Theorem 2 of [1] states:

Suppose $C$ is an additive self-orthogonal sub-code of $\textrm{GF}(4)^n$, containing $2^{n-k}$ vectors, such that there are no vectors of weight $<d$ in $C^\perp/C$. Then any eigenspace of $\phi^{-1}(C)$ is an additive quantum-error-correcting code with parameters $[[n, k, d]]$.

where here $\phi: \mathbb{Z}_2^{2n} \rightarrow \textrm{GF}(4)^n$ is the map between the binary representation of $n$-fold Pauli operators and their associated codeword, and $C$ is self-orthogonal if $C \subseteq C^\perp$ where $C^\perp$ is the dual of $C$.

This tells us that each additive self-orthogonal $\textrm{GF}(4)^n$ classical code represents a $[[n, k, d]]$ quantum code.

My question is whether the reverse is also true, that is: is every $[[n, k, d]]$ quantum code represented by an additive self-orthogonal $\textrm{GF}(4)^n$ classical code?

Or equivalently: Are there any $[[n, k, d]]$ quantum codes that are not represented by an additive self-orthogonal $\textrm{GF}(4)^n$ classical code?

[1]: Calderbank, A. Robert, et al. "Quantum error correction via codes over GF (4)." IEEE Transactions on Information Theory 44.4 (1998): 1369-1387.

  • $\begingroup$ Aren't the stabilizer codes such as Toric codes or color codes self orthogonal? there is an isomorphism between both!! $\endgroup$
    – Trect
    Jun 20 '18 at 18:08
  • $\begingroup$ Sorry, I don't understand your point. I am looking for a quantum code that is not self-orthogonal, not examples of those that are. $\endgroup$ Jun 21 '18 at 7:42
  • $\begingroup$ What is the question exactly? As far as I have understood in the question you are trying to find quantum codes that represent classical code? $\endgroup$ Jun 21 '18 at 11:07
  • $\begingroup$ No, I am trying to find out if all quantum codes (on qubits) have equivalent classical codes. For clarity, I have highlighted the exact question and added another rephrasing. $\endgroup$ Jun 21 '18 at 12:23

The additive self-orthogonal constraint on the classical codes in order to create stabilizer quantum codes is needed due to the fact that the stabilizer generators must commute between them in order to create a valid code space. When creating quantum codes from classical codes, the commutation relationship for the stabilizers is equivalent to having a self-orthogonal classical code.

However, quantum codes can be constructed from non-self-orthogonal classical codes over $GF(4)^n$ by means of entanglement-assistance. In this constructions, an arbitrary classical code is selected, and by adding some Bell pairs in the qubit system, commutation between the stabilizers is obtained.

This entanglement-assisted paradigm for constructing QECCs from any classical code is presented in arXiv:1610.04013, which is based on the paper "Correcting Quantum Errors with Entanglement" published in Science by Brun, Devetak and Hsieh.


Your question can in parts be seen as a notational issue.

The notation $[[n,k,d]]_D$ is often (but not always) reserved for codes is of stabilizer type. As the paper by Calderbank et al shows, qubit stabilizer codes are equivalent to additive self-orthogonal GF(4)^n classical codes. This construction generalizes, see Refs. Ketkar et al. and Ashikhmin and Knill. Here, the dimension of the code is $D^k$ for quDits.

Some authors use $((n,K,d))_D$ to denote (stabilizer and non-stabilizer) codes that have dimension $K$. Note that $K$ is then not necessarily a power of $D$.

Rains et al. were the first to construct a $((5,6,2))$ code that is of non-stabilizer type, and which is provably better than any stabilizer code on five qubits: in comparison, the best one has parameters $[[5,2,2]]$, and is thus of dimension $2^2 = 4 < 6$. You'll find more examples for non-additive quantum codes in Yu et al., Smolin et al., and Grassl and Beth.


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.