Multivariate trace inequalities. David Sutter, Mario Berta, Marco Tomamichel

Transcription

1 Multivariate trace inequalities David Sutter, Mario Berta, Marco Tomamichel

2 What are trace inequalities and why we should care. Main difference between classical and quantum world are complementarity and entanglement Quantum mechanical observables may not be simultaneously measurable (complementarity) Mathematically this means that operators do not need to commute A and B commute if [A, B] := AB BA = 0

3 What are trace inequalities and why we should care. Main difference between classical and quantum world are complementarity and entanglement Quantum mechanical observables may not be simultaneously measurable (complementarity) Mathematically this means that operators do not need to commute A and B commute if [A, B] := AB BA = 0 To understand QM we need to comprehend the behavior of functions involving matrices that do not commute Trace inequalities allow us to do that

4 What are trace inequalities and why we should care. Main difference between classical and quantum world are complementarity and entanglement Quantum mechanical observables may not be simultaneously measurable (complementarity) Mathematically this means that operators do not need to commute A and B commute if [A, B] := AB BA = 0 To understand QM we need to comprehend the behavior of functions involving matrices that do not commute Trace inequalities allow us to do that. Trace inequalities are powerful (mathematical) tools in proofs

5 Golden-Thompson (GT) inequality (965) Golden-Thompson: Let H and H be Hermitian. Then tr e H +H tr e H e H

6 Golden-Thompson (GT) inequality (965) Golden-Thompson: Let H and H be Hermitian. Then tr e H +H tr e H e H Not so easy to prove If [H, H ] = 0 then equality holds (trivial) Incredibly useful (wherever matrix exponentials occur) Statistical physics (bound partition function) [Golden-65 & Thompson-65] Random matrix theory (tail bounds via Laplace method) [Ahlswede-Winter-0] Information theory (entropy inequalities) [Lieb-Ruskai-73] Control theory, dynamical systems, Does not extend to n matrices (at least not in an obvious way)

7 GT inequality for more than two matrices tr e H +H tr e H e H Extensions to three matrices are not immediate tr e H +H +H 3 tr e H e H e H 3

8 GT inequality for more than two matrices tr e H +H tr e H e H Extensions to three matrices are not immediate tr e H +H +H 3 tr e H e H e H 3 tr e H +H +H 3 tr e H e H e H 3 e H

9 GT inequality for more than two matrices tr e H +H tr e H e H Extensions to three matrices are not immediate tr e H +H +H 3 tr e H e H e H 3 tr e H +H +H 3 tr e H e H e H 3 e H Lieb s triple matrix inequality (973) tr e H +H +H 3 0 dλ tr e H ( e H + λ ) e H 3 ( e H + λ ) Equivalent to many other interesting statements Lieb s concavity theorem: A tr exp(h + log A) is concave Strong subadditivity of quantum entropy (SSA): H(AB) + H(BC) H(ABC) H(B) 0

10 GT inequality for more than two matrices tr e H +H tr e H e H Extensions to three matrices are not immediate tr e H +H +H 3 tr e H e H e H 3 tr e H +H +H 3 tr e H e H e H 3 e H Lieb s triple matrix inequality (973) tr e H +H +H 3 0 dλ tr e H ( e H + λ ) e H 3 ( e H + λ ) Equivalent to many other interesting statements Lieb s concavity theorem: A tr exp(h + log A) is concave Strong subadditivity of quantum entropy (SSA): H(AB) + H(BC) H(ABC) H(B) 0 Open problem: extensions of GT for more than 3 matrices?

11 Outline for the rest of the talk. Understanding GT better (intuitive proof based on pinching). Extending GT to n matrices 3. Tightening the result (using interpolation theory) 4. Application: entropy inequalities via extended GT

12 The spectral pinching method Question: How do we force matrices to commute, changing them as little as possible?

13 The spectral pinching method Question: How do we force matrices to commute, changing them as little as possible? Any positive definite matrix A can be written (spectral decomposition) as A = λp λ λ spec(a) The pinching map with respect to A is P A : X λ spec(a) P λ X P λ

