9
$\begingroup$

Is it possible to encode a single logical qubit in 3 physical qubits so that the resulting code has distance 2?

In other words, does a $[\![3,1,2]\!]$ code exist?

Comments:

  • No $[\![3,1,2]\!]$ stabilizer code exists. This simple argument is given for example below equation (14) in https://arxiv.org/abs/quant-ph/0512170. In general it shows that no $[\![2n+1,2n-1,2]\!]$ stabilizer code exists.

  • No $[\![3,1,3]\!]$, $[\![3,2,2]\!]$, or $[\![2,1,2]\!]$ code exists; this follows from the quantum singleton bound $$ n-k\geq 2(d-1). $$

  • There is a well known $[\![4,2,2]\!]$ stabilizer code with stabilizer $ XXXX,ZZZZ $. This is the smallest known quantum code with $ d \geq 2 $.

$\endgroup$

1 Answer 1

5
+50
$\begingroup$

No, there is no $[\![3,1,2]\!]$ code.

Background

As in classical error correction, existence of quantum codes can often be ruled out using a range of inequalities that codes must satisfy. Indeed, the question already makes use of the singleton bound to rule out existence of similar codes. Below, I will use equations and inequalities in theorem $10$ on page $7$ in Quantum shadow enumerators. The paper derives the quantum analogue of the shadow of a classical binary code and uses it to extend the results in Quantum MacWilliams Identities which in turn derives quantum versions of MacWilliams identities relating the weight distribution of a code $C$ and its dual $C^\perp$.

No $[\![3,1,2]\!]$ code exists

Theorem $10$ lists a few equations and ineqalities that relate weights$^1$ $A_i$ and $B_i$ and shadow weights$^2$ $S_i$ for $i=0,\dots,n$ to each other and to the parameters of the code $[\![n,k,d]\!]=(\!(n,K=2^k,d)\!)$. If the code exists, then the relations are satisfied. Suppose then that there is a code $[\![3,1,2]\!]=(\!(3,2,2)\!)$. We will use equations and inequalities in theorem $10$ to derive a contradiction. It is sufficient to compute $A_0,\dots,A_3$, $B_0$, $B_1$ and $S_0$ since at this point we will obtain a contradiction with one of the inequalities - namely, $S_0\ge 0$ - proving that no $[\![3,1,2]\!]$ code exists.

First, we compute Krawtchouk polynomials $P_i(x):=P_i(x,n=3)$

$$ \begin{align} P_0(x)&=1\\ P_1(x)&=9-4x. \end{align}\tag1 $$

We won't need $P_2(x)$ and $P_3(x)$. Next, we use the polynomials to write down equations for $B_0$, $B_1$ and $S_0$ in terms of $A_i$

$$ \begin{align} 8B_0 &= A_0+A_1+A_2+A_3\\ 8B_1 &= 9A_0+5A_1+A_2-3A_3\\ 8S_0 &= A_0-A_1+A_2-A_3. \end{align}\tag2 $$

Now, by the first relation in theorem $10$, we have $A_0=K^2=4$ and by the fourth one $2B_0=A_0$ and $2B_1=A_1$. Substituting into $(2)$, we obtain

$$ \begin{align} A_1+A_2+A_3&=12\\ A_1+A_2-3A_3&=-36 \end{align}\tag3 $$

so $A_1+A_2=0$ and $A_3=12$. However, from the second relation in theorem $10$, we know that $A_1\ge 0$ and $A_2\ge 0$, so $A_1=A_2=0$. But then $S_0=-1$ and we arrive at a contradiction with the last relation in theorem $10$, namely $S_0\ge 0$. Therefore, no code $[\![3,1,2]\!]$ exists.


$^1$ For definition of $A_i$ and $B_i$ see discussion on page $2$ in Quantum shadow enumerators.
$^2$ For definition of $S_i$ see theorem $4$ on page $4$ in Quantum shadow enumerators and preceding discussion.

$\endgroup$
1
  • 1
    $\begingroup$ Nice! The shadow enumerators paper, below theorem 10, mentions the bounds in the GF(4) paper arxiv.org/abs/quant-ph/9608006. The GF(4) paper has a table, made from the same type of inequalities used in your answer, with bounds on maximum distance given $ n $ and $ k $ (the table was originally meant for additive codes, but it turns out the bounds are valid for all codes, except where they are marked by $ \beta $ in which case the bound is +1 for the non-additive case). In particular, the table states that every $ [[3,1,d]] $ code is $ d=1 $, confirming what you have worked out above. $\endgroup$ Aug 5 at 12:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.