During a description of zero-dimensional self-dual $\text{GF}(4)$ quantum codes in "On self-dual quantum codes, graphs, and Boolean functions" by L.E. Danielsen, it states:

A zero-dimensional stabilizer code with high distance represents a single quantum state which is robust to error, sometimes called a stabilizer state. Codes of higher dimension can be constructed from zero-dimensional quantum codes...

My question has two parts:

Firstly, I am confused by what is meant here by a "single quantum state". To my understanding, the passage seems to confuse the encoded 0-dimensional qubit state (which is robust to Pauli error), i.e. a single "qubit" state, with the single "stabilizer" state which is represented by the generators of the code (which is not robust to error, as a single Hadamard on any qubit would be sufficient to take it to a completely different stabilizer state). Is this the case, or am I misunderstanding something here?

Secondly, what form do these constructions take in practice? Is this simply achieved by not enforcing one of the stabilizer's generators, or are there other more general methods? Furthermore, if there are multiple construction methods, what are their advantages or disadvantages (both in the codes they create and/or the complexity of the construction)?


1 Answer 1


For the first part of the question, I think that with "single quantum state", the author is referring to the encoded quantum state, which will be an state formed by $n$-qubits. Such state is called stabilizer state because the encoding operation takes the input state, in this case $|0\rangle^{\otimes n}$, and takes it to a state in the codespace defined by the stabilizers. This code space is defined to be the $+1$ simultaneous eigenspace of the stabilizer operators, and in general has dimension $2^k$, so as in this case $k=0$, the dimension of this codespace will be $1$. Consequently, the author is trying to state that such zero-dimensional stabilizer codes refer to the basis of such $1$ dimensional subspace, which will be just a single quantum state.

Also the author says that such state is robust to errors because an stabilizer error correction code is being applied to the state, and so errors will be correctable. Obviously a Hadamard gate would change the encoded state, but I think that with robust, the author is trying to say that the state can be corrected after the appearance of Pauli errors, that is $\{X,Y,Z\}^{\otimes n}$ operators applied to the state. This robustness obviously comes from the fact that an error-correction code is being applied to the $n$-ancilla qubit input state.

Then this codes would be obtainded by finding such encoding operator that would take the input $n$-state to the $+1$ simultaneous eigenspace that the $n$ stabilizer generators define (Note that in general ${n-k}$ generators are needed and that here $k=0$). I don't know what do you exactly want to say with the frase "enforcing one of the stabilizer's generators". To find out the encoder unitary, most of the times, combinations of so-called Clifford gates are used, there are differents methods and algorithms in literature that are useful for finding the exact combination.

  • $\begingroup$ Of course! I was forgetting that by "robust to errors" they were referring to the ability to detect (and correct) some set of local errors (such as the application of a Hadamard) through parity checks on the state. However, I don't understand your answer to the second question. And for clarification, by "not enforcing one of the stabilizer's generators" I am referring to simply creating a $n$-qubit state that is stabilized by $n-1$ generators, thus defining a logical qubit subspace defined by some logical $X$ and $Z$ operators. $\endgroup$ Jun 11, 2018 at 8:43
  • $\begingroup$ I wrote wrong the number of generators needed, I edited it in the answer. What I was trying to say is that if you use $n-1$ then it would not be a zero dimensional stabilizer code, as that way $k=1$. Afterwards I just say that the way to construct a quantum circuit that takes an state $|0\rangle^{\otimes n}$ to the codespace is done by selecting a combination of Clifford gates, that is Hadamard, Phase and CNOT gates. Algorithms to create the correct combinations of such gates exist, see en.wikipedia.org/wiki/Gottesman%E2%80%93Knill_theorem . $\endgroup$ Jun 11, 2018 at 8:50

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