# How to complete this teleportation circuit? How to create a copy of $|\psi〉$?

This is a quantum circuit. M represents the act of making a measurement on the first two qubits. The circuit is supposed to transfer the state $$|\psi\rangle = a |0\rangle + b |1\rangle$$ ($$a, b \in \Bbb C$$) from the first qubit to the third qubit, involving a classical operation to select a quantum gate. Depending on the measurement outcome of the first two qubits, what quantum gate should be applied in place of ‘...’, such that the transfer is successful? The measurement results of M and the quantum gate to be applied in each case.

For this circuit a purely classical and a purely quantum circuit is possible or not?

• You're asking two questions here. The first is a basic question about teleportation. I suggest reading the wikipedia article. The second part has essentially been asked here before. – DaftWullie Mar 19 '19 at 9:23

A trick you can use, in order to figure out the necessary fixups, is to see what happens when you condition the output display on the measurement results. Like this:

See how the conditioned states are all pointing in different directions? Your goal is to fix that by, for each of the measurement cases, introducing controlled operations that make them all match. For example, if we decide that we want everything to match the 00 output, then the 11 output is fixed by introducing a Y fixup:

See how the first and last controlled displays now show an indicator pointing in the same direction? The CCY made them match. Find operations that do the same thing for the other two, and you have a valid teleportation circuit. Once you have that, try to simplify the fixups.

Some rules of thumb for finding these fixups, which continue to work in more complicated cases:

• Most fixups are Pauli operations (X, Y, Z). Try those first.
• It is common for fixups to have one classical control instead of many classical controls. Do the cases with a single ON measurement result first, and see if controlling only on that one ON bit instead of the entire measurement result just happens to also work for all the other cases.