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Families of Schmidt-number witnesses for high dimensional quantum states

  • Xian Shi
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  • College of Information Science and Technology, Beijing University of Chemical Technology, Beijing 100029, China

Received date: 2024-03-22

  Revised date: 2024-05-07

  Accepted date: 2024-05-09

  Online published: 2024-07-04

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© 2024 Institute of Theoretical Physics CAS, Chinese Physical Society and IOP Publishing

Abstract

Higher dimensional entangled states demonstrate significant advantages in quantum information processing tasks. The Schmidt number is a quantity of the entanglement dimension of a bipartite state. Here we build families of k-positive maps from the symmetric information complete positive operator-valued measurements and mutually unbiased bases, and we also present the Schmidt number witnesses, correspondingly. At last, based on the witnesses obtained from mutually unbiased bases, we show the distance between a bipartite state and the set of states with a Schmidt number less than k.

Cite this article

Xian Shi . Families of Schmidt-number witnesses for high dimensional quantum states[J]. Communications in Theoretical Physics, 2024 , 76(8) : 085103 . DOI: 10.1088/1572-9494/ad48fb

1. Introduction

Entanglement is one of the most fundamental features in quantum mechanics compared to classical physics [1, 2]. It also plays a critical role in quantum information and quantum computation theory, such as quantum cryptography [3], teleportation [4], and superdense coding [5].
Since the start of quantum information theory, a lot of effort has been devoted to the problems of distinguishing whether a state is separable or entangled [616]. A commonly used method to certify the entanglement of a state is to build an effective entanglement witness. There are many ways to construct the entanglement witness [17, 18]. In 2018, Chruscinski et al showed a method to construct entanglement witnesses from mutually unbiased bases (MUB) [19]. Whereafter, the authors in [2022] generalized the method to build entanglement witnesses with other classes of positive operator-valued measurements (POVM). Quantifying the entanglement of a bipartite state is the other important problem in entanglement theory [2327]. One of the most essential entanglement measures is the Schmidt number (SN) [28], this quantity indicates the lowest dimension of the system needed to generate the entanglement. Furthermore, genuine high dimensional entanglement plays an important role in many quantum information tasks, such as quantum communication [29], quantum control [30] and universal quantum computation [31, 32].
However, like most entanglement measures, it is hard to obtain the SN of a generic entangled state. Recently, the method to bound the SN of an entangled state has attracted much attention from relevant researchers [3338]. In [33], Bavaresco et al proposed a method to bound the dimension of an entangled state. Recently, Liu et al presented the results of the SN of a given state based on its covariance matrix [35, 37]. Tavakoli and Morelli showed the bound of the SN of a given state with the help of MUBs and symmetric information complete (SIC) POVMs [38]. Similar to entanglement, a straightforward method to certify the dimension of an entangled state is by constructing the k-positive maps [28] or SN witnesses [28, 34, 39]. However, there are few results obtained on constructing the SN witnesses with the help of certain POVMs.
In this manuscript, we will present the methods to construct k-positive maps with the use of SIC-POVMs and MUBs, which generalizes the methods of [19]. We also give the corresponding SN witness. Moreover, we compare the k-positive maps here with those in [28, 40]. Lastly, we present the lower bounds of the distance between a bipartite mixed state and the set of states with a Schmidt number less than k based on the SN witnesses, which are built from MUBs.

2. Preliminary knowledge

In this manuscript, the quantum systems we considered here are finite dimensions. Next, we denote ${ \mathcal D }\,({{ \mathcal H }}_{{AB}})$ as the set consisting of the states of ${{ \mathcal H }}_{{AB}},$
$\begin{eqnarray*}{ \mathcal D }\,({{ \mathcal H }}_{{AB}})=\{{\rho }_{{AB}}| {\rho }_{{AB}}\geqslant 0,\mathrm{tr}{\rho }_{{AB}}=1\}.\end{eqnarray*}$
And we denote $| {\psi }_{d}\rangle =\tfrac{1}{\sqrt{d}}{\sum }_{i=0}^{d-1}| {ii}\rangle $ as the maximally entangled states of ${{ \mathcal H }}_{{AB}}$ with $\mathrm{Dim}\,({{ \mathcal H }}_{A})=\mathrm{Dim}\,({{ \mathcal H }}_{B})=d.$
In this section, we will first recall the knowledge of the Schmidt number for a bipartite mixed state, then we will recall the definitions and properties of SICs and MUBs, correspondingly. Lastly, we will present the definition of the distance to the set of states with a Schmidt number less than k, Dk( · ).

2.1. Schmidt number

Assume ∣ψAB = ∑ijcijij⟩ is a pure state in ${{ \mathcal H }}_{{AB}}$ with $\mathrm{Dim}\,({{ \mathcal H }}_{A})={d}_{A}$ and $\mathrm{Dim}\,({{ \mathcal H }}_{B})={d}_{B}.$ There always exists orthonormal bases $\{| \tilde{i}{\rangle }_{A}\}$ and $\{| \tilde{i}{\rangle }_{B}\}$ in ${{ \mathcal H }}_{A}$ and ${{ \mathcal H }}_{B},$ respectively such that
$\begin{eqnarray*}| \psi {\rangle }_{{AB}}=\displaystyle \sum _{i=1}^{k}\,\sqrt{{\lambda }_{i}}| \tilde{i}\tilde{i}\rangle ,\end{eqnarray*}$
where λi > 0 and ${\sum }_{i}\,{\lambda }_{i}^{2}=1.$ Here the nonzero number k is called the Schmidt number of ∣ψ⟩ [28], i.e., $\mathrm{SN}\,(| \psi \rangle )=k.$ The Schmidt number of a mixed state ρAB is defined as follows [28];
$\begin{eqnarray}\mathrm{SN}\,(\rho )=\displaystyle \mathop{\min }\limits_{\rho ={\sum }_{i}\,{p}_{i}\,| {\psi }_{i}\rangle \,\langle {\psi }_{i}| }\;\mathop{\max }\limits_{i}\;\mathrm{SR}\,(| {\psi }_{i}\rangle ),\end{eqnarray}$
where the minimization takes over all of the decompositions of ρAB = ∑ipiψi⟩⟨ψi∣. The Schmidt number is an entanglement monotone, and it can be seen as a key quantity of the power of entanglement resources.
Through the definition of Schmidt number, one can classify the states of ${{ \mathcal H }}_{{AB}}$ as follows. Let
$\begin{eqnarray}{S}_{k}=\{\rho | \mathrm{SN}\,(\rho )\leqslant k\},\end{eqnarray}$
due to the definition of Sk, we have SkSk+1, and S1 is the set of separable states.
Similar to the method of entanglement detection, we recall the following tools; k-positive maps and SN witnesses, on detecting whether a state ρABSk. Assume a linear map Λk on ${ \mathcal B }\,({ \mathcal H })$ is k-positive, if Ik ⨂ Λk is a positive map on ${ \mathcal B }\,({{ \mathcal C }}^{k}\otimes { \mathcal H })$, here Ik is the identity map on ${ \mathcal B }\,({{ \mathcal C }}^{k}).$ Assume Wk is Hermite, if
$\begin{eqnarray*}\begin{array}{l}\mathrm{tr}\,({W}_{k}\,{\rho }_{k})\geqslant 0\quad \forall {\rho }_{k}\in {S}_{k},\\ \mathrm{tr}\,({W}_{k}\,\rho )\lt 0\quad \exists \rho \in { \mathcal D }\,({{ \mathcal H }}_{{AB}}),\end{array}\end{eqnarray*}$
then we call Wk an SN-(k+1) witness.
Lastly, we recall the following result on the Schmidt number of a bipartite state obtained in [28],

