Presentation is loading. Please wait.

Presentation is loading. Please wait.

Quantum Foundations Lecture 15

Similar presentations


Presentation on theme: "Quantum Foundations Lecture 15"β€” Presentation transcript:

1 Quantum Foundations Lecture 15
April 2, 2018 Dr. Matthew Leifer HSC112

2 Announcements Schmid College Academic Advising:
Tuesday April 3, 4:30pm-6:30pm AF209A (Prof. Leifer) Wednesday April 4, 4:30pm-6:30pm Henley Hall Basement (Prof. Dressel) Adam Becker is returning to Chapman: Book event and signing at 1888 center: Monday April 16. RSVP required Assignments First Draft due on Blackboard April 11. Peer review until April 16. Discussion in class April 16. Final Version due May 2. Homework 3 due April 11. I like lunch invitations

3 Application: Minimum Error Discrimination
Alice has a preparation device that prepares the system in either the state 𝜌 or the state 𝜎. She chooses each with 50/50 probability and sends the system to Bob. Bob makes a measurement on the system and has to guess whether 𝜌 or 𝜎 was prepared. What is his maximum probability of success and what measurement should he make?

4 Classical Case Let’s look at the classical case first. There is a variable that can take 𝑑 possible values 𝑗=1,2,β‹―,𝑑. Alice prepares the probability distributions 𝒑 or 𝒒 with 50/50 probability. Bob sees the value of 𝑗 and has to guess whether 𝒑 or 𝒒 was prepared. Bob decides on a subset 𝐸 𝒑 βŠ†{1,2,β‹―,𝑑}. If π‘—βˆˆ 𝐸 𝒑 he guesses 𝒑. If it is in the complement 𝐸 𝒒 = 1,2,β‹―,𝑑 \ 𝐸 𝒑 , he guesses 𝒒.

5 Classical Case Therefore,
𝑝 succ ≀ {𝑗| 𝑝 𝑗 β‰₯ π‘ž 𝑗 } ( 𝑝 𝑗 βˆ’ π‘ž 𝑗 ) = {𝑗| 𝑝 𝑗 β‰₯ π‘ž 𝑗 } ( 𝑝 𝑗 βˆ’ π‘ž 𝑗 ) + {𝑗| 𝑝 𝑗 < π‘ž 𝑗 } ( π‘ž 𝑗 βˆ’ 𝑝 𝑗 ) = 𝑗=1 𝑑 𝑝 𝑗 βˆ’ π‘ž 𝑗 = 𝐷 𝑐 (𝒑,𝒒) Where 𝐷 𝑐 𝒑,𝒒 = 1 2 𝑗=1 𝑑 𝑝 𝑗 βˆ’ π‘ž 𝑗 is called the variational distance.

6 Quantum Case Theorem (Helstrom, Holevo): The optimal success probability in the quantum case is given by 𝑝 succ = 𝐷 π‘ž (𝜌,𝜎) where 𝐷 π‘ž 𝜌,𝜎 = 1 2 Tr πœŒβˆ’πœŽ is known as the trace distance, and the matrix norm is given by 𝑀 = 𝑀 † 𝑀 .

7 𝑀 = 𝑀 † 𝑀 = 𝑀 2 because 𝑀 is Hermitian.
Quantum Case Lemma: If 𝑀 is a Hermitian operator then Tr 𝑀 = 𝑗 | πœ† 𝑗 | where πœ† 𝑗 are the eigenvalues of 𝑀. Proof: Let 𝑀= 𝑗 πœ† 𝑗 | πœ™ 𝑗 ⟩⟨ πœ™ 𝑗 | be the spectral decomposition of 𝑀 and write 𝑀= {𝑗| πœ† 𝑗 β‰₯0} πœ† 𝑗 | πœ™ 𝑗 ⟩⟨ πœ™ 𝑗 | βˆ’ {π‘˜| πœ† π‘˜ <0} πœ† π‘˜ | πœ™ π‘˜ ⟩⟨ πœ™ π‘˜ | Then, 𝑀 = 𝑀 † 𝑀 = 𝑀 2 because 𝑀 is Hermitian.

