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In quantum error correction, two stabilizer codes are considered to be equivalent if they differ by a local Clifford transformation or a qubit permutation. Represented as a tableau, these set of transformations establish an equivalence class between different tableaus that represent equivalent codes.

The literature suggests that, using this notion of code equivalence, there is a unique $[\![5,1,3]\!]$ code with 4 cyclic stabilizer generators $g_1 = XZZX1,~g_2 = 1XZZX, \dots $ and logical basis $X_L = XXXXX$, $Z_L = ZZZZZ$. I have no problem in generating a tableau for such a representation of the code, and I can also verify that it is indeed a $d=3$ code by checking that the (Knill-Laflamme) error-correcting conditions are satisfied for a standard symmetric depolarizing channel.

However, I have also been able to find a $[\![5,1,3]\!]$ code (i.e. I can satisfy the same Knill-Laflamme equations as in the previous code) represented by a tableau in which both $X_L$ and $Z_L$ are weight 3 Pauli strings. To me it doesn't seem that such a code can be mapped to the standard $[\![5,1,3]\!]$ code that I've described in the previous paragraph using only local Clifford transformations. The reason is that I don't see a way in which the weight of the logical operators can be changed using these transformations.

My question(s) are thus the following:

  1. In which sense is the $[\![5,1,3]\!]$ code unique?

  2. Is it true that two different tableaus can only represent equivalent codes if all the weights of their Pauli strings are the same?

I'm aware of https://physics.stackexchange.com/questions/283008/equivalent-quantum-codes but it doesn't answer my question.

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  • $\begingroup$ It might help if you showed us your alternative code... (Also, remember that stabilizers and logical operators can be re-written as products of each other. So $X_L g_1$ is just as good a way of writing $X_L$ (and is a weight 3 Pauli string).) $\endgroup$
    – DaftWullie
    Mar 1, 2023 at 13:11
  • $\begingroup$ Sure, but what is the circuit that maps the tableau with $X_L$ to a new tableau where the old $X_L$ gets replaced by $X_L g_1$ ? Understanding how it's done or whether it is even necessary to begin with might be the root of my confusion. $\endgroup$
    – Jan Olle
    Mar 1, 2023 at 13:29
  • $\begingroup$ You don't need a circuit. It is the same code. $\endgroup$
    – DaftWullie
    Mar 1, 2023 at 13:55

1 Answer 1

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TL;DR: 1. The code is unique under the stated equivalence. 2. No.

Weight of a logical operator is ill-defined

The crux of the issue is that the weight of a logical operator is ill-defined. Logical operator is not a single Pauli operator in the $n$-qubit Pauli group $\mathcal{P}_n$, but a coset $X_L\mathcal{S}$ of the stabilizer group $\mathcal{S}$ where $X_L$ is some representative of the logical operator.

For example, in the $[\![5,1,3]\!]$ code folks typically choose $X_L=XXXXX$, but $g_1=XZZXI\in\mathcal{S}$, so $X_Lg_1\equiv IYYIX\in X_L\mathcal{S}$ is another valid representative of the logical operator $X_L\mathcal{S}$. Clearly, the two representatives have different weight.

A more dramatic example occurs in an odd distance $d$ planar surface code where people typically represent $X_L$ as a weight $d$ chain of physical $X$ operators along a straight line connecting two boundaries. However, an alternative representative is the operator that applies $X$ to all $d^2$ data qubits of the code.

Weights in a tableau are not fixed under choice of generators

It is true that weights of the generators listed in a tableau remain unchanged under local Cliffords and qubit permutations. However, they do not remain fixed under multiplication of generators. Consequently, it is in fact not true that two different tableaus can only represent equivalent codes if all the weights of their Pauli strings are the same.

As a simple counterexample, consider the following two tableaus representing the group stabilizing $\mathrm{span}(|000\rangle,|001\rangle)$ $$ \begin{array}{ccc} ZII\\ IZI \end{array}\quad\quad\quad \begin{array}{ccc} ZZI\\ IZI \end{array} $$ The issue is that if $h,g_1,g_2,\dots,g_k$ are independent generators of a group $\mathcal{G}$ then so are $h,hg_1,hg_2,\dots,hg_k$ and the weights of $g_i$ and $hg_i$ are generally not the same.

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