Clifford gates are transversal What exactly does this transversal mean? What is the difference between non-Clifford gates and Clifford gates? Why is it simple for Clifford gates to implement transversal for quantum fault tolerance? Please explain the definitions, differences, and functions of the two in detail. I don't understand them very well, thank you.
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2$\begingroup$ Welcome to QC stackexchange! Your question is not very narrow and would require several pages of answer. I would suggest that you read up on fault-tolerance & transversality in the book by Nielsen & Chuang. The role that Clifford gates play in fault-tolerant quantum computing is more subtle and can be treated in an answer here. $\endgroup$– Markus HeinrichCommented Dec 8, 2021 at 11:11
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$\begingroup$ please have a look at quantumcomputing.stackexchange.com/help/how-to-ask. Questions should be laser-focused on individual issues. You can ask separate questions in separate posts. Feel free to edit your post to make it more focused $\endgroup$– glS ♦Commented Dec 8, 2021 at 11:41
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1 Answer
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Transversal and Clifford are not as closely linked as your question would seem to imply.
Transversal gates are those for which an error-correcting code can achieve the transformation on a logical qubit by applying that gate on each of the physical qubits. For example, in a 7-qubit code, if you can achieve logical Hadamard by applying Hadamard on each of the 7 physical qubits, that gate is transversal. These gates are the simplest to implement because they are automatically fault-tolerant: if an error occurs on one physical qubit, there is no action that can propagate it onto a different physical qubit in the same code because no two qubits in the block every interact. They are also particularly short gate sequences (just one time step), which has the knock-on effect of making the corresponding fault-tolerant threshold particularly high.
Now it happens that for some of the most common quantum error correcting codes, the gates that you can apply transversally coincide with the Clifford gates. Indeed, iirc, this is true for all self-dual CSS codes. But this is just a particular choice of error correcting code. There are other choices where it is not the Clifford gates that can be implemented transversally. In that sense, there's nothing particularly special about Clifford.
answered Dec 8, 2021 at 11:25
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$\begingroup$ Also, it might be added that although the set of transversal gates for a code may not necessarily coincide with the Clifford gates, it will never be a universal gate set. $\endgroup$– JSdJCommented Dec 8, 2021 at 17:44
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$\begingroup$ This is known as the Eastin-Knill Theorem. Wikipedia has a good explanation (en.wikipedia.org/wiki/Eastin%E2%80%93Knill_theorem). The original paper is worth reading for as well (arxiv.org/abs/0811.4262) $\endgroup$– hollCommented Dec 13, 2021 at 14:56
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1$\begingroup$ It seems to me that the original paper shows that the set of transversal gates always forms a finite group (a discrete subset of the group of unitary gates closed under products and inverses). Is that correct? $\endgroup$ Commented Mar 1, 2022 at 0:26
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2$\begingroup$ @IanGershonTeixeira Yes. And in contrast, you know that a universal set does not form a finite group. $\endgroup$ Commented Mar 1, 2022 at 7:43
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1$\begingroup$ In some other places (for example here quantumcomputing.stackexchange.com/questions/24399/… and quantumcomputing.stackexchange.com/questions/15301/… ) it says that for phase gate $ P $ to be transversal you need the code to be double even. In your answer you say that Clifford is transversal for all self dual CSS codes. Is that because every self dual CSS code is doubly even? $\endgroup$ Commented Jun 1, 2022 at 19:50