# Given 2 unknown qubits, which series of gates can put them in an equal superposition of $\vert 00 \rangle$ and $\vert 11 \rangle$?

You have 2 qubits, the states of which are unknown to you. They are either in the state $$\vert 0 \rangle$$ or $$\vert 1\rangle$$. Which sequences of gates can be applied so as to put the system in the state $$\frac{1}{\sqrt{2}}$$($$\vert 00 \rangle$$ + $$\vert 11\rangle$$)?

• I think more information is required, e.g. do you have some black-box procedure allowing you to prepare this unknown state as many times as you would like? If so, some variational minimization of some parameterized quantum circuit over a Hamiltonian whose ground state is the bell state would work. – Arthur-1 Apr 23 at 19:37

• If not, and you're only allowed to use unitary gates, it's impossible. Unitary gates are reversible, so you cannot map several different states to the same $$\frac{1}{\sqrt{2}}(|00 \rangle + |11\rangle)$$ state. If there was such a transformation, applying its adjoint to the $$\frac{1}{\sqrt{2}}(|00 \rangle + |11\rangle)$$ state would lead to only one state, not both of your states.
• Yes, that would work: set two auxiliary qubits to $(|00\rangle + |11\rangle)/\sqrt{2}$, then SWAP the auxiliary qubits with the input qubits. You are also correct that uncomputation is impossible without the knowledge of the input state (which reflects what Mariia said about reversibility of unitaries). – Adam Zalcman Apr 24 at 6:12
If you can use a reset, then applying this operation sets both qubits to $$|0\rangle$$. Then apply Hadamard gate on the first qubit and CNOT gate on both qubits. In the end you have your desired Bell state.