## How to prepare a quantum circuit for $\frac{1}{\sqrt{3}}(|00\rangle+|01\rangle+|10\rangle)$ starting from $|00\rangle$

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How do I prepare a quantum circuit for $$\frac{1}{\sqrt{3}}(|00\rangle+|01\rangle+|10\rangle)$$ state starting from the $$|00\rangle$$ state?

I have no clue how to do it. I tried with controlled Hadamard gate but of no use here. Can someone help? Furthermore, can it be constructed with only one CNOT and any number of single-qubit gates?

– Danylo Y – 2020-07-07T07:50:41.403

Step 1: Find the Schmidt decomposition $$|\psi\rangle=\sum_i\alpha_i|u_i\rangle|v_i\rangle$$. I won't do this completely here, but $$\alpha_0=\sqrt{\frac{3+\sqrt{5}}{6}},\qquad |u_0\rangle=|v_0\rangle=\frac{1}{\sqrt{10-2\sqrt{5}}}(2|0\rangle+(\sqrt{5}-1)|1\rangle)$$ should be enough to confirm you're going in the right direction.
Step 2: Express this as $$(U_A\otimes U_B)(\alpha_0|00\rangle+\alpha_1|11\rangle)$$.
Step 3: Write down the circuit. First apply $$U|0\rangle\rightarrow \alpha_0|0\rangle+\alpha_1|1\rangle$$ on the first qubit. Then apply $$CNOT$$ gate with first qubit as control and second qubit as target. Then apply $$U_A\otimes U_B$$.