Some remarkable differences between quantum and classical - PowerPoint PPT Presentation

Quantum bit (Qubit) ✻ 1 1 = | 0 ′ � 2 | 0 � + 2 | 1 � √ √ | 1 � � ✒ � ✒ � � Basis 1: � p (0) = 1 � 2 � � {| 0 � , | 1 �} � � � � � � Basis 2: � ✲ � { 1 1 2 | 1 � = | 0 ′ � , ❅ √ 2 | 0 � + √ ❅ | 0 � ❅ 1 1 2 | 1 � = | 1 ′ �} 2 | 0 � − √ √ ❅ ❅ ❅ p (0 ′ ) = 1 ❅ | 1 ′ � ❅ ❘ Pure state � = (generalized) probability distribution Mika Hirvensalo Some remarkable ... 23 of 43

Quantum gate Example: W : C 2 → C 2 1 | 0 � + 1 W | 0 � = √ √ | 1 � 2 2 1 | 0 � − 1 √ √ W | 1 � = | 1 � 2 2 is unitary (Hadamard-Walsh transform) Mika Hirvensalo Some remarkable ... 24 of 43

Interference / Walsh transform once Mika Hirvensalo Some remarkable ... 25 of 43

Interference / Walsh transform once | 0 � | 0 � ✓ ❙ ✓ ❙ 1 1 √ √ ✓ ❙ 2 2 ✓ ❙ ✴ ✓ ✇ ❙ 1 1 | 0 � | 1 � 2 | 0 � + 2 | 1 � √ √ Mika Hirvensalo Some remarkable ... 25 of 43

Interference / Walsh transform twice | 0 � | 0 � ✓ ❙ ✓ ❙ 1 1 √ √ ✓ ❙ 2 2 ✓ ❙ ✴ ✓ ❙ ✇ 1 1 | 0 � | 1 � 2 | 0 � + 2 | 1 � √ √ ✁ ❆ ✁ ❆ ✁ ❆ ✁ ❆ 1 1 1 − 1 √ √ √ √ ✁ ❆ ✁ ❆ 2 2 2 2 ✁ ❆ ✁ ❆ ✁ ☛ ❆ ❯ ☛ ✁ ❯ ❆ 2 | 0 � + 1 1 | 0 � | 1 � | 0 � | 1 � 2 | 1 � + 1 2 | 0 � − 1 2 | 1 � = | 0 � Mika Hirvensalo Some remarkable ... 26 of 43

Interference / Walsh transform twice | 0 � | 0 � ✓ ❙ ✓ ❙ 1 1 √ √ ✓ ❙ 2 2 ✓ ❙ ✴ ✓ ❙ ✇ 1 1 | 0 � | 1 � 2 | 0 � + 2 | 1 � √ √ ✁ ❆ ✁ ❆ ✁ ❆ ✁ ❆ 1 1 1 − 1 √ √ √ √ ✁ ❆ ✁ ❆ 2 2 2 2 ✁ ❆ ✁ ❆ ✁ ☛ ❯ ❆ ✁ ☛ ❯ ❆ 1 2 | 0 � + 1 | 0 � | 1 � | 0 � | 1 � 2 | 1 � + 1 2 | 0 � − 1 2 | 1 � = | 0 � ❅ ■ � ✒ ❅ � ❅ � Constructive interference Mika Hirvensalo Some remarkable ... 26 of 43

Interference / Walsh transform twice | 0 � | 0 � ✓ ❙ ✓ ❙ 1 1 √ √ ✓ ❙ 2 2 ✓ ❙ ✓ ✴ ❙ ✇ 1 1 | 0 � | 1 � 2 | 0 � + 2 | 1 � √ √ ✁ ❆ ✁ ❆ ✁ ❆ ✁ ❆ 1 1 1 − 1 √ √ √ √ ✁ ❆ ✁ ❆ 2 2 2 2 ✁ ❆ ✁ ❆ ☛ ✁ ❯ ❆ ✁ ☛ ❯ ❆ 1 2 | 0 � + 1 | 0 � | 1 � | 0 � | 1 � 2 | 1 � + 1 2 | 0 � − 1 2 | 1 � = | 0 � ❅ ■ ❅ ■ � ✒ ✒ � ❅ � ❅ � ❅ � ❅ � Constructive Destructive interference interference Mika Hirvensalo Some remarkable ... 26 of 43

