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How can we construct a single qubit gate $U = \mathrm{e}^{\frac{\mathrm{i}\pi}{4}}*\exp(−\frac{\mathrm{i}\pi}{4} Y)$ from $S$, $H$ (Hadamard), and Pauli gates?

I already know that final answer will be:

$$SHSHS3 = SHSHS^\dagger$$

I just don't know the process that led to this answer.

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There may be a simpler way of doing this, but this certainly works.

First find the matrix representation of $U$ by multiplying out the terms. Remember that $e^{i \theta Y} = \cos(\theta) I + i \sin(\theta)Y$. The final result (thank you Sympy) is

$$ U = \frac{1}{2} \begin{bmatrix} 1 + i & -1 - i \\ 1 + i & 1 + i \\ \end{bmatrix} $$

Now all single qubit transforms can be translated mechanically into the form $e^{i\gamma} R_Z(\phi) R_X(\theta)R_Z(\lambda)$. Qiskit provides a method to do that:

from qiskit.quantum_info import OneQubitEulerDecomposer

decomposer = OneQubitEulerDecomposer('ZXZ')
phi, theta, lam, gamma = decomposer.angles_and_phase(U)

You learn that $\theta = \phi = \frac{\pi}{2}$, $\lambda = \frac{-\pi}{2}$, and $\gamma = \frac{\pi}4$.

Now $R_Z(\phi) = R_Z(\pi/2)$ is just $S$, and $R_Z(\lambda) = R_Z(-\pi/2)$ is just $S^\dagger$. Any rotation around the X axis can be expressed as a rotation around the Z axis preceded and followed by an H. So $R_X(\theta) = H R_Z(\theta) H = H S H$

We have a leftover global phase $\gamma$ which we can ignore.

Putting the pieces together you get $S HSH SSS$.

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I go about this in quite a different way (and get a different result). You're trying to make $$ e^{-i\pi Y/4}=\frac{1}{\sqrt{2}}(I-iY). $$ Start by considering Hadamard: $$ H=\frac{X+Z}{\sqrt{2}}. $$ We can use Pauli relations to write $Z=-iXY$. Hence, $$ H=X\frac{I-iY}{\sqrt{2}}. $$ Thus, $e^{i\pi Y/4}=XH$. This is exactly what you want up to a global phase (and global phases don't matter).

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