14 The spectral pinching method Question: How do we force matrices to commute, changing them as little as possible? Any positive definite matrix A can be written (spectral decomposition) as A = λp λ λ spec(a) The pinching map with respect to A is P A : X λ spec(a) P λ X P λ Properties of pinching maps:. [P A (X ), A] = 0 for all X 0. tr P A (X )A = tr AX for all X 0 3. P A (X ) spec(a) X for all X 0 trace is cyclic, i.e., tr AB = tr BA

15 The spectral pinching method Question: How do we force matrices to commute, changing them as little as possible? Any positive definite matrix A can be written (spectral decomposition) as A = λp λ λ spec(a) The pinching map with respect to A is P A : X λ spec(a) P λ X P λ Properties of pinching maps:. [P A (X ), A] = 0 for all X 0. tr P A (X )A = tr AX for all X 0 3. P A (X ) spec(a) X for all X 0 trace is cyclic, i.e., tr AB = tr BA Operator inequality A B A B 0

16 An intuitive proof of the GT inequality Golden-Thompson: Let H and H be Hermitian. Then tr e H +H tr e H e H Any Hermitian matrix H can be written as log A for some positive definite matrix A Let H k := log A k A k = e H k for k {, }

17 An intuitive proof of the GT inequality Golden-Thompson: Let H and H be Hermitian. Then tr e H +H tr e H e H Any Hermitian matrix H can be written as log A for some positive definite matrix A Let H k := log A k A k = e H k for k {, } Let A and A be positive definite matrices. Then tr exp(log A + log A ) tra A

18 An intuitive proof of the GT inequality (con t) To show: tr exp(log A + log A ) tra A

19 An intuitive proof of the GT inequality (con t) To show: tr exp(log A + log A ) tra A log tr exp(log A + log A ) = m log tr exp(log A m + log A m ) trace is multiplicative under tensor products, i.e., tr B m = (tr B) m

20 An intuitive proof of the GT inequality (con t) To show: tr exp(log A + log A ) tra A log tr exp(log A + log A ) = m log tr exp(log A m + log A m ) m log tr exp ( log A m + log P A m(a m ) ) log poly(m) + m Pinching property 3: P A (X ) spec(a) X spec(a m ) = ( ) m+d d = poly(m) log( ) is operator monotone, i.e. X Y log X log Y tr exp( ) is operator monotone If spec(a) = {λ, λ } then spec(a ) = {λ, λ λ, λ λ, λ }

21 An intuitive proof of the GT inequality (con t) To show: tr exp(log A + log A ) tra A log tr exp(log A + log A ) = m log tr exp(log A m + log A m ) m log tr exp ( log A m + log P A m(a m ) ) log poly(m) + m = log tr A m m P A m (A m ) + log poly(m) m Pinching property : [P A (X ), A] = 0 log A + log B = log AB if [A, B] = 0

22 An intuitive proof of the GT inequality (con t) To show: tr exp(log A + log A ) tra A log tr exp(log A + log A ) = m log tr exp(log A m + log A m ) m log tr exp ( log A m + log P A m(a m ) ) log poly(m) + m = log tr A m m = m P A m log tr A m A m + (A m ) + log poly(m) m log poly(m) m Pinching property : tr P A (X )A = tr AX

23 An intuitive proof of the GT inequality (con t) To show: tr exp(log A + log A ) tra A log tr exp(log A + log A ) = m log tr exp(log A m + log A m ) m log tr exp ( log A m + log P A m(a m ) ) log poly(m) + m = log tr A m m = m P A m log tr A m A m + = log tr A A + log poly(m) m log poly(m) m (A m ) + log poly(m) m trace is multiplicative under tensor products

24 An intuitive proof of the GT inequality (con t) To show: tr exp(log A + log A ) tra A log tr exp(log A + log A ) = m log tr exp(log A m + log A m ) m log tr exp ( log A m + log P A m(a m ) ) log poly(m) + m = log tr A m m = m P A m log tr A m A m + = log tr A A + log poly(m) m log poly(m) m (A m ) + log poly(m) m = log tr A A log poly(m) lim m m = 0

25 An intuitive proof of the GT inequality (con t) To show: tr exp(log A + log A ) tra A log tr exp(log A + log A ) = m log tr exp(log A m + log A m ) m log tr exp ( log A m + log P A m(a m ) ) log poly(m) + m = log tr A m m = m P A m log tr A m A m + = log tr A A + log poly(m) m log poly(m) m (A m ) + log poly(m) m = log tr A A Why should this be intuitive?

