I'm still learning Quantum Information right now, especially with Quantum Teleportation (QT). I have found an interesting article suggesting a Simplified Protocol of QT that doesn't need classical channel. Here's the link for the article : https://www.scirp.org/journal/paperinformation.aspx?paperid=86656 enter image description here

This protocol is more satisfying and make sense for me. However, it makes me wonder, why classical channel should be used in the first place, if this protocol is more simple and efficient?

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    $\begingroup$ Compare equations 23 and 25 in the paper. As far as I can tell, in this "simplified" protocol, the state arrives on the same qubit it started on. It's not being teleported anywhere. $\endgroup$
    – DaftWullie
    Commented Nov 10, 2021 at 15:27
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    $\begingroup$ This looks like the teleportation protocol with postselection on the 00 outcome for $b_1$ and $b_2$. $\endgroup$
    – Condo
    Commented Nov 10, 2021 at 18:37
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    $\begingroup$ This is a garbage article from the garbage publisher en.wikipedia.org/wiki/Scientific_Research_Publishing $\endgroup$
    – Danylo Y
    Commented Nov 10, 2021 at 18:51

3 Answers 3


This "protocol" is completely wrong. The author is simply preparing the state $|\psi\rangle$ on Bob's side and leaving it there untouched. Meanwhile, Alice prepares a maximally entangled state, and applies a CNOT and a Hadamard to it, which maps it into the state $|00\rangle$.

It is not possible to teleport without a channel.

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    $\begingroup$ Yes, the paper is just bad. It does a trivial analysis of the wrong thing then calls it teleportation. $\endgroup$ Commented Nov 10, 2021 at 17:40

No. If no classical information is transmitted from Alice to Bob, then Bob's state is completely unaffected by anything that happens on Alice's side. This means, in particular, that no information whatsoever can be sent using such protocol.

More precisely, if $\rho$ is the initial bipartite (presumably entangled) state, then the only thing Bob has access to is the partial state $\rho^B\equiv \operatorname{Tr}_A(\rho)$. This partial state is unaffected by any operation Alice can perform, as easily seen observing that $\operatorname{Tr}_A((\Phi\otimes I)\rho)=\operatorname{Tr}_A(\rho)$ for any CPTP channel $\Phi$. Thus, in particular, no teleportation is possible.


First things first. The quantum teleportation protocol requires classical bits to be sent from Alice to Bob. Without these bits, Bob cannot apply the proper corrections to actually obtain the desired "target" state $|\psi\rangle$, this is apparent by understanding what's going on in the teleportation protocol.

Additionally, the requirement of a classical channel in the teleportation protocol eliminates the notion that quantum teleportation could be used to send information faster than the speed of light, which should be impossible by other physical laws, see the no-communication theorem.

From quickly glancing at the article and this "simplified" teleportation protocol, it seems that the author is not proposing that you can do teleportation without sending any classical bits (at least I hope that is not their claim), but rather, they seem to be analyzing the quality or fidelity of the teleportation protocol in the case where you don't send the classical bits. This is all fine, but it should be stressed that this protocol will only work a small proportion of the time.


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