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10/30/09 Quantum Computing Devices Quantum Bits A quantum bit, qubit, is a two-level quantum system A two dimensional Hilbert space H2 is a quantum qubit H2 has the basis B = { 0 , 1 } so-called computational basis States 0 , 1 are called basis states € € 1 10/30/09 A general state of a single quantum bit is a vector ω 0 0 + ω1 1 2 2 ω 0 + ω1 = 1 • Having unit length €Observation of a quantum bit in such a state will give 0 or 1 as an outcome with probabilities 2 ω 0 , ω1 2 € An operation on a qubit, called unary quantum gate, is a unitary mapping U: H2 → H2 Unary quantum gate defines a linear operation, such that the corresponding matrix is unitary • a* stands for the complex conjugate number a 0 → a 0 + b1 1 →c 0 + d1 a b a* and * c d b c * 1 0 = d * 0 1 € 2 10/30/09 Let use the coordinate representation T T 0 = (1,0) 1 = (0,1) The unitary matrix defines an action 0 1 M¬ = 1 0 € The unitary quantum gate defined by M¬ is called a quantum-not gate € M¬ 0 = 1 , M¬ 1 = 0 € M¬ 0 = 1 , M¬ 1 = 0 Can € be written using the XOR operation M¬ x = x ⊕ 1 0 ⊕1 = 1 1⊕1 = 0 € € € 3 10/30/09 Another quantum gate 1+ i M¬ = 2 1− i 2 1+ i 1− i 0 + 1 2 2 1− i 1+ i M¬ 1 = 0 + 1 2 2 1− i 2 1+ i 2 M¬ 0 = 2 2 1+ i 1− i 1 = = 2 2 2 € € 0 and 1€ with a probability 1/2, because M¬ ⋅ M¬ = M¬ Is called square root of the not-gate € Gate W2 = € defined by W2 1 2 1 2 1 2 1 − 2 1 1 0 + 1 2 2 1 1 W2 1 = 0 − 1 2 2 W2 0 = W2 is called Walsh matrix, Hadamard matrix € or Hamarad-Walsh matrix • Quantum gates a linear 1 1 1 1 1 1 1 W 2 ( 0 + 1 ) = 2 W 2 0 + 2 W 2 1 = 2 2 ( 0 + 1 ) + 2 2 ( 0 − 1 ) = 0 2 € 4 10/30/09 Let use the coordinate representation 0 = 1 T (1,1) 2 1 = 1 T (1,−1) 2 What is the unitary matrix representation € of the quantum-not gate in this basis? M¬ 0 = 1 , M¬ 1 = 0 € The closure Relation a basis set { x1 , x 2 ,…, x n } The identity operator can be written as Given n ∑x i xi = I € i=1 € n n n a = I a = ∑ x i x i a = ∑ x i x i a = ∑ω i x i i=1 i=1 i=1 € 5 10/30/09 One quantum bit I= 0 0 + 1 1 I= 1 1 0 0 + 0 0 1 1 1 I = 0 1 I = 0 0 0 0 + 0 0 1 0 1 € Representation of Operators A = IAI = ∑ x i x i A∑ x j x j i j A = ∑ xi A x j xi x j i, j x1 A x1 x 2 A x1 x n A x1 € x1 A x 2 x2 A x2 x1 A x n xn A xn € 6 10/30/09 Matrix representation of an operator with respect to a certain basis u1 A u1 u2 A u1 un A u1 u1 A u2 u2 A u2 u1 A un un A un € Operators in two dimensional space Convention with respect to the computational basis 0 A0 A = 1A0 0 A1 1 A1 € 7 10/30/09 Z operator Z 0 = 0, Z = Z = Z 1 = −1 0 Z 1 0 (Z 0 ) = 1 Z 1 1 (Z 0 ) 0Z0 1Z 0 0 (Z 1 ) 1 (Z 1 ) − 0 1 1 0 = − 1 1 0 −1 00 10 € Unitary Transformation Change xi of basis to ui u1 x1 U = u2 x1 u1 x 2 u2 x 2 a′ = U a A′ = UAU * € 8 10/30/09 Let use the coordinate representation T T 0 = (1,0) 1 = (0,1) The unitary matrix defines an action 0 1 M¬ = 1 0 € The new basis + = € 1 