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

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?

• Let me note that the 2nd and 3rd qubits are not in the state |00⟩+|01⟩+|10⟩+|11⟩ , If you trace out the first and fourth qubits, they're in the completely mixed state. If you don't trace them out, they're part of an entangled system and have no definite state of their own. I've added an answer which may contain a hint for your homework (if such homework exists). Feb 18, 2020 at 13:24
• Have you considered teleportation? Feb 18, 2020 at 23:01

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.
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.