The role of the square root of phase in defining the Pauli group

The role of the square root of phase in defining the Pauli group Ask Question Asked 1 month ago Modified yesterday Viewed 152 times 1 Consider a d 𝑑 dimensional qudit. The computational basis states are |0⟩,…,|d−1⟩ | 0 ⟩ , … , | 𝑑 − 1 ⟩ . The generalized X,Z 𝑋 , 𝑍 operators are defined by X|j⟩=|j+1modd⟩ 𝑋 | 𝑗 ⟩ = | 𝑗 + 1 mod 𝑑 ⟩ and Z|j⟩= ω j |j⟩ 𝑍 | 𝑗 ⟩ = 𝜔 𝑗 | 𝑗 ⟩ , where ω= e 2πi/d 𝜔 = 𝑒 2 𝜋 𝑖 / 𝑑 . Let Z d =Z/dZ 𝑍 𝑑 = 𝑍 / 𝑑 𝑍 , and τ= ω − − √ 𝜏 = 𝜔 . Now when d 𝑑 is even, there are two definitions of the Pauli group in the quantum computing literature: (i) P (d) :={ ω j X k Z l :j,k,l∈ Z d } 𝑃 ( 𝑑 ) := { 𝜔 𝑗 𝑋 𝑘 𝑍 𝑙 : 𝑗 , 𝑘 , 𝑙 ∈ 𝑍 𝑑 } , and (ii) P ¯ ¯ ¯ ¯ (d) :={ τ j X k Z l :k,l∈ Z d ,j∈ Z 2d } 𝑃 ¯ ( 𝑑 ) := { 𝜏 𝑗 𝑋 𝑘 𝑍 𝑙 : 𝑘 , 𝑙 ∈ 𝑍 𝑑 , 𝑗 ∈ 𝑍 2 𝑑 } . I would say that most papers use the definition in (ii). I have myself previously used definition (i), as the phase didn't matter in what I was doing. However, I would like to find a clean answer why the definition in (ii) is preferred and used so widely. What exactly goes wrong if we don't include the square root of the phase? Some papers like https://arxiv.org/pdf/2305.13178 go one step further and define the Pauli group to include all phases in U(1) U ( 1 ) . A footnote on the bottom of Page 5 of this paper comments that the Clifford groups obtained with or without this continuous extension remains the same. Is it the case then that the Clifford group defined as the normalizer of the Pauli group does not change, irrespective of which definition of the Pauli group we use? clifford-grouppauli-group Share Improve this question Follow asked Feb 9 at 19:04 Rahul Sarkar 1414 4 bronze badges If you define P x,z = τ xz X x Z z 𝑃 𝑥 , 𝑧 = 𝜏 𝑥 𝑧 𝑋 𝑥 𝑍 𝑧 , you get the nice properties P † x,z = P −x,−z 𝑃 𝑥 , 𝑧 † = 𝑃 − 𝑥 , − 𝑧 and P x P y = τ xΩy P x+y 𝑃 𝑥 𝑃 𝑦 = 𝜏 𝑥 Ω 𝑦 𝑃 𝑥 + 𝑦 , where Ω Ω is the symplectic matrix. –  Ethan Davies Commented Feb 9 at 19:09 @EthanDavies thank you! is this the only reason? why is this desirable, and are there serious consequences anywhere if this does not hold? –  Rahul Sarkar Commented Feb 9 at 20:59 Add a comment 1 Answer Sorted by: Highest score (default) Date modified (newest first) Date created (oldest first) 0 I think the main reason is because physicists want Y 𝑌 to be hermitian like X 𝑋 and Z 𝑍 . It also makes <X,Y,Z> < 𝑋 , 𝑌 , 𝑍 > a generating set of the Pauli group. Even if you pick Y=XZ 𝑌 = 𝑋 𝑍 your Pauli and Clifford groups wouldn't change at all. The first can always be generated by <τI,X,Z> < 𝜏 𝐼 , 𝑋 , 𝑍 > as you note. The usual S 𝑆 gate still maps this group to itself, as it maps X to the product of all generators. For qubits for example, whether you call this product Y 𝑌 or iY 𝑖 𝑌 makes no difference. In fact some older papers use this convention instead: https://arxiv.org/abs/quant-ph/9702029. Share Improve this answer Follow answered Feb 13 at 22:55 user35159 4742 2 silver badges 7 7 bronze badges "Clifford groups wouldn't change at all" - is this true? Note that when d=2, if I use definition (i) for the Pauli group, then the S = (1 0; 0 i) gate is not in the normalizer, but it is in the normalizer if I use definition (ii). So it seems that adding extra phases to the Pauli group, does help to extend the Clifford group. –  Rahul Sarkar Commented Feb 14 at 18:39 Add a comment Your Answer Sign up or log in Sign up using Google Sign up using Email and Password Post as a guest Name Email Required, but never shown Post Your Answer By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy. Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions clifford-grouppauli-group See similar questions with these tags. The Overflow Blog Open source for awkward robots Domain expertise still wanted: the latest trends in AI-assisted knowledge for... 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