Necessary conditions on effective quantum entanglement catalysts

Abstract

Quantum catalytic transformations play important roles in the transformation of quantum entangled states under local operations and classical communications (LOCC). The key problems in catalytic transformations are the existence and the bounds on the catalytic states. We present the necessary conditions of catalytic states based on a set of points given by the Schmidt coefficients of the entangled source and target states. The lower bounds on the dimensions of the catalytic states are also investigated. Moreover, we give a detailed protocol of quantum mixed state transformation under entanglement-assisted LOCC.

Appendix

1.1 A. Proof of Proposition 1

Proof

If the conclusion is not true, one would have $q_1r_{s+1}\le q_lr_s$, $q_{l+1}r_s\le q_dr_{s-1}$ or $q_1r_{s+1}\le q_dr_{s-1}$, which imply that the largest $d(s-1)+l$ elements of ${\mathbf {q}} \otimes {\mathbf {r}}$ are $q_1r_1,\ldots ,q_dr_1$, $q_1r_2,\ldots ,q_dr_2$, $\ldots$, $q_1r_s,\ldots ,q_lr_s$. The summation of these largest $d(s-1)+l$ elements of ${\mathbf {q}} \otimes {\mathbf {r}}$ gives

$$\begin{aligned} \sum _{ i = 1 }^{d(s-1)+l}( {\mathbf {q}} \otimes {\mathbf {r}} ) _ { i }&=\sum _{ i = 1}^d\sum _{j=1}^{s-1}q_ir_j+\sum _{t=1}^lq_tr_s\\&=\sum _{ i = 1 }^{s-1}r_i+r_s\sum _{j= 1}^lq_j. \end{aligned}$$

Meanwhile, the largest $d(s-1)+l$ elements of ${\mathbf {p}} \otimes {\mathbf {r}}$ are not less than the sum of the terms $p_1r_1,\ldots ,p_dr_1$, $p_1r_2,\ldots ,p_dr_2$, $\ldots$, $p_1r_s,\ldots ,p_lr_s$,

$$\begin{aligned} \sum _{ i = 1 }^{d(s-1)+l}( {\mathbf {p}} \otimes {\mathbf {r}} ) _ { i } ^ { \downarrow }&\ge \sum _{ i = 1}^d\sum _{j=1}^{s-1}p_ir_j+\sum _{t=1}^lp_tr_s\\&=\sum _{ i = 1 }^{s-1}r_i+r_s\sum _{j= 1}^lp_j. \end{aligned}$$

On the other hand, since $l\in {\mathcal {L}}$ and $\sum _{j= 1}^lp_j >\sum _{j= 1}^lq_j$, we have

$$\begin{aligned} \sum _{ i = 1 }^{d(s-1)+l}( {\mathbf {p}} \otimes {\mathbf {r}} ) _ { i } ^ { \downarrow }\geqslant&\sum _{ i = 1 }^{s-1}r_i+r_s\sum _{j= 1}^lp_j \\ >&\sum _{ i = 1 }^{s-1}r_i+r_s\sum _{j= 1}^lq_j\\ =&\sum _{ i = 1 }^{d(s-1)+l}( {\mathbf {q}} \otimes {\mathbf {r}} ) _ { i } ^ { \downarrow } \end{aligned}$$

for any $s=2,3,\ldots ,k-1$, which contradicts the assumption that ${\mathbf {p}} \otimes {\mathbf {r}} \prec {\mathbf {q}} \otimes {\mathbf {r}}$.

Next we prove the first inequality in (6). Since $l\in {\mathcal {D}}_1$, we have $p_1+\cdots +p_l>q_1+\cdots +q_l$ and $p_1+\cdots +p_{l+1}\le q_1+\cdots +q_{l+1}$, which give rise to $p_{l+1}<q_{l+1}$. Taking into account $p_d>q_d$, we have $\frac{p_d}{p_{l+1}}>\frac{q_d}{q_{l+1}}$ as ${\mathbf {p}}$ and ${\mathbf {q}}$ are solvable incomparable.

