Notation: $|\text{qubit}_{1}, ..., \text{qubit}_{N}\rangle$.

The goal of quantum teleportation is to send quantum information using classical bits.

A source transmits a state $|\psi\rangle_{A_{0}} = \alpha|0\rangle + \beta|1\rangle$ to Alice. Alice has no knowledge of what this state is and is forbidden from measuring this state as the laws of quantum mechanics result in the collapse of the state.

At this point, quantum teleportation protocol dictates the introduction of a Bell state, say $|\Phi_{+}\rangle = \frac{1}{\sqrt{2}}[|00\rangle + |11\rangle]$, that tensors with $|\psi\rangle_{A_{0}}$ to give

$|\psi\rangle_{A_{0}}|\Psi_{+}\rangle = \frac{1}{\sqrt{2}}[\alpha (|000\rangle + |011\rangle) + \beta(|100\rangle + |111\rangle)]$.

What is the physical (or mathematical) motivation behind which the above Bell state is introduced?


1 Answer 1


The motivation for introducing the Bell state is simple: it gets the job done! This of course raises the question: what properties of the Bell state make it suitable for use as a resource in quantum teleportation? Perhaps the most intuitive answer to this question is proivded by state-channel duality.

The simplest statement of state-channel duality is as a mapping from operators to states defined by $$ |i\rangle\langle k|\mapsto|i\rangle|k\rangle\tag1 $$ and extended by linearity$^1$. But then $$ I=\sum_i|i\rangle\langle i|\mapsto\sum_i|i\rangle|i\rangle=\sqrt{d}|\Phi_+\rangle\tag2 $$ where $d$ is the dimension of the Hilbert space. Thus, the Bell state corresponds to the identity channel under state-channel duality. But identity is exactly the channel we seek to realize in quantum teleportation.

Equation $(2)$ has an attractive visual representation in string diagrams and tensor networks where it corresponds to a deformation wherein a wire is bent so that its input becomes its second output, see e.g. equation $(30)$ on page $8$ in this paper.

$^1$ There are two levels of state-channel duality. The first maps linear operators to states. The second maps superoperators to linear operators. In the present context, the former is sufficient.


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