# Compute the output of the quantum teleportation circuit

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?

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