Quantum information and quantum computing

Transcription

1 Middle East Technical University, Department of Physics January 7, 009

2 Outline Measurement 1 Measurement 3 Single qubit gates Multiple qubit gates 4 Distinguishability 5

3 What s measurement? Quantum measurement A quantum measurement on a system is described by a measurement operator M m. The probability that the result m occurs is p(m) = ψ M mm m ψ The probabilities for all the outcomes in the problem must add up to 1. p(m) = ψ M mm m ψ = 1 m m

4 What s measurement? Quantum measurement A quantum measurement on a system is described by a measurement operator M m. The probability that the result m occurs is p(m) = ψ M mm m ψ The probabilities for all the outcomes in the problem must add up to 1. p(m) = ψ M mm m ψ = 1 m m

5 What s measurement? Quantum measurement A quantum measurement on a system is described by a measurement operator M m. The probability that the result m occurs is p(m) = ψ M mm m ψ The probabilities for all the outcomes in the problem must add up to 1. p(m) = ψ M mm m ψ = 1 m m

6 Quantum measurement on qubit Two measurement operators M 0 = 0 0 and M 1 = 1 1 Operators are idempotent and satisfy the completeness relation. M i = M i i = 0, 1 I = M 0 M 0 + M 1 M 1 = M 0 + M 1 For the qubit ψ = a a 1 1 The state after measurement : M i ψ a i = a i a i i = i Probabilities : p(i) = ψ M i M i ψ = a i

7 Quantum measurement on qubit Two measurement operators M 0 = 0 0 and M 1 = 1 1 Operators are idempotent and satisfy the completeness relation. M i = M i i = 0, 1 I = M 0 M 0 + M 1 M 1 = M 0 + M 1 For the qubit ψ = a a 1 1 The state after measurement : M i ψ a i = a i a i i = i Probabilities : p(i) = ψ M i M i ψ = a i

8 Quantum measurement on qubit Two measurement operators M 0 = 0 0 and M 1 = 1 1 Operators are idempotent and satisfy the completeness relation. M i = M i i = 0, 1 I = M 0 M 0 + M 1 M 1 = M 0 + M 1 For the qubit ψ = a a 1 1 The state after measurement : M i ψ a i = a i a i i = i Probabilities : p(i) = ψ M i M i ψ = a i

9 Measurement Single qubit gates Multiple qubit gates The X gate or NOT gate : 0 1, 1 0 In matrix notation For a general qubit X X ( ) ( ) α = β ( ) β α

10 Measurement Single qubit gates Multiple qubit gates The X gate or NOT gate : 0 1, 1 0 In matrix notation For a general qubit X X ( ) ( ) α = β ( ) β α

11 Measurement Single qubit gates Multiple qubit gates The X gate or NOT gate : 0 1, 1 0 In matrix notation For a general qubit X X ( ) ( ) α = β ( ) β α

12 Measurement Single qubit gates Multiple qubit gates The iy gate : 0 1, 1 0 In matrix notation For a general qubit iy ( 0 ) X(α 0 ) + (β 1 ) = α 1 ) + β 0

13 Measurement Single qubit gates Multiple qubit gates The Z gate : 0 0, 1 1 In matrix notation For a general qubit Z ( ) Z(α 0 + β 1 ) = α 0 β 1

14 Measurement Single qubit gates Multiple qubit gates The Hadamard gate Square-root of the NOT gate H 1 ( Takes 0 and 1 to halfway inbetween ) H 0 = 1 ( ) H 1 = 1 ( 0 1 ) H is not a NOT gate but instead H = I

15 Properties of gates Measurement Single qubit gates Multiple qubit gates Infinitely many matrices infinitely many gates A gate must be unitary. An arbitrary quantum gate may be decomposed from a smaller set of gates.

16 Multiple qubit gates Measurement Single qubit gates Multiple qubit gates The controlled-not (CNOT) gate Action on a two-component system : In other words A, B A, A B where A B = A + B (mod ) In matrix notation (with respect to 00, 01, 10 and 11 ) U CN = Just like a single qubit gate U CN is unitary.

