Can teleportation violate the no-cloning theorem with 3 entangled qubits?

Question

I saw the How to prove teleportation does not violate no-cloning theorem? question.

But I want to address this issue from a different angle. What if instead of entangling 2 qubits, we would have entangled 3 qubits in a "same entangle without phase". Then we send the 2nd entangled qubit to Bob, and the 3rd entangled qubit to Carol.

Alice could then send the classical bits (the measurements results) to both Bob and Carol. Then both Bob and Carol apply Not gate and/or Z gate. Hence now the same qubit is located both at Bob's and at Carol's.

Am I wrong?

Answer 1

There have been a number of questions on this stack exchange (including one by me), all of which indicate confusion about three entangled qubits. It turns out that three entangled qubits behave very differently from two entangled qubits. The very existence of the third qubit, even if it is never looked at or modified, is enough to affect the results.

Quirk has a good example of teleportation. You can see that the sent qubit (first wire) and the received qubit (fifth wire) are in perfect sync.

Now cause the sixth qubit to also be entangled. Circuit. Even though we never look at the sixth wire, the teleportation is broken.

Answer 2

I imagine that you're thinking of a state that looks like $$ |\Psi\rangle=\frac{1}{\sqrt{2}}(|000\rangle+|111\rangle). $$ If you were to try and teleport a basis state such as $|0\rangle$ or $|1\rangle$ through this state, Bob and Charlie would correctly receive the states. (The quickest way to see this is that Alice teleporting a state $|\psi\rangle$ is equivalent to projecting the first qubit of $|\Psi\rangle$ in the state $|\psi^\star\rangle$ (the complex conjugate of the state), up to normalisation. So, $|0\rangle\mapsto|00\rangle$.

However, now imagine Alice trying to teleport a state such as $|+\rangle=(|0\rangle+|1\rangle)/\sqrt{2}$. It will arrive as $$ \langle +|\otimes I\otimes I\cdot |\Psi\rangle\sim \frac{1}{\sqrt{2}}(|00\rangle+|11\rangle). $$ This is certainly not the desired state $|++\rangle$. So, it does not achieve perfect cloning.