Assume ρ is a bipartite state on ${{ \mathcal H }}_{{AB}}.$ ρ has a Schmidt number of at least $k+1$ if, and only if, there exists a k-positive linear map ${{\rm{\Lambda }}}_{k}$ such that

$\begin{eqnarray}(I\otimes {{\rm{\Lambda }}}_{k})\,(\rho )\,\ngeqq 0.\end{eqnarray}$
The linear Hermiticity-preserving map Λ is k-positive if and only if
$\begin{eqnarray}(I\otimes {\rm{\Lambda }})\,(| {\psi }_{k}\rangle \,\langle {\psi }_{k}| )\geqslant 0,\end{eqnarray}$
where $| {\psi }_{k}\rangle $ are arbitrary maximally entangled states with a Schmidt number of $k.$ Finally, if Λ is k-positive, then ${{\rm{\Lambda }}}^{\dagger }$ defined by $\mathrm{tr}{A}^{\dagger }{\rm{\Lambda }}(B)=\mathrm{tr}{{\rm{\Lambda }}}^{\dagger }\,({A}^{\dagger })\,B$ for all A and B, is also k-positive.

In [28, 40], the authors showed a family of positive maps Λp(X) with the following form
$\begin{eqnarray}{{\rm{\Lambda }}}_{p}\,(X)=\mathrm{tr}\,(X)\,{\mathbb{I}}-{pX},\end{eqnarray}$
where X is a linear operator of ${ \mathcal H }$. When $\tfrac{1}{k+1}\lt p\leqslant \tfrac{1}{k}$, Λp is k-positive.

2.2. SICs and MUBs

Assume ${{ \mathcal H }}_{d}$ is a Hilbert space with dimension d, $\{{E}_{i}=\tfrac{1}{d}| {\phi }_{i}\rangle \,\langle {\phi }_{i}| | i=1,2,\cdots ,{d}^{2}\}$ is a POVM of ${{ \mathcal H }}_{d},$ here ∣φi⟩ are pure states with
$\begin{eqnarray}| \,\langle {\phi }_{j}| {\phi }_{k}\rangle \,{| }^{2}=\displaystyle \frac{1}{d+1},\hspace{4mm}\forall j\ne k,\end{eqnarray}$
then ${\{{E}_{i}\}}_{i=1}^{{d}^{2}}$ is an SIC-POVM. The existence of SIC-POVMs in arbitrary dimensions is still an open problem [41], readers who are interested in the problem can refer to [4244].
Assume ρ is a state of $({{ \mathcal H }}_{d}),$ {Eii = 1, 2, ⋯ ,d2} is an SIC-POVM, then
$\begin{eqnarray}\displaystyle \sum _{j=1}^{{d}^{2}}\,| \mathrm{tr}\,({E}_{j}\,\rho )\,{| }^{2}=\displaystyle \frac{\mathrm{tr}{\rho }^{2}\,+\,1}{d\,+\,{d}^{2}},\end{eqnarray}$
which is showed in [45].
Next we recall the definition of MUBs. Assume $\{| {e}_{i}^{l}\rangle \,| i=1,2,\cdots ,d\}\,{}_{l=1}^{L}$ are L orthonormal bases, and
$\begin{eqnarray*}\begin{array}{l}| \,\langle {e}_{j}^{m}| \;{f}_{k}^{{m}^{{\prime} }}\,\rangle \,{| }^{2}=\displaystyle \frac{1}{d}\hspace{5mm}\forall j,k\qquad m\ne {m}^{{\prime} },\\ | \,\langle {e}_{j}^{m}| \;{f}_{k}^{m}\,\rangle \,{| }^{2}={\delta }_{{jk}},\end{array}\end{eqnarray*}$
then they are MUBs. For any space with dimension d, there exists at most d + 1 MUBs. If the upper bound is reached, the set of MUBs is called a complete set. It is well known that the complete sets of MUBs exist when the dimension of the Hilbert space is a number with prime power. However, the existence of the complete sets of MUBs is unknown for arbitrary dimensional systems, even if the dimension is 6 [41].
Let $\{{Q}_{i}^{(\alpha )}=| {e}_{i}^{(\alpha )\,}\rangle \,\langle {e}_{i}^{(\alpha )\,}| | i=1,2,\cdots ,d\}\,{}_{\alpha =1}^{L}$ are L MUBs and ρ is a state in ${ \mathcal D }\,({{ \mathcal H }}_{d})$, then
$\begin{eqnarray}\displaystyle \sum _{\alpha =1}^{L}\,\displaystyle \sum _{i=1}^{d}\,| \mathrm{tr}\,(\rho {Q}_{i}^{(\alpha )})\,{| }^{2}\leqslant \mathrm{tr}\,({\rho }^{2})+\displaystyle \frac{L-1}{d},\end{eqnarray}$
the above inequality is obtained in [46].
Lastly, we present the distance to the set Sk for a bipartite state, Dk( · ). Assume ρAB is a bipartite mixed state, its distance to the set Sk in terms of the Frobenius norm is defined as
$\begin{eqnarray}{D}_{k}\,(\rho )=\mathop{\min }\limits_{\sigma \in {S}_{k}}\;\parallel \rho -\sigma {\parallel }_{F},\end{eqnarray}$
where the minimum takes over all the state in Sk.

3. k-positive maps based on SICs and MUBs

Assume ${{ \mathcal H }}_{d}$ is a Hilbert space with dimension d, ${ \mathcal M }={\left\{{P}_{i}=\tfrac{1}{d}| {\phi }_{i}\rangle \,\langle {\phi }_{i}| \}\,\right\}}_{i=1}^{{d}^{2}}$ and is an SIC-POVM. Next, we present a class of k-positive maps Λ( · ) with the help of the SIC-POVM ${ \mathcal M }$. Let ${ \mathcal O }$ be an orthogonal rotation in ${{\mathbb{R}}}^{d}$ around the axis ${n}_{* }=\tfrac{(1,1,\cdots ,1)}{\sqrt{d}}$, that is, ${ \mathcal O }{n}_{* }={n}_{* },$
$\begin{eqnarray}{\rm{\Lambda }}(X)=\displaystyle \frac{{{\mathbb{I}}}_{d}}{d}\mathrm{tr}\,(X)-h\displaystyle \sum _{g,l=1}^{{d}^{2}}\,{{ \mathcal O }}_{{gl}}\,\mathrm{tr}\,[(X-\displaystyle \frac{{{\mathbb{I}}}_{d}}{d}\mathrm{tr}\,(X))\,{P}_{l}]\,{P}_{g},\end{eqnarray}$
where $h=\sqrt{\tfrac{{d}^{4}+{d}^{3}}{({kd}-1)\,({kd}+k-2)}},$ Pl and Pk take over all the elements in an SIC-POVM ${ \mathcal M }.$

The map ${\rm{\Lambda }}(\cdot )$ defined in (10) is k-positive.

