Maximal noiseless code rates for collective rotation channels on qudits

Abstract

We study noiseless subsystems on collective rotation channels of qudits, i.e., quantum channels with operators in the set $\mathcal {E}(d,n) = \{ U^{\otimes n}: U \in {\mathrm {SU}}(d)\}.$ This is done by analyzing the decomposition of the algebra $\mathcal {A}(d,n)$ generated by $\mathcal {E}(d,n)$. We summarize the results for the channels on qubits ($d=2$) and obtain the maximum dimension of the noiseless subsystem that can be used as the quantum error correction code for the channel. Then we extend our results to general d. In particular, it is shown that the code rate, i.e., the number of protected qudits over the number of physical qudits, always approaches 1 for a suitable noiseless subsystem. Moreover, one can determine the maximum dimension of the noiseless subsystem by solving a non-trivial discrete optimization problem. The maximum dimension of the noiseless subsystem for $d = 3$ (qutrits) is explicitly determined by a combination of mathematical analysis and the symbolic software Mathematica.

Appendix: Proofs of Theorems

1.1 Proof of Theorem 2

By the Frobenius formula [8, 10, 28],

$$\begin{aligned} f(p_1,\ldots ,p_d) = \frac{(dk)! \prod _{\ell =1}^{d-1} \ell !}{\prod _{\ell =1}^d (k+d-\ell )!} = f_1(k)f_2(k), \end{aligned}$$

where

$$\begin{aligned} f_1(k) = \frac{(dk)!}{(k!)^d} \qquad \hbox { and } \qquad f_2(k) = \frac{ \prod _{\ell =1}^{d-1} \ell ! }{\prod _{\ell =1}^d ((k+1)\cdots (k+d-\ell ))}. \end{aligned}$$

Now, by Stirling’s formula, we obtain

$$\begin{aligned} \lim _{k\rightarrow \infty } \frac{\log _d f_1(k)}{dk}= & {} \lim _{k\rightarrow \infty } \frac{\ln (dk)! - d \ln k!}{(\ln d)dk}\nonumber \\= & {} \lim _{k\rightarrow \infty } \frac{ dk(\ln d) + O(\ln (dk)) - O(\ln k )}{(\ln d) dk} = 1, \end{aligned}$$

and

$$\begin{aligned} \lim _{k\rightarrow \infty } \frac{\log _d(f_2(k))}{dk} = \lim _{k\rightarrow \infty } \frac{\log _d\left( \prod _{\ell =1}^{d-1} \ell !\right) - \sum _{i=1}^d \sum _{j=1}^{d-i} \log _d(k+j)}{dk} = 0. \end{aligned}$$

The result follows. $\square$

1.2 Proof of Theorem 3

Consider the following terms:

$$\begin{aligned} \begin{array}{rcll} f(p_1,p_2,p_3)&{}-&{}f(p_1+1,p_2-1,p_3)\qquad &{}(1)\\ f(p_1,p_2,p_3)&{}-&{}f(p_1,p_2-1,p_3+1)\qquad &{}(2)\\ f(p_1,p_2,p_3)&{}-&{}f(p_1-1,p_2+1,p_3)\qquad &{}(3)\\ f(p_1,p_2,p_3)&{}-&{}f(p_1,p_2+1,p_3-1)\qquad &{}(4)\\ f(p_1,p_2,p_3)&{}-&{}f(p_1+1,p_2,p_3-1)\qquad &{}(5)\\ f(p_1,p_2,p_3)&{}-&{}f(p_1-1,p_2,p_3+1)\qquad &{}(6) \end{array} \end{aligned}$$

Clearly, if $f(p_1,p_2,p_3)$ attains the maximum, then each of the above terms is nonnegative.

(a) When $n = 3k$, there are three cases (a.1) $p_2 > k$, (a.2) $p_2<k$ and (a.3) $p_2 = k$. We use Mathematica [13, see supplementary Mathematica file] to show the following.

