Are Quantum States Exponentially Long Vectors?

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Are Quantum States Exponentially Long Vectors?
?
?

• Scott Aaronson
• (who did and will have an affiliation)
• (did IAS will Waterloo)

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The Computer Science Picture of Reality
?
Quantum computing challenges this picture Thats
why everyone should care about it, whether or not
quantum factoring machines are ever built
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As far as I am concerned, the QC model consists
of exponentially-long vectors (possible
configurations) and some uniform (or simple)
operations (computation steps) on such vectors
The key point is that the associated complexity
measure postulates that each such operation can
be effected at unit cost (or unit time). My main
concern is with this postulate. My own intuition
is that the cost of such an operation or of
maintaining such vectors should be linearly
related to the amount of non-degeneracy of
these vectors, where the non-degeneracy may
vary from a constant to linear in the length of
the vector (depending on the vector). Needless
to say, I am not suggesting a concrete definition
of non-degeneracy, I am merely conjecturing
that such exists and that it captures the
inherent cost of the computation. Oded Goldreich
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My Two-Pronged Response

• (1) Its not easy to explain current experiments
  (let alone future ones!), if you dont think that
  quantum states are exponentially long
  vectorsA. 2004, Multilinear Formulas and
  Skepticism of Quantum Computing
• (2) But its not that badA. 2004, Limitations
  of Quantum Advice and One-Way Communication

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Prong (1)Quantum states are exponentially long
vectors
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How Good Is The Evidence for QM?

1. Interference Stability of e- orbits,
   double-slit, etc.
2. Entanglement Bell inequality, GHZ experiments
3. Schrödinger cats C60 double-slit experiment,
   superconductivity, quantum Hall effect, etc.

C60
Arndt et al., Nature 401680-682 (1999)
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Intuition Once we accept ?? and ?? into our
set of possible states, were almost forced to
accept ????? and ?????? as well But we might
restrict ourselves to tree states n-qubit states
obtainable from 0? and 1? by a polynomial
number of linear combinations and tensor products
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Definition The tree size of a pure state ?? is
the minimum number of linear combinations and
tensor products in any tree representing ??
Question Can we show that quantum states arising
in real (or at least doable) experiments have
superpolynomial tree size?
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Main Result
If
is a uniform superposition
over the codewords of a binary linear code, then
C? has tree size n?(log n), with high
probability if the generator matrix is chosen
uniformly at random from

• Even to approximate C? takes tree size n?(log
  n)
• Can derandomize using Reed-Solomon codes
• Yields the first function f0,1n?? with a
  superpolynomial gap between formula size and
  multilinear formula size

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Conjectures

• n?(log n) tree size lower bounds can be improved
  to exponentialCan show this under the
  restriction that linear combinations are
  manifestly orthogonal

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Conjectures (cont)

• 2D and 3D cluster states (lattices of spins
  with pairwise interactions) have large (2?(n)?)
  tree size
• Would mean that states with enormous tree sizes
  have already been seen in the lab, if we believe
  the condensed-matter physicists (e.g. Ghosh et
  al., Entangled quantum state of magnetic
  dipoles, Nature 42548-51, 2003)
• Can show that a 1D cluster state of n spins has
  tree size O(n4)
• If a quantum computer is in a tree state at
  every time step, then that computer has a
  nontrivial classical simulationCan show its
  simulable in the third level of PH

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Prong (2)Its not that bad
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Quantum Advice
Nielsen Chuang We know that many systems in
Nature prefer to sit in highly entangled states
of many systems might it be possible to exploit
this preference to obtain extra computational
power?
BQP/qpoly Class of languages decidable by
polynomial-size, bounded-error quantum circuits,
given a polynomial-size quantum advice state ?n?
that depends only on the input length n
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Obvious Challenge Prove an oracle separation
between BQP/poly and BQP/qpoly
Harry Buhrman Hey Scottwhy not try for an
unrelativized separation? After all, if quantum
states are like 2n-bit classical strings, then
maybe BQP/qpoly ? NEEEEE/poly!
Maybe BQP/qpoly even contains NP!
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Result
Like BPP but with no gap
BQP/qpoly ? PP/poly
Proof based on new communication resultGiven
f0,1n?0,1m?0,1 (partial or total), D1(f)
O(m Q1(f) logQ1(f))D1(f) deterministic 1-way
communication complexity of fQ1(f)
bounded-error quantum 1-way complexity
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Alices Classical Message
Bob, if you use the maximally mixed state in
place of my quantum message, then y1 is the
lexicographically first input for which youll
output the wrong answer with probability at least
1/3. But if you condition on succeeding on y1,
then y2 is the next input for which youll output
the wrong answer with probability at least
1/3. But if you condition on succeeding on y1
and y2, then y3 is the
y1
y2
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Technicality We assume Alices quantum message
was boosted, so that the error probability is
negligible
Claim Alice only needs to send TO(Q) inputs
y1,,yT, where Q size of her quantum message
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BQP/qpoly ? PP/poly
Bob is the PP algorithm
Alice is the advisor
But why can Bobs procedure be implemented in
PP? A. 2005 PostBQP PP, where PostBQP BQP
with postselected measurement outcomes (The
PostBQP ? PP direction is an easy modification of
Adleman et al.s proof that BQP ? PP)
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Remarks

• Contrast with PQP/qpoly Everything, and Razs
  recent result that QIP/qpoly Everything
• Using similar techniques, I get that for every
  k, there exists a language in PP that does not
  have quantum circuits of nk, not even with
  quantum advice
• A second limitation of quantum advice (A. 2004)
  there exists an oracle relative to which NP ?
  BQP/qpoly

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Conjectures

• The question of whether classical advice can
  always replace quantum advice (i.e. BQP/poly
  BQP/qpoly) is not independent of ZF set theory
• Similarly for whether classical proofs can
  replace quantum proofs (i.e. QCMA QMA)
• QMA/qpoly ? PP/poly
• Randomized and quantum one-way communication
  complexities are asymptotically equal for all
  total Boolean functions f

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Summary

• The state of n particles is an exponentially
  long vector. Welcome to Quantum World!
• But for most purposes, its no worse than a
  probability distribution being an exponentially
  long vector
• If youre still not happy, suggest a Sure/Shor
  separator