17 Symbolic representation Single qubit gates Multiple qubit gates A> A> (control qubit) B> B + A> (target qubit) Symbolic representation of the CNOT gate. If the control qubit is zero, nothing happens to the target qubit. If the control qubit is one, the other is swapped. The CNOT gate may be generalized to multiple qubits. If the control qubit (topmost) is zero, nothing happens to the other qubits. If the control qubit is one, the gate U is applied to the other qubits

18 Symbolic representation Single qubit gates Multiple qubit gates A> A> (control qubit) B> B + A> (target qubit) Symbolic representation of the CNOT gate. If the control qubit is zero, nothing happens to the target qubit. If the control qubit is one, the other is swapped. The CNOT gate may be generalized to multiple qubits. If the control qubit (topmost) is zero, nothing happens to the other qubits. If the control qubit is one, the gate U is applied to the other qubits

19 Symbolic representation Single qubit gates Multiple qubit gates A> A> (control qubit) B> B + A> (target qubit) Symbolic representation of the CNOT gate. If the control qubit is zero, nothing happens to the target qubit. If the control qubit is one, the other is swapped. The CNOT gate may be generalized to multiple qubits. If the control qubit (topmost) is zero, nothing happens to the other qubits. If the control qubit is one, the gate U is applied to the other qubits

20 Symbolic representation Single qubit gates Multiple qubit gates U Symbolic representation of the CNOT gate. If the control qubit is zero, nothing happens to the target qubit. If the control qubit is one, the other is swapped. The CNOT gate may be generalized to multiple qubits. If the control qubit (topmost) is zero, nothing happens to the other qubits. If the control qubit is one, the gate U is applied to the other qubits

21 Symbolic representation Single qubit gates Multiple qubit gates U Symbolic representation of the CNOT gate. If the control qubit is zero, nothing happens to the target qubit. If the control qubit is one, the other is swapped. The CNOT gate may be generalized to multiple qubits. If the control qubit (topmost) is zero, nothing happens to the other qubits. If the control qubit is one, the gate U is applied to the other qubits

22 Symbolic representation Single qubit gates Multiple qubit gates U Symbolic representation of the CNOT gate. If the control qubit is zero, nothing happens to the target qubit. If the control qubit is one, the other is swapped. The CNOT gate may be generalized to multiple qubits. If the control qubit (topmost) is zero, nothing happens to the other qubits. If the control qubit is one, the gate U is applied to the other qubits

23 Single qubit gates Multiple qubit gates Symbolic representation an example Series of CNOT gates. Read from left to right Swaps qubits a, b a, a b a a b), a b = b, a b b, a b b = b, a

24 Distinguishability Measurement Single qubit gates Multiple qubit gates Let s imagine a game between two individuals Alice and Bob : Alice Chooses a state ψ i (i=1,, n) Gives state to Bob without telling him which one she has chosen Bob Knows the entire set of states Makes a measurement to find out the state Can he find out the state?

25 Distinguishability Scenario 1 : Orthonormal states If { ψ i } are orhonormal, Define operator M j ψ j ψ j, one for each state {M i } satisfy completeness relations If the state that Bob receives is ψ, then for i = j p(j) = ψ M j ψ = 1 Thus ψ = ψ j with certainty. Bob CAN determine the state!!!

26 Distinguishability Scenario 1 : Orthonormal states If { ψ i } are orhonormal, Define operator M j ψ j ψ j, one for each state {M i } satisfy completeness relations If the state that Bob receives is ψ, then for i = j p(j) = ψ M j ψ = 1 Thus ψ = ψ j with certainty. Bob CAN determine the state!!!

27 Distinguishability Scenario 1 : Orthonormal states If { ψ i } are orhonormal, Define operator M j ψ j ψ j, one for each state {M i } satisfy completeness relations If the state that Bob receives is ψ, then for i = j p(j) = ψ M j ψ = 1 Thus ψ = ψ j with certainty. Bob CAN determine the state!!!

28 Distinguishability Scenario 1 : Orthonormal states If { ψ i } are orhonormal, Define operator M j ψ j ψ j, one for each state {M i } satisfy completeness relations If the state that Bob receives is ψ, then for i = j p(j) = ψ M j ψ = 1 Thus ψ = ψ j with certainty. Bob CAN determine the state!!!

29 Distinguishability Scenario 1 : Orthonormal states If { ψ i } are orhonormal, Define operator M j ψ j ψ j, one for each state {M i } satisfy completeness relations If the state that Bob receives is ψ, then for i = j p(j) = ψ M j ψ = 1 Thus ψ = ψ j with certainty. Bob CAN determine the state!!!

30 Distinguishability Scenario : Non-orthonormal states If { ψ i } are not orhonormal, Suppose a determining measurement exists, M i. Assume a two-state system : ψ 1 and ψ ψ 1 M 1 M 1 ψ 1 = 1 and ψ M M ψ = 1 Decompose ψ = α ψ 1 + β φ, where φ is orthonormal to ψ 1. ψ M M ψ = β φ M M φ β < 1 Contradiction! Bob CAN NOT determine the state!!!