T (1,1) 2 − = 1 T (1,−1) 2 € +0 +1 U = −1 −0 1 1 1 U= = W2 2 1 −1 Hadarmad € matrix (also called H) + , − The basis is called Hadamard basis € 9 10/30/09 U = U −1 = U * 1 1 1 0 11 1 UM¬U = 2 1 −11 01 −1 1 0 UM¬U = 0 −1 € Quantum Register A system of two quantm bits is a fourdimensional Hilbert space H 4 = H 2 ⊗ H 2 With the orthonormal basis {0 0 , 0 1,1 0 ,1 1} € 0 0 = 00 , 0 1 = 01 , 1 0 = 10 , 1 1 = 11 We € € write: A state of a two-qubit system is a unitlength vector ω 0 00 + ω1 01 + ω 2 10 + ω 3 11 2 2 2 2 ω 0 + ω1 + ω 2 + ω 3 = 1 € 10 10/30/09 ω 0 00 + ω1 01 + ω 2 10 + ω 3 11 Observation of two-qubit system gives, 00, 01, 10, 11 with probabilities 2 2 2 2 ω 0 , ω1 , ω 2 , ω 3 € Observation of one qubit (out of two) give 0 and one with probabilities for the first and probabilities for the second € 2 2 2 ω 0 + ω1 , ω 2 + ω 3 2 2 2 2 2 ( ω 0 + ω 2 , ω1 + ω 3 ) € The tensor product of vectors does not commute 0 1 ≠1 0 0 0 1 ≠ 0 0 1 0 0 We use linear ordering from left to right to address the qubits individually € 11 10/30/09 state z ∈ H4 of a two-qubit system is decomposable if z can be written as a product of states in H2 A z=x⊗y A state that is not decomposable is entangled € The 1 state 2 ( 00 + 01 + 10 + 11 ) Is decomposable because € 1 00 + 01 + 10 + 11 ) = ( 2 1 1 1 = (0 0 + 0 1 + 1 0 + 1 1)= 0 + 1 )⋅ ( (0 + 1) 2 2 2 € 1 1 1 1 1 1 1 = ⊗ 2 1 2 1 2 1 1 (1) Ψ ( 2) ⊗ Ψ (2) ω (1) ω 00 0 ω0 (1) (2) (2) ω (1) ω 01 ω ω ω 1,2 = 0(1) ⊗ 0(2) = 0(1) 1(2) = Ψ( ) = ω10 ω1 ω1 ω1 ω 0 (1) (2) ω11 ω1 ω1 € € 12 10/30/09 1 state 2 ( 00 + 11 ) Is entangled, to prove it we assume the contrary 1 ( 00 + 11 ) = (a0 0 + a1 1 )(b0 0 + b1 1 ) = 2 The € = a0b0 00 + a0b1 01 + a1b0 10 + a1b1 11 → a0b0 = 1 2 a0b1 = 0 a1b0 = 0 1 a1b1 = 2 contradiction € 1 If two qubits are entangled state 2 ( 00 + 11 ) then observing one of them will give 0 or 1, both with probability 1/2 It is not possible to observe € different values on the qubits Experiments have shown that this correlation can remain even if the qubits are separated more than 10 km Opportunities for quantum communication (teleportation) • Such as state is also called EPR-pair (Einstein, Podlsky, Rosen, regarded distant correlation as a source of paradox for quantum physics) 13 10/30/09 A 1 ( 00 + 11 ) 2 • is called EPR-pair (Einstein, Podolsky and Rosen) pair of bits in the state If both qubits are run through a Hadamard gate, the resulting€state is again 1 ( 00 + 11 ) 2 GHZ (Greenberger-Horne-Zeilinger) state state is as well entangled GHZ = M = 3, 1 0 2 ( ⊗M +1 ⊗M ), M > 2 € 1 ( 000 + 111 ) 2 € A binary quantum gate is a unitary mapping H4 → H4 We use the coordinate representation T T T 00 = (1,0,0,0) , 01 = (0,1,0,0) , 10 = (0,0,1,0) , 11 = (0,0,0,1) T € 14 10/30/09 Gate Mcnot M cnot 1 0 = 0 0 0 1 0 0 0 0 0 1 0 0 1 0 M cnot 00 = 00 , M cnot 01 = 01 , M cnot 10 = 11 , M cnot 11 = 10 € Is called controlled not, since the