· Geometry Classical Quantum Set A 2Rn Random variable X on R n with prob. density function f X...

Quantum analogues of geometric inequalities

for Information Theory

Anna Vershynina

June 17, 2016Workshop on Mathematical Many-Body Theory and its Applications

Bilbao, Spain

based on a joint work with Robert Koenig and Stefan Huber

Technical University of Munich, Germany

Outline of the talk

Introduction

Geometric inequalities in information theory

Classical vs quantum inequalities

Quantum inequalities

Application to the quantum Ornstein-Uhlenbeck semigroup

Optimizing entropy rates of the quantum attenuator semigroup

Entropy power inequality

Concavity of the entropy power of diffusion semigroup

Quantum isoperimetric inequality

Geometric inequalities

Brunn-Minkowski inequality

A B A +B

A+B = a+ b|a ∈ A, b ∈ B

Rn = R2

vol(A)1/n vol(B)1/n+ vol(A+B)1/n≤

Isoperimetric inequality

A area(A) ≤ 14π length(∂A)2

Geometry vs Information theory

Set A ⊂ Ω Random variable X on Ω

with prob. mass function px = Pr[X = x]

volume vol(A) entropy power 2H(X)

Shannon entropy H(X) = −∑px log px


Set A ⊂ Ω Random variable X on Ω

with prob. mass function px = Pr[X = x]

volume vol(A) entropy power 2H(X)

Shannon entropy H(X) = −∑px log px

Consider a set An = A× · · · ×A ∈ Ωn Consider n i.i.d. with PXn = PX · · · · · PXhas the following property:

∀ε > 0 ∃Mn,ε ⊂ Ωn s.t.

and Pr[Xn ∈Mn,ε] ≥ 1− ε

|Mn,ε| ∼(2H(X)

)nvol(An) = (vol(A))n


Set A ∈ Rn Random variable X on Rnwith prob. density function fX

volume vol(A) entropy power e2H(X)/n

entropy H(X) = −∫Rn fX(x) log fX(x)dx

addition A+B convolution: X + Y has a density function

fX+Y (x) =∫fX(x− z)fY (z)dz

fX fY fX+Y

Geometric inequalitiesBrunn-Minkowski inequality

vol(A)1/n + vol(B)1/n ≤ vol(A+B)1/n

Shannon‘s entropy power inequality [‘48]

e2H(X)/n + e2H(Y )/n ≤ e2H(X+Y )/n




e2H(X)/n + e2H(Y )/n ≤ e2H(X+Y )/n


area(A) ≤ 14π length(∂A)2

length(∂A) = limε→0area(A+Bε(0))−area(A)

ε“length“ limε→0

e2H(X+√εZ)/n−e2H(X)/d

ε

+ =A

Bε(0)

A+Bε(0)

fX f√εZ fX+√εZ

ε




e2H(X)/n + e2H(Y )/n ≤ e2H(X+Y )/n






ε

limε→0e2H(X+

√εZ)/n−e2H(X)/n

ε ≥ limε→0e2H(X)/n+e2H(


ε = 2π e




e2H(X)/n + e2H(Y )/n ≤ e2H(X+Y )/n






ε

limε→0e2H(X+




ε = 2π e

ddε

∣∣ε=0

e2H(X+√εZ)/n = 2

ne2H(X)/n d

dε

∣∣ε=0

H(X +√εZ)

Fisher information

J(X) = 2 ddε

∣∣ε=0

H(X +√εZ)




e2H(X)/n + e2H(Y )/n ≤ e2H(X+Y )/n






ε

limε→0e2H(X+




ε = 2π e

ddε

∣∣ε=0

e2H(X+√εZ)/n = 2

ne2H(X)/n d

dε

∣∣ε=0

H(X +√εZ)

Fisher information

J(X) = 2 ddε

∣∣ε=0

H(X +√εZ)

Isoperimetric inequality for entropies

1nJ(X)e2H(X)/n ≥ 2π e

Geometric analogue of Fisher information

x2

x1

A ε

ε

A+ ε ~x1

A+ ε ~x2

D(A1, A2) =“difference“ between A1 and A2

“Fisher Information“= d2

dε2

∣∣∣ε=0

∑2j=1D(A,A+ ε ~xj)

A

A

Geometric analogue of Fisher information

x2

x1

A ε

ε

A+ ε ~x1

A+ ε ~x2

D(A1, A2) =“difference“ between A1 and A2

“Fisher Information“= d2

dε2

∣∣∣ε=0

∑2j=1D(A,A+ ε ~xj)

A

A

J(X) =∑nj=1

∂2

∂ε2

∣∣∣ε=0

D(f ||f ε ~xj )

distribution translated in ~xj direction by ε

de Bruijn‘s identity

= 2 ddε

∣∣ε=0

H(X +√εZ)

Geometry Classical Quantum

Set A ∈ Rn Random variable X on Rn

with prob. density function fX


H(X) = −∫Rn fX(x) log fX(x)dx

addition A+B convolution: X + Y


ρ - n-mode state.[Qj , Pk] = −[Pk, Qj ] = iδj,kI[Qj , Qk] = [Pj , Pk] = 0 1 ≤ j, k,≤ n

S(ρ) = −Tr(ρ log ρ)

N(ρ) = expS(ρ)/n










N(ρ) = expS(ρ)/n

ρX + ρY ??










