# Why is measurement needed in teleportation?

One of the postulates of QC is, that measurements in every circuit can be postponed or never performed in a circuit while achieving the same functionality of the circuit (at least this the way I understand it). Given this, why does Alice have to do any measurement at all? Isn't a much better / simpler circuit is, not sending any classical / quantum information to Bob at all and nevertheless Bob can achieve the original vector state.

What's the point at all of sending any classical bits? I mean, imagine today I show you an option that we can communicate with other without a need of sending anything at all (provided we got some prior entangled qubit), just by working remotely, isn't it much better? Than requiring the transformation of either qubits or bits.

• Maybe turn the question the other way around. Why do you think it's possible to achieve without performing a measurement? Specify a protocol, calculate the corresponding mathematics of how it works, and see if it does. If you genuinely believe you have a protocol that works without the measurement, show us, and we'll try to figure out what's going on. – DaftWullie Feb 28 at 8:38

To answer this, contrast quantum teleportation with the swap gate. Ignoring everything in the middle, the effect of quantum teleportation is to get from state $$|\phi00\rangle$$ to $$|00\phi\rangle$$. This can obviously be accomplished with a simple application of the swap gate. So, why do we care about quantum teleportation?

You'd never run the quantum teleportation protocol on a single quantum computer, except as an academic exercise. Quantum teleportation is a way of getting from $$|\phi00\rangle$$ to $$|00\phi\rangle$$ when the first two qbits exist on one quantum computer, and the third exists on another. Any communication between the computers is classical, and all operations are local to the qbits on that computer (except for the initial process of entangling the qbits). Given these restrictions, quantum teleportation is the simplest protocol to get from $$|\phi00\rangle$$ to $$|00\phi\rangle$$. It enables us to transmit a qbit state with perfect fidelity using just two bits of classical information.

• Thanks . I see what you say, but I have two questions. One , you use the swap and I see somewhere people using a much more complicated circuit for implemention. Why is that? You say it can be accomplished so easily just with a swap gate. Second question, the first entangled qubit transferred to Alice , does require quantum communication, right? We can't get away with it, is it correct? – bilanush Feb 28 at 17:45
• You can use the swap gate when all the qbits are on a single quantum computer. If the qbits are on different quantum computers, you have to use quantum teleportation. The entangled qbit is indeed transferred from Alice to Bob, but this qbit is "disposable"; it doesn't matter if the transmission fails, because you can just generate & send another. Thus we can use existing optical transmission methods to send entangled qbits and eventually have one received by Bob. – ahelwer Feb 28 at 18:29
• No, I asked about an implemention without teleportation (ok, so it's on the same computer). But my question is why they used a much more complicated circuit. I am not taking about telportation now as I got your answer. I am asking a different question. You say it can be performed easily by the swap. Why do they use many gates to perform it? Concerning your second part, do you basically mean, that it doesn't have any information it's just the bell state qubit so it doesn't matter? – bilanush Feb 28 at 18:54
• Without looking at the circuit you're talking about I can't really say why they did it that way, but it's possibly because they wanted to do it in terms of basic operations rather than introducing the swap gate. You are correct re: the bell state qbit. This is the reason we use quantum teleportation rather than just sending the qbit we want to teleport over an optical network, because optical networks have very high error rates so we risk losing the qbit. – ahelwer Feb 28 at 19:18
• The swap gate is not usually implemented as a gate in itself, but rather a series of more primitive gates; that method is one such way. Basically given that Alice is sending a qbit to Bob (the entangled qbit), we might question why Alice doesn't just send the qbit she wants to teleport to Bob instead. The answer is that the method of transmission is very error-prone, so Alice risks losing the qbit she wants to teleport. Thus we send the disposable entangled qbits instead, until one goes through. From there the quantum teleportation protocol is very solid. – ahelwer Mar 1 at 18:17

measurements in every circuit can be postponed or never performed in a circuit while achieving the same functionality of the circuit

That's correct. But if the circuit involves two parties, this process will introduce quantum operations between the two parties. It will require a quantum communication channel, so that the qubits can be shuttled back and forth as needed in order to perform these operations.

Isn't a much better / simpler circuit is, not sending any classical / quantum information to Bob at all and nevertheless Bob can achieve the original vector state.

You're sending information either way. Teleportation allows you to send the quantum information at leisure ahead of time (while establishing entanglement), and then in the heat of the moment when time is sensitive you can use cheaper/faster/better classical communication instead of having to deal with expensive/brittle/slow quantum communication.