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
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
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
Geometry vs Information theory
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
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
Isoperimetric inequality
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
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
Isoperimetric inequality
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
limε→0e2H(X+
√εZ)/n−e2H(X)/n
ε ≥ limε→0e2H(X)/n+e2H(
√εZ)/n−e2H(X)/n
ε = 2π e
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
Isoperimetric inequality
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
limε→0e2H(X+
√εZ)/n−e2H(X)/n
ε ≥ limε→0e2H(X)/n+e2H(
√εZ)/n−e2H(X)/n
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)
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
Isoperimetric inequality
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
limε→0e2H(X+
√εZ)/n−e2H(X)/n
ε ≥ limε→0e2H(X)/n+e2H(
√εZ)/n−e2H(X)/n
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
volume vol(A) entropy power e2H(X)/n
H(X) = −∫Rn fX(x) log fX(x)dx
addition A+B convolution: X + Y
fX+Y (x) =∫fX(x− z)fY (z)dz
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
Geometry Classical Quantum
Set A ∈ Rn Random variable X on Rn
with prob. density function fX
volume vol(A) entropy power e2H(X)/n
H(X) = −∫Rn fX(x) log fX(x)dx
addition A+B convolution: X + Y
fX+Y (x) =∫fX(x− z)fY (z)dz
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
ρX + ρY ??
Geometry Classical Quantum
Set A ∈ Rn Random variable X on Rn
with prob. density function fX
volume vol(A) entropy power e2H(X)/n
H(X) = −∫Rn fX(x) log fX(x)dx
addition A+B convolution: X + Y
fX+Y (x) =∫fX(x− z)fY (z)dz
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
ρXλY = Tr2
(Uλ(ρX ⊗ ρY )U†λ
)beamsplitter with transmissivity λ ∈ [0, 1]
Geometry Classical Quantum
Set A ∈ Rn Random variable X on Rn
with prob. density function fX
volume vol(A) entropy power e2H(X)/n
H(X) = −∫Rn fX(x) log fX(x)dx
addition A+B convolution: X + Y
fX+Y (x) =∫fX(x− z)fY (z)dz
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
ρXλY = Tr2
(Uλ(ρX ⊗ ρY )U†λ
)beamsplitter with transmissivity λ ∈ [0, 1]
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]
Brunn-Minkowski inequality
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
Geometry Classical Quantum
Set A ∈ Rn Random variable X on Rn
with prob. density function fX
volume vol(A) entropy power e2H(X)/n
H(X) = −∫Rn fX(x) log fX(x)dx
addition A+B convolution: X + Y
fX+Y (x) =∫fX(x− z)fY (z)dz
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
ρXλY = Tr2
(Uλ(ρX ⊗ ρY )U†λ
)beamsplitter with transmissivity λ ∈ [0, 1]
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]
Brunn-Minkowski inequality
vol(A)1/n + vol(B)1/n
Shannon entropy power inequality
e2H(X)/d + e2H(Y )/d ≤ e2H(X+Y )/d
Isoperimetric inequality
area(A) ≤ 14π length(∂A)2
Isoperimetric inequality for entropies
1nJ(X)e2H(X)/n ≥ 2π e
≤ vol(A+B)1/n
???
Geometry Classical Quantum
Set A ∈ Rn Random variable X on Rn
with prob. density function fX
volume vol(A) entropy power e2H(X)/n
H(X) = −∫Rn fX(x) log fX(x)dx
addition A+B convolution: X + Y
fX+Y (x) =∫fX(x− z)fY (z)dz
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
Brunn-Minkowski inequality
vol(A)1/n + vol(B)1/n
Shannon entropy power inequality
e2H(X)/d + e2H(Y )/d ≤ e2H(X+Y )/d
Isoperimetric inequality
area(A) ≤ 14π length(∂A)2
Isoperimetric inequality for entropies
1nJ(X)e2H(X)/n ≥ 2π e
≤ vol(A+B)1/n
f ?t ρ =∫f(ξ)W (
√tξ)ρW †(
√tξ)dξ
classical-quantum
Geometry Classical Quantum
Set A ∈ Rn Random variable X on Rn
with prob. density function fX
volume vol(A) entropy power e2H(X)/n
H(X) = −∫Rn fX(x) log fX(x)dx
addition A+B convolution: X + Y
fX+Y (x) =∫fX(x− z)fY (z)dz
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
Brunn-Minkowski inequality
vol(A)1/n + vol(B)1/n
Shannon entropy power inequality
e2H(X)/d + e2H(Y )/d ≤ e2H(X+Y )/d
Isoperimetric inequality
area(A) ≤ 14π length(∂A)2
Isoperimetric inequality for entropies
1nJ(X)e2H(X)/n ≥ 2π e
≤ 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
Isoperimetric inequality for entropies
1nJ(ρ)eS(ρ)/n ≥ 4π e
Classical vs Quantum information theory
Entropy
ρ - n-mode state.[Qj , Pk] = −[Pk, Qj ] = iδj,kI[Qj , Qk] = [Pj , Pk] = 0 1 ≤ j, k,≤ n
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
ρ - n-mode state.[Qj , Pk] = −[Pk, Qj ] = iδj,kI[Qj , Qk] = [Pj , Pk] = 0 1 ≤ j, k,≤ n
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
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
Weyl operators translate position and momentum operators
W (~θ)QjW (~θ) = Qj + θI W (~θ)PjW (~θ) = Pj + θI
Quantum Fisher information
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.
