Quantum Teleportation: Preparation

Quantum teleportation is a protocol that enables the transmission of an unknown quantum state from a sender (Alice) to a receiver (Bob) without the physical transfer of the particle carrying the state.

It is an incredibly interesting physical phenomenon and has been implemented successfully across hundreds of kilometers by various scientific teams around the world!

This process relies on two primary resources: a pre-shared pair of entangled qubits and a so-called "classical communication channel", which really just means Alice has to communicate some simple information to Bob during the protocol (for instance via phone).

Let's define the setup:

1. Qubit Q (Alice's Source Qubit): This is Alice's qubit, carrying the unknown quantum state she wishes to teleport. This state can be generally written as

   $$|\psi\rangle_Q = \alpha|0\rangle_Q + \beta|1\rangle_Q,$$

   where $|\alpha|^2 + |\beta|^2 = 1$, and have used the subscript $Q$ on $\ket{\psi}_{Q}$ to help keep track of which qubit this is in the protocol. This is the state to be teleported.

   Alice does not necessarily know the values of $\alpha$ and $\beta$.

   In column vector notation (basis $\{|0\rangle, |1\rangle\}$), this state is $\ket{\psi}_{Q} =$ $\begin{bmatrix} \alpha \\ \beta \end{bmatrix}_{Q}$.

2. Qubit A (Alice's Entangled Qubit): This qubit is also in Alice's possession and is one half of an entangled Bell state.

3. Qubit B (Bob's Entangled Qubit): This qubit is held by Bob and is the other half of the entangled pair.

The standard entangled pair utilized for teleportation is the following Bell state:

$$|\Phi^+\rangle_{AB} = \frac{1}{\sqrt{2}}(|00\rangle_{AB} + |11\rangle_{AB})$$

In the computational basis $\{|00\rangle, |01\rangle, |10\rangle, |11\rangle\}$, we recall that the column vector for this Bell state is:

$$|\Phi^+\rangle_{AB} = \frac{1}{\sqrt{2}}\begin{bmatrix} 1 \\ 0 \\ 0 \\ 1 \end{bmatrix}_{AB}.$$

Note that we have used the subscript $AB$ to denote that this state is shared between Alice and Bob, i.e. Alice has one of the qubits, and Bob has the other.

Thus, the total initial state $\ket{\Psi_{0}}$ of the 3-qubit system (ordered $Q$, $A$, $B$) is formed by the Kronecker product of Alice's unknown qubit state and the shared Bell pair:

$$|\Psi_0\rangle = |\psi\rangle_Q \otimes |\Phi^+\rangle_{AB}$$

To be clear, Alice has two qubits in her lab, the source qubit (the state to be teleported), and her half of the entangled pair. Bob has one qubit in his lab, which is his half of the entangled pair.

Bra-ket Expansion of the Initial State:

Let's expand $|\Psi_0\rangle$:

$|\Psi_0\rangle = (\alpha|0\rangle_Q + \beta|1\rangle_Q) \otimes \left( \frac{1}{\sqrt{2}}(|00\rangle_{AB} + |11\rangle_{AB}) \right)$

Using the distributive property of the Kronecker product:

$|\Psi_0\rangle = \frac{1}{\sqrt{2}} \left( \alpha|0\rangle_Q \otimes (|00\rangle_{AB} + |11\rangle_{AB}) + \beta|1\rangle_Q \otimes (|00\rangle_{AB} + |11\rangle_{AB}) \right)$

$|\Psi_0\rangle = \frac{1}{\sqrt{2}} \left( \alpha|0\rangle_Q|00\rangle_{AB} + \alpha|0\rangle_Q|11\rangle_{AB} + \beta|1\rangle_Q|00\rangle_{AB} + \beta|1\rangle_Q|11\rangle_{AB} \right)$

Or, removing the subscripts:

$$|\Psi_0\rangle = \frac{1}{\sqrt{2}} (\alpha|000\rangle + \alpha|011\rangle + \beta|100\rangle + \beta|111\rangle)$$

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