The Quantum Pontiff

QIP 2009 Day 5 Liveblogging

The last day of QIP in Santa Fe. Also note that Joe has posted some nice notes on additivity on his blog: part I and part II. Oh, and QIP next year will be in Zurich, and QIP the year after next will be in Singapore.

Lluis Masanes, “Towards device-independent security in QKD”

Paper: arXiv:0807.2158. I cam in late, so was a bit lost. But basically Lluis talked about secret key distillation under weaker assumptions than just assuming quantum theory. It seems that one an do secret key distillation from accessing the correlations that violate Bell inequalities under the assumption of no signaling.

Amnon Ta-Shma, “Short seed extractors against quantum storage”

Amnon talked about the paper arXiv:0808.1994.

Jop Briet (speaker), Harry Buhrman, and Ben Toner, “A generalized Grothendieck inequality and entanglement in XOR games”

Dude I need LaTeX support in scienceblogs. Jop talked about a new Grothendieck’s type inequality (see the paper arXiv:0901.2009 for the definition.) This can then be used to show that the amount of entanglement required to maximally violate a Bell inequality must depend on the number of measurements (not just the number of measurement outcomes.)

Dejan Dukaric, Manuel Forster, Severin Winkler, and Stefan Wolf , “On non-locality distillation”

At this point my brain was full.

Gilles Brassard, Louis Salvail, and Alain Tapp, “Key distribution and oblivious transfer à la Merkle”

Robert König, Renato Renner, and Christian Schaffner, “The operational meaning of min- and max-entropy”

Matthias Christandl, Dejan Dukaric, Robert König, and Renato Renner, “Postselection-technique with applications to quantum cryptography and the parallel repetition problem”

Robert König (speaker), Ben Reichardt, and Guifre Vidal, “Exact entanglement renormalization for string-net models”

Here Robert described his work on Levin and Wen’s string net models. The main result is that the Levin and Wen models can be thought about as arising from the requirement of scale invariant. In particular they show how these models have ground states which are efficiently describable by multi-scale entanglement renormalization ansatz states. Paper here: arXiv:0806.4583. I like this paper a lot.

Aram Harrow and Richard Low (speaker), “Efficient quantum tensor product expanders and k-designs”

It would be nice to efficiently implement random unitaries sampled uniformly with respect to Haar measure. This is hard. However often we want to construct methods for sampling random unitaries which give things like random unitaries. Here they talk about the case where one wants to sample with respect to Haar measure which matches the first k moments of the Haar measure (a k-design). Aram and Richard present ways to do this efficiently by efficiently constructing a “k-copy tensor product expander.” Paper: arXiv:0811.2597.

Bill Rosgen, “Distinguishability of random unitary channels”

Comments

  1. #1 davie
    January 16, 2009

    To make an mathematical expression for your blog, go to the Wikipedia sandbox (http://en.wikipedia.org/wiki/Wikipedia:Sandbox). Go to the Edit page and follow these directions: http://en.wikipedia.org/wiki/Help:Formula save the page, copy-and-paste the result to your blog.

  2. #2 Ben
    January 16, 2009

    Quantum eavesdropping à la Merkle (Gilles Brassard): As there doesn’t seem to be an arxiv paper yet, let me write some comments from the talk.

    Merkle’s idea for public-key cryptography: Given a one-way function f: [N] -> [N], Alice and Bob each choose sqrt{N} inputs and publish f applied to those inputs. With good probability, their lists have overlap. They use the preimage of that element as their secret key. Note that their effort is sqrt{N}, but an adversary needs to use N time to invert the function to figure out the key. Thus there is a quadratic effort gap. (Not as good as the suppose exponential effort gap for, e.g., Diffie-Hellman, but has a weaker assumption and is perhaps still useful.)

    Quantumly, this doesn’t work, since the eavesdropper can apply Grover search to invert the permutation in time sqrt{N}, so there is no effort gap. However, if we also allow Bob to be quantum, then, when Alice publishes (classically) a list of t=N^{1/3} encodings, Bob can find a preimage in time sqrt{N/t} = N^{1/3}. The adversary’s required effort is still sqrt{N}, so there is a (3/2)-power effort gap.

    A similar algorithm works for oblivious transfer (recall: Bob should get exactly one of Alice’s strings a or b, and Alice shouldn’t know which). Alice now publishes two columns with N^{1/3} rows. Roughly, one column holds a and the other b; each row can be hidden by the same random mask; and then f is applied. (I missed the details of the construction but this seems to work.) Bob chooses which string/column he wants to learn and searches for the preimage of some entry in that column, in time N^{1/3}. He then asks Alice for the mask for that row to get the output bit.

    This construction has a (3/2)-power gap classically, with an appropriate choice of t. It seems like the best attack Bob can make to learn both preimages for some row takes time N^{4/9} (?), giving a (4/3)-power gap. But this gap has not been proved.

    More open questions: The classical quadratic effort gap is tight (just proved in 2008 by Barak and Mahmoody-Ghidary). Is 3/2 tight quantumly? Are there fully classical protocols secure against quantum attack. He said that they have a candidate.

