2
$\begingroup$

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?

enter image description here

$\endgroup$
  • 2
    $\begingroup$ What have you tried? How far have you got? How are answers to previous, very similar questions not helping? $\endgroup$ – DaftWullie Jan 23 at 14:01
  • 1
    $\begingroup$ 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 at 14:15
  • $\begingroup$ This question was mentioned to me in the exam as I wrote above @MartinVesely $\endgroup$ – Ba. Taj Jan 24 at 16:04
5
$\begingroup$

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

|improve this answer|||||
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.