① ① ① ① ① ① ① ① ① ① ① ① ① n quantum bits Tensor product notation: | ❛ � ⊗ | ❜ � = | ❛ � | ❜ � = | ❛ ❜ � . Mika Hirvensalo Some remarkable ... 27 of 43

① ① ① ① ① ① ① ① ① ① ① ① ① n quantum bits Tensor product notation: | ❛ � ⊗ | ❜ � = | ❛ � | ❜ � = | ❛ ❜ � . For example, | 0 � | 0 � = | 00 � , | 0 � | 1 � = | 01 � , etc. Mika Hirvensalo Some remarkable ... 27 of 43

① ① ① ① ① ① ① ① n quantum bits Tensor product notation: | ❛ � ⊗ | ❜ � = | ❛ � | ❜ � = | ❛ ❜ � . For example, | 0 � | 0 � = | 00 � , | 0 � | 1 � = | 01 � , etc. � c ① | ① � (2 n -dimensional Hilbert space), General state ① ∈{ 0 , 1 } n � | c ① | 2 = 1 where ① ∈{ 0 , 1 } n Mika Hirvensalo Some remarkable ... 27 of 43

n quantum bits Tensor product notation: | ❛ � ⊗ | ❜ � = | ❛ � | ❜ � = | ❛ ❜ � . For example, | 0 � | 0 � = | 00 � , | 0 � | 1 � = | 01 � , etc. � c ① | ① � (2 n -dimensional Hilbert space), General state ① ∈{ 0 , 1 } n � | c ① | 2 = 1 where ① ∈{ 0 , 1 } n If U f | ① � | 0 � = | ① � | f ( ① ) � can be realized, then � � 1 1 U f √ | ① � | 0 � = √ | ① � | f ( ① ) � 2 n 2 n ① ∈{ 0 , 1 } n ① ∈{ 0 , 1 } n (Quantum parallelism) Mika Hirvensalo Some remarkable ... 27 of 43

① ① ① ① n quantum bits If U f | ① � | 0 � = | ① � | f ( ① ) � can be realized, then � � 1 1 U f √ | ① � | 0 � = √ | ① � | f ( ① ) � 2 n 2 n ① ∈{ 0 , 1 } n ① ∈{ 0 , 1 } n (Quantum parallelism) Mika Hirvensalo Some remarkable ... 28 of 43

① ① n quantum bits If U f | ① � | 0 � = | ① � | f ( ① ) � can be realized, then � � 1 1 U f √ | ① � | 0 � = √ | ① � | f ( ① ) � 2 n 2 n ① ∈{ 0 , 1 } n ① ∈{ 0 , 1 } n (Quantum parallelism) � � 2 � � 1 = 1 P ( | ① � | f ( ① ) � ) = � � √ 2 n 2 n Mika Hirvensalo Some remarkable ... 28 of 43

n quantum bits If U f | ① � | 0 � = | ① � | f ( ① ) � can be realized, then � � 1 1 U f √ | ① � | 0 � = √ | ① � | f ( ① ) � 2 n 2 n ① ∈{ 0 , 1 } n ① ∈{ 0 , 1 } n (Quantum parallelism) � � 2 � � 1 = 1 P ( | ① � | f ( ① ) � ) = � � √ 2 n 2 n Observation “collapses” the system into | ① � | f ( ① ) � (Projection postulate) Mika Hirvensalo Some remarkable ... 28 of 43

n quantum bits If U f | ① � | 0 � = | ① � | f ( ① ) � can be realized, then � � 1 1 U f √ | ① � | 0 � = √ | ① � | f ( ① ) � 2 n 2 n ① ∈{ 0 , 1 } n ① ∈{ 0 , 1 } n (Quantum parallelism) � � 2 � � 1 = 1 P ( | ① � | f ( ① ) � ) = � � √ 2 n 2 n Observation “collapses” the system into | ① � | f ( ① ) � (Projection postulate) Direct method offers no advantage over probabilistic guessing! Mika Hirvensalo Some remarkable ... 28 of 43

Nondeterministic Computing Mika Hirvensalo Some remarkable ... 29 of 43

Nondeterministic Computing Good computational paths should be supported Mika Hirvensalo Some remarkable ... 29 of 43

Nondeterministic Computing Good computational paths should be supported Seems impossible in classical computing Mika Hirvensalo Some remarkable ... 29 of 43

Nondeterministic Computing Good computational paths should be supported Seems impossible in classical computing Sometimes possible in quantum computing Mika Hirvensalo Some remarkable ... 29 of 43

