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I am currently trying to gain a fuller understanding of the meaning of various spin states and their relation to the direction of measurement by a Stern-Gerlach device. I came across two spin-${1 \over 2}$ states, the first of which given by:

$$\left|\psi_1\right\rangle={1 \over 2} \left|+z\right\rangle+{i\sqrt3 \over 2}\left|-z\right\rangle $$

I know that if a spin-${1 \over 2}$ particle is prepared spin up along an axis specified by:

$$\hat{\boldsymbol{n}}=\sin{\theta}\cos{\phi}\hat{\boldsymbol{i}}+\sin{\theta}\sin{\phi}\hat{\boldsymbol{j}}+\cos{\theta}\hat{\boldsymbol{k}}$$

Then its spin state is given by:

$$\left|+n\right\rangle=\cos{{\theta \over 2}}\left|+z\right\rangle+e^{i\phi}\sin{{\theta \over 2}}\left|-z\right\rangle$$

By inspection, then, I determined that for $\left|\psi_1\right\rangle$ possible values for $\theta$ and $\phi$ are $\theta = {2\pi \over 3}$ and $\phi={\pi \over 2}$. Knowing these angles I then determined the direction in which $\left|\psi_1\right\rangle$ is spin up. This made sense.

The second state I encountered, however, is given by:

$$\left|\psi_2\right\rangle={-i \over 2}\left|+z\right\rangle+{\sqrt3 \over 2}\left|-z\right\rangle$$

According to the expression I gave above for $\left|+n\right\rangle$, however, in order for the amplitude for $\left|+z\right\rangle$ to be complex, the argument passed to the cosine function must be complex. In turn, this means $\theta$ must be complex. This doesn't make sense to me, because my understanding is that $\hat{\boldsymbol{n}}$ is a vector in ordinary 3-dimentional space.

This lead me to believe I am determining $\theta$ incorrectly. If I am making a mistake in determining the values for $\theta$ and $\phi$, I am curious to know what it is. If my method is correct, then why am I coming up with a complex angle for the spin direction of $\left|\psi_2\right\rangle$?

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  • $\begingroup$ The expression for the $|+n\rangle$ state also includes a global phase (which is usually not mentioned explicitly). You could assume a global phase of $\pi/2$ for $|\psi_2\rangle$, i.e. multiply it with $i$ and find $|\psi_2\rangle \equiv |\psi_1\rangle$ $\endgroup$ – jayann Jan 15 '15 at 19:18
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Recall that two ket vectors which differ by a global phase represent identical physical states. In this case, $|\psi_2\rangle = i|\psi_1\rangle$, and the states are therefore equivalent. One can always use this global phase freedom to make the coefficient of $\lvert+z\rangle$ real, and then a parametrisation in terms of the angles $\theta$ and $\phi$ becomes straightforward.

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