Quantum gate: Phase shift I dont undestand how to apply a phase shift gate to a qubit. By example how to map $|\psi_0\rangle = \cos (30^\circ) |0\rangle + \sin (30^\circ) |1\rangle$ to $|\psi_1\rangle = \cos(-15^\circ) |0\rangle + \sin(-15^\circ) |1\rangle$
 A: A phase gate will not map between the two vectors you give.  A phase gate changes the phase of the $\left|1\right>$ component, which is not what you want since for your example all components are real.
Your two vectors lie in the X-Z plane of the Bloch sphere.  To map from your first vector to your second vector, you need to rotate about the Y axis.  The following unitary does the job, with $\theta=(-15)-30=-45$.
$$
\left[
\begin{array}[rr]
\textrm{cos}(\theta)
&
-\textrm{sin}(\theta)
\\
\textrm{sin}(\theta)
&
\textrm{cos}(\theta)
\end{array}
\right]
$$
A: So, you have two vectors. Let $|0\rangle = \begin{pmatrix}
        1 \\
        0 \\
        \end{pmatrix}$ and $|1\rangle = \begin{pmatrix}
        0 \\
        1 \\
        \end{pmatrix}$. So, your initial vector is $\begin{pmatrix}
        cos \frac{\pi}{6} \\
        sin \frac{\pi}{6} \\
        \end{pmatrix}$ and final vector is $\begin{pmatrix}
        cos \frac{-\pi}{12} \\
        sin \frac{-\pi}{12} \\
        \end{pmatrix}$. The phase difference between these two vectors, $\theta$ is $cos^{-1} \left[ \frac{\begin{pmatrix}
        cos \frac{\pi}{6} \\
        sin \frac{\pi}{6} \\
        \end{pmatrix} . \begin{pmatrix}
        cos \frac{-\pi}{12} \\
        sin \frac{-\pi}{12} \\
        \end{pmatrix}}{|\begin{pmatrix}
        cos \frac{\pi}{6} \\
        sin \frac{\pi}{6} \\
        \end{pmatrix}| | \begin{pmatrix}
        cos \frac{-\pi}{12} \\
        sin \frac{-\pi}{12} \\
        \end{pmatrix}|} \right]$. Evaluate $\theta$ and plug in the value in \begin{pmatrix}
        1 & 0 \\
        0 & e^{i\theta} \\
        \end{pmatrix}. The resulting matrix will be your gate.