Here we place the proof of theorem 2 in the appendix.
Based on lemma 1, we can provide a class of witnesses on detecting whether a bipartite state is in S k through the k-positive map defined in (10),
$\begin{eqnarray}{W}_{k}=\displaystyle \frac{h\,+\,d}{{d}^{2}}{{\mathbb{I}}}_{d}\otimes {{\mathbb{I}}}_{d}-h\displaystyle \sum _{{gl}}\,{{ \mathcal O }}_{{gl}}\,\overline{{P}_{l}\,}\otimes {P}_{g},\end{eqnarray}$
where $h=\sqrt{\tfrac{{d}^{4}+{d}^{3}}{({kd}-1)\,({kd}+k-2)}}.$
Next, we present a class of k-positive maps Λ( · ) based on the MUBs. Let $\{| {e}_{i}^{\alpha }\rangle \,| \alpha =1,2,\cdots ,L\}\,{}_{i=1}^{d}$ be the MUBs, ${{ \mathcal O }}^{(\alpha )}$ be L orthogonal rotation in ${{\mathbb{R}}}^{d}$ around the axis ${n}_{* }=\tfrac{1}{d}\,(1,1,\cdots ,1),$ that is, ${{ \mathcal O }}^{(\alpha )\,}{n}_{* }={n}_{* }.$ Based on the MUBs, we also present a set of k-positive maps,
$\begin{eqnarray}\begin{array}{l}{{\rm{\Theta }}}_{k}\,(X)=\displaystyle \frac{{{\mathbb{I}}}_{d}}{d}\mathrm{tr}\,(X)-{h}_{s}\,\displaystyle \sum _{\alpha =1}^{L}\,\displaystyle \sum _{g,l=1}^{d}\,{{ \mathcal O }}_{{gl}}^{(\alpha )\,}\\ \quad \mathrm{tr}\,[(X-\displaystyle \frac{{{\mathbb{I}}}_{d}}{d}\mathrm{tr}\,(X))\,{Q}_{l}^{(\alpha )}]\,{Q}_{g}^{(\alpha )},\end{array}\end{eqnarray}$
where ${h}_{s}=\sqrt{\tfrac{1}{({dk}-1)\,({Lk}-L+d-1)}},$ ${Q}_{l}^{(\alpha )}=| {e}_{i}^{\alpha }\rangle \,\langle {e}_{i}^{\alpha }| $, $\{| {e}_{i}^{\alpha }\rangle \,| \alpha \,=1,2,\cdots ,L\}\,{}_{i=1}^{d}$ are L MUBs.

${{\rm{\Theta }}}_{k}\,(\cdot )$ defined in (12) are k-positive.

Here we place the proof of theorem 3 in the appendix.
Based on lemma 1, we can provide a class of witnesses on detecting whether a bipartite state is in Sk through the k-positive map defined in (12),
$\begin{eqnarray}{W}_{k}=\displaystyle \frac{1+{{Lh}}_{s}}{d}{{\mathbb{I}}}_{d}\otimes {{\mathbb{I}}}_{d}-{h}_{s}\,\displaystyle \sum _{g,l=1}^{d}\,\displaystyle \sum _{\alpha =1}^{L}\,{{ \mathcal O }}_{{gl}}^{(\alpha )\,}\overline{{Q}_{l}^{(\alpha )\,}}\otimes {Q}_{g}^{(\alpha )},\end{eqnarray}$
where ${h}_{s}=\sqrt{\tfrac{1}{({dk}-1)\,({Lk}-L+d-1)}}.$

When k = 1, ${h}_{s}=\tfrac{1}{d-1},$

$\begin{eqnarray}\begin{array}{l}{W}_{k}=\displaystyle \frac{d+L-1}{d\,(d-1)}{{\mathbb{I}}}_{d}\otimes {{\mathbb{I}}}_{d}\\ \quad -\displaystyle \frac{1}{d-1}\displaystyle \sum _{g,l=1}^{d}\,\displaystyle \sum _{\alpha =1}^{L}\,{{ \mathcal O }}_{{gl}}^{(\alpha )\,}\overline{{Q}_{l}^{(\alpha )\,}}\otimes {Q}_{g}^{(\alpha )},\end{array}\end{eqnarray}$
which is the entanglement witnesses shown in [19].

When $L=d+1,$ and ${ \mathcal O }={\mathbb{I}},$ the k-positive map (12) can be written as

$\begin{eqnarray}{{\rm{\Omega }}}_{k}\,(X)=\displaystyle \frac{{{\mathbb{I}}}_{d}}{d}\,(1+{h}_{s})\,\mathrm{tr}\,(X)-{h}_{s}\,X,\end{eqnarray}$
where ${h}_{s}=\sqrt{\tfrac{1}{({dk}-1)\,({kd}+k-2)}}$, and
$\begin{eqnarray*}\displaystyle \frac{1+{h}_{s}}{{{dh}}_{s}}=\displaystyle \frac{\tfrac{1}{{h}_{s}}+1}{d}=\displaystyle \frac{\sqrt{({dk}-1)\,({kd}+k-2)}+1}{d},\end{eqnarray*}$
as $\tfrac{\sqrt{({dk}-1)\,({kd}+k-2)}+1}{d}\in [k,k+1),$ we have this class of k-positive maps constructed from complete sets of MUBs belonging to the family of (7) in [28].

Next, we investigate the witness (13) for the following states.

Assume

$\begin{eqnarray*}{\rho }_{v}=v| {\psi }_{d}\rangle \,\langle {\psi }_{d}| +\displaystyle \frac{1-v}{{d}^{2}}{{\mathbb{I}}}_{{d}^{2}},\end{eqnarray*}$
where $| {\psi }_{d}\rangle =\tfrac{1}{\sqrt{d}}{\sum }_{i}| {ii}\rangle $ is the maximally entangled state and $v\in (0,1]$.

By the witness (13), we present a sufficient condition of ${SN}\,({\rho }_{v})\gt k$:

$\begin{eqnarray*}v\gt \displaystyle \frac{\sqrt{({dk}-1)\,({Lk}-L+d-1)}+L-1}{{dL}-1}.\end{eqnarray*}$
Here we place the computation in the appendix.

