How Much Information Is In A Quantum State?

 How Much Information Is In A Quantum State? Scott Aaronson MIT

Computer Scientist / Physicist Nonaggression Pact • You accept that, for this talk: • Polynomial vs. exponential is the basic dichotomy of the universe • “For all x” means “for all x” In return, I will not inflict the following computational complexity classes on you: #P AM AWPP BQP BQP/qpoly MA NP P/poly PH PostBQP PP PSPACE QCMA QIP QMA SZK YQP

So, how much information is in a quantum state? An infinite amount, of course, if you want to specify the state exactly… Life is too short for infinite precision

A More Serious Point In general, a state of n possibly-entangled qubits takes ~2n bits to specify, evenapproximately Spin-½ particles To a computer scientist, this is arguably 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  Not something I just made up!“As seen in Science & Nature” Well-known problem: To do tomography on an entangled state of n spins, you need ~cn measurements Current record: 8 spins / ~656,000 experiments (!) This is a conceptual problem—not just a practical one!

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?

Today we’ll 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” • Setting the stage: Holevo’s Theorem and random access codes • Describing a state by postselected measurements [A. 2004] • “Pretty good tomography” using far fewer measurements [A. 2006] • - Numerical simulation [A.-Dechter, in progress] • Encoding quantum states as ground states of simple Hamiltonians [A.-Drucker 2009]

 Alice Bob Theorem [Holevo 1973]: By sending an n-qubit state , Alice can communicate no more than n classical bits to Bob (or 2n bits assuming prior entanglement) How can that be? Well, Bob has to measure , and measuring makes most of the wavefunction go poof… Lesson: “The linearity of QM helps tame the exponentiality of QM”

The Absent-Minded Advisor Problem Can you give your graduate student a quantum state  with n qubits (or 10n, or n3, …)—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… Suppose Alice wants to describe an n-qubit state  to Bob, well enough that for any 2-outcome measurement E, Bob can estimate Tr(E) Then she’ll need to send ~cn bits, in the worst case. But… suppose Bob only needs to be able to estimate Tr(E) for every measurement E in a finite set S. Theorem (A. 2004): In that case, it suffices for Alice to send ~n log n  log|S| bits

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

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 his guess, by repeatedly telling him a measurement EtS on which his guess t-1 badly fails Bob lets t be the state obtained by starting from t-1, then performing Et and postselecting on the right outcome

Claim: After only O(n) of these learning steps, Bob gets a state T such that Tr(ET)Tr(E) for all measurements ES. Proof Sketch: For simplicity, assume =|| is pure and Tr(E) is ≤1/n2 or 1-1/n2 for all ES. Let p be the probability that E1,E2,…,ET all yield the desired outcomes. By assumption, p is at most (say) (2/3)T On the other hand, if Bob had made the lucky guess 0=||, then p would’ve been at least (say) 0.9 But we can decompose I as an equal mixture of | and 2n-1 other states orthogonal to |! Hence p  0.9/2n 0.9/2n ≤ (2/3)T  T=O(n) Conclusion: Alice can describe  to Bob by telling him its behavior on E1,E2,…,ET. This takes O(n log|S|) bits

Discussion • We’ve shown that for any n-qubit state  and set S of observables, one can “compress” the measurement data {Tr(E)}ES into a classical string x of only Õ(nlog|S|) bits • Just two tiny problems… • Computing x seems astronomically hard • Given x, estimating Tr(E)also seems astronomically hard • I’ll now state the “Quantum Occam’s Razor Theorem,” which at least addresses the first problem…

Quantum Occam’s Razor Theorem • Let  be an unknown quantum state of n spins • Suppose you just want to be able to estimate Tr(E) for mostmeasurements E drawn from some probability measure D • Then it suffices to do the following, for some m=O(n): • Choose E1,…,Em independently from D • Go into your lab and estimate Tr(Ei) for each 1≤i≤m • Find any “hypothesis state”  such that Tr(Ei)Tr(Ei) for all 1≤i≤m

Theorem [A. 2006]: Provided (C a constant) and “Quantum states are PAC-learnable” for all i, you’ll be guaranteed that with probability at least 1- over the choice of E1,…,Em. Proof combines Ambainis et al.’s result on the impossibility of quantum compression with some power tools from classical computational learning theory

Remark 1: To do this “pretty good tomography,” you don’t need any prior assumptions about ! (No Bayesian nuthin’...) • Removes a lot of conceptual problems... • Instead, you assume a distribution D over measurements • Might be preferable—after all, you can control which measurements to apply, but not what  is Remark 2: Given the measurement data Tr(E1),…,Tr(Em), finding a hypothesis state  consistent with it could still be an exponentially hard computational problem Semidefinite / convex programming in 2n dimensions But this seems unavoidable: even finding a classical hypothesis consistent with data is conjectured to be hard!

Numerical Simulation[A.-Dechter, in progress] We implemented the “pretty-good tomography” algorithm in MATLAB, using a fast convex programming method developed specifically for this application [Hazan 2008] We then tested it (on simulated data) using MIT’s computing cluster 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

Recap: Given an unknown n-qubit entangled quantum state , and a set S of two-outcome measurements… Learning theorem: “Any hypothesis state  consistent with a small number of sample points behaves like  on most measurements in S” Postselection theorem: “A particular state T (produced by postselection) behaves like  on all measurements in S” Dream theorem: “Any state  that passes a small number of tests behaves like  on all measurements in S” [A.-Drucker 2009]: The dream theorem holds Caveat:  will have more qubits than , and in general be a very different state Proof combines Quantum Occam’s Razor Theorem with a new classical result about “isolatability” of functions

A “Practical” Implication It’s the year 2500. Everyone and her grandfather has a personal quantum computer. You’re a software vendor who sells “magic initial states” that extend quantum computers’ problem-solving abilities. Amazingly, you only need one kind of state in your store: ground states of 1D nearest-neighbor Hamiltonians! UNIVERSAL RESOURCE STATE Reason: Finding ground states of 1D spin systems is known to be “universal” for quantum constraint satisfaction problems[Aharonov-Gottesman-Irani-Kempe 2007], building on [Kitaev 1999]

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 a 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 “Experimental demonstration” would be nice too

www.scottaaronson.com(/papers /talks /blog) Postselection theorem: quant-ph/0402095 Learning theorem: quant-ph/0608142 Ground state theorem, numerical simulations: “in preparation”

How Much Information Is In A Quantum State?