N(ρ) = expS(ρ)/n

ρXλY = Tr2

(Uλ(ρX ⊗ ρY )U†λ

)beamsplitter with transmissivity λ ∈ [0, 1]










N(ρ) = expS(ρ)/n

ρXλY = Tr2



Quantum entropy power inequality

λeS(X)/n + (1− λ)eS(Y )/n ≤ eS(XλY )/n

[Koenig, Smith 14] λ = 1/2

[de Palma et al 15]


vol(A)1/n + vol(B)1/n

Shannon entropy power inequality

e2H(X)/d + e2H(Y )/d ≤ e2H(X+Y )/d

≤ vol(A+B)1/n










N(ρ) = expS(ρ)/n

ρXλY = Tr2



Quantum entropy power inequality

λeS(X)/n + (1− λ)eS(Y )/n ≤ eS(XλY )/n

[Koenig, Smith 14] λ = 1/2

[de Palma et al 15]




e2H(X)/d + e2H(Y )/d ≤ e2H(X+Y )/d





≤ vol(A+B)1/n

???










N(ρ) = expS(ρ)/n




e2H(X)/d + e2H(Y )/d ≤ e2H(X+Y )/d





≤ vol(A+B)1/n

f ?t ρ =∫f(ξ)W (

√tξ)ρW †(

√tξ)dξ

classical-quantum










N(ρ) = expS(ρ)/n




e2H(X)/d + e2H(Y )/d ≤ e2H(X+Y )/d





≤ vol(A+B)1/n

f ?t ρ =∫f(ξ)W (

√tξ)ρW †(

√tξ)dξ

classical-quantum

Classical-quantum entropy power inequality

teH(f)/n + eS(ρ)/n ≤ eS(f?tρ)/n t ≥ 0


1nJ(ρ)eS(ρ)/n ≥ 4π e

Classical vs Quantum information theory

Entropy


H(X) = H(f) = −∫f(x) log f(x)dx S(ρ) = −Tr(ρ log ρ)

Entropy Power N(f) = exp2H(f)/n N(ρ) = expS(ρ)/n

Relative entropy D(f ||g) =∫f(x) log f(x)

g(x)dx D(ρ||σ) = Tr(ρ log ρ− ρ log σ)

Classical QuantumX - Rn-valued r.v. with a prob. density function fX .

Shannon von Neumann

Classical vs Quantum information theory

Entropy


H(X) = H(f) = −∫f(x) log f(x)dx S(ρ) = −Tr(ρ log ρ)

Entropy Power N(f) = exp2H(f)/n N(ρ) = expS(ρ)/n

Relative entropy D(f ||g) =∫f(x) log f(x)

g(x)dx D(ρ||σ) = Tr(ρ log ρ− ρ log σ)

Fisher information matrix For ~θ ∈ Rn, define

f (θ)(x) = f(x− θ)

J(f (θ))∣∣θ=θ0

=(

∂2

∂θi∂θj

∣∣θ=θ0

D(f (θ0)||f (θ)

))2ni,j=1

Fisher information

J(f) = Tr(J(f (θ))

∣∣θ=θ0

)

Classical QuantumX - Rn-valued r.v. with a prob. density function fX .

Shannon von Neumann

Quantum Fisher information

W (~θ) = ei√2π ~θ·(σ ~R),

ρ(θ) = W (θ)ρW †(θ)

~R = (Q1, P1, . . . , Qn, Pn), σ =

(0 1−1 0

)⊕n.

For ~θ ∈ R2n the Weyl displacement operators are defined as

where

Translated state is


W (~θ) = ei√2π ~θ·(σ ~R),

ρ(θ) = W (θ)ρW †(θ)

~R = (Q1, P1, . . . , Qn, Pn), σ =

(0 1−1 0

)⊕n.


where

Translated state is

Weyl operators translate position and momentum operators

W (~θ)QjW (~θ) = Qj + θI W (~θ)PjW (~θ) = Pj + θI


W (~θ) = ei√2π ~θ·(σ ~R),

ρ(θ) = W (θ)ρW †(θ)

J(ρ(θ))∣∣θ=θ0

=(

∂2

∂θi∂θj

∣∣θ=θ0

D(ρ(θ0)||ρ(θ)

))2ni,j=1

J(ρ) = Tr(J(ρ(θ))

∣∣θ=θ0

)

~R = (Q1, P1, . . . , Qn, Pn), σ =

(0 1−1 0

)⊕n.


where

Translated state is

Weyl operators translate position and momentum operators

W (~θ)QjW (~θ) = Qj + θI

Fisher information matrix

Fisher information

W (~θ)PjW (~θ) = Pj + θI

Inequalities

Classical

(fX , fY )→ fX+Y (z) =∫fX(z − x)fY (x)dx

Quantum

(f, ρ)→ f ?t ρ =∫f(ξ)W (

√tξ)ρW †(

√tξ)dξt = 1 [Werner ‘84]

Inequalities

Classical


the Fisher information ineq.