8 Quantum Case 𝑀 = 𝑀 2 = {𝑗| πœ† 𝑗 β‰₯0} πœ† 𝑗 | πœ™ 𝑗 ⟩⟨ πœ™ 𝑗 | βˆ’ {π‘˜| πœ† π‘˜ <0} πœ† π‘˜ | πœ™ π‘˜ ⟩⟨ πœ™ π‘˜ | {𝑙| πœ† 𝑙 β‰₯0} πœ† 𝑙 | πœ™ 𝑙 ⟩⟨ πœ™ 𝑙 | βˆ’ {π‘š |πœ† π‘š <0} πœ† π‘š | πœ™ π‘š ⟩⟨ πœ™ π‘š | = {𝑗| πœ† 𝑗 β‰₯0} {𝑙| πœ† 𝑙 β‰₯0} πœ† 𝑗 πœ† 𝑙 πœ™ 𝑗 πœ™ 𝑗 πœ™ 𝑙 ⟩⟨ πœ™ 𝑙 | + {π‘˜| πœ† π‘˜ <0} {π‘š| πœ† π‘š <0} πœ† π‘˜ πœ† π‘š πœ™ π‘˜ πœ™ π‘˜ πœ™ π‘š ⟩⟨ πœ™ π‘š | Note: cross terms are zero because the subspaces with πœ† 𝑗 β‰₯0 and πœ† 𝑗 <0 are orthogonal. 𝑀 = {𝑗| πœ† 𝑗 β‰₯0} {𝑙| πœ† 𝑙 β‰₯0} πœ† 𝑗 πœ† 𝑙 𝛿 𝑗𝑙 πœ™ 𝑗 ⟨ πœ™ 𝑙 | + {π‘˜| πœ† π‘˜ <0} {π‘š| πœ† π‘š <0} πœ† π‘˜ πœ† π‘š 𝛿 π‘˜π‘š πœ™ π‘˜ ⟨ πœ™ π‘š | = {𝑗| πœ† 𝑗 β‰₯0} πœ† 𝑗 πœ™ 𝑗 πœ™ 𝑗 + {π‘˜| πœ† π‘˜ <0} πœ† π‘˜ πœ™ π‘˜ πœ™ π‘˜ = 𝑗 πœ† 𝑗 πœ™ 𝑗 πœ™ 𝑗 = 𝑗 πœ† 𝑗 πœ™ 𝑗 πœ™ 𝑗 And hence Tr 𝑀 = 𝑗 | πœ† 𝑗 | .

9 Quantum Case Now consider Bob’s strategy. He has to choose a two-outcome POVM: 𝐸 𝜌 , 𝐸 𝜎 =πΌβˆ’ 𝐸 𝜌 , such that if he gets the outcome 𝐸 𝜌 he will guess 𝜌 and if he gets the outcome 𝐸 𝜎 he will guess 𝜎. His success probability is 𝑝 succ =Prob 𝜌 is prepared Prob( 𝐸 𝜌 |𝜌)+Prob 𝜎 is prepared Prob( 𝐸 𝜎 |𝜎) = 1 2 Tr 𝐸 𝜌 𝜌 +Tr( 𝐸 𝜎 𝜎) = 1 2 Tr 𝐸 𝜌 𝜌 +Tr πΌβˆ’ 𝐸 𝜌 𝜎 = 1 2 Tr 𝜎 +Tr 𝐸 𝜌 (πœŒβˆ’πœŽ) = Tr 𝐸 𝜌 (πœŒβˆ’πœŽ)

10 𝐸 𝜌 = 𝑃 + = {𝑗| πœ† 𝑗 β‰₯0} | πœ™ 𝑗 ⟩⟨ πœ™ 𝑗 |
Quantum Case Now let πœŒβˆ’πœŽ= 𝑗 πœ† 𝑗 πœ™ 𝑗 πœ™ 𝑗 be the spectral decomposition of πœŒβˆ’πœŽ. Tr 𝐸 𝜌 (πœŒβˆ’πœŽ) =Tr 𝐸 𝜌 𝑗 πœ† 𝑗 πœ™ 𝑗 πœ™ 𝑗 = 𝑗 πœ† 𝑗 Tr 𝐸 𝜌 | πœ™ 𝑗 ⟩⟨ πœ™ 𝑗 | = 𝑗 πœ† 𝑗 ⟨ πœ™ 𝑗 | 𝐸 𝜌 | πœ™ 𝑗 ⟩ Now, 0≀ πœ™ 𝑗 𝐸 𝜌 πœ™ 𝑗 ≀1, so this is clearly maximized if we can choose πœ™ 𝑗 𝐸 𝜌 πœ™ 𝑗 =1 for πœ† 𝑗 β‰₯0 and πœ™ 𝑗 𝐸 𝜌 πœ™ 𝑗 =0 for πœ† 𝑗 <0. This can be achieved if we choose 𝐸 𝜌 = 𝑃 + = {𝑗| πœ† 𝑗 β‰₯0} | πœ™ 𝑗 ⟩⟨ πœ™ 𝑗 |