4. Applications

In this section, we will present the distance between the state and the set Sk based on the witness obtained in the last section, the method here is based on [47].
Assume ρ is a bipartite mixed state, Yk is a witness of the states with Schmidt number k of ${{ \mathcal H }}_{d}\otimes {{ \mathcal H }}_{d}$, let $a=\tfrac{\mathrm{tr}\,({Y}_{k})}{{d}^{2}}$, $b=\sqrt{\mathrm{tr}\,({Y}_{k}^{\dagger }\,{Y}_{k})-\tfrac{{\left(\mathrm{tr}{Y}_{k}\,\right)}^{2}}{{d}^{2}}},$ ${V}_{k}=\tfrac{{Y}_{k}-a{\mathbb{I}}\otimes {\mathbb{I}}}{b}$,
$\begin{eqnarray}\begin{array}{rcl}{D}_{k}\,({\rho }_{{AB}}) & = & \mathop{\min }\limits_{\sigma \in {S}_{k}}\;\mathop{\max }\limits_{\parallel W{\parallel }_{F}=1}\;\mathrm{tr}\,(W\,({\rho }_{{AB}}-{\sigma }_{{AB}}))\\ & \geqslant & | \mathrm{tr}\,[{V}_{k}\,(\rho -\omega )]\,| \\ & = & | \mathrm{tr}\,[\displaystyle \frac{{Y}_{k}}{b}\,(\rho -\omega )-\displaystyle \frac{a}{b}\,(\rho -\omega )]\,| \\ & = & | \mathrm{tr}\,[\displaystyle \frac{{Y}_{k}}{b}\,(\rho -\omega )]\,| \\ & \geqslant & -\displaystyle \frac{1}{b}\mathrm{tr}{Y}_{k}\,\rho ,\end{array}\end{eqnarray}$
in the first inequality, ω is the optimal state in Sk, Vk is a Hermite operator and ∥VkF = 1. The last inequality is due to ωSk, $\mathrm{tr}{Y}_{k}\,\omega \geqslant 0.$
Next, we apply the witnesses obtained in (13) to show the distance between a bipartite state and the set Sk relying on (16).

Assume ${\rho }_{{AB}}$ is a bipartite mixed state on ${{ \mathcal H }}_{d}\otimes {{ \mathcal H }}_{d}$, let $\{| {e}_{i}^{\alpha }\rangle \,| \alpha =1,2,\cdots ,m\}\,{}_{i=1}^{L}$ be the MUBs of the system ${{ \mathcal H }}_{d},$ then

$\begin{eqnarray}{D}_{k}\,(\rho )\geqslant -\displaystyle \frac{1}{\sqrt{{h}_{s}^{2}\,({Ld}-L)}}\mathrm{tr}{W}_{k}\,\rho ,\end{eqnarray}$
where ${h}_{s}=\sqrt{\tfrac{1}{({dk}-1)\,({Lk}-L+d-1)}},$ Wk is defined in (13).

Due to (16), to obtain the lower bound of ${D}_{k}\,(\rho )$, we only need to compute the value of b. Let ${Z}_{k}={\sum }_{g,l=1}^{d}\,{\sum }_{\alpha =1}^{L}\,{{ \mathcal O }}_{{gl}}^{(\alpha )\,}\overline{{Q}_{l}^{(\alpha )\,}}\otimes {Q}_{g}^{(\alpha )},$

$\begin{eqnarray}\begin{array}{l}\mathrm{tr}{Z}_{k}\\ =\displaystyle \sum _{g,l=1}^{d}\,\displaystyle \sum _{\alpha =1}^{L}\,{{ \mathcal O }}_{{gl}}^{(\alpha )\,}\mathrm{tr}\overline{{Q}_{l}^{(\alpha )\,}}\otimes {Q}_{g}^{(\alpha )},\\ =\displaystyle \sum _{g,l=1}^{d}\,\displaystyle \sum _{\alpha =1}^{L}\,{{ \mathcal O }}_{{gl}}^{(\alpha )}\\ ={Ld},\end{array}\end{eqnarray}$
$\begin{eqnarray}\begin{array}{l}\mathrm{tr}{Z}_{k}^{\dagger }\,{Z}_{k}\\ \quad =\displaystyle \sum _{g,l,m,n\,=\,1}^{d}\,\displaystyle \sum _{\alpha ,\beta \,=\,1}^{L}\,{{ \mathcal O }}_{{gl}}^{(\alpha )\,}{{ \mathcal O }}_{{mn}}^{(\beta )\,}\mathrm{tr}\overline{{Q}_{l}^{(\alpha )\,}}\overline{{Q}_{n}^{(\beta )\,}}\otimes {Q}_{g}^{(\alpha )\,}{Q}_{m}^{(\beta )}\\ \quad =\,{Ld}+{L}^{2}-L,\end{array}\end{eqnarray}$
then
$\begin{eqnarray}\begin{array}{rcl}b & = & \sqrt{\mathrm{tr}\,({W}_{k}^{\dagger }\,{W}_{k})-\displaystyle \frac{{\left(\mathrm{tr}{W}_{k}\,\right)}^{2}}{{d}^{2}}}\\ & = & \sqrt{{h}_{s}^{2}\,({Ld}-L)},\end{array}\end{eqnarray}$
hence
$\begin{eqnarray}{D}_{k}\,({\rho }_{{AB}})\geqslant -\displaystyle \frac{1}{\sqrt{{h}_{s}^{2}\,({Ld}-L)}}\mathrm{tr}{W}_{k}\,\rho .\end{eqnarray}$

Assume ${\rho }_{v}$ is a state mixed with the maximally entangled state with the white noise,

$\begin{eqnarray}{\rho }_{v}=v| {\psi }_{d}\rangle \,\langle {\psi }_{d}| +\displaystyle \frac{1-v}{{d}^{2}}{{\mathbb{I}}}_{{d}^{2}},\end{eqnarray}$
where $| {\psi }_{d}\rangle =\tfrac{1}{\sqrt{d}}{\sum }_{i}| {ii}\rangle $ is the maximally entangled state and $v\in (0,1].$

Let $\{| {e}_{i}^{\alpha }\rangle \,| \alpha =1,2,\cdots ,L\}\,{}_{i=1}^{d}$ be the MUBs,

