Profile picture
Frédéric Grosshans @fgrosshans
, 8 tweets, 3 min read Read on Twitter
Now Dominik Hangleiter talks about getting rid of the anticoncentration conjecture in random universal circuits
#LTQI
Dominik Hangleiter: the random circuit model uses a √n×√n lattice of qubits with gates taken out of {CZ,√X,√Y, T}

#LTQI
Dominik Hangleiter: More abstractly, the unitary U outputing X is chosen at random in some family. We need the anticonconetration conjecture.

Which families of U satisfies this ? At least the 2-designs do it.
Paley-Zygmund : Pr(Z≥αE(Z))≥(1-α)²E(Z)²/E(Z²)
#LTQI
Dominik Hangleiter:
A unitary k-design is an ensemble such that ∀X∈B(H)^⊗k E_design(U^⊗k X U⁺^⊗k)=E_Haar(U^⊗k X U⁺^⊗k)
and a similar def. for states.

For an ε-approximate design, the expectation values are within a (1±ε) factor of Haar’s
#LTQI
Domink Hangleiter shows that, if U is a k-design, |ψ⟩∈ℋ, then U|ψ⟩ is a state k-design by computing the expectaton values.

Using PAley-Zygmund, for an ε-design
Pr(P_U(X)≥α(1-ε)/N) ≥ (1-α)³(1-ε)²/2(1+ε)

#LTQI
Dominik Hangleiter now needs random circuits to be ϵ-approximate designs
It’s known (thm by BHH) that G-local PRQC (parallel random local quantum random circuits) are ε-approx 2 designs in a depth ∼n log(1/ε)
⇒ G-local PRQC anticoncentrate at depth n log(1/ε)
#LTQI
Domink Hangleiter: This is worse than Google setting, which is of depth √n, but it is expected in this 1D set-up
#LTQI
Dominik Hangleiter: If one exactly synthesizes the family of IQP {SWAP, exp(-iπ/8 X⊗X)} ⇒ approx, of U are ♯P hard to simulate to ∼¼ in linear depth

The idea is to use IQP results
#LTQI
Missing some Tweet in this thread?
You can try to force a refresh.

Like this thread? Get email updates or save it to PDF!

Subscribe to Frédéric Grosshans
Profile picture

Get real-time email alerts when new unrolls are available from this author!

This content may be removed anytime!

Twitter may remove this content at anytime, convert it as a PDF, save and print for later use!

Try unrolling a thread yourself!

how to unroll video

1) Follow Thread Reader App on Twitter so you can easily mention us!

2) Go to a Twitter thread (series of Tweets by the same owner) and mention us with a keyword "unroll" @threadreaderapp unroll

You can practice here first or read more on our help page!

Did Thread Reader help you today?

Support us! We are indie developers!


This site is made by just three indie developers on a laptop doing marketing, support and development! Read more about the story.

Become a Premium Member and get exclusive features!

Premium member ($3.00/month or $30.00/year)

Too expensive? Make a small donation by buying us coffee ($5) or help with server cost ($10)

Donate via Paypal Become our Patreon

Thank you for your support!