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.