Nondeterministic Computing Good computational paths should be supported Seems impossible in classical computing Sometimes possible in quantum computing The “efficiency” of quantum computing is based on interference Mika Hirvensalo Some remarkable ... 29 of 43

Quantum algorithms Interference should favour the good computation paths Difficult to control in algorithm design Mika Hirvensalo Some remarkable ... 30 of 43

Quantum algorithms Interference should favour the good computation paths Difficult to control in algorithm design Main methods Quantum Fourier transform Grover iteration Adiabatic quantum computing Quantum random walks Mika Hirvensalo Some remarkable ... 30 of 43

Quantum Fourier transform Discrete Fourier transform on coefficients of c 0 | 00 . . . 0 � + c 1 | 00 . . . 1 � + . . . + c 2 n − 1 | 11 . . . 1 � , Can be implemented in time Poly( n ) (instead of 2 n ) Exponential advantage for problems with periodic structure Main ingredient in Shor’s factoring algorithm Mika Hirvensalo Some remarkable ... 31 of 43

Grover iteration Basic idea: Using k calls of function f , the superposition � � 1 √ | 00 . . . 0 � + | 00 . . . 1 � + . . . + | 11 . . . 1 � , 2 n coefficients of such vectors | ① � for which f ( ① ) = 1 can be k increased to ≈ C · 2 n , hence the probability of seeing such √ an element becomes ≈ | C | 2 k 2 2 n . Provides a quadratic advantage over classical algorithms Works on all search problems Mika Hirvensalo Some remarkable ... 32 of 43

EPR Paradox EPR pair 1 √ ( | 00 � + | 11 � ) 2 (entangled state), perfect correlation Mika Hirvensalo Some remarkable ... 33 of 43

EPR Paradox EPR pair 1 √ ( | 00 � + | 11 � ) 2 (entangled state), perfect correlation; cf. with 1 2( | 00 � + | 01 � + | 10 � + | 11 � ) Mika Hirvensalo Some remarkable ... 33 of 43

EPR Paradox EPR pair 1 √ ( | 00 � + | 11 � ) 2 (entangled state), perfect correlation; cf. with 1 1 ( | 0 � + | 1 � ) 1 2( | 00 � + | 01 � + | 10 � + | 11 � ) = √ √ ( | 0 � + | 1 � ) 2 2 (decomposable state). Mika Hirvensalo Some remarkable ... 33 of 43

Compound Systems Correlation over distance also possible in classical mechanics: Probability distribution 1 2 [00] + 1 2 [11] Mika Hirvensalo Some remarkable ... 34 of 43

Compound Systems Correlation over distance also possible in classical mechanics: Probability distribution 1 2 [00] + 1 2 [11] But 1 | 00 � + 1 √ √ | 11 � 2 2 violates a Bell inequality . Mika Hirvensalo Some remarkable ... 34 of 43

John Bell John Steward Bell (1928–1990) Mika Hirvensalo Some remarkable ... 35 of 43

EPR Paradox Einstein, Podolsky, Rosen: Can Quantum-Mechanical Description of Physical Reality Be Considered Com- plete? Physical Review 47, 777–780 (1935) Niels Bohr (1885–1962) & Albert Einstein (1879–1955) Mika Hirvensalo Some remarkable ... 36 of 43

EPR Paradox (Bohm formulation) Mika Hirvensalo Some remarkable ... 37 of 43

EPR Paradox (Bohm formulation) Einstein: The physical world is local and realistic Mika Hirvensalo Some remarkable ... 37 of 43

EPR Paradox (Bohm formulation) Einstein: The physical world is local and realistic 1 1 Assume distant qubits in state 2 | 00 � + 2 | 11 � √ √ Mika Hirvensalo Some remarkable ... 37 of 43

EPR Paradox (Bohm formulation) Einstein: The physical world is local and realistic 1 1 Assume distant qubits in state 2 | 00 � + 2 | 11 � √ √ Quantum mechanics: neither qubit has definite pre-observation value Mika Hirvensalo Some remarkable ... 37 of 43

EPR Paradox (Bohm formulation) Einstein: The physical world is local and realistic 1 1 Assume distant qubits in state 2 | 00 � + 2 | 11 � √ √ Quantum mechanics: neither qubit has definite pre-observation value Observe the first qubit Mika Hirvensalo Some remarkable ... 37 of 43