For ~θ ∈ R2n the Weyl displacement operators are defined as
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
(fX , fY )→ fX+Y (z) =∫fX(z − x)fY (x)dx
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
the Fisher information ineq.
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
(fX , fY )→ fX+Y (z) =∫fX(z − x)fY (x)dx
the Fisher information ineq.
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
the Fisher information ineq.
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
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
Entropy power inequality
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
Isoperimetric inequality
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
Entropy power inequality
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
Isoperimetric inequality
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
for 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
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),
for ω =√tωc + ωq
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
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
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),
for ω =√tωc + ωq
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
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
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),
for ω =√tωc + ωq
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
Quantum Fisher information isoperimetric inequality
ddt
∣∣∣∣t=0
[12nJ(fZ ?t ρ)
]−1≥ 1
Recall that for a Gaussian r.v. Z we have etL(ρ) = fZ ?t ρ
Quantum Fisher information isoperimetric inequality
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→∞.
Recall that for a Gaussian r.v. Z we have etL(ρ) = fZ ?t ρ
Quantum Fisher information isoperimetric inequality
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→∞.
Recall that for a Gaussian r.v. Z we have etL(ρ) = fZ ?t ρ
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
Concavity of the quantum entropy power
d2
dt2
∣∣∣∣t=0
N(fZ ?t ρ) ≤ 0
Recall that for a Gaussian r.v. Z we have etL(ρ) = fZ ?t ρ
Concavity of the quantum entropy power
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
Entropy power inequality
N(f ?t ρ) ≥ tN(f) +N(ρ)
Entropy power inequality
N(etL(ρ)) ≥ N(ρ) + t 2πe
Entropy power inequality
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
Entropy power inequality
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
EPI in equivalent to δ ≤ 1
Entropy power inequality
N(f ?t ρ) ≥ tN(f) +N(ρ)
Proof:
Classical heat diffusion semigroup on R2n Cc = 12
∑2nj=1
∂2
∂~ξ2j
exp(H(etCc(f))/d
)= (2πe)t+O(1)
exp(S(etL(ρ))/d
)= (2πe)t+O(1)
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
EPI in equivalent to δ ≤ 1
Step 1 lims→∞ δ(s) = 1
(in the limit t→∞)
Entropy power inequality
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
EPI in equivalent to δ ≤ 1
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
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→∞.
