Probability of a qubit state

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Everything I've read says that you can get the probability of a qubit state by squaring the state's component in the amplitude vector.

For instance $[1/\sqrt{2}, 1/\sqrt{2}]$ and $[1/\sqrt{2}, -1/\sqrt{2}]$ both have a 50% chance of being true because squaring $1\sqrt{2}$ and $-1\sqrt{2}$ both result in 0.5.

Squaring doesn't seem to be enough though, due to this vector: $[0,i]$

That ought to come out to be 100% probability of being true, but when you square $i$ you get -1, so it comes out to -100% probability of being true.

It seems like you might need to absolute value the result of the square, but I haven't seen anyone see that before.

Is that the correct way to get probability from amplitude?

Edit:

It also seems like there is another wrinkle! In the amplitude vector $[1/\sqrt{2}, -0.5 -0.5i]$, when i square the $|1\rangle$ state, i get $0.5i$ which definitely needs some more work before being a probability.

Is the right answer that instead of squaring, you need to multiply by the complex conjugate?

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You have rediscovered the fact that $f(x,y)=xy$ is an inner product on $\mathbb{R}$ but not an inner product on $\mathbb{C}$. The usual inner product on $\mathbb{C}$ is $f(x,y)=x\overline{y}$, and $f(x,x)=|x|^2$.