I know that the $T$ gate is equivalent to a $\frac{\pi}{4}$ rotation around the $Z$-axis, but what about $X$ and $Y$?
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$\begingroup$ Not entirely sure if this is what you're asking, but you can't write the $T$ gate only as a product of $X$ and $Y$ gates. For a simple proof, notice that $\det(X)=\det(Y) = -1$, but $\det(T) = e^{\pi i/4}$. The determinant is multiplicative, but we cannot get $e^{\pi i/4}$ by multiplying copies of $(-1)$. $\endgroup$– Franklin Pezzuti DyerMar 13 at 21:33
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1$\begingroup$ @FranklinPezzutiDyer the determinant is not everything. The restriction is stricter than that. Even if you want to get $T$ up to a global phase it is still not possible from $X$ and $Y$ only (they generate the Pauli group which does not include anything close to $T$). $\endgroup$– MauricioMar 13 at 22:29
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2 Answers
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A $T$ gate is a rotation of $\pi/4$ around the $Z$-axis of the Bloch sphere. Meaning that $T$ is equivalent to $R_Z(\pi/4)$ up to a global phase (where $R_Z(\theta)=\exp(-i\theta Z/2 )$ is the rotation operator around the $Z$ axis).
You cannot get a $T$ gate from $X$ and $Y$ only. Multiplications of $X,Y,Z$ form a closed group (Pauli group), that does not include $T$.
Nevertheless you can write a rotation in $Z$-axis as multiples of rotations of around the axes $X$ and $Y$ axes: $$R_X(\pi/2)R_Y(\pi/4)R_X(-\pi/2)=R_Z(\pi/4).$$
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As already mentioned by Mauricio in his answer, you can express the $T$ gate as a combination of rotations around the $X$ and the $Y$ axes (up to a global phase). In particular, this is how you can see this transformation by using the Qiskit transpiler:
from qiskit import QuantumCircuit, transpile
qc = QuantumCircuit(1)
qc.t(0)
qc = transpile(qc, basis_gates=['rx', 'ry'])
qc.draw('mpl')
answered Mar 14 at 8:46