Why doesn't the contextual QRAC contradict Nayak's bound?

Why doesn't the contextual QRAC contradict Nayak's bound? Ask Question Asked 2 years ago Modified today Viewed 100 times 5 A QRAC encodes n 𝑛 bits in m 𝑚 qubits so that one of the bits can be retrieved. Let p>0.5 𝑝 > 0.5 be the success probability of the receiver correctly obtaining any one of the n 𝑛 bits of their choice by performing the appropriate measurement. Nayak's bound states that such an (n,m,p) ( 𝑛 , 𝑚 , 𝑝 ) -QRAC must satisfy m≥(1−H(p))n 𝑚 ≥ ( 1 − 𝐻 ( 𝑝 ) ) 𝑛 , where H(·) is the entropy function. So for any constant p 𝑝 , the "compression ratio" ( m/n 𝑚 / 𝑛 ) that can be obtained is at least (1−H(p)) ( 1 − 𝐻 ( 𝑝 ) ) . That is we can only get a constant (in n 𝑛 ) factor space savings. This paper introduces a Quantum Random Access code that claims (in section 7) to be able to store ≈0.5⋅ 3 u ≈ 0.5 ⋅ 3 𝑢 bits in C u 2 1.5 u 𝐶 𝑢 2 1.5 𝑢 qubits where C 𝐶 is a constant ≈243 ≈ 243 , where the success probability is 0.999. I'm confused how this doesn't violate Nayak's bound since the space savings is not a constant factor, yet the success probability is constant. contextuality Share Improve this question Follow edited Mar 30, 2024 at 11:48 glS♦ 28k7 7 gold badges 43 43 silver badges 140 140 bronze badges asked Mar 6, 2024 at 11:09 shashvat 9676 6 silver badges 16 16 bronze badges Add a comment 1 Answer Sorted by: Highest score (default) Date modified (newest first) Date created (oldest first) 0 I just came across the question, and I don't quite understand, but I had a follow up to Nayak's bound and QRAC preparation. I am assuming that Nayak's bounds apply to pure states as in quantum states that are preparable. So in the paper linked in the original post the idea is to used mixed states, which would mean 4-qubit pure states and then sample them in a way that is able to produce a biased measurement according to a specific basis. From my limited understanding of mixed states, it means we can purify the pure state into a pure state of higher dimensions, which is kind of hacking Hayak's bound right? In the sense that the left hand side is actually greater than the 4-qubit pure states being used. So Nayak's bounds still hold? The purified state is of the higher degree as stated by the bound. Share Improve this answer Follow answered Feb 19 at 15:21 timmy1691 111 1 bronze badge 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 contextuality See similar questions with these tags. The Overflow Blog After all the hype, was 2025 really the year of AI agents? 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