How Much Information Is In Entangled Quantum States?

1. | How Much Information Is In Entangled Quantum States? Scott Aaronson MIT

2. The Problem In quantum mechanics, a state of n entangled particles requires at least 2n complex numbers to specify To a computer scientist, this is probably the central fact about quantum mechanics But why should we worry about it?

3. 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!

4. Answer 2: Quantum Computing Skepticism Levin Goldreich ‘t Hooft Davies Wolfram 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 But is it really an exponential amount?

5. Let’s tame the exponential beast Idea: “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”

6. 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 HG be an exponentially-large subgroup. Then the following highly-entangled state (if you have it) can be used to decide whether a given element xG is in H or not, and to prove statements of the form xH: [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 trick: Measure the first qubit in

7. 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

8. On the Bright Side… 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 | Theorem (A. 2004): In that case, it suffices for Alice to send Bob only ~n log n  log|S| classical bits

9. | ALL MEASUREMENTS PERFORMABLE USING ≤n2 QUANTUM GATES ALL MEASUREMENTS

10. How does the theorem work? 1 2 3 I 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 EtS 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

11. Just two tiny problems with this compression theorem… • Computing the classical “compressed representation” of quantum state seems astronomically hard • 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…

12. Quantum Occam’s Razor Theorem • 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): • Choose m measurements independently from D • Go into your lab and estimate acceptance probabilities of all of them on | • Find any “hypothesis state” approximately consistent with all measurement outcomes “Quantum states are PAC-learnable”

13. 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

15. 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

16. Summary In 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