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In the following answer to a recent question, it is claimed that when we concatenate two error correcting codes of distances $d_1$ and $d_2$, the greatest distance that we can prove without assuming any structure of the code is $d_1d_2/2$ while if we know something about the structure, e.g. that it's a CSS code, the distance is at least $d_1d_2$.

Are there any known examples of pairs of codes for which the combined distance is less than $d_1d_2$? For preference, the codes should be on qubits and expressed within the stabilizer formalism, if that is even possible. The smaller the better!

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  • $\begingroup$ Do you want an example only for error-correcting code? Otherwise, a simple example would be concatenating $[\![4,2,2]\!]$ code with $[\![4,2,2]\!]$ yeilds $[\![8,2,2]\!]$ code. $\endgroup$
    – FDGod
    Dec 5, 2023 at 21:50
  • $\begingroup$ that would do, if you could provide me with some details (it's not a code I know) $\endgroup$
    – DaftWullie
    Dec 6, 2023 at 8:45

1 Answer 1

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$[\![4,2,2 ]\!]$ code

$[\![4,2,2 ]\!]$ is the smallest quantum error-detecting code which can detect a single qubit error. Let's work with the following generators-

$$\begin{align} g_1 &= XZZX \,,\tag{1.1}\\ g_2 &= YXXY \,.\tag{1.2} \end{align}$$

Let's choose the logical operators- $$\begin{align} \overline{X_1} &= XIYY \,, \tag{2.1}\\ \overline{X_2} &= XIXZ \,,\tag{2.2}\\ \overline{Z_1} &= YZYI \,,\tag{2.3}\\ \overline{Z_2} &= IXZZ \,.\tag{2.4} \end{align}$$


Concatenating $[\![4,2,2 ]\!]$ with $[\![4,2,2 ]\!]$

Let $\mathcal{C}_1 \equiv [\![4,2,2 ]\!] \equiv [\![n_1,k_1,d_1 ]\!]$ & $\mathcal{C}_2 \equiv [\![4,2,2 ]\!] \equiv [\![n_2,k_2,d_2 ]\!]\,.$

Let's concateate $\mathcal{C}_1$ & $\mathcal{C}_2$. Let the resultant code be $\mathcal{C} \equiv [\![n,k,d ]\!]\,.$

Let's first encode two qubits into four qubits, which are then partitioned into two blocks. Block 1 contains qubits $i$ and $ii$, and block 2 contains qubits $iii$ and $iv$.

Now, encode the $j^{\text{th}}$ block, say $B(j)$ using code $\mathcal{C}_2$.

The qubits in block $B(1)$ will be labelled $1,2,3,4$. The qubits in the block $B(2)$ will be labelled $5,6,7,8$.

Thus, final code $\mathcal{C}$ has $n=8$.

Each block $B(j)$ gets the copy of generators of $\mathcal{C}_2$. Hence, we can write first four generators for $\mathcal{C}\,,$ where block $B(1)$ have

$$\begin{align} g_1 &= X_1Z_2Z_3X_4 \,,\tag{3.1}\\ g_2 &= Y_1X_2X_3Y_4 \,.\tag{3.2} \end{align}$$

and block $B(2)$ have

$$\begin{align} g_3 &= X_5Z_6Z_7X_8 \,,\tag{3.3}\\ g_4 &= Y_5X_6X_7Y_8 \,.\tag{3.4} \end{align}$$

The two remaining generators are the images of $\mathcal{C}_1$ under $\mathcal{C}_2$. By doing some algebra (using the logical operators given in Eqs.$(2.1)$ to $(2.4)$), you can see that $\mathcal{C}_2$ maps the generators of $\mathcal{C}_1$ as follows-

$$\begin{align} g_5 &= X_iZ_{ii}Z_{iii}X_{iv} = X_1 X_2 X_3 X_4 Z_5 Z_6 Z_7 Z_8\,,\tag{3.5}\\ g_6 &= Y_iX_{ii}X_{iii}Y_{iv} = Y_1 Z_2 X_3 X_4 I_5 X_6 I_7 Y_8\,.\tag{3.6} \end{align}$$

The number of generators, $r = n-k = 6$, hence $k=2$, as expected-

encoding

We can check that all single-qubit errors anticommute with at least one generator of $\mathcal{C}$.