$\begin{eqnarray}\begin{array}{l}{\tilde{W}}_{k}=\displaystyle \frac{d+L-1}{d\,(d-1)}{{\mathbb{I}}}_{d}\otimes {{\mathbb{I}}}_{d}\\ \quad -\displaystyle \frac{1}{d-1}\displaystyle \sum _{g,l=1}^{d}\,\displaystyle \sum _{\alpha =1}^{L}\,{{ \mathcal O }}_{{gl}}^{(\alpha )\,}\overline{{Q}_{l}^{(\alpha )\,}}\otimes {Q}_{g}^{(\alpha )},\end{array}\end{eqnarray}$
where ${Q}_{l}^{(\alpha )}=| {e}_{l}^{\alpha }\rangle \,\langle {e}_{i}^{\alpha }| ,$ ${{ \mathcal O }}_{{gl}}^{(\alpha )}={\delta }_{{gl}},$ $\alpha =1,2,\cdots ,L$. In Example 1, we showed
$\begin{eqnarray}\mathrm{tr}\,({\tilde{W}}_{k}\,{\rho }_{v})=\alpha -\displaystyle \frac{{h}_{s}}{d}+\displaystyle \frac{{h}_{s}}{d}v-{{vh}}_{s}\,L,\end{eqnarray}$
where $\alpha =\tfrac{1+{{Lh}}_{s}}{d},$ then by the theorem 6, we have
$\begin{eqnarray}{D}_{k}\,(\rho )\geqslant -\displaystyle \frac{{h}_{c}+L-1\,+\,v-{dvL}}{d\sqrt{{Ld}-L}},\end{eqnarray}$
where ${h}_{c}=\sqrt{({dk}-1)\,({Lk}-L+d-1)}.$

In fact, the method to compute the lower bound of the distance between a given state and the set Sk in (16) is universal, it can be used in the scenarios of generic valid SN witnesses.

5. Conclusion

Here we have presented families of k-positive maps in arbitrary dimensional systems based on the SIC-POVMs and MUBs. By using the k-positive maps, we also built the SN witnesses correspondingly. Then we compared the k-positive maps built from MUBs and the existing results. When k = 1, we found that the conclusion degrades into the map obtained in [19]. Lastly, we presented a definition of the distance between a bipartite state and the set Sk. Moreover, we showed a lower bound of the distance based on the SN witnesses constructed from MUBs. Due to the important roles that higher dimensional systems played, our results can provide a reference for future work on the study of entanglement theory.

Acknowledgments

XS was supported by the National Natural Science Foundation of China (Grant No. 12301580) and the Funds of the College of Information Science and Technology, Beijing University of Chemical Technology (Grant No. 0104/11170044115).

Appendix

Theorem 2: The map Λ( · ) defined in (10) is k-positive.

Due to the lemma 1, when we prove

$\begin{eqnarray}(I\otimes {\rm{\Lambda }})\,(| {\psi }_{k}\rangle \,\langle {\psi }_{k}| )\geqslant 0,\end{eqnarray}$
for all maximally entangled states $| {\psi }_{k}\rangle $ with a Schmidt number of k, then we finish the proof. Next, we denote B as a maximal ball centered at the maximally mixed state $\eta =\tfrac{1}{d}{\mathbb{I}}$ inscribed in ${ \mathcal D }\,({{ \mathcal H }}_{d})$ [48], then $\rho \in B\subset { \mathcal D }\,({{ \mathcal H }}_{d})$ if and only if
$\begin{eqnarray*}\mathrm{tr}{\rho }^{2}\leqslant \displaystyle \frac{1}{d-1}.\end{eqnarray*}$
Here we will utilize the above fact to prove (A1).

Assume $| {\psi }_{k}\rangle =(U\otimes V)\,{\sum }_{i=0}^{k-1}\,\sqrt{\tfrac{1}{k}}| {ii}\rangle $, here U and V are arbitrary unitary operators of ${{ \mathcal H }}_{A}$ and ${{ \mathcal H }}_{B}$, respectively, then

$\begin{eqnarray}\begin{array}{l}\mathrm{tr}{\left[\,(I\otimes {\rm{\Lambda }})\,(U\otimes V)\,(| {\psi }_{k}\rangle \,\langle {\psi }_{k}| )\,({U}^{\dagger }\otimes {V}^{\dagger })\right]}^{2}\\ \quad =\mathrm{tr}\displaystyle \frac{1}{{k}^{2}}{\,\left[\left(I\otimes {\rm{\Lambda }})\,(U\otimes V)\,(\displaystyle \sum _{i=0}^{k-1}\,\displaystyle \sum _{j=0}^{k-1}\,| {ii}\rangle \,\langle {jj}| )\,({U}^{\dagger }\otimes {V}^{\dagger }\right)\,\right]}^{2}\\ \quad =\mathrm{tr}\displaystyle \frac{1}{{k}^{2}}{\,\left[\displaystyle \sum _{i,j=0}^{k-1}\,U| i\rangle \,\langle \;j\,| {U}^{\dagger }\otimes {\rm{\Lambda }}(V| i\rangle \,\langle \;j\,| {V}^{\dagger })\,\right]}^{2}\\ \quad =\mathrm{tr}\displaystyle \frac{1}{{k}^{2}}\,[\displaystyle \sum _{i,j,m=0}^{k-1}\,U| i\rangle \,\langle m| {U}^{\dagger }\otimes {\rm{\Lambda }}(V| i\rangle \,\langle \;j\,| {V}^{\dagger }){\rm{\Lambda }}(V| \;j\,\rangle \,\langle m| {V}^{\dagger })]\\ \quad =\displaystyle \frac{1}{{k}^{2}}\mathrm{tr}\displaystyle \sum _{i,j=0}^{k-1}{\rm{\Lambda }}(V| i\rangle \,\langle \;j\,| {V}^{\dagger }){\rm{\Lambda }}(V| \;j\,\rangle \,\langle i| {V}^{\dagger }).\end{array}\end{eqnarray}$
Next, we compute ${\rm{\Lambda }}(| i\rangle \,\langle \;j\,| ),$ when i = j,
$\begin{eqnarray}{\rm{\Lambda }}(V| i\rangle \,\langle i| {V}^{\dagger })=\displaystyle \frac{{{\mathbb{I}}}_{d}}{d}-h\displaystyle \sum _{g,l=1}^{{d}^{2}}\,{{ \mathcal O }}_{{gl}}\,\mathrm{tr}\,[(V| i\rangle \,\langle i| {V}^{\dagger }-\displaystyle \frac{{{\mathbb{I}}}_{d}}{d})\,{P}_{l}]\,{P}_{g},\end{eqnarray}$
when $i\ne j$,
$\begin{eqnarray}{\rm{\Lambda }}(V| i\rangle \,\langle \;j\,| {V}^{\dagger })=-h\displaystyle \sum _{g,l=1}^{{d}^{2}}\,{{ \mathcal O }}_{{gl}}\,\mathrm{tr}\,[(V| i\rangle \,\langle \;j\,| {V}^{\dagger })\,{P}_{l}]\,{P}_{g}.\end{eqnarray}$
$\begin{eqnarray}\begin{array}{l}(A2)=\displaystyle \frac{1}{{kd}}+\displaystyle \frac{{h}^{2}}{{k}^{2}}\displaystyle \sum _{i=0}^{k-1}\displaystyle \sum _{g,l,s,t\,=\,1}^{{d}^{2}}{{ \mathcal O }}_{{gl}}{{ \mathcal O }}_{{ts}}\mathrm{tr}[(V| i\rangle \langle i| {V}^{\dagger }-\displaystyle \frac{{{\mathbb{I}}}_{d}}{d}){P}_{l}]\mathrm{tr}[(V| i\rangle \langle i| {V}^{\dagger }-\displaystyle \frac{{{\mathbb{I}}}_{d}}{d}){P}_{s}]\mathrm{tr}({P}_{g}{P}_{t})\\ \quad -\displaystyle \frac{2h}{{k}^{2}}\displaystyle \sum _{i=0}^{k-1}\mathrm{tr}\displaystyle \sum _{g,l=1}^{{d}^{2}}{{ \mathcal O }}_{{gl}}\mathrm{tr}[(V| i\rangle \langle i| {V}^{\dagger }-\displaystyle \frac{{{\mathbb{I}}}_{d}}{d}){P}_{l}]\mathrm{tr}({P}_{g})\\ \quad +\displaystyle \frac{{h}^{2}}{{k}^{2}}\displaystyle \sum _{i\ne j}\displaystyle \sum _{g,l,s,t\,=\,1}^{{d}^{2}}{{ \mathcal O }}_{{gl}}{{ \mathcal O }}_{{st}}\mathrm{tr}[V| i\rangle \langle j| {V}^{\dagger }{P}_{l}]\mathrm{tr}[V| j\rangle \langle i| {V}^{\dagger }{P}_{t}]\mathrm{tr}({P}_{g}{P}_{s})\\ \quad =\displaystyle \frac{1}{{kd}}+\displaystyle \frac{{h}^{2}}{{d}^{2}{k}^{2}}\displaystyle \sum _{i=0}^{k-1}\displaystyle \sum _{l\,=\,1}^{{d}^{2}}| \mathrm{tr}[(V| i\rangle \langle i| {V}^{\dagger }-\displaystyle \frac{{{\mathbb{I}}}_{d}}{d}){P}_{l}]{| }^{2}\\ \quad +\displaystyle \frac{{h}^{2}}{{k}^{2}{d}^{2}}\displaystyle \sum _{i\ne j}\displaystyle \sum _{l\,=\,1}^{{d}^{2}}\mathrm{tr}({P}_{l}V| i\rangle \langle j| {V}^{\dagger })\mathrm{tr}({P}_{l}V| j\rangle \langle i| {V}^{\dagger })\\ \quad \leqslant \displaystyle \frac{1}{{kd}}+\displaystyle \frac{{h}^{2}(d-1)}{{d}^{4}(d+1)k}+\displaystyle \frac{{h}^{2}(k-1)}{{d}^{4}k}\\ \quad =\displaystyle \frac{{d}^{4}\,+\,{d}^{3}\,+\,{h}^{2}{kd}+{{kh}}^{2}-2{h}^{2}}{{{kd}}^{4}(d+1)}\\ \quad =\displaystyle \frac{1}{{dk}-1},\end{array}\end{eqnarray}$
in the last inequality, we apply lemma 8 in the appendix and the following derived in [45]
$\begin{eqnarray}\displaystyle \sum _{j=1}^{{d}^{2}}\,| \mathrm{tr}\,({P}_{j}\,\rho )\,{| }^{2}=\displaystyle \frac{\mathrm{tr}{\rho }^{2}+1}{d+{d}^{2}},\end{eqnarray}$
hence we have ${\rm{\Psi }}\,(\cdot )$ is k-positive.□

