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?


  • 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 $.


1 Answer 1


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


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.

  • 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

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