• (a.1) If $(p_1,p_2,p_3) = (k+a+ b,k+a,k-2a-b)$ with $a > 0$ and $b \ge 0$, the sum of (1) and (2) is negative and so f cannot attain the maximum in this case.

• (a.2) If $(p_1,p_2,p_3) = (k+2a+b,k-a,k-a-b)$ with $a > 0$ and $b \ge 0$, then either (3) or (4) is negative and so f cannot attain the maximum in this case.

• (a.3) Let $(p_1,p_2,p_3) = (k+r,k,k-r)$ and define $r_0 \equiv \frac{1}{2} \left( -3+\sqrt{3+3 k+\sqrt{12+20 k+9 k^2}}\right)$. Then (5) and (6) are both nonnegative if and only if $\left\lceil r_0 \right\rceil \le r \le \left\lceil r_0\right\rceil +1$.

Moreover, $r_0$ in (a.3) is not an integer for all $k\ge 1$. Thus, $f(k+r,k,k-r)$ will attain its maximum when $r = \left\lceil r_0 \right\rceil$. Therefore, f has a unique maximum $f(k+r,k,k-r)$ with $r = \left\lceil r_0 \right\rceil$ when $n = 3k$ as stated.

(b) When $n = 3k+1$, we also have three cases (b.1) $p_2>k$, (b.2) $p_2<k$ and (b.3) $p_2=k$. We use Mathematica [13] to show the following.

• (b.1) If $(p_1,p_2,p_3) = (k+1+a+b,k+a,k-2a-b)$ with $a > 0$ and $b \ge 0$, the sum of (1) and (2) is negative and so f cannot attain the maximum in this case.

• (b.2) If $(p_1,p_2,p_3) = (k+1+2a+b,k-a,k-a-b)$ with $a > 0$ and $b \ge 0$, then either (3) or (4) is negative and so f cannot attain the maximum in this case.

• (b.3) Let $(p_1,p_2,p_3) = (k+1+r,k,k-r)$ and define $r_3 \equiv \frac{1}{4}(-8 + \sqrt{40+24k})$. Then (5) and (6) are both nonnegative if and only if $\lceil r_3 \rceil \le r \le \lceil r_3 \rceil +1$.

There is a subtlety that does not exist for the case (a). We show that $r_3$ is a positive integer if and only if $k = 1 + 8 q + 6 q^2$ or $9+16q+6q^2$, for some integer $q\ge 0$. In this case, the term (5) is equal to zero when $(p_1,p_2,p_3) = (k+1+r,k,k-r)$. Therefore, $f(k+1+r,k,k-r)=f(k+2+r,k,k-r-1)$ and f has a maximum value at $(p_1,p_2,p_3)=(k+1+r,k,k-r)$ and $(k+2+r,k,k-r-1)$.

(c) When $n = 3k+2$, we use Mathematica [13] to show the following.

• (c.1) If $(p_1,p_2,p_3) = (k+a+b,k+a,k+2-2a-b)$ with $a > 1$ and $b \ge 0$, the sum of (1) and (2) is negative and so f cannot attain the maximum in this case.

• (c.2) If $(p_1,p_2,p_3) = (k+2+a+b,k-a,k-a-b)$ with $a \ge 1$ and $b > 0$, the sum of (3) and (4) is negative and f cannot attain the maximum in this case.

The cases (c.1) and (c.2) show that the maximum of $f(p_1,p_2,p_3)$ can occur only at $p_2=k$ or $k+1$. We show by Mathematica [13] that

• (c.3) For $r \ge 0$, $f(k+2+r,k,k-r)\le f(k+2+\left\lceil r_1 \right\rceil ,k,k-\left\lceil r_1 \right\rceil )$ and $f(k+1+r,k+1,k-r)\le f(k+1+\left\lceil r_3 \right\rceil ,k+1,k-\left\lceil r_3 \right\rceil )$.

