Are Quantum States Exponentially Long Vectors?

Scott Aaronson(who did and will have an affiliation)

(did: IAS will: Waterloo)

Distributions over n-bit strings 2n-bit strings

The Computer Science Picture of Reality

Quantum computing challenges this picture

That’s why everyone should care about it, whether or not quantum factoring machines are ever built

+ details

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

(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 Computing”]

(2) But it’s not that bad

[A. 2004, “Limitations of Quantum Advice and One-Way Communication”]

My Two-Pronged Response

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.

C60

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?

DIV

IDIN

G L

INE

Not precision in amplitudes:n

2

10

Not entanglement across hundreds of particles:

2

10nn

Not a combination of the two:

nn

2

10

2

10

2

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

+

|01 |12

++

|01 |11 |02 |12

1

2

1

2

1

2

1

2

2

7

3

7

Definition: The tree size of a pure state | is the minimum number of linear combinations and tensor products in any tree representing |

,1,0

nx

xxfObservation: If

then the tree size of | equals the multilinear formula size of f up to a constant factor

Question: Can we show that quantum states arising in real (or at least doable) experiments have superpolynomial tree size?

Main Result

• 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

If

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

Cx

xC

C1

is a uniform superposition

nn 3/1

2F

Conjectures

• States arising in Shor’s algorithm also have superpolynomial tree size

Can show this assuming a number-theoretic conjecture:

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

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 condensed-matter 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 simulationCan show it’s simulable in the third level of PH

Prong (2)It’s not that bad

Quantum Advice

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

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?”

Maybe BQP/qpoly even contains NP!

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!

Result:

BQP/qpoly PP/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 fQ1(f) = bounded-error quantum 1-way complexity

Corollary: Can’t show BQP/poly BQP/qpoly

without also showing PP P/poly

Like BPP but with no gap

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 …

y1

y2

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 where |1 is

Q

iiiQ

2

12

1

Alice’s “true” quantum message. Therefore Bob also succeeds with probability (1/2Q).

BQP/qpoly PP/poly

Alice is the “advisor”

Bob is the PP algorithm

But why can Bob’s 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)

Remarks

• 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

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

Summary

• 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 separator