Bounds on the number of mutually unbiased entangled bases

Abstract

We provide several bounds on the maximum size of MU k-Schmidt bases in \(\mathbb {C}^{d}\otimes \mathbb {C}^{d'}\). We first give some upper bounds on the maximum size of MU k-Schmidt bases in \(\mathbb {C}^{d}\otimes \mathbb {C}^{d'}\) by conversation law. Then we construct two maximally entangled mutually unbiased (MU) bases in the space \(\mathbb {C}^{2}\otimes \mathbb {C}^{3}\), which is the first example of maximally entangled MU bases in \(\mathbb {C}^d\otimes \mathbb {C}^{d'}\) when \(d\not \mid d'\). By applying a general recursive construction to this example, we are able to obtain two maximally entangled MU bases in \(\mathbb {C}^{d}\otimes \mathbb {C}^{d'}\) for infinitely many \(d,d'\) such that d is not a divisor of \(d'\). We also give some applications of the two maximally entangled MU bases in \(\mathbb {C}^{2}\otimes \mathbb {C}^{3}\). Further, we present an efficient method of constructing MU k-Schmidt bases. It solves an open problem proposed in [Y. F. Han et al., Quantum Inf. Process. 17, 58 (2018)]. Our work improves all previous results on maximally entangled MU bases.

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Notes

  1. 1.

    Maximally entangled MU bases are called MUMEBs in [9,10,11,12]; k-Schmidt bases are called SEBks in [13, 14]; MU k-Schmidt bases are called MUSEBks in [15].

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Acknowledgements

We appreciate the anonymous referee’s suggestion that points out an early version of Theorem 4. FS and XZ were supported by NSFC under Grant No. 11771419, the Fundamental Research Funds for the Central Universities, and Anhui Initiative in Quantum Information Technologies under Grant No. AHY150200. LC and YS were supported by the NNSF of China (Grant No. 11871089), and the Fundamental Research Funds for the Central Universities (Grant No. ZG216S2005).

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Appendices

Appendix A:The proof of Lemma 10

Proof

We write U and V in the matrix form as follows:

$$\begin{aligned} U= \left( \begin{matrix} {\frac{1}{\sqrt{2}}} &{} \quad \frac{1}{\sqrt{2}} &{} \quad 0 &{} \quad 0 &{} \quad 0 &{} \quad 0 \\ 0 &{} \quad 0 &{} \quad \frac{1}{\sqrt{2}} &{} \quad \frac{1}{\sqrt{2}} &{} \quad 0 &{} \quad 0 \\ 0 &{} \quad 0 &{}\quad 0 &{}\quad 0 &{} \quad \frac{1}{\sqrt{2}} &{} \quad \frac{1}{\sqrt{2}} \\ 0 &{} \quad 0 &{} \quad 0 &{} \quad 0 &{} \quad \frac{1}{\sqrt{2}} &{} \quad -\frac{1}{\sqrt{2}} \\ {\frac{1}{\sqrt{2}}} &{} \quad -\frac{1}{\sqrt{2}} &{} \quad 0 &{} \quad 0 &{} \quad 0 &{} \quad 0 \\ 0 &{} \quad 0 &{} \quad \frac{1}{\sqrt{2}} &{} \quad -\frac{1}{\sqrt{2}} &{} \quad 0 &{} \quad 0 \end{matrix} \right) , \end{aligned}$$
(A1)

and

$$\begin{aligned} V&:= \left( \begin{matrix} |v_1\rangle&|v_2\rangle&|v_3\rangle&|v_4\rangle&|v_5\rangle&|v_6\rangle \end{matrix} \right) \nonumber \\&= \left( \begin{matrix} {\frac{1}{\sqrt{6}}} &{} \quad {\frac{1}{\sqrt{6}}} &{} \quad 0 &{} \quad 0 &{} \quad {\frac{\sqrt{2}}{\sqrt{6}}} &{} \quad -{\frac{\sqrt{2}}{\sqrt{6}}} \\ {\frac{\sqrt{2}}{\sqrt{6}}} &{} \quad -{\frac{\sqrt{2}}{\sqrt{6}}} &{} \quad {\frac{1}{\sqrt{6}}} &{} \quad {\frac{1}{\sqrt{6}}} &{} \quad 0 &{} \quad 0 \\ 0 &{} \quad 0 &{} \quad {\frac{\sqrt{2}}{\sqrt{6}}} &{} \quad -{\frac{\sqrt{2}}{\sqrt{6}}} &{} \quad {\frac{1}{\sqrt{6}}} &{} \quad {\frac{1}{\sqrt{6}}} \\ -{\frac{\sqrt{2}}{\sqrt{6}}}i &{} \quad -{\frac{\sqrt{2}}{\sqrt{6}}}i &{} \quad 0 &{} \quad 0 &{} \quad {\frac{i}{\sqrt{6}}} &{} \quad -{\frac{i}{\sqrt{6}}} \\ {\frac{i}{\sqrt{6}}} &{} \quad -\frac{i}{\sqrt{6}} &{} \quad -\frac{\sqrt{2}}{\sqrt{6}}i &{} \quad -\frac{\sqrt{2}}{\sqrt{6}}i &{} \quad 0 &{} \quad 0 \\ 0 &{}\quad 0 &{} \quad {\frac{i}{\sqrt{6}}} &{} \quad -\frac{i}{\sqrt{6}} &{} \quad -\frac{\sqrt{2}}{\sqrt{6}}i &{} \quad -\frac{\sqrt{2}}{\sqrt{6}}i \end{matrix} \right) . \end{aligned}$$
(A2)

