Further entangling two ebits to obtain $|0000\rangle + |1111\rangle$ state

Question

I have two ebits $e1$ and $e2$ and my 4-qubit system is in the state:

$$(|00\rangle + |11\rangle) (|00\rangle + |11\rangle)$$

Ignoring the normalizing constants for now. As mentioned, I need to obtain $|0000\rangle + |1111\rangle$.

I have the additional restriction that I can only use single gates on all the qubits, and controlled gates (CNOT, etc) only on the second and third qubit (otherwise I could just use inverse operations to disentangle all qubits, and proceed to get what I want in same way we create the GHZ state).

My idea to do this was to entangle second and third qubits, using CNOT and H gates, but it doesn't work because the 2nd and third qubits are in the state $|00\rangle + |01\rangle + |10\rangle + |11\rangle$, and I only know how to entangle if we were starting in $|00\rangle$ (or even if we're in $|0\rangle(|0\rangle + |1\rangle)$, which is what we get after H gate).

Is this possible to do?

Answer 1

If you can only use unitary control gates on the second and third qubits, you cannot change the density matrix on the first and last qubits except by single qubit operations. This density matrix is the completely mixed state: $$\frac{1}{4}\left(\begin{array}{cc}1&0\\0&1\end{array}\right)\otimes \left(\begin{array}{cc}1&0\\0&1\end{array}\right),$$ and unitary single qubit operations don't change it. In the 4-qubit state you want, the first and fourth qubits are not in the completely mixed state.

Are you allowed to use measurements on the second and third qubits?

Answer 2

If you allow for measurements and communication of measurement outcomes, and joint operations on qubits 2 and 3, parties $2$ and $3$ can prepare the desired state locally and teleport it to 1 and 4.

Answer 3

I think this is the quantum circuit that you want, first produce two pairs of Bell states, then undo the second pair coherently, and finally apply two CNOT two flip the last two qubits.