Understanding the Pauli-Y gate in the Bloch sphere I'm having some trouble understanding the Bloch representation of qubits in some cases. 
The canonical representation $\cos(\psi/2) |0\rangle + \sin(\psi/2)e^{i\theta}|1\rangle$ has the first coefficient, $\cos(\psi/2)$, always a nonnegative real number.
But the Pauli-Y gate is defined to operate on the computational basis $|0\rangle$ and $|1\rangle$ with the matrix
$$
Y = \begin{bmatrix}
        0 & -i \\ i & 0
    \end{bmatrix},
$$
which when applied to $|1\rangle$ yields $-i|0\rangle$. Now this is not in the canonical representation. My questions is: should we normalize the phase to get $|0\rangle$? and does this means that $-i|0\rangle=|0\rangle$?
 A: The state vector
$$ |\Psi\rangle=\cos \psi/2 |0\rangle + \sin \psi/2 ~e^{i\theta} |1\rangle  =  \begin{pmatrix}
      \cos \psi/2          \\
         e^{i\theta} \sin \psi/2
    \end{pmatrix}$$
defines a pure state density matrix through its projection operator,
$$\bbox[yellow]{
|\Psi\rangle \langle \Psi | =  \begin{pmatrix}
      \cos^2 \psi/2             &  \sin \psi/2 ~ \cos\psi/2 ~e^{-i\theta} \\
         \sin \psi/2 ~ \cos\psi/2 ~e^{i\theta} &  \sin^2 \psi/2
    \end{pmatrix}=\rho  }~.
$$ 
Note the manifest invariance under over-all rephasing of $|\Psi\rangle$. 
The general principles' expression of this idempotent hermitean density matrix is also, evidently,   $$
\rho=\frac{1}{2}(1\!\! 1 +  \hat n  \cdot \vec \sigma) ,
 $$
with  $\hat n = (\sin \psi \cos \theta, \; \sin \psi  \sin \theta, \; \cos \psi)^T. $
It is now obvious that $-i|0\rangle$ corresponds to the same point of the sphere as $|0\rangle$, its rephasing by an over-all angle angle of $\pi/2$.  
On the Bloch sphere, Y has sent (rotated by π/2) the x-axis (1,0,0) to the z-axis (0,0,1).