Theorem 3: Θk( · ) defined in (12) are k-positive.

Here we apply a similar method of the proof of theorem 2. Assume $| {\psi }_{k}\rangle =(U\otimes V)\,{\sum }_{i=0}^{k-1}\,\sqrt{\tfrac{1}{k}}| {ii}\rangle $, where U and V are arbitrary unitary operators of ${{ \mathcal H }}_{A}$ and ${{ \mathcal H }}_{B}$, respectively, if $\mathrm{tr}{\left[\,({I}_{k}\otimes {{\rm{\Theta }}}_{k})\,(| {\psi }_{k}\rangle \,\langle {\psi }_{k}| )\right]}^{2}\leqslant \tfrac{1}{{dk}-1}$, then we finish the proof [48].

$\begin{eqnarray}\begin{array}{l}\mathrm{tr}{\left[\,(I\otimes {{\rm{\Theta }}}_{k})\,(U\otimes V)\,(| {\psi }_{k}\rangle \,\langle {\psi }_{k}| )\,({U}^{\dagger }\otimes {V}^{\dagger })\right]}^{2}\\ \quad =\displaystyle \frac{1}{{k}^{2}}\mathrm{tr}\displaystyle \sum _{i,j=0}^{k-1}\,{{\rm{\Theta }}}_{k}\,(V| i\rangle \,\langle \;j\,| {V}^{\dagger })\,{{\rm{\Theta }}}_{k}\,(V| \;j\,\rangle \,\langle i| {V}^{\dagger }).\end{array}\end{eqnarray}$
Next we compute ${{\rm{\Theta }}}_{k}\,(V| i\rangle \,\langle \;j\,| {V}^{\dagger }),$ when $i=j,$
$\begin{eqnarray}\begin{array}{l}{{\rm{\Theta }}}_{k}\,(V| i\rangle \,\langle i| {V}^{\dagger })\\ \quad =\displaystyle \frac{{{\mathbb{I}}}_{d}}{d}-{h}_{s}\,\displaystyle \sum _{\alpha =1}^{L}\,\displaystyle \sum _{g,l=1}^{d}\,{{ \mathcal O }}_{{gl}}^{(\alpha )\,}\mathrm{tr}\,[(V| i\rangle \,\langle i| {V}^{\dagger }-\displaystyle \frac{{{\mathbb{I}}}_{d}}{d})\,{Q}_{l}^{(\alpha )}]\,{Q}_{g}^{(\alpha )},\end{array}\end{eqnarray}$
when $i\ne j$,
$\begin{eqnarray}\begin{array}{l}{{\rm{\Theta }}}_{k}\,(V| i\rangle \,\langle \;j\,| {V}^{\dagger })\\ \quad =-{h}_{s}\,\displaystyle \sum _{\alpha =1}^{L}\,\displaystyle \sum _{g,l=1}^{d}\,{{ \mathcal O }}_{{gl}}^{(\alpha )\,}\mathrm{tr}\,[(V| i\rangle \,\langle \;j\,| {V}^{\dagger })\,{Q}_{l}^{(\alpha )}]\,{Q}_{g}^{(\alpha )},\end{array}\end{eqnarray}$
then based on (A8) and (A9), we have
$\begin{eqnarray*}\begin{array}{l}(A7)\leqslant \displaystyle \frac{1}{{k}^{2}}\,\left[\displaystyle \frac{k}{d}+{h}_{s}^{2}\,\displaystyle \sum _{i=1}^{k}\,\displaystyle \sum _{\alpha =1}^{L}\,\displaystyle \sum _{n=1}^{d}\,| \mathrm{tr}\,\left(V| i\rangle \,\langle i| {V}^{\dagger }-\displaystyle \frac{{{\mathbb{I}}}_{d}}{d}\left.,\Space{0ex}{1.24em}{0ex}\right)\right.\right.\\ \,\times \,{Q}_{n}^{(\alpha )\,}{| }^{2}+{h}_{s}^{2}\,\displaystyle \frac{{Lk}\,(k-1)}{d}\Space{0ex}{3.0ex}{0ex}]\\ \quad \leqslant \displaystyle \frac{1}{{k}^{2}}\,\Space{0ex}{1.0ex}{0ex}[\displaystyle \frac{k}{d}+{{kh}}_{s}^{2}\,\Space{0ex}{1.0ex}{0ex}(1-\displaystyle \frac{1}{d}\Space{0ex}{1.0ex}{0ex})+{h}_{s}^{2}\,\displaystyle \frac{{Lk}\,(k-1)}{d}\left.=\displaystyle \frac{1}{{dk}-1}\Space{0ex}{1.55em}{0ex}\right].\end{array}\end{eqnarray*}$
In the inequalities, we have used lemma 9 in the appendix the following derived in [46]
$\begin{eqnarray}\displaystyle \sum _{\alpha =1}^{L}\,\displaystyle \sum _{l\,=\,1}^{d}\,| \mathrm{tr}\,(\rho {Q}_{l}^{(\alpha )})\,{| }^{2}\leqslant \mathrm{tr}\,({\rho }^{2})+\displaystyle \frac{L-1}{d}.\end{eqnarray}$