Let

$$\begin{aligned} X= \left( \begin{matrix} |x_1\rangle&|x_2\rangle&|x_3\rangle&|x_4\rangle&|x_5\rangle&|x_6\rangle \end{matrix} \right) , \end{aligned}$$
(A3)

where \(|x_j\rangle \)’s are vectors in \(\mathbb {C}^6\), and \(|x_j\rangle \ne |v_1\rangle ~\forall j\). Suppose U and X are two MUBs which can be extended to four MUBs in \(\mathbb {C}^6\). If \(|x_1\rangle =|v_2\rangle \), then we define

$$\begin{aligned} P=\mathop {\mathrm{diag}}(1,-1,1,1,-1,1),\quad Q=\left( \begin{matrix} 0 &{} \quad 1 &{} \quad 0 &{} \quad 0 &{} \quad 0 &{} \quad 0 \\ 1 &{} \quad 0 &{} \quad 0 &{} \quad 0 &{} \quad 0 &{} \quad 0 \\ 0 &{}\quad 0 &{} \quad 0 &{} \quad -1 &{} \quad 0 &{} \quad 0 \\ 0 &{} \quad 0 &{} \quad -1 &{} \quad 0 &{} \quad 0 &{} \quad 0 \\ 0 &{} \quad 0 &{} \quad 0 &{} \quad 0 &{} \quad 1 &{} \quad 0 \\ 0 &{} \quad 0 &{} \quad 0 &{} \quad 0 &{} \quad 0 &{} \quad 1 \end{matrix} \right) . \end{aligned}$$
(A4)

One can verify \(PUQ=U\), and thus PUQ and PX are two MUBs which can be extended to four MUBs in \(\mathbb {C}^6\). Since \(P|x_1\rangle =P|v_2\rangle =|v_1\rangle \), we have a contradiction with the hypothesis that X cannot have \(|v_1\rangle \). So we have shown that \(|x_1\rangle \) cannot be \(|v_2\rangle \). Then up to a column permutation of X, we have \(|x_j\rangle \ne |v_2\rangle ~\forall j\). Using the same idea, it suffices to exclude two cases, namely \(|x_1\rangle =|v_3\rangle \) and \(|x_1\rangle =|v_5\rangle \), in order to prove that X cannot have any column vector of V. If \(|x_1\rangle =|v_3\rangle \), then we define

$$\begin{aligned} P=\left( \begin{matrix} 0 &{} \quad 0 &{} \quad 0 &{} \quad 0 &{} \quad 0 &{} \quad 1 \\ 0 &{}\quad 0 &{} \quad 0 &{} \quad 0 &{} \quad -1 &{} \quad 0 \\ 0 &{} \quad 0 &{} \quad 0 &{} \quad 1 &{} \quad 0 &{} \quad 0 \\ 0 &{} \quad 0 &{} \quad 1 &{} \quad 0 &{} \quad 0 &{} \quad 0 \\ 0 &{} \quad -1 &{} \quad 0 &{}\quad 0 &{} \quad 0 &{} \quad 0 \\ 1 &{} \quad 0 &{} \quad 0 &{} \quad 0 &{} \quad 0 &{} \quad 0 \end{matrix} \right) ,\quad Q=\left( \begin{matrix} 0 &{} \quad 0 &{} \quad 0 &{} \quad -1 &{} \quad 0 &{} \quad 0 \\ 0 &{} \quad 0 &{} \quad 1 &{} \quad 0 &{} \quad 0 &{} \quad 0 \\ 0 &{} \quad 1 &{}\quad 0 &{} \quad 0 &{} \quad 0 &{} \quad 0 \\ -1 &{} \quad 0 &{} \quad 0 &{} \quad 0 &{} \quad 0 &{} \quad 0 \\ 0 &{} \quad 0 &{} \quad 0 &{} \quad 0 &{} \quad 1 &{} \quad 0 \\ 0 &{} \quad 0 &{} \quad 0 &{} \quad 0 &{} \quad 0 &{} \quad -1 \end{matrix} \right) . \end{aligned}$$
(A5)

