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Property on Clifford gates and unitary 2-design: wrong reference in the paper. I am looking for a proof

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I am currently reading this paper , and on page 9, Eq (5), they state that: $$ \mathbb{E}_C[(C \otimes C^*) \rho_{AB} (C \otimes C^*)^{\dagger}]=f^n \Phi^{\otimes n}_{AB} + \frac{1-f^n}{4^n-1} (\mathbb{I}^{\otimes n}_{AB}-\Phi^{\otimes n}_{AB}) $$ , where: $C$ is a Clifford gate. $C^*$ is the complex conjugate of this Clifford (conjugate performed in the computational basis to my understanding). $\mathbb{E}_C$ is an average over the Clifford group. $\rho_{AB}$ is a $2n$ qubit density matrix ( $n

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    Property on Clifford gates and unitary 2-design: wrong reference in the paper. I am looking for a proof Ask Question Asked today Modified today Viewed 12 times 2 I am currently reading this paper, and on page 9, Eq (5), they state that: E C [(C⊗ C ∗ ) ρ AB (C⊗ C ∗ ) † ]= f n Φ ⊗n AB + 1− f n 4 n −1 ( I ⊗n AB − Φ ⊗n AB ) 𝐸 𝐶 [ ( 𝐶 ⊗ 𝐶 ∗ ) 𝜌 𝐴 𝐵 ( 𝐶 ⊗ 𝐶 ∗ ) † ] = 𝑓 𝑛 Φ 𝐴 𝐵 ⊗ 𝑛 + 1 − 𝑓 𝑛 4 𝑛 − 1 ( 𝐼 𝐴 𝐵 ⊗ 𝑛 − Φ 𝐴 𝐵 ⊗ 𝑛 ) , where: C 𝐶 is a Clifford gate. C ∗ 𝐶 ∗ is the complex conjugate of this Clifford (conjugate performed in the computational basis to my understanding). E C 𝐸 𝐶 is an average over the Clifford group. ρ AB 𝜌 𝐴 𝐵 is a 2n 2 𝑛 qubit density matrix ( n 𝑛 qubits in A, n 𝑛 qubits in B). Physically it represent the total quantum state of the n 𝑛 pairs before distillation. Φ AB Φ 𝐴 𝐵 is a 2-qubit "perfect" Bell pair. f=Tr[ Φ ⊗n AB ρ AB ] 1/n 𝑓 = 𝑇 𝑟 [ Φ 𝐴 𝐵 ⊗ 𝑛 𝜌 𝐴 𝐵 ] 1 / 𝑛 : it is the initial "single-pair" fidelity with respect to n 𝑛 perfect Bell pairs. They claim that this derivation is done in the reference [ 83 , Theorem 7.25], but I opened the reference and I cannot find the theorem. I suspect they made a mistake in the reference. Where can I find such a proof? Note: it could possibly be a very natural implication from the definition of unitary 2-designs, but I just learnt the concept so I'm not very familiar with it. It could also be "non trivial" to be shown. In any case I'm looking for a reference showing this which could replace the one they cite. clifford-groupt-designs Share Improve this question Follow asked 2 hours ago Marco Fellous-Asiani 2,6002 2 gold badges 19 19 silver badges 50 50 bronze badges To be honest, the discussion after example 7.25 in the linked reference can be considered the proof, but I think such proof is not appropriate if you're not familiar with these topics. There are many ways to prove that result, for instance, using the representation theory of the Clifford group (Schur's lemma). For another point of view, check out Werner's state en.wikipedia.org/wiki/Werner_state –  David Commented 1 hour ago Add a comment Know someone who can answer? Share a link to this question via email, Twitter, or Facebook. 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-groupt-designs See similar questions with these tags. The Overflow Blog Prevent agentic identity theft Related 6 Proof for Cardinality of the Clifford Group 11 Review paper on depth, qubits and T 𝑇 gates number on Clifford+T decomposition for various "typical" algorithms 11 Clifford gates are transversal What exactly does this transversal mean? 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    Mar 31, 2026
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