Lecture 4 - Quantum Circuit Model, Entanglement, Superdense Coding 02.03.2020

Home , Superdense coding

Ozye˘ginUniversity¨ Spring 2020 CS480/CS580 Quantum Computing

Scribe: Eren Balatkan Lecturer: Ozlem¨ Salehi K¨oken

4 Quantum Circuit Model

In this section, we will be talking about the quantum circuit model.

4.1 1-Qubit Gates

Not (X) : Corresponds to a 180 rotation around X - Axis

0 1 Z |0i X = 1 0

X |0i = |1i Y |0i

X |0i X |1i = |0i

X(α |0i + β |1i) = β |0i + α |1i |1i

Y-Gate (Y) : Corresponds to a 180 rotation around Y - Axis

0 −i Z |0i Y = i 0

Y |0i = i |1i

Y |0i

X |0i Y |1i = −i |0i

Y (α |0i + β |1i) = −βi |0i + α |1i |1i

1 Z-Gate (Z) : Corresponds to a 180 rotation around Z - Axis

1 0 Z |0i Z = 0 −1

Z |0i = |0i Y |0i

X |0i Z |1i = − |1i

Z(α |0i + β |1i) = α |0i + −β |1i |1i

Rotation Gates

Let |Ψi be a quantum state and let µ be a unitary operator. The action of µ on |Ψi can be thought as a rotation on the Bloch sphere

−iθX/2 Rx(θ) = e

−iθY/2 Ry(θ) = e

−iθZ/2 Rz(θ) = e

If A2 = I, eiAx = cos(x)I + isin(x)A

 θ θ  cos −isin θ θ  2 2  Rx(θ) = cos I + isin X =   2 2 θ θ  −isin cos  2 2  θ θ  cos −sin θ θ  2 2  Ry(θ) = cos I + isin Y =   2 2 θ θ  sin cos  2 2

θ θ R (θ) = e−iθZ/2 = cos I − Z z 2 2  θ   θ  cos 0 isin 0 2 2 =   −    θ   θ   0 cos   0 −isin  2 2  θ θ  cos − isin 0 2 2 =    θ θ   0 cos + isin  2 2 e−iθ/2 0 = 0 eiθ/2

2 Example:

Eﬀect of Rz(θ) is to change the angle ϕ to ϕ + θ which is a rotation around z-axis

iα Theorem 1. Suppose µ is a 1-Qubit unitary gate. Then there exists α, β, γ and δ such that µ = e Rz(β)Ry(γ)Rz(δ)

Theorem 2. Any unitary µ can be expressed as µ = eiαAXBXC where ABC = I, A,B,C are unitary

S-Gate T-Gate

1 0 1 0 S = T = 0 i 0 eπi/4

S = Rz(π/2) T = Rz(π/4) S2 = Z T 2 = S

Hadamard Gate (H)

! 1 1 √1 √1 H = √1 = 2 2 2 1 −1 √1 − √1 2 2 ! ! √1 √1 1 √1 |+i = H |0i = 2 2 = 2 = √1 |0i + √1 |1i √1 − √1 0 √1 2 2 2 2 2 ! ! √1 √1 0 √1 |−i = H |1i = 2 2 = 2 = √1 |0i − √1 |1i √1 − √1 1 − √1 2 2 2 2 2

3 1 1 H |+i = H √ |0i + √ |1i 2 2 1 1 1 1 1 1 = √ √ |0i + √ |1i + √ √ |0i − √ |1i 2 2 2 2 2 2 1 1 1 1 = |0i + |1i + |0i − |1i 2 2 2 2 = |0i

H2 = I

(H ⊗ H)(|00i) = H⊗2 |00i 1 1 = √ (|0i + |1i) ⊗ √ (|0i + |1i) 2 2 1 1 1 1 = |00i + |01i + |10i + |11i 2 2 2 2

1 n H⊗n |0...0i = √ P2 −1 |xi where x is written in binary 2n x=0 1 1 1 1 H⊗3 |000i = √ P7 |xi = √ |000i + √ |001i + ... + √ |111i 8 x=0 8 8 8

Hadamard corresponds to a rotation of π around √1 , 0, √1 . When restricted to real amplitudes, Hadamard 2 2 is a reﬂection matrix in 2D plane over angle π/8

|0i

|+i

π/8 |1i

4 4.2 Multi-Qubit Gates

Controlled Not (CNOT)