11 Quantum Case So we have Tr 𝐸 𝜌 (πœŒβˆ’πœŽ) ≀Tr 𝑃 + (πœŒβˆ’πœŽ)
= 𝑗 πœ† 𝑗 ⟨ πœ™ 𝑗 | 𝑃 + | πœ™ 𝑗 ⟩ = 𝑗 {π‘˜| πœ† π‘˜ β‰₯0} πœ† 𝑗 ⟨ πœ™ 𝑗 | πœ™ π‘˜ ⟩⟨ πœ™ π‘˜ | πœ™ 𝑗 ⟩ = {𝑗| πœ† 𝑗 β‰₯0} | πœ† 𝑗 | However, 𝜌 and 𝜎 are both density matrices, so Tr πœŒβˆ’πœŽ =Tr 𝜌 βˆ’Tr 𝜎 =1βˆ’1=0 Therefore 𝑗 πœ† 𝑗 = {𝑗| πœ† 𝑗 β‰₯0} | πœ† 𝑗 | βˆ’ {𝑗| πœ† 𝑗 <0} | πœ† 𝑗 | =0 or {𝑗| πœ† 𝑗 β‰₯0} | πœ† 𝑗 | = {𝑗| πœ† 𝑗 <0} | πœ† 𝑗 |

12 Quantum Case Hence, {𝑗| πœ† 𝑗 β‰₯0} | πœ† 𝑗 | = {𝑗| πœ† 𝑗 β‰₯0} | πœ† 𝑗 | + {𝑗| πœ† 𝑗 <0} | πœ† 𝑗 | = 1 2 𝑗 | πœ† 𝑗 | Now, if we apply the lemma, this gives {𝑗| πœ† 𝑗 β‰₯0} | πœ† 𝑗 | = 1 2 Tr πœŒβˆ’πœŽ = 𝐷 π‘ž (𝜌,𝜎) Putting it all together gives 𝑝 succ ≀ 𝐷 π‘ž (𝜌,𝜎) with equality achieved if Bob chooses 𝐸 𝜌 = 𝑃 + , i.e. the projector onto the positive eigenspace of πœŒβˆ’πœŽ.

13 𝜌= 𝑗 𝑝 𝑗 |π‘—βŸ©βŸ¨π‘—| and 𝜎= 𝑗 π‘ž 𝑗 |π‘—βŸ©βŸ¨π‘—|
Special Cases Note that if 𝜌 and 𝜎 are diagonal in the same basis 𝜌= 𝑗 𝑝 𝑗 |π‘—βŸ©βŸ¨π‘—| and 𝜎= 𝑗 π‘ž 𝑗 |π‘—βŸ©βŸ¨π‘—| then the eigenvalues of πœŒβˆ’πœŽ are 𝑝 𝑗 βˆ’ π‘ž 𝑗 and we get 𝐷 π‘ž 𝜌,𝜎 = 1 2 𝑗 𝑝 𝑗 βˆ’ π‘ž 𝑗 = 𝐷 𝑐 (𝒑,𝒒) recovering the classical result. If 𝜌=|πœ“βŸ©βŸ¨πœ“| and 𝜎=|πœ™βŸ©βŸ¨πœ™| are both pure states then (you will prove on Hwk. 4) 𝐷 π‘ž 𝜌,𝜎 = 1βˆ’ πœ™ πœ“ 2 Therefore, pure states are perfectly distinguishable iff πœ™ πœ“ =0.

14 8.vi) The Lindblad Equation

15 Continuous Time Dynamics

16 The View from the Larger Church

17 The View from the Larger Church

18 Deriving the Lindblad Equation

19 Deriving the Lindblad Equation

20 Deriving the Lindblad Equation

21 Example: Decoherence

22 9) Ontological Models The aim of this section is to investigate the possibility of constructing a realist theory (known as an ontological model) that can reproduce the predictions of quantum theory. We start with a simple toy-model that reproduces many of the apparently puzzling phenomena we have studied so far: The Spekkens’ toy theory. These phenomena are naturally explained if there is a restriction on the amount of information we can have about the ontic state (an β€œepistemic restriction” or β€œepistriction”) and the quantum state is epistemic. After this we will present the general definition of an ontological model and prove a number of no-go theorems that imply that a realist theory underlying quantum theory cannot be like this.

23 9) Ontological Models Good references for this section include:
David Jennings and Matthew Leifer, β€œNo Return to Classical Reality”, Contemporary Physics, vol. 57, iss. 1, pp (2015) preprint: Robert W. Spekkens, β€œQuasi-Quantization: Classical Statistical Theories with an Epistemic Restriction”, in β€œQuantum Theory: Informational Foundations and Foils”, Giulio Chiribella and Robert W. Spekkens (eds.), pp , Springer (2015) preprint: Robert W. Spekkens, β€œContextuality for preparations, transformations, and unsharp measurements”, Physical Review A, vol (2005). Preprint: J. S. Bell, β€œSpeakable and Unspeakable in Quantum Mechanics”, 2nd edition, Cambridge University Press (2004). Matthew Leifer, β€œIs the Quantum State Real? An Extended Review of ψ-ontology Theorems”, Quanta, vol. 3, no. 1, pp (2014).

24 9) Ontological Models Ontological Models Epistricted Theories
Definitions Examples Excess Baggage Contextuality Ξ¨-ontology Bell’s Theorem The Colbeck-Renner Theorem


Download ppt "Quantum Foundations Lecture 15"

Similar presentations


Ads by Google