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arXiv:0809
.3972v3
[quant-ph
]6Oct2008
A Counterexample to Additivity of Minimum Output Entropy
M. B. Hastings1
1Center for Nonlinear Studies and Theoretical Division,Los
Alamos National Laboratory, Los Alamos, NM, 87545
We present a random construction of a pair of channels which
gives, with non-zero probabilityfor sufficiently large dimensions,
a counterexample to the minimum output entropy conjecture. Asshown
by Shor[4], this implies a violation of the additivity conjecture
for the classical capacity ofquantum channels. The violation of the
minimum output entropy conjecture is relatively small.
PACS numbers:
The field of information theory started with the studyof the
capacity of a classical noisy channel to send clas-sical
information[1]. One fundamental property is addi-tivity: the
asymptotic rate at which information can betransmitted noiselessly
over a channel can be computedfrom a single-letter formula for the
information trans-mitted in one use of the channel. The only
advantage ofcorrelating the symbols across multiple channels (or
the
same channel multiple times) is the ability to do channelcoding.
Channel coding reduces the probability of er-ror, but at the cost
of sending a shorter message so thatthe asymptotic amount of
information that can be sentremains the same.
With the development of quantum information the-ory, it becomes
natural to ask about the ability to sendquantum information over
quantum channels, or clas-sical information over quantum channels.
In the firstcase, even the operationaldefinition of additivity was
re-cently shown to be false in dramatic fashion: there existtwo
channels, each of which has zero capacity to sendquantum
information no matter how many times it is
used, but which can be used in tandem to send
quantuminformation[2].
The question of additivity for sending classical infor-mation
over a quantum channel is known as the additivityconjecture. The
Holevo capacity[3] for sending classicalinformation over a quantum
channel E is defined to bethe maximum over density matrices i and
probabilities
pi of
H
i
piE(i)
i
piHE(i)
, (1)
where H() = Tr( ln()) is the von Neumann entropy.Shor[4] showed
that the additivity conjecture for the
capacity of quantum channels is equivalent to anotherconjecture,
the minimum output entropy conjecture[5].Given a channel E, we
define the minimum output en-tropy Hmin by
Hmin(E) = min|H(E(||)). (2)The minimum output entropy conjecture
is that for allchannels E1 and E2, we have
Hmin(E1 E2) = Hmin(E1) + Hmin(E2). (3)
In this letter we give a random construction that leads,with
high probability, to a counterexample to the min-imum output
entropy conjecture. Our construction issimilar to the constructions
Winter and Hayden used toshow violation of the maximal p-norm
multiplicativityconjecture for all p > 1[6, 7, 8]. For p = 1,
this violationwould imply violation of the minimum output
entropyconjecture; however, the counterexample found in [7] re-
quires a matrix size which diverges as p 1. We use dif-ferent
system and environment sizes (note that D > D, l2i /L
2 is very likely to be close to 1/D, sothat we apply unitaries
with roughly equal probability.The only reason in what follows for
not choosing all theprobabilities equal is that the choice we made
will allowus to appeal to certain exact results on random
bipartitestates later.
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We also define the conjugate channel
EC() =Di=1
Dj=1
liljL2
Tr
Ui Uj
|ij|, (7)
As shown in [10]
Hmin(E) = Hmin(EC). (8)In the O(...) notation that follows, we
will take
1
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Similarly, we can expand
S = i
pi ln(pi) ln(D) Di
p2i/2 + ... (18)
Using Eq. (17,18), we find that the probability of having
S = ln(D)S is roughly O(N)O(D2) exp[(ND)DS].Using -nets, these
estimates (17,18) give some motiva-
tion for the construction we are using, but just fail to givea
good enough bound on their own: define an -net withdistance d cmm
of
O(N2)O(D2)DNDpND exp[(N D)D(p 1/D)] exp[O(D2)ln(N) (N
D)D2c2MM(ln(N)/(N D))/
By picking cMM large enough, we can make this probabil-ity
arbitrarily small for sufficiently large D and N/D.
The next lemma is the crucial step.
Lemma 4. Consider a given choice of Ui and li whichgive a EC
such that PEC 1/2. Suppose there exists astate |0, such that
EC(|00|) has given eigenvalues
p1,...,pD. Let Pnear denote the probability that, for arandomly
chosen state|, the density matrixEC(||)has eigenvalues q1,...,qD
which obey
|qi(ypi+(1y)(1/D))| poly(D)O(
ln(N)/(N D))(21)
for some y 1/2. Then,
Pnear exp(O(N))(1/2 1/poly(D)), (22)
where the power of D in the polynomial in (22) can bemade
arbitrarily large by an appropriate choice of thepolynomial in
(21).
Proof. Consider a random state . We can write
| =
1 x2|0 + x|, (23)
With probability exp(O(N)), we have x2 1/2.Since is random, the
probability distribution of |
is that of a random state with|0
= 0. One way to
generate such a random state | with this property is tochoose a
random state | and set
| = 11 |0||2
1 |00|
. (24)
If we choose a random state |, then with probabil-ity at least
1/2, the state EC(||) is close to maxi-mally mixed. Further, for
any given i, j, the probabil-ity that |0|UiUj || is greater than
O(
ln(D)/N) is
1/poly(D), and the polynomial poly(D) can be chosento be any
given power of D by appropriate choice of theconstant hidden in
the
Onotation for
O(ln(D)/N).
Therefore,
P r
Tr|EC(|0|)|
poly(D)
ln(D)/N
1/poly(D),
(25)with any desired power of D in the polynomial on
theright-hand side.
Then, since
Pr| | O(ln(D)/N) 1/poly(D), (26)
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we find that
Pr
Tr|EC(|0|)|
poly(D)
ln(D)/N
1/poly(D), (27)
with again any desired power in the polynomial.The probability
that EC(||) is close to maximally
mixed is at least 1/2, and so by (26) the probability that
the eigenvalues r1,...,rD ofEC(||) obey|ri 1/D| cMM
ln(N)/(N D) + poly(D)(ln(D)/N)
poly(D)O(
ln(N)/N) (28)
is at least 1/2 1/poly(D). Lety = 1 x2. (29)
Thus, since
EC(||) = ( 1 x2)EC(|00|) + x2EC(||)(30)+x1
x2)
EC(
|
0
|) + h.c.,
using Eq. (27) we find that for given x, the probabilitythat a
randomly chosen | gives a state with eigenvaluesq1,...,qD such
that
|qi(ypi+(1y)(1/D))| poly(D)O(
ln(N)/N) (31)
is 1/2 1/poly(D). Combining this result with theexp(O(N))
probability of x2 1/2, the claim of thelemma follows.
Lemma 5. If the unitary matrices Ui are chosen atrandom from the
Haar measure, and the li are chosenrandomly as described above,
then the probability thatHmin(EC) is less than ln(D) Smax is less
than one
for sufficiently largeN, for appropriate choice of c1 andp1. The
N required depends on D.
Proof. Let Pbad denote the probability that Hmin(EC)