Let ${ \mathcal M }=\{{M}_{j}=\tfrac{1}{d}| {\phi }_{j}\rangle \,\langle {\phi }_{j}| \}$ be an SIC-POVM in a d dimensional system ${ \mathcal H }$, and $\{| i\rangle \,| i=1,2,\cdots ,d\}$ is any orthonormal base of ${ \mathcal H }$, when $i\ne j,$

$\begin{eqnarray*}\displaystyle \sum _{j=1}^{{d}^{2}}\,{\mathrm{tr}}^{2}\,({M}_{j}| i\rangle \,\langle \;j\,| )=\displaystyle \frac{1}{{d}^{2}}.\end{eqnarray*}$

Here we apply a similar method of the proof of proposition 5 in [45]. Let

$\begin{eqnarray}| {\rm{\Psi }}\rangle =\displaystyle \frac{1}{{d}^{3/2}}\displaystyle \sum _{j=1}^{{d}^{2}}\,| {\phi }_{j}\rangle \,| {\phi }_{j}^{* }\rangle ,\end{eqnarray}$
$\begin{eqnarray}| {{\rm{\Phi }}}_{k}\rangle =\displaystyle \frac{\sqrt{d\,+\,1}}{{d}^{3/2}}\displaystyle \sum _{j=1}^{{d}^{2}}\,{\omega }^{k\,(\,j-1)\,}| {\phi }_{j}\rangle \,| {\phi }_{j}^{* }\rangle ,\end{eqnarray}$
where $k=1,2,\cdots ,{d}^{2}-1,$ and $| {\phi }^{* }\rangle $ is a vector such that its components are conjugate to the corresponding components of $| \phi \rangle $. The d2 vectors (A11), (A12) construct an orthonormal basis of the space ${ \mathcal H }\otimes { \mathcal H },$ hence,
$\begin{eqnarray}| i\rangle \,\langle \;j\,| \otimes {\mathbb{I}}| {\rm{\Psi }}\rangle =\displaystyle \sum _{k\,=\,1}^{{d}^{2}-1}\,{a}_{k}| {{\rm{\Phi }}}_{k}\rangle ,\end{eqnarray}$
then
$\begin{eqnarray}\begin{array}{rcl}{a}_{k} & = & \langle {{\rm{\Phi }}}_{k}| | i\rangle \,\langle \;j\,| \otimes {\mathbb{I}}| {\rm{\Psi }}\rangle \\ & = & \displaystyle \frac{\sqrt{d\,+\,1}}{{d}^{3}}\displaystyle \sum _{i,j\,=\,1}^{{d}^{2}}\,{\omega }^{-q\,(i-1)\,}\langle {\phi }_{i}| | i\rangle \,\langle \;j\,| | {\phi }_{j}\rangle \,\langle {\phi }_{j}| | {\phi }_{i}\rangle \\ & = & \displaystyle \frac{\sqrt{d\,+\,1}}{d}\displaystyle \sum _{i=1}^{{d}^{2}}\,{\omega }^{-q\,(i-1)\,}{p}_{i},\end{array}\end{eqnarray}$
where ${p}_{i}=\tfrac{\langle {\phi }_{i}| | i\rangle \,\langle \;j\,| | {\phi }_{i}\rangle }{d}.$ Next,
$\begin{eqnarray}\langle {\rm{\Psi }}| {\left(\,| i\rangle \,\langle \;j\,| \otimes {\mathbb{I}}\right)}^{2}| {\rm{\Psi }}\rangle =\displaystyle \frac{1}{d}\mathrm{tr}\,(| i\rangle \,\langle \;j\,| i\rangle \,\langle \;j\,| )=0.\end{eqnarray}$
Through (A13),
$\begin{eqnarray}\displaystyle \sum _{k\,=\,1}^{{d}^{2}-1}\,{a}_{k}^{* }{a}_{k}=(d+1)\,\displaystyle \sum _{j=1}^{{d}^{2}}\,{p}_{j}^{2}-\displaystyle \frac{d\,+\,1}{{d}^{2}},\end{eqnarray}$
that is,
$\begin{eqnarray*}\displaystyle \sum _{j=1}^{{d}^{2}}\,{p}_{j}^{2}=\displaystyle \frac{1}{{d}^{2}}.\end{eqnarray*}$

Let $\{| {e}_{i}^{\alpha }\rangle \,| i=1,2,\cdots ,m\}\,{}_{\alpha =1}^{L}$ be the MUBs in a d dimensional system ${ \mathcal H }$, and ${{ \mathcal N }}^{(\alpha )}=\{{Q}_{i}^{(\alpha )}=| {e}_{i}^{\alpha }\rangle \,\langle {e}_{i}^{\alpha }| | i\,=1,2,\cdots ,m\}\,{}_{\alpha =1}^{L}$. Assume $\{| i\rangle \,| i=1,2,\cdots ,d\}$ is any orthonormal base of ${ \mathcal H }$, when $i\ne j,$ let $\rho =| i\rangle \,\langle \;j\,| ,$

$\begin{eqnarray*}\displaystyle \sum _{\alpha }^{L}\,\displaystyle \sum _{i=1}^{{d}^{2}}\,{\mathrm{tr}}^{2}\,({Q}_{i}^{(\alpha )\,}\rho )\leqslant \displaystyle \frac{L}{d}.\end{eqnarray*}$

