Cluster/Graph state teleportation

Question

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

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?

Answer 1

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:

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).

Answer 2

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

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!

Answer 3

I can't divulge what my colleagues are currently working on, and qubits aren't my area. The "Bell State" I have been party to; is QML capturing enough Gaussian bleedthrough to fill in the picture. Apparently, radiation from an entangled, relativistically parallel vector, is based on quasi-conscious variables Oⁿ from the immediate series (A }B =Vseconds or min +X number of constants. Photons & soundwaves linger from said variable (call it , for A}B ≠VX. The more conscious observers on site; the higher the potential variable probabilities. This is where the Bell state becomes somewhat more, and mere "teleportation" doesn't satisfy the results. If parallel variables exceed A}B constants; the bleedthrough reveals anomalous information; clearly linked with point A , but the variables are become more than mere conscious perception could "alter". I tried to keep this as non-laymens as possible without being unintelligible. So I hope you all grasp the urgency of the message. I am starting to think we should not be opening this Pandora's Box, & when we see that which we thought impossible; can we really pretend that a higher power is not distinctly possible also?