26 Extension of GT to n matrices Same proof technique can be applied (pinch iteratively) Fact: For any A > 0 a probability measure µ on R such that P A (X ) = µ(dt)a it XA it Note that A it is a unitary that commutes with A

27 Extension of GT to n matrices Same proof technique can be applied (pinch iteratively) Fact: For any A > 0 a probability measure µ on R such that P A (X ) = µ(dt)a it XA it Note that A it is a unitary that commutes with A For three matrices we find tr e H +H +H 3 sup tr e H e +it H e H 3 e it H t R Same is true for n matrices (each additional matrix gives an additional pair of unitaries)

28 Extension of GT to n matrices Same proof technique can be applied (pinch iteratively) Fact: For any A > 0 a probability measure µ on R such that P A (X ) = µ(dt)a it XA it Note that A it is a unitary that commutes with A For three matrices we find tr e H +H +H 3 sup tr e H e +it H e H 3 e it H t R Same is true for n matrices (each additional matrix gives an additional pair of unitaries) Example: n = 4 tr e H +H +H 3 +H 4 sup tr e H e +it H e +it H 3 e H 4 e it H 3 e it H t,t R

29 Extension of GT to n matrices Same proof technique can be applied (pinch iteratively) Fact: For any A > 0 a probability measure µ on R such that P A (X ) = µ(dt)a it XA it Note that A it is a unitary that commutes with A For three matrices we find tr e H +H +H 3 sup tr e H e +it H e H 3 e it H t R Same is true for n matrices (each additional matrix gives an additional pair of unitaries) Can we replace the supremum Example: n = 4 by something independent of H k? tr e H +H +H 3 +H 4 sup tr e H e +it H e +it H 3 e H 4 e it H 3 e it H t,t R

30 Extension of GT to n matrices (con t) n matrix extension of GT: Let p, n N and consider a collection {H k } n k= of Hermitian matrices. Then ( n ) log exp n H k dt β 0 (t) log p exp( p ( + it)h k) k= k= where β 0 (t) := π ( cosh(πt) + ) β 0(t) t

31 Extension of GT to n matrices (con t) n matrix extension of GT: Let p, n N and consider a collection {H k } n k= of Hermitian matrices. Then ( n ) log exp n H k dt β 0 (t) log p exp( p ( + it)h k) k= k= Let n = 3 and p = tr e H +H +H 3 = 0 dt β 0 (t) tr e H e +it H e H 3 e it H dλ tr e H ( e H + λ ) e H 3 ( e H + λ ) Reproduces Lieb s triple matrix inequality Proof uses complex interpolation theory (Stein-Hirschman see [Junge-Renner-S-Wilde-Winter-5]) Complex interpolation theory has been used in QIT recently, e.g., [Beigi-3], [Dupuis-4], [Wilde-5]

32 Applications Approximate quantum Markov chains Strengthened strong subadditivity of entropy

33 Approximate quantum Markov chains A B C Definition: A density matrix ρ ABC is a quantum Markov chain (QMC) if there exists a recovery map R B BC such that ρ ABC = (I A R B BC )(ρ AB )

34 Approximate quantum Markov chains A B C Definition: A density matrix ρ ABC is a quantum Markov chain (QMC) if there exists a recovery map R B BC such that ρ ABC = (I A R B BC )(ρ AB ) Theorem [Petz-88]: ρ ABC is a QMC iff I (A : C B) = 0 with R B BC : X B ρ BC (ρ B X Bρ B id C )ρ BC

35 Approximate quantum Markov chains A B C Definition: A density matrix ρ ABC is a quantum Markov chain (QMC) if there exists a recovery map R B BC such that ρ ABC = (I A R B BC )(ρ AB ) Theorem [Petz-88]: ρ ABC is a QMC iff I (A : C B) = 0 with R B BC : X B ρ BC (ρ B X Bρ B id C )ρ BC Question: What about states such that I (A : C B) ɛ?

