How Much Information Is In Entangled Quantum States? Scott Aaronson MIT |

How Much Information Is In Entangled Quantum States?

Scott AaronsonMIT

|

The Problem

In quantum mechanics, a state of n entangled particles

requires at least 2n complex numbers to specify

nxx x

1,0

To a computer scientist, this is probably the central fact about quantum mechanics

But why should we worry about it?

Answer 1: Quantum State Tomography

Task: Given lots of copies of an unknown quantum state, produce an approximate classical description of it

Central problem: To do tomography on an entangled state of n particles, you need ~cn measurements

Innsbruck group: 8 particles / ~656,000 experiments!

Answer 2: Quantum Computing Skepticism

Some physicists and computer scientists believe quantum computers will be impossible for a fundamental reason

For many of them, the problem is that a quantum computer would “manipulate an exponential amount of information” using only polynomial resources

Levin Goldreich ‘t Hooft Davies Wolfram

But is it really an exponential amount?

Let’s tame the exponential beastIdea: “Shrink quantum states down to reasonable

size” by viewing them operationally

Analogy: A probability distribution over n-bit strings also takes ~2n bits to specify. But that fact seems to be “more about the map than the territory”

Holevo’s Theorem (1973): By sending an n-qubit quantum state, Alice can transmit no more than n classical bits to Bob

This talk: Limitations on the information content of quantum states [A. 2004], [A. 2006], [A. Drucker 2010]

Lesson: “The linearity of QM helps tame the exponentiality of QM”

First, where does the exponentiality of quantum states manifest itself?

Quantum Proofs and Advice: Let G be a finite (but exponentially-large) group G, and let HG be an exponentially-large subgroup. Then the following highly-entangled state (if you have it) can be used to decide whether a given element xG is in H or not, and to prove statements of the form xH:

The trick: Measure the first qubit in

[A.-Kuperberg 2007]: For the Group Non-Membership problem, there might also be short classical proofs that are quickly verifiable by a QC

But at least “relative to a quantum oracle,” there are also cases where quantum proofs are provably exponentially more compact than classical proofs

Outstanding challenge to show such a separation relative to a classical oracle

The Absent-Minded Advisor Problem

Can you give your graduate student a quantum state | with ~n qubits, such that by measuring | in a suitable basis, the student can learn your answer to any one yes-or-no question of size n?

NO [Ambainis, Nayak, Ta-Shma, Vazirani 1999]Indeed, quantum communication is no better

than classical for this problem as n

On the Bright Side…

Theorem (A. 2004): In that case, it suffices for Alice to send Bob only

~n log n log|S|classical bits

Suppose Alice wants to describe an n-qubit quantum state | to Bob … well enough that, for any 2-outcome measurement M in some finite set S, Bob can estimate the probability that M accepts |

|ALL MEASUREMENTSALL MEASUREMENTS PERFORMABLE

USING ≤n2 QUANTUM GATES

How does the theorem work?

Alice is trying to describe the quantum state to Bob

In the beginning, Bob knows nothing about , so he guesses it’s the maximally mixed state 0=I

Then Alice helps Bob improve, by repeatedly telling him a measurement EtS on which his current guess t-1 badly fails

Bob lets t be the state obtained by starting from t-1, then performing Et and postselecting on getting the right outcome

I123

Just two tiny problems with this compression theorem…

1.Computing the classical “compressed representation” of quantum state seems astronomically hard

2.Given the compressed representation, computing the probability some measurement on the state accepts also seems astronomically hard

The “Quantum Occam’s Razor Theorem” [A. 2006] at least addresses the first problem…

Let | be an unknown entangled state of n particles

Suppose you just want to be able to estimate the acceptance probabilities of most measurements E drawn from some probability distribution D

Then it suffices to do the following, for some m=O(n):1.Choose m measurements independently from D

2.Go into your lab and estimate acceptance probabilities of all of them on |

3.Find any “hypothesis state” approximately consistent with all measurement outcomes

Quantum Occam’s Razor Theorem

“Quantum states are PAC-learnable”

Numerical Simulation[A.-Dechter 2008]

We implemented this “pretty-good quantum state tomography” method in MATLAB, using a fast convex programming method developed specifically for this application [Hazan 2008]

We then tested it on simulated data

We studied how the number of sample measurements m needed for accurate predictions scales with the number of qubits n, for n≤10

Result of experiment: My theorem appears to be true

Combining My Postselection and Quantum Learning Results

[A.-Drucker 2010]: Let | be an n-qubit state and let T be a complexity bound. Then there exists a local Hamiltonian H on poly(n,T) qubits, such that any 2-outcome measurement on | performable by a circuit of size T can be simulated in poly(n,T) time by a suitable measurement M’ on the unique ground state of H

Application: Trusted quantum advice is equivalent to trusted classical advice + untrusted quantum advice

SummaryIn many natural scenarios, the “exponentiality” of quantum states is an illusion

That is, there’s a short (though possibly cryptic) classical string that specifies how the quantum state behaves, on any measurement you could actually perform

Applications: Pretty-good quantum state tomography, characterization of quantum computers with “magic initial states”…

Biggest open problem: Find special classes of quantum states that can be learned in a computationally efficient way

[Aaronson-Gottesman, in preparation] Learnability of stabilizer states