• (c.4) For $r\ge 0$ and $(p_1,p_2,p_3)=(k+2+r,k,k-r)$, (3) is non-positive if and only if $r\ge r_4$ and (4) is non-positive if and only if $r\le r_2$.

By (c.3), the maximum of f occurs at either $(p_1,p_2,p_3) = (k+2+\left\lceil r_1 \right\rceil ,k,k-\left\lceil r_1 \right\rceil )$ or $(k+1+\left\lceil r_3 \right\rceil ,k+1,k-\left\lceil r_3 \right\rceil )$. Since $0< r_3-r_1=\frac{1}{4}\left( 2-\sqrt{60+24k}+\sqrt{40+24k}\right) <\dfrac{1}{2}$, we have $\left\lceil r_1 \right\rceil \le \left\lceil r_3 \right\rceil \le \left\lceil r_1 \right\rceil +1$. Furthermore, by (c.3) and (c.4),

$$\begin{aligned}&f(k+2+\left\lceil r_1 \right\rceil ,k,k-\left\lceil r_1 \right\rceil )\le f \left( k+1+\left\lceil r_3 \right\rceil ,k+1,k-\left\lceil r_3 \right\rceil \right) \nonumber \\&\hbox {if }\left\lceil r_1 \right\rceil \le r_2\ \hbox {or}\ \left\lceil r_1 \right\rceil \ge r_4 \end{aligned}$$

(9)

Conversely, suppose $r_2\le \left\lceil r_1 \right\rceil \le r_4$. Then both (3) and (4) are nonnegative. Hence,

$$\begin{aligned} f(k+2+\left\lceil r_1 \right\rceil ,k,k-\left\lceil r_1 \right\rceil )\ge & {} f\left( k+1+\left\lceil r_1 \right\rceil ,k+1,k-\left\lceil r_1 \right\rceil \right) \quad \hbox {and} \nonumber \\ f\left( k+2+\left\lceil r_1 \right\rceil ,k,k-\left\lceil r_1 \right\rceil \right)\ge & {} f\left( k+2+\left\lceil r_1 \right\rceil ,k+1,k-\left\lceil r_1 \right\rceil -1\right) . \end{aligned}$$

Since $\left\lceil r_1 \right\rceil \le \left\lceil r_3 \right\rceil \le \left\lceil r_1 \right\rceil +1$, it follows that

$$\begin{aligned}&f\left( k+2+\left\lceil r_1 \right\rceil ,k,k-\left\lceil r_1 \right\rceil \right) \ge f\left( k+1+\left\lceil r_3 \right\rceil ,k+1,k- \left\lceil r_3 \right\rceil \right) \nonumber \\&\hbox {if}\quad r_2\le \left\lceil r_1 \right\rceil \le r_4. \end{aligned}$$

(10)

By (9) and (10), f has a unique maximum if neither $r_2$ nor $r_4$ is an integer.

Suppose $r_2$ is an integer for some k. Then we have

$$\begin{aligned} k=\left\{ \begin{array}{ll} \frac{4}{3} + 9 q + 6 q^2&{}\quad \hbox {if }r_2=3q ,\\ 5 + 13 q + 6 q^2&{}\quad \hbox {if }r_2=3q+1 , \\ 10 + 17 q + 6 q^2&{}\quad \hbox {if }r_2=3q+2 . \end{array} \right. \end{aligned}$$

Thus, $r_2$ is an integer for some integer k if and only if $k=5 + 13 q + 6 q^2$ or $10 + 17 q + 6 q^2$ for some integer $q\ge 0$. For $k=5 + 13 q + 6 q^2$, $r_1=\frac{1}{4} \left( -10 + \sqrt{180 + 312 q + 144 q^2}\right)$. We have

$$\begin{aligned} 3+12q = -10 + \sqrt{(13+12q)^2}\le & {} -10 + \sqrt{180+312 q + 144 q^2}\nonumber \\\le & {} -10 + \sqrt{(14+12q)^2} = 4+12q. \end{aligned}$$