Let

$\begin{eqnarray}| \psi \rangle =\displaystyle \frac{1}{d}\displaystyle \sum _{i=1}^{d}\,| {e}_{i}^{l}\rangle \,| {e}_{i}^{l}{\rangle }^{* },\end{eqnarray}$
$\begin{eqnarray}| {\phi }_{\alpha ,k}\rangle =\displaystyle \frac{1}{\sqrt{d}}\displaystyle \sum _{i=1}^{d}\,{\omega }^{k\,(i-1)\,}| {e}_{i}^{\alpha }\rangle \,| {e}_{i}^{\alpha }{\rangle }^{* }.\end{eqnarray}$
Here l is a given constant, $\omega ={{\rm{e}}}^{2\pi {\rm{i}}/d}$, $k=1,2,\cdots ,d-1$, and $\alpha =1,2,\cdots ,L$. The above $L\,(d-1)+1$ states are orthogonal to each other, so we can assume $\{| \psi \rangle ,| {\phi }_{\alpha ,k}\rangle $, $| {\gamma }_{x}\rangle \,| $ $k=1,2,\cdots ,\,d\,-\,1,\alpha \,=\,1,2,\cdots ,L,$ $m=1,2,\cdots ,\,{d}^{2}-L\,(d-1)-1\}$ is an orthonormal basis of the system ${{ \mathcal H }}_{d}\otimes {{ \mathcal H }}_{d}.$

Next assume

$\begin{eqnarray}\rho \otimes {\mathbb{I}}| \psi \rangle =\displaystyle \sum _{\alpha =1}^{L}\,\displaystyle \sum _{k\,=\,1}^{d-1}\,{\rho }_{\alpha ,k}| {\phi }_{\alpha ,k}\rangle +\displaystyle \sum _{x\,=\,1}^{{d}^{2}-L\,(d-1)-1}\,{c}_{x}| {\gamma }_{x}\rangle ,\end{eqnarray}$
where ${\rho }_{\alpha ,k}=\tfrac{1}{d}{\sum }_{i=1}^{d}\,{\omega }^{-k\,(i-1)\,}\langle {e}_{i}^{\alpha }| \rho | {e}_{i}^{\alpha }\rangle ,$ then
$\begin{eqnarray*}\begin{array}{l}\displaystyle \sum _{\alpha =1}^{L}\,\displaystyle \sum _{k=1}^{d-1}\,| {\rho }_{\alpha ,k}{| }^{2}=\displaystyle \frac{1}{{d}^{2}}\displaystyle \sum _{\alpha =1}^{L}\,\displaystyle \sum _{k=1}^{d-1}\,\displaystyle \sum _{i,j=1}^{d}\,{\omega }^{-k\,(\,j-i)\,}\langle {e}_{i}^{\alpha }| \rho | {e}_{i}^{\alpha }\rangle \,\langle {e}_{j}^{\alpha }| \rho | {e}_{j}^{\alpha }\rangle \\ \quad =\displaystyle \frac{1}{d}\displaystyle \sum _{\alpha =1}^{L}\,\displaystyle \sum _{i,j=1}^{d}\,\,(d{\delta }_{{ij}}-1)\,\langle {e}_{i}^{\alpha }| \rho | {e}_{i}^{\alpha }\rangle \,\langle {e}_{j}^{\alpha }| \rho | {e}_{j}^{\alpha }\rangle \\ \quad =\displaystyle \frac{1}{d}\displaystyle \sum _{\alpha =1}^{L}\,\displaystyle \sum _{i=1}^{d}\,{\mathrm{tr}}^{2}\,({Q}_{i}^{(\alpha )\,}\rho )-\displaystyle \frac{L}{{d}^{2}}.\end{array}\end{eqnarray*}$
Hence,
$\begin{eqnarray*}\begin{array}{l}\langle \phi | \rho \otimes {\mathbb{I}}| \phi \rangle =0\geqslant \displaystyle \sum _{\alpha =1}^{L}\,\displaystyle \sum _{k=1}^{d-1}\,| {\rho }_{\alpha ,k}{| }^{2}\\ \Space{0ex}{0.74em}{0ex}\quad =\displaystyle \frac{1}{d}\displaystyle \sum _{\alpha =1}^{L}\,\displaystyle \sum _{i=1}^{d}\,{\mathrm{tr}}^{2}\,({Q}_{i}^{(\alpha )\,}\rho )-\displaystyle \frac{L}{{d}^{2}},\end{array}\end{eqnarray*}$
that is,
$\begin{eqnarray*}{\mathrm{tr}}^{2}\,({Q}_{i}^{(\alpha )\,}\rho )\leqslant \displaystyle \frac{L}{d}.\end{eqnarray*}$

Example 1: Here we first denote $\alpha =\tfrac{1+{{Lh}}_{s}}{d},$ $\Psi$d = ∣ψd⟩⟨ψd∣, then
$\begin{eqnarray}\begin{array}{l}\mathrm{tr}{W}_{k}\,{\rho }_{v}\\ \quad =[(\alpha {\mathbb{I}}-{h}_{s}\,\displaystyle \sum _{g,l=1}^{d}\,\displaystyle \sum _{\alpha =1}^{L}\,{{ \mathcal O }}_{{gl}}^{(\alpha )\,}\overline{{Q}_{l}^{(\alpha )\,}}\otimes {Q}_{g}^{(\alpha )})\\ \quad \times \,(v{{\rm{\Psi }}}_{d}+(1-v)\,\displaystyle \frac{1}{{d}^{2}}{{\mathbb{I}}}_{{d}^{2}})]\\ \quad =\,(1-v)\,\alpha -\displaystyle \frac{{h}_{s}\,(1-v)}{d}+\alpha v\\ \quad -{{vh}}_{s}\,\displaystyle \sum _{g,l=1}^{d}\,\displaystyle \sum _{\alpha =1}^{L}\,\mathrm{tr}\,[{{ \mathcal O }}_{{gl}}^{(\alpha )\,}\overline{{Q}_{l}^{(\alpha )\,}}\otimes {Q}_{g}^{(\alpha )\,}{{\rm{\Psi }}}_{d}]\\ \quad =\alpha -\displaystyle \frac{{h}_{s}}{d}+\displaystyle \frac{{h}_{s}}{d}v-{{vh}}_{s}\,\displaystyle \sum _{g,\alpha }\,\displaystyle \frac{{{ \mathcal O }}_{{gg}}^{(\alpha )}}{d}.\end{array}\end{eqnarray}$
Here in the second equality, we apply the following fact (AI)ψd = (IAT)ψd. Let ${{ \mathcal O }}_{{gg}}^{(\alpha )}=1,\forall g,\alpha ,$ then
$\begin{eqnarray}(A20)=\alpha -\displaystyle \frac{{h}_{s}}{d}+\displaystyle \frac{{h}_{s}}{d}v-{{vh}}_{s}\,L,\end{eqnarray}$
that is, when $v\geqslant \tfrac{1/{h}_{s}+L-1}{d{L}-1},$ Schmidt number of ρv is bigger than k.
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