# Why when we have GHZ, then we cannot send a qubit using teleportation technique? [closed]

Suppose the following example

Alice has a qubit $|A\rangle$ and Bob has a qubit $|B\rangle$, then by teleportation technique we can send an unknown qubit state (i.e. $|\psi\rangle = \alpha |0\rangle + \beta |1\rangle )$ Alice to Bob.

Suppose another example

Suppose that there are three parity: Alice, Bob, and Dina. All these guys have an entangled qubits that are entangled. That is, Alice has a qubit $|A\rangle$, Bob has a qubit $|B\rangle$ and Dina has a qubit $|C\rangle$. They are entangled as $\frac{1}{\sqrt{2}}(|000\rangle + |111\rangle)$. (It is understood that the general case of entangled qubits is known as GHZ-state). Now to the question.

The Question: Can we send an unknown qubit from Alice to Bob and Dina where they (Bob and Dina) can retrieve the unknown qubit using the teleportation technique?

I said: Yes. Professor says: No, I said: Why? Professor: Because no-cloning theorem! I said: Professor, the three qubits are entangled (I want to emphasize that the entangled qubit are different from those that are non-entangled qubits) Professor: We cannot do this in quantum mechanics, it is impossible. (The discussion is over)

My opinion

I don't really understand, so I decided to put this question here to see if there is any further information about this problem. Therefore, the question for this side is

Could someone tell me why GZA-state doesn't work as EPR-state?

That is, if we have an entangled two-qubits in different places (EPR-state), then we are able to send an unknown qubit from differen places. But if we have three entangled qubits or more that are entangled (GHZ-state), then we cannot send an unknown qubit; because of no-cloning theorem.

my argument

I just don't see any replication of qubit (since no-cloning theorem prevents us to construct a copy of a qubit) instead the only thing that is replicated is the 2-classical bits that is sent from Alice to Bob and Dian, so where is exactly the no-cloning theorem here?

Thank you.

See the figure for three qubits that are entangled in different places. ## closed as off-topic by Norbert Schuch, ACuriousMind♦, Gert, Daniel Griscom, JamalSFeb 12 '16 at 16:27

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• Have you tried working this out yourself? I mean have you tried to perform the teleportation protocol on Alice and Bob's pairs and see what happens with Dina's part? – Martin Feb 8 '16 at 12:13
• @Martin Yes, I tried, it works exactly right and you cannot retrieve the qubit. But I'm thinking that suppose that we have 3 qubits that are entangled, then what is different between 3 entangled qubits and 3 non-entangled qubits? Is the power of 3 entangled qubits are the same as 3 entangled qubits? For me, It seems by non-cloning theorem the power is the same. Hence, this is exactly why I feel there must be something here that I don't know either in the qubit or the entangled-qubit. – YOUSEFY Feb 13 '16 at 18:53
• Three entangled qubits share some correlations between them. It is a good (yet different) question to ask what you can do with this correlation - mind, there is a lot between "nothing" (separable product state) and "teleportation" (maximally entangled bipartite system). Multipartite entanglement is difficult, yet certain things that you can do are known. For example, here are multi-party games where you can have a better strategy using entangled particles: arxiv.org/abs/1510.09210 or dare.uva.nl/document/2/97125 (these are random hits) – Martin Feb 13 '16 at 20:58

A three-party GHZ state can't be used for teleportation because extra copies of a qubit's value act like de-facto measurements. Charlie's copy of the entanglement's 'secret value' prevents some crucial interference from happening when Alice and Bob interact.

You can still transfer information from Alice to Bob using a GHZ state and classical communication... it's just that you need Charlie to help out. For example, he could use LOCC erasure to destroy his share of the entanglement.

Alternatively, Alice can tweak the teleportation procedure to use a GHZ state and classical communication to send a qubit to both Bob and Charlie. But they'll end up with entangled copies of the qubit, instead of independent copies. (If they ended up with independent copies, we'd have violated the no-cloning theorem.) Here's the circuit that does that: I suggest working out, on paper, what actually happens when you try to teleport with a GHZ state as if it was an EPR pair (i.e. if you drop the last CNOT onto the bottom qubit in the above diagram). It's analogous to what happens when you put a detector on one of the arms of a Mach-Zehnder interferometer: there's extra state space, allowing the various paths to end up in different final states instead of the same state, meaning no interference occurs, and thus no "magic".

• since each qbit exists once at one moment, how one can check that the copy is right ? – user46925 Feb 10 '16 at 0:37
• @igael I don't understand your question. What do you mean "the copy is right"? Are you asking how do we verify that Bob and Charlie ended up with a pair of qubits in the state $\alpha \left|00\right\rangle + \beta \left|11\right\rangle$ if Alice started with the state $\alpha \left|0\right\rangle + \beta \left|1\right\rangle$ and GHZ-teleported it to both of them? Just have them get together and repeat the process a bunch of times and do joint measurements and check predictions based on the system being in that state. – Craig Gidney Feb 10 '16 at 1:54
• @Strilanc Thank you for your answer. After working out in a paper, I found the last state is: $$\frac12(|00\rangle_{\psi,A} \times (\alpha |00\rangle_{B,D}+ \beta|00\rangle_{B,D}) + |01\rangle_{\psi,A} \times (\alpha |11\rangle_{B,D}+ \beta |11\rangle_{B,D}) + |10\rangle_{\psi,A} \times (\alpha |00\rangle_{B,D}-\beta|00\rangle_{B,D}) + |11\rangle_{\psi,A} \times (\alpha |11\rangle_{B,D}-\beta|11\rangle_{B,D}))$$. Now, can you tell me what does it mean to have this state $(\alpha |00\rangle_{B,D}-\beta|00\rangle_{B,D})$ physically and mathematically? For example, Is $\alpha$ equal $\beta$? – YOUSEFY Feb 13 '16 at 18:45
• @YOUSEFY It's just equal to $(\alpha - \beta) \left| 00 \right\rangle$. But I think you might have made a mistake if you ended up with a state that interferes the $\alpha$ and $\beta$ amplitudes. You might have accidentally duplicated the controlled-Z that erases Alice's copy, assuming it was associated with Bob and Charlie instead of with Alice? Here is the GHZ teleport-entangled-copy-to-two-receivers circuit: imgur.com/sGB7Qpc – Craig Gidney Feb 13 '16 at 19:21
• Maybe a intuitive but not strict picture can say something. If ER=EPR, then there is a strange ER bridge among A, B, D. Only when we combine BD, then A has a bridge connecting to BD, otherwise A is not connected with B and D (Just as the entanglement of GHZ state, AB are not entangled, but A is entangled with BD). So the teleportation is only valid from A to BD(1 copy), but not A to B and D(2 copies) since there is no such a geometrical structure allowing this. – XXDD Feb 14 '16 at 3:53

No-cloning theorem says that you cannot "clone" the state. As in there is no way to make perfect copies of a quantum state. In the simple teleportation case with two parties the Alice's state is transferred to Bob by using the EPR pair. There is no copying. But if Alice were to try and send her state to two people(Bob and Dina) then that would mean that two copies of the state would exist after the teleportation is done. This is a violation of no-cloning theorem.