Counterexamples to the maximal p-norm multiplicativity conjecture

Patrick Hayden (McGill University)

|| ||N(½)p

C&QIC, Santa Fe 2008

A challenge to the physicists

John Pierce [1973]: I think that I have never met a physicist

who understood information theory. I wish that physicists would stop talking about reformulating information theory and would give us a general expression for the capacity of a channel with quantum effects taken into account rather than a number of special cases.

Sending classical information

through noisy quantum channels

Physical model of a noisy channel:(Trace-preserving, completely positive map)

HSW noisy coding theorem: In the limit of many uses, the optimalrate at which Alice can send bits reliably to Bob through N is given by the (regularization of the) formula

where the maximization is over some family of input/output states.

m Encoding( state)

Decoding(measurement)

m’

Sending classical information

through noisy quantum channels

Physical model of a noisy channel:(Trace-preserving, completely positive map)

m Encoding( state)

Decoding(measurement)

m’

HSW noisy coding theorem: In the limit of many uses, the optimalrate at which Alice can send bits reliably to Bob through N is given by the (regularization of the) formula

The additivity conjecture:These two formulas are equal

where

Sustained, heroic, and so far inconclusive efforts by: Datta, Eisert, Fukuda, Holevo, King, Ruskai,

Schumacher, Shirokov, Shor, Werner...

Why do they care so much?

The additivity conjecture:These two formulas are equal

where

Operational interpretation: •Alice doesn’t need to entangle her inputs across multiple uses of the channel.• Codewords look like ¾x1

¾x2 L ¾xn

QMAC solution pre-QIP 2005

Interpretation: Alice and Bob treat each others’ actions as noise. Independent decoding.No-go theorem for use of quantum side information.

[Yard/Devetak/H 05 v1]

QMAC solution post-QIP 2005

Interpretation: Charlie decodes Alice’s quantum data first and uses itto help him decode Bob’s. (Or vice-versa.)Go theorem for use of quantum side information.

[Yard/Devetak/H 05 v2]

Capacity formulas matter

Fair question to throw at the speaker if you’re getting bored in any quantum Shannon theory talk: “Can you describe an effective procedure

for calculating this capacity you claim to have determined?”

If we can’t write down a tractable formula for thesolution to a capacity problem, then we don’t fully understand the structure of the optimal codes.

Lesson:

An (Almost) Equivalent Form:

Minimum Entropy Outputs

• H() = - Tr[ log ] (von Neumann entropy of the density operator )

• N, N1 and N2 are quantum channels. (CPTP)

Notation:

• Hmin(N) = min H(N()) is the minimum output entropy of N.

Conjecture:

The minimum entropy output state for the product channel N1 N2 is attained by a product state input 1 2.

[King-Ruskai 99]

Maximal p-norm multiplicativity conjecture

Conjecture:

The minimum entropy output state for the product channel N1 N2 is attained by a product-state input 1 2.

Maximal p-norm multiplicativity conjecture

Conjecture:

The minimum entropy output state for the product channel N1 N2 is attained by a product-state input 1 2.

Renyi entropy (1 < p ):

(Recover von Neumann entropy as p 1.)

Norm? What norm?

[Amosov-Holevo-Werner 00]

Partial results: Additivity holds if...

One channel is Unitary A unital qubit channel A generalized depolarizing channel A generalized dephasing channel Entanglement-breaking A very noisy channel

Complements of these channels

[Amosov, Devetak, Eisert, Fujiwara, Hashizume, Holevo, King, Matsumoto, Nathanson, Ruskai, Shor, Wolf, Werner]

[See Holevo ICM 2006]

But...

2002: Additivity fails for p > 4.79... [Holevo-Werner]

2007: Additivity fails for p > 2.

[Winter]

Counterexamples for 1<p<2!

