Download The Learnability of Quantum States

How Much Information Is In
Entangled Quantum States?
|
Scott Aaronson
MIT
The Problem
In quantum mechanics, a state of n entangled particles
requires at least 2n complex numbers to specify
 

x0,1
x
x
n
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
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?
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”
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[A.-Kuperberg
2007]:
For
the
Group
Nonlarge subgroup. Then the following highly-entangled state (if
Membership
therewhether
might also
be short
you have
it) can be problem,
used to decide
a given
element
that
are quickly
verifiable
a QCxH:
xGclassical
is in H orproofs
not, and
to prove
statements
of thebyform
But at least “relative to a quantum oracle,” there
are also cases where quantum proofs are provably
exponentially more compact than
proofsthe
Theclassical
trick: Measure
qubit in
Outstanding challenge to show suchfirst
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…
Suppose Alice wants to describe an n-qubit quantum
state | to Bob … well enough that, for any 2outcome 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
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, 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
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…
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
“Quantum
states
are
from some probability
distribution
D
PAC-learnable”
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
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
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