How to calculate distance of $k=0$ stabilizer code?

Question

This could be seen as a followup to the question "How to calculate the distance of stabilizer code?". Summarizing the accepted answer : distance is the minimum weight of the set $$E = \bigl\{e : e \not \in S, e \in \mathrm{Nor}(P_N,S)/(\pm I) \bigr\}$$ where $S$ is the stabilizer group (generated by $K_n$'s in the previous question), and $\mathrm{Nor}(P_N,S)$ is its normalizer in the Pauli group of order $2^{2N+1}$ (where $N$=number of qubits; using real version of group here).

My question is the following: does this hold for $k=0$ stabilizer codes? I suspect that it doesn't always hold but can't find a reference for it... it does seem to work for most cases, but some simple counter examples are also easy to find : take the GHZ state $\tfrac{1}{\sqrt 2}\bigl(\lvert00\rangle + \lvert11\rangle\bigr)$, with $K_1=X_1X_2$ and $K_2=Z_1Z_2$. In this case, $\mathrm{Nor}(P,S)=\pm S$, so the set $E$ is empty. Something is obviously broken in this process: I think that the distance should be 2. What's going on here?

Answer 1

Note that in the case $k = 0$, the stabiliser 'code' is a $2^0 = 1$ dimensional subspace of the Hilbert space, which is to say that it consists of a single stabiliser state. This will have somewhat peverse effects on the features such as the 'distance' of the code.

The "code distance" is ultimately defined in terms of the minimum weight of a Pauli operator $E$ which is not 'detectable' (by which I mean, distinguishable from the identity) according to the Knill–Laflamme conditions: $$ \langle \psi_j \rvert E \lvert \psi_k \rangle = C_E \delta_{j,k} $$ where $\lvert \psi_j \rangle, \lvert \psi_k \rangle$ are states in the code. In the case of a 1-dimensional subspace, there is only a single state $\lvert \psi \rangle =: \lvert \psi_0 \rangle$. Thus we would take $j,k \in \{ 0 \}$, so that the $\delta_{j,k}$ term is always equal to $1$. But that means that by simply defining $C_E = \langle \psi \rvert E \lvert \psi \rangle$, the Knill–Laflamme condition is always satisfied. Thus, the 'distance' of the code is defined for a $k = 0$ stabiliser code as the minimum over the empty set.

Using the less abstract approach for stabiliser codes, of considering weights of Pauli operators which are in the normaliser of the code, bear in mind that we're talking then of operators which map the code-space to itself, but are not proportional to a member ofthe stabiliser group. But for $k = 0$ operators which map the state $\lvert \psi \rangle$ to itself are necessarily proportional to stabilisers, so no such operator exists. Again, we are considering the minimum weight over an empty set of operators.

According to your conventions, it could possibly be sensible to talk about the distance as being infinite; but in practise it would be better to say that the distance is undefined.

Answer 2

In the classic paper https://arxiv.org/pdf/quant-ph/9608006.pdf, on page 10, the distance of an $[n,0]$ code is defined as the smallest non-zero weight of any stabilizer in the code. The physical interpretation for this definition given is, "An $[[n, 0, d]]$ code is a quantum state such that, when subjected to a decoherence of $[(d − 1)/2]$ coordinates, it is possible to determine exactly which coordinates were decohered."