Proving an amplitude bound for an $n$-qudit state whose stabilizer fidelity is maximized at $|0^n\rangle$

Proving an amplitude bound for an n 𝑛 -qudit state whose stabilizer fidelity is maximized at | 0 n ⟩ | 0 𝑛 ⟩ Ask Question Asked 3 days ago Modified 3 days ago Viewed 69 times 2 Let |ψ⟩∈( C d ) ⊗n | 𝜓 ⟩ ∈ ( 𝐶 𝑑 ) ⊗ 𝑛 be an n 𝑛 -qudit state. Suppose the stabilizer fidelity of |ψ⟩ | 𝜓 ⟩ (i.e., the maximum fidelity |⟨ψ|ϕ⟩ | 2 | ⟨ 𝜓 | 𝜙 ⟩ | 2 over all n 𝑛 -qudit stabilizer states |ϕ⟩ | 𝜙 ⟩ ) is achieved by the all-zero state |ϕ⟩=| 0 n ⟩ | 𝜙 ⟩ = | 0 𝑛 ⟩ . I am trying to prove that there exists a constant c∈(0,1) 𝑐 ∈ ( 0 , 1 ) such that the following inequality holds: 1 d−1 ∑ k=1 d−1 |⟨ψ|k 0 n−1 ⟩ | 2 ≤c|⟨ψ| 0 n ⟩ | 2 . 1 𝑑 − 1 ∑ 𝑘 = 1 𝑑 − 1 | ⟨ 𝜓 | 𝑘 0 𝑛 − 1 ⟩ | 2 ≤ 𝑐 | ⟨ 𝜓 | 0 𝑛 ⟩ | 2 . Note that the inequality holds for c=1 𝑐 = 1 trivially because |⟨ψ|k 0 n−1 ⟩ | 2 ≤|⟨ψ| 0 n ⟩ | 2 | ⟨ 𝜓 | 𝑘 0 𝑛 − 1 ⟩ | 2 ≤ | ⟨ 𝜓 | 0 𝑛 ⟩ | 2 . How can I rigorously establish this upper bound? Any hints or references to similar properties of the stabilizer states would be greatly appreciated. quantum-statestabilizer-codefidelity Share Improve this question Follow edited Mar 13 at 2:50 asked Mar 13 at 2:48 Boye 212 2 bronze badges New contributor 1 I guess you are assuming that ψ 𝜓 is non-stabilizer? Because otherwise |+0⟩ | + 0 ⟩ gives you a counterexample. –  Markus Heinrich Commented Mar 13 at 7:31 @MarkusHeinrich I suppose the point is that a stabilizer state such as your example does not satisfy the property that the largest inner product with a stabilizer state is achieved by |00..0⟩ | 00..0 ⟩ . –  DaftWullie Commented Mar 13 at 9:00 1 @MarkusHeinrich Thank you for the comment! Actually, \ketψ=\ket+0 \ket 𝜓 = \ket + 0 does not satisfy the premise of the problem, so it wouldn't serve as a counterexample. Like DaftWullie said, \ketψ=\ket+0 \ket 𝜓 = \ket + 0 doesn't satisfy that its maximum fidelity over all stabilizer states is achieved by \ket 0 n \ket 0 𝑛 . To clarify my question: I am looking to prove that there exists a universal constant c∈(0,1) 𝑐 ∈ ( 0 , 1 ) such that the inequality holds for any n 𝑛 and any \ketψ \ket 𝜓 , provided that \ketψ \ket 𝜓 satisfies the condition that \ket 0 n \ket 0 𝑛 is its closest stabilizer state. –  Boye Commented Mar 13 at 9:38 @DaftWullie, Boye of course, thanks –  Markus Heinrich Commented 13 hours 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 quantum-statestabilizer-codefidelity See similar questions with these tags. 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