Timeline for Equivalent $[\![n,k,d]\!]$ codes and transversal gates
Current License: CC BY-SA 4.0
18 events
| when toggle format | what | by | license | comment | |
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| Nov 1, 2022 at 23:12 | comment | added | Ian Gershon Teixeira | @AdamZalcman hey just seeing this. I think everything in your answer is still true for the new definition so its probably ok in this case, but you're right that's it's always good to just ask a new question! | |
| Nov 1, 2022 at 23:09 | comment | added | Ian Gershon Teixeira | @squiggles if I asked a new question about codes with the same parameters and both implementing transversal Clifford but which are not equivalent would you be willing to post this example as an answer? | |
| Oct 21, 2022 at 16:16 | comment | added | squiggles | They're both doubly even and self-dual. In each, the X- and Z-stabilizers have identical support (modulo on the throwaway fixed padding qubits). They both implement the full Clifford group transversally. | |
| Oct 21, 2022 at 15:21 | comment | added | Adam Zalcman | You shouldn't change the meaning of a question after answers are posted as this will confuse future readers. You can of course post any number of related questions you like :-) | |
| Oct 21, 2022 at 13:27 | comment | added | Ian Gershon Teixeira | @AdamZalcman good point! I updated the definition of equivalence | |
| Oct 21, 2022 at 12:43 | comment | added | Ian Gershon Teixeira | @squiggles are those CSS codes just doubly even or are they also self-dual? We need both properties to know that the full transversal gate group for those codes is the Clifford group (see quantumcomputing.stackexchange.com/a/15305/19675 ). Comparing the two codes you describe is an interesting idea in its own right but I'm just trying to make sure I understand how it relates back to the new question raised in the comments about finding two distinct Clifford transversal codes with the same parameters. Or were you just suggesting that as another answer to the title question? | |
| Oct 20, 2022 at 7:52 | comment | added | squiggles | To get rid of the permutation issue, one could just use any distance color code on a hexagonal lattice and the same distance color code on a "squares and octagons lattice". Then pad the hexagonal lattice with qubits fixed to zero to formally give them identical [[n,k,d]] params. Different size stabs --> not locally equivalent. Both doubly even. | |
| Oct 20, 2022 at 7:08 | comment | added | Adam Zalcman | I think Paulis (and CNOT) are transversal in all vanilla surface codes regardless of distance. That said, there are variations of the surface code with more transversal gates (e.g. IIRC, Cliffords are transversal in the folded surface code and CCZ in 3D surface code). | |
| Oct 20, 2022 at 7:01 | comment | added | Adam Zalcman | I suppose the above example is disappointing. I think what it tells us is that your definition of equivalence may need some rethinking. You should probably allow all non-entangling unitaries (not just product unitaries), so that merely swapping qubits isn't sufficient to break equivalence. N.B. check out the definition of equivalence of classical codes where bit permutations are explicitly mentioned (see e.g. p.39 in "The Theory of Error-Correcting Codes" by MacWilliams and Sloane). | |
| Oct 20, 2022 at 7:00 | comment | added | Adam Zalcman | Hehe, I see. Well, a very simple example would be the following two $[\![14, 2, 3]\!]$ CSS codes. First code: use Steane code to encode the first logical qubit into physical qubits $1\dots 7$ and the second logical qubit into $8\dots 14$. Second code: use Steane code to encode the first logical qubit into physical qubits $1, 3, 5, 7, 9, 11, 13$ and the second logical qubit into $2, 4, 6, 8, 10, 12, 14$. In both codes, Cliffords are of course transversal. | |
| Oct 19, 2022 at 18:42 | comment | added | Ian Gershon Teixeira | Ya, putting all my cards on the table here the counterexample I was definitely fishing for was two self dual doubly even CSS codes with the same parameters but that are not equivalent by local unitaries. That I would be very very very interested to see. If you can find that example I'll write you a thank you haiku. And just to clarify the common knowledge surface code thing, are saying that it is known that the $ 3 \times 3 $ rotated surface code has only Paulis transversal? Or are you saying more generally that its known all surface codes have only Paulis transversal? | |
| Oct 19, 2022 at 18:32 | comment | added | Adam Zalcman | Regarding the transversal gates of the surface code, I'm relying here on what seems to be fairly common knowledge, but admittedly, I don't know how to prove it at the moment (it would make a great QCSE question!). That's why I included the other examples and suggestions. BTW, it seems that a simple yet rigorous non-equivalence proof might be constructed for two color codes or two CSS codes deriving from self-dual doubly even classical codes (since in those cases the transversal group coincides with the Cliffords and we can appeal to Eastin-Knill theorem to rule out other transversal gates). | |
| Oct 19, 2022 at 18:29 | comment | added | Adam Zalcman | Unfortunately, that answer had an error (It is true that $UgU^\dagger\in\mathcal{S}$ for all stabilizers $g$ implies that $U$ is a logical operator, but I used the converse which is false; we only have that $UgU^\dagger = U_L\oplus U_N$ where unitary $U_L$ maps the logical subspace to itself; this is actually rather important since it's why codes with transversal non-Cliffords exist). I rewrote it to fix the issue, so please have another look there. Apologies for the mistake. | |
| Oct 19, 2022 at 13:41 | vote | accept | Ian Gershon Teixeira | ||
| Oct 19, 2022 at 13:37 | comment | added | Ian Gershon Teixeira | I of course remember your very thorough proof that the only transversal gates of the 9-qubit Shor code are the transversal Paulis quantumcomputing.stackexchange.com/questions/27157/…. Could you say more about why the $ 3 \times 3 $ rotated surface code also only has transversal Paulis as its transversal gates? | |
| Oct 19, 2022 at 8:48 | history | edited | Adam Zalcman | CC BY-SA 4.0 |
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| Oct 19, 2022 at 8:43 | history | edited | Adam Zalcman | CC BY-SA 4.0 |
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| Oct 19, 2022 at 8:34 | history | answered | Adam Zalcman | CC BY-SA 4.0 |