# Why is super-dense coding called the inverse of quantum teleportation?

I am new to quantum computation and I recently came across the statement that super-dense coding can be called the inverse of quantum teleportation

• For me, teleportation is more physically fundamental than superdense coding. tt seems superdence coding is just to carry 2 classical bits by applying 4 different unitary operators. But teleportation, 2 classical bits seem to deliver an infinite amount of information. – XXDD Nov 5 '18 at 12:10

In quantum teleportation, one starts with an entangled state shared between two parties, and (after some messing at the sender's side), two classical bits are transmitted from one party to the other so that the net effect is a quantum state is sent from the first party to the second without sending any quantum data.

In superdense coding, the parties start with an entangled state shared between two parties, and (after some messing at the sender's side), a quantum state is sent from one party to the other so that the net effect is two classical bits are sent from the first party to the second.

Hopefully I've written that in such a way that it conveys the symmetry between the two settings. Where I say "quantum state", I specifically mean a single qubit in an unknown state.

Basically quantum teleportation is in facto the determinate side of super-dense coding. In superdense coding we fit two classical bits of information using fairly a single qubit. On the other hand, quantum teleportation uses two classical bits of information to send a single qubit that is in an unknown quantum state. I suggest you to check IBM Q documentation for more details. Also, the βNo Cloning Theoremβ states that you cannot prevailingly clone a qubit in an unknown quantum state. Letβs delve a deeper into this to see why this is accurate.

Given the sum |πβ© + |πβ© is a vector. The scalar product πΌ|πβ© is a vector and πΌ is a complex number πΌ β β.