A straightforward computing yields that \(PUQ=U\), and \((-i)P|v_3\rangle =|v_1\rangle \). So PUQ and \((-i)PX\) are two MUBs that can be extended to four MUBs in \(\mathbb {C}^6\). Since \((-i)P|x_1\rangle =(-i)P|v_3\rangle =|v_1\rangle \), we have a contradiction with the hypothesis that X cannot have \(|v_1\rangle \). Hence we exclude \(|x_1\rangle =|v_3\rangle \). If \(|x_1\rangle =|v_5\rangle \) we may choose

$$\begin{aligned} P=\left( \begin{matrix} 0 &{} \quad 0 &{} \quad 1 &{} \quad 0 &{} \quad 0 &{} \quad 0 \\ 1 &{} \quad 0 &{} \quad 0 &{} \quad 0 &{} \quad 0 &{} \quad 0 \\ 0 &{} \quad 1 &{} \quad 0 &{} \quad 0 &{} \quad 0 &{} \quad 0 \\ 0 &{} \quad 0 &{} \quad 0 &{} \quad 0 &{} \quad 0 &{} \quad 1 \\ 0 &{} \quad 0 &{} \quad 0 &{} \quad 1 &{} \quad 0 &{} \quad 0 \\ 0 &{} \quad 0 &{} \quad 0 &{} \quad 0 &{} \quad 1 &{} \quad 0 \end{matrix} \right) ,\quad Q=\left( \begin{matrix} 0 &{} \quad 0 &{} \quad 1 &{} \quad 0 &{} \quad 0 &{} \quad 0 \\ 0 &{} \quad 0 &{} \quad 0 &{} \quad 1 &{} \quad 0 &{} \quad 0 \\ 0 &{} \quad 0 &{} \quad 0 &{} \quad 0 &{} \quad 1 &{} \quad 0 \\ 0 &{} \quad 0 &{} \quad 0 &{} \quad 0 &{} \quad 0 &{} \quad 1 \\ 1 &{} \quad 0 &{} \quad 0 &{} \quad 0 &{} \quad 0 &{} \quad 0 \\ 0 &{} \quad 1 &{} \quad 0 &{} \quad 0 &{} \quad 0 &{} \quad 0 \end{matrix} \right) . \end{aligned}$$
(A6)

A straightforward computing yields that \(PUQ=U\), and \(P|v_5\rangle =|v_1\rangle \). Similarly we can exclude \(|x_1\rangle =|v_5\rangle \). Therefore, X cannot have any column vector of V. This completes the proof. \(\square \)

Appendix B:\(M_3(6,6)\ge 3\)

The following three maximally entangled MU bases in \(\mathbb {C}^{3}\otimes \mathbb {C}^{3}\) are constructed from [8].

