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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?

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  • $\begingroup$ 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. $\endgroup$ Jul 30, 2022 at 16:35

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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.

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    $\begingroup$ 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 :) $\endgroup$ Jul 30, 2022 at 18:46

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