We use proof by contradiction. If $\frac{q_d}{q_{l+1}}\ge \frac{r_k}{r_{k-1}}$, i.e., $q_dr_{k-1}\ge q_{l+1}r_k$, then the smallest $d-l$ elements of ${\mathbf {q}} \otimes {\mathbf {r}}$ are $q_{l+1}r_k,q_{l+2}r_k,\ldots ,q_dr_k$. Moreover, the sum of the smallest $d-l$ elements of ${\mathbf {p}} \otimes {\mathbf {r}}$ is not larger than the sum of $p_{l+1}r_k,p_{l+2}r_k,\ldots ,p_dr_k$. Because of $l\in {\mathcal {L}}$, we get $p_{l+1}+\cdots +p_d<q_{l+1}+\cdots +q_d$. Therefore, we have the following relation,

$$\begin{aligned}&\sum _ { i = dk-d+l+1 } ^ { dk } ( {\mathbf {p}} \otimes {\mathbf {r}} ) _ { i } ^ { \downarrow }\leqslant \sum _{ i = l+1 }^dp_ir_k\\&\quad <\sum _{ i = l+1 }^dq_ir_k= \sum _ { i = dk-d+l+1 } ^ { dk} ( {\mathbf {q}} \otimes {\mathbf {r}} ) _ { i } ^ { \downarrow }, \end{aligned}$$

which contradicts the assumption that ${\mathbf {p}} \otimes {\mathbf {r}} \prec {\mathbf {q}} \otimes {\mathbf {r}}$, and proves the first inequality in (6).

Concerning the second inequality in (6), we use again the proof by contradiction. If $\frac{q_1}{q_l}\le \frac{r_1}{r_2}$, the largest l elements of ${\mathbf {q}} \otimes {\mathbf {r}}$ are $q_1r_1,q_2r_1,\ldots ,q_lr_1$. Since the sum of the largest l elements of ${\mathbf {p}} \otimes {\mathbf {r}}$ is not less than the sum of $p_1r_1,p_2r_1,\ldots ,p_lr_1$ for $l\in {\mathcal {L}}$, we have $p_1+\cdots +p_l>q_1+\cdots +q_l$. Then we have the relation,

$$\begin{aligned}&\sum _ { i = 1 } ^l( {\mathbf {p}} \otimes {\mathbf {r}} ) _ { i } ^ { \downarrow }\geqslant \sum _{ i = 1 }^lp_ir_1\\&\quad >\sum _{ i = 1 }^lq_ir_1= \sum _ { i = 1} ^ l ( {\mathbf {q}} \otimes {\mathbf {r}} ) _ { i } ^ { \downarrow }, \end{aligned}$$

which contradicts the assumption that ${\mathbf {p}} \otimes {\mathbf {r}} \prec {\mathbf {q}} \otimes {\mathbf {r}}$ and proves the second inequality in (6). $\square$

1.2 B. A remark on the relationship between Proposition 1 and Theorem 1 in [23]

In the proof of Theorem 1 in [23], it is mentioned that “$q_{m} r_{v^{\prime }} \geqslant q_{1} r_{v^{\prime }+1}$ implies that the first $\left( v^{\prime }-1\right) d+m$ elements of $({\mathbf {q}} \otimes {\mathbf {r}})^{\downarrow }$ consist of $q_{x} r_{y}$, $\forall x \in \{1,2, \ldots , d\}$ and $\forall y \in \left\{ 1,2, \ldots , v^{\prime }-1\right\}$, along with $q_{x^{\prime }} r_{v^{\prime }}$ $\forall x^{\prime } \in \{1,2, \ldots , m\}$.” Generally this could be not true because some elements of $q_{x} r_{y}$, $x \in \{1,2, \ldots , d\}$, $y \in \left\{ 1,2, \ldots , v^{\prime }-1\right\}$, along with $q_{x^{\prime }} r_{v^{\prime }}$ $\forall x^{\prime } \in \{1,2, \ldots , m\}$ might be less than some elements of $q_{x^{\prime }} r_{v^{\prime }}$, $x^{\prime } \in \{1,2, \ldots , d\}$, $y^{\prime } \in \{ v^{\prime }+1, \ldots , k\}$, along with $q_{x^{\prime }} r_{v^{\prime }}$ $\forall x^{\prime } \in \{m+1, \ldots , d\}$, when $q_{d} r_{v^{\prime }-1} < q_{m+1} r_{v^{\prime }}$ or $q_{d} r_{v^{\prime }-1} < q_{1} r_{v^{\prime }+1}$. This implies that the first $\left( v^{\prime }-1\right) d+m$ elements of $({\mathbf {q}} \otimes {\mathbf {r}})^{\downarrow }$ do not consist of $q_{x} r_{y}$, $\forall x \in \{1,2, \ldots , d\}$ and $\forall y \in \left\{ 1,2, \ldots , v^{\prime }-1\right\}$, along with $q_{x^{\prime }} r_{v^{\prime }}$ $\forall x^{\prime } \in \{1,2, \ldots , m\}$.