$$\begin{aligned} \mathcal {S}_{1}&=\left\{ \frac{1}{3\sqrt{3}}\left( \begin{matrix} w+2 &{} \quad w+2 &{} \quad w+2\\ 2w^{2}+1 &{} \quad w+2 &{} \quad w^{2}+2w\\ w^{2}+2w &{} \quad w+2 &{} \quad 2w^{2}+1 \end{matrix}\right) , \ \frac{1}{3\sqrt{3}}\left( \begin{matrix} w+2 &{} \quad w^{2}+2w &{} \quad 2w^{2}+1\\ 2w^{2}+1 &{}\quad w^{2}+2w &{} \quad w+2\\ w^{2}+2w &{}\quad w^{2}+2w &{}\quad w^{2}+2w \end{matrix} \right) ,\right. \\&\quad \frac{1}{3\sqrt{3}}\left( \begin{matrix} w+2 &{}\quad 2w^{2}+1 &{}\quad w^{2}+2w\\ 2w^{2}+1 &{} \quad 2w^{2}+1 &{}\quad 2w^{2}+1 \\ w^{2}+2w &{}\quad 2w^{2}+1 &{} \quad w+2 \end{matrix} \right) ,\\&\quad \frac{1}{3\sqrt{3}}\left( \begin{matrix} w+2 &{} \quad w+2 &{}\quad w+2\\ w+2 &{} \quad w^{2}+2w &{}\quad 2w^{2}+1\\ w+2 &{} \quad 2w^{2}+1 &{}\quad w^{2}+2w \end{matrix}\right) , \frac{1}{3\sqrt{3}}\left( \begin{matrix} w+2 &{}\quad w^{2}+2w &{}\quad 2w^{2}+1\\ w+2 &{}\quad 2w^{2}+1 &{}\quad w^{2}+2w\\ w+2 &{}\quad w+2 &{}\quad w+2 \end{matrix} \right) , \\&\quad \frac{1}{3\sqrt{3}}\left( \begin{matrix} w+2 &{}\quad 2w^{2}+1 &{}\quad w^{2}+2w\\ w+2 &{}\quad w+2 &{}\quad w+2 \\ w+2 &{}\quad w^{2}+2w &{}\quad 2w^{2}+1 \end{matrix} \right) ,\\&\quad \frac{1}{3\sqrt{3}}\left( \begin{matrix} w+2 &{}\quad w+2 &{}\quad w+2\\ w^{2}+2w &{}\quad 2w^{2}+1 &{}\quad w+2\\ 2w^{2}+1 &{}\quad w^{2}+2w &{}\quad w+2 \end{matrix}\right) , \ \frac{1}{3\sqrt{3}}\left( \begin{matrix} w+2 &{}\quad w^{2}+2w &{}\quad 2w^{2}+1 \\ w^{2}+2w &{}\quad w+2 &{}\quad 2w^{2}+1\\ 2w^{2}+1 &{}\quad 2w^{2}+1 &{}\quad 2w^{2}+1 \end{matrix} \right) ,\\&\quad \left. \frac{1}{3\sqrt{3}}\left( \begin{matrix} w+2 &{}\quad 2w^{2}+1 &{}\quad w^{2}+2w\\ w^{2}+2w &{}\quad w^{2}+2w &{}\quad w^{2}+2w\\ 2w^{2}+1 &{}\quad w+2 &{}\quad w^{2}+2w \end{matrix} \right) \right\} ,\\ \mathcal {S}_{2}&=\left\{ \frac{1}{3\sqrt{3}}\left( \begin{matrix} 0 &{} \quad 3w^{2} &{} \quad 0\\ 3w &{} \quad 0 &{} \quad 0\\ 0 &{} \quad 0 &{} \quad 3 \end{matrix}\right) , \ \frac{1}{3\sqrt{3}}\left( \begin{matrix} 0 &{} \quad 3 &{} \quad 0\\ 3w &{} \quad 0 &{} \quad 0\\ 0 &{} \quad 0 &{} \quad 3w^{2} \end{matrix} \right) , \ \frac{1}{3\sqrt{3}}\left( \begin{matrix} 0 &{} \quad 3w &{} \quad 0\\ 3w &{} \quad 0 &{} \quad 0\\ 0 &{} \quad 0 &{} \quad 3w \end{matrix} \right) \right. ,\\&\quad \frac{1}{3\sqrt{3}}\left( \begin{matrix} 3w^{2} &{} \quad 0 &{} \quad 0\\ 0 &{} \quad 0 &{} \quad 3w\\ 0 &{} \quad 3 &{} \quad 0 \end{matrix}\right) , \ \frac{1}{3\sqrt{3}}\left( \begin{matrix} 3w^{2}&{} \quad 0 &{} \quad 0\\ 0 &{} \quad 0 &{} \quad 3\\ 0 &{}\quad 3w &{} \quad 0 \end{matrix} \right) ,\\&\quad \frac{1}{3\sqrt{3}}\left( \begin{matrix} 3w^{2} &{} \quad 0 &{} \quad 0\\ 0 &{} \quad 0 &{} \quad 3w^{2}\\ 0 &{} \quad 3w^{2} &{} \quad 0 \end{matrix} \right) , \frac{1}{3\sqrt{3}}\left( \begin{matrix} 0 &{} \quad 0 &{} \quad 3w^{2}\\ 0 &{} \quad 3w &{} \quad 0\\ 3 &{} \quad 0 &{} \quad 0 \end{matrix}\right) , \ \frac{1}{3\sqrt{3}}\left( \begin{matrix} 0 &{}0 &{}3w\\ 0 &{}3w^{2} &{} 0\\ 3 &{}0 &{} 0 \end{matrix} \right) ,\\&\quad \left. \frac{1}{3\sqrt{3}}\left( \begin{matrix} 0 &{} \quad 0 &{} \quad 3\\ 0 &{}\quad 3 &{}\quad 0\\ 3 &{}\quad 0 &{} \quad 0 \end{matrix} \right) \right\} ,\\ \mathcal {S}_{3}&=\left\{ \frac{1}{\sqrt{3}}\left( \begin{matrix} 1 &{} \quad 0 &{} \quad 0\\ 0 &{} \quad 1 &{} \quad 0\\ 0 &{} \quad 0 &{} \quad 1 \end{matrix}\right) , \ \frac{1}{\sqrt{3}}\left( \begin{matrix} 1 &{} \quad 0 &{} \quad 0\\ 0 &{} \quad w &{} \quad 0\\ 0 &{} \quad 0 &{} \quad w^{2} \end{matrix} \right) , \ \frac{1}{\sqrt{3}}\left( \begin{matrix} 1 &{} \quad 0 &{} \quad 0\\ 0 &{} \quad w^{2} &{} \quad 0\\ 0 &{} \quad 0 &{} \quad w \end{matrix} \right) ,\right. \\&\quad \frac{1}{\sqrt{3}}\left( \begin{matrix} 0 &{} \quad 0 &{} \quad 1\\ 1 &{} \quad 0 &{} \quad 0\\ 0 &{} \quad 1 &{} \quad 0 \end{matrix}\right) , \ \frac{1}{\sqrt{3}}\left( \begin{matrix} 0 &{} \quad 0 &{} \quad w^{2}\\ 1 &{} \quad 0 &{} \quad 0\\ 0 &{} \quad w &{} \quad 0 \end{matrix} \right) , \frac{1}{\sqrt{3}}\left( \begin{matrix} 0 &{} \quad 0 &{} \quad w\\ 1 &{} \quad 0 &{} \quad 0\\ 0 &{} \quad w^{2} &{} \quad 0 \end{matrix} \right) ,\\&\quad \left. \frac{1}{\sqrt{3}}\left( \begin{matrix} 0 &{} \quad 1 &{} \quad 0\\ 0 &{} \quad 0 &{} \quad 1\\ 1 &{} \quad 0 &{} \quad 0 \end{matrix}\right) , \ \frac{1}{\sqrt{3}}\left( \begin{matrix} 0 &{}\quad w &{} \quad 0\\ 0 &{}\quad 0 &{} \quad w^{2}\\ 1 &{}\quad 0 &{} \quad 0 \end{matrix} \right) , \ \frac{1}{\sqrt{3}}\left( \begin{matrix} 0 &{}w^{2} &{}0\\ 0 &{}0 &{}w\\ 1 &{}0 &{} 0 \end{matrix} \right) \right\} . \end{aligned}$$

