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For a given teleportation process as depicted in the figure, how one can say that teleporting the qubit state $|q\rangle$ has not cloned at the end of Bob's measurement?

enter image description here

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Firstly, carefully read through the formal presentation of the protocol as described on Wikipedia. Secondly, there's nothing to prove as such here. It is evident from the teleportation protocol itself.

Yes, at the conclusion of the teleportation protocol, Bob's qubit certainly will take on the same state as Alice's qubit's initial state i.e. $|q\rangle$. However, Alice's qubit will now exist as an inseparable part of an entangled Bell state. You haven't really been able to create a "copy" of Alice's qubit, in the sense that Alice's qubit's initial state is "destroyed" in the process and is no longer $|q\rangle$.

In (crude) analogical terms, in the quantum teleportation protocol you "cut and paste" the state $|q\rangle$ rather than "copying and pasting". So there's no violation of the No cloning theorem!

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Cloning means the generation of $|q\rangle|q\rangle$ from $|q\rangle|0\rangle$. This is not what happens in teleportation. Teleportation is kind of a swap operation, i.e. something like $|q\rangle|0\rangle \to |0\rangle|q\rangle$.

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I.e., if we have the entire initial state is written as follows

$$| q \rangle \otimes | \beta_{00} \rangle =(\alpha | 0\rangle+\beta | 1 \rangle ) \otimes \frac{1}{\sqrt{2}}( |00\rangle+| 11 \rangle)$$ $$= \frac{1}{\sqrt{2}}(\alpha | 000\rangle+\alpha | 011 \rangle+\beta | 100 \rangle+\beta | 111 \rangle ),$$

then, in step three, we obtain the state (the system state collapses to one of four possible result after measurement),

$$|\psi \rangle \equiv\{ |00\rangle \frac{\alpha | 0 \rangle+\beta | 1\rangle}{2}+| 01 \rangle \frac{\alpha | 1 \rangle+\beta | 0\rangle}{2}+| 10 \rangle \frac{\alpha | 0 \rangle-\beta | 1\rangle}{2}+| 11 \rangle \frac{\alpha | 1 \rangle-\beta | 0\rangle}{2} \},$$

where the right two-qubit is for Alice and the superposed qubits with corresponding probabilities $\alpha$ and $\beta$ are for Bob

hence we can no longer write

$$| q \rangle_A \otimes | \text{something} \rangle_B,$$

namely, Alice's $| q \rangle_A$ dissapeared.

enter image description here

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  • $\begingroup$ What do you mean by after the measurement, we obtain the state |ψ⟩≡{|00⟩α|0⟩+β|1⟩2+|01⟩α|1⟩+β|0⟩2+|10⟩α|0⟩−β|1⟩2+|11⟩α|1⟩−β|0⟩2}? Is this correct? You don't have it written in the Bell basis vectors. And your measurement is supposed to collapse your vector to either one of the four Bell basis terms. So why do you say it's obtained after measuring? $\endgroup$ – bilanush Mar 3 at 8:04
  • $\begingroup$ After "measurement" you do not get the state you wrote down: $$|\psi \rangle = \{ |00\rangle \frac{\alpha | 0 \rangle+\beta | 1\rangle}{2}+| 01 \rangle \frac{\alpha | 1 \rangle+\beta | 0\rangle}{2}+| 10 \rangle \frac{\alpha | 0 \rangle-\beta | 1\rangle}{2}+| 11 \rangle \frac{\alpha | 1 \rangle-\beta | 0\rangle}{2} \}.$$ That is incorrect. Please re-read the teleportation protocol from the Wikipedia page I linked in my answer. $\endgroup$ – Sanchayan Dutta Mar 3 at 8:29
  • $\begingroup$ I re-edited it! $\endgroup$ – Student404Mus Mar 3 at 9:07
  • $\begingroup$ @bilanush Is the idea that I cannot re-write the state as a product of the initial state implies that the latter had broken? $\endgroup$ – Student404Mus Mar 3 at 9:11
  • $\begingroup$ Yes. I believe so. But what you wrote even after editing looks wrong. You are supposed to get in the left side bell basis vectors . $\endgroup$ – bilanush Mar 3 at 9:56

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