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

Question

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:

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

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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.

Answer 1

The circuit is a Hadamard + a Controlled - Hadamard gate.

Note that $ S $ gate is $ P $ gate in your notation.

Answer 2

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.