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?


1 Answer 1


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 $U_A\otimes U_B$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.