The geometry of Bloch space in the context of quantum random access codes

Abstract

We study the communication protocol known as a quantum random access code (QRAC) which encodes n classical bits into m qubits ($m<n$) with a probability of recovering any of the initial n bits of at least $p>\tfrac{1}{2}$. Such a code is denoted by (n, m, p)-QRAC. If cooperation is allowed through a shared random string, we call it a QRAC with shared randomness. We prove that for any (n, m, p)-QRAC with shared randomness the parameter p is upper bounded by $\tfrac{1}{2}+\tfrac{1}{2}\sqrt{\tfrac{2^{m-1}}{n}}$. For $m=2$, this gives a new bound of $p\le \tfrac{1}{2}+\tfrac{1}{\sqrt{2n}}$ confirming a conjecture by Imamichi and Raymond (AQIS’18). Our bound implies that the previously known analytical constructions of $(3,2,\tfrac{1}{2}+\tfrac{1}{\sqrt{6}})$- , $(4,2,\tfrac{1}{2}+\tfrac{1}{2\sqrt{2}})$- and $(6,2,\tfrac{1}{2}+\tfrac{1}{2\sqrt{3}})$-QRACs are optimal. To obtain our bound, we investigate the geometry of quantum states in the Bloch vector representation and make use of a geometric interpretation of the fact that any two quantum states have a nonnegative overlap.

Appendices

Appendix A: Proof of observation 6

Observation 6 Let $\beta \in \varvec{\beta }(Q_N) \backslash \{ 0\}$. Then,

$$\begin{aligned} \tfrac{\beta }{1-N\lambda _{\max }}, \tfrac{\beta }{1-N\lambda _{\min }} \in \partial \varvec{\beta }(Q_N) \end{aligned}$$

(25)

where $\lambda _{\min }$ and $\lambda _{\max }$ is the smallest and largest eigenvalue of $\varvec{\rho }(\beta )$, respectively. In particular, $-\beta \in \partial \varvec{\beta }(Q_N)$ if and only $\lambda _{\max }= r_N R_N$.

Proof

We list the eigenvalues of $\varvec{\rho }(\beta )$ as $\lambda _{\text {max}} \ge \cdots \ge \lambda _{\text {min}} \ge 0$. Then, the eigenvalues of $\tfrac{1}{2} \sum _{i=1}^{N^2-1} \beta _i \varvec{\sigma }_i$ can be listed as

$$\begin{aligned} \lambda _{\max }-\tfrac{1}{N} \ge \cdots \ge \lambda _{\min } - \tfrac{1}{N}. \end{aligned}$$

(42)

Let $\beta '=\gamma \beta$ for some $\gamma \in {\mathbb {R}}$. If $\gamma \ge 0$ The eigenvalues of $\varvec{\rho }(\gamma \beta )$ can be listed as

$$\begin{aligned} \tfrac{1}{N}(1-\gamma )+\gamma \lambda _{\max } \ge \cdots \ge \tfrac{1}{N}(1-\gamma )+\gamma \lambda _{\min } \ge 0 \end{aligned}$$

(43)

and hence $\beta ' \in \partial \varvec{\beta }(Q_N)$ is in the boundary of Bloch space if $\gamma =\tfrac{1}{1-N\lambda _{\min }}$. Similarly if $\gamma <0$ then $\beta ' \in \partial \varvec{\beta }(Q_N)$ if $\gamma =\tfrac{1}{1-N\lambda _{\max }}$.

$\square$

Appendix B: Proof of lemma 9

Lemma 9 For any set of vectors, $\{\mu _i \}_{i=1}^n\subseteq \mathbb {R}^{N}$, the inequality

$$\begin{aligned} \sum _{x \in \{0,1 \}^n}\left| \left| \sum _{i\in [n]} (-1)^{x_i} \mu _i\right| \right| \le 2^n \sqrt{\sum _{i\in [n]} \left| \left| \mu _i\right| \right| ^2} \end{aligned}$$

(32)

holds with equality if and only if the $\mu _i$’s are orthogonal.

Proof

The inequality in (32) holds if all the $\mu _i$’s are 0, so we assume they are not all 0. We can interpret the sum on the left hand side of (32) as the inner product of

$$\begin{aligned} y_1=(1, ... ,1) \in {\mathbb {R}}^{\{0,1 \}^n} \cong {\mathbb {R}}^{2^n} \end{aligned}$$

(44)

and a vector $y_2 \in {\mathbb {R}}^{\{0,1 \}^n}$, whose entry corresponding to bit string $x\in {\mathbb {R}}^{\{0,1 \}^n}$ is given by

$$\begin{aligned} (y_2)_x=\left| \left| \sum _{i\in [n]}(-1)^{x_{i}} \mu _{i}\right| \right| . \end{aligned}$$

(45)

