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

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?

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.