Transcript The Learnability of Quantum States

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…
1
0
0
C 2
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, even approximately
 

x
 x
x0,1n
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
ALL MEASUREMENTS
USING ≤n2 QUANTUM GATES
How does the theorem work?
I 321
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…
1. Computing x seems astronomically hard
2. 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 most
measurements E drawn from some probability measure D
Then it suffices to do the following, for some m=O(n):
1. Choose E1,…,Em independently from D
2. Go into your lab and estimate Tr(Ei) for each 1≤i≤m
3. Find any “hypothesis state”  such that Tr(Ei)Tr(Ei)
for all 1≤i≤m
Theorem [A. 2006]: Provided
C n
1
2 1
m  4  4 log  log  (C a constant)
 


and
T rEi   T rEi   
2
states
for all i, you’ll “Quantum
be guaranteed that
7
are
PAC-learnable”
Pr  TrE   TrE      1   ,
E~D
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”