Applying the Cauchy–Schwarz inequality, we get that the left hand side of (32) is upper bounded by

$$\begin{aligned} \left| \left| y_1\right| \right| \left| \left| y_2\right| \right| = \sqrt{2^n}\sqrt{\sum _{x \in \{0,1 \}^n}\left| \left| \sum _{i\in [n]} (-1)^{x_i} \mu _i\right| \right| ^2}. \end{aligned}$$

(46)

We claim now that

$$\begin{aligned} \sum _{x \in \{0,1 \}^n}\left| \left| \sum _{i\in [n]} (-1)^{x_i} \mu _i\right| \right| ^2=2^n \sum _{i \in [n]} \left| \left| \mu _i\right| \right| ^2. \end{aligned}$$

(47)

This can be proved by induction on n. First, Eq. (47) holds for $n=1$ since

$$\begin{aligned} \left| \left| \mu _1\right| \right| ^2+\left| \left| -\mu _1\right| \right| ^2=2\left| \left| \mu _1\right| \right| ^2. \end{aligned}$$

(48)

Assume that Eq. (47) holds for $n=k$ and consider the case when $n=k+1$. By explicitly carrying out the sum over $x_{k+1} \in \{0,1 \}$ on the left hand side of Eq. (47) we get

$$\begin{aligned}&\sum _{x \in \{0,1 \}^{k}} \left[ \left| \left| \big ( (-1)^{x_1} \mu _1 + ... + (-1)^{x_k} \mu _k \big ) + \mu _{k+1} \right| \right| ^2 \right. \nonumber \\&\left. + \left| \left| \big ( (-1)^{x_1} \mu _1 + ... + (-1)^{x_k} \mu _k \big ) - \mu _{k+1} \right| \right| ^2 \right] . \end{aligned}$$

(49)

Applying the parallelogram identity, i.e.,

$$\begin{aligned} \left| \left| u_1+u_2\right| \right| ^2+ \left| \left| u_1-u_2\right| \right| ^2=2(\left| \left| u_1\right| \right| ^2+\left| \left| u_2\right| \right| ^2) , \end{aligned}$$

(50)

the expression in (49) equals

$$\begin{aligned} 2 \sum _{x \in \{0,1 \}^{k}}\big ( \left| \left| \textstyle \sum _{i \in [k]} (-1)^{x_i} \mu _i \right| \right| ^2+\left| \left| \mu _{k+1}\right| \right| ^2 \big ). \end{aligned}$$

(51)

Finally, applying the induction hypothesis, we complete the inductive step as follows:

$$\begin{aligned} 2\big ( 2^k \sum _{i \in [k]} \left| \left| \mu _i\right| \right| ^2+2^k \left| \left| \mu _{k+1}\right| \right| ^2 \big )=2^{k+1}\sum _{i \in [k+1]}\left| \left| \mu _{i}\right| \right| ^2. \end{aligned}$$

(52)

Now, by inserting (47) in (46) we can conclude (32).

Recall that $\left\langle y_1 ,y_2\right\rangle = \left| \left| y_1\right| \right| \left| \left| y_2\right| \right|$ if and only if $y_2 =k y_1$ for some $k\in {\mathbb {R}}$. Hence, the bound in (32) holds with equality if and only if the quantity $(y_2)_x$ in Eq. (45) is equal to some constant c independent of $x\in \{0,1\}^n$. In other words, for all $x \in \{0,1 \}^n$ we must have that

$$\begin{aligned} \sum _{i\ne j \in [n]} (-1)^{x_i+x_j}\left\langle \mu _i ,\mu _j\right\rangle =c^2-\sum _{i\in [n]} \left| \left| \mu _i\right| \right| ^2, \end{aligned}$$

(53)

where the right hand side is constant. Now, fix $m\in [n]$. The left hand side of (53) can be rewritten as

$$\begin{aligned} \sum _{i\ne j \in [n]\backslash \{ m\}}(-1)^{x_i+x_j}\left\langle \mu _i,\mu _j\right\rangle +2\sum _{i\in [n]\backslash \{ m\}}(-1)^{x_m+x_i}\left\langle \mu _m,\mu _i\right\rangle \end{aligned}$$

(54)

This must be invariant upon the interchange $x_m \rightarrow \overline{x_m}$. Since $(-1)^{x_m}=-(-1)^{\overline{x_m}}$, we have that

$$\begin{aligned} \sum _{i\in [n]\backslash \{ m\}}(-1)^{x_m+x_i}\left\langle \mu _m,\mu _i\right\rangle =0. \end{aligned}$$

(55)

Now fix $m'\in [n]\backslash \{m \}$ and rewrite (55) as

$$\begin{aligned} (-1)^{x_m+x_{m'}}\left\langle \mu _m,\mu _{m'}\right\rangle +\sum _{i\in [n]\backslash \{ m,m'\}}(-1)^{x_m+x_i}\left\langle \mu _m,\mu _i\right\rangle =0. \end{aligned}$$

(56)