  1 0 0 0 CNOT |10i = |11i  0 1 0 0  CNOT =    0 0 0 1  CNOT |11i = |10i 0 0 1 0 CNOT |01i = |01i

CNOT |00i = |00i

Example:

Let |Ψi = √1 |00i + √1 |10i = √1 |0i + √1 |1i ⊗ |0i 2 2 2 2

CNOT (|Ψi = √1 |00i + √1 |11i) 2 2

Controlled Z-Gate (CZ)

  1 0 0 0 CZ |00i = |00i  0 1 0 0  CZ =    0 0 1 0  CZ |01i = |01i 0 0 0 −1 CZ |10i = |10i

CZ |11i = − |11i

Toﬀoli Gate (CCX)

  1 0 0 0 0 0 0 0 CCX |000i = |000i 0 1 0 0 0 0 0 0   0 0 1 0 0 0 0 0   CCX |101i = |101i 0 0 0 1 0 0 0 0 CCX =   0 0 0 0 1 0 0 0   CCX |110i = |111i 0 0 0 0 0 1 0 0   0 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 CCX |111i = |110i

5 Example:

C4NOT : Apply NOT to target if 4 control qubits are in state |1i ?

Input 1 Input 2 Input 3 Input 4

Ancilla 1

Ancilla 2

Target

Example:

Apply NOT to target if control qubit is on state |0i ?

Input 1 X X

Target

6 5 Basic Quantum Protocols

5.1 Entanglement

Suppose that we have a quantum system in |Ψi = |00i. Let’s apply the following operations.

Qubit 1 H

Qubit 2

1 1 (H ⊗ I)(|00i) = √ |0i + √ |1i ⊗ |0i 2 2 1 1 = √ |00i + √ |10i 2 2

CNOT √1 |00i + √1 |10i) = √1 |00i + √1 |11i 2 2 2 2

If we measure ﬁrst qubit and observe |0i, then the second qubit collapses to |0i. Similarly, if we measure the second qubit and observe |1i, then the second qubit collapses to |1i. This is true even if the qubits are seperated from each other. This is called entanglement.

Bell States

•| Φ+i = √1 |00i + √1 |11i 2 2

•| Φ−i = √1 |00i − √1 |11i 2 2

•| Ψ+i = √1 |01i + √1 |10i 2 2

•| Ψ−i = √1 |01i − √1 |10i 2 2 The states above are known as the Bell states and they are entangled states. Entangled states can not be written as a tensor product of two subsystems.

7 5.2 Superdense Coding

Suppose that Alice wants to send Bob two classical bits of information. Alice will send only a single qubit to Bob to achieve this task. • Initially, Alice and Bob have an entangled pair of qubits in state √1 |00i + √1 |11i. 2 2 • Alice has the ﬁrst qubit and Bob has the second qubit. Alice wants to send two bits of classical information, ab, to Bob. • If a=1, Alice applies Z-Gate to her qubit. • If b=1, Alice applies X and sends to Bob.

ab Operation Result ———– ———– ———– 00 - √1 |00i + √1 |11i 2 2 01 X √1 |10i + √1 |01i 2 2 10 Z √1 |00i − √1 |11i 2 2 11 XZ √1 |10i − √1 |01i 2 2

• Now Bob has both qubits. • Bob applies CNOT where ﬁrst qubit is control and second qubit is target. • Bob applies Hadamard to ﬁrst qubit • He makes the measuerment and observes quantum state |abi

1’st Case - |00i

1 1 |Ψi = CNOT √ |00i + √ |11i 2 2 1 1 = √ |00i + √ |10i 2 2

1 1 1 1 1 1 (H ⊗ I) |Ψi = √ √ |0i + √ |1i |0i + √ √ |0i − √ |1i |0i 2 2 2 2 2 2 1 1 1 1 = √ |00i + √ |10i + √ |00i − √ |10i = |00i 2 2 2 2 2’nd Case - |01i

1 1 |Ψi = CNOT √ |10i + √ |01i 2 2 1 1 = √ |11i + √ |01i 2 2

1 1 1 1 1 1 (H ⊗ I) |Ψi = √ √ |0i − √ |1i |1i + √ √ |0i + √ |1i |1i 2 2 2 2 2 2 1 1 1 1 = √ |01i − √ |11i + √ |01i − √ |11i = |01i 2 2 2 2