I would like to calculate the state after a transformation using the Hadamard gate on a complex state. I get stuck mid-calculation, most likely due to not being able to dealing with the global state. Anybody who can tell me what the part on the question mark should be (and/or which other errors I made)?
$H {|0\rangle + i|1\rangle\over\sqrt 2} \equiv {1\over\sqrt 2}\begin{bmatrix}1 & 1 \\ 1 & -1\end{bmatrix} {1\over\sqrt 2} \begin{bmatrix} 1 \\ i \end{bmatrix} = {1\over2} \begin{bmatrix}1+i\\1-i\end{bmatrix} = ? = {1\over\sqrt 2} \begin{bmatrix}1\\-i\end{bmatrix} \equiv {|0\rangle - i|1\rangle\over\sqrt 2} $
Update trying to use @DaftWullie his answer: $H {|0\rangle + i|1\rangle\over\sqrt 2} \equiv {1\over\sqrt 2}\begin{bmatrix}1 & 1 \\ 1 & -1\end{bmatrix} {1\over\sqrt 2} \begin{bmatrix} 1 \\ i \end{bmatrix} = {1\over2} \begin{bmatrix}1+i\\1-i\end{bmatrix} \neq {1\over\sqrt 2} \begin{bmatrix}1+i\\1-i\end{bmatrix} = \begin{bmatrix}{1\over\sqrt 2}+{1\over\sqrt 2}i\\{1\over\sqrt 2}-{1\over\sqrt 2}i\end{bmatrix} = \begin{bmatrix}\cos(\pi/4)+i \cdot \sin(\pi/4)\\ (\cos(\pi/4)+i \cdot \sin(\pi/4))\cdot(\cos(\pi/2-i\cdot \sin(\pi/2))\end{bmatrix} = \begin{bmatrix}e^{i\pi/4}\\e^{-i\pi/2}e^{i\pi/4}\end{bmatrix}=\\ e^{i\pi/4}\begin{bmatrix}1\\e^{-i\pi/2}\end{bmatrix}= e^{i}e^{\pi/4}\begin{bmatrix}1\\e^{-i\pi/2}\end{bmatrix}\equiv e^{\pi/4}\begin{bmatrix}1\\e^{-i\pi/2}\end{bmatrix} = {1\over\sqrt 2}\begin{bmatrix}1\\-i\end{bmatrix} \equiv {|0\rangle - i|1\rangle\over\sqrt 2} $
Here I still get partly stuck, as I was expecting to calculate using ${1\over 2}$ instead of ${1\over\sqrt 2}$. I see that this too falls into the category of "multiplies the whole vector has no observable consequence", but I wonder if I can calculate this "cleaner" (or did I simply make a mistake?).
Also, how do I indicate removing the global phase in an equation? Do I use the equivalence symbol? An equal symbol with a footnote above it?