For all 1 < p < 2, there exist channels N1 and N2 to Cd such that:• Hp

min(N1) , Hpmin(N2) log d - O(1)

• Hpmin(N1 N2) p log d + O(1)

Additivity would have implied:

Hpmin(N1 N2) 2 log d -

O(1)Near p=1, minimum output entropy of N1 N2

not significantly greater than that of N1 or N2 alone!

Intuition: Channels that look very noisy (nearly depolarizing)need not be anywhere near depolarizing on entangled input.

2p

1

The counterexamples

U|0

N()R

SA

B TRASHN N()S A

Fix dimensions |R|<<|S|, |A|=|B| and choose U at randomaccording to Haar measure. Demonstrate resulting channelsviolate Renyi additivity with non-zero probability.

Two things to prove:i) Product channel has low minimum output entropy.ii) Individual channels have high minimum output entropies.

N N has low output entropy

The key identity:

N N has low output entropy

The key identity (v1):

The key identity (v2):

U|0

N()R

SA

B TRASH

Easy calculation:

This is BIG if |R| is small! (Compare 1/|A|2 for maximally mixed state.)Choose |R| ~ |A|p-1.

N and N have high output entropy

U|0 N()R

SA

B TRASH|

If U is selected at random, what can be said about U||0?

U||0 is highly entangled between A and B: Hp( N() ) log|A| - O(1)

(Compare maximally mixed state: log|A|.)

N N()S A|

[Lubkin, Lloyd, Page, Foong & Kanno, Sanchez-Ruiz, Sen…]

Is this true simultaneously for all | S with a typical U?

i.e. Is min| S Hp( N() ) log|A| - O(1) ?

Concentration of measure

Sn

LEVY: Given an -Lipschitz function f : Sn ! R with median M, the probability that, for a random x 2R Sn , f (x) is further than from M is bounded above by exp (-n2 C/2) from some C > 0.

An

An < exp[-n g()]for some g() indep. of n

f (x)=x1

Just need a Lipschitz constant: Choosing f the map from | to Hp(N()), can take 2 |A|p-1.

Pr[ Hp(N()) < log|A|- const - ] ~ exp( - const 2|A|3-p )

Connect the dotsU (S |0 ½ A B 1) Choose a fine net F of states on the

unit sphere of S |0.2) P( Not all states in UF highly entangled )

· |F| P( One state isn’t )3) Highly entangled for sufficiently

fine N implies same for all states in S.

THEOREM: If |R|~|A|p-1, then |S| ~ |A|3-p and w.h.p. as|A| ,

min| S Hp( N() ) log|A| - O(1).

N and N have high minimum output entropy.

Done!

For all 1 < p < 2, there exist channels N1 and N2 to Cd such that:• Hp

min(N1) , Hpmin(N2) log d - O(1)

• Hpmin(N1 N2) p log d + O(1)

Additivity would have implied:

Hpmin(N1 N2) 2 log d -

O(1)Near p=1, minimum output entropy of N1 N2

not significantly greater than that of N1 or N2 alone!

What about von Neumann (p=1)???

Method fails: recall |R|~|A|p-1. Constants depend on p and blow up.

Artifact of the analysis or does the conjecture

survive at p=1?

|R|=3 |A|=|B|=24(NN)()

What about von Neumann (p=1)???

Method fails: recall |R|~|A|p-1. Constants depend on p blow up.

Artifact or does the conjecturesurvive at p=1?

Hp for p > 1 very sensitive to a single large eigenvalue, but H1 is not.

Do some calculating

Contribution from eigenvalue ~1/|R|

Contribution from all the others

For Hp, p > 1, first term dominates but second term dominates H1

H1((N N)()) = 2 log|A| - O(1)is BIG not small

No additivity violations. To be sure, can anyone calculate the O(1) terms?

Summary

Additivity fails for 1 < p < 2. Closes main approach to additivity for capacity itself.

Further developments: Winter tightened Lipschitz bound, showing

same examples work for 1 < p < Dupuis showed orthogonal group can

replace unitary group: N1 = N2

Cubitt, Harrow, Leung, Montanaro & Winter have found violations for 0 p 0.12