So

$$\begin{aligned} \frac{3}{4} + 3q = \frac{1}{4} (3+12q) \le r_1 \le \frac{1}{4}(4+12q) = 1 + 3q \quad \Longrightarrow \quad \left\lceil r_1 \right\rceil =3q+1=r_2. \end{aligned}$$

For $k=10 + 17 q + 6 q^2$, $r_1=\frac{1}{4} \left( -10 + \sqrt{300 + 408 q + 144 q^2}\right)$. We have

$$\begin{aligned} 7+12q = -10 + \sqrt{(17+12q)^2}\le & {} -10 + \sqrt{300 + 408 q + 144 q^2}\nonumber \\\le & {} -10 + \sqrt{(18+3q)^2} = 8 + 12q. \end{aligned}$$

So

$$\begin{aligned} \frac{7}{4} + 3q = \frac{1}{4}(7+12q) \le r_1 \le \frac{1}{4}(8+12q) = 2 + 3q \quad \Longrightarrow \quad \left\lceil r_1 \right\rceil =3q+2=r_2. \end{aligned}$$

The proof for the case when $r_4$ is an integer is similar. $\square$

1.3 Proof of Theorem 4

We first prove the following Lemma.

Lemma 1

Let $r_0, r_1,r_2,r_3$ and $r_4$ be the numbers defined in Theorem 3 and let

$$\begin{aligned} {\hat{r}}_0= & {} \frac{1}{2} \left( -3+\sqrt{3+3 (k+1)+\sqrt{12 + 20 (k+1) + 9 (k+1)^2}}\right) \nonumber \\= & {} \frac{1}{2} \left( -3+\sqrt{6+3 k+\sqrt{41 + 38 k + 9 k^2}}\right) . \end{aligned}$$

Then

$$\begin{aligned} r_1< r_3< r_0< {\hat{r}}_0 < r_1 +1 < r_3 + 1. \end{aligned}$$

(11)

Hence, $\{\left\lceil r_0 \right\rceil ,\left\lceil {\hat{r}}_0 \right\rceil \} \subseteq \{\left\lceil r_3\right\rceil , \left\lceil r_3\right\rceil + 1\}$. Furthermore, $\left\lceil r_0\right\rceil = \left\lceil r_3\right\rceil + 1$ if $r_3$ is an integer and $\lceil {\hat{r}}_0 \rceil = \left\lceil r_1\right\rceil + 1$ if $r_2 \le \lceil r_1 \rceil \le r_4$.

Proof

With the help of Mathematica [13], we can show that

$$\begin{aligned}&{\left( r_0-r_3\right) \left( 1+\sqrt{10+6 k}+\sqrt{3+3 k+\sqrt{12+k (20+9 k)}}\right) }\nonumber \\&\qquad \times \left( 6+3 k+\sqrt{12+k (20+9 k)}+2 \sqrt{3+3 k+\sqrt{12+k (20+9 k)}}\right) \nonumber \\&\qquad \times \left( 3+k+\sqrt{12+k (20+9 k)}\right. \nonumber \\&\qquad \left. +\,\sqrt{12+k (20+9 k)} \sqrt{3+3 k+\sqrt{12+k (20+9 k)}}\right) \nonumber \\&\quad = 2 \left[ 3 \left( 13+4 \sqrt{12+k (20+9 k)}+8 \sqrt{3+3 k+\sqrt{12+k (20+9 k)}}\right) \right. \nonumber \\&\qquad +\,k \left\{ 10 \left( 11+2 \sqrt{12+k (20+9 k)}+4 \sqrt{3+3 k+\sqrt{12+k (20+9 k)}}\right) \right. \nonumber \\&\qquad \left. \left. +\,k \left( 95+27 k+9 \sqrt{12+k (20+9 k)}+18 \sqrt{3+3 k+\sqrt{12+k (20+9 k)}}\right) \right\} \right] \nonumber \\&\quad >0,\\&{({\hat{r}}_0- r_0)\left( \sqrt{3+3 k+\sqrt{12+k (20+9 k)}}+\sqrt{6+3 k+\sqrt{41+k (38+9 k)}}\right) } \nonumber \\&\qquad \times \left( 3+\sqrt{12+k (20+9 k)}+\sqrt{41+k (38+9 k)}\right) \nonumber \\&\quad =19+9 k+3 \sqrt{41+k (38+9 k)}>0. \end{aligned}$$

