There are a variety of ways in which the teleportation protocol can be written mathematically. Some are very compact but abstract, and some are verbose but straightforward and concrete. The presentation below is of the latter form: verbose, but has the benefit of showing each quantum state simply and directly. Later sections review more compact notations. The teleportation protocol begins with a quantum state or qubit |\psi\rangle, in Alice's possession, that she wants to convey to Bob. This qubit can be written generally, in
bra–ket notation, as: :|\psi\rangle_C = \alpha |0\rangle_C + \beta|1\rangle_C. The subscript
C above is used only to distinguish this state from
A and
B, below. Next, the protocol requires that Alice and Bob share a maximally entangled state. This state is fixed in advance, by mutual agreement between Alice and Bob, and can be any one of the four
Bell states shown. It does not matter which one. :|\Phi^+\rangle_{AB} = \frac{1}{\sqrt{2}} (|0\rangle_A \otimes |0\rangle_{B} + |1\rangle_A \otimes |1\rangle_{B}), :|\Psi^+\rangle_{AB} = \frac{1}{\sqrt{2}} (|0\rangle_A \otimes |1\rangle_{B} + |1\rangle_A \otimes |0\rangle_{B}), :|\Psi^-\rangle_{AB} = \frac{1}{\sqrt{2}} (|0\rangle_A \otimes |1\rangle_{B} - |1\rangle_A \otimes |0\rangle_{B}). :|\Phi^-\rangle_{AB} = \frac{1}{\sqrt{2}} (|0\rangle_A \otimes |0\rangle_{B} - |1\rangle_A \otimes |1\rangle_{B}), In the following, assume that Alice and Bob share the state |\Phi^+\rangle_{AB}. Alice obtains one of the particles in the pair, with the other going to Bob. (This is implemented by preparing the particles together and shooting them to Alice and Bob from a common source.) The subscripts
A and
B in the entangled state refer to Alice's or Bob's particle. At this point, Alice has two particles (
C, the one she wants to teleport, and
A, one of the entangled pair), and Bob has one particle,
B. In the total system, the state of these three particles is given by : |\psi\rangle_C\otimes|\Phi^+\rangle_{AB} = (\alpha |0\rangle_C + \beta|1\rangle_C)\otimes\frac{1}{\sqrt{2}} (|0\rangle_A \otimes |0\rangle_B + |1\rangle_A \otimes |1\rangle_B) . Alice will then make a local measurement in the Bell basis (i.e. the four Bell states) on the two particles in her possession. To make the result of her measurement clear, it is best to write the state of Alice's two qubits as superpositions of the Bell basis. This is done by using the following general identities, which are easily verified: :|0\rangle \otimes |0\rangle = \frac{1}{\sqrt{2}} (|\Phi^+\rangle + |\Phi^-\rangle), :|0\rangle \otimes |1\rangle = \frac{1}{\sqrt{2}} (|\Psi^+\rangle + |\Psi^-\rangle), :|1\rangle \otimes |0\rangle = \frac{1}{\sqrt{2}} (|\Psi^+\rangle - |\Psi^-\rangle), and :|1\rangle \otimes |1\rangle = \frac{1}{\sqrt{2}} (|\Phi^+\rangle - |\Phi^-\rangle). After expanding the expression for \begin{align} \end{align} , one applies these identities to the qubits with
A and
C subscripts. In particular, \alpha \frac{1}{\sqrt{2}} |0\rangle_C \otimes |0\rangle_A \otimes |0\rangle_B = \alpha \frac{1}{2}(|\Phi^+\rangle_{CA} + |\Phi^-\rangle_{CA}) \otimes |0\rangle_B, and the other terms follow similarly. Combining similar terms, the total three particle state of
A,
B and
C together becomes the following four-term superposition: : \begin{align} \frac{1}{2} \Big \lbrack \ & |\Phi^+\rangle_{CA} \otimes (\alpha |0\rangle_B + \beta|1\rangle_B) \ + \ |\Phi^-\rangle_{CA} \otimes (\alpha |0\rangle_B - \beta|1\rangle_B) \\ \ + \ & |\Psi^+\rangle_{CA} \otimes (\alpha |1\rangle_B + \beta|0\rangle_B) \ + \ |\Psi^-\rangle_{CA} \otimes (\alpha |1\rangle_B - \beta|0\rangle_B) \Big \rbrack . \\ \end{align} Note that all three particles are still in the same total state since no operations have been performed. Rather, the above is just a change of basis on Alice's part of the system. This change has moved the entanglement from particles A and B to particles C and A. The actual teleportation occurs when Alice measures her two qubits (C and A) in the Bell basis gate followed by a
