# Prepare state $|00\rangle+|1+\rangle$ using Clifford gates and the T-gate

I am looking for a quantum circuit which maps state $$|00\rangle$$ to $$|\psi\rangle=\frac{1}{\sqrt{2}} |00\rangle+\frac{1}{\sqrt{2}}|1+\rangle$$.

The circuit should only apply quantum gates from the Clifford group (specifically, $$CNOT$$, $$H$$, $$P$$) and the $$T$$ gate:

$$CNOT = \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \end{bmatrix}, \quad H = \frac{1}{\sqrt{2}}\begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix}, \quad P = \begin{bmatrix} 1 & 0 \\ 0 & i \end{bmatrix}, \quad T = \begin{bmatrix} 1 & 0 \\ 0 & e^{i \pi / 4} \end{bmatrix}$$

### My Thoughts

Because these gates are universal for quantum computation (as stated here), I know that a circuit which approximates $$|\psi\rangle$$ must exist. I am hoping that I can produce $$|\psi\rangle$$ exactly, but I was not able to find the corresponding circuit.

I already figured out the the circuit needs to apply $$T$$ at least one, as $$|\psi\rangle$$ has no stabilizers from the Pauli group (determined by brute-force), and any state produced by a Clifford circuit would have stabilizers from the Pauli group.

• Do you allow any gate from the Clifford group or only these specific generators $(CNOT, H, P)$?
– JSdJ
Dec 5, 2020 at 13:48
• @JSDJ I clarified now that I only want to allow these specific generators. Dec 5, 2020 at 17:39 Note that $$S$$ gate is $$P$$ gate in your notation. This circuit is also a proof that you need two T gates to produce the state you want. Because it makes clear that, given the state, you can apply operations that undo the $$\sqrt{X} = HPH$$ and the CNOT leaving behind two T states.