Isoperimetric Inequality for Entropies1nJ(ρ)N(ρ) ≥ 4πe
Optimality: let ρ = ωn be a Gaussian thermal state with a mean photon number n
Then S(ωn) = g(n) = (n + 1) log(n + 1)− n logn
Isoperimetric Inequality for Entropies1nJ(ρ)N(ρ) ≥ 4πe
Optimality: let ρ = 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 the state evolves as etL(ωn) = ωnt where nt = n + 2πt
In particular, by de Bruijn identity
J(ωn) = 2 ddt
∣∣∣∣t=0
S(etL(ωn)) = 4π log(n+1n
)
Isoperimetric Inequality for Entropies1nJ(ρ)N(ρ) ≥ 4πe
Optimality: let ρ = 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 the state evolves as etL(ωn) = ωnt where nt = n + 2πt
In particular, by de Bruijn identity
J(ωn) = 2 ddt
∣∣∣∣t=0
S(etL(ωn)) = 4π log(n+1n
)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
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
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
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
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
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
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
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
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))
Quantum Ornstein-Uhlenbeck semigroup
D(ρ‖σµ,λ) = −S(ρ)− (log ν)n− log(1− ν)
We have
nt = e−(µ2−λ2)tn + (1− e−(µ2−λ2)t)n∞ n∞ = tr(σµ,λn) = λ2
µ2−λ2
And soddt
∣∣∣t=0
nt = −(µ2 − λ2) (n− n∞)
Therefore,
− ddt
∣∣∣t=0
D(etLµ,λ(ρ)‖σµ,λ) = µ2
2 J−(ρ) + λ2
2 J+(ρ)− (log ν)(µ2 − λ2)n + λ2 log ν
Lemma−ζD(ρ‖σµ,λ)− d
dt
∣∣∣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))
n = Tr(ρn)
Proof
Quantum Ornstein-Uhlenbeck semigroup
D(ρ‖σµ,λ) = −S(ρ)− (log ν)n− log(1− ν)
We have
nt = e−(µ2−λ2)tn + (1− e−(µ2−λ2)t)n∞ n∞ = tr(σµ,λn) = λ2
µ2−λ2
And soddt
∣∣∣t=0
nt = −(µ2 − λ2) (n− n∞)
Therefore,
− ddt
∣∣∣t=0
D(etLµ,λ(ρ)‖σµ,λ) = µ2
2 J−(ρ) + λ2
2 J+(ρ)− (log ν)(µ2 − λ2)n + λ2 log ν
Lemma−ζD(ρ‖σµ,λ)− d
dt
∣∣∣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))
For ζ = µ2 − λ2 we obtain the desired equality
n = Tr(ρn)
Proof
Quantum Ornstein-Uhlenbeck semigroup
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)
Lemma−ζD(ρ‖σµ,λ)− d
dt
∣∣∣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))
Quantum Ornstein-Uhlenbeck semigroup
Theorem−ζD(ρ‖σµ,λ)− d
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)
Lemma−ζD(ρ‖σµ,λ)− d
dt
∣∣∣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))
Proof
Quantum Ornstein-Uhlenbeck semigroup
Theorem−ζD(ρ‖σµ,λ)− d
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
−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)
for ζ = µ2 − λ2, n = Tr(ρn)
Lemma−ζD(ρ‖σµ,λ)− d
dt
∣∣∣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))
Proof
Quantum Ornstein-Uhlenbeck semigroup
Theorem−ζD(ρ‖σµ,λ)− d
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
S(ρ) ≥ −2πAJ−(ρ)− 2πAJ+(ρ) + 2 + log(4πA)
Gaussian optimality: J−(ρ) ≥ J−(ωn) = −2n log(1 + 1/n)
for ζ = µ2 − λ2, n = Tr(ρn)
With the choice of A = λ2
4π(µ2−λ2) we obtain the desired bound
D(ρ‖σµ,λ) = −S(ρ)− (log ν)n− log(1− ν)
Lemma−ζD(ρ‖σµ,λ)− d
dt
∣∣∣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))
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)
Fast convergence of qOU semigroup µ2 = 2, λ2 = 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) ≤ −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)
Proof From the last Theorem we have
Fast convergence of qOU semigroup µ2 = 2, λ2 = 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.
We had that
−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)
ddt
∣∣∣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)
Proof From the last Theorem we have
Fast convergence of qOU semigroup µ2 = 2, λ2 = 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.
We had that
−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)
From [Buscemi et al.16] we have J+(ρ) ≥ 2
ddt
∣∣∣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)
Proof From the last Theorem we have
Fast convergence of qOU semigroup µ2 = 2, λ2 = 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.
We had that
−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)
From [Buscemi et al.16] we have J+(ρ) ≥ 2
From [De Palma et al. 16] we have
−ζD(ρ‖σµ,λ)− ddt
∣∣∣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) ≤ −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)
Proof From the last Theorem we have
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
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
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
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
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
≤ 14π4
∫| 5 h
1/2X |2dx− n−
∫hX log hY dx
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
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 ) ≤ 14π4
∫| 5 h
1/2X |2dx− n−
∫hX log hY dx
14πJ(X) +πE|X|2
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
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 ) ≤ 14πJ(X)− n+ πE|X|2
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
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 ) ≤ 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
)
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
)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)
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
)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)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))
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
)Step 1 (Correspondence to classical problem)
Let p = (p0, p1, . . . , ) be a prob. distribution
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))
H(p) := −∑∞k=0 pk log pk
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
)Step 1 (Correspondence to classical problem)
Let p = (p0, p1, . . . , ) be a prob. distribution
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))
Step 2 (Classical problem)
infp:Ep[N ]≤nddt
∣∣∣t=0
H(etC−(p)) = −n log(1 + 1n )
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
)Step 1 (Correspondence to classical problem)
Let p = (p0, p1, . . . , ) be a prob. distribution
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))
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
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