36 Approximate quantum Markov chains A B C Definition: A density matrix ρ ABC is a quantum Markov chain (QMC) if there exists a recovery map R B BC such that ρ ABC = (I A R B BC )(ρ AB ) Theorem [Petz-88]: ρ ABC is a QMC iff I (A : C B) = 0 with R B BC : X B ρ BC (ρ B X Bρ B id C )ρ BC Question: What about states such that I (A : C B) ɛ? Theorem [Fawzi-Renner-4]: For any ρ ABC there exists R B BC such that I (A : C B) ρ log F ( ρ ABC, R B BC (ρ AB ) ) 0

37 Why the classical case is easy Theorem [Fawzi-Renner-4]: For any ρ ABC there exists R B BC such that I (A : C B) ρ log F ( ρ ABC, R B BC (ρ AB ) ) 0 Suppose A, B, and C are classical (i.e., ρ AB, ρ BC, and ρ B are diagonal) I (A : C B) ρ = D ( ρ ABC exp(log ρ AB + log ρ BC log ρ B ) ) = D ( ρ ABC ρ BC (ρ B ρ ABρ = D ( id ρ ABC R B BC (ρ AB ) )B C )ρ ) BC If [X, Y ] := XY YX = 0, then log XY = log X + log Y

38 Why the classical case is easy Theorem [Fawzi-Renner-4]: For any ρ ABC there exists R B BC such that I (A : C B) ρ log F ( ρ ABC, R B BC (ρ AB ) ) 0 Suppose A, B, and C are classical (i.e., ρ AB, ρ BC, and ρ B are diagonal) I (A : C B) ρ = D ( ρ ABC exp(log ρ AB + log ρ BC log ρ B ) ) = D ( ρ ABC ρ BC (ρ B ρ ABρ = D ( id ρ ABC R B BC (ρ AB ) )B C )ρ ) BC If A, B, and C are quantum, the density matrices ρ AB, ρ BC, and ρ B are not diagonal and do not commute

39 Why the classical case is easy Theorem [Fawzi-Renner-4]: For any ρ ABC there exists R B BC such that I (A : C B) ρ log F ( ρ ABC, R B BC (ρ AB ) ) 0 Suppose A, B, and C are classical (i.e., ρ AB, ρ BC, and ρ B are diagonal) I (A : C B) ρ = D ( ρ ABC exp(log ρ AB + log ρ BC log ρ B ) ) = D ( ρ ABC ρ BC (ρ B ρ ABρ = D ( id ρ ABC R B BC (ρ AB ) )B C )ρ ) BC If A, B, and C are quantum, the density matrices ρ AB, ρ BC, and ρ B are not diagonal and do not commute D(ρ σ) log F (ρ, σ)

40 Details about Fawzi-Renner-4 Theorem [Fawzi-Renner-4]: For any ρ ABC there exists R B BC such that I (A : C B) ρ log F ( ρ ABC, R B BC (ρ AB ) ) 0 Measured relative entropy: D M (ρ σ) := sup M D ( M(ρ) M(σ) ). D M (ρ σ) log F (ρ, σ). D M (ρ σ) = D(ρ σ) iff [ρ, σ] = 0 There are several generalizations and improvements of the Fawzi-Renner bound (see QIP 06) Open question: a bound that is tight in the classical case with an explicit and universal recovery map?

41 Application: Strenghtened strong subadditivity Variational formula for relative entropy [Petz-88]: D(ρ σ) = sup tr ρ log ω + tr exp(log σ + log ω) ω>0 Variational formula for measured relative entropy [Berta-Fawzi-Tomamichel-5]: D M (ρ σ) = sup tr ρ log ω + trσω ω>0

42 Application: Strenghtened strong subadditivity Variational formula for relative entropy [Petz-88]: D(ρ σ) = sup tr ρ log ω + tr exp(log σ + log ω) ω>0 Variational formula for measured relative entropy [Berta-Fawzi-Tomamichel-5]: D M (ρ σ) = sup tr ρ log ω + trσω ω>0 I (A : C B) ρ = D ( ρ ABC exp(log ρ AB + log ρ BC log ρ B ) ) Follows by definition D(ρ σ) := trρ log ρ trρ log σ

