How to create a Bell state with asymmetric amplitudes using single-qubit and CNOT gates?

Question

Is there a systematic way - in terms of a quantum circuit with single qubit and CNOT gates - to create a bell state with asymmetric amplitudes, e.g., $$ \alpha |00\rangle + \beta|11\rangle $$ where $\alpha, \beta$ are arbitrary complex values satisfying the condition of normalization?

Answer 1

Start with a two-qubit system: $$|00\rangle$$ An arbitrary single-qubit state can be achieved by applying the unitary $$ U = e^{i\alpha}R_z(\theta)R_y(\gamma)R_z(\delta) $$ to one of the qubits. Since we are starting in the $|0\rangle$ state it is sufficient to set $\alpha = \delta = 0$, yielding $$ U|0\rangle = \cos\frac{\gamma}{2}|0\rangle + e^{i\theta}\sin\frac{\gamma}{2}|1\rangle $$

We can now identify $cos\frac{\gamma}{2} \equiv \alpha$ and $e^{i\theta}\sin\frac{\gamma}{2} \equiv \beta$, yielding

$$U|00\rangle = (\alpha|0\rangle+\beta|1\rangle)|0\rangle=\alpha|00\rangle+\beta|10\rangle$$

Perform a CNOT gate on the second qubit controlled by the first one: $$\alpha|00\rangle+\beta|11\rangle$$

An optional $X$ gate can bring the state to "the other" bell state, $$ \alpha|01\rangle+\beta|10\rangle $$

All in all, you simply have to replace the $H$ gate to create the amplitude you want and then proceed as usual, leaving the $X$ gate as an option in the end.