For three MUBs in \(\mathbb {C}^{2}\), we choose

$$\begin{aligned} \mathcal {T}_{1}=\{(0,1),(1,0)\}, \ \mathcal {T}_{2}=\left\{ \frac{1}{\sqrt{2}}(1,1),\frac{1}{\sqrt{2}}(1,-1)\right\} , \ \mathcal {T}_{3}=\left\{ \frac{1}{\sqrt{2}}(1,i),\frac{1}{\sqrt{2}}(1,-i)\right\} . \end{aligned}$$

Then \(\{\mathcal {S}_{1}\otimes \mathcal {T}_{1}, \mathcal {S}_{2}\otimes \mathcal {T}_{2},\mathcal {S}_{3}\otimes \mathcal {T}_{3}\}\) is a set of three maximally entangled MU bases in \(\mathbb {C}^{3}\otimes \mathbb {C}^{6}\) by Eq. (23), and \(\{\mathcal {T}_{1}^{\mathrm {T}}\otimes \mathcal {S}_{1}\otimes \mathcal {T}_{1}, \mathcal {T}_{2}^{\mathrm {T}}\otimes \mathcal {S}_{2}\otimes \mathcal {T}_{2},\mathcal {T}_{3}^{\mathrm {T}}\otimes \mathcal {S}_{3}\otimes \mathcal {T}_{3}\}\) is a set of three MU 3-Schmidt bases in \(\mathbb {C}^{6}\otimes \mathbb {C}^{6}\) by Eq. (24).

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Shi, F., Shen, Y., Chen, L. et al. Bounds on the number of mutually unbiased entangled bases. Quantum Inf Process 19, 383 (2020). https://doi.org/10.1007/s11128-020-02890-4

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Keywords

  • Mutually unbiased bases
  • Maximally entangled MU bases
  • MU k-Schmidt bases