Hadamard operation. In the outputs, a and b take on values of 0 or 1. :|\Phi^+\rangle_{CA}, |\Phi^-\rangle_{CA}, |\Psi^+\rangle_{CA}, |\Psi^-\rangle_{CA}. Equivalently, the measurement may be done in the computational basis, \{|0\rangle,|1\rangle\} , by mapping each Bell state uniquely to one of \{|0\rangle \otimes|0\rangle, |0\rangle \otimes|1\rangle, |1\rangle \otimes|0\rangle, |1\rangle \otimes|1\rangle\} with the quantum circuit in the figure to the right. The result of Alice's (local) measurement is a collection of two classical bits (00, 01, 10 or 11) related to one of the following four states (with equal probability of 1/4), after the three-particle state has
collapsed into one of the states: • |\Phi^+\rangle_{CA} \otimes (\alpha |0\rangle_B + \beta|1\rangle_B) • |\Phi^-\rangle_{CA} \otimes (\alpha |0\rangle_B - \beta|1\rangle_B) • |\Psi^+\rangle_{CA} \otimes (\alpha |1\rangle_B + \beta|0\rangle_B) • |\Psi^-\rangle_{CA} \otimes (\alpha |1\rangle_B - \beta|0\rangle_B) Alice's two particles are now entangled to each other, in one of the four
Bell states, and the entanglement originally shared between Alice's and Bob's particles is now broken. Bob's particle takes on one of the four superposition states shown above. Note how Bob's qubit is now in a state that resembles the state to be teleported. The four possible states for Bob's qubit are unitary images of the state to be teleported. The result of Alice's Bell measurement tells her which of the above four states the system is in. She can now send her result to Bob through a classical channel. Two classical bits can communicate which of the four results she obtained. After Bob receives the message from Alice, he will know which of the four states his particle is in. Using this information, he performs a unitary operation on his particle to transform it to the desired state \alpha |0\rangle_B + \beta|1\rangle_B: • If Alice indicates her result is |\Phi^+\rangle_{CA}, Bob knows his qubit is already in the desired state and does nothing. This amounts to the trivial unitary operation, the identity operator. • If the message indicates |\Phi^-\rangle_{CA}, Bob would send his qubit through the unitary
quantum gate given by the
Pauli matrix :\sigma_3 = \begin{bmatrix} 1 & 0 \\ 0 & -1\end{bmatrix} to recover the state. • If Alice's message corresponds to |\Psi^+\rangle_{CA}, Bob applies the gate :\sigma_1 = \begin{bmatrix} 0 & 1 \\ 1 & 0\end{bmatrix} to his qubit. • Finally, for the remaining case, the appropriate gate is given by : \sigma_3 \sigma_1 = - \sigma_1 \sigma_3 = i \sigma_2 = \begin{bmatrix} 0 & 1 \\ -1 & 0\end{bmatrix}. Teleportation is thus achieved. The above-mentioned three gates correspond to rotations of π radians (180°) about appropriate axes (X, Y and Z) in the
Bloch sphere picture of a qubit. Some remarks: • After this operation, Bob's qubit will take on the state |\psi\rangle_B= \alpha |0\rangle_B + \beta|1\rangle_B, and Alice's qubit becomes a part of an entangled state. Teleportation does not result in the copying of qubits, and hence is consistent with the
no-cloning theorem. • There is no transfer of matter or energy involved. Alice's particle has not been physically moved to Bob; only its state has been transferred. The term "teleportation", coined by Bennett, Brassard, Crépeau, Jozsa, Peres and Wootters, reflects the indistinguishability of quantum mechanical particles. • For every qubit teleported, Alice needs to send Bob two classical bits of information. These two classical bits do not carry complete information about the qubit being teleported. If an eavesdropper intercepts the two bits, she may know exactly what Bob needs to do in order to recover the desired state. However, this information is useless if she cannot interact with the entangled particle in Bob's possession. == Certifying quantum teleportation ==