31 Distinguishability Scenario : Non-orthonormal states If { ψ i } are not orhonormal, Suppose a determining measurement exists, M i. Assume a two-state system : ψ 1 and ψ ψ 1 M 1 M 1 ψ 1 = 1 and ψ M M ψ = 1 Decompose ψ = α ψ 1 + β φ, where φ is orthonormal to ψ 1. ψ M M ψ = β φ M M φ β < 1 Contradiction! Bob CAN NOT determine the state!!!

32 Distinguishability Scenario : Non-orthonormal states If { ψ i } are not orhonormal, Suppose a determining measurement exists, M i. Assume a two-state system : ψ 1 and ψ ψ 1 M 1 M 1 ψ 1 = 1 and ψ M M ψ = 1 Decompose ψ = α ψ 1 + β φ, where φ is orthonormal to ψ 1. ψ M M ψ = β φ M M φ β < 1 Contradiction! Bob CAN NOT determine the state!!!

33 Distinguishability Scenario : Non-orthonormal states If { ψ i } are not orhonormal, Suppose a determining measurement exists, M i. Assume a two-state system : ψ 1 and ψ ψ 1 M 1 M 1 ψ 1 = 1 and ψ M M ψ = 1 Decompose ψ = α ψ 1 + β φ, where φ is orthonormal to ψ 1. ψ M M ψ = β φ M M φ β < 1 Contradiction! Bob CAN NOT determine the state!!!

34 Distinguishability Scenario : Non-orthonormal states If { ψ i } are not orhonormal, Suppose a determining measurement exists, M i. Assume a two-state system : ψ 1 and ψ ψ 1 M 1 M 1 ψ 1 = 1 and ψ M M ψ = 1 Decompose ψ = α ψ 1 + β φ, where φ is orthonormal to ψ 1. ψ M M ψ = β φ M M φ β < 1 Contradiction! Bob CAN NOT determine the state!!!

35 Distinguishability Scenario : Non-orthonormal states If { ψ i } are not orhonormal, Suppose a determining measurement exists, M i. Assume a two-state system : ψ 1 and ψ ψ 1 M 1 M 1 ψ 1 = 1 and ψ M M ψ = 1 Decompose ψ = α ψ 1 + β φ, where φ is orthonormal to ψ 1. ψ M M ψ = β φ M M φ β < 1 Contradiction! Bob CAN NOT determine the state!!!

36 Entangled states Measurement Distinguishability An entangled state is a joint state which cannot be described by the tensor product of two states. Consider the two qubit state ψ = 1 ( ) For this state there are no single qubits a and b such that ψ = a b

37 Distinguishability Entangled states superdense coding Superdense coding Superdense coding is a technique for sending two bits of classical information using only one qubit. Alice(A) sends Bob(B) a qubit and B recovers the information that A sends by conducting a measurement. Because information cannot be recovered reliably from nonorthogonal states, A cannot do better than sending one bit of classical information per qubit. However assume additionally that A and B share pieces of an entangled state made up of two qubits. ψ = 1 ( 0 A 0 B + 1 A 1 B )

38 Superdense coding Measurement Distinguishability Before sending her qubit A has a choice of applying Identity X the NOT gate ψ ψ 1 = 1 ( 0 A 0 B + 1 A 1 B ) ψ ψ = 1 ( 1 A 0 B + 0 A 1 B ) iy gate Z ψ ψ 3 = 1 ( 1 A 0 B + 0 A 1 B ) ψ ψ 4 = 1 ( 0 A 0 B + 1 A 1 B )

39 Superdense coding Measurement Distinguishability Before sending her qubit A has a choice of applying Identity X the NOT gate ψ ψ 1 = 1 ( 0 A 0 B + 1 A 1 B ) ψ ψ = 1 ( 1 A 0 B + 0 A 1 B ) iy gate Z ψ ψ 3 = 1 ( 1 A 0 B + 0 A 1 B ) ψ ψ 4 = 1 ( 0 A 0 B + 1 A 1 B )

40 Superdense coding Measurement Distinguishability Before sending her qubit A has a choice of applying Identity X the NOT gate ψ ψ 1 = 1 ( 0 A 0 B + 1 A 1 B ) ψ ψ = 1 ( 1 A 0 B + 0 A 1 B ) iy gate Z ψ ψ 3 = 1 ( 1 A 0 B + 0 A 1 B ) ψ ψ 4 = 1 ( 0 A 0 B + 1 A 1 B )

