Quantum Hypothesis Testing, Summaries of Physics

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Quantum Hypothesis Testing

Non-Commutative Chernoff and Hoeffding bounds

Koenraad M.R. Audenaert

Quantum Information at

February 16, 2007

Impressum

Based on joint work with:

• J. Calsamiglia, R. Munoz-Tapia, E. Bagan, Ll. Masanes, A. Acin (Barcelona)
• F. Verstraete (Vienna)
• M. Nussbaum (Cornell) and A. Szkola (Leipzig).

Read more about it in:

• KA et al. et F. Verstraete, “The Quantum Chernoff Bound”, quant-ph/0610027, to appear in PRL.
• M. Nussbaum and A. Szkola, quant-ph/0607216.
• ... and in a forthcoming full-length paper.

Distinguishing coins

• What I can do is: throw the coin n times and see how much heads come up.
• What I know about the coins is:
  • With coin H 0 , heads come up k times in n throws with probability

Pk =

n k

1 / 2 n.

• With coin H 1 this probability is

Qk =

n k

qHk qTn −k.

• Say, in an actual experiment, heads come up k times out of n.
• Maximum Likelihood (ML) Decision rule: if Pk > Qk, decide H 0 , else H 1.

ML Decision rule

(^00 20 40 60 80 100 )

k

Pk

N=

(0.55,0.45)

(0.9,0.1)

(0.5,0.5)

α (^) β

Error probabilities

(^00 20 40 60 80 100 )

k

Pk

N=

(0.55,0.45)

(0.9,0.1)

(0.5,0.5)

α (^) β

Answer 1

• Answer to Question 1: total error minimised by ML decision rule.
• What about Question 2? How big must n be to get “negligible” error?
• Depends on definition of “negligible”.
• Let’s look at how total error behaves in terms of n.

Answer 2

• Total Error Probability goes roughly as exp(−nR).
• Exponent R is the error rate (error exponent).
• We can take R as a qualitative answer to Question 2.
• It quantifies how well we’re doing, given p and q: efficiency of the decision rule
• In turn quantifies how alike p and q are: gives a distance measure on distributions
• Well, almost...

We need the Asymptotic Error Rate

(^010 20 30 40 50 60 70 80 90 )

N

R = −log(Pe)/N

(0.5,0.5) vs (0.9,0.1)

(0.5,0.5) vs (0.55,0.45)

Asymptotic Error Rate

• Asymptotic error rate hard to calculate directly: large n
• H. Chernoff (1952): Simple formula for asymptotic error rate:

n^ lim→∞ −^

n

log Pe = − log Q(p, q), where Q is defined as Q(p, q) = (^0) ≤infs≤ 1

i

psi q^1 i −s.

• The quantity − log Q is called the Chernoff Distance (Divergence, Bound).
• It is a measure of distinguishability between distributions.
• − log Q((0. 5 , 0 .5), (0. 9 , 0 .1)) = 0. 0488
• − log Q((0. 5 , 0 .5), (0. 55 , 0 .45)) = 0. 000545

Asymptotic Error Rate

(^010 20 30 40 50 60 70 80 90 )

N

R = −log(Pe)/N

(0.5,0.5) vs (0.9,0.1)

(0.5,0.5) vs (0.55,0.45)

Optimal Measurement

• Quantum version of n throws = ρ⊗n^ vs σ⊗n^ ∈ H⊗n.
• Measurement = POVM {E 0 , E 1 } on H⊗n, with 0 ≤ E 0 , E 1 ≤ 1 1 and E 0 +E 1 = 1 1.
• Decide on H 0 if outcome is ‘0’ (E 0 ), otherwise H 1.
• Type-I error: αn = Tr[E 0 σ⊗n], Type-II error: βn = Tr[E 1 ρ⊗n].
• Total error: Pe,n = (αn + βn)/ 2 (assuming equal priors).
• Optimal measurement: minimise Pe over all E 0 , E 1

Pe,min,n = min 0 ≤E 1 ≤ 11

Tr[( 11 − E 1 ) σ⊗n^ + E 1 ρ⊗n]/ 2

= (1 − max 0 ≤E 1 ≤ 11

Tr[E 1 (σ⊗n^ − ρ⊗n)])/ 2.

• Solution is based on the positive part of an operator/matrix.

The Positive Part

• The positive part H+ of a Hermitian matrix H is obtained by setting its negative eigenvalues equal to 0.
• In terms of the matrix absolute value: H+ = (H + |H|)/ 2.
• If P is the projector on (the support of) H+, we can write H+ = P H.
• For all Hermitian H, one has H+ ≥ H, and H+ ≥ 0.
• Variational expression for Tr H+: Tr H+ = maxQ Tr QH, where the maximisation is over all Hermitian projectors Q, and the optimum is achieved in Q = P , the projector on H+.
• Variant: maximise Tr QH over all positive contractions Q ( 0 ≤ Q ≤ 1 1). Same answer.

The Quantum Chernoff Bound

• Again, Pe goes down exponentially with n, with asymptotical rate

n^ lim→∞ −^

n log(1 − T (ρ⊗n, σ⊗n))

• Can we find a closed-form expression in the sense of Chernoff?
• Long-standing open problem.
• Ogawa and Hayashi (2004): three candidate expressions, based on the quantities

ψ 1 (s) = min{Tr[ρσs/^2 ρ−sσs/^2 ], Tr[σρ(1−s)/^2 σ−(1−s)ρ(1−s)/^2 ]} ψ 2 (s) = Tr[ρsσ^1 −s] ψ 3 (s) = Tr[exp((1 − s) log ρ + s log σ)], each of which reduces to

k p s kq 1 −s k for commuting^ ρ^ and^ σ.

Candidate #2 is an upper bound

• Nussbaum and Szkola (’06) proved that candidate #2,

− log min 0 ≤s≤ 1 Tr[ρsσ^1 −s],

is an upper bound to the error rate.

• Proof is based on a very special mapping of pairs of d-dim. states to pairs of d^2 -dim. probability vectors:

ρ = U ΛU ∗, σ = V M V ∗^7 → p = vec(ΛW ), q = vec(W M ), where W is an entrywise positive matrix s.t. Tr[ρσ] =

i,j(ΛW M^ )i,j.

• Can this bound be achieved? Is it also a lower bound?
• If so, this solves the problem completely!