and

$$\begin{aligned}&{\left( r_1 +1- {\hat{r}}_0 \right) \left( \sqrt{15+6 k}+\sqrt{6+3 k+\sqrt{41+k (38+9 k)}}\right) }\nonumber \\&\qquad \times \left( 9+3 k+\sqrt{41+k (38+9 k)}\right) \nonumber \\&\quad =20+8 k >0. \end{aligned}$$

Hence, Eq. (11) follows. $\square$

Now we are ready to prove the theorem. Notice that by Theorem 3,

$$\begin{aligned} f^*(n) = \left\{ \begin{array}{ll} f(k+\lceil r_0\rceil ,k,k-\lceil r_0\rceil ) &{} \hbox {if } n = 3k, \\ f(k+ 1 +\lceil r_3\rceil ,k,k-\lceil r_3\rceil ) &{} \hbox {if } n = 3k+1, \\ f(k+1+\lceil r_3\rceil ,k+1,k-\lceil r_3\rceil ) &{} \hbox {if } n = 3k+2 \hbox { with } \left\lceil r_1\right\rceil \le r_2\ \hbox { or } \left\lceil r_1\right\rceil \ge r_4, \\ f(k+2+\lceil r_1\rceil ,k,k-\lceil r_1\rceil ) &{} \hbox {if } n = 3k+2 \hbox { with } r_2 \le \left\lceil r_1\right\rceil \le r_4,\\ f(k+\lceil {\hat{r}}_0\rceil ,k,k-\lceil {\hat{r}}_0\rceil ) &{} \hbox {if } n = 3k+3. \\ \end{array} \right. \end{aligned}$$

Furthermore, $f(k+ 1 +\lceil r_3\rceil ,k,k-\lceil r_3\rceil ) = f(k+ 2 +\lceil r_3\rceil ,k,k-1-\lceil r_3\rceil )$ if $r_3$ is an integer.

From $n = 3k$ to $n+1 = 3k+1$, suppose $(k+\lceil r_0\rceil ,k,k-\lceil r_0\rceil ) = (p_1,p_2,p_3)$. By Lemma 1, $\left\lceil r_0 \right\rceil = \left\lceil r_3\right\rceil$ or $\left\lceil r_3\right\rceil + 1$. Then

$$\begin{aligned} (k+1+\lceil r_3\rceil ,k,k-\lceil r_3\rceil ) = \left\{ \begin{array}{ll} (p_1+1,p_2,p_3) &{} \hbox {if } \lceil r_0 \rceil = \lceil r_3 \rceil , \\ (p_1,p_2,p_3+1) &{} \hbox {if } \lceil r_0 \rceil = \lceil r_3 \rceil +1. \end{array} \right. \end{aligned}$$

From $n = 3k+1$ to $n+1 = 3k+2$, suppose $(k+1+\lceil r_3\rceil ,k,k-\lceil r_3\rceil ) = (p_1,p_2,p_3)$. By Lemma 1, $\left\lceil r_3 \right\rceil = \left\lceil r_1\right\rceil$ or $\left\lceil r_1\right\rceil + 1$. Then

$$\begin{aligned} (k+1+\lceil r_3\rceil ,k+1,k-\lceil r_3\rceil )= & {} (p_1,p_2+1,p_3) \quad \hbox {and} \\ (k+2+\lceil r_1\rceil ,k,k-\lceil r_1\rceil )= & {} \left\{ \begin{array}{ll} (p_1+1,p_2,p_3) &{} \hbox {if } \lceil r_3 \rceil = \lceil r_1 \rceil , \\ (p_1,p_2,p_3+1) &{} \hbox {if } \lceil r_3 \rceil = \lceil r_1 \rceil +1. \end{array} \right. \end{aligned}$$