From Proposition 1, we have that the first $\left( s-1\right) d+l$ elements of $({\mathbf {q}} \otimes {\mathbf {r}})^{\downarrow }$ consist of $q_{x} r_{y}$, $\forall x \in \{1,2, \ldots , d\}$ and $\forall y \in \left\{ 1,2, \ldots , s-1\right\}$, along with $q_{x^{\prime }} r_{s}$ $\forall x^{\prime } \in \{1,2, \ldots , l\}$ when $q_{l} r_{s} \geqslant q_{1} r_{s+1}$, $q_{d} r_{s-1} \geqslant q_{l+1} r_{s}$ and $q_{d} r_{s-1} \geqslant q_{1} r_{s+1}$, which is true for $\forall l \in \mathcal {L}$. Hence, Proposition 1 fills such loopholes in Theorem 1 in [23] and also gives finer characterizations of ${\mathbf {r}}$ by increasing the number of constraints on ${\mathbf {r}}$.

1.3 C. Proof of corollary 2

Proof

By the definition of $M_{\mathcal{L}}$, we have:

$$\begin{aligned} \begin{aligned} \sum _{i=M_{\mathcal{L}}+1}^{d}p_i\leqslant \sum _{i=M_{\mathcal{L}}+1}^{d}q_i, \\ \sum _{i=M_{\mathcal{L}}+2}^{d}p_i>\sum _{i=M_{\mathcal{L}}+2}^{d}q_i. \end{aligned} \end{aligned}$$

This implies that $p_{M_{\mathcal{L}}+1}<q_{M_{\mathcal{L}}+1}$. Combining with $p_d\geqslant q_d$, we get

$$\begin{aligned} \frac{p_d}{p_{M_{\mathcal{L}}+1}}>\frac{q_d}{q_{M_{\mathcal{L}}+1}}. \end{aligned}$$

If $\frac{q_d}{q_{M_{\mathcal{L}}+1}}\geqslant \frac{r_k}{r_{k-1}}$, we have the following partial order the vector ${\mathbf {q}} \otimes {\mathbf {r}}$,

$$\begin{aligned} q_dr_k\leqslant q_{d-1}r_k\leqslant \dots \leqslant q_{M_{\mathcal{L}}+1}r_k\leqslant q_dr_{k-1}. \end{aligned}$$

(12)

For $\frac{p_d}{p_{M_{\mathcal{L}}+1}}>\frac{q_d}{q_{M_{\mathcal{L}}+1}}$, we have the following partial order the vector ${\mathbf {p}} \otimes {\mathbf {r}}$,

$$\begin{aligned} p_dr_k\leqslant p_{d-1}r_k\leqslant \dots \leqslant p_{M_{\mathcal{L}}+1}r_k\leqslant p_dr_{k-1} \end{aligned}$$

(13)

Combining (12) and (13), we have that the last $d-M_{\mathcal{L}}$ elements of $( {\mathbf {p}} \otimes {\mathbf {r}} ) ^ { \downarrow }$ and $( {\mathbf {q}} \otimes {\mathbf {r}} ) ^ { \downarrow }$ are $p_dr_k$, $p_{d-1}r_k$, $\dots$, $p_{M_{\mathcal{L}}+1}r_k$ and $q_dr_k$, $q_{d-1}r_k$, $\dots$, $q_{M_{\mathcal{L}}+1}r_k$, respectively. Therefore, we have

$$\begin{aligned}&\sum _{i=dk-d+M_{\mathcal{L}}+1}^{dk}( {\mathbf {p}} \otimes {\mathbf {r}} )_i ^ {\downarrow }=r_k\sum _{i=M_{\mathcal{L}}+1}^{d}p_i\\&\quad \leqslant r_k\sum _{i=M_{\mathcal{L}}+1}^{d}q_i=\sum _{i=dk-d+M_{\mathcal{L}}+1}^{dk}( {\mathbf {q}} \otimes {\mathbf {r}} )_i ^ {\downarrow }, \end{aligned}$$

which implies that

$$\begin{aligned} \sum _{i=1}^{dk-d+M_{\mathcal{L}}}( {\mathbf {p}} \otimes {\mathbf {r}} )_i ^ {\downarrow }> \sum _{i=1}^{dk-d+M_{\mathcal{L}}}( {\mathbf {p}} \otimes {\mathbf {r}} )_i ^ {\downarrow }, \end{aligned}$$

and contradicts the assumption that ${\mathbf {p}} \otimes {\mathbf {r}} \prec {\mathbf {q}} \otimes {\mathbf {r}}$. $\square$