# General Proof of the Statement that You Need $1$ Ebit and $2$ Bits to Teleport $1$ Qubit?

I understand from the standard teleportation protocol that 1 ebit is used up in teleporting 1 qubit and thus, cannot be used again -- and thus, we need 1 fresh ebit of shared entanglement between Alice and Bob per each qubit that Alice wants to teleport to Bob. Furthermore, the teleported state could have a phase-flip or a bit-flip error and thus, the $$2$$ bit syndrome of Alice's measurements need to be sent to Bob for Bob to recover the correct qubit.

However, is there a general proof of this statement apart from the heuristic argument? If so, is there a fundamental no-go theorem of quantum mechanics behind this restriction? For example, the fact that you need $$2$$ classical bits to be sent in order to teleport $$1$$ qubit appears to be less than optimal, i.e., even if it were possible to teleport a qubit using only $$1$$ classical bit (along with the $$1$$ ebit, of course), the no-communication theorem wouldn't be violated because one can only extract $$1$$ bit of classical information from $$1$$ qubit. Is there some other basic tenet of quantum mechanics that would be violated if this were to be possible?

• Hmm, one thing I can think of is that if the qubit being teleported is entangled to another qubit that Bob has then superdense coding could be used to send $2$ bits of information and thus, if the teleportation only required $1$ bit of communication then we'd be in trouble vis-a-vis no-communication theorem. Jul 30, 2022 at 16:35

To prove that teleportation of a single qubit cannot be accomplished with less than 2 classical bits, you can assume the opposite to show that No-Signalling was violated (as your comment suggests).

Recall the two protocols that we know we can perform:

1. TP (Teleportation): Alice transmits $$1\text{ qubit}$$ to Bob by physically sending $$2\text{ cbits}$$, consuming $$1 \text{ ebit}$$:

$$\text{TP:}\qquad1\text{ ebit} + 2 \text{ cbits} \rightarrow 1 \text{ qubit}\tag{1}$$

1. SD (Superdense Coding): Alice transmits $$2\text{ cbits}$$ to Bob by physically sending $$1\text{ qubit}$$ and consuming $$1 \text{ ebit}$$ in the process: $$\text{SD:}\qquad1\text{ ebit} + 1 \text{ qubit} \rightarrow 2 \text{ cbits} \tag{2}$$

Now suppose there is a version of Teleportation (call it SuperTP) that transmits one qubit to bob but sends $$x < 2$$ cbits. This allows us to do the following: We will perform SD, but when it comes time to physically send 1 qubit, we will instead perform SuperTP to transmit that single qubit worth of information using $$x <2$$ cbits:

$$\text{SuperTP:}\qquad1\text{ ebit} + x \text{ cbits} \rightarrow 1 \text{ qubit}\tag{3}$$

Substituting SuperTP for the physically transmitted qubit on the LHS of Eq. (2), we can describe a composition of the SuperTP and SD protocols as

$$\text{SD[SuperTP]:}\qquad2\text{ ebits} + x \text{ cbits} \rightarrow 2 \text{ cbits} \tag{4}$$

Now it is clear that any $$x<2$$ will violate the No-Signalling principle, and so SuperTP cannot exist. In this sense, Teleportation with cbits and qubits is optimal; it uses the lower bound of $$x=2$$ cbits that does not violate No-Signalling. Furthermore, we have not assumed anything about the optimality of SD for this proof: A more efficient protocol for SD would only drive the lower bound for $$x$$ up. Yet, the existence of TP in Eq. (1) allows us to do this exact same procedure to show that SD is optimal, by creating a composition TP[SuperSD] (you might find it worthwhile to work through that). In this sense, there is a duality between SD and TP where each one's existence implies the other's optimality.

• Thanks for your answer! Can you also comment on the optimality of the number ebits? Again, heuristically, I get it why we need 1 ebit for 1 qubit to be teleported (because any teleportation scheme would need a "phase kickback" step that would require Alice to make measurements and that would use up the entanglement) but I don't know what's the proper formal argument. Thanks again :) Jul 30, 2022 at 18:46