This must be invariant upon the interchange $x_{m'}\rightarrow \overline{x_{m'}}$ so we get $\left\langle \mu _m,\mu _{m'}\right\rangle =0$. This completes the proof. $\square$

Appendix C: Proof of corollary 12

Corollary 12 For $m>1$ there exists $n_{\max }(m) \le 4^m-1$ such that any $(n\ge n_{\max }(m),m,p)$-QRAC fulfills

$$\begin{aligned} p<\tfrac{1}{2}+\tfrac{1}{2}\sqrt{\tfrac{2^{m-1}}{n}}. \end{aligned}$$

(39)

Proof

The average success probability of an (n, m)-QRAC reaches the bound in Theorem 10 if and only if it is given by the following Bloch vector configuration:

• 1. A set $\{\nu _i\}_{i=1}^n$ of orthogonal unit vectors such that $\pm \sqrt{r_{2^m}R_{2^m}}\nu _i \in \partial \varvec{\beta }(Q_{2^m})$. The POVM for measuring the jth bit is then associated with the pair $\pm \sqrt{r_{2^m}R_{2^m}}\nu _j$ in the sense of (31).

• 2. For the encodings, $2^{n-1}$ pairs of orthogonal pure state Bloch vectors, $\{\beta _x, \beta _{{{\overline{x}}}} \}$, where

  $$\begin{aligned} \tfrac{1}{2}(\beta _x-\beta _{{{\overline{x}}}})=\tfrac{1}{\sqrt{n}}\sum _{i\in [n]} (-1)^{x_i} \nu _i=: V_x, \end{aligned}$$

  (57)

  with $\{\nu _i\}_{i=1}^n$ given as in 1.

Let $V:=\text {span} \big \{V_x \mid x\in \{ 0,1\}^n \big \}$. Assume, for some (n, m)-QRAC, that 1 and 2 above are fulfilled. This implies that its average success probability is $\tfrac{1}{2}+\tfrac{1}{2}\sqrt{\tfrac{2^{m-1}}{n}}$. In view of Eqs. (17), (31) and (57), one can use the decomposition

$$\begin{aligned} \beta _{x,{{\overline{x}}}}= \pm \tfrac{1}{2}(\beta _x-\beta _{{{\overline{x}}}})+\tfrac{1}{2}(\beta _x+\beta _{{{\overline{x}}}}), \end{aligned}$$

(58)

to calculate the probability of correctly decoding the ith bit of x to be $x_i$ as

$$\begin{aligned} p_{i,x}&=\tfrac{1}{r_{2^m}R_{2^m}}\text {Tr}\left[ \varvec{\rho }\left( \tfrac{1}{2}(\beta _x-\beta _{{{\overline{x}}}})+\tfrac{1}{2}(\beta _x+\beta _{{{\overline{x}}}})\right) \varvec{\rho }\left( \sqrt{r_{2^m}R_{2^m}}\nu _i\right) \right] \end{aligned}$$

(59)

$$\begin{aligned}&=\tfrac{1}{r_{2^m}R_{2^m}}\left( \tfrac{1}{2^{m}}+\tfrac{\sqrt{r_{2^m}R_{2^m}}}{2}\Big \langle \tfrac{1}{2}(\beta _x-\beta _{{{\overline{x}}}})+\tfrac{1}{2}(\beta _x+\beta _{{{\overline{x}}}}),\nu _i\Big \rangle \right) \end{aligned}$$

(60)

$$\begin{aligned}&=\tfrac{1}{2}+\tfrac{1}{2}\tfrac{1}{\sqrt{r_{2^m}R_{2^m}}}\left( \tfrac{1}{\sqrt{n}}+\Big \langle \tfrac{1}{2}(\beta _x+\beta _{{{\overline{x}}}}),\nu _i\Big \rangle \right) \end{aligned}$$

(61)

$$\begin{aligned}&= \tfrac{1}{2}+\tfrac{1}{2}\sqrt{\tfrac{2^{m-1}}{n}}\big ( 1 +\sqrt{n}\Big \langle \tfrac{1}{2}(\beta _x+\beta _{{{\overline{x}}}}),\nu _i\Big \rangle \big ). \end{aligned}$$

(62)

As noted, the average of this is $\tfrac{1}{2}+\tfrac{1}{2}\sqrt{\tfrac{2^{m-1}}{n}}$. It follows that the worst case success probability reaches $\tfrac{1}{2}+\tfrac{1}{2}\sqrt{\tfrac{2^{m-1}}{n}}$ only if, we additionally have that for all $x\in \{0,1 \}^n$, $\tfrac{1}{2}(\beta _x+\beta _{{{\overline{x}}}})\in V^{\perp }$. Unless $m=1$, we have that $\tfrac{1}{2}(\beta _x+\beta _{{{\overline{x}}}})\ne 0$, i.e., for $m>1$, $V^{\perp }$ must be of non-vanishing dimension. From this, we conclude the desired. $\square$