Sender and receiver use the teleportation protocol, where the sender teleports a quantum state $\left| \varphi \right>=\alpha\left| 0 \right> + \beta \left|1\right>$ to the receiver.

I want to implement this protocol, and then find the output $\left| ABC \right>$ of the quantum circuit when the measurement of the teleportation protocol in the left side is $\left| 11\right>$.

In other words: when we measure $|11\rangle$, how to show that the state $|\varphi\rangle$ was really teleported?

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    $\begingroup$ What have you tried? How far have you got? How are answers to previous, very similar questions not helping? $\endgroup$ – DaftWullie Jan 23 '20 at 14:01
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    $\begingroup$ I am a little bit lost in your question. Could you please add results you have and how they differ from expectations? $\endgroup$ – Martin Vesely Jan 23 '20 at 14:15
  • $\begingroup$ This question was mentioned to me in the exam as I wrote above @MartinVesely $\endgroup$ – Ba. Taj Jan 24 '20 at 16:04

Since your circuit is teleportation, $|C\rangle =|\varphi\rangle$ and since you measured $|11\rangle$ on $|AB\rangle$ the answer is $|ABC\rangle = |11\rangle|\varphi\rangle$.

Now, let look why this is true.

Firstly Hadamard and CNOT gate on second and third qubit prepares entangled Bell state $|\beta_{00}\rangle = \frac{1}{\sqrt{2}}(|00\rangle + |11\rangle)$.

Now the circuit is in state

$$ |\varphi\rangle|\beta_{00}\rangle = \frac{1}{\sqrt{2}}(\alpha|0\rangle+\beta|1\rangle)(|00\rangle + |11\rangle) = \frac{1}{\sqrt{2}}[\alpha|0\rangle(|00\rangle + |11\rangle) + \beta|1\rangle(|00\rangle + |11\rangle)] $$

Then you apply CNOT controlled by first qubit and targeting second qubit. This will negate second qubit in case the first qubit is in state $|1\rangle$. This means that only part $\beta|1\rangle(|00\rangle + |11\rangle)$ is influenced.

Now, state of the circuit is changed to

$$ \frac{1}{\sqrt{2}}[\alpha|0\rangle(|00\rangle + |11\rangle) + \beta|1\rangle(|10\rangle + |01\rangle)] $$

Application of Hadamard gate on the first qubit change the state further to

$$ \frac{1}{2}[\alpha(|0\rangle + |1\rangle)(|00\rangle + |11\rangle) + \beta(|0\rangle - |1\rangle)(|10\rangle + |01\rangle)] $$

because $H|0\rangle = \frac{1}{\sqrt{2}}(|0\rangle + |1\rangle)$ and $H|1\rangle = \frac{1}{\sqrt{2}}(|0\rangle - |1\rangle)$.

Since you will measure first and second qubit, it is convenient to rearange the state to separate first two qubits. So, you can rewrite the state as

$$ \frac{1}{2} \big( |00\rangle(\alpha|0\rangle + \beta|1\rangle) + |01\rangle (\alpha|1\rangle + \beta|0\rangle) + |10\rangle (\alpha|0\rangle - \beta|1\rangle) + |11\rangle (\alpha|1\rangle - \beta|0\rangle) \big) $$

In your case you measured $|11\rangle$ on first and second qubit. This means that the third qubit is in state

$$ (\alpha|1\rangle - \beta|0\rangle) $$

Since both first and second qubits are in state $|1\rangle$ both CNOT gates after measurement will be activated. The first one change the state of third qubit to

$$ (\alpha|0\rangle - \beta|1\rangle) $$

Next two Hadamards together with CNOT implements controlled $Z$ gate which change a phase to oposite in case input qubit is in state $|1\rangle$. This leads to final state of third qubit

$$ (\alpha|0\rangle + \beta|1\rangle) $$

Hence, you can see that state from first qubit was teleported to third qubit.

Note: based on Nielsen and Chuang, pg. 27 and expanded


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