  3. #3 joe nahhas
    January 29, 2009

    Einstein’s Nemesis: DI Her Eclipsing Binary Stars Solution
    The problem that the 100,000 PHD Physicists could not solve

    This is the solution to the “Quarter of a century” Smithsonian-NASA Posted motion puzzle that Einstein and the 100,000 space-time physicists including 109 years of Nobel prize winner physics and physicists and 400 years of astronomy and Astrophysicists could not solve and solved here and dedicated to Drs Edward Guinan and Frank Maloney
    Of Villanova University Pennsylvania who posted this motion puzzle and started the search collections of stars with motion that can not be explained by any published physics
    For 350 years Physicists Astrophysicists and Mathematicians and all others including Newton and Kepler themselves missed the time-dependent Newton’s equation and time dependent Kepler’s equation that accounts for Quantum – relativistic effects and it explains these effects as visual effects. Here it is

    Universal- Mechanics

    All there is in the Universe is objects of mass m moving in space (x, y, z) at a location
    r = r (x, y, z). The state of any object in the Universe can be expressed as the product

    S = m r; State = mass x location

    P = d S/d t = m (d r/dt) + (dm/dt) r = Total moment

    = change of location + change of mass

    = m v + m’ r; v = velocity = d r/d t; m’ = mass change rate

    F = d P/d t = d²S/dt² = Force = m (d²r/dt²) +2(dm/d t) (d r/d t) + (d²m/dt²) r

    = m γ + 2m’v +m”r; γ = acceleration; m” = mass acceleration rate

    In polar coordinates system

    r = r r(1) ;v = r’ r(1) + r θ’ θ(1) ; γ = (r” – rθ’²)r(1) + (2r’θ’ + rθ”)θ(1)

    F = m[(r”-rθ’²)r(1) + (2r’θ’ + rθ”)θ(1)] + 2m'[r’r(1) + rθ’θ(1)] + (m”r) r(1)

    F = [d²(m r)/dt² – (m r)θ’²]r(1) + (1/mr)[d(m²r²θ’)/d t]θ(1) = [-GmM/r²]r(1)

    d² (m r)/dt² – (m r) θ’² = -GmM/r²; d (m²r²θ’)/d t = 0

    Let m =constant: M=constant

    d²r/dt² – r θ’²=-GM/r² —— I

    d(r²θ’)/d t = 0 —————–II

    r²θ’=h = constant ————– II
    r = 1/u; r’ = -u’/u² = – r²u’ = – r²θ'(d u/d θ) = -h (d u/d θ)
    d (r²θ’)/d t = 2rr’θ’ + r²θ” = 0 r” = – h d/d t (du/d θ) = – h θ'(d²u/d θ²) = – (h²/r²)(d²u/dθ²)
    [- (h²/r²) (d²u/dθ²)] – r [(h/r²)²] = -GM/r²
    2(r’/r) = – (θ”/θ’) = 2[λ + ỉ ω (t)] – h²u² (d²u/dθ²) – h²u³ = -GMu²
    d²u/dθ² + u = GM/h²
    r(θ, t) = r (θ, 0) Exp [λ + ỉ ω (t)] u(θ,0) = GM/h² + Acosθ; r (θ, 0) = 1/(GM/h² + Acosθ)
    r ( θ, 0) = h²/GM/[1 + (Ah²/Gm)cosθ]
    r(θ,0) = a(1-ε²)/(1+εcosθ) ; h²/GM = a(1-ε²); ε = Ah²/GM

    r(0,t)= Exp[λ(r) + ỉ ω (r)]t; Exp = Exponential

    r = r(θ , t)=r(θ,0)r(0,t)=[a(1-ε²)/(1+εcosθ)]{Exp[λ(r) + ì ω(r)]t} Nahhas’ Solution

    If λ(r) ≈ 0; then:

    r (θ, t) = [(1-ε²)/(1+εcosθ)]{Exp[ỉ ω(r)t]

    θ'(r, t) = θ'[r(θ,0), 0] Exp{-2ỉ[ω(r)t]}

    h = 2π a b/T; b=a√ (1-ε²); a = mean distance value; ε = eccentricity
    h = 2πa²√ (1-ε²); r (0, 0) = a (1-ε)

    θ’ (0,0) = h/r²(0,0) = 2π[√(1-ε²)]/T(1-ε)²
    θ’ (0,t) = θ'(0,0)Exp(-2ỉwt)={2π[√(1-ε²)]/T(1-ε)²} Exp (-2iwt)

    θ'(0,t) = θ'(0,0) [cosine 2(wt) – ỉ sine 2(wt)] = θ'(0,0) [1- 2sine² (wt) – ỉ sin 2(wt)]
    θ'(0,t) = θ'(0,t)(x) + θ'(0,t)(y); θ'(0,t)(x) = θ'(0,0)[ 1- 2sine² (wt)]
    θ'(0,t)(x) – θ'(0,0) = – 2θ'(0,0)sine²(wt) = – 2θ'(0,0)(v/c)² v/c=sine wt; c=light speed

    Δ θ’ = [θ'(0, t) – θ'(0, 0)] = -4π {[√ (1-ε) ²]/T (1-ε) ²} (v/c) ²} radians/second
    {(180/π=degrees) x (36526=century)

    Δ θ’ = [-720×36526/ T (days)] {[√ (1-ε) ²]/ (1-ε) ²}(v/c) = 1.04°/century

    This is the T-Rex equation that is going to demolished Einstein’s space-jail of time

    The circumference of an ellipse: 2πa (1 – ε²/4 + 3/16(ε²)²—) ≈ 2πa (1-ε²/4); R =a (1-ε²/4)
    v (m) = √ [GM²/ (m + M) a (1-ε²/4)] ≈ √ [GM/a (1-ε²/4)]; m<

    v = v (center of mass); v is the sum of orbital/rotational velocities = v(cm) for DI Her
    Let m = mass of primary; M = mass of secondary

    v (m) = primary speed; v(M) = secondary speed = √[Gm²/(m+M)a(1-ε²/4)]
    v (cm) = [m v(m) + M v(M)]/(m + M) All rights reserved. joenahhas1958@yahoo.com

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