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

2

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?

Anjishnu Adhikari

Posted 2020-07-07T06:35:56.723

Reputation: 21

Answers

6

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 $U_A\otimes U_B$.

DaftWullie

Posted 2020-07-07T06:35:56.723

Reputation: 35 722