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.

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).