Suppose Alice starts with two qubits in the Bell state (1/√2)|00⟩+(1/√2)|11⟩ and teleports these qubits to Bob by applying the quantum teleportation protocol to each qubit separately.
(a) What is the state of the two qubits received by Bob?
To determine the state of the two qubits received by Bob after applying the quantum teleportation protocol, we need to understand how the protocol works.
The quantum teleportation protocol involves three qubits: the original qubit to be teleported (in this case, the two qubits in the Bell state), an entangled pair of qubits shared between Alice and Bob, and an additional qubit that Alice prepares and sends to Bob.
The protocol proceeds as follows:
1. Alice applies a Controlled NOT (CNOT) gate to her qubit (the least significant qubit of the Bell state) and the original qubit she wants to teleport. Then, she applies a Hadamard gate to her qubit.
2. Alice measures both of her qubits and obtains two classical bits of information.
3. Alice sends the classical bits to Bob through classical communication.
4. Based on the classical bits received from Alice, Bob applies a series of operations to his qubits to reconstruct the state.
Now let's go through each step and determine the state of the qubits received by Bob.
Step 1: Applying CNOT and Hadamard gates
With the Bell state (1/√2)|00⟩ + (1/√2)|11⟩ as the input, Alice applies a CNOT gate to the first qubit (her qubit) and the corresponding qubit she wants to teleport. This results in the following state:
(1/√2)|00⟩ + (1/√2)|11⟩ → (1/√2)|00⟩ + (1/√2)|10⟩
Next, Alice applies a Hadamard gate to her qubit, resulting in:
(1/√2)|00⟩ + (1/√2)|10⟩ → (1/2)(|00⟩ + |10⟩) + (1/2)(|00⟩ - |10⟩)
Step 2: Measuring qubits
Alice measures both of her qubits. Suppose she measures the qubits and obtains the first classical bit, b1, from her own qubit, and the second classical bit, b2, from the qubit being teleported.
Step 3: Sending classical bits
Alice sends the classical bits, b1 and b2, to Bob through classical communication.
Step 4: Bob's operations
Based on the classical bits received from Alice, Bob performs the following operations on his qubits:
If b2 = 0, Bob does nothing.
If b2 = 1, Bob applies a Pauli-X gate (bit flip) to his qubit.
If b1 = 0, Bob does nothing.
If b1 = 1, Bob applies a Pauli-Z gate (phase flip) to his qubit.
Taking all the possibilities into account, the final state of the qubits received by Bob can be expressed as a linear combination of basis states:
|ψ⟩ = (1/2)(|00⟩ + |10⟩) + (1/2)(|00⟩ - |10⟩) (if b1 = 0 and b2 = 0)
|ψ⟩ = (1/2)(|01⟩ + |11⟩) + (1/2)(|01⟩ - |11⟩) (if b1 = 0 and b2 = 1)
|ψ⟩ = (1/2)(|00⟩ - |10⟩) + (1/2)(|00⟩ + |10⟩) (if b1 = 1 and b2 = 0)
|ψ⟩ = (1/2)(|01⟩ - |11⟩) + (1/2)(|01⟩ + |11⟩) (if b1 = 1 and b2 = 1)
Simplifying each expression, we can rewrite the final state as:
|ψ⟩ = (1/2)(|00⟩ + |01⟩ + |10⟩ + |11⟩) [if b1 = 0 and b2 = 0]
|ψ⟩ = (1/2)(|00⟩ - |01⟩ + |10⟩ - |11⟩) [if b1 = 0 and b2 = 1]
|ψ⟩ = (1/2)(|00⟩ + |01⟩ - |10⟩ - |11⟩) [if b1 = 1 and b2 = 0]
|ψ⟩ = (1/2)(|00⟩ - |01⟩ - |10⟩ + |11⟩) [if b1 = 1 and b2 = 1]
Therefore, depending on the measurement outcomes of Alice's qubits (b1 and b2), the state of the two qubits received by Bob will be a superposition of the basis states |00⟩, |01⟩, |10⟩, and |11⟩.