EPR Paradox (Bohm formulation) Einstein: The physical world is local and realistic 1 1 Assume distant qubits in state 2 | 00 � + 2 | 11 � √ √ Quantum mechanics: neither qubit has definite pre-observation value Observe the first qubit ⇒ The value of the second qubit is known certainly Mika Hirvensalo Some remarkable ... 37 of 43

EPR Paradox (Bohm formulation) Einstein: The physical world is local and realistic 1 1 Assume distant qubits in state 2 | 00 � + 2 | 11 � √ √ Quantum mechanics: neither qubit has definite pre-observation value Observe the first qubit ⇒ The value of the second qubit is known certainly (without “touching” or “disturbing” it) Mika Hirvensalo Some remarkable ... 37 of 43

EPR Paradox (Bohm formulation) Einstein: The physical world is local and realistic 1 1 Assume distant qubits in state 2 | 00 � + 2 | 11 � √ √ Quantum mechanics: neither qubit has definite pre-observation value Observe the first qubit ⇒ The value of the second qubit is known certainly (without “touching” or “disturbing” it) ⇒ The value if the second qubit is “an element of reality” Mika Hirvensalo Some remarkable ... 37 of 43

EPR Paradox (Bohm formulation) Einstein: The physical world is local and realistic 1 1 Assume distant qubits in state 2 | 00 � + 2 | 11 � √ √ Quantum mechanics: neither qubit has definite pre-observation value Observe the first qubit ⇒ The value of the second qubit is known certainly (without “touching” or “disturbing” it) ⇒ The value if the second qubit is “an element of reality” ⇒ Quantum mechanics is an incomplete theory Mika Hirvensalo Some remarkable ... 37 of 43

Bell Inequalities Itamar Pitowsky: Quantum Probability – Quantum Logic, Springer (1989) Mika Hirvensalo Some remarkable ... 38 of 43

Bell Inequalities Itamar Pitowsky: Quantum Probability – Quantum Logic, Springer (1989) Ballot box of 100 balls Mika Hirvensalo Some remarkable ... 38 of 43

Bell Inequalities Itamar Pitowsky: Quantum Probability – Quantum Logic, Springer (1989) Ballot box of 100 balls Each red or blue, wooden or plastic Mika Hirvensalo Some remarkable ... 38 of 43

Bell Inequalities Itamar Pitowsky: Quantum Probability – Quantum Logic, Springer (1989) Ballot box of 100 balls Each red or blue, wooden or plastic 80 red, 60 wooden Mika Hirvensalo Some remarkable ... 38 of 43

Bell Inequalities Itamar Pitowsky: Quantum Probability – Quantum Logic, Springer (1989) Ballot box of 100 balls Each red or blue, wooden or plastic 80 red, 60 wooden 30 red and wooden? Mika Hirvensalo Some remarkable ... 38 of 43

Bell Inequalities Itamar Pitowsky: Quantum Probability – Quantum Logic, Springer (1989) Ballot box of 100 balls Each red or blue, wooden or plastic 80 red, 60 wooden 30 red and wooden? Then 80+60-30=110 are red or wooden. No way! Mika Hirvensalo Some remarkable ... 38 of 43

Bell Inequalities Itamar Pitowsky: Quantum Probability – Quantum Logic, Springer (1989) Ballot box of 100 balls Each red or blue, wooden or plastic 80 red, 60 wooden 30 red and wooden? Then 80+60-30=110 are red or wooden. No way! In other words: (0 . 8 , 0 . 6 , 0 . 3) does not express probabilities ( p 1 , p 2 , p 12 ) of two events and their intersection. Mika Hirvensalo Some remarkable ... 38 of 43

Bell Inequalities Itamar Pitowsky: Quantum Probability – Quantum Logic, Springer (1989) Ballot box of 100 balls Each red or blue, wooden or plastic 80 red, 60 wooden 30 red and wooden? Then 80+60-30=110 are red or wooden. No way! In other words: (0 . 8 , 0 . 6 , 0 . 3) does not express probabilities ( p 1 , p 2 , p 12 ) of two events and their intersection. Reason: P (1 ∨ 2) = p 1 + p 2 − p 12 is a probability, too. Mika Hirvensalo Some remarkable ... 38 of 43

Bell Inequalities Lemma ( p 1 , p 2 , p 12 ) is an “eligible” probability vector if and only if 0 ≤ p 12 ≤ p 1 , p 2 ≤ 1 and 0 ≤ p 1 + p 2 − p 12 ≤ 1 These are Bell inequalities! Mika Hirvensalo Some remarkable ... 39 of 43