# Resource cost and noise effects in quantum teleportation of multible (entangled) qbits

Suppose you have n qubits that are in an unknown state (may be entangled, etc).

Can you teleport this state by teleporting each qubit individually (using a Bell state and a classical channel)?

If not, how many classical bits and what kind of Bell states are needed? Do we suffer an exponential blowup? Does adding small amount of noise (i.e. imperfect hardware) have a large impact on these costs?

This can be seen by observing that teleportation of a qubit implements the identity channel on the qubit, i.e., it maps any state $|\alpha\rangle$ to itself (while preserving the phase!). More precisely, teleportation (like any physical map) is a completely positive trace preserving (CPTP) map which is of the form $\mathcal E(\rho)=\rho$, i.e., it also preserves coherences. Several independent teleportations thus implement the identity on $N$ qubits, i.e., they preserve any states (including entangled ones).
On the other hand, if there is noise (i.e., the entangled states are not perfect) you will need to use an encoding/decoding scheme. For general noise (however small it is), this will only work asymptotically, i.e., you will need to teleport a large number $N$ of qubits using $cN$ imperfect Bell pairs (with $c>1$ some constant), and in the limit $N\rightarrow\infty$, the teleportation will work perfectly given $c$ is not larger than the quantum capacity of the channel.
• @KevinKostlan Well, the main answer follows basically from the linearity of teleportation: If $|i\rangle|j\rangle$ is mapped to $|i\rangle|j\rangle$ by teleportation (=two parallel independent teleportations of the two qubits), so is every superposition. In fact, it would be extremely weird if this would not happen: What would then be the action of teleporting two qubits independently if we apply it to an entangled state? – Norbert Schuch Dec 29 '14 at 2:25