Where does the imaginary unit $i$ come from in representing spin vector along y axis? I am currently reading Leonard Susskind's - "Quantum Mechanics - The Theoretical Minimum". On Page 38 of the book, the writer described representing spin vectors along the $x$- and $y$-axes using the spin up and down state vectors as basis vectors. I followed up to the part when he described the spin state along $x$-axis as:
$$|r\rangle =  \frac{1}{\sqrt2}|u\rangle +\frac{1}{\sqrt2}|d\rangle$$ (where the letters $r$, $u$ and $d$ represented spin along right, up and down directions respectively)
A similar representation of the spin vector along negative direction of x-axis was described with opposite signs. But then later the writer described spin vector along y-axis as :
$$|o\rangle =  \frac{1}{\sqrt2}|u\rangle +\frac{i}{\sqrt2}|d\rangle$$  (where the letters $o$, $u$ and $d$ represented spin along out, up and down directions respectively).
I failed to understand the reason for using $i$ in the numerator of second component of the spin vector, even though the conditions used to derive it were similar to the one used for deriving spin vector along x-axis. It would be great if someone could explain it.
 A: Leonard Susskind's book is a very introductory book and therefore just half-justifies why the imaginary unit appears there. Actually it comes from a "workaround", the expressions of the $x$ and $y$ components of spin angular momentum is not direct. In the same fashion then, I'm gonna give my answer. You know that the probability of measuring spin up when the system is in the state $\vert o \rangle$ is given by
$$\mathcal{P}_u(\vert o \rangle)=\langle o\vert u\rangle\langle u\vert o\rangle=\frac{1}{2}$$
and same goes for spin down. Also, it is stated in the book that the sum of probabilities should be one. For the $x$ axis, the author has chosen the components to be $\pm\frac{1}{\sqrt{2}}$. For the $y$ axis, he does the same reasoning for the $1/2$ probability, but notice that if you choose $\pm\frac{1}{\sqrt{2}}$ to accomplish probability conditions again, you will end up with the same state than for the $x$ axis, thus this is not valid. Thinking like the book, you have to find some other way of expressing the states $\vert o \rangle $ and $\vert i \rangle $ so that these probabilities relations hold. There is no other way to do it than using the imaginary unit, for it has unitary magnitude, and the probability associated with the component that has the imaginary unit would be
$$\mathcal{P}=\frac{i}{\sqrt{2}}\left(\frac{-i}{\sqrt{2}}\right)=\frac{1}{2}$$
Notice that actually, you can choose the imaginary unit to be in the position you want, even use it in the $x$ axis state.
