0
$\begingroup$

I came across somewhere about the circuit diagram that depicts the teleportation of a 4-qubit cluster state. Here it isenter image description here

Let me tell what i understood.

  1. The qubits on the first two wires give the state $\dfrac{1}{\sqrt{2}}(|0\rangle|0\rangle+|1\rangle|1\rangle)$, which is a bell stae. This state i guess is teleported with the help of the 4 qubits cluster state, anyways ( i am not sure).

  2. The qubits on wires 3,4,5,6 after Hadamards and controlled phase give the cluster state $$|a\rangle=\frac{1}{2}(|0000\rangle+|0011\rangle+|1100\rangle-|1111\rangle)$$

Now after this i am not able to understand the circuit, apart from the measurement. For example why is Hadamard on the 3rd wire?

Is there any specific way to make this circuit, any sequential procedure? Can somebody explain?

$\endgroup$
2
  • $\begingroup$ It would help to know where you came across this so that we can understand what the authors were actually trying to achieve. $\endgroup$
    – DaftWullie
    Jun 23 '20 at 7:11
  • $\begingroup$ The authors just wanted to do some simulation on an IBM simulator. link.springer.com/article/10.1007/s11128-020-2586-x $\endgroup$
    – Upstart
    Jun 23 '20 at 7:35
2
$\begingroup$

The calculation of the resulting state of the described circuit:

After applying the same circuit identities described in this answer (and here) to the connected question we will obtain a "simplified" circuit:

enter image description here

Here are the links to the initial, first intermediate, second intermediate and final circuits presented in the Quirk. Whenever it was possible (by commuting relations of the gates) and convenient the places of the gates were changed. For example adjacent CNOTs on different pairs of qubits can change their positions. In the first part discussed here combined with the Bell state we will have the following state:

$$| \psi_1 \rangle = \frac{1}{2\sqrt{2}} \big(|00\rangle + |11\rangle\big)\big(| 0000 \rangle + | 0011 \rangle + | 1100 \rangle - | 1111 \rangle \big)$$

After the following "cascade-like" sequence consisted of $4$ CNOTs:

$$| \psi_2 \rangle = \frac{1}{2\sqrt{2}} \Big(| 000000 \rangle + | 000011 \rangle + | 001111 \rangle - | 001100 \rangle + \\ + | 111111 \rangle + | 111100 \rangle + | 110000 \rangle - | 110011 \rangle \Big)$$

After the final two CNOTs:

$$| \psi_3 \rangle = \frac{1}{2\sqrt{2}} \Big(| 000000 \rangle + | 110011 \rangle + | 111111 \rangle - | 001100 \rangle + \\ + | 001111 \rangle + | 111100 \rangle + | 110000 \rangle - | 000011 \rangle \Big)$$

After the final two Hadamard gates:

$$| \psi_4 \rangle = \frac{1}{2\sqrt{2}} \Big(| ++0000 \rangle + | --0011 \rangle + | --1111 \rangle - | ++1100 \rangle + \\ + | ++1111 \rangle + | --1100 \rangle + | --0000 \rangle - | ++0011 \rangle \Big)$$

We can simplify this state:

$$| \psi_4 \rangle = \frac{1}{2\sqrt{2}} \Big(| 000000 \rangle + | 110000 \rangle - | 010011 \rangle - | 100011 \rangle + \\ + | 001111 \rangle + | 111111 \rangle - | 011100 \rangle - | 101100 \rangle \Big)$$

Regrouping the terms:

$$| \psi_4 \rangle = \frac{1}{2\sqrt{2}} \Big(\big(| 0000 \rangle + | 1100 \rangle - | 0111 \rangle - | 1011 \rangle \big) |00\rangle + \\ + \big(| 0011 \rangle + | 1111 \rangle - | 0100 \rangle - | 1000\rangle \big) |11\rangle \Big)$$

This is the final state that should be measured (only the last two qubits). This final result can be checked from Quirk's output that can be found from the initial version of the circuit (the order of the qubits in the ket notation is reversed there).

$\endgroup$
6
  • $\begingroup$ Okay i understood the whole circuit wince you simplified it very much. But where does the telepoetation comes in? $\endgroup$
    – Upstart
    Jun 23 '20 at 22:26
  • $\begingroup$ @Upstart, if my calculations are right then I'm not sure if this circuit does any teleportation. $\endgroup$ Jun 23 '20 at 22:35
  • $\begingroup$ Can you have look at the article that i gave the link for..it has something to do with teleprtation using this circuit $\endgroup$
    – Upstart
    Jun 23 '20 at 22:40
  • $\begingroup$ @Upstart, I have looked at the paper (not carefully). If I understand right they tried to teleport Bell state from the first two qubits to the last two qubits, right? It didn't happen in the calculations mentioned in the answer. $\endgroup$ Jun 23 '20 at 22:47
  • $\begingroup$ So are there any steps that need to be added for the qubits to get trasferred? $\endgroup$
    – Upstart
    Jun 23 '20 at 22:55
1
$\begingroup$

I think it is worth trying to understand the circuit that the authors really want to implement: enter image description here

Here they produce the Bell state they want to teleport onto the final two qubits, and cluster state (personally, I wouldn't call it a cluster state after they've added the two extra hadamards). Then they do two single-qubit teleportation protocols to make the state arrive where they want them to. Hopefully my circuit has made this structure fairly clear.

So, now you need to understand how they've converted this into the circuit they've implemented. Firstly, instead of operations controlled off classical measurement results, they've just implemented quantumly controlled $X$ and $Z$ gates, and dropped the measurements on the four qubits. So, the output will still be Bell state on the last two qubits even without measuring the other four. Then, they add measurements on the last two qubits in an attempt to verify what they have produced is what they wanted. So, the important output is the output of the last two qubits, not the first four!

$\endgroup$
4
  • $\begingroup$ Actually, sir I understood the earlier answer more clearly. But how does it occur to someone, that we can switch the gates while just having a look at the circuit, because when i did this from the original paper, it got all messed up. Is there some simpler way to construct these circuits. I mean what if I wanted to do GHZ state teleportaion? First of all can we do this? $\endgroup$
    – Upstart
    Jun 23 '20 at 19:38
  • $\begingroup$ Okay, but how to make sure that you transfer the first two qubits onto the last two qubits? Is there any method that we need to follow? $\endgroup$
    – Upstart
    Jun 28 '20 at 11:04
  • $\begingroup$ I don’t know because I’ve not been back to read the paper that the one you cite is based upon. $\endgroup$
    – DaftWullie
    Jun 28 '20 at 12:05
  • $\begingroup$ Okay..but thanks any way. I tried but to no avail. $\endgroup$
    – Upstart
    Jun 28 '20 at 12:08

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.