Are Quantum States Exponentially Long Vectors?

download report

Transcript Are Quantum States Exponentially Long Vectors?

Are Quantum States
Exponentially Long Vectors?
over n-bit strings
2n-bit strings
Scott Aaronson
(who did and will have an affiliation)
(did: IAS
will: Waterloo)
The Computer Science
Picture of Reality
+ details
Quantum computing challenges this picture
That’s why everyone should care about it,
whether or not quantum factoring machines
are ever built
As far as I am concern[ed], 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 capture[s] the inherent cost of the computation.
—Oded Goldreich
My Two-Pronged Response
(1) It’s not easy to explain current experiments
(let alone future ones!), if you don’t think that quantum
states are exponentially long vectors
[A. 2004, “Multilinear Formulas and Skepticism of Quantum
(2) But it’s not that bad
[A. 2004, “Limitations of Quantum Advice and One-Way
Prong (1)
Quantum states are exponentially long vectors
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.
Arndt et al., Nature
401:680-682 (1999)
Exactly what property separates
the Sure States we know we can
create, from the Shor States that
suffice for factoring?
Not precision in amplitudes:
 0 1
Not entanglement across hundreds of particles:
Not a combination of the two:
1  0  1
2 
 0 1
 
Intuition: Once we accept | and | into our set of
possible states, we’re 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
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?
Observation: If
 
 f x  x ,
then the tree size of | equals the multilinear
formula size of f up to a constant factor
C 
Main Result
 x 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
1/ 3
uniformly at random from F n
• Even to approximate |C takes tree size n(log n)
• Can derandomize using Reed-Solomon codes
• Yields the first function f:{0,1}n with a
superpolynomial gap between formula size and
multilinear formula size
• n(log n) tree size lower bounds can be improved
to exponential
Can show this under the restriction that linear combinations
are “manifestly orthogonal”
• States arising in Shor’s algorithm also have
superpolynomial tree size
Can show this assuming a number-theoretic conjecture:
Conjectures (con’t)
• 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 condensedmatter physicists (e.g. Ghosh et al., “Entangled quantum
state of magnetic dipoles,” Nature 425:48-51, 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 simulation
Can show it’s simulable in the third level of PH
Prong (2)
It’s not that bad
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
Obvious Challenge: Prove an oracle
separation between BQP/poly and BQP/qpoly
Harry Buhrman: Hey Scott—why 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!
Like BPP but with
no gap
BQP/qpoly  PP/poly
Corollary: Can’t show
BQP/poly  BQP/qpoly without
also showing PP  P/poly
Proof based on new communication result:
Given f:{0,1}n{0,1}m{0,1} (partial or total),
D1(f) = O(m Q1(f) logQ1(f))
D1(f) = deterministic 1-way communication complexity of f
Q1(f) = bounded-error quantum 1-way complexity
Alice’s 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 you’ll 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
you’ll output the wrong answer with
probability at least 1/3.
But if you condition on succeeding on
y1 and y2, then y3 is the …
Technicality: We assume Alice’s quantum
message was boosted, so that the error
probability is negligible
Claim: Alice only needs to send T=O(Q) inputs
y1,…,yT, where Q = size of her quantum message
Proof Sketch: Bob succeeds on y1,…,yT
simultaneously with probability at most (2/3)T.
But we can decompose the Q-qubit maximally
mixed state as 1 2
where |1 is
i i
Q 
i 1
Alice’s “true” quantum message. Therefore Bob
also succeeds with probability (1/2Q).
BQP/qpoly  PP/poly
Alice is the
Bob is
the PP
But why can Bob’s procedure be implemented in
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)
• Contrast with PQP/qpoly = Everything, and Raz’s
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
• 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
• The state of n particles is an exponentially long
vector. Welcome to Quantum World!
• But for most purposes, it’s no worse than a
probability distribution being an exponentially
long vector
• If you’re still not happy, suggest a Sure/Shor