Further, suppose $r_3$ is an integer, then $\lceil r_3 \rceil = \lceil r_1 \rceil$, and hence, we must have $r_2 \le \lceil r_1 \rceil \le r_4$ and $f^*(3k+2) = f(k+2+\lceil r_1\rceil ,k,k-\lceil r_1\rceil )$. If $(p_1,p_2,p_3) = (k+2+\lceil r_3\rceil ,k,k-1-\lceil r_3\rceil )$, then

$$\begin{aligned} (k+2+\lceil r_1\rceil ,k,k-\lceil r_1\rceil ) = (p_1,p_2,p_3+1). \end{aligned}$$

Furthermore, if $f^*(3k+1) = f(p_1^1,p_2^1,p_3^1) = f(p_1^2,p_2^2,p_3^2)$ with $(p_1^1,p_2^1,p_3^1) = (k+1+\lceil r_3\rceil ,k,k-\lceil r_3\rceil )$ and $(p_1^2,p_2^2,p_3^2) = (k+2+\lceil r_3\rceil ,k,k-1-\lceil r_3\rceil )$, then $f^*(3k+2) = f(p_1^2,p_2^1,p_3^1)$.

Now from $n = 3k+2$ to $n+1 = 3k+3$, suppose $\left\lceil r_1\right\rceil \le r_2$ or $\left\lceil r_1\right\rceil \ge r_4$ and $(k+1+\lceil r_3\rceil ,k+1,k-\lceil r_3\rceil ) = (p_1,p_2,p_3)$. By Lemma 1, $\left\lceil {\hat{r}}_0 \right\rceil = \left\lceil r_3\right\rceil$ or $\left\lceil r_3\right\rceil + 1$. Then

$$\begin{aligned} (k+1 + \lceil {\hat{r}}_0 \rceil ,k+1,k+1-\lceil {\hat{r}}_0\rceil ) = \left\{ \begin{array}{ll} (p_1,p_2,p_3+1) &{} \hbox {if } \lceil {\hat{r}}_0 \rceil = \lceil r_3\rceil , \\ (p_1+1,p_2,p_3) &{} \hbox {if } \lceil {\hat{r}}_0 \rceil = \lceil r_3 \rceil + 1. \end{array}\right. \end{aligned}$$

Now suppose $r_2 \le \lceil r_1 \rceil \le r_4$ and $(k+2+\lceil r_1\rceil ,k,k-\lceil r_1\rceil ) = (p_1,p_2,p_3)$. By Lemma 1, $\lceil {\hat{r}}_0\rceil = \lceil r_1\rceil + 1$. Then

$$\begin{aligned} (k+1 + \lceil {\hat{r}}_0 \rceil ,k+1,k+1-\lceil {\hat{r}}_0\rceil ) = (p_1,p_2+1,p_3). \end{aligned}$$

Furthermore, if $f^*(3k+2) = f(p_1^1,p_2^1,p_3^1) = f(p_1^2,p_2^2,p_3^2)$ with $(p_1^1,p_2^1,p_3^1) = (k+1+\lceil r_3\rceil ,k+1,k-\lceil r_3\rceil )$ and $(p_1^2,p_2^2,p_3^2) = (k+2+\lceil r_1\rceil ,k,k-\lceil r_1\rceil )$, then $f^*(3k+3) = f(p_1^2,p_2^1,p_3^1)$. Thus, the result follows. $\square$