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I guess code distance is the same as "Hamming distance" which is related to the overlap between different "codewords" in the codespace?

More technically, my understanding is that a stabilizer quantum-error-correcting code (QECC) having code distance $d$ implies that any operator $\mathcal{E}$ capable of producing a logical error must have support (i.e., act nontrivially) on at least $d$ qubits? This implies that every logical operator acts nontrivially on at least $d$ qubits.

I'm curious if there are other (equivalent) definitions. Specifically, I'm wondering if the minimum number of errors applies equally to deletion / erasure errors (on at most $d-1$ qubits). However, I guess deletion errors aren't the same as logical errors (i.e., they don't change between codewords), so you can correct at most $t = (d-1)/2$ of them, so that's the maximum number of deletion errors that can be tolerated?

If so, then I would expect the following statement to hold: Given a QECC with code distance $d$ and nominal logical operators $\{ \overline{X}, \overline{Y}, \overline{Z}\}$, then for any region $C$ that contains $\left| C \right| = d-1$ qubits (or more likely $\left| C \right| = t = (d-1)/2$ qubits), there exist elements $ \mathcal{S}_x $, $\mathcal{S}_y$, $\mathcal{S}_z$ of the stabilizer group such that $\{ \mathcal{S}_x \overline{X} , ~\mathcal{S}_y \overline{Y} , ~ \mathcal{S}_z \overline{Z} \}$ are also valid logical operators, each of which acts trivially on every qubit in $C$?

As a bonus question: If this definition is correct, are we guaranteed that the stabilizer elements $ \mathcal{S}_x $, $\mathcal{S}_y$, and $\mathcal{S}_z$ are unique (assuming that $\left| C\right|$ is maximal for code distance $d$)? Do the numbers $N$ and $k$ of physical and logical qubits matter for this?

Also if I just need to open Nielsen and Chuang to a particular section, feel free to let me know! I'll have a copy soon. Also, if you have a random fun fact relating code distance to something else, feel free to comment it, since I'm generally interested.

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  • $\begingroup$ Can I just clarify what you mean by deletion/erasure errors? Do you mean what I'm imagining - a qubit gets removed from the system, and we can spot that it has been removed, and we know which qubit was removed? $\endgroup$
    – DaftWullie
    Commented Feb 3, 2023 at 7:16
  • $\begingroup$ If so, there's some mention of a relevant result here: quantumcomputing.stackexchange.com/a/4802/1837 (I'm sure I went through more of the details in an answer at some point, but cannot immediately lay my hands on it) $\endgroup$
    – DaftWullie
    Commented Feb 3, 2023 at 7:21
  • $\begingroup$ @DaftWullie yes, I mean to allow for both of these cases (we know which qubit was traced out / removed, OR we just know that some qubit was traced out / removed). I'll take a look at the "≥5" proof and see if any aspects of that answer my questions! $\endgroup$ Commented Feb 3, 2023 at 23:06

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I'm curious if there are other (equivalent) definitions. Specifically, I'm wondering if the minimum number of errors applies equally to deletion / erasure errors (on at most 𝑑−1 qubits). However, I guess deletion errors aren't the same as logical errors (i.e., they don't change between codewords), so you can correct at most 𝑡=(𝑑−1)/2 of them, so that's the maximum number of deletion errors that can be tolerated?

For the toric code, it's relatively easy to see how many erasure errors can be corrected. You can understand this by looking at how one can correct erasure errors. In the following figure a) (from https://arxiv.org/abs/0904.3556) erasure errors happen to data qubits indicated by the 2 red crosses.

enter image description here

As shown in b) one erasure error can be overcome by 'merging' two weight 4 stabilizers to form a weight 6 stabilizer. In the example shown in the figure, four weight 4 stabilizers are merged to form two weight 6 stabilizers.

A logical operator on the toric code is any cycle going around the torus (donut). Therefore the logical operator can be measured if there is at least one cycle that avoids all deleted qubits.

If you delete all qubits that form a cycle, you can't measure the logical operator anymore. The smallest cycle one can form consists of $d$ data qubits, and therefore we can correct upto $d-1$ erasure errors in the toric code.

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  • $\begingroup$ Shouldn't this be "we can correct up to d-1 erasure errors"? Since d erasure errors cannot be corrected. $\endgroup$ Commented Feb 19, 2023 at 9:38
  • $\begingroup$ yes, I edited it now $\endgroup$
    – Peter-Jan
    Commented Feb 20, 2023 at 7:39

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