Just starting out with QM here. I believe I'm missing something simple but can't get my head around it.
Consider the state:
$$ \left | \psi \right> = \frac{1}{\sqrt 2}\left|0\right> + \frac{i}{\sqrt 2}\left|1\right> $$
This should be normalised, since:
$$ \sqrt{|\alpha|^2 + |\beta|^2} = 1 $$
But when I try to calculate $|\left<\psi | \psi \right>|^2$
I get: $$ \left<\psi | \psi \right> = \frac{1}{\sqrt 2}*\frac{1}{\sqrt 2}\left<0|0\right> + \frac{i}{\sqrt 2}*\frac{i}{\sqrt 2}\left<1|1\right> $$ $$ \left<\psi | \psi \right> = \frac{1}{2}-\frac{1}{2}=0 $$ $$ |\left<\psi | \psi \right>|^2=0 $$
I would have expected $\left<\psi | \psi \right>$ to be 1 since it's normalised. By interpretation, I'm measuring the probability of the state in the SAME state, I would expect to get 1 as well.
Am I doing something wrong?
My reference article which I'm learning from is here by the way: https://qiskit.org/textbook/ch-states/representing-qubit-states.html#2.-The-Rules-of-Measurement-
If
$$ \left| \psi \right>=\frac{1}{\sqrt{2}} \left| 0 \right> +\frac{i}{\sqrt{2}} \left| 1 \right> $$
Then, taking the adjoint
$$\left< \psi \right|=\frac{1}{\sqrt{2}} \left< 0 \right| +\frac{-i}{\sqrt{2}} \left< 1 \right| $$
So that, multiplying out and using $\left< 0 \middle| 1 \right>=0$, we have
$$ \left< \psi \middle| \psi \right>=\frac{1}{\sqrt{2}}\frac{1}{\sqrt{2}} \left< 0\middle| 0\right>+\frac{i}{\sqrt{2}}\frac{(-i)}{\sqrt{2}}\left< 1\middle| 1\right>=\frac{1}{2}+\frac{1}{2}=1$$