Quantum

(f, ρ)→ f ?t ρ =∫f(ξ)W (

√tξ)ρW †(

√tξ)dξ

for λ ∈ [0, 1]

J(√

λX +√

1− λY)≤ λJ(X) + (1− λ)J(Y )

Stam inequality

J (X + Y )−1 ≥ J(X)−1 + J(Y )−1


for ω =√tωc + ωq

ω2J(f ?t ρ) ≤ ω2cJ(f) + ω2

qJ(ρ)

Stam inequality

J(f ?t ρ)−1 ≥ tJ(f)−1 + J(ρ)−1

t = 1 [Werner ‘84]

Inequalities

Classical



Quantum

(f, ρ)→ f ?t ρ =∫f(ξ)W (

√tξ)ρW †(

√tξ)dξ

Define Liouvillean L(ρ) = −π∑2nj=1[Rj , [Rj , ρ]].

for λ ∈ [0, 1]

J(√

λX +√

1− λY)≤ λJ(X) + (1− λ)J(Y )

Stam inequality

J (X + Y )−1 ≥ J(X)−1 + J(Y )−1



ω2J(f ?t ρ) ≤ ω2cJ(f) + ω2

qJ(ρ)

Stam inequality

J(f ?t ρ)−1 ≥ tJ(f)−1 + J(ρ)−1

ρ(t) = etL(ρ) : ddtρ(t) = L(ρ(t))

etL(ρ) = 1(2π)n

∫e−‖

~ξ‖2/2W (√t~ξ) ρW †(

√t~ξ)d~ξ

For Z - Gaussian r.v. : fZ(~ξ) = (2π)−ne−|ξ|2/2

fZ ?t ρ = etL(ρ)

Z - Gaussian r.v. : fZ(~ξ) = (2π)−n/2e−|ξ|2/2

t = 1 [Werner ‘84]

Consider

Note that

Inequalities

de Bruijn identity

J(X +

√tZ)

= 2 ∂∂tH

(X +

√tZ)

J(ρ) = 2 ddtS

(etL(ρ)

)∣∣t=0

Classical Quantum


N(X + Y ) ≥ N(X) +N(Y ) N(f ?t ρ) ≥ tN(f) +N(ρ)

Concavity of the entropy power

d2

dε2

∣∣∣∣ε=0

N(X +√εZ) ≤ 0 d2

dt2

∣∣∣∣t=0

N(etL(ρ)) ≤ 0

Fisher information isoperimetric inequality

ddε

∣∣∣∣ε=0

[1n J(X +

√εZ)

]−1 ≥ 1 ddt

∣∣∣∣t=0

[12nJ(etL(ρ))

]−1≥ 1


1nJ(X)N(X) ≥ 2πe 1

nJ(ρ)N(ρ) ≥ 4πe

etL(ρ) = fZ ?t ρ

[Koenig, Smith ‘14]

Inequalities

de Bruijn identity

J(X +

√tZ)

= 2 ∂∂tH

(X +

√tZ)

J(ρ) = 2 ddtS

(etL(ρ)

)∣∣t=0

Classical Quantum


N(X + Y ) ≥ N(X) +N(Y ) N(f ?t ρ) ≥ tN(f) +N(ρ)

Concavity of the entropy power

d2

dε2

∣∣∣∣ε=0

N(X +√εZ) ≤ 0 d2

dt2

∣∣∣∣t=0

N(etL(ρ)) ≤ 0

Fisher information isoperimetric inequality

ddε

∣∣∣∣ε=0

[1n J(X +

√εZ)

]−1 ≥ 1 ddt

∣∣∣∣t=0

[12nJ(etL(ρ))

]−1≥ 1


1nJ(X)N(X) ≥ 2πe 1

nJ(ρ)N(ρ) ≥ 4πe

Tight for Gaussian states:ωn a 1-mode G. thermal st.with a mean-photon number n

J(ωn)N(ωn) = 4π(n+1n

)nlog(n+1n

)n+1→4πe, as n→∞

etL(ρ) = fZ ?t ρ

[Koenig, Smith ‘14]

Quantum Inequalities

Rescaling

fX ?t ρ = f√tX ?1 ρ

If fX is a prob. density function of a r.v. X, then

here the r.v.√tX is given by f√tX(ζ) = f(ζ/

√t)/(√t)2n

AdditionfX ?1 (fY ?1 ρ) = fX+Y ?1 ρ

where (fX , fY )→ fX+Y (z) =∫fX(z − x)fY (x)dx

Translation Let ωc, ωq > 0, and t ≥ 0. Then for all θ ∈ R2n

(f ?t ρ)(ωθ) = f (ωcθ) ?t ρ(ωqθ) ω =

√tωc + ωq

Quantum Inequalities

Data Processing Inequality

D (f ?t ρ‖ g ?t σ) ≤ D (f‖ g) +D (ρ‖σ)