41 Superdense coding Measurement Distinguishability Before sending her qubit A has a choice of applying Identity X the NOT gate ψ ψ 1 = 1 ( 0 A 0 B + 1 A 1 B ) ψ ψ = 1 ( 1 A 0 B + 0 A 1 B ) iy gate Z ψ ψ 3 = 1 ( 1 A 0 B + 0 A 1 B ) ψ ψ 4 = 1 ( 0 A 0 B + 1 A 1 B )

42 Superdense coding Measurement Distinguishability Before sending her qubit A has a choice of applying Identity X the NOT gate ψ ψ 1 = 1 ( 0 A 0 B + 1 A 1 B ) ψ ψ = 1 ( 1 A 0 B + 0 A 1 B ) iy gate Z ψ ψ 3 = 1 ( 1 A 0 B + 0 A 1 B ) ψ ψ 4 = 1 ( 0 A 0 B + 1 A 1 B )

43 Superdense coding Measurement Distinguishability States ψ 1, ψ, ψ 3 and ψ 4 are orthogonal. They are called Bell basis, Bell states or EPR pairs (Einstein, Podolsky and Rosen). B can reliably measure which of the four states A has sent. If someone intercepts, all they get is a partial entangled state which is useless by itself.

44 Bell states through gates Distinguishability x y H β xy > Transforms the four states 00, 01, 10 and 11 into the Bell states. Example : 00 1 The Hadamard gate : 00 1 ( ) The CNOT gate : 1 ( ) 1 ( ) All four states 00 1 ( ) 01 1 ( ) 10 1 ( ) 11 1 ( )

45 Bell states through gates Distinguishability x y H β xy > Transforms the four states 00, 01, 10 and 11 into the Bell states. Example : 00 1 The Hadamard gate : 00 1 ( ) The CNOT gate : 1 ( ) 1 ( ) All four states 00 1 ( ) 01 1 ( ) 10 1 ( ) 11 1 ( )

46 Bell states through gates Distinguishability x y H β xy > Transforms the four states 00, 01, 10 and 11 into the Bell states. Example : 00 1 The Hadamard gate : 00 1 ( ) The CNOT gate : 1 ( ) 1 ( ) All four states 00 1 ( ) 01 1 ( ) 10 1 ( ) 11 1 ( )

47 Bell states through gates Distinguishability x y H β xy > Transforms the four states 00, 01, 10 and 11 into the Bell states. Example : 00 1 The Hadamard gate : 00 1 ( ) The CNOT gate : 1 ( ) 1 ( ) All four states 00 1 ( ) 01 1 ( ) 10 1 ( ) 11 1 ( )

48 is a technique for moving states around even in the absence of a quantum communications channel A and B initially share an EPR pair. Then A wants to deliver a qubit ψ = α 0 + β 1 to B. IMPORTANT! She DOES NOT know what the qubit is! A does the following :

49 Prepare input : A single qubit times an entangled state. ψ 0 = ψ A φ AB = 1 [α 0 A ( 0 A 0 B + 1 A 1 B ) + β 1 A ( 0 A 0 B + 1 A 1 B )] Send A s qubits through a CNOT gate ψ 1 = 1 [α 0 A ( 0 A 0 B + 1 A 1 B )+β 1 A ( 1 A 0 B + 0 A 1 B )]

50 Prepare input : A single qubit times an entangled state. ψ 0 = ψ A φ AB = 1 [α 0 A ( 0 A 0 B + 1 A 1 B ) + β 1 A ( 0 A 0 B + 1 A 1 B )] Send A s qubits through a CNOT gate ψ 1 = 1 [α 0 A ( 0 A 0 B + 1 A 1 B )+β 1 A ( 1 A 0 B + 0 A 1 B )]

51 Send first qubit through a Hadamard gate ψ = 1 [α( 0 A + 1 A )( 0 A 0 B + 1 A 1 B ) + β( 0 A 1 A )( 1 A 0 B = 1 [ 0 A0 A (α 0 B + β 1 B ) + 0 A 1 A (α 1 B + β 0 B )+ 1 A 0 A (α 0 B β 1 B ) + 1 A 1 A (α 1 B β 1 B )] Notice that now B s particles resemble the one that A initially had. A is now done preparing the output using gates and is now going to conduct a measurement.

52 A makes a measurement and gets one of the four possible states 00, 01, 10 and 11. A then sends this result classically to B. A s outcome B s state B must 00 α 0 + β 1 Do nothing 01 α 1 + β 0 Apply an X gate 10 α 0 β 1 Apply a Z gate 11 α 1 β 0 Apply ZX Deliver quantum information using just a single classical transfer!