However, weight two errors like $Z_1Y_2$ commutes with all generators and $Z_1 Y_2 \in N(S) \setminus S\,.$ Hence, $d=2$.


Conclusion

Concatenating $[\![4,2,2 ]\!]$ & $[\![4,2,2 ]\!]$ results in the $[\![8,2,2 ]\!]$ code, and thus satisfying your required condition, $d < d_1 d_2\,.$


Note$^1$

The result about distances after code concatenation is much more general than what you have linked.

Concatenating two quantum stabilizer codes $[\![n_1,k_1,d_1]\!]$ (as outer layer) and $[\![n_2,k_2,d_2]\!]$(as inner layer) gives you a resultant code $[\![n,k,d]\!]$ where

Case(1):

  • $n = n_1 n_2\,,$
  • $k = k_1 k_2 \,,$
  • $d \geq d_1 d_2\,,$

holds true if $n_1 (\text{mod } {k_2}) \neq 0\,.$

Case(2):

  • $n = \frac{n_1 n_2}{k_2}\,,$
  • $k = k_1 \,,$
  • $d \geq \frac{d_1 d_2}{k_2}\,,$

holds true if $n_1(\text{mod } {k_2}) =0\,.$

So, it's preferable to pick codes for which $k_2$ does not divide $n_1$.


1: Reference for the note$\to$ Theorems 3.5 & 3.6 in Frank Gaitan (2008), Quantum Error Correction and Fault Tolerant Quantum Computing, Taylor & Francis Group.

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    $\begingroup$ Do you have a reference for your note? $\endgroup$
    – DaftWullie
    Dec 7, 2023 at 7:25
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    $\begingroup$ @DaftWullie Theorems 3.5 & 3.6 in Frank Gaitan (2008), Quantum Error Correction and Fault Tolerant Quantum Computing, Taylor & Francis Group. $\endgroup$
    – FDGod
    Dec 7, 2023 at 7:51
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    $\begingroup$ hmmm I can't seem to lay my hands easily on that book, so I'm trying to reconstruct the result for myself. I think I see it, however, are you missing a case: where $n_1$ and $k_2$ share a common factor that's not $1,k_2$? I assume you get a similar bound of $n=\frac{n_1n_2}{\text{gcd}(n_1,k_2)}$ and $d\geq \frac{d_1d_2}{\text{gcd}(n_1,k_2)}$? $\endgroup$
    – DaftWullie
    Dec 8, 2023 at 10:41
  • $\begingroup$ @DaftWullie This is interesting. Can you elaborate, what concatenation procedure you are using here? I tried few random examples, and it seems that then also get- $$k = \frac{k_1 k_2}{\text{gcd}(n_1, k_2)} $$ My concern, in this case, is that- can you get generators for $\mathcal{C}$ with this type of concatenation procedure, starting with generators for $\mathcal{C}_1$ and $\mathcal{C}_2$? Whereas getting generators for $\mathcal{C}$ with a concatenation procedure where $n = n_1 n_2$ is fairly straightforward. $\endgroup$
    – FDGod
    Dec 9, 2023 at 2:37
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    $\begingroup$ I think the construction works exactly as you'd expect it to. Let $n_1=pr$ and $k_2=sr$ where $r=\text{gcd}(n_1,k_2)$. For you first code, you've got a stabilizer which you can block into sections of length $r$. You can collect $s$ of these together and each Pauli operator in those $sr$ qubits translates into a logical operator of $n_2$ qubits. There may be one last logical qubit that you have to pad a bit to make up the numbers. $\endgroup$
    – DaftWullie
    Dec 11, 2023 at 7:38

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