Rescaling

fX ?t ρ = f√tX ?1 ρ

If fX is a prob. density function of a r.v. X, then

here the r.v.√tX is given by f√tX(ζ) = f(ζ/

√t)/(√t)2n

AdditionfX ?1 (fY ?1 ρ) = fX+Y ?1 ρ

where (fX , fY )→ fX+Y (z) =∫fX(z − x)fY (x)dx

Translation Let ωc, ωq > 0, and t ≥ 0. Then for all θ ∈ R2n

(f ?t ρ)(ωθ) = f (ωcθ) ?t ρ(ωqθ) ω =

√tωc + ωq

Stam Inequality

TheoremLet ωc, ωq ∈ R, and t ≥ 0. Then ω2J(f ?t ρ) ≤ ω2

cJ(f) + ω2qJ(ρ)

In particular,J(f ?t ρ)−1 − J(ρ)−1 − tJ(f)−1 ≥ 0


Stam Inequality


cJ(f) + ω2qJ(ρ)


Proof

By Data processing inequality

Tr(J(f (ωc

~θ) ?t ρ(ωq~θ)); ~θ)

∣∣∣~θ= ~θ0

)≤ Tr

(J(f (ωc

~θ); ~θ)∣∣∣~θ= ~θ0

+ J(ρ(ωq~θ); ~θ)

∣∣∣∣~θ= ~θ0

)Since, (f ?t ρ)

(ω~θ)= f (ωc

~θ) ?t ρ(ωq~θ),


Tr

(J(f (ωc

~θ) ?t ρ(ωq~θ)); ~θ)

∣∣∣∣~θ=~0

)= Tr

(J(f ?t ρ(ω

~θ); ~θ)

∣∣∣∣~θ=~0

)= ω2J(f ?t ρ)

Stam Inequality


cJ(f) + ω2qJ(ρ)


Proof


Tr(J(f (ωc

~θ) ?t ρ(ωq~θ)); ~θ)

∣∣∣~θ= ~θ0

)≤ Tr

(J(f (ωc

~θ); ~θ)∣∣∣~θ= ~θ0

+ J(ρ(ωq~θ); ~θ)

∣∣∣∣~θ= ~θ0

)Since, (f ?t ρ)

(ω~θ)= f (ωc

~θ) ?t ρ(ωq~θ),


Tr

(J(f (ωc

~θ) ?t ρ(ωq~θ)); ~θ)

∣∣∣∣~θ=~0

)= Tr

(J(f ?t ρ(ω

~θ); ~θ)

∣∣∣∣~θ=~0

)= ω2J(f ?t ρ)

Since, J(f (ωc~θ); ~θ) = ω2

cJ(f (~θ); ~θ), and J(ρ(ωq

~θ); ~θ) = ω2qJ(ρ(

~θ); ~θ)

Tr

(J(f (ωc

~θ); ~θ)

∣∣∣∣~θ=~0

)+ Tr

(J(ρ(ωq

~θ); ~θ)

∣∣∣∣~θ=~0

)= ω2

cJ(f) + ω2qJ(ρ)

Stam Inequality


cJ(f) + ω2qJ(ρ)


Proof


Tr(J(f (ωc

~θ) ?t ρ(ωq~θ)); ~θ)

∣∣∣~θ= ~θ0

)≤ Tr

(J(f (ωc

~θ); ~θ)∣∣∣~θ= ~θ0

+ J(ρ(ωq~θ); ~θ)

∣∣∣∣~θ= ~θ0

)Since, (f ?t ρ)

(ω~θ)= f (ωc

~θ) ?t ρ(ωq~θ),


Tr

(J(f (ωc

~θ) ?t ρ(ωq~θ)); ~θ)

∣∣∣∣~θ=~0

)= Tr

(J(f ?t ρ(ω

~θ); ~θ)

∣∣∣∣~θ=~0

)= ω2J(f ?t ρ)

Since, J(f (ωc~θ); ~θ) = ω2

cJ(f (~θ); ~θ), and J(ρ(ωq

~θ); ~θ) = ω2qJ(ρ(

~θ); ~θ)

Tr

(J(f (ωc

~θ); ~θ)

∣∣∣∣~θ=~0

)+ Tr

(J(ρ(ωq

~θ); ~θ)

∣∣∣∣~θ=~0

)= ω2

cJ(f) + ω2qJ(ρ)

Taking ωc =√tJ(f)−1

J(ρ)−1+tJ(f)−1 and ωq = J(ρ)−1

J(ρ)−1+tJ(f)−1 , we obtain J(f ?t ρ)−1 − J(ρ)−1 − tJ(f)−1 ≥ 0

Quantum Fisher information isoperimetric inequality

ddt

∣∣∣∣t=0

[12nJ(etL(ρ))

]−1≥ 1


ddt

∣∣∣∣t=0

[12nJ(fZ ?t ρ)

]−1≥ 1

Recall that for a Gaussian r.v. Z we have etL(ρ) = fZ ?t ρ


Proof:

ddt

∣∣∣∣t=0

[12nJ(fZ ?t ρ)

]−1≥ 1

Take f = fZ in q. Stam inequality:

1t

J(fZ ?t ρ)−1 − J(ρ)−1

≥ J(fZ)−1 = (2n)−1

Take a limit t→∞.