43 Application: Strenghtened strong subadditivity Variational formula for relative entropy [Petz-88]: D(ρ σ) = sup tr ρ log ω + tr exp(log σ + log ω) ω>0 Variational formula for measured relative entropy [Berta-Fawzi-Tomamichel-5]: D M (ρ σ) = sup tr ρ log ω + trσω ω>0 I (A : C B) ρ = D ( ρ ABC exp(log ρ AB + log ρ BC log ρ B ) ) = sup trρ ABC log ω + tr exp(log ρ AB + log ρ BC log ρ B + log ω) ω>0 Variational formula for relative entropy

44 Application: Strenghtened strong subadditivity Variational formula for relative entropy [Petz-88]: D(ρ σ) = sup tr ρ log ω + tr exp(log σ + log ω) ω>0 Variational formula for measured relative entropy [Berta-Fawzi-Tomamichel-5]: D M (ρ σ) = sup tr ρ log ω + trσω ω>0 I (A : C B) ρ = D ( ρ ABC exp(log ρ AB + log ρ BC log ρ B ) ) = sup trρ ABC log ω + tr exp(log ρ AB + log ρ BC log ρ B + log ω) ω>0 sup trρ ABC log ω+ ω>0 dtβ 0 (t)trρ +it BC 4 matrix extension of GT (n = 4 and p = ) ) (ρ +it B ρ AB ρ it B id C ρ it BC ω

45 Application: Strenghtened strong subadditivity Variational formula for relative entropy [Petz-88]: D(ρ σ) = sup tr ρ log ω + tr exp(log σ + log ω) ω>0 Variational formula for measured relative entropy [Berta-Fawzi-Tomamichel-5]: D M (ρ σ) = sup tr ρ log ω + trσω ω>0 I (A : C B) ρ = D ( ρ ABC exp(log ρ AB + log ρ BC log ρ B ) ) = sup trρ ABC log ω + tr exp(log ρ AB + log ρ BC log ρ B + log ω) ω>0 sup trρ ABC log ω+ ω>0 dtβ 0 (t)trρ +it BC ) (ρ +it B ρ AB ρ it B id C ρ it BC ω = D M (ρ ABC R B BC (ρ AB )), Variational formula for meas. rel. entropy with R B BC ( ) = ) dtβ 0(t)ρ +it BC (ρ +it B ( )ρ it B id C ρ it BC

46 Application: Strenghtened strong subadditivity Variational formula for relative entropy [Petz-88]: D(ρ σ) = sup tr ρ log ω + tr exp(log σ + log ω) ω>0 Variational formula for measured relative entropy [Berta-Fawzi-Tomamichel-5]: D M (ρ σ) = sup tr ρ log ω + trσω ω>0 I (A : C B) ρ = D ( ρ ABC exp(log ρ AB + log ρ BC log ρ B ) ) = sup trρ ABC log ω + tr exp(log ρ AB + log ρ BC log ρ B + log ω) ω>0 sup trρ ABC log ω+ ω>0 dtβ 0 (t)trρ +it BC ) (ρ +it B ρ AB ρ it B id C = D M (ρ ABC R B BC (ρ AB )), with R B BC ( ) = ) dtβ 0(t)ρ +it BC (ρ +it B ( )ρ it B id C ρ it BC ω ρ it BC

47 Strenghtened strong subadditivity (con t) We just saw that Theorem: I (A : C B) ρ D M ( ρabc R B BC (ρ AB ) ) for R B BC ( ) = dtβ 0 (t)ρ +it BC ) (ρ +it B ( )ρ it B id C ρ it BC

48 Strenghtened strong subadditivity (con t) We just saw that Theorem: I (A : C B) ρ D M ( ρabc R B BC (ρ AB ) ) for R B BC ( ) = dtβ 0 (t)ρ +it BC Tight for commutative case ) (ρ +it B ( )ρ it B id C ρ it BC Explicit recovery map that is universal (only depends on ρ BC ) Proof based (only) on 4 matrix extension of GT Can be generalized to monotonicity of relative entropy Improves Fawzi-Renner and its follow up papers

49 Conclusions arxiv: Commun. Math. Phys. 06 If matrices do not commute things get complicated Trace inequalities are powerful tools expressing relations between matrices that do not commute Spectral pinching method is an intuitive approach to prove matrix (trace) inequalities Applications: Strengthening of strong subadditivity (FR bound) Hopefully many more (random matrix theory? other entropy inequalities?,...)

50 Conclusions arxiv: Commun. Math. Phys. 06 If matrices do not commute things get complicated Trace inequalities are powerful tools expressing relations between matrices that do not commute Spectral pinching method is an intuitive approach to prove matrix (trace) inequalities Applications: Strengthening of strong subadditivity (FR bound) Hopefully many more (random matrix theory? other entropy inequalities?,...) Thank you

51 More trace inequalities Let A and B be positive definite matrices and q R + A q := exp(q log A) is well-defined Araki-Lieb-Thirring: Let r [0, ] tr(b r/ A r B r/ ) q r tr(b / AB / ) q If r the inequality holds in the opposite direction

52 More trace inequalities Let A and B be positive definite matrices and q R + A q := exp(q log A) is well-defined Araki-Lieb-Thirring: Let r [0, ] tr(b r/ A r B r/ ) q r tr(b / AB / ) q If r the inequality holds in the opposite direction Implies the GT inequality via Lie-Trotter formula lim r 0 ( n k= C r k ) r ( n ) = exp log C k k= For q = this gives tr exp(log A + log B) trab

53 More trace inequalities Let A and B be positive definite matrices and q R + A q := exp(q log A) is well-defined Araki-Lieb-Thirring: Let r [0, ] tr(b r/ A r B r/ ) q r tr(b / AB / ) q If r the inequality holds in the opposite direction Implies the GT inequality via Lie-Trotter formula lim r 0 ( n k= C r k ) r ( n ) = exp log C k k= For q = this gives tr exp(log A + log B) trab Exercise: Prove ALT via the spectral pinching method We can prove extensions to n matrices via pinching or/and interpolation theory

54 Summary of results n matrix extension of ALT: Let p, r (0, ], n N, and consider a collection {A k } n k= of positive semi-definite matrices. Then log n r A r k n dt β r (t) log A +it k p β r (t) = k= sin(πr) r(cosh(πt)+cos(πr)) k= p is a probability distribution on R β 0(t) β 4 (t) β (t) β 3 4 (t) t

55 Summary of results n matrix extension of ALT: Let p, r (0, ], n N, and consider a collection {A k } n k= of positive semi-definite matrices. Then log n r A r k n dt β r (t) log A +it k p β r (t) = k= sin(πr) r(cosh(πt)+cos(πr)) k= p is a probability distribution on R Proof uses Stein-Hirschman interpolation theorem Using Lie-Trotter (i.e. r 0) we get as a corollary n matrix extension of GT: Let p, n N and consider a collection {H k } n k= of Hermitian matrices. Then ( n ) log exp n H k dt β 0 (t) log k= p exp( p ( + it)h k) k=

56 Stein-Hirschman operator interpolation theorem Strengthening of the Hadamard three lines theorem see [Junge-Renner-S-Wilde-Winter-5] S := {z C : 0 < Re(z) < } L (H) is the space of bounded linear operators acting on H Let G : S L(H) be uniformly bounded on S holomorphic on S continuous on the boundary S Let θ (0, ) and p θ = θ p 0 + θ p where p 0, p [, ] log G(θ) pθ ( ) dt β θ (t) log G(it) θ p 0 + β θ (t) log G( + it) θ p with R β θ (t) := sin(πθ) θ [cosh(πt) + cos(πθ)]

57 Proof of n matrix extension of ALT Choose G(z) = n k= Az k is bounded on S, holomorphic on S and continuous on S Let θ = r, p 0 = and p = p log G( + it) θ p = r log n log G(it) θ p 0 k= A+it k p = ( r) log n k= Ait k = 0 = r log n log G(θ) pθ = log n k= Ar k p r k= Ar k r p

58 Proof of n matrix extension of ALT Choose G(z) = n k= Az k is bounded on S, holomorphic on S and continuous on S Let θ = r, p 0 = and p = p log G( + it) θ p = r log n log G(it) θ p 0 k= A+it k p = ( r) log n k= Ait k = 0 = r log n log G(θ) pθ = log n k= Ar k p r Now we apply Stein-Hirschman log n r A r k p k= n β r (t) log k= k= Ar k r A +it k p p