second qubit (target qubit) is flipped if and only if the first (control qubit) is 1 Tensor product, Kornecker product of matrices b11 b12 a1s a2s b b ,B = 21 22 ars bt1 bt 2 a11B a12 B a1sB a21B a22 B a2sB A⊗B= ar1B ar 2 B arsB a11 a12 a a A = 21 22 ar1 ar 2 b1u b2u btu € 15 10/30/09 If M1 and M2 are 2 x 2 matrices that describe unitary quantum gates, then it is easy to verify that the joint actions of M1 of the first qubis and M2 on the second are described by M1 ⊗ M2 This generalize to quantum systems of any size If matrices M1 and M2 define unitary mappings on Hilbert soace Hn and Hm, then nm x nm matrix M1 ⊗ M2 defines a unitary mapping on the space Hn ⊗ Hm Let M1=M2=W2 be the Hadamard matrix 1 1 1 1 1 1 −1 1 −1 W4 = W2 ⊗ W2 = 2 1 1 −1 −1 1 −1 −1 1 W 4 x 0 x1 = € 1 ( 0 + (−1)x0 1 )( 0 + (−1)x1 1 ) 2 1 = ( 00 + (−1) x1 01 + (−1) x 0 10 + (−1) x 0 +x1 11 ) 2 x 0 , x1 ∈ {0,1} € 16 10/30/09 result of action of W4 in our example is decomposable. Why? H 4 = H 2 ⊗ H 2 Input state is decomposable x 0 x1 = x 0 , x1 = x 0 x1 = x 0 ⊗ x1 € x 0 , x1 ∈ {0,1} W4 is decomposable The € W4 = W2 ⊗ W2 € € Mcnot cannot be expressed as a tensor product of 2 x 2 matrices Consider 00 (…….operator is linear) U t (ω1 x1 + ω 2 x 2 + …ω n x n ) = ω1U t x1 + ω 2U t x 2 + …ω nU t x n 1 1 (0 + 1)0 = ( 00 + 10 € 2 2 1 1 M cnot ( 00 + 10 ) = (M cnot 00 + M cnot 10 ) 2 2 1 1 M cnot ( 00 + 10 ) = ( 00 + 11 2 2 W2 0 0 = € € 17 10/30/09 action of Mcnot turns a decomposable state into an entangled state The ..it cannot be a tensor product of two unary operators M cnot M cnot 1 ( 00 + 11 ) = 2 1 ( 00 + 11 ) = 2 1 (M cnot 00 + M cnot 11 ) 2 1 1 ( 00 + 10 = (0 + 1)0 2 2 1 W 2 ( 0 + 1 ) 0 = 0 0 = 00 2 € 18 10/30/09 By a quantum register of length m we understand an ordered system of m qubits State is the m-fold tensor product H2m = H2 ⊗ ⊗ H2 The basis states are { x m x ∈ {0,1} } x = x1 … x m € We can also say that the basis states of m-qubit register are {a € a ∈ {0,1,…,2 m −1}} € A peculiar feature feature of exponential packing density associated with quantum system of m bits corresponds to c 0 0 + c1 1 + + c 2 m −1 2 m −1 2 2 2 c 0 + c1 + + c 2 m −1 = 1 € A general description of an m two-state quantum system requires 2m complex numbers Time evolution is described by unitary matrix of 2m x 2m 19 10/30/09 Time evolution is described by unitary matrix of 2m x 2m Exponential in the physical size of the system Effective deterministic simulation difficult Difficult to simulate on a probabilistic computer (Interference) A set of n elements can be identified with vectors of an ortonormal basis of n-dimensional complex vector space Hn A state is a unit-length vector in the state space 2 with probability ω i The state space of a compound system, consisting of two subsystems is the tensor product € of the subsystem state space State transformations are length-preserving mappings (Unitary mappings) 20