Proof:

ddt

∣∣∣∣t=0

[12nJ(fZ ?t ρ)

]−1≥ 1

Take f = fZ in q. Stam inequality:

1t

J(fZ ?t ρ)−1 − J(ρ)−1

≥ J(fZ)−1 = (2n)−1

Take a limit t→∞.


Optimality: for n = 1, let ωn be a Gaussian thermal state with a mean-photon number n

Then S(ωn) = g(n) = (n + 1) log(n + 1)− n logn

Under L: etL(ωn) = ωnt with nt = n + 2πt

Therefore, J(ωn) = 2 ddt

∣∣∣∣t=0

S(etL(ωn)) = 4π log(n+1n

)J(etL(ωn)) = 4π log

(1 + 1

n+2πt

)ddt

∣∣∣∣t=0

[12nJ(etL(ρ))

]−1= 1

n(n+1) log−2(1 + 1

n

)→ 1 as n→∞

Concavity of the quantum entropy power

d2

dt2

∣∣∣∣t=0

N(etL(ρ)) ≤ 0


d2

dt2

∣∣∣∣t=0

N(fZ ?t ρ) ≤ 0



d2

dt2

∣∣∣∣t=0

N(etL(ρ)) ≤ 0

Proofde Bruijn identity: J(ρ) = 2 d

dtS(etL(ρ)

)∣∣t=0

Therefore:

d2

dt2

∣∣∣∣t=0

N(etL(ρ)) = N(ρ)[

12nJ(ρ)

]2+ 1

2nddtJ(etLρ)

∣∣∣∣t=0

ddtN(etL(ρ)) = 1

nN(etL(ρ)) ddtS(etL(ρ))

d2

dt2N(etL(ρ)) = 1n2N(etL(ρ))

(ddtS(etL(ρ))

)2+ 1

nN(etL(ρ)) d2

dt2S(etL(ρ))

From quantum Fisher information isoperimetric inequality:

12nJ(ρ)2 + d

dtJ(etLρ)

∣∣∣∣t=0

≤ 0


N(f ?t ρ) ≥ tN(f) +N(ρ)


N(etL(ρ)) ≥ N(ρ) + t 2πe


N(f ?t ρ) ≥ tN(f) +N(ρ)

Proof: Consider δ := expS(ρ)/n+t expH(f)/nexpS(f?tρ)/n

EPI in equivalent to δ ≤ 1


N(f ?t ρ) ≥ tN(f) +N(ρ)

Proof:

Classical heat diffusion semigroup on R2n Cc = 12

∑2nj=1

∂2

∂~ξ2j

Define

µ→ EA(µ) := exp(S(eµL(ρ))/n

)ν → EB(ν) := exp

(H(eνCc(f))/n

)ξ → EC(ξ) := exp

(S(eξL(f ?t ρ)))/n

)Consider

µ(s) = EA(µ(s)) , µ(0) = 0

ν(s) = EB(ν(s)) , ν(0) = 0

ξ(s) := µ(s) + tν(s)

Define

δ(s) := EA(µ(s))+tEB(ν(s))EC(ξ(s))

EPI is equivalent to δ(0) ≤ 1

Consider δ := expS(ρ)/n+t expH(f)/nexpS(f?tρ)/n



N(f ?t ρ) ≥ tN(f) +N(ρ)

Proof:


∑2nj=1

∂2

∂~ξ2j

exp(H(etCc(f))/d

)= (2πe)t+O(1)

exp(S(etL(ρ))/d

)= (2πe)t+O(1)

Define



(H(eνCc(f))/n



)Consider

µ(s) = EA(µ(s)) , µ(0) = 0

ν(s) = EB(ν(s)) , ν(0) = 0

ξ(s) := µ(s) + tν(s)

Define





Step 1 lims→∞ δ(s) = 1

(in the limit t→∞)


N(f ?t ρ) ≥ tN(f) +N(ρ)

Proof:


∑2nj=1

∂2

∂~ξ2j

Define



(H(eνCc(f))/n



)Consider

µ(s) = EA(µ(s)) , µ(0) = 0

ν(s) = EB(ν(s)) , ν(0) = 0

ξ(s) := µ(s) + tν(s)

Define





Step 1 lims→∞ δ(s) = 1 Step 2 δ(s) ≥ 0 for all s ≥ 0

eξL(f ?t ρ) = eνCc(f) ?t eµL(ρ)

Note that

Isoperimetric Inequality for Entropies1nJ(ρ)N(ρ) ≥ 4πe


Proof:from de Bruijn identity:

ddt

∣∣∣∣t=0

N(ρ(t)) = 12nJ(ρ)N(ρ)

denoting ρ(t) = etL(ρ), t ≥ 0, the entropy power inequality yields:

1t [N(ρ(t))−N(ρ)] ≥ 2πe

take a limit t→∞.


Optimality: let ρ = ωn be a Gaussian thermal state with a mean photon number n





Under L the state evolves as etL(ωn) = ωnt where nt = n + 2πt

In particular, by de Bruijn identity

J(ωn) = 2 ddt

∣∣∣∣t=0


)




Under L the state evolves as etL(ωn) = ωnt where nt = n + 2πt

In particular, by de Bruijn identity

J(ωn) = 2 ddt

∣∣∣∣t=0


)N(ωn) = exp(S(ωn)/1) = (n+1)n+1

nn

Also

J(ωn)N(ωn) = 4π(n+1n

)nlog(n+1n

)n+1 →n→∞

4πe

Therefore

Quantum Ornstein-Uhlenbeck semigroup

Quantum Attenuator:L−(ρ) = aρa† − 1

2a†a, ρ

Quantum Amplifier L+(ρ) = a†ρa− 12aa

†, ρ

n = 1

Let ρ = ωn be a Gaussian thermal state with a mean photon number n. Then ρ(t) = ωn(t)

Under L−n−(t) = e−tn

Under L+

n+(t) = etn + et − 1



2a†a, ρ


†, ρ

n = 1



Under L+

n+(t) = etn + et − 1

One-parameter group of CPTP maps eLµ,λt≥0, generated by

Lµ,λ = µ2L− + λ2L+ for µ > λ > 0

Then ωn(t) is

nt = e−(µ2−λ2)tn + (1− e−(µ2−λ2)t)n∞ n∞ = tr(σµ,λn) = λ2

µ2−λ2



2a†a, ρ


†, ρ

n = 1



Under L+

n+(t) = etn + et − 1


Lµ,λ = µ2L− + λ2L+ for µ > λ > 0

Then ωn(t) is


µ2−λ2

Conjecture

D(etLµ,λ(ρ)‖σµ,λ) ≤ e−(µ2−λ2)tD(ρ‖σµ,λ) for all t ≥ 0

Fixed point of Lµ,λ is σµ,λ = (1− ν)∑∞n=0 ν

n|n〉〈n|, with ν = λ2/µ2



2a†a, ρ


†, ρ

n = 1



Under L+

n+(t) = etn + et − 1


Lµ,λ = µ2L− + λ2L+ for µ > λ > 0

Then ωn(t) is


µ2−λ2

Conjecture

D(etLµ,λ(ρ)‖σµ,λ) ≤ e−(µ2−λ2)tD(ρ‖σµ,λ) for all t ≥ 0

To this end we will focus on −ζD(ρ‖σµ,λ)− ddt

∣∣∣t=0

D(etLµ,λ(ρ)‖σµ,λ) for ζ = µ2 − λ2

Fixed point of Lµ,λ is σµ,λ = (1− ν)∑∞n=0 ν

n|n〉〈n|, with ν = λ2/µ2

Quantum Ornstein-Uhlenbeck semigroupLemma

−ζD(ρ‖σµ,λ)− ddt

∣∣∣t=0

D(etLµ,λ(ρ)‖σµ,λ) = µ2

2 J−(ρ) + λ2

2 J+(ρ) + ζS(ρ) + λ2 log ν + ζ log(1− ν)

J±(ρ) := 2 ddtS(etL±(ρ)) J(ρ) = 2π(J−(ρ) + J+(ρ))


D(ρ‖σµ,λ) = −S(ρ)− (log ν)n− log(1− ν)

We have


µ2−λ2

And soddt

∣∣∣t=0

nt = −(µ2 − λ2) (n− n∞)

Therefore,

− ddt

∣∣∣t=0


2 J−(ρ) + λ2

2 J+(ρ)− (log ν)(µ2 − λ2)n + λ2 log ν

Lemma−ζD(ρ‖σµ,λ)− d

dt

∣∣∣t=0


2 J−(ρ) + λ2



n = Tr(ρn)

Proof



We have


µ2−λ2

And soddt

∣∣∣t=0

nt = −(µ2 − λ2) (n− n∞)

Therefore,

− ddt

∣∣∣t=0


2 J−(ρ) + λ2

2 J+(ρ)− (log ν)(µ2 − λ2)n + λ2 log ν


dt

∣∣∣t=0


2 J−(ρ) + λ2



For ζ = µ2 − λ2 we obtain the desired equality

n = Tr(ρn)

Proof


Theorem−ζD(ρ‖σµ,λ)− d

dt

∣∣∣t=0

D(etLµ,λ(ρ)‖σµ,λ) ≥ −ζn log(1 + 1/n) + δ for ζ = µ2 − λ2, n = Tr(ρn)


dt

∣∣∣t=0


2 J−(ρ) + λ2





dt

∣∣∣t=0

D(etLµ,λ(ρ)‖σµ,λ) ≥ −ζn log(1 + 1/n) + δ

From Isoperimetric inequality for entropies

−S(ρ) ≤ AJ(ρ)− (2 + log(4πA))

for A > 0

for ζ = µ2 − λ2, n = Tr(ρn)


dt

∣∣∣t=0


2 J−(ρ) + λ2



Proof



dt

∣∣∣t=0



−S(ρ) ≤ AJ(ρ)− (2 + log(4πA))

for A > 0

−S(ρ) ≤ log

14πeJ(ρ)

using log x ≤ x− 1

log(

14πeJ(ρ)

)= log

(AJ(ρ)4πeA

)= log

(1

4πeA

)+ log

(AJ(ρ)

)≤ AJ(ρ)− 2− log(4πA)



dt

∣∣∣t=0


2 J−(ρ) + λ2



Proof



dt

∣∣∣t=0



−S(ρ) ≤ AJ(ρ)− (2 + log(4πA))

for A > 0

S(ρ) ≥ −2πAJ−(ρ)− 2πAJ+(ρ) + 2 + log(4πA)

Gaussian optimality: J−(ρ) ≥ J−(ωn) = −2n log(1 + 1/n)


With the choice of A = λ2

4π(µ2−λ2) we obtain the desired bound



dt

∣∣∣t=0


2 J−(ρ) + λ2



Proof

Fast convergence of qOU semigroup µ2 = 2, λ2 = 1

Exampleddt

∣∣∣t=0

D(etL√

2,1(ρ)‖σ√2,1) ≤ −D(ρ‖σ√2,1) for all states ρ s.t. tr(ρn) . 0.67

or S(ρ) & 2.4In particular,D(etL

√2,1(ρ)‖σ√2,1) ≤ e−tD(ρ‖σ√2,1)


Example

−ζD(ρ‖σ√2,1)− ddt

∣∣∣t=0

D(etL√

2,1(ρ)‖σ√2,1) ≥ −n log(1 + 1/n) + 2− 2 log(2)

The RHS is monotonically decreasing, and for n . 0.67, is non-negative.

ddt

∣∣∣t=0

D(etL√



√2,1(ρ)‖σ√2,1) ≤ e−tD(ρ‖σ√2,1)

Proof From the last Theorem we have


Example


∣∣∣t=0

D(etL√

2,1(ρ)‖σ√2,1) ≥ −n log(1 + 1/n) + 2− 2 log(2)


We had that


∣∣∣t=0


2 J−(ρ) + λ2


ddt

∣∣∣t=0

D(etL√



√2,1(ρ)‖σ√2,1) ≤ e−tD(ρ‖σ√2,1)



Example


∣∣∣t=0

D(etL√

2,1(ρ)‖σ√2,1) ≥ −n log(1 + 1/n) + 2− 2 log(2)


We had that


∣∣∣t=0


2 J−(ρ) + λ2


From [Buscemi et al.16] we have J+(ρ) ≥ 2

ddt

∣∣∣t=0

D(etL√



√2,1(ρ)‖σ√2,1) ≤ e−tD(ρ‖σ√2,1)



Example


∣∣∣t=0

D(etL√

2,1(ρ)‖σ√2,1) ≥ −n log(1 + 1/n) + 2− 2 log(2)


We had that


∣∣∣t=0


2 J−(ρ) + λ2


From [Buscemi et al.16] we have J+(ρ) ≥ 2

From [De Palma et al. 16] we have


∣∣∣t=0

D(etLµ,λ(ρ)‖σµ,λ) ≥ F (S0) + λ2 + λ2 log ν + ζ log(1− ν)

Also, for S0 & 2.06, RHS is non-negative

ddt

∣∣∣t=0

D(etL√



√2,1(ρ)‖σ√2,1) ≤ e−tD(ρ‖σ√2,1)


Function F (S) is monotonically increasing for S & 0.5 for µ2 = 2, λ2 = 1

Optimality of the convergence rate

ConjectureD(etLµ,λ(ρ)‖σµ,λ) ≤ e−(µ2−λ2)tD(ρ‖σµ,λ) for all t ≥ 0

Optimality of the convergence rate

ConjectureD(etLµ,λ(ρ)‖σµ,λ) ≤ e−(µ2−λ2)tD(ρ‖σµ,λ) for all t ≥ 0

Tightness:for any Gaussian state

ddt

∣∣∣t=0

D(etLµ,λ(ρ)‖σµ,λ) ≤ −(µ2 − λ2)D(ρ||σµ,λ)

Moreover, for any ζ > µ2 − λ2 there exists a Gaussian state ρ s.t.

ddt

∣∣∣t=0

D(etLµ,λ(ρ)‖σµ,λ) > −ζD(ρ||σµ,λ)

Log-Sobolev inequality

∫|f |2 log |f |2e−π|x|2dx ≤ 1

π

∫| 5 f |2e−π|x|2dx for

∫|f |2e−π|x|2dx = 1

Let g(x) = f(x)e−π|x|2/2. Then

Gross‘ logarithmic Sobolev inequality is[Carlen ‘91]

∫|g|2 log |g|2dx ≤ 1

π

∫| 5 g|2dx− n with

∫|g|2dx = 1


∫|f |2 log |f |2e−π|x|2dx ≤ 1

π


∫|f |2e−π|x|2dx = 1



∫|g|2 log |g|2dx ≤ 1

π


∫|g|2dx = 1

Let r.v. X has density hX(x) = |g(x)|2 and r.v. Y has density hY (x) = e−π|x|2

. Then

H(X|Y )

∫hX log hXdx−

∫hX log hY dx


∫|f |2 log |f |2e−π|x|2dx ≤ 1

π


∫|f |2e−π|x|2dx = 1



∫|g|2 log |g|2dx ≤ 1

π


∫|g|2dx = 1


. Then

H(X|Y )

∫hX log hXdx−

∫hX log hY dx

≤ 14π4

∫| 5 h

1/2X |2dx− n−

∫hX log hY dx


∫|f |2 log |f |2e−π|x|2dx ≤ 1

π


∫|f |2e−π|x|2dx = 1



∫|g|2 log |g|2dx ≤ 1

π


∫|g|2dx = 1


. Then

H(X|Y ) ≤ 14π4

∫| 5 h

1/2X |2dx− n−

∫hX log hY dx

14πJ(X) +πE|X|2


∫|f |2 log |f |2e−π|x|2dx ≤ 1

π


∫|f |2e−π|x|2dx = 1



∫|g|2 log |g|2dx ≤ 1

π


∫|g|2dx = 1


. Then

H(X|Y ) ≤ 14πJ(X)− n+ πE|X|2


∫|f |2 log |f |2e−π|x|2dx ≤ 1

π


∫|f |2e−π|x|2dx = 1



∫|g|2 log |g|2dx ≤ 1

π


∫|g|2dx = 1


. Then

H(X|Y ) ≤ 14πJ(X)− n+ πE|X|2

D(ρ‖σµ,λ) = −S(ρ)− (log ν)n− log(1− ν) ≤ AJ(ρ)− 2− log(1− ν)− log(4πA) + n log 1ν

Quantum case:

ν = λ2/µ2 < 1

A > 0

Gaussian optimality for energy-constrained entropy rates

Quantum Attenuator: L−(ρ) = aρa† − 12a†a, ρ

Theoreminfρ:Tr(nρ)≤n

ddt

∣∣t=0

S(etL−(ρ)) = −n log(1 + 1

n

)




ddt

∣∣t=0

S(etL−(ρ)) = −n log(1 + 1

n

)Step 1 (Correspondence to classical problem)

Let p = (p0, p1, . . . , ) be a prob. distribution

For ρ =∑n pn|n〉〈n| we have

ρ(t) = etL−(ρ) =∑n pn(t)|n〉〈n|

with pn(t) = −npn(t) + (n+ 1)pn+1(t)




ddt

∣∣t=0

S(etL−(ρ)) = −n log(1 + 1

n



For ρ =∑n pn|n〉〈n| we have

ρ(t) = etL−(ρ) =∑n pn(t)|n〉〈n|

with pn(t) = −npn(t) + (n+ 1)pn+1(t)Denote (C−(p))n = −npn + (n+ 1)pn+1

Then p(t) = etC−(p) - pure-death processTheorem

infρ:Tr(nρ)≤nddt

∣∣∣t=0

S(etL−(ρ)) = infp:Ep[N ]≤nddt

∣∣∣t=0

H(etC−(p))




ddt

∣∣t=0

S(etL−(ρ)) = −n log(1 + 1

n



Denote (C−(p))n = −npn + (n+ 1)pn+1



∣∣∣t=0


∣∣∣t=0

H(etC−(p))

H(p) := −∑∞k=0 pk log pk




ddt

∣∣t=0

S(etL−(ρ)) = −n log(1 + 1

n






∣∣∣t=0


∣∣∣t=0

H(etC−(p))

Step 2 (Classical problem)

infp:Ep[N ]≤nddt

∣∣∣t=0

H(etC−(p)) = −n log(1 + 1n )




ddt

∣∣t=0

S(etL−(ρ)) = −n log(1 + 1

n






∣∣∣t=0


∣∣∣t=0

H(etC−(p))

Step 2 (Classical problem)

infp:Ep[N ]≤nddt

∣∣∣t=0

H(etC−(p)) = −n log(1 + 1n )

Step 3 (Gaussian optimality) Let ωn be a Gaussian thermal state with a mean-photon number nThen

ddt

∣∣∣t=0

S(etL−(ωn)) = −n log(1 + 1

n

)

Thank you!

References

S. Huber, R. Koenig, A. Vershynina. Geometric inequalities from phase-space translations. In preparation

R. Koenig and G. Smith. The entropy power inequality for quantum systems. Infor- mation Theory, IEEETransactions on, 60(3):1536-1548, March 2014.

G. De Palma, D. Trevisan, and V. Giovannetti. Gaussian states minimize the output entropy of the one-modequantum attenuator. 2016. arXiv:1605.00441

F. Buscemi, S. Das, and M. M. Wilde. Approximate reversibility in the context of entropy gain, information gain, andcomplete positivity. 2016. arXiv:1601.01207

G. De Palma, A. Mari, S. Lloyd, and V. Giovannetti. Multimode quantum entropy power inequality. Phys. Rev. A,91:032320, Mar 2015

R. Werner. Quantum harmonic analysis on phase space. Journal of Mathematical Physics, 25(5):1404-1411, 1984

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