i$-th column of $\widetilde{X}$, and $y_i$ the $i$-th column of $\widetilde{Y}$. Then $\widetilde{X}\widetilde{Y}^*=\sum_{i=1}^{r}x_i y_i^*$, so by (48), $C=\sum_{i=1}^{r}x_i y_i^*$, which is (41).

For (42), read off the entries of $G$:

By (49) the two matrices both equal $G$, so $x_i^*x_j=y_i^*y_j=g_{ij}$, which is (42).

Finally,

shows (43). This ends the proof. □

Proof of Theorem A

Proof. The strategy is as follows: By (27), the left-hand side of (4) equals $\left\lVert\Phi(C)\right\rVert^2$. Expanding $C$ as a balanced rank-$r$ decomposition we can write $\Phi(C)=\sum_i\Phi(x_i y_i^*)$. Its squared norm $\left\lVert\Phi(C)\right\rVert^2$ can be written as a double sum of diagonal and off-diagonal terms, which we bound separately from above. The resulting upper bound plus an additional sum of squares expression then equals the right-hand side of (4), proving the assertion.

Step 1: balanced decomposition

Let $N:=d_1d_2$. Apply the balanced decomposition in Lemma 7.1 to obtain vectors $x_1,\ldots,x_r$, $y_1,\ldots,y_r\in\mathbb{C}^N$ and a Hermitian positive semidefinite matrix $G=(g_{ij})_{i,j=1}^r$ such that

where

We also define $\delta_i:=g_{ii}=\lVert x_i\rVert^2=\lVert y_i\rVert^2\ge0$.

By linearity of $\Phi$,

Then,

Our aim in steps 2 and 3 in then to bound the diagonal and off-diagonal terms appearing in $\left\lVert\Phi(C)\right\rVert^2$.

Step 2: diagonal terms

By Lemma 6.4, (53), and the definition of $\delta_i$,

Step 3: cross terms

Fix $i\ne j$. By Corollary 6.7, respectively the form given in Eq. (40),

By Cauchy–Schwarz in the direct-sum Hilbert space,

Now consider the rank-one operator $y_j y_i^*$ appearing on the rhs of (59). By (7), (8), and then the definition of $\delta_i$ in (53),

By Lemma 6.4 therefore

The same computation for $x_jx_i^*$ gives

Combining (59), (62), and (63),

These inequalities (57) and (64) finish the upper bound on the lhs of (4).

Step 4: a combined upper bound

Now expand the squared norm of $\Phi(C)$. Using (56),

Using the upper bounds of (57) and (64),

Above, we have regrouped the terms involving $\delta_i$ by $\sum_i\delta_i^2+2\sum_{i<j}\delta_i\delta_j=(\sum_i\delta_i)^2$.

Step 5: express $\|C\|_F^2$ through the common Gram matrix

As in the balanced rank-$r$ decomposition of Lemma 7.1, let $\widetilde X$ have columns $x_1\cdots x_r$ and $\widetilde Y$ have columns $y_1\cdots y_r$. Then

where $G=(g_{ij})_{ij}$. Therefore

Since $G$ is Hermitian, $g_{ji}=\overline{g_{ij}}$. Hence

Thus

Step 6: the remaining term is a sum of squares

Taking the modulus squared of (54) gives $|\tau|^2=|\operatorname{tr}(C)|^2/r^2$, and hence

Let us now show that (4) holds. To this end, subtract its lhs from its rhs. Using (70), (71) and the upper bound on $\lVert\Phi(C)\rVert^{2}$ of (66), we have

The last expression can then be regrouped using the Lagrange identity [Lemma 6.3],

This is a clearly non-negative expression, and thus proves inequality (4). This ends the proof.

Back to distillability

We can now answer the first question.

Problem 1.1 (Problem 5 of [9]). Decide whether the Werner state $\varrho(4,-\frac{1}{2})$ of two ququarts, $d=4$, defined in Eq. (2), is two-copy distillable.

The answer is:

Proof. Theorem A with $d_{1}=d_{2}=4$ and $r=2$ states that every $C\in M_{16}(\mathbb{C})$ of rank at most two satisfies $\lVert\operatorname{tr}_{1}(C)\rVert_{F}^{2}+\lVert\operatorname{tr}_{2}(C)\rVert_{F}^{2}\le2\lVert C\rVert_{F}^{2}+\frac{1}{2}\lvert\operatorname{tr}(C)\rvert^{2}$. This is condition (3) of Theorem 2.1 at $d=4$, $\alpha=-\frac{1}{2}$. It follows that $\varrho(4,-\frac{1}{2})$ is two-copy undistillable. This ends the proof. □

We can also answer the second main question:

Problem 1.2. For which $d$ and $\alpha$ is $\varrho(d,\alpha)$ two-copy distillable?

The answer is:

Proof. We proceed by cases:

Case 1: $-\frac{1}{d}\le\alpha$. $\varrho(d,\alpha)$ is separable [1] and thus two-copy undistillable.

For $\alpha<-\frac{1}{d}$, $\varrho(d,\alpha)$ is NPT [1] and we distinguish two further cases:

Case 2: $-\frac{1}{2}\le\alpha<-\frac{1}{d}$. Note that this is only relevant for $d\ge3$. Let $C\in M_{d^{2}}(\mathbb{C})$ have rank at most two, put $t:=-\alpha\in(0,\frac{1}{2}]$,

By Theorem A with $d_{1}=d_{2}=d$ and $r=2$, $\lVert\Phi(C)\rVert^{2}\le2a+b/2$, so it suffices to show that

for all $t:=-\alpha\in(0,\frac{1}{2}]$. This indeed holds due to,

Above we used that the first factor is nonnegative due to $t\le\frac{1}{2}$. The second factor is nonnegative due to (23), $b\le2a$, which gives $bt\le2at\le a\le2a$, i.e. $\frac{a}{t}\ge\frac{b}{2}$. Hence (76) holds. Combining (76) with Theorem A gives

which is condition (3) of Theorem 2.1. Hence $\varrho(d,\alpha)$ is two-copy undistillable for $-\frac{1}{2}\le\alpha$.

Case 3: $\alpha<-\frac{1}{2}$. Then $\varrho(d,\alpha)$ is one-copy distillable [2], and one-copy distillability trivially implies distillability at every larger number of copies; hence $\varrho(d,\alpha)$ is two-copy distillable. This ends the proof. □

As a consequence, the one and two-copy distillability regions of $\varrho(d,\alpha)$ coincide.

{sup}1` Department of Mathematical Sciences, University of Copenhagen, Copenhagen, 2100, Denmark

Email address: [email protected]

{sup}2` Division of Quantum Information, Institute of Informatics, Faculty of Mathematics, Physics and Informatics, University of Gdańsk, Wita Stwosza 57, 80-308 Gdańsk, Poland

Email address: [email protected]

{sup}3` MTA-ELTE "Momentum" Integrable Quantum Dynamics Research Group, ELTE Eötvös Loránd University, Pázmány Péter sétány 1/A, 1117 Budapest, Hungary

{sup}4` Holographic Quantum Field Theory Research Group, HUN-REN Wigner Research Centre for Physics, Konkoly-Thege Miklós u